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{"url":"https:\/\/www.groundai.com\/project\/the-reduced-hartree-fock-model-for-short-range-quantum-crystals-with-defects\/","text":"Short-range quantum crystals with defects\n\n# The reduced Hartree-Fock model for short-range quantum crystals with defects\n\nSalma Lahbabi CNRS & Laboratoire de Math\u00e9matiques (UMR 8088), Universit\u00e9 de Cergy-Pontoise\n95000 Cergy-Pontoise Cedex, France\nCERMICS, \u00c9cole Nationale des Ponts et Chauss\u00e9es (Paristech)\n& INRIA (Micmac Project), 6-8 Av. Blaise Pascal, 77455 Champs-sur-Marne, France\nAugust 30, 2019\n###### Abstract.\n\nIn this article, we consider quantum crystals with defects in the reduced Hartree-Fock framework. The nuclei are supposed to be classical particles arranged around a reference periodic configuration. The perturbation is assumed to be small in amplitude, but need not be localized in a specific region of space or have any spatial invariance. Assuming Yukawa interactions, we prove the existence of an electronic ground state, solution of the self-consistent field equation. Next, by studying precisely the decay properties of this solution for local defects, we are able to expand the density of states of the nonlinear Hamiltonian of a system with a random perturbation of Anderson-Bernoulli type, in the limit of low concentration of defects. One important step in the proof of our results is the analysis of the dielectric response of the crystal to an effective charge perturbation.\n\n## 1. Introduction\n\nIn solid state physics and materials science, the presence of defects in materials induces many interesting properties, such as Anderson localization and leads to many applications such as doped semi-conductors The mathematical modeling and the numerical simulation of the electronic structure of these materials is a challenging task, as we are in the presence of infinitely many interacting particles.\n\nThe purpose of this paper is to construct the state of the quantum electrons of a mean-field crystal, in which the nuclei are classical particles arranged around a reference periodic configuration. We work with the assumption that the nuclear distribution is close to a chosen periodic arrangement locally, but the perturbation need not be localized in a specific region of space and it also need not have any spatial invariance. To our knowledge, this is the first result of this kind for Hartree-Fock type models for quantum crystals, with short-range interactions. By studying precisely the behavior of our solution, we are then able to expand the density of states of the Hamiltonian of the system in the presence of a random perturbation of Anderson-Bernoulli type, in the limit of low concentration of defects, that is when the Bernoulli parameter tends to zero. The state of the random crystal and the mean-field Hamiltonian were recently constructed in\u00a0[8]. Our small- expansion is the nonlinear equivalent of a previous result by Klopp\u00a0[19] in the linear case.\n\nThe mean-field model we consider in this paper is the reduced Hartree-Fock model\u00a0[31], also called the Hartree model in the physics literature. It is obtained from the generalized Hartree-Fock model\u00a0[25] by removing the exchange term. As the Coulomb interaction is long-range, it is a difficult mathematical question to describe infinite systems interacting through the Coulomb potential. In the following, we assume that all the particles interact through Yukawa potential of parameter . In fact, we can assume any reasonable short-range potential, but we concentrate on the Yukawa interaction in dimension for simplicity. We consider systems composed of infinitely many classical nuclei distributed over the whole space and infinitely many electrons.\n\nWe start by recalling the definition of the reduced Hartree-Fock (rHF) model for a finite system composed of a set of nuclei having a density of charge and electrons. The electrons are described by the -body wave-function (called a Slater determinant)\n\n \u03c8(x1,\u22ef,xN)=1\u221aN!det(\u03c6j(xi)),\n\nwhere the functions satisfy . The rHF equations then read\n\n \u23a7\u23aa \u23aa \u23aa\u23a8\u23aa \u23aa \u23aa\u23a9H\u03c6i=\u03bbi\u03c6iH=\u221212\u0394+V\u2212\u0394V+m2V=\u2223\u2223Sd\u22121\u2223\u2223(\u03c1\u03c8\u2212\u03bdnuc)\u22001\u2264i\u2264N, (1)\n\nwhere and are the smallest eigenvalues of the operator , assuming that . Here, is the Lebesgue measure of the unit sphere (, , ). The existence of a solution of\u00a0(1) is due to Lieb and Simon\u00a0[26].\n\nIn order to describe infinite systems, it is more convenient to reformulate the rHF problem in terms of the one-particle density matrix formalism\u00a0[24]. In this formalism, the state of the electrons is described by the orthogonal projector of rank and the equations\u00a0(1) can be recast as\n\n \u23a7\u23aa \u23aa \u23aa \u23aa\u23a8\u23aa \u23aa \u23aa \u23aa\u23a9\u03b3=1(H\u2264\u03f5F)H=\u221212\u0394+V\u2212\u0394V+m2V=\u2223\u2223Sd\u22121\u2223\u2223(\u03c1\u03b3\u2212\u03bdnuc), (2)\n\nwhere formally and the Fermi level is any real number in the gap .\n\nFor infinite systems, the rHF equation is still given by\u00a0(2), but is now an infinite rank operator as there are infinitely many electrons in the system. The operator needs to be locally trace class for the electronic density to be well-defined in .\n\nThe rHF equation\u00a0(2) was solved for periodic nuclear densities\n\n \u03bdnuc=\u03bdper=\u2211k\u2208R\u03b7(\u22c5\u2212k)\n\nby Catto, Le Bris and Lions in\u00a0[10], and periodic nuclear densities with local perturbations\n\n \u03bdnuc=\u2211k\u2208R\u03b7(\u22c5\u2212k)+\u03bd\n\nwere studied by Canc\u00e8s, Deleurence and Lewin in\u00a0[7]. We have denoted by the underlying discrete periodic lattice. The corresponding Hamiltonians are denoted by and . Stochastic distributions,\n\n \u03bdnuc(\u03c9,\u22c5)=\u2211k\u2208R\u03b7(\u22c5\u2212k)+\u2211k\u2208Rqk(\u03c9)\u03c7(\u22c5\u2212k)\n\nfor instance, were treated in\u00a0[8].\n\nOur present work follows on from\u00a0[7, 6, 8]. We are going to solve the equation\u00a0(2) in the particular case where\n\n \u03bdnuc=\u03bdper+\u03bd, (3)\n\nwhere is a periodic nuclear distribution so that the corresponding background crystal is an insulator (the mean-field Hamiltonian has a gap around ), and is a small enough arbitrary perturbation of the background crystal. The perturbation needs to be small in amplitude locally, but must not be local or have any spatial invariance.\n\nThe rHF model is an approximation of the -body Schr\u00f6dinger model, for which there is no well-defined formulation for infinite systems so far. The only available result is the existence of the thermodynamic limit of the energy: the energy per unit volume of the system confined to a box, with suitable boundary conditions, converges when the size of the box grows to infinity. The first theorem of this form for Coulomb interacting systems is due to Lieb and Lebowitz in\u00a0[22]. In this latter work, nuclei are considered as quantum particle and rotational invariance plays a crucial role. For quantum systems in which the nuclei are classical particles, the thermodynamic limit was proved for perfect crystals by Fefferman\u00a0[12] (a recent proof has been proposed in\u00a0[17]) and for stationary stochastic systems by Blanc and Lewin\u00a0[4]. Similar results for Yukawa interacting systems are simpler than for the Coulomb case and follow from the work of Ruelle and Fisher\u00a0[13] for perfect crystals and Veniaminov\u00a0[32] for stationary stochastic systems. Unfortunately, very little is known about the limiting quantum state in both cases.\n\nFor (orbital-free) Thomas-Fermi like theories, the periodic model was studied in\u00a0[26, 9], the case of crystals with local defects was studied in\u00a0[5] and stochastic systems were investigated in\u00a0[3]. To the best of our knowledge, the only works dealing with systems with arbitrary distributed nuclei are\u00a0[9, 2] for Thomas-Fermi type models.\n\nAs mentioned before, our work is the first one to consider this kind of systems in the framework of Hartree-Fock type models. Our results concern small perturbations of perfect crystals interacting through short-range Yukawa potential. Extending these results to more general geometries and for the long-range Coulomb interaction are important questions that we hope to address in the future.\n\nAfter having found solutions of\u00a0(2) for any (small enough) , we study the properties of this solution for local perturbations . This enables us to investigate small random perturbations of perfect crystals. Precisely, we consider nuclear distributions\n\n \u03bdnuc(\u03c9,x)=\u03bdper(x)+\u2211k\u2208Rqk(\u03c9)\u03c7(x\u2212k),\n\nwhere are i.i.d. Bernoulli variables of parameter and is a compactly supported function which is small enough in . We are interested in the properties of the system in the limit of low concentration of defects, that is when the parameter goes to zero. We prove that the density of states of the mean-field Hamiltonian , which describes the collective behavior of the electrons, admits an expansion of the form\n\n np=n0+J\u2211j=1\u03bcjpj+O(pJ+1). (4)\n\nHere, is the density of states of the unperturbed Hamiltonian and is a function of the spectral shift function for the pair of operators and , the latter being the mean-field Hamiltonian of the system with only one local defect constructed in\u00a0[7]. We give in Theorem\u00a02.7 a precise meaning of .\n\nIn\u00a0[19], Klopp considers the empirical linear Anderson-Bernoulli model\n\n H=\u221212\u0394+V0+VwithV(\u03c9,x)=\u2211k\u2208Rqk(\u03c9)\u03b7(x\u2212k),\n\nwhere is a linear periodic potential and an exponentially decaying potential. He proves that the density of states of the Hamiltonian admits an asymptotic expansion similar to\u00a0(4). The case where is distributed following a Poisson law instead of Bernoulli is dealt with in\u00a0[20]. Our proof of\u00a0(4) follows the same lines as the one of Klopp. The main difficulty here is to understand the decay properties of the mean-field potential solution of the self-consistent equations\u00a0(2). For this reason, we dedicate an important part of this paper to the study of these decay properties. In Theorem\u00a02.3 below, we show that for a compactly supported perturbation , the difference decays faster than any polynomial far from the support of the perturbation . Moreover, we show that the potential generated by two defects that are far enough is close to the sum of the potentials generated by each defect alone.\n\nThe article is organized as follow. In Section\u00a02, we present the main results of the paper. We start by recalling the reduced Hartree-Fock model for perfect crystals and perfect crystals with local defects in Section\u00a02.1. In Section\u00a02.2, we state the existence of solutions to the self-consistent equations\u00a0(2) for given by\u00a0(3). We also explain that our solution is in some sense the minimizer of the energy of the system. We also prove a thermodynamic limit, namely, the ground state of the system with the perturbation confined to a box converges, when the size of the box goes to infinity, to the ground state of the system with the perturbation . In Section\u00a02.3, we prove decay estimates for the mean-field density and potential. In Section\u00a02.4, we present the expansion of the density of states of the mean-field Hamiltonian. The proofs of all these results are provided in Sections\u00a0456 and\u00a07. In Section\u00a03, we study the dielectric response of a perfect crystal to a variation of the effective charge distribution, which plays a key role in this paper.\n\nAcknowledgement. I thoroughly thank \u00c9ric Canc\u00e8s and Mathieu Lewin for their precious help and advices. The research leading to these results has received funding from the European Research Council under the European Community\u2019s Seventh Framework Programme (FP7\/2007\u20132013 Grant Agreement MNIQS no. 258023).\n\n## 2. Statement of the main results\n\n### 2.1. The rHF model for crystals with and without local defects\n\nIn defect-free materials, the nuclei and electrons are arranged according to a discrete periodic lattice of , in the sense that both the nuclear density and the electronic density are -periodic functions. For simplicity, we take in the following. The reduced Hartree-Fock model for perfect crystals has been rigorously derived from the reduced Hartree-Fock model for finite molecular systems by means of thermodynamic limit procedure in\u00a0[10, 7] in the case of Coulomb interaction. The same results for Yukawa interaction are obtained with similar arguments. The self-consistent equation\u00a0(2) then reads\n\n \u23a7\u23aa \u23aa \u23aa \u23aa\u23a8\u23aa \u23aa \u23aa \u23aa\u23a9\u03b30=1(Hper\u2264\u03f5F)Hper=\u221212\u0394+Vper\u2212\u0394Vper+m2Vper=\u2223\u2223Sd\u22121\u2223\u2223(\u03c1\u03b30\u2212\u03bdper). (5)\n\nIt has been proved in\u00a0[10, 7] that\u00a0(5) admits a unique solution which is the unique minimizer of the periodic rHF energy functional.\n\nMost of our results below hold only for insulators (or semi-conductors). We therefore make the assumption that\n\n Hper\u00a0has a spectral gap around\u00a0\u03f5F. (6)\n\nThe rHF model for crystals with local defects was introduced and studied in\u00a0[7]. A solution of the rHF equation\u00a0(2) is constructed using a variational method. One advantage of this method is that there is no need to assume that the perturbation is small in amplitude. The idea is to find a minimizer of the infinite energy of the system by minimizing the energy difference between the perturbed state and the perfect crystal. The ground state density matrix can thus be decomposed as\n\n \u03b3=\u03b30+Q\u03bd, (7)\n\nwhere is a minimizer of the energy functional\n\n E\u03bd(Q)=Tr\u03b30((Hper\u2212\u03f5F)Q)+12Dm(\u03c1Q\u2212\u03bd,\u03c1Q\u2212\u03bd) (8)\n\non the convex set\n\n K={Q\u2217=Q,\u2212\u03b30\u2264Q\u22641\u2212\u03b30,(\u2212\u0394+1)12Q\u2208S2(L2(Rd)),(\u2212\u0394+1)12Q\u00b1\u00b1(\u2212\u0394+1)12\u2208S1(L2(Rd))}, (9)\n\nwhere , and . We use the notation to denote the Schatten class. In particular is the set of Hilbert-Schmidt operators. The second term of\u00a0(8) accounts for the interaction energy and is defined for any charge densities by\n\n Dm(f,g) =\u2223\u2223Sd\u22121\u2223\u2223\u222bRd\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u02c6f(p)\u02c6g(p)|p|2+m2dp=\u222bRd\u222bRdf(x)Ym(x\u2212y)g(y)dxdy,\n\nwhere is the Fourier transform of . The Yukawa kernel , the inverse Fourier transform of , is given by\n\n Ym(x)=\u23a7\u23aa \u23aa\u23a8\u23aa \u23aa\u23a9m\u22121e\u2212m|x|ifd=1,K0(m|x|)ifd=2,|x|\u22121e\u2212m|x|ifd=3,\n\nwhere is the modified Bessel function of the second type\u00a0[23]. It has been proved in\u00a0[7] that the energy functional\u00a0(8) is convex and that all its minimizers share the same density . These minimizers are of the form\n\n \u23a7\u23aa \u23aa \u23aa \u23aa\u23a8\u23aa \u23aa \u23aa \u23aa\u23a9\u03b3=1(H\u2264\u03f5F)+\u03b4H=\u221212\u0394+V\u2212\u0394V+m2V=\u2223\u2223Sd\u22121\u2223\u2223(\u03c1\u03b3\u2212\u03bdper\u2212\u03bd), (10)\n\nwhere . If is small enough in the -norm, then .\n\nOne of the purposes of this article is to find decay estimates of the potential solution of\u00a0(10) that are necessary in the study of the Anderson-Bernoulli random perturbations of crystals.\n\n### 2.2. Existence of ground states\n\nIn this section, we state our results concerning the electronic state of a perturbed crystal. The host crystal is characterized by a periodic nuclear density such that the gap assumption\u00a0(6) holds. The perturbation is given by a distribution . The total nuclear distribution is then\n\n \u03bdnuc=\u03bdper+\u03bd.\n\nIn Theorem\u00a02.1 below, we show that if is small enough in the -norm, then the rHF equation\u00a0(2) admits a solution . This solution is unique in a neighborhood of . The proof consists in formulating the problem in terms of the density and using a fixed point technique, in the spirit of\u00a0[15].\n\n###### Theorem 2.1 (Existence of a ground state).\n\nThere exists and such that for any satisfying , there is a unique solution to the self-consistent equation\n\n \u23a7\u23aa \u23aa \u23aa \u23aa\u23a8\u23aa \u23aa \u23aa \u23aa\u23a9\u03b3=1(H\u2264\u03f5F)H=\u221212\u0394+V\u2212\u0394V+m2V=\u2223\u2223Sd\u22121\u2223\u2223(\u03c1\u03b3\u2212\u03bd\u2212\u03bdper) (11)\n\nsatisfying\n\n \u2225\u2225\u03c1\u03b3\u2212\u03c1\u03b30\u2225\u2225L2\\rm unif\u00a0\u2264C\u2225\u03bd\u2225L2\\rm unif\u00a0. (12)\n\nWe denote this solution by , the response electronic density by and the defect mean-field potential by .\n\nFor a local defect such that , equation\u00a0(11) admits a unique solution which coincides with the ground state solution of\u00a0(7) constructed in\u00a0[7]. Indeed, the solution given in Theorem\u00a02.1 is a solution of the defect problem\u00a0(10). Moreover, in the proof of Theorem\u00a02.1, we prove that has a gap around , thus necessarily in\u00a0(10). As all the solutions of\u00a0(10) share the same density,\u00a0(10) (thus\u00a0(11)) admits a unique solution.\n\nThe ground state constructed in Theorem\u00a02.1 is in fact the unique minimizer of the \"infinite\" rHF energy functional. Indeed, following ideas of\u00a0[16], we can define the relative energy of the system with nuclear distribution by subtracting the \"infinite\" energy of from the \"infinite\" energy of a test state :\n\n Erel\u03bd(\u03b3):=Tr\u03b3\u03bd((H\u2212\u03f5F)(\u03b3\u2212\u03b3\u03bd))+12Dm(\u03c1\u03b3\u2212\u03c1\u03b3\u03bd,\u03c1\u03b3\u2212\u03c1\u03b3\u03bd).\n\nThis energy is well-defined for states such that is finite rank and smooth enough for instance, but one can extend it to states in a set similar to in\u00a0(9). The minimum of the energy is attained for . Moreover, as has a gap around , is strictly convex and is its unique minimizer.\n\nIn the following theorem, we show that if we confine the defect to a box of finite size, then the ground state of the system defined by the theory of local defects presented in Section\u00a02.1 converges, when the size of the box goes to infinity, to the ground state of the system with the defect defined in Theorem\u00a02.1. We denote by .\n\n###### Theorem 2.2 (Thermodynamic limit).\n\nThere exists such that for any satisfying , the sequence converges in to as .\n\n### 2.3. Decay estimates\n\nIn this section, we prove some decay estimates of the mean-field potential and the mean-field density , which will be particularly important to understand the system in the presence of rare perturbations in the next section.\n\nTheorem\u00a02.3 below is crucial in the proof of Theorem\u00a02.7. Indeed, we will need uniform decay estimates for compactly supported defects, with growing supports and uniform local norms.\n\n###### Theorem 2.3 (Decay rate of the mean-field potential and density).\n\nThere exists and such that for any satisfying , we have for\n\n \u2225V\u03bd\u2225H2\\rm unif\u00a0(Rd\u2216CR(\u03bd))+\u2225\u03c1\u03bd\u2225L2\\rm unif\u00a0(Rd\u2216CR(\u03bd))\u2264Ce\u2212C\u2032(logR)2\u2225\u03bd\u2225L2\\rm unif\u00a0(Rd), (13)\n\nwhere .\n\n###### Remark 2.4.\n\nUsing the same techniques as in the proof of Theorem\u00a02.3, we can prove (see\u00a0[21]) that there exists and such that for any satisfying and , we have for\n\n \u2225V\u03bd\u2225H2(Rd\u2216CR(\u03bd))+\u2225\u03c1\u03bd\u2225L2(Rd\u2216CR(\u03bd))\u2264Ce\u2212C\u2032(logR)2\u2225\u03bd\u2225L2(Rd). (14)\n\nEstimate\u00a0(14) gives a decay rate of the solution of the rHF equation for crystals with local defects, far from the support of the defect. In particular, it shows that . This decay is due to the short-range character of the Yukawa interaction. In the Coulomb case, it has been proved in\u00a0[6] that for anisotropic materials, .\n\nThe decay rate of and proved in Theorem\u00a02.3 is faster than the decay of any polynomial, but is not exponential, which we think should be the optimal rate.\n\nProposition\u00a02.5 below is an important intermediary result in the proof of Theorem\u00a02.2. It says that the mean-field density and potential on a compact set depend mainly on the nuclear distribution in a neighborhood of this compact set.\n\n###### Proposition 2.5 (The mean-field potential and density depend locally on \u03bd).\n\nThere exists such that for any there exists such that for any satisfying and any we have\n\n \u2225\u2225V\u03bd\u2212V\u03bdL\u2225\u2225H2\\rm unif\u00a0(B(0,L\/4\u03b2))+\u2225\u2225\u03c1\u03bd\u2212\u03c1\u03bdL\u2225\u2225L2\\rm unif\u00a0(B(0,L\/4\u03b2))\u2264CL\u03b2\u2225\u03bd\u2225L2\\rm unif\u00a0,\n\nwhere .\n\nIn the same way, we obtain the following result which will be very useful in the proof of Theorem\u00a02.7. We prove that the potential generated by two defects that are far enough is close to the sum of the potentials generated by each defect alone in the sense of\n\n###### Proposition 2.6.\n\nThere exists such that for any , there exists such that for any satisfying and , we have\n\n \u2225\u2225V\u03bd1+\u03bd2\u2212V\u03bd2\u2225\u2225H2\\rm unif%\u00a0(CR\/4\u03b2(\u03bd2))+\u2225\u2225\u03c1\u03bd1+\u03bd2\u2212\u03c1\u03bd2\u2225\u2225L2\\rm unif\u00a0(CR\/4\u03b2(\u03bd2)) \u2264CR\u03b2(\u2225\u03bd1\u2225L2\\rm unif\u00a0+\u2225\u03bd2\u2225L2\\rm unif\u00a0).\n###### Proof.\n\nThe proof is the same as the one of Proposition\u00a02.5 with and . \u220e\n\n### 2.4. Asymptotic expansion of the density of states\n\nIn this section, we use our previous results to study a particular case of random materials. In the so-called statistically homogeneous materials, the particles are randomly distributed over the space with a certain spatial invariance. More precisely, the nuclear distribution (thus the electronic density) is stationary in the sense\n\n \u03bdnuc(\u03c4k(\u03c9),x)=\u03bdnuc(\u03c9,x+k),\n\nwhere is an ergodic group action of on the probability set (see Figure\u00a01).\n\nOne famous example of such distributions is the Anderson model\n\n \u03bdnuc(\u03c9,x)=\u2211k\u2208Zdqk(\u03c9)\u03c7(x\u2212k),\n\nwhere, typically, and the \u2019s are i.i.d. random variables. The reduced Hartree-Fock model for statistically homogeneous materials was introduced in\u00a0[8]. The state of the electrons is described by a random self-adjoint operator acting on such that almost surely. The rHF equation is then\n\n \u23a7\u23aa \u23aa \u23aa \u23aa\u23a8\u23aa \u23aa \u23aa \u23aa\u23a9\u03b3(\u03c9)=1(H(\u03c9)\u2264\u03f5F)+\u03b4(\u03c9)H(\u03c9)=\u221212\u0394+V(\u03c9,\u22c5)\u2212\u0394V(\u03c9,\u22c5)+m2V(\u03c9,\u22c5)=\u2223\u2223Sd\u22121\u2223\u2223(\u03c1\u03b3(\u03c9)\u2212\u03bd(\u03c9,\u22c5))almost surely, (15)\n\nwhere almost surely. The solutions of\u00a0(15) turn out to be the minimizers of the energy functional\n\n E\u2013\u2013\u03bdnuc(\u03b3)=Tr\u2013\u2013\u2013((\u221212\u0394\u2212\u03f5F)\u03b3)+D\u2013\u2013m(\u03c1\u03b3\u2212\u03bdnuc,\u03c1\u03b3\u2212\u03bdnuc),\n\nwhere and\n\n D\u2013\u2013m(f,g)=E(\u222bRd\u222b\u0393f(x)Ym(x\u2212y)g(y)dxdy).\n\nHere, denotes the semi-open unit cube. Thanks to the convexity of , it has been proved in\u00a0[8] that the minimizers of share the same density. Therefore, the Hamiltonian solution of\u00a0(15) is uniquely defined.\n\nIn this paper, we are interested in the particular case of random perturbation of perfect crystals\n\n \u03bdnuc(\u03c9,x)=\u03bdper(x)+\u03bdp(\u03c9,x)\n\nin the limit of low concentration of defects. We restrict our study to Anderson-Bernoulli type perturbations, that is, we suppose that at each site of , there is a probability to see a local defect , independently of what is happening in the other sites. More precisely, we consider the probability space endowed with the measure and the ergodic group action . The defect distribution we consider is then given by\n\n \u03bdp(\u03c9,x)=\u2211k\u2208Zdqk(\u03c9)\u03c7(x\u2212k)\n\nwhere is the coordinates of and with . The \u2019s are i.i.d. Bernoulli variables of parameter . If , then almost surely and\u00a0(15) admits a unique solution. For almost every , this solution coincides with the solution of\u00a0(11) constructed in Theorem\u00a02.1. For convenience, we will from now on use the notation\n\n H0=Hper\u2212\u03f5F,\n\nwhere we recall that is the Fermi level. We introduce the mean-field Hamiltonian corresponding to the system with the defect\n\n Hp=H0+V\u03bdpwithV\u03bdp(\u03c9,x)=Ym\u2217(\u03c1\u03bdp\u2212\u03bdp).\n\nAs is stationary with respect to the ergodic group and uniformly bounded in , then by\u00a0[27, Theorem 5.20], there exists a deterministic positive measure , the density of states of , such that for any in the Schwartz space\n\n \u222bR\u03c6(x)np(dx)=Tr\u2013\u2013\u2013(\u03c6(Hp)).\n\nFor , we define the self-consistent operator corresponding to the system with the defects in\n\n HK=H0+VK,\n\nwhere\n\n VK=Ym\u2217(\u03c1K\u2212\u03bdK),\u03bdK=\u2211k\u2208K\u03c7(\u22c5\u2212k)and\u03c1K=\u03c1\u03bdK.\n\nIf , we denote by the spectral shift function\u00a0[33] for the pair of operators and . It is the tempered distribution in satisfying, for any ,\n\n Tr(\u03c6(HK)\u2212\u03c6(H0))=\u222bR\u03beK(x)\u03c6\u2032(x)dx=\u2212\u222bR\u03be\u2032K(x)\u03c6(x)dx.\n\nIn Theorem\u00a0(2.7) below, we give the asymptotic expansion of the density of states in terms of powers of the Bernoulli parameter .\n\n###### Theorem 2.7.\n\nFor such that and such that , we define the tempered distribution by\n\n \u03bcK(x)=\u22121|K|\u2211K\u2032\u2282K(\u22121)\u2223\u2223K\u2216K\u2032\u2223\u2223\u03be\u2032K\u2032(x).\n\nThere exists such that if , then\n\n1. for , is a well-defined convergent series in .\n\n2. for , there exists , independent of such that for any ,\n\nwhere is the density of states of the unperturbed Hamiltonian and .\n\nIn Theorem\u00a02.7, we only present the expansion of the density of states until the second order . The proof of the expansion up to any order should follow the same lines and techniques used here.\n\nA result similar to Theorem\u00a02.7 was obtained in\u00a0[19] in the linear case. Materials with low concentration of defects were studied by Le Bris and Anantharaman\u00a0[1]. in the framework of stochastic homogenization.\n\nThe proof of Theorem\u00a02.7 follows essentially the proof of\u00a0[19, Theorem 1.1]. It uses the decay of the potential related to each local defect. In\u00a0[19, Theorem 1.1], the linear potential is assumed to decay exponentially. In our nonlinear model, the decay estimates established in Section\u00a02.3 play a crucial role in the proof.\n\nThe rest of the paper is devoted to the proofs of the results presented in this section. In the next section, we study the dielectric response of the crystal to an effective charge perturbation. The results of Section\u00a03 will be used in later sections.\n\n## 3. Dielectric response for Yukawa interaction\n\nIn this section, we study the dielectric response of the electronic ground state of a crystal to a small effective charge perturbation . This means more precisely that we expand the formula\n\n Qf=1(H0+f\u2217Ym\u22640)\u22121(H0\u22640)\n\nin powers of (for small enough) and state important properties of the first order term. The higher order term will be dealt with later in Lemma\u00a04.1. For Coulomb interactions and local perturbation , where is the Coulomb space, this study has been carried out in\u00a0[6] in dimension .\n\nThe results of this section can be used in the linear model or the mean-field framework. In the reduced Hartree-Fock model we consider in this paper, the effective charge perturbation is , where is the electronic density of the response of the crystal to the nuclear perturbation defined in Theorem\u00a02.1. Expanding (formally) in powers of and using the resolvent formula leads to considering the following operator\n\n Q1,f=12i\u03c0\u222eC1z\u2212H0f\u2217Ym1z\u2212H0dz,\n\nwhere is a smooth curve in the complex plane enclosing the whole spectrum of below (see Figure\u00a02).\n\nBy the residue Theorem, the operator does not depend on the particular curve chosen as above. We recall that is bounded with relative bound . Thus is bounded below by the Rellich-Kato theorem\u00a0[28, Theorem X.12]. Theorem\u00a03.1 below studies the properties of the dielectric response operator and the operator , which will play an important role in the resolution of the self-consistent equation\u00a0(11). In particular, it gives the functional spaces on which and are well-defined for both local and extended charge densities. It also says that is local in the sense that its off-diagonal components decay faster than any polynomial. We consider , endowed with the scalar product\n\n \u27e8f,g\u27e9H\u22121=1(2\u03c0)d\u222bRd\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u02c6f(p)\u02c6g(p)|p|2+m2dp.\n###### Theorem 3.1 (Properties of the dielectric response).\n\nWe have\n\n1. The operator\n\n L:H\u22121(Rd)\u2192H\u22121(Rd)f\u21a6\u2212\u03c1Q1,f,\n\nis well-defined, bounded, non-negative and self-adjoint. Hence is invertible and bicontinuous.\n\n2. The operator is bounded from to and is a well-defined, bounded operator from into itself.\n\n3. The operator\n\n L:L2\\rm unif\u00a0(Rd)\u2192L2\\rm unif\u00a0(Rd)f\u21a6\u2212\u03c1Q1,f,\n\nis well-defined and bounded. The operator is invertible on and its inverse is bounded.\n\n4. There exist and such that for any such that , we have\n\n \u2225\u2225\u22251\u0393+j11+L1\u0393+k\u2225\u2225\u2225B\u2264Ce\u2212C\u2032(log|k\u2212j|)2. (16)\n###### Proof.\n\nThe proof consists in the following 6 steps. In the whole paper and are constants whose value might change from one line to the other.\n\n#### Step 1\n\nProof of (i). The proof is similar to the one of\u00a0[6, Proposition 2], with the Yukawa kernel , instead of the Coulomb kernel. In the Yukawa case, plays the role of the Coulomb space. The proof of\u00a0[6, Proposition 2] can easily be adapted to our case. We skip the details for the sake of brevity.\n\n#### Step 2\n\nProof of (ii). Let . Then and\n\n (17)\n\nTherefore, by\u00a0[6, Proposition 1], , where has been defined in\u00a0(9), and . Arguing by duality, we have for any ,\n\n Tr(Q1,fW)=\u222bRd\u03c1Q1,fW. (18)\n\nBesides, by the Kato-Seiler-Simon inequality\u00a0[30, Theorem 4.1] for\n\n \u2200p\u22652,\u2225f(\u2212i\u2207)g(x)\u2225S2\u2264(2\u03c0)\u2212dp\u2225f\u2225Lp\u2225g\u2225Lp (19)\n\nand the fact that\n\n (z\u2212H0)\u22121(1\u2212\u0394)\u00a0is uniformly bounded on % the contour\u00a0C, (20)\n\nwe have\n\nand\n\n \u2223\u2223Tr(Q1,fW)\u2223\u2223=\u2223\u2223\u222312i\u03c0\u222eCTr(1z\u2212H0Ym\u2217f1z\u2212H0W)dz\u2223\u2223\u2223 \u2264C\u2225Ym\u2217f\u2225L2\u2225W\u2225L2. (21)\n\nThe bound\u00a0(20) follows from the following lemma.\n\n###### Lemma 3.2.\n\nLet . Then there exists , depending only on the -norm of , such that for any , we have\n\n \u2225\u2225(\u2212\u0394+1)(\u2212\u0394+W\u2212z)\u22121\u2225\u2225B\u2264C1+|z|d(z,\u03c3(\u2212\u0394+W)).\n\nIn particular, if is a compact set of , then is uniformly bounded on .\n\nThe proof of Lemma\u00a03.2 is elementary, it can be read in\u00a0[21]. In view of\u00a0(17),\u00a0(18) and\u00a0(21), it follows that\n\n \u2223\u2223\u2223\u222bRd(Lf)W\u2223\u2223\u2223 \u2264C\u2225f\u2225H\u22121\u2225W\u2225L2.\n\nWe deduce that\n\n \u2225Lf\u2225L2\u2264C\u2225f\u2225H\u22121.\n\nWe now prove that is bounded on . Let and such that . Then, . As is bounded from into itself, we have\n\n \u2225f\u2225H\u22121\u2264C\u2225g\u2225H\u22121\u2264C\u2225g\u2225L2.\n\nTherefore, as is continuous from to ,\n\n \u2225f\u2225L2 =\u2225g\u2212Lf\u2225L2\u2264\u2225g\u2225L2+\u2225Lf\u2225L2\u2264\u2225g\u2225L2+C\u2225f\u2225H\u22121\u2264C\u2225g\u2225L2,\n\nwhich concludes the proof of (ii).\n\n#### Step 3\n\nProof of the first part of (iii): is well-defined and bounded on . First, we consider a bounded operator and prove that is locally trace class. For and , we have by\u00a0(20) and the Kato-Simon-Seiler inequality\u00a0(19) that is trace class and that there exists independent of such that\n\n \u2223\u2223\u2223Tr(\u03c71z\u2212H0A1z\u2212H0\u03c7)\u2223\u2223\u2223 \u2264\u2225\u2225\u2225\u03c71z\u2212H0A1z\u2212H0\u03c7\u2225\u2225\u2225S1 \u2264\u2225\u2225\u2225\u03c71z\u2212H0\u2225\u2225\u2225S2\u2225A\u2225B\u2225\u2225\u22251z\u2212H0\u03c7\u2225\u2225\u2225S2\u2264C\u2225A\u2225B\u2225\u03c7\u22252L2.\n\nIt follows that the operator is locally trace class and that its density is in . 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\section{Introduction}
\noindent
The precision determinations of $\alpha_s(M_Z^2)$, the mass of the charm quark $m_c$ and the
parton distribution functions from the world data on deep-inelastic scattering (DIS) require
the heavy flavor corrections to 3-loop order \cite{Alekhin:2012vu}. Here the structure function
$F_2(x,Q^2)$ provides the highest precision. As has been shown in \cite{Buza:1995ie} at scales
$Q^2/m_c^2 \raisebox{-0.07cm 10$ the asymptotic representation of the heavy flavor Wilson coefficients
provides a representation on the per cent level.~\footnote{The corresponding scales are much
higher in case of the structure function $F_L(x,Q^2)$ \cite{Buza:1995ie}, for which the
3-loop heavy flavor corrections for general values of $N$ have been calculated in
\cite{Blumlein:2006mh}.} They are given in terms of convolutions of massive operator matrix
elements (OMEs) and the massless Wilson coefficients \cite{Vermaseren:2005qc}. A series
of 3-loop Mellin-moments for $F_2(x,Q^2)$ and transversity and the OMEs describing the
transition matrix elements in the variable flavor number scheme (VFNS)
\cite{Buza:1996wv,Bierenbaum:2009zt}
have been calculated in 2009 in Refs.~\cite{Bierenbaum:2009mv,Blumlein:2009rg} projecting the
respective tensor quantities onto massive tadpoles which could be computed using {\tt MATAD}
\cite{Steinhauser:2000ry}.
A program to compute the massive 3-loop Wilson coefficients at general values of $N$ and
their analytic continuation to $N \in \mathbb{C}$ started thereafter. In the unpolarized case,
eight Wilson coefficients/OMEs contribute. All logarithmic contributions
\cite{Bierenbaum:2010jp} are available since they rely on the the 2-loop results
\cite{Buza:1995ie,Bierenbaum:2007qe} up to $O(\alpha_s^2 \varepsilon)$ \cite{Bierenbaum:2008yu}.
Two of the eight Wilson coefficients resp. OMEs, $L_{qg,Q}^{(3)}$ and $L_{qq,Q}^{(3),\rm PS}$,
were calculated in \cite{Ablinger:2010ty}. We studied the contributions to specific color factors,
such as $O(N_F T_F^2 C_{A,F})$, which are completely known now
\cite{Ablinger:2010ty,Blumlein:2012vq}.
Further investigations are devoted to diagrams with two fermion lines with finite equal
\cite{Ablinger:2012ej} or unequal mass \cite{Ablinger:2011pb,BW13}. Genuine 3-loop topologies of
the ladder- and V-graph type have been studied in \cite{Ablinger:2012sm,Ablinger:2012qm}. These
calculations were accompanied by mathematical and computer-algebraic developments. In course of
this systematic use is made of higher hypergeometric functions, Mellin-Barnes techniques,
and modern summation theory \cite{SIGMA}. The latter are encoded in the packages {\tt Sigma,
EvaluateMultiSums} and {\tt SumProduction} \cite{CODE1}. Extensions of the harmonic sums
\cite{HSUM} and polylogarithms \cite{Remiddi:1999ew} to generalized harmonic sums
\cite{Moch:2001zr,Ablinger:2013cf} and the associated iterated integrals, the cyclotomic and
generalized cyclotomic sums and integrals \cite{Ablinger:2011te} were developed. Most recently
iterated integrals over root-valued letters were systematized. These functions and their relations
were encoded in the package {\tt HarmonicSums}, \cite{Ablinger:2013cf,Ablinger:2013hcp}, see also
\cite{Ablinger:2013jta}. All these developments were necessary to perform the present
calculations. They are, however, of much wider use.
In this note we report on progress being obtained during the last year.
\section{3-Loop OMEs with Two Fermion Lines of Equal Mass}
\noindent
A subset of graphs contributing to the 3-loop massive Wilson coefficients contains two fermion
lines with equal mass, characterized by the color factor $T_F^2 C_{F,A}$. These graphs may contain
new types of sums, which, to a wider extent also emerge in the V-topologies, see
Section~\ref{sec:BV}.
These are weighted inverse binomial sums. An example is given by the diagram in Figure~\ref{FIG3}.
\begin{figure}[t]
\begin{center}
\includegraphics[scale=0.5]{Johannes_Bluemlein1.eps}
\end{center}
\caption[]{An example for a graph with two massive fermion lines}\label{FIG3}
\end{figure}
The diagram is given by
\begin{eqnarray}
I(N) &=& \frac{1 + (-1)^N}{2} \Biggl\{
\frac{1}{45 \varepsilon^2 (N+1)}
- \frac{1}{\varepsilon} \left[\frac{S_1(N)}{90(N+1)}
+\frac{47 N^3 + 20 N^2 -67 N +40}{1800(N-1)N(N+1)^2}\right]
\nonumber\\
&&
+\frac{105 N^3 - 175 N^2 + 56 N + 96}{13440(N+1)^2(2N-3)(2N-1)4^N}\binom{2N}{N}\left[
\sum_{j=1}^N \frac{4^j S_1(j)}{\binom{2j}{j} j^2} - \sum_{j=1}^N \frac{4^j}{\binom{2j}{j}
j^3} - 7 \zeta_3\right]
\nonumber\\ &&
+ \frac{5264 N^3 -2409 N^2 -12770 N +3528}{100800(N+1)^2(2N-3)(2N-1)} S_1(N)
+ \frac{S_1^2(N)+S_2(N) +3 \zeta_2}{360(N+1)}
\nonumber\\ &&
+ \frac{S_3(N) - S_{2,1}(N) + 7 \zeta_3}{420(N+1)}
+ \frac{Q_0(N)}{2268000(N-1)^2N^2(N+1)^3(2N-3)(2N-1)}
\Biggr\}~.
\nonumber
\end{eqnarray}
Here and in the following $Q_i$ denote polynomials in $N$.
The terms $\propto 1/(2N-3), 1/(2N-1)$ deserve special attention. It can be shown that both are
removable poles in $I(N)$. It is generally expected that in QCD the rightmost singularity is
located at $N=1$. All basic topologies of this type contributing to the OME $A_{gg}^{(3)}$ have
been calculated.
\section{3-Loop OMEs with Two Fermion Lines of Different Mass}
\noindent
From the level of the 3-loop correction onwards, also graphs with two fermion lines of different
mass contribute. They require an extension of the renormalization programme of
Ref.~\cite{Bierenbaum:2009mv}. It turns out that the equal mass case is better included alongside
with the case of two different masses $m_c$ and $m_b$. The very close values of the charm and bottom
quark masses do not allow to treat charm massless at the scale $\mu^2 = m_b^2$ and one has
to deal with a two-mass scenario. Yet $\xi = m^2_c/m_b^2 \sim 1/10$ allows an expansion in $\xi$.
For the fixed moments $N = 2,4,6$ the calculation of all OMEs has been performed in
\cite{Ablinger:2011pb,BW13} after
mapping them to tadpoles and using the code {\tt qexp} \cite{QEXP}. First results were
derived for general values of $N$.
It is needless to say that also the matching conditions in the variable flavor scheme require
these new and no other expressions to stay in accordance with the renormalization group
equations inside the correct framework of perturbative QCD. Moreover, the matching scales
may vary considerably for different observables \cite{Blumlein:1998sh}.
\restylefloat{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[scale=0.8]{Johannes_Bluemlein2.eps}
\caption[]{Ladder graph with operator insertion.}\label{FIG1}
\end{center}
\end{figure}
\section{Ladder Graphs}
\noindent
First results have been obtained in the calculation of ladder graphs in the massive case, which
belong to the genuine 3-loop topologies \cite{Ablinger:2012qm}. Here the class of functions
appearing in intermediate and final results extends to generalized harmonic sums,
cf.~\cite{Ablinger:2013cf}. Let us consider the diagram in Figure~\ref{FIG1}.
The corresponding scalar graph yields
\begin{center}
\includegraphics[angle=0,width=\textwidth]{Johannes_Bluemlein3.eps}
\end{center}
It can be calculated with an extension of the method of hyperlogaritms \cite{Brown:2008um}
to the case of massive graphs with operator insertion \cite{Ablinger:2012qm} and is of weight {\sf
w = 5}. One notices the
emergence of terms growing individually like $\propto 2^N$, which would potentially imply an
instability at large $N$. However, the asymptotic expansion of the function $\hat{I}_4(N)$
shows that the corresponding terms cancel. In case of this and more involved topologies
both in the sum-representation and likewise also in that by iterated integrals the individual
entities of the representation, despite spanning the algebraic basis, partly act together forming
the physical structures. Individually they may not reflect the properties of the complete diagram.
\section{Massive Benz and V-Topologies}
\label{sec:BV}
\noindent
The method of hyperlogarithms is also suited to compute non-divergent diagrams of other
massive topologies such as Benz-diagrams and the V-topology. This has been done in
\cite{Ablinger:2012sm}.
\begin{figure}[t]
\begin{center}
\includegraphics[scale=0.85]{Johannes_Bluemlein4.eps}~~~
\includegraphics[scale=0.80]{Johannes_Bluemlein5.eps}
\end{center}
\caption{An example of a diagram with Benz subtopology and a diagram of the
V-topology}\label{FIG2}.
\end{figure}
The diagram shown in Figure~\ref{FIG2} (left) results in
\begin{eqnarray}
I(N)&=&
\frac {1} {(N+1) (N+2)}
\Biggl\{
\frac{2 \left(1-13 (-1)^N+(-1)^N \textcolor{red}{2^{3+N}}+N-7 (-1)^N N+3 (-1)^N
\textcolor{red}{2^{1+N}}N\right) }{(1+N) (2+N)} \zeta_3
\nonumber\\&&
+\frac{1}{(2+N)} S_3
+\frac{(-1)^N}{2 (2+N)} S_1^3
-\frac{(-1)^N (3+2 N)}{2 (1+N)^2 (2+N)} S_2
+\frac{5 (-1)^N}{2} S_2^2
\nonumber\\&&
+\frac{(-1)^N (3+2 N)}{2 (1+N)^2(2+N)} S_1^2
-\frac{(-1)^N}{2} S_2 S_1^2
+\frac{3 (-1)^N (4+3 N)}{(1+N) (2+N)} S_3
+3 (-1)^N S_4
+\frac{2}{(2+N)} S_{-2,1}
\nonumber\\&&
+{2 (-1)^N } \zeta_3 S_1 \left(2\right)
+\frac{2 (-1)^N (3+N) }{(1+N) (2+N)} S_{2,1}
-{12 (-1)^N } S_1 \zeta_3
\nonumber\\&&
+\frac{(-1)^N (5+7 N) }{2 (1+N) (2+N)} S_1 S_2
+{3 (-1)^N } S_1 S_3
+{4 (-1)^N } S_{2,1} S_1
-{4 (-1)^N} S_{3,1}
\nonumber\\&&
-\frac{4 \left((-1)^N \textcolor{red}{2^{2+N}}-3 \textcolor{red}{(-2)^N} N+3 (-1)^N
\textcolor{red}{2^{1+N}}
N\right)}{(1+N)
(2+N)} \textcolor{green}{S_{1,2}\left(\frac{1}{2},1\right)}
-{5 (-1)^N } S_{2,1,1}
\nonumber\\&&
+\frac{2 \left(-(-1)^N \textcolor{red}{2^{2+N}}-13 \textcolor{red}{(-2)^N} N+5 (-1)^N
\textcolor{red}{2^{1+N}}
N\right)}{(1+N) (2+N)} \textcolor{green}{S_{1,1,1}\left(\frac{1}{2},1,1\right)}
\nonumber\\&&
-{2 (-1)^N } \textcolor{green}{S_{1,1,2}\left(2,\frac{1}{2},1\right)}
-{(-1)^N} \textcolor{green}{S_{1,1,1,1}\left(2,\frac{1}{2},1,1\right)}
\Biggr\}~.
\nonumber
\end{eqnarray}
Also in this case the asymptotic expansion is regular. The corresponding representation in
$x$-space leads to generalized harmonic polylogarithms. In the case of the massive V-topology,
cf. Figure~\ref{FIG2} (right), further extensions arise. Here finite nested binomial and inverse
binomial sums weighted with generalized harmonic sums contribute. In $x$-space root-valued
letters contribute to the alphabet, extending those of the harmonic polylogarithms by 30 letters
in the case of the given graph. An example of a contributing sum is
\begin{eqnarray}
\sum_{i=1}^N \frac{1}{(i+1) \displaystyle \binom{2i}{i}}\sum_{j=1}^i \binom{2j}{j} \frac{1}{j}
S_2(j) &=& \int_0^1 dx \frac{x^N-1}{x-1} \Biggl[\frac{x}{2} \left(H^*_{\sf w_8,w_8,1,0}(x)
- \zeta_2 H^*_{\sf w_8,w_8}(x)\right)
\nonumber\\ &&
- \frac{x}{\sqrt{x-1/4}}\left(H^*_{\sf w_8,1,0}(x)
- \zeta_2 H^*_{\sf w_8}(x)\right)\Biggr]
\nonumber\\ &&
+\frac{2}{3} \zeta_3 \int_0^1 dx
\frac{\left(\tfrac{x}{4}\right)^N-1}{x-4} \left[\frac{x}{2} H^*_{\sf w_3}(x) -
\frac{x}{\sqrt{1-x}}\right],
\nonumber
\end{eqnarray}
with the letters ${\sf w_3, w_8}$ given by
\begin{eqnarray}
{\sf w_3} = \frac{1}{x \sqrt{1-x}},~~~~~~{\sf w_8} = \frac{1}{x \sqrt{x - 1/4}}.
\nonumber
\end{eqnarray}
Here the harmonic polylogarithms $H^*$ are defined as iterated integrals w.r.t. the point
$x=1$. In the case of the scalar integral of diagram Figure~\ref{FIG2} (right) potential
divergencies
$\propto 8^N, 4^N$
cancel, while the one $\propto 2^N$ remains. It is expected to cancel for the physical graphs.
\section{$\mathbf{O(\alpha_s^2)}$ Charged Current Corrections}
\noindent
Charged current data on heavy flavor production will improve the sea-quark densities. Therefore,
here the $O(\alpha_s^2)$ QCD corrections are desirable. In the present analyses \cite{Alekhin:2012ig}
the $O(\alpha_s)$ contributions, cf.~\cite{Gluck:1996ve,Blumlein:2011zu}, are used. Since the
charged current HERA data are located in the high $Q^2$ region, the asymptotic form of the
$O(\alpha_s^2)$ corrections yields a sufficient representation. It has been studied in Ref.
\cite{Buza:1997mg} before. Recently these corrections have been derived independently in
\cite{BHP13} giving the representations both in Mellin and $x$-space, extending the former analysis
and correcting some errors.
\section{Calculation of OMEs containing Benz graphs}
\vspace{1mm}\noindent
Recently we have calculated the massive 3-loop OMEs $A_{qq,Q}^{(3),\rm NS}$ and
$A_{qq,Q}^{(3),\rm NS,TR}$ for general values of $N$ and obtained the Wilson coefficient
$L_{qq,Q}^{(3),\rm NS}$, cf.~\cite{Bierenbaum:2009zt,Blumlein:2009rg}. The corresponding class
of graphs contains also massive Benz diagrams. An extension of the code {\tt Reduze~2}
\cite{Studerus:2009ye,vonManteuffel:2012np} to graphs with local operator insertions allowed to
reduce the corresponding integrals to master integrals, which have been calculated using
hypergeometric, Mellin-Barnes and advanced summation techniques \cite{SIGMA}. In course of this
we have also computed the complete 2-loop anomalous dimensions for transversity
$\gamma^{\pm,(1)}_{\rm NS,TR}$ \cite{TRANS2} and the contributions $\propto T_F$ of the 3-loop
anomalous dimensions $\gamma^{\pm,(2)}_{\rm NS}$ and $\gamma^{\pm,(2)}_{\rm NS,TR}$ in an ab
initio calculation. In the first case we confirm the results of
\cite{Larin:1993vu,Larin:1996wd,Retey:2000nq,Blumlein:2004xt,Moch:2004pa} and in the second
case our earlier moments \cite{Blumlein:2009rg} and the results in \cite{GRACEY,Bagaev:2012bw}.
Details of this calculation are given in \cite{NS1}. The calculation of further massive OMEs is
underway.
\section{Conclusions}
\noindent
Recently progress has been made towards the complete calculation of the 3-loop heavy
flavor corrections to DIS in the region $Q^2 \gg m^2$, including the matrix elements
needed in the variable flavor number scheme at general values of $N$. The $O(n_f T_F^2 C_{F,A})$
contributions have been completed. The gluonic $O(T_F^2)$ terms are currently calculated,
after all principal topologies have been solved. The renormalization in the 2-mass case has been
performed and for all OMEs the moments $N = 2,4,6$ were calculated. Also the setup for a VFNS in
case {\it both} charm and bottom become massless, has been derived.
No hierarchy exists for these terms individually.
This scheme is different from the former single mass VFNS. Diagrams of ladder-, V- and
Benz-topologies containing no singularities in $\varepsilon$ can be systematically calculated.
Here new functions occur, including a larger number of root-letters in iterated
integrals. All logarithmic contributions to the asymptotic heavy flavor Wilson coefficients
have been determined \cite{Bierenbaum:2010jp}.
After the two Wilson coefficients $L_{qq,Q}^{(3), \rm ps}$ and $L_{qg,Q}^{(3)}$ had
been computed in \cite{Ablinger:2010ty} we have calculated $L_{qq,Q}^{(3),\rm NS}$ and
$A_{qq,Q}^{(3),\rm NS,TR}$ as well as the associated 2- and 3-loop anomalous dimensions.
The calculation of further Wilson coefficients is underway.
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Bangkok-based Backyard Travel is presenting a new way to explore China's Yunnan Province, by train. The tour will call at some of the province's most popular destinations including Kunming, Dali and Lijiang as well as the Jade Dragon Mountain, Erhai Lake and the Three Pagodas at the foot of the Changshan Mountain.?
The six-day, five-night trip begins in Kunming, the provincial capital province with an important visit to the Yunnan Ethnic Minorities Museum. With more than 20 of China's 56 ethnic minorities living in the Yunnan, the museum is an essential stop on the rail tour.
From Kunming the journey moves by overnight train to Dali, home to the Bai hill tribes and several scenic regions including Erhai Lake and the Changshang Mountains. Dali's old town is a maze of white-walled homes fronted by beautiful porches which the local Bai community takes care of with great pride.
Lijiang is the next stop after a 2.5 hour train ride. Lijiang is known for its labyrinth of traditional redwood and brick houses topped with grey slate roofs interlaced with canals and stone bridges. Populated mainly by the Naxi people, Lijiang is lovingly referred to as the 'Big Inkpot' due to its network of canals that spread as if originating from an overturned pot of ink. While in Lijiang, travelers can visit the Mu Palace (named after Imperial Chinese warlords) and spend time in the old part of town to experience the local markets.
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Vacations By Rail is introducing six new rail tours in 2014 including independent and escorted tours in the USA, Canada, Europe and South America; they've also enhanced some of their most popular existing packages. The new vacations range from eight to 11 days.
The new 11-day Memphis to New Orleans Rail-River Journey, for instance, goes from Chicago to Memphis aboard Amtrak's City of New Orleans, and from Memphis to New Orleans via steamboat with ports of call including Vicksburg, Natchez, St. Francisville, Baton Rouge and Plantation Road.
The new 10-day Rail through the Canadian Rockies escorted rail tour features an all-daylight journey aboard the Rocky Mountaineer between Vancouver and Banff and a classic overnight rail trip between Jasper and Vancouver aboard VIA Rail's Canadian—and visits to Banff, Yoho, and Jasper National Parks, Lake Louise and Vancouver.
The new 10-day Best of Western Europe includes three nights in both London and Amsterdam and two nights in Paris with city orientation tours and transportation in each city; and, rail travel aboard the high-speed Eurostar and Thalys trains. Two new escorted tours, one in England and one in Italy, round out new offerings in Europe in 2014.
The eight-day escorted tour to Lima, Cusco, and Machu Picchu on the new Highlights of Peru vacation travels aboard the Vistadome train to Machu Picchu, a visit to the Sacred Valley, and sightseeing in Lima and Cusco.
Some of the additions and modifications to popular rail vacations include the addition of an excursion to Martha's Vineyard on the seasonal Fall Colors, Scenic Trains & Martha's Vineyard tour, and a revamped independent journey aboard Eastern Canada's premier Le Massif de Charlevoix train on the expanded eight-day Le Massif de Charlevoix & Quebec vacation.
Eurail is offering an Early Bird deal which offers non-European residents free travel days on Eurail Global Passes purchased from Feb. 1 through March 31.
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The "Early Bird" campaign offers two extra travel days for the 15-day Eurail Global Pass, three extra travel days for the 21-day Eurail Global Pass and five extra travel days for a one month Eurail Global continuous Pass.
Last year's Early Bird promotion proved very successful, boosting sales in March of the Global Pass by 96 percent. The opportunity to receive extra travel days motivated nearly twice as many customers to buy the Eurail Global Pass.
Generating more than 55 percent of total sales revenues, the Eurail Global Pass continues to be the most popular pass in the Eurail portfolio. With its 24 participating countries, it covers nearly most of Europe (including free or discounted travel on some major shipping lines) and gives travelers the largest variety of countries to choose from.
Rail riders can now book Italy's Trenitalia via AccesRail thanks to a new agreement signed by the two companies. The deal utilizes AccesRail's expertise to enable Trenitalia to make its train tickets available on the sales platforms of all major airlines of the world. To the travel agent, this means that they can now easily book air and rail segments in the airline primary screen display of all important GDSs.
And to the customer this means that they can now travel with air and rail segments on the same itinerary. For example, passengers landing in Rome will already have their Trenitalia ticket in-hand and can more easily board the train en route to their chosen destination.
Trenitalia's Le Frecce trains transport passengers throughout Italy. This intermodality, joining airlines and rail, allows airlines to extend their network by accessing secondary destinations with no or smaller airports, and railways are able to increase the visibility of their rail options while maximizing the load factor of their trains.
AccesRail, a division of ACP Rail International, specializes in intermodal travel and GDS distribution, making it possible for its partner railways to interact with airlines in this way around the world. Air-rail intermodal solutions have been successfully implemented for numerous railways including Renfe, Deutsche Bahn (DB), Norwegian State Railways (NSB), Swedish Railways (SJ), East Japan Railways, J R Kyushu, Dutch and Belgium Railways (NS/SNCB) and Thalys. | {
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<?php
namespace Grav\Plugin;
use Grav\Common\Plugin;
use RocketTheme\Toolbox\Event\Event;
use Grav\Plugin\SocialFeed\Twig\PostExtension;
use Grav\Plugin\SocialFeed\Manager\PostManager;
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* Class SocialFeedPlugin.
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class SocialFeedPlugin extends Plugin
{
/**
* @return array
*
* The getSubscribedEvents() gives the core a list of events
* that the plugin wants to listen to. The key of each
* array section is the event that the plugin listens to
* and the value (in the form of an array) contains the
* callable (or function) as well as the priority. The
* higher the number the higher the priority.
*/
public static function getSubscribedEvents()
{
return [
'onPluginsInitialized' => ['onPluginsInitialized', 0],
];
}
/**
* Initialize the plugin.
*/
public function onPluginsInitialized()
{
// Don't proceed if we are in the admin plugin
if ($this->isAdmin()) {
return;
}
require_once __DIR__.'/vendor/autoload.php';
$uri = $this->grav['uri'];
if (false !== strpos($uri->path(), '/social-posts')) {
return $this->getPosts();
} else {
$this->enable([
'onTwigExtensions' => ['onTwigExtensions', 0],
]);
}
}
/**
* Return posts as json.
*
* @return Json
*/
public function getPosts()
{
$manager = new PostManager();
header('Content-Type: application/json');
echo json_encode($manager->getPosts($this->grav['uri']->query(null, true)));
die;
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/**
* Add Twig Extensions.
*/
public function onTwigExtensions()
{
$this->grav['twig']->twig->addExtension(new PostExtension());
}
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La Binche-Chimay-Binche 2015, nota anche come Mémorial Frank Vandenbroucke, diciannovesima edizione della corsa, valida come evento dell'UCI Europe Tour 2015 categoria 1.1, si svolse il 6 ottobre 2015 per un percorso di 194,5 km, con partenza ed arrivo a Binche, in Belgio. Fu vinta dall'olandese Ramon Sinkeldam, al traguardo in 4h24'24" alla media di 44,14 km/h, precedendo l'altro olandese Pim Ligthart e il belga Tom Van Asbroeck, piazzatosi terzo.
Dei 140 ciclisti iscritti furono in 135 a partire e in 54 a completare la gara.
Squadre partecipanti
Ordine d'arrivo (Top 10)
Collegamenti esterni
Binche-Chimay-Binche
Ciclismo nel 2015 | {
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} | 4,822 |
{"url":"http:\/\/one4life.nl\/v9r23cg\/e77cb6-advanced-number-theory-pdf","text":"Advanced Number Theory When a person thinks of algebra, they typically think of a process used to solve polynomial equations. If Ais not equal to the zero ideal f0g, then the generator gis the smallest positive integer belonging to A. Probability theory is the most directly relevant mathematical background, and it is assumed that the reader has a working knowledge of measure-theory-based probability theory. It does not assume prior familiarity with abstract algebra. Advanced Number Theory Adithya B., Brian L., William W., Daniel X. 0000003821 00000 n 0000003019 00000 n 0000001711 00000 n 0000000016 00000 n 0000003058 00000 n number of lattice points in the rectangle, which is just p 1 2 q 1 2 , as desired. endstream Problems in Advanced Convex Number Theory G. Cartan, B. Kronecker, L. Huygens and O. Heaviside Abstract Let us assume there exists a finite, 0000000962 00000 n \u2026 The result was a broadly based international gathering of leading number theorists who reported on recent advances in both classical analytic number theory as well as in related parts of number theory and algebraic geometry. 1791 15 View Math62708.pdf from PLN 50 at Harvard University. Notions of primality and divisibility are indeed quite clas-sical, and the ancients even knew a great deal about some relatively stream We are really very thankful to Mr. Anwar Khan for providing these notes and appreciates his effort to publish these notes on MathCity.org It covers the complete syllabus of Advanced Analysis paper of MSc Mathematics. It is now being made available in book form with an appendix\u2013an English translation of Siegel\u2019s paper \u201cBerechnung von Zetafunktionen an ganzzahligen Stellen\u201d One of the unique characteristics of these notes is the careful choice of topics and its importance in the theory of numbers. There is nothing original to me in the notes. Also some approaches to number theory start with inversion, and define division using inversion without discussing how it relates to integer division, which is another reason $$\/$$ is often avoided. %\ufffd\ufffd\ufffd\ufffd download 1 file . This lecture note is an elementary introduction to number theory \u2026 If you think about it, it is hard to give a satisfactory de nition of any area of mathematics that would make much sense to someone who has not taken one or more courses in it. Author: Richard A. Mollin Publisher: CRC Press ISBN: 1420083295 Size: 78.12 MB Format: PDF, Docs Category : Computers Languages : en Pages : 440 View: 5533 Get Book. 1 In some sense three-dimensional CS was the rst and most important example of a topological quantum eld theory. Finite continued fractions 17 9. CRC Press, 2009. 0000001176 00000 n startxref The ideals that are listed in Example 4 are all generated by a single number g. We next show that all ideals of Z have this property. is known as the father of analytic number theory. SINGLE PAGE PROCESSED JP2 ZIP download. The freedom is given in the last two chapters because of the advanced nature of the topics that are presented. Primes and factorization 12 7. Introduction to Number Theory Lecture Notes Adam Boocher (2014-5), edited by Andrew Ranicki (2015-6) December 4, 2015 1 Introduction (21.9.2015) These notes will cover all material presented during class. 1791 0 obj <> endobj %%EOF Better titles for this book would be \"A Second Course in Number Theory\" or \"Introduction to quadratic forms and quadratic fields\". These are the notes of the course MTH6128, Number Theory, which I taught at Queen Mary, University of London, in the spring semester of 2009. endobj G68ia\ufffd1J\ufffd'\ufffdm\ufffd\u0216\ufffd\ufffd%\ufffd\ufffd\ufffd!$r.J\ufffdL\ufffd\ufffd\\\u0205\ufffdc\ufffd%D\ufffd\ufffd2\ufffdAr\ufffdr\ufffd\ufffd\ufffd\ufffd\ufffd\u06a2c\ufffd;y)\ufffdR!o \ufffdk\ufffd\ufffd(m\u0592g. people call number theory are related, in fact deeply and increasingly so over time. 0000001440 00000 n endobj The natural numbers 1 2. Dive into this fun collection to play with numbers like never before, and start unlocking the connections that are the foundation of Number Theory. << \/Names 1175 0 R \/OpenAction 986 0 R \/PageLabels << \/Nums [ 0 << \/P (1) >> 1 << \/P (2) >> 2 << \/P (3) >> 3 << \/P (4) >> 4 << \/P (5) >> 5 << \/P (6) >> 6 << \/P (7) >> 7 << \/P (8) >> 8 << \/P (9) >> 9 << \/P (10) >> 10 << \/P (11) >> 11 << \/P (12) >> 12 << \/P (13) >> 13 << \/P (14) >> 14 << \/P (15) >> 15 << \/P (16) >> 16 << \/P (17) >> 17 << \/P (18) >> 18 << \/P (19) >> 19 << \/P (20) >> 20 << \/P (21) >> 21 << \/P (22) >> 22 << \/P (23) >> 23 << \/P (24) >> 24 << \/P (25) >> 25 << \/P (26) >> 26 << \/P (27) >> 27 << \/P (28) >> 28 << \/P (29) >> 29 << \/P (30) >> 30 << \/P (31) >> 31 << \/P (32) >> 32 << \/P (33) >> 33 << \/P (34) >> 34 << \/P (35) >> 35 << \/P (36) >> 36 << \/P (37) >> 37 << \/P (38) >> 38 << \/P (39) >> 39 << \/P (40) >> ] >> \/PageMode \/FullScreen \/Pages 1156 0 R \/Type \/Catalog >> Number Theory .-WACLAW SIERPINSKI \"250 Problems in Elementary Number Theory\" presents problems and their solutions in five specific areas of this branch of mathe\u00ad matics: divisibility of numbers, relatively prime numbers, arithmetic progressions, prime and composite numbers, and Diophantic equations. 4 Number Theory I: Prime Numbers Number theory is the mathematical study of the natural numbers, the positive whole numbers such as 2, 17, and 123. Congruences 9 6. shed light on analytic number theory, a subject that is rarely seen or approached by undergraduate students. The ramification theory needed to understand the properties of conductors from the point of view of the Herbrand distribution is given in C.J. 0000003302 00000 n 9\/9 \u00a71Quadratic Residues When we are working with natural numbers, it is easy to tell if a number has a square root - just check if it is a perfect square. Advanced Number Theory: This course will focus on local fields and Galois cohomology. Moreno, Advanced Analytic Number Theory [127]. The only prerequisite is Math 602; students may or may not have taken Math 620. x\ufffdcbd\ufffdgb8$\ufffdo fE\ufffd\u0132\ufffd~@\ufffd\ufffd,H\ufffd1w?\ufffd8$\ufffdB\ufffd\ufffd\ufffd3\ufffd0\ufffdcbL\ufffdR\ufffd\ufffd8JS\ufffd}\ufffd@\ufffd\ufffd\ufffd2 \ufffd\ufffd Exploring one of the most dynamic areas of mathematics, Advanced Number Theory with Applications covers a wide range of algebraic, analytic, combinatorial, cryptographic, and geometric aspects of number theory. Book Description: Exploring one of the most dynamic areas of mathematics, Advanced Number Theory with Applications covers a wide range of algebraic, analytic, combinatorial, cryptographic, and geometric aspects of number theory. View Adv prac 1.pdf from MATH 042 at Chitkara University. 0 download 1 file . These lectures have been compiled from a variety of sources, mainly from the recommended books: Congruences modulo a prime 14 8. 974 0 obj stream <<9EA8F97A12C8AB4BAA909E26D8DDAFEF>]>> Advanced Calculus & Number Theory #322 . This chapter lays the foundations for our study of the theory of numbers by weaving together the themes of prime numbers, integer factorization, and the distribution of primes. 0000002596 00000 n However, it is much harder when we are working with modular arithmetic. The Euclidean Algorithm and the method of back-substitution 4 4. x\ufffdbbqda\ufffdc\ufffde@ ^\ufffdrL` )\u069d\ufffd\ufffd\ufffd% \u8d8f%\ufffdX\ufffd\ufffd9?\ufffd\ufffd\ufffd\ufffd\ufffd\\\u042e\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdP\ufffd\ufffd\ufffd\ufffd\u01170\ufffd\ufffd10\ufffdp\ufffd\ufffdw\ufffd\ufffd\ufffda\ufffd0dX\ufffd\ufffdA#\ufffd\ufffd\ufffd\ufffdsLn:\ufffd$\ufffdHW\ufffd\ufffd]6~\ufffd\u0392\ufffd*\ufffd\ufffd\ufffd\ufffd. The integers 3 3. Adithya B., Brian L., William W., Daniel X. \u201cAdvanced Analytic Number Theory\u201d was \ufb01rst published by the Tata Insti-tute of Fundamental Research in their Lecture Notes series in 1961. 0000000611 00000 n trailer It is not a very advanced book in the sense that required background is only a one-semester course in number theory. xref 0000002175 00000 n Advanced Analysis: Handwritten Notes These notes are provided by Mr. Anwar Khan. In nite continued fractions 19 10. Chapter 1. Basic Number Theory 1 1. The definitions and elementary properties of the absolute Weil group of a number \u2026 It \u2026 If numbers aren't beautiful, we don't know what is. 0000003564 00000 n 0000005851 00000 n In Section 1.1, we rigorously prove that the Theory of Statistics c 2000\u20132020 James E. Gentle Unit Content Objective Performance Indicator Performance Task State Standards Code: Functions of Several Variables The students will be - Desc ribe the level surface of - in-class problems 2.5.11.A; x\ufffd\ufffd[Ks7\ufffd\ufffdW\ufffd\ufffd^V\\$\ufffd\ufffdL\ufffd3yT\ufffd\ufffdz\ufffd\ufffdvrPd5QcKEN\ufffd\ufffd\ufffdV\/\u06f2\ufffd\ufffd\"7\u0771\ufffd\ufffd\ufffdH\ufffd#\ufffd%\ufffda\ufffd#*9\ufffd2\ufffd,\ufffdB)x*\u0150ul!X\ufffd\ufffd Example: $$2\\times 3+4(5^{ \u2026 Despite their ubiquity and apparent sim-plicity, the natural integers are chock-full of beautiful ideas and open problems. endobj Cohn_Harvey_-_Advanced_Number_Theory Identifier-ark ark:\/13960\/t76t4pz82 Isbn 048664023X Lccn 80065862 Ocr ABBYY FineReader 11.0 Openlibrary OL4120501M Openlibrary_edition ... PDF download. One \u2026 Now, substituting this equality back into our initial expression, we get that p q q p = ( 1) (p 1)(q 1) 4, as desired. 1805 0 obj <>stream \ufffd Text: Elementary Number Theory, Charles vanden Eynden, 2nd edition, Waveland Press, ISBN 1-57766-445-0 (McGraw-Hill ISBN 0-07232-571-2 is the same edition). Proof. - 440 pages. Theory of Numbers Lecture Notes. Chapter 1 covers this theory at a fairly rapid pace. number theory, postulates a very precise answer to the question of how the prime numbers are distributed. << \/Linearized 1 \/L 286356 \/H [ 4574 570 ] \/O 977 \/E 53044 \/N 40 \/T 280252 >> There is, in addition, a section of 0000003900 00000 n Theorem 1.8 Every ideal Ais generated by a unique nonnegative number g, that is A= gZ = fngjn2Zg. Advanced Number Theory with Applications by Richard A. Mollin CRC Press, Taylor & Francis Groups 2010 ISBN: 978-1-4200-8328-6 Fan Junjie Bertrand Centre for Strategic Infocomm Technologies 30 May 2011 1 Overview of Book This is the sequel to the introductory text 'Fundamental Number Theory with Applications' written by a Number theory (or arithmetic or higher arithmetic in older usage) is a branch of pure mathematics devoted primarily to the study of the integers and integer-valued functions.German mathematician Carl Friedrich Gauss (1777\u20131855) said, \"Mathematics is the queen of the sciences\u2014and number theory is the queen of mathematics.\" Student Inquiries | \u0627\u0633\u062a\u0641\u0633\u0627\u0631\u0627\u062a \u0627\u0644\u0637\u0644\u0627\u0628: registration@zuj.edu.jo: registration@zuj.edu.jo The tabular method 7 5. Start Divisibility. 972 0 obj Modern Number theory has evolved through several stages in the past two millennia. Al-Zaytoonah University of Jordan P.O.Box 130 Amman 11733 Jordan Telephone: 00962-6-4291511 00962-6-4291511 Fax: 00962-6-4291432. Number Theory Warmups. Advanced students, mathematicians and number theorists will welcome this stimulating treatment of advanced number theory, which approaches the complex topic of algebraic number theory from a historical standpoint, taking pains to show the reader how concepts, definitions and theories have evolved during the last two centuries. Course Outline Number theory is essentially the study of the natural numbers 1,2,3,...and their properties. We will follow convention, and reserve the \\(\/$$ symbol for integer division. Email: president@zuj.edu.jo. %PDF-1.5 eld theory. 973 0 obj Lesson 13: Advanced Number Theory September 202015\/40 << \/Type \/ObjStm \/Length 1696 \/Filter \/FlateDecode \/N 96 \/First 974 >> %PDF-1.4 %\ufffd\ufffd\ufffd\ufffd 975 0 obj 11\/15\/2020 Advanced Number Theory Advanced Number Theory Click on a question number to \u2026 A primary focus of number theory is the study of prime numbers, which can be See the contents of the notes given below to see the topics covered by these notes. The course was designed by Su-san McKay, and developed by Stephen Donkin, Ian Chiswell, Charles Leedham- << \/Type \/XRef \/Length 79 \/Filter \/FlateDecode \/DecodeParms << \/Columns 4 \/Predictor 12 >> \/W [ 1 2 1 ] \/Index [ 972 204 ] \/Info 106 0 R \/Root 974 0 R \/Size 1176 \/Prev 280253 \/ID [<705a870930ca72e764245e092d083645>] >> It is recommended for graduate students in algebra, including those interested in number theory and in algebraic geometry from an algebraic point of view. 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Rst and most important example of a process used to solve polynomial equations # 322 in the rectangle, is.","date":"2021-08-01 00:17:29","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.46575927734375, \"perplexity\": 2007.5545197026772}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-31\/segments\/1627046154127.53\/warc\/CC-MAIN-20210731234924-20210801024924-00373.warc.gz\"}"} | null | null |
/*
* File: queue.h
* -------------
* This file exports the <code>Queue</code> class, a collection
* in which values are ordinarily processed in a first-in/first-out
* (FIFO) order.
*
* @version 2016/09/24
* - refactored to use collections.h utility functions
* - added iterators begin(), end()
* @version 2016/09/22
* - optimized equals, ==, != to avoid deep-copy
* @version 2016/08/10
* - added constructor support for std initializer_list usage, such as {1, 2, 3}
* @version 2016/08/04
* - fixed operator >> to not throw errors
* @version 2015/07/05
* - using global hashing functions rather than global variables
* @version 2014/11/13
* - added comparison operators ==, !=, <, etc.
* - added add() method as synonym for enqueue()
* - added remove() method as synonym for dequeue()
* - added template hashCode function
* - optimized some functions (operator <<, toStlQueue) to avoid making unneeded deep copy
* @version 2014/10/10
* - removed dependency on 'using namespace' statement
* - removed usage of __foreach macro
*/
#ifndef _queue_h
#define _queue_h
#include <deque>
#include <initializer_list>
#include <iterator>
#include <queue>
#include "collections.h"
#include "error.h"
#include "hashcode.h"
#include "vector.h"
/*
* Class: Queue<ValueType>
* -----------------------
* This class models a linear structure called a <b><i>queue</i></b>
* in which values are added at one end and removed from the other.
* This discipline gives rise to a first-in/first-out behavior (FIFO)
* that is the defining feature of queues.
*/
template <typename ValueType>
class Queue {
public:
/*
* Constructor: Queue
* Usage: Queue<ValueType> queue;
* ------------------------------
* Initializes a new empty queue.
*/
Queue();
/*
* Constructor: Queue
* Usage: Queue<ValueType> queue {1, 2, 3};
* ----------------------------------------
* Initializes a new queue that stores the given elements from front-back.
*/
Queue(std::initializer_list<ValueType> list);
/*
* Destructor: ~Queue
* ------------------
* Frees any heap storage associated with this queue.
*/
virtual ~Queue();
/*
* Method: add
* Usage: queue.add(value);
* ------------------------
* Adds <code>value</code> to the end of the queue.
* A synonym for the enqueue method.
*/
void add(const ValueType& value);
/*
* Method: back
* Usage: ValueType last = queue.back();
* -------------------------------------
* Returns the last value in the queue by reference.
*/
const ValueType& back() const;
/*
* Method: clear
* Usage: queue.clear();
* ---------------------
* Removes all elements from the queue.
*/
void clear();
/*
* Method: dequeue
* Usage: ValueType first = queue.dequeue();
* -----------------------------------------
* Removes and returns the first item in the queue.
*/
ValueType dequeue();
/*
* Method: enqueue
* Usage: queue.enqueue(value);
* ----------------------------
* Adds <code>value</code> to the end of the queue.
*/
void enqueue(const ValueType& value);
/*
* Method: equals
* Usage: if (queue.equals(queue2)) ...
* ------------------------------------
* Compares two queues for equality.
* Returns <code>true</code> if this queue contains exactly the same
* values as the given other queue.
* Identical in behavior to the == operator.
*/
bool equals(const Queue<ValueType>& queue2) const;
/*
* Method: front
* Usage: ValueType first = queue.front();
* ---------------------------------------
* Returns the first value in the queue by reference.
*/
const ValueType& front() const;
/*
* Method: isEmpty
* Usage: if (queue.isEmpty()) ...
* -------------------------------
* Returns <code>true</code> if the queue contains no elements.
*/
bool isEmpty() const;
/*
* Method: peek
* Usage: ValueType first = queue.peek();
* --------------------------------------
* Returns the first value in the queue, without removing it. For
* compatibility with the STL classes, this method is also exported
* under the name <code>front</code>, in which case it returns the
* value by reference.
*/
const ValueType& peek() const;
/*
* Method: remove
* Usage: ValueType first = queue.remove();
* ----------------------------------------
* Removes and returns the first item in the queue.
* A synonym for the dequeue method.
*/
ValueType remove();
/*
* Method: size
* Usage: int n = queue.size();
* ----------------------------
* Returns the number of values in the queue.
*/
int size() const;
/*
* Returns an STL deque object with the same elements as this Queue.
*/
std::queue<ValueType> toStlDeque() const;
/*
* Returns an STL queue object with the same elements as this Queue.
*/
std::queue<ValueType> toStlQueue() const;
/*
* Method: toString
* Usage: string str = queue.toString();
* -------------------------------------
* Converts the queue to a printable string representation.
*/
std::string toString() const;
/*
* Operator: ==
* Usage: queue1 == queue2
* -------------------
* Returns <code>true</code> if <code>queue1</code> and <code>queue2</code>
* contain the same elements.
*/
bool operator ==(const Queue& queue2) const;
/*
* Operator: !=
* Usage: queue1 != queue2
* -------------------
* Returns <code>true</code> if <code>queue1</code> and <code>queue2</code>
* do not contain the same elements.
*/
bool operator !=(const Queue& queue2) const;
/*
* Operators: <, >, <=, >=
* Usage: queue1 < queue2 ...
* --------------------------
* Relational operators to compare two queues.
* The <, >, <=, >= operators require that the ValueType has a < operator
* so that the elements can be compared pairwise.
*/
bool operator <(const Queue& queue2) const;
bool operator <=(const Queue& queue2) const;
bool operator >(const Queue& queue2) const;
bool operator >=(const Queue& queue2) const;
template <typename T>
friend int hashCode(const Queue<T>& s);
template <typename T>
friend std::ostream& operator <<(std::ostream& os, const Queue<T>& queue);
/* Private section */
/**********************************************************************/
/* Note: Everything below this point in the file is logically part */
/* of the implementation and should not be of interest to clients. */
/**********************************************************************/
/*
* Implementation notes: Queue data structure
* ------------------------------------------
* The Queue class is implemented using a ring buffer.
*/
private:
/* Instance variables */
Vector<ValueType> ringBuffer;
int count;
int capacity;
int head;
int tail;
/* Private functions */
void expandRingBufferCapacity();
int queueCompare(const Queue& queue2) const;
/*
* Iterator support
* ----------------
* The classes in the StanfordCPPLib collection implement input
* iterators so that they work symmetrically with respect to the
* corresponding STL classes.
*/
class iterator : public std::iterator<std::input_iterator_tag, ValueType> {
public:
iterator(const Queue* gp, int index) {
this->gp = gp;
this->index = index;
}
iterator(const iterator& it) {
this->gp = it.gp;
this->index = it.index;
}
iterator& operator ++() {
index = (index + 1) % gp->capacity;
return *this;
}
iterator operator ++(int) {
iterator copy(*this);
operator++();
return copy;
}
bool operator ==(const iterator& rhs) {
return gp == rhs.gp && index == rhs.index;
}
bool operator !=(const iterator& rhs) {
return !(*this == rhs);
}
const ValueType& operator *() {
return gp->ringBuffer[index];
}
ValueType* operator ->() {
return &gp->ringBuffer[index];
}
private:
const Queue* gp;
int index;
};
public:
iterator begin() const {
return iterator(this, /* index */ head);
}
iterator end() const {
return iterator(this, /* index */ tail);
}
};
/*
* Implementation notes: Queue data structure
* ------------------------------------------
* The array-based queue stores the elements in successive index
* positions in a vector, just as a stack does. What makes the
* queue structure more complex is the need to avoid shifting
* elements as the queue expands and contracts. In the array
* model, this goal is achieved by keeping track of both the
* head and tail indices. The tail index increases by one each
* time an element is enqueued, and the head index increases by
* one each time an element is dequeued. Each index therefore
* marches toward the end of the allocated vector and will
* eventually reach the end. Rather than allocate new memory,
* this implementation lets each index wrap around back to the
* beginning as if the ends of the array of elements were joined
* to form a circle. This representation is called a ring buffer.
*/
static const int INITIAL_CAPACITY = 10;
/*
* Implementation notes: Queue constructor
* ---------------------------------------
* The constructor must allocate the array storage for the queue
* elements and initialize the fields of the object.
*/
template <typename ValueType>
Queue<ValueType>::Queue() {
clear();
}
template <typename ValueType>
Queue<ValueType>::Queue(std::initializer_list<ValueType> list) {
clear();
for (const ValueType& value : list) {
add(value);
}
}
/*
* Implementation notes: ~Queue destructor
* ---------------------------------------
* All of the dynamic memory is allocated in the Vector class,
* so no work is required at this level.
*/
template <typename ValueType>
Queue<ValueType>::~Queue() {
// empty
}
template <typename ValueType>
void Queue<ValueType>::add(const ValueType& value) {
enqueue(value);
}
template <typename ValueType>
const ValueType& Queue<ValueType>::back() const {
if (count == 0) {
error("Queue::back: Attempting to read back of an empty queue");
}
return ringBuffer[(tail + capacity - 1) % capacity];
}
template <typename ValueType>
void Queue<ValueType>::clear() {
capacity = INITIAL_CAPACITY;
ringBuffer = Vector<ValueType>(capacity);
head = 0;
tail = 0;
count = 0;
}
/*
* Implementation notes: dequeue, peek
* -----------------------------------
* These methods must check for an empty queue and report an error
* if there is no first element.
*/
template <typename ValueType>
ValueType Queue<ValueType>::dequeue() {
if (count == 0) {
error("Queue::dequeue: Attempting to dequeue an empty queue");
}
ValueType result = ringBuffer[head];
head = (head + 1) % capacity;
count--;
return result;
}
template <typename ValueType>
void Queue<ValueType>::enqueue(const ValueType& value) {
if (count >= capacity - 1) {
expandRingBufferCapacity();
}
ringBuffer[tail] = value;
tail = (tail + 1) % capacity;
count++;
}
template <typename ValueType>
bool Queue<ValueType>::equals(const Queue<ValueType>& queue2) const {
return stanfordcpplib::collections::equals(*this, queue2);
}
template <typename ValueType>
const ValueType& Queue<ValueType>::front() const {
if (count == 0) {
error("Queue::front: Attempting to read front of an empty queue");
}
return ringBuffer[head];
}
template <typename ValueType>
bool Queue<ValueType>::isEmpty() const {
return count == 0;
}
template <typename ValueType>
const ValueType& Queue<ValueType>::peek() const {
if (count == 0) {
error("Queue::peek: Attempting to peek at an empty queue");
}
return ringBuffer.get(head);
}
template <typename ValueType>
ValueType Queue<ValueType>::remove() {
// this isEmpty check is also done in dequeue(), but we repeat it
// here so that the possible error message will be more descriptive.
if (isEmpty()) {
error("Queue::remove: Attempting to remove from an empty queue");
}
return dequeue();
}
template <typename ValueType>
int Queue<ValueType>::size() const {
return count;
}
template <typename ValueType>
std::queue<ValueType> Queue<ValueType>::toStlDeque() const {
std::deque<ValueType> result;
for (int i = 0; i < count; i++) {
result.push_back(ringBuffer[(head + i) % capacity]);
}
return result;
}
template <typename ValueType>
std::queue<ValueType> Queue<ValueType>::toStlQueue() const {
std::queue<ValueType> result;
for (int i = 0; i < count; i++) {
result.push(ringBuffer[(head + i) % capacity]);
}
return result;
}
template <typename ValueType>
std::string Queue<ValueType>::toString() const {
std::ostringstream os;
os << *this;
return os.str();
}
/*
* Implementation notes: expandRingBufferCapacity
* ----------------------------------------------
* This private method doubles the capacity of the ringBuffer vector.
* Note that this implementation also shifts all the elements back to
* the beginning of the vector.
*/
template <typename ValueType>
void Queue<ValueType>::expandRingBufferCapacity() {
Vector<ValueType> copy = ringBuffer;
ringBuffer = Vector<ValueType>(2 * capacity);
for (int i = 0; i < count; i++) {
ringBuffer[i] = copy[(head + i) % capacity];
}
head = 0;
tail = count;
capacity *= 2;
}
template <typename ValueType>
int Queue<ValueType>::queueCompare(const Queue& queue2) const {
if (this == &queue2) {
return 0;
}
for (int i1 = 0, i2 = 0;
i1 < count && i2 < queue2.count;
i1++, i2++) {
if (ringBuffer[(head + i1) % capacity] < queue2.ringBuffer[(queue2.head + i2) % queue2.capacity]) {
return -1;
} else if (queue2.ringBuffer[(queue2.head + i2) % queue2.capacity] < ringBuffer[(head + i1) % capacity]) {
return 1;
}
}
if (count < queue2.count) {
return -1;
} else if (count > queue2.count) {
return 1;
} else {
return 0;
}
}
template <typename ValueType>
bool Queue<ValueType>::operator ==(const Queue& queue2) const {
return equals(queue2);
}
template <typename ValueType>
bool Queue<ValueType>::operator !=(const Queue& queue2) const {
return !equals(queue2);
}
template <typename ValueType>
bool Queue<ValueType>::operator <(const Queue& queue2) const {
return queueCompare(queue2) < 0;
}
template <typename ValueType>
bool Queue<ValueType>::operator <=(const Queue& queue2) const {
return queueCompare(queue2) <= 0;
}
template <typename ValueType>
bool Queue<ValueType>::operator >(const Queue& queue2) const {
return queueCompare(queue2) > 0;
}
template <typename ValueType>
bool Queue<ValueType>::operator >=(const Queue& queue2) const {
return queueCompare(queue2) >= 0;
}
template <typename ValueType>
std::ostream& operator <<(std::ostream& os, const Queue<ValueType>& queue) {
os << "{";
if (!queue.isEmpty()) {
writeGenericValue(os, queue.ringBuffer[queue.head], /* forceQuotes */ true);
for (int i = 1; i < queue.count; i++) {
os << ", ";
writeGenericValue(os, queue.ringBuffer[(queue.head + i) % queue.capacity], /* forceQuotes */ true);
}
}
os << "}";
return os;
}
template <typename ValueType>
std::istream& operator >>(std::istream& is, Queue<ValueType>& queue) {
ValueType element;
return stanfordcpplib::collections::readCollection(is, queue, element, /* descriptor */ "Queue::operator >>");
}
/*
* Template hash function for queues.
* Requires the element type in the queue to have a hashCode function.
*/
template <typename T>
int hashCode(const Queue<T>& q) {
int code = hashSeed();
for (int i = 0; i < q.count; i++) {
code = hashMultiplier() * code + hashCode(q.ringBuffer[(q.head + i) % q.capacity]);
}
return int(code & hashMask());
}
#include "private/init.h" // ensure that Stanford C++ lib is initialized
#endif // _queue_h
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,263 |
Émeringes és un municipi francès situat al departament del Roine i a la regió d'Alvèrnia-Roine-Alps. L'any 2007 tenia 226 habitants.
Demografia
Població
El 2007 la població de fet d'Émeringes era de 226 persones. Hi havia 88 famílies de les quals 20 eren unipersonals (12 homes vivint sols i 8 dones vivint soles), 32 parelles sense fills, 32 parelles amb fills i 4 famílies monoparentals amb fills.
La població ha evolucionat segons el següent gràfic:
Habitants censats
Habitatges
El 2007 hi havia 124 habitatges, 94 eren l'habitatge principal de la família, 24 eren segones residències i 6 estaven desocupats. 111 eren cases i 13 eren apartaments. Dels 94 habitatges principals, 61 estaven ocupats pels seus propietaris, 25 estaven llogats i ocupats pels llogaters i 9 estaven cedits a títol gratuït; 5 tenien dues cambres, 12 en tenien tres, 28 en tenien quatre i 50 en tenien cinc o més. 80 habitatges disposaven pel capbaix d'una plaça de pàrquing. A 35 habitatges hi havia un automòbil i a 51 n'hi havia dos o més.
Piràmide de població
La piràmide de població per edats i sexe el 2009 era:
Economia
El 2007 la població en edat de treballar era de 153 persones, 125 eren actives i 28 eren inactives. De les 125 persones actives 120 estaven ocupades (66 homes i 54 dones) i 5 estaven aturades (1 home i 4 dones). De les 28 persones inactives 17 estaven jubilades, 3 estaven estudiant i 8 estaven classificades com a «altres inactius».
Ingressos
El 2009 a Émeringes hi havia 92 unitats fiscals que integraven 225 persones, la mediana anual d'ingressos fiscals per persona era de 18.399 €.
Activitats econòmiques
Dels 10 establiments que hi havia el 2007, 5 eren d'empreses de comerç i reparació d'automòbils, 2 d'empreses d'hostatgeria i restauració, 1 d'una empresa de serveis i 2 d'empreses classificades com a «altres activitats de serveis».
Dels 2 establiments de servei als particulars que hi havia el 2009, 1 era un taller de reparació d'automòbils i de material agrícola i 1 saló de bellesa.
L'any 2000 a Émeringes hi havia 20 explotacions agrícoles que ocupaven un total de 126 hectàrees.
Equipaments sanitaris i escolars
El 2009 hi havia una escola elemental integrada dins d'un grup escolar amb les comunes properes formant una escola dispersa.
Poblacions més properes
El següent diagrama mostra les poblacions més properes.
Referències
Résumé statistique Fitxa resum de dades estadístiques d'Émeringes a l'INSEE.
Évolution et structure de la population Fitxa amb el detall de dades d'Émeringes a l'INSEE
France par commune Dades detallades de tots els municipis de França accessibles a partir del mapa.
Municipis del Roine | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,304 |
News Segment
Posted on February 16, 2007 by Skid in Needs Reviewed, Uncategorized // 0 Comments
WILL ROCK STAR SUPERNOVE TAKE OFF?:
In the world of professional bull riding, the goal is eight seconds. In the world of Rock Star Supernova, the reality is eight minutes.
That's how much phone time Supernova drummer Tommy Lee gave this music journalist during a recent interview. The music industry standard is the "15-minute phoner," although 20-minute interviews are fairly commonplace as well.
Yet, very few musicians are as busy as Lee. The 44-year-old native of Athens, Greece is a successful author, reality TV star and legendary rock musician. He also maintains the type of challenging social calendar that would put Paris Hilton or Lindsay Lohan to shame.
For the past 25 years, Lee's main claim to fame has been as the heart and soul of Motley Crue. His newer fans, however, know him primarily for his involvement in Rock Star Supernova, a reality TV-spawned supergroup that performs Wednesday at Arco Arena in Sacramento and Thursday at Oracle Arena in Oakland.
To an outsider, it seems like Lee never slows down. That perception, the drummer says, is pretty much right on the money.
"If I am sitting still, that's when I get in trouble," he says from Los Angeles. "I just like to stay busy and stay creative. If I'm not playing music, I'm playing with my kids or something."
Even a marathon man enjoys some new scenery now and again. Indeed, the key to Lee's vitality has been change. Before he tires of one thing, he's off to the next — be it co-authoring a book ("Tommyland"), starting up a new band (the short-lived Methods of Mayhem), starring in a reality TV show (NBC's "Tommy Lee Goes to College") or joining the other original members of Motley Crue for a lengthy reunion tour.
It was after the latter that Lee found himself really in need of a change. From early 2005 to mid-2006, he logged in more than 130 dates with his fellow Crue men, vocalist Vince Neil, bassist Nikki Sixx and guitarist Mick Mars.
He would find what he was looking for when he was asked to take part in the second season of the CBS reality TV show "Rock Star." The first season, during which singers competed to front the band INXS, proved tremendously popular with viewers.
For the second go-round, vocalists battled to lead a newly formed supergroup featuring Lee, Walnut Creek bassist Jason Newsted (formerly of Metallica) and guitarist Gilby Clarke (formerly of Guns N' Roses). The show was co-hosted by Dave Navarro, the ex-Jane's Addiction guitarist whose band Panic Channel will open Rock Star Supernova's concerts in Oakland and Sacramento.
The series ran from July to September last year and was a hit with television audiences. Critics were less enthralled. They complained that the sanitized talent show format, in which the singers were judged by the band members, was a very unnatural and unorthodox way to start a new group.
Most of the competitors seemed straight out of the "American Idol" fold, only with a lot more leather and hairspray, which left many wondering how an established rock vocalist would have fared on the show. Indeed, Lee admits that the man he's backed for most of his career — Crue's Neil — likely wouldn't have made the cut.
"That's really not the kind of guy we were looking for," he says. "We were looking for something new and fresh — something 2007."
It turned out that they were looking for Canadian singer Lukas Rossi, a choice that surprised many viewers. Rossi, a former cook at a Hooters restaurant, was considered by some to be a huge underdog among the final four contestants. Yet, Lee says he had a hunch that Rossi would win the prize basically from the first time he heard him sing.
"Lucas busted out this original piece of music called `Headspin,' " Lee recalls. "Right then, I said, `You know what? That might be our guy right there.' "
The clincher, as far as Lee was concerned, was when Rossi was able to make a certain cover song sound appealing.
"Lucas took a Bon Jovi song and actually made it sound great," the drummer marvels. "I was blown away."
As reward for his efforts, Rossi was able to record a CD with Lee, Clarke and Newsted titled "Rock Star Supernova," which was released in November. The disc has thus far been a commercial disappointment — failing even to break into the top 100 of the Billboard album chart.
The band, obviously, is hoping for better results from the road show.
The tour opener, on New Year's Eve at the Joint in Las Vegas, received negative reviews from the press. One review that ran on Canada.com called it "wildly uneven" and chided the band for its NC-17 stage antics, the kind that Motley Crue built a career on. Lee, not surprisingly, has a different take on the debut bash.
"It was awesome," he says. "What a way to bring in the new year with a new band. And Vegas, obviously, is retarded. After about three or four days of that, we finally got out of there alive."
Newsted, however, didn't even make it onstage for opening night. A little more than a month after the "Supernova" TV series ended, the bassist injured his left shoulder and right arm while trying to catch a falling bass amp. The injuries will probably sideline him for the entire tour.
That's a bummer for local fans who were hoping to see Newsted play in the Bay Area. Performing in his place will be the Black Crowes' Johnny Colt. The good news, according to Lee, is that the current band is still an appealing marriage of musical styles.
"Gilby comes from a hard rock, bluesy kind of school. I come from an anything-goes type of school," he says. "You got Johnny, who comes from hard rock, but is also a bluesy player. Then you got Lucas and he's into the Radioheads and Coldplays. It's really the ultimate mix."
Courtesy of www.insidebayarea.com | {
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The 2010 Italian motorcycle Grand Prix was the fourth round of the 2010 Grand Prix motorcycle racing season. It took place on the weekend of 4–6 June 2010 at the Mugello Circuit. Seven-time MotoGP world champion Valentino Rossi suffered a displaced compound fracture of his right tibia in free practice, after losing control of his Yamaha in one of the circuit's fast corners. The injury saw him lose any hope of retaining his 2008 and 2009 crowns.
Dani Pedrosa dominated the MotoGP event aboard his Honda, reaching the finish line well clear of chasing championship leader Jorge Lorenzo in second. Andrea Dovizioso took third, while ex-champion Casey Stoner took his Ducati to fourth on the final lap. The 125cc race saw the first victory of future multiple MotoGP world champion Marc Márquez.
MotoGP classification
Moto2 classification
125 cc classification
Championship standings after the race (MotoGP)
Below are the standings for the top five riders and constructors after round four has concluded.
Riders' Championship standings
Constructors' Championship standings
Only the top five positions are included for both sets of standings.
References
Italian motorcycle Grand Prix
Italian
Motorcycle Grand Prix
June 2010 sports events in Italy | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,219 |
_To C and the children,
And with love and gratitude to my family and friends_
# Contents
Cover
Title Page
Dedication
Introduction
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
Chapter 17
Chapter 18
Chapter 19
Chapter 20
Chapter 21
Chapter 22
Chapter 23
Chapter 24
Chapter 25
Chapter 26
Chapter 27
Chapter 28
Chapter 29
Chapter 30
Chapter 31
Chapter 32
Additional Reading
Acknowledgments
About the Author
Reader's Guide
Copyright
_I am—yet what I am, none cares or knows;
My friends forsake me like a memory lost:
I am the self-consumer of my woes:—
They rise and vanish in oblivion's host,
Like shadows in love's frenzied stifled throes:—
And yet I am, and live—like vapours tossed_
_Into the nothingness of scorn and noise,—
Into the living sea of wakeful dreams,
Where there is neither sense of life or joys,
But the vast shipwreck of my life's esteems;
Even the dearest, that I love the best
Are strange—nay, rather stranger than the rest._
—JOHN CLARE (C. 1840)
# Introduction
_You have to begin to lose your memory, if only in bits and pieces, to realize that memory is what makes our lives. Life without memory is no life at all. Our memory is our coherence, our reason, our feeling, even our action. Without it, we are nothing._
—LUIS BUÑUEL
_It is singular how soon we lose the impression of what ceases to be constantly before us. A year impairs, a luster obliterates. There is little distinct left without an effort of memory, then indeed the lights are rekindled for a moment—but who can be sure that the Imagination is not the torch-bearer?_
—LORD BYRON
THE PROCESS OF WRITING THIS BOOK, THE PHYSICAL act of putting it together from diaries, scribbled notes, books about the mind, and concentrated bouts of introspection, has proven an illuminating exercise for me, demonstrating just what it is that dementia takes away. (Answer: everything; every last thing we reassure ourselves that nothing could take away from us.) The way the brain works, the supercomputer folded modestly into every human head, marshaling its forces, making connections, prompting and synthesizing, is dazzling and extraordinary and yet seems every day perfectly unexceptional and ordinary to us. There's nothing we take more for granted. In recording the decline of somebody with dementia, and seeing her preoccupations grow narrower and narrower, and her intellectual pathways block off, I've found myself preoccupied with unexpected things, more and wider things, my mind disappearing down all kinds of unforeseen alleys, which has been exhilarating but also poignant. I'm left feeling a profound gratitude to the life of the mind, how associative it is and how rich, in its leading on from one thing to another, into that whole interior landscape of yoked-together and often incongruous thoughts that adds up to a self. This book has turned out to be as much about the unraveling of a caregiver as it is about the person cared for, but its starting point was wanting to write about Alzheimer's and about life with an Alzheimer's sufferer, my mother-in-law, Nancy.
We spent many years looking after Nancy at one remove from us, a responsibility made more stressful by distance, and then at closer range—in a big Victorian house in a remote part of Scotland, with Nancy and her disabled husband, Morris, living with us and our three children. The house was not an ordinary one and, in a way I didn't anticipate, became another character in the story. It was an imposing, drafty mansion on a wild, near-treeless headland. We moved there specifically to attempt an extended family; when that failed, we had little choice but to leave. The official gloss put upon this exit is of the "phases of life" sort: job done, time to go. The private verdict is soaked marrow-deep in defeat.
I'm aware that in many ways this is a story about privilege. We could afford (could convince the bank we could afford) the big house and the part-time help, and when push came to shove (and it did, literally), my in-laws could come up with the fees for a good nursing home. But there are monetary consequences to caregiving, above and beyond the obvious weekly bills, and there has been a real financial hangover that we're still working through, brought on by months and years of having no choice but to put work second.
_Well, so what_ , you may be thinking. _You took in your husband's parents. Boo-hoo. Big deal._ Across other, more populous continents, three-generation households are the norm, after all (the Asian three-generational photograph is lodged reprovingly in my brain), and they will likely become more commonplace here, as the care crisis bites harder. It's pretty clear that it _will_ bite. The world seems to be in the grip of a dementia epidemic. Here in Britain, there are 820,000 people who've been diagnosed with dementia, two-thirds of them women, and the figure is rising sharply. In the United States, it's more than ten times that number. Of these, according to the Alzheimer's Association, 5.3 million have Alzheimer's disease. There are estimated to be more than 35 million dementia sufferers across the globe, with over 65 million forecast for 2030 and more than 115 million for 2050: the figures near doubling _every twenty years._ That's why the phrase "dementia time bomb" is beginning to be used. The devastating extra sting of dementia is that, unlike heart disease and cancer, it doesn't shorten life. It's a cruelly lengthy business. The changes in the brain can begin twenty years before a formal diagnosis, and the average life expectancy afterward is eight years.
Alzheimer's disease is only one of many varieties of dementia, though by far the commonest. Over 60 percent of diagnosed dementia sufferers have Alzheimer's disease. Back in 2002, BBC News reported that more than 40 percent of UK home caregivers of someone with Alzheimer's had been forced to give up work in order to look after the person. In the United States, 10 million people act as caregivers to someone with dementia and millions more offer support. About the same percentage of American caregivers are not employed, and two-thirds of those who can manage to hold down a job report major disruption to the workweek. I quote these statistics as a roundabout way of answering my own question: Why write this book at all? There were several reasons. One of these was to share in my own revelation, hard-earned, that Alzheimer's isn't just about memory loss; that _memory loss_ isn't just about memory loss, but leads to disintegration. I wanted also to kick the system ineffectually in the shins; to give a glimpse into the dementia abyss; to show that for every "client" in the statistics there are one, two, four, six others (aka the family) whose lives are blighted in addition; in short, to give a little insight into the reality that ensues from the apparently noble idea (the noble, and for the country's financial bottom line, far preferable idea) that the elderly ill should stay at home whenever possible.
Question: Do governments understand just how dehumanizing Alzheimer's is? (A rhetorical question. Answer: no, or they wouldn't withhold good drug treatments or limit research programs on grounds of cost.) Question: Does anybody who hasn't been through it understand just how dehumanizing caregiving can be? (A rhetorical question. Answer: no, or there would be proper nursing home provision and it would be free.) As things stand in the United Kingdom, dementia patients in nursing homes, unlike cancer patients in hospitals, are regarded as "social care clients" and charged hotel rates, and if they have savings and houses must give them up to pay the bills. We British may regard ourselves as two steps ahead of the United States in the matter of universal rights to health care, but when it comes to dementia, the two systems are very alike. Medicaid will step in and pay for residence in a nursing home only if the ill person's own assets have dwindled away almost to nothing, and it's pretty much an identical situation in the United Kingdom. Once the money runs out, the ill person's house is likely to be sold to pay for care, unless a spouse or dependent is still living in it. Even if American houses can be placed out of risk in the short term, certainly they are at risk after the owner with dementia has died, via _estate recovery_ (the rebate of nursing home fees in arrears to Medicaid—a policy that's pursued energetically in most states). Advice about loopholes in the system that allow a family to hang on to a loved one's home long term has grown into an industry, and almost every American Web site that talks about costs and rights to do with dementia suggests consulting an attorney. It's a system that's good for lawyers: in other words, bad law.
There's also a selfish answer to the why-write-the-book question. I'm one of those who have found work incompatible with caregiving, even work that I have always done at home, sitting at a table by a window, or slouched uncomfortably on a sofa, laptop at a precarious angle, mediating children's interruptions—work that you might assume would be ideal in the circumstances. It's more than economics in my case. Writing is more in the way of a compulsion. It may even be a psychiatric disorder. If days pass dryly—that is, without sentences being made and remade—I find that I begin to drift into the arena of the unwell. Throughout my years of caring for Nancy, the drive was there to produce something salable, but other than the occasional article, the content wouldn't follow the impulse. Following an early career producing sensible nonfiction and then a long hiatus while having and raising children, I was supposed to be cutting loose and writing a novel—and, on the face of it, I was immensely productive, almost manically so. I wrote two and a half novels. I wrote them in a rush, thinking, I can make some money at this (almost a guarantee of failure). The two finished ones were bad, superficial, studded with frustrations like cloves in an orange. The half is still a half, stopped, stalled. The muse left me. She did it quite abruptly, though things had been sticky between us for a while. After that, all I could seem to write about with any passion or conviction was my mother-in-law. Writing about her was sustaining through the dark days of creative roadblock. It was, to be blunt, a way of not cracking up.
This might also be the moment to tell you that names in the account that follows have been changed. Nancy is beyond minding or even registering the fact that she's the subject of what you might call an unauthorized biography, and changing names provides only a tissue-paper-thin layer of anonymity, but it feels right, nonetheless.
A lot of what follows is taken from unedited diaries, which accounts for the use of the present tense and also for the emotional rawness of some passages. While filling the diaries, I used some of the entries in a newspaper piece I wrote about Nancy. It was straightforward and at moments graphic about her problems (and ours), and this didn't go down well with online commentators. Their chief complaint had to do with my having written intrusively about my mother-in-law without her consent. Even by then Nancy was long past the point of being able to consent to anything; she found the choice of Weetabix or cornflakes baffling enough. Intellectual competence aside, the argument remains that whatever the truth about rights, it's in bad taste to write in such unsparing detail about another's decline. The daughter of former British prime minister Margaret Thatcher has been pelted with rebukes since disclosing her mother's dementia. Her critics insist that the disease should be "kept in the family," which is only a short hop from suggesting that it's stigmatizing and shameful. Tony Robinson, the actor who played Baldrick on the popular BBC show _Blackadder_ , was accused of something similar when he let UK Channel 4 make a documentary about his mother's last weeks. His response was robust: no, quite the opposite; he was proud of the program. There's a public service element to allowing media access, even if it might appear to the viewer to be cloaked in voyeurism. Those of us who have loved ones embarked on the dementia journey—and it _is_ a journey, with clearly defined stages—publicize the details of their decline not despite our love, but in large part because of it. In some cases, the love is shared with the nation at large. In the case of Ronald Reagan, the announcement of his Alzheimer's disease by means of an open letter in 1994 (a poignant and brave last message that marked the end of his public life) prompted a whole new national surge of affection for the former U.S. president.
SCIENCE STILL ISN'T sure precisely what triggers Alzheimer's, though things are moving so fast that the mystery may be solved by the time you get to read this. (In fact, the pattern in the last few years has been that they move fast and get nowhere much.) What's uncontroversial is that Alzheimer's brains show the presence of two weird and provocative things: (1) a wild overproduction of beta-amyloid, a naturally produced and usually soluble protein, contributing to sticky blobs called _plaques_ and (2) the knotting and snagging of the tau protein that forms the "rungs" in the communication ladders within brain cells into _tangles._ The race is still on to determine what the definitive cause is.
An adult brain has about 100,000 million nerve cells, individual _neurons_ that each look rather like the branching root of a tuber pulled out of the ground—tubers of different shapes according to flavor. A good analogy, put forward by the Oxford professor Susan Greenfield, is to think of the brain as the Amazon rain forest inside your head. In the Amazon rain forest's 2.7 million square miles, she says, there are about 100,000 million trees. Imagine all that foliage condensed into the size of a cauliflower within your skull: 100,000 million tiny trees, making a dense neuron forest. Our memories and our thoughts travel through the forest as encoded electrical signals. The "roots" of the neuron are called _dendrites_ (from the Greek for _treelike)_ , and its stalk (trunk) is called an _axon._ The information comes in to the neuron via the dendrites, into the _soma_ (cell body)—that's the front door—and then goes out the back door, travels up the axon, along parallel lines of communication called _microtubules_ , and out the other end at branches called _synaptic terminals._ This information moves, in tiny leaps, from axon to dendrite, from one neuron to the next. How does it do that? For a while there were two camps of conjecture, spark versus soup. The sparkers, who believed in an electrical leap, lost out in the end to the soupers, who thought that the constituency of the soup was key. The spaces at which the crossing is made are called _synapses_ , though they're more like ports than spaces—ports at which clusters of _neurotransmitters_ are waiting as a chemical transport system. Subsequent research has shown that there are indeed electrical as well as chemical synapses in the brain, though the electrical ones are heavily outnumbered. The number of dendrites and synapses varies hugely according to the neuron's function, but on average a neuron is thought to have around 7,000 synaptic terminals. Multiply that by 100,000 million and the mind begins to boggle.
In photographic comparison, a normal brain resembles a freshly peeled chestnut, pale and fat and glistening, and a brain with advanced Alzheimer's disease looks rather like a walnut, shrunken and shriveled with bits apparently eaten away. The disease takes place as a physical invasion, involving the progressive destruction of the neuron forest. Under the microscope, the damage is theatrically obvious: there are plaques—fuzzy, rust-colored accretions of protein fragments—which interfere with the transport network, and tangles, which look rather like strands that have grown over the neurons, like bindweed in a garden, though in fact they're a distortion of the neuron wall itself, its microtubules having collapsed into knots. As cells wither and die, gaps form in the tissues, leaving characteristic holes. American researchers working with the new generation of scanners, and thus able for the first time to look into the brains of living Alzheimer's patients, have found that the disease starts in or adjacent to the hippocampus (the memory-processing zone) and moves farther into the limbic system (our emotional nerve center); around eighteen months later, it has crept into the frontal lobe (site of the thinking, reflecting self). The disease always starts in the same place and takes the same general route, but proceeds unevenly in its spread. Some sections of the brain will be decimated, but neighboring ones might be unaffected and normal. It's rather like a forest fire in which clumps of blackened stumps sit adjacent to trees that seem oblivious to the disaster, untouched, their green canopies intact.
The term _dementia_ (from _de mentis_ , "out of the mind") was coined in 1801 in the asylums of Paris. Today it is used to mean brain failure, and in just the same way that heart failure is a condition caused by a whole host of problems, brain failure has many sponsors. One in fourteen UK citizens over sixty-five has some form of dementia and one in six over eighty, but for UK citizens reaching the age of sixty-five in 2010, the risk of developing dementia is one in three. Almost one in six Americans aged sixty-five will go on to develop dementia, and more than one in five aged eighty-five. And that's the trouble with it, in terms of PR. It's an old person's disease, by and large, and elderly ill people aren't easy to "sell." The issue is confused by our muddle about what's normal in old age—the idea that senility is an ordinary part of the human condition, that it is aging itself made manifest, and thus can't be cured. Progress is slow.
Research funds aren't generous, despite the fact that currently dementia costs the United Kingdom about £23 billion a year and the United States a staggering $148 billion just to deal with damage limitation and long-term care. Unpaid caregivers, their lives transformed into a round-the-clock vigil, are saving the British government about £12.4 billion. In just one year (2008), the economic value of unpaid caregiving in the United States was estimated to be $94 billion. Two-thirds of UK citizens with late-onset dementia are living in a family home; about 70 percent in the United States. Both figures are probably higher when undiagnosed cases are taken into account.
In the United Kingdom, only £61 is spent on research per Alzheimer's victim, though the amount is £295 per patient for cancer. In the United States in 2008, $5.6 billion was spent on cancer research, but only $0.4 billion on dementia science. Cancer has higher cultural status, even, perversely, a twisted, dark kind of glamour. Plucky young people get it, pop stars battle it, pretty wives and dashing young husbands die of it, and their pictures are spread across the newspapers. Cancer is a disease that journalists get and write about on the premise that if life hands you lemons, make lemonade. People with dementia don't write about it much because writing isn't something they do—or wasn't, until recently, when the very-early-diagnosed patient lobby sprang into being and people like the writer Terry Pratchett began speaking out. The much-loved author of the _Discworld_ novels, a man who's sold 55 million books worldwide, allowed a BBC TV crew to follow him for twelve months. The resulting television program _(Living with Alzheimer's)_ charted unsentimentally the beginnings of his decline, his defeat by the attempt to tie a knot in his tie, his having to pause in giving a reading because he found that a "shadow" was falling repeatedly across the page. This is the kind of cultural event that introduces people to the idea that dementia has something to do with them. It will be a long road. In general, the Alzheimer's demographic and its symptoms have meant that it's very low caste—something that, even now, we associate with decay and the cabbage-and-disinfectant scent of the geriatric ward.
There are widespread misconceptions about the disease. Uncertainty is the midwife of misconception. The trouble is, nobody knows for sure what triggers Alzheimer's. All we can hope for is that keeping fit, doing crosswords, and eating well will spare us. They don't, necessarily. The illness of writer and philosopher Iris Murdoch attracted so much interest because people were amazed that someone like that could fall prey to Alzheimer's, someone so clever, articulate, affluent. We live in an age-defying, mortality-denying culture. We don't believe in ourselves as elderly. We're interested in cancer and the carcinogenic because those are words that might turn out to apply to the thirty-eight-year-old as much as the seventy-eight-year-old; cancer afflicts the young and rich and fit. If Alzheimer's equals old age, then that's something we'll deal with later... though we'll be fine, because we drink soy milk and do Sudoku and play tennis on the weekend. The most widespread misconception is that dementia's a good way to go: "They're in their own little world and pretty happy" the misconception goes, and "they've no idea they're going to die of it right up to the very end, which doesn't sound too bad to me." Very occasionally and exceptionally, in the online Alzheimer's community, sweet-tempered-to-the-last is reported; the slow-fade sweetie who was never any trouble and died smiling in bed before indignity could take hold. But that isn't the norm. That hasn't been Nancy's fate, alas.
IF I HAD to pick one catchall descriptor for Nancy's life in the last few years it would be _misery_. Profound misery, unceasing and insoluble. She knows that something is wrong, very wrong, but what is it? She's had a series of terrible daily encounters with herself and her environment that might have come directly from an amnesiac thriller: waking to find she has aged fifty years overnight, that her parents have disappeared, that she doesn't know the woman in the mirror, nor the people who claim to be her husband and children, and has never seen the rooms and furnishings that everyone around her claims insistently are her home. Time has slipped, gone seriously askew. Every day for her is spent in an ongoing quest to put things right. The trouble is, she can't seem to concentrate on the question or on possible clues to it. She can't navigate the problem. When she left us for the nursing home, she was daily engaged in a very protracted, slow-motion form of panic. It's been over eight years now since the formal diagnosis and eleven years at least since symptoms began, but even after all this time, she's only at stage 6 of the disease. Stage 7 looms, the cruelest and last phase, with its loss of continence, motor control, speech, and ability to swallow. Eventually her lungs will forget how to breathe, her heart forget how to beat, and her quest will come to an end.
I have thought, said, and probably even written in here somewhere that Nancy has lost her self. That at least is the impression anyone who knew Nancy twenty years ago would have if they spent a weekend with her. The things that made her herself are all but gone now, I say, but what does it mean to say that? Obviously she is still herself, isn't she? She isn't anyone else. It's just that the self is changed. Disease has changed it, or else, in some vaguely science-fictional manner, overlaid it with something new. But what exactly is the self, anyway? Must it have unity, continuity, in order to be authentic? Does it exist beyond and beneath the health or otherwise of 100,000 million neurons? Is there something else that encapsulates the self, something extra, indefinable, that we call the soul? If, as some philosophers of the mind argue, being conscious can't be said to be without content, that it has to do with being aware of not only your own person but also your past and future, your place in the world, your culture and context, your hopes and fears, then where does that leave Nancy? John Locke may have come up with the notion of "consciousness" specifically to spike Descartes's idea that we are thinking all the time, even when sleeping, but Locke also thought that we are only ourselves in having our memories, and defined personhood accordingly. Locke's definition, being antique, is easy to forgive. It's surprising, though, to find much more recent definitions that agree broadly with his. As late as 1973 an American philosopher named Mary Anne Warren demanded of _persons_ that they be conscious, rational, capable of abstract thought, able to communicate, able to exercise free will, and have self-awareness. Under this severity, nobody with brain damage is a person, and Alzheimer's, so often misreferred to as a mental illness, involves a catastrophic form of brain damage.
Materialists would contend that there is no soul, that we are only a kind of organic machine, our notion of a unique self misguided. It's difficult not to be convinced by this idea, seeing Nancy's selfhood warp and flicker and wane as the disease colonizes her. It's not good—not even for privileged bystanders, counting their blessings—to see a self under attack. We prefer to think of our _selves_ as something original in the world, inviolate, independent of our physical bodies. The idea that we are biochemistry, and that's all, that thoughts and feelings are produced by neurons, that neurons can die and our selves die with them... that's a deeply undermining idea. It's far more comforting to contend that Nancy's soul, her essential self, remains intact beyond the reach of her struggle to think and express herself, and will be liberated and restored by immortality. I try hard to believe this when I see her, alone in the dayroom in the nursing home, sitting rubbing her hands together and muttering. I can't help wondering what she's thinking. _Is_ she thinking? Is she having a dialogue with her disease, negotiating with it in some way, aware of the great buried store of memory, her past, her self, glimpsed under the tangles of Alzheimer's like a ruined house under the suffocating grip of ivy?
Now that she's at one remove from us again, it's easy to love her, and where love falters, guilt is primed and ready to fill its place.
# Chapter 1
_Three Years Ago_
NANCY IS STANDING AT THE WINDOW AGAIN, THE ONE with the spectacular view, worrying about how the oil tanker will manage to get out of the bay. She is making her anxious hands, rubbing each palm against the back of the other in turn, brisk and rhythmical.
"I just don't think it will get out of the space, it's too big," she says, rubbing harder, her eyes full of concern. She is wearing all the cardigans she could find in her bedroom, in layers, having insisted on doing up all the buttons on each and tucking each sleeve under at the wrist. Her mood has improved since breakfast time, when she woke with the now-characteristic belief that she was newly discharged from hospital into the care of strangers: "But where is my family? Are they coming for me?"
"We are your family, honey," I soothe.
She laughs disdainfully, shaking her head. "Either you're a liar, or I'm going mad." Most mornings, there are tears. Tears and confusion. Dressing is hard. She wants to do it herself, but bras and trousers go on backward. If we don't get to her quickly enough, she wanders the halls in her underwear. She looks younger than seventy-nine, everybody says so, and this is especially evident in her near-naked wanderings. Physically she's amazingly good for her age: unstooping at five foot seven, well proportioned other than a mild potbelly, determinedly upright. Her legs are strong and shapely. She can walk for miles, has thick silvery hair cut in a bob (it was sandy colored once, set into soft curls at the local salon once a month) and a charming smile, her pale face barely lined, though her blue eyes are rheumy now and her nose growing hooky. She's acquired a prickly white beard under her chin, which my husband, Chris, shaves off every now and then. She won't always let him at it. She can be protective of it, sitting stroking it in her chair. Some days it horrifies her. "Who put this here? Where did this come from? Take it away!" Or she thinks it's a wound, a scab. "I must have tripped and fallen. But it's getting better now." Nancy's at a good-days-and-bad-days stage of Alzheimer's, and on bad days she accuses Morris of having given her the stubble, perhaps because she recognizes that beards are properly the province of men.
She returns to her little sitting room, her coal fire, her husband, and sits in her pale blue winged armchair. She asks, now, if it's hers and if she can sit there. She hasn't had it long enough to remember it. Only the very-long-term memory is functioning. Morris is sitting in the chair beside her, is always sitting in the chair beside her. His is electrically powered, tips back, is upholstered in orange tapestry. He was stout once and, with his square face, mischievous dark eyes, dark hair combed over, and mustache, resembled a rather better-looking Oliver Hardy, and was just as likely to suffer fools gladly. He's mellowed. He appears to have shrunk, in all dimensions.
I've known Morris and Nancy for twenty-two years. When I first met them, brought home by Chris from university, I thought them old-fashioned, thrifty (furnishings and appliances had remained unchanged over decades), sociable, hardworking, right-wing. They were _Daily Mail_ readers, natural conservatives, but generous about our student leftiness. I don't recall anything much in the way of ideological standoff. They were all hospitality, bailed us out when we got into financial hot water, let us stay with them on an indefinite basis when work plans went awry. Despite finding our postgraduate ideas about office jobs and steady security highly provoking (we didn't fancy either of these much), they were nothing but kind. Kind but unforthcoming, opinion withheld. This has been a pattern in our relationships.
Nancy and Morris moved here with the rest of us this summer. We have a lot of latitude in where we live. Latitude and longitude. Chris is an internationally known-in-his-own-niche expert on a specific use of new technology, and he consults widely, mostly from his home office, though there are bouts of meetings and flying. We have two teenage girls—Millie, sixteen, who's tall and dark like her mother, and Caitlin, fourteen, who shares her father's ash-blond coloring—and a boy called Jack, ten, a senior at primary school, tall and lanky and Italianate, with a scruffy dark shock of hair.
Moving, it turns out, isn't good for Alzheimer's patients. Leaving behind the familiar, having to adapt to the new. Nancy's disorientation is ongoing. "I don't know where I am," she sobs, "I don't know what I'm supposed to be doing."
I've been reading about memory. In cases of transient global amnesia (total but temporary memory loss), people ask over and over where they are and what they should do, how they got there, what they should do, _what should they do now?_ Doing is a big preoccupation. They don't ask what might seem to be the obvious question: Who am I? That doesn't seem to be a question the self asks of the self. Instead, it looks for clues from context: where, how, what.
Chris and I have different responses to her anxiety. He takes her hand and is tender, explaining that they weren't coping, she and his father, and have come to live with us. I go for a jollier approach. "Well, lucky for you you're retired now and you can sit in this chair by the fire and eat biscuits and watch the afternoon film on the telly," I say. "Not like poor old me, I've got washing to see to, dogs to walk and vacuuming, the dinner to sort out, and you should see Jack's bedroom." Jack is proving dedicated to the acquisition of stuff, particularly electronic stuff (gadgets, dead laptops), as well as guns, swords, and lighters. Sometimes I worry about where these interests might lead.
"Oh, poor you, having to do all that," Nancy says, fleetingly lucid, playing along, and I'm embarrassed at being caught out talking to her in this nice-nurse fashion. But the moment passes and she's back at the window. "Look at all that water." Her voice is astonished.
"Yes. We live here, out on the peninsula; the sea's all around us. Do you remember coming here with us to live? We came last month. Do you remember?"
"Edinburgh," she says under her breath.
"You used to live in Edinburgh, years ago. But then you moved up to Speyside, near our old house. Do you remember the bungalow? By the river?" She looks blank. "And now you live here, with us."
She looks at me, grim-faced. "That's all very well, but they laugh at me, you know. Not you, I'm not talking about you, but the others. They look me up and down in the street and I can see that they're thinking, Who the hell does she think she _is?"_ Paranoia, an Alzheimer's marker, is just beginning to get its grip on her. But she's been lovely to the children all summer, which is reassuring. Her face lights up when they go into her sitting room. She pats her knee, like she used to; Millie's five foot ten and can't help laughing. "Now come and tell me all about it," Nancy says. About what, she doesn't specify. The girls are good with her, as Morris is always telling me. They're patient, tolerant, don't rise to verbal bait. They do things at Granny's pace, taking her arm in theirs. "Come on, Gran. Let's go and make Granddad some tea," talking her through the operation step by step. "Put the tea bags in the pot now. In the pot, not the mug. That's it. Right. Hot water next, can you manage the kettle okay? That's the kettle. Yes. Here, let me."
Morris prefers television to conversation, or indeed anything, and it's been this way for a long time. Depressed and immobile, he is master of the remote and flicks between channels with a desperate air. It's like he can't look away. Things are too awful in his present to contemplate them squarely. Because he's so focused on his television day, Nancy's life is frequently lonely. She can't follow a television program any longer. She's more interested in being with me, because—when running the household, at least—I appear to be doing things. She's less keen on me when I'm writing or reading. "The men just sit there," she tells me scornfully, unable to distinguish between one kind of sitting and another: one at his desk on his laptop and phone, consulting and earning, and the other in the armchair next to her, absorbed fifteen hours a day by the flickering screen. She follows me around. She wonders half a dozen times a day where _the friends_ are, and if they are coming.
"I don't want the friends to know I've been ill," she says, as we pick tomatoes in the greenhouse. She eats the ones she picks or puts them slyly in her pocket, thinking I haven't seen. Or just picks the dried-out leaves from the plants and puts those in the basket, smoothing them carefully. Then she takes them out again. "I don't think these are ready," she'll tell me, trying to fix them back on the trusses.
The friends—imaginary friends—visit us sometimes, and she has days when she worries about how they'll get here and how they'll get home. In truth, her real friends have long deserted her, had deserted the two of them long before their move north. _Desertion_ is a strong word; the truth is the process wasn't so premeditated—it was a more gradual loss of attentiveness, a social slippage, the kind that happens when people get sick and have little to talk about other than their problems. Three from their old circle telephone from time to time, but it's us they want to speak to, for reports.
"I need to say good-bye," Nancy insists, twisting her handkerchief. "I need to see the friends off."
"Don't worry," Chris says, trying to ease her agitation. "They've gone already. I saw them leave earlier." And then, seeing her expression, he adds, "But they said to tell you they'd had a lovely day."
"Gone already? But they didn't say good-bye."
"They did, don't you remember? I think you might have been asleep."
"They haven't gone."
"They have. I saw them; they left on the bus."
She looks indignant, draws her shoulders up tight. _"They didn't come on a bus."_
She appears to be having hallucinations. These are new, have arrived quite abruptly, and it occurs to me that our moving her here has aggravated the decline somehow, has accelerated it. Guilt is something I'm going to get used to, but for now it's fresh and new. I take Nancy into the drawing room and we look through a stack of interiors magazines, me commenting and Nancy cooing. My laptop's open on the table, and my attention is 80 percent diverted while I trawl the Internet for answers. Temporal lobe damage, it seems, can cause autobiographical hallucinations. Does she see the friends striding toward her across the lawn, looking just as they did twenty years ago or more?
Sometimes I think I can see them myself. The house doesn't feel haunted—some big old houses do, but this one doesn't—though there have been sightings, I'm told, in years past, of Victorians paused on the stairs, their eyes oblivious to the present. The first day we were here and went to the pub for supper, a fisherman propping up the bar asked how we were getting on with the spooks. I haven't seen anything or heard spectral footsteps, but the whole property is soaked in what I can only describe as pastfulness. It's pastful, and sometimes, even though I know it's just this, I've half believed there are women in rustling silk frocks in that part of the wood that was once the rose garden, have half heard brief melodious laughter in the paddock that was once a tennis court. Who are these people, the friends Nancy talks about? It's occurred to me that the altered perceptions of Alzheimer's might allow people to see ghosts.
The house sits out in near seclusion at the neck of the headland, at the point where the neck joins the shoulder of a second, bigger peninsula, two miles from a village, fifteen miles from a small town, and far, far away from everything. It's a great, four-square Victorian house with sash windows, crenellations, and crowstep gables, its overgrown walled garden framed in lichen-covered stone. It's the kind of house that, while not grand enough for Manderley or Gothic enough for Walter Scott or English enough for Jane Austen, might serve as the scene of a death at the vicarage in Agatha Christie. It sticks up high on the low, gently undulating profile of this wind-scoured green promontory like a church, the sea rushing up the cliff faces around it. Building upward in this climate is an act of faith, almost of defiance. The architectural vernacular hereabouts favors single-story longhouses, long and low and hugging the ground, though a good many of these have been weathered into rubble, with kit-built bungalows parked alongside. After the longhouses fell from fashion, the local style favored one-and-a-half-story cottages, high enough to be provided with an upper floor snuggled into the eaves, low enough to brace themselves against the weather. The eighteenth-century terraced housing that lines the two principal streets of the village is fully double story, held in a self-protective loop around a deep harbor.
The house and neighboring farm were once one property, and together they owned all the land that can be seen from the single third-floor window, the attic window that leads out onto a precarious half balcony. The original farmhouse is two hundred yards down the hill, across the lane, enfolded by its barns and cattle courtyard into a wind-resistant square. The building of the big house in 1860 marked the achievement of wealth and status, a move up from the cottage to a grander residence on higher ground, one gleefully elaborate in its luxurious details. All that remains of the estate are the four garden acres inside the high wall, the privacy-giving wall, marking off the domestic world from the working one, separating peasant from gentry, keeping the bullocks and harvest workers in the adjoining fields out of the sight of the strolling, tea-taking, tennis-playing manor dwellers within.
The house layout is ideal for an extended family. The kitchen has two doors: one into Morris and Nancy's sitting room, and one into the rear corridor, where their bedroom and private bathroom were converted from two former maids' rooms. Off their sitting room in the other direction is a small lobby, which leads into their private daytime bathroom. So Nancy and Morris have, in effect, their own suite of rooms, with only the kitchen shared, and even that is two-family friendly, having two stoves and two full-size tables along its double length. The original thinking was that Nancy and Morris would self-cater, up to a point and with our assistance. They were keen, Morris said, to have as much independence as possible. They brought what remained of their marital past in packing crates, everything that had survived successive years of downsizing: their 1960s crockery and pastel-colored kitchenware; tarnished silver cutlery with worn bone handles; old pillows, duvets, blankets, marital linens smelling of cedar wood; boxes of clothes and miscellaneous items dating back forty years; old toiletries, socks, lamp shades; wallets and watches, belts and business paper.
With the exception of a daily excursion into the conservatory for coffee, their world has shrunk into this little sitting room by the kitchen: its two armchairs, a 1960s coffee table, a partner's desk, a television, a dresser laden with ornaments—unused steak knives and ancient paperwork idling in its drawers—and a bookcase scantily furnished with photograph albums, thrillers, _Reader's Digests_ , the _RAC Guide to Great Days Out_ , 1970s cookbooks whose pages are stuck together with cake mix.
Nancy's Alzheimer's seems to advance in phases, as if we're mining underground, into the unknown, toward obliteration. Currently we've hit the seam of lost prepositions. Morris gets exasperated with her lapses, her confusion, her failure to recognize what's ordinary and deal with it in the old ordinary way. He has his own health problems: replacement hips that have worn out, poor circulation, numbed legs and feet. Walking is a struggle, and in the past he's relied on his wife to be his legs. She has trouble with this role now.
"No, no!" we hear him shouting. "The cup! The cup! In front of the book! No, not under it, in front of it! Now put the spoon in it. In it! In it! Not behind it! That's a book! Not a book, a cup! Oh, for god's sake, woman!" She can't seem to distinguish between cup and book. Parietal lobe damage is responsible for this, apparently; for failing to match objects with words in that apparently simple but sneakily complex two-hander we call recognition. But telling Morris so and asking him to be less irritable makes no difference. Occasionally Nancy gets fed up with being yelled at and gets her coat and handbag. On one such day she finds me in the kitchen making soup.
"Excuse me." A plaintive little voice. She can't any longer remember my name. "Excuse me, lady. I think I should tell you that I am going to have to find other accommodations." This formal way of speaking is new. Perhaps it stems from uncertainty: her being a stranger in a strange land, needing the help of good Samaritans and needing to be polite to them. If you're unsure who anybody is, or indeed who _you_ are, come to that—their rank, your rank, what your relationship might be—then you're likely to be deferential. Either that or bolshie, asserting your position. Bolshie will come later.
When Nancy's upset, distraction's the only way out. Everything else, and especially _reasoning_ , only escalates and intensifies the trouble. I take her outside, where flowers and butterflies and birds and trees do the job like nothing else, all upset forgotten. We go down to the road, down the long driveway between looming dark hedges of fuchsia, and stand between the entrance pillars and admire the view. She runs an appreciative finger over the house name, indented in brass set into the stone, and I'm shocked to find that she can't read the word that's the house name. She can't recognize the letters and, even when I tell her what they are, can't vocalize them into a run of sounds. She's interested in them, though, as in something half remembered, on the tip of her tongue, running her fingers over the brass a second time, frowning and with concentration. Shocked, I go back to Chris in his office. "Your mother can't read; she can't read anymore," I tell him. It's stunning because it's so absolute, so concrete a loss. Parietal lobe damage is to blame again, it seems, in that zone of the brain where visual impressions are organized and reading and writing are ordered and understood. I read about this on the Internet, which has become my personal guide, dementia caregivers' network, MD, and hospital rolled into one handy package.
It isn't, any of it, a linear progression. Damage, or at least the symptoms of damage, can appear to waver like flickering wiring. Some days Nancy has vocabulary, some days not. She's wandering the house looking for her shoes, and when I ask if I can help, she looks down at the floor, offering me a lifted socked foot. "The things, the things that go on the... that go on the things. I want to. I want the things that go on the end." Perhaps this is a sign of parietal lobe damage again, failing to match word and object, or perhaps it has to do with the plaques/tangles invading Broca's area, a patch on the left side of the frontal lobe that was named after Pierre-Paul Broca (1824–1880), who had a patient in 1861 who could say only "tan." It's the zone charged specifically with talking. It's fascinating, this physical loss of abilities in the departments of self, but in tracking Nancy's neuronal failure, I face self-accusations of ghoulishness.
Random stream-of-consciousness nonsense has become a feature in the mornings. Miscellaneous phrases from the past, from the long-term memory, fall out of the box in random order.
"I'm so glad you're here," she says, "because I was worried about that."
"About what?"
She looks at me appraisingly, as if making a decision about whether she can confide, before launching in. "It's been a long time, and I didn't always do it that way, oh no, don't you believe him when he puts it off, because I can tell you, it's all the other way, really, to be quite truthful, and he knows it is, and I could strangle him sometimes, but the woman said I was to go that way, so I went, and it wasn't there. Did I tell you that? I said that before and you haven't got it. I know that. I do know that. I'm not really as stupid as I look, but she says—oh the things I could tell you about her, but I won't because you shouldn't—and I have got to find the thing now or I won't hear the end of it."
"Her?" I ask.
"The woman," Nancy says, rolling her eyes.
"But it's just you and me here," I say. "We're the only ones."
"No, no, no," Nancy says briskly. "Not you. The other woman."
TAKING ON NANCY'S care, full time, seven days, twenty-four hours, has been... I wish I could find a better word than _shock_. It's been a shock. The thesaurus offers "trauma," but that isn't remotely it. It hasn't been a "blow" or an "upset," a "bombshell" or a "jolt." It's more like the kind of experience that leaves you staring into space openmouthed. _How on earth did I get here?_ you think. _And how am I going to extricate myself?_ There's no adequate preparation for the physical demands, the physical hour-after-hourness of full-time caregiving. It hadn't occurred to me that I would need to dress and undress her, for example, and get her toileted and into the shower, and would find myself, in consequence of this, adopting the nice-nurseish patter that theoretically I hate. "Righto, Nancy, let's get you sorted for bed, shall we? Cardigan first." When I get her into her nightie and take her trousers off, her feet are bluish: white and blue and mauve, her toenails thickened, opaque and yellowed like smokers' fingers, her shins crocodile-skinned. _Proximity_. That's the key word. Up close and disturbingly personal. There's emptiness behind her eyes, something missing that used to be there. It's sinister. It seems sometimes, in fanciful moments, that it's Nancy who's missing, though her body continues to live and breathe and walk around in the world, redundantly.
I HAVE A new role, a new identity. Mothering somebody's mother, and being thanked for it effusively. Nancy comes into the kitchen when I'm cooking and wants to help. I find something for her to do and then she bursts into tears.
"Oh no. What on earth is it?" I put my arm round her and she cries harder.
"It's just that you're so-o g-good to me," she blubbers. "You're so good and kind and you do everything for me. I wish I could do something for you. Tell me what I can do. I want to give you something. A present. Will you take my money out of the bank and get yourself a present?"
"There's no need, really. I don't need anything. Really," I tell her.
She goes back into her sitting room.
"Oh god, what is it now?" I hear Morris asking.
What exactly is my new relationship with my in-laws? I am their housekeeper, something approximating their parent, their perpetual hostess, but also a servant. I send Morris a pot of Earl Grey and a warm Victoria sponge, feeling as if I have visitors and need to provide afternoon tea, and in return he gives Jack a penny and says, "Here, give this to the waitress."
We begin to integrate ourselves a little into peninsula society. First into commerce, then into other people's kitchens. Professions here are often of the multiple kind. Paul, the gas fitter, installs an eight-burner stove in place of the inherited curly-plate electric, then makes new stable doors for the yard, and is turning out to be a very nifty tiler. Though tradesmen aren't easy to find. At the end of the week I scissor the local paper, cutting out announcements for the pinboard. The newspaper's being read everywhere we go on publication day, by shopkeepers, office and health workers, people at the wholesalers and in boatyards, people in tea shops. Ordinary routine comes to a halt. There's a piece about new Neolithic finds made farther up the coast. Someone has been shooting seals and the public is appealed to for tip-offs. Wrecks have been plundered by treasure seekers, and a diver's brought up dead. A man's airlifted from an uninhabited island, injured while birding. A skipper's been charged with being drunk in charge of a boat. There's been another suicide, someone who came from England on holiday and leapt off our cliffs to his death. There's been a country dance, and intoxicated teenagers hospitalized. All this is absorbing enough, but I'm more interested in the advertising. The advertisements are a godsend. Not every trader has a shop or even a sign, and lots of the smaller businesses are done anonymously from home. Thus it is that we find ourselves in a barn one morning, choosing tiles, while being watched intently by heifers.
We take afternoon walks on the beach, going down in the car so that Morris can come. He can't make it over the strip of pebbles, nor manage the low grassy dune, so he sits in the car with the door open, watching and smoking. I take the dogs to the water's edge and throw sticks toward America, the retriever plunging in after them and the Jack Russell barking at him from the shallows. Chris walks his mother up and down the length of the sand, Nancy holding on to his arm and striding along. She's happy, just for this moment, radiant, smiling into the sun. Sometimes a change in the weather is enough to restore our optimism, and this seems truer for those with Alzheimer's than the rest of us. Nancy's world is re-created every minute. She lives in the moment, and therein lies the problem. The minute we get back indoors, she's lobbying to go out for a walk. The walk she's just had is rendered down into an idea, one that persists and nags at her. Perhaps the best thing for Alzheimer's sufferers might be nomadism of a kind. A permanent ambling trek in talkative company, with pauses only for meals and to sleep, would make her happy, I think. Everything, every moment, would be new, and everybody in her party would be on a more equal footing of constant change.
It's our wedding anniversary at the end of the month and Chris and I go to the village restaurant to celebrate, leaving the children in charge. We eat crab cakes, a fish and crustacean stew, a lemon tart with marmalade ice cream, delighting in everything but preoccupied with home, two mobile phones winking on the tablecloth.
"The fish is wonderfully fresh," Chris tells the owner. "Is it caught in the bay here?"
"Actually no," the owner says. "We can't get the quality here. All the good stuff goes south. All our fish comes down from Shetland."
# Chapter 2
_Inspiration may be a form of super-consciousness, or perhaps of subconsciousness, I wouldn't know. But I am sure it is the antithesis of self-consciousness_.
—AARON COPLAND
WE HAVE ALWAYS HAD A TASTE FOR SEMIREMOTENESS. The short-drive-for-milk, long-drive-for-olives model is one Chris and I have honed and perfected. We thought we knew all about backwaters. But this, the peninsula, is a backwater in a different manner. Living here makes you question the terms of your disengagement with the Big World, something that thus far has been merely instinctive. In this society, almost as far north as it's possible to be in Britain, the Big World becomes a catchall for everything that is wrong with life. The Big World is referred to as _south_. The question of whether, for instance, children will have to go south for work is much discussed, in worried terms. South is corrupt, spoiled, feared. We have a little kingdom here, a far-flung corner that prides itself on difference. It's a Roman outpost. The barbarians are talked about partly with pity and partly with scorn. It's Shakespeare: "This other Eden, demi-paradise, / this fortress built by Nature for herself / against infection and the hand of war. / This happy breed of men, this little world, / this precious stone set in the silver sea."
It's easy to assume that there'll be a natural camaraderie between those who choose the edge and not the center. It isn't always true. The physical edge is easy to achieve—you just take up your bed and hand it to the furniture removal company. But what about the metaphysical edge? Edge dwellers are no more likely to be readers, to be articulate, to be interesting people with fulfilled creative impulses—all the usual stuff we hope for in our neighbors—than anyone else. They're just as likely to watch bad television and talk about it. They're just as likely to be unhappy. More so, probably. Unhappiness has driven a good number of the edge dwellers edgeward; unhappiness that morphs into reclusiveness. Utopianism brings others, and intense sociability, an aptitude for running things, starting things, galvanizing all of us disparate souls into community.
Aside from the spatial demands of a two-family setup, Chris and I came here for the usual material reasons (big house, smallish price), for the particularly privileged reason that we work at home and can choose our location, but also, certainly in my case, looking for a new relationship: one with the Sublime. I came looking for inspiration (for work, yes, but also for life) as something concrete (the quality of the view), something engulfing and omnipresent (the quality and shape of clouds)—all of which is just outside the door and there at whim. It wasn't really about clouds, of course, or views. It's always about engaging with elementals. Mail-order catalogs find their sales rise if they choose backdrops of beaches, meadows, hillsides, riverbanks, and forests. Our visitors' book is bulging with remarks about the spirituality of the location. If these things are emblematic, then we on the peninsula are emblem rich, emblem saturated. We have house, meadow, wood and wall, a vast panorama of sea and sky, steep drops and long beaches; we have weather and the tides at our disposal. Those with a lot of geography in their lives are envied. A person at one with geography is admired—the more extreme the geography, the more extremely.
The landscape here is people dwarfing. A succession of headlands rise vast cliffed. The Sublime is here if it's anywhere. Wordsworth is its chief prophet, in my library at least. I have tried to look at my surroundings with his eyes, feel bolstered in his near-supernatural manner. Look at this, from "Tintern Abbey":
_And I have felt
A presence that disturbs me with the joy
Of elevated thoughts; a sense sublime
Of something far more deeply interfused,
Whose dwelling is the light of setting suns,
And the round ocean and the living air,
And the blue sky, and in the mind of man:
A motion and a spirit, that impels
All thinking things, all objects of all thought,
And rolls through all things_.
We expect this of wild places, because we are all Romantics in our way. We still live in the Romantic age, the age of will and the individual, seeking some anthropomorphic godlike power in immensity, perhaps, as we park the car at the cliff-top walk and stand, coat blowing, looking out at nothingness, everythingness, somethingness: whatever it is that rolls through all things, it's ill defined. Or something more mysterious. Here's Wordsworth in _The Prelude:_
_... and I would stand...
Beneath some rock, listening to notes that are
The ghostly language of the ancient earth,
Or make their dim abode in distant winds.
Thence did I drink the visionary power_.
Visionary power. That's what I came looking for. For writing, but also as a corrective to the person that caring for the in-laws might make me; a balance, a bulwark, a reasserted sense of perspective. But what is it, this _spirit that impels all thinking things?_ It might be God, it might be Gaia, or it might be the effect that immensity, despite its inanimate nonconsciousness, has on the mind, and that's what I take the Sublime properly to be. Whatever the case, people come to live here with just the same kind of impulses, expectations, needs. The wilderness knows nothing about us. Self-reinvention isn't only possible here; it's provoked, nurtured, made flesh.
The Sublime, it turns out, is disappointingly elusive. Which is to say that I don't often feel it, almost never; I more often find I'm feeling nothing, and I know I'm not alone in this. There are things that stop us from being properly present in the landscape, that stop me, at any rate. The self intrudes on the Sublime; the past and future intrude, and worse, much worse, the banal considerations of domesticity, the _lists_. The lists follow me out to the cliff tops and onto the beach. I go there hoping for, at the least, the experience of a kind of self-dilution, but instead the strip of sand becomes just another venue for the things that bothered me at home. In addition to which, I'm cold, it's too windy to breathe easily, my ears hurt. My feet are wet from the thick dew of the dunes I had to cross to get here, my ankle turning painfully on the smooth pink landslip of pebbles. I'm distracted. The risk, now, is that the landscape will become peopled by my own mind. The sea, its restlessness. The wind, its stubbornness. The gulls, their superficiality. The Romantic poets went out into the Sublime, sublimity bagging, as a response to the encroaching materialism of their times, the Industrial Revolution and the mystery-extinguishing age of science. I go in order to feel stronger, strong enough to deal with Nancy and Morris and their constant neediness. Wordsworth would have enjoyed the peninsula. He'd have been out all day, returning unhungry, unthirsty—he was possessed of extraordinary unworldly stamina—and would have settled in a fireside chair to write about it, barely aware of numb fingers, cold ears, wet socks. The experience would have suffused him with his reportedly "lofty thoughts"; it would have convinced him that despite the "dreary intercourse of daily life... all which we behold [i]s full of blessings." Which is how we all want to feel.
Wild and desolate beauty is, it turns out, not a backdrop to life that works for Alzheimer's. Nancy seems barely to register the landscape, and when it's pointed out to her—a stunning lighting effect over the hills, a sunset over the bay—she seems not to notice, or at least not to see it properly. Her admiration comes across as forced, something done to humor me. Perhaps beauty, aesthetic sense, is lodged deep in the memory, and for Nancy that memory is lost. Perhaps beauty is something we're taught and must remember, and now that Nancy's lost her bearings intellectually, she's also lost the idea that a golden hill is preferable to a cloudy one, a red sun prettier than a yellow one. Perhaps our aesthetic sense is as much autobiographical as innate. Mistakenly I'd thought that the Sublime would make itself felt in Nancy's circumscribed life, that its primitive kind of language of symbols and feelings (the feeling, simultaneously, of being nothing in the universe and at its dead center) would speak to her, digging deeper in than language can, or that at the very least she'd find the wildness stimulating. I thought she'd love the cliffs and the views, the walks and the seabird colonies, the seal families pulled out on the estuary rocks. But Nancy sees the hills and headlands, seas and skies, as a backdrop to nothing happening, an absence, a stage set where no play will take place. Only people interest her now. Alzheimer's sufferers like cities, bustle, noise, a person-made world. Or at least this one does.
We made the decision to take Morris and Nancy in some six months before it happened. We decided that the answer to the problems proliferating with their living alone was that we should find a much bigger house, one with an annex or a cottage, a project Nancy and Morris would make a contribution to, an agreement put on a formal footing with a solicitor. It was either that or get them urgently into residential care, and Morris was miserable about that prospect (this is an understatement—throat cutting was mentioned, as I recall). But all the contenders that came up in our own area were way out of our league. They were the kind of houses that had their own brochure. Nevertheless we went to see some of them. The ones we could almost (but not quite) afford were in grotty surroundings, by main roads or cheek by jowl with stalag-style chicken farms, encircled by council estates and dead cars. Realizing that this was the reality of our budget took a lot of legwork and a lot of driving. We drove a long way and made Nancy very carsick. Then we spent two months almost buying a ruined farmhouse, drawing up plans for the conversion of outbuildings, but the projected costs spiraled out of control and we abandoned the plan.
Faced with a dead end, I cast the Internet net wider, including anything big in any location. And that is how I came to see the house, on a Web site, and send the fateful e-mail to Chris, who was working in the next room.
The answer flashed back.
"Far too remote. How would I get to meetings? Need to be practical about this." And then hard on its heels, another e-mail. "Had a look at the flights situation. Possible, if not exactly cheap. And I could get a little boat for weekends. Tempting. But the running costs will be horrendous."
"Well, I could turn part of it into a bed-and-breakfast," I replied. "There's a separate apartment, up a separate stairway."
WE MADE THE family visit to the house on April 1 and a second visit the following afternoon, invited to tea by the charming eighty-year-old owners, last of the line. Not literally the last of the line, in fact, but the last who could countenance _living_ here, a long way from proper jobs and department stores. The house set its trap with care. It was a perfect spring day, warm, with barely a breath of wind. The beach down the lane shone out yellow and blue. Spring birds were all atwitter. Children romped around the garden, their distant shouts brought closer by the reverberating of voices off old stone. Away from the formal lawn, down the drive toward the sea, wilder areas of garden beckoned, tall grasses mown into paths, and a secretive wood, where sycamores stunted by wind, venturing only tentatively above the line of coping stones, huddled, heads down, arms linked above their heads like a rugby scrum. Pools of sunlight fell among them. I sat on a mossy bench and the sun was warm on my face. A tame turkey sat at the base of a tree looking back. Once the heart is lost, the head can only throw in the towel.
Perhaps this should be known as the Lichen Peninsula. Lichen's everywhere in pale green mats, curly fingered, densely layered. The air smells different here. Linen fresh, ozoney, briny, undercut by something earthy and sappy. When the tide is out and the sun is warm, there's a rank drying seaweed note. In summer—and summer is short, sweet, cherished—the air is full of dry grass aromas, sweet hay scents mixed in with the brine, and the light, sea bounced, is dazzling, jabbing in unprotected eyes. Everything seems vividly colored. There's a soaring pale wash of blue above, with a quality about it that's nostalgic: the kind of soft and summery depth that childhood skies had once, the kind that small airplanes leave trails in. The grass is the brightest kind of green. The sea is clear and painterly: the royal blue and azure and turquoise marking shallowness on clean sand, the dark green and brown patches indicating depth and weed. There are three beaches within a five-minute walk, all different: estuary and pebble and sand. The sand beach is closest, just down from the house, and the pebble beach is at right angles to it, round the corner of the headland. The estuary, on the road toward the village, is huge and golden and puddley. Comical oyster catchers stride briskly about on drinking straw legs, then stand together crouched over, round-shouldered in black coats like old men in a bar.
I'd thought that Nancy might respond to the history of the house. It was a foolish thought. But I've always liked buildings with a strong sense of identity. Houses that don't need you, their character already made and set by other, more interesting people who pushed their experiences, their thoughts, into the stone of walls and wood of floors, the faded wallpapers and paneled doors. That's what original features have always signified, to me at least. It's relaxing to feel yourself peripheral to another era, a ghost from the future in a house where the past is still present. I had a peculiar idea that Nancy would respond to this. Her early life was spent at a castle—a real one, with acres of lawns and walled warm corners where pineapples and peaches were grown under glass. Her father was head gardener at a great estate, one that's now a hotel, wedding venue, and conference center with depressingly corporate Web pages. Added to which, Nancy's early married life and her child-raising years were spent in Victorian city surroundings. I thought she'd feel at home.
On our second day, excitedly, I take her on a tour of the outbuildings. The main yard has an L shape of them, incorporating tractor shed, coach house, garage, stables, a quaint row of low outhouses bordering the drying green beyond. The gardens are charming, though romantically gone to ruin, with wide herbaceous borders, extensive shrubbery areas with paths behind, and elephantine hummocks of _Escallonia_ and _Hebe_. Generations of family dogs and cats are buried in the wood and in the vegetable garden, with headstones and names and dates.
I take Morris on the tour, too, and we move at his slow, stick-aided pace round the grounds. Morris had been typically gung-ho about the move. He was going to learn to sea fish, and go sailing with Chris. He was going to get one of those electric buggies. He was going to plan and oversee the planting of the kitchen garden. But the truth is that he's no longer good with outdoors. Outdoors taunts him with everything that he's lost. His life today consists of the achievement of selfhood through television. Pictorial absorption. Mind meld. A domestic annihilation that invokes Nancy's presence, perhaps, in healthier and younger days, when all the adult comings and goings of television pictures mirrored their own busy lives, their powers, their choices, and was restfully vicarious. Now it's as if he disappears down a wormhole out of the present. Nancy's presence is preferred, the two of them driving the spaceship together, like they did in their heyday at the family house, when they converted a tiny study into a private TV room, two armchairs and a television squeezed into a pod. But now that she's ill, Nancy isn't content any longer to sit in her armchair all day with the TV on, and why should she be? There has to be more to life. Even she, standing at the doorway that leads from moderate to severe Alzheimer's disease, can see that.
Morris wants to be indoors, and Nancy wants to be out. She comes into the garden with me half a dozen times a day, and every time we go I point out the view. Ordinarily she'll say "Oh, ye-es," drawing the word out as if impressed, but her attention flickering. I'm not convinced she really sees it. So I persist. "Look at the next headland, Nancy—do you see the lighthouse?"
"It's wonderful. Look at that! And really not very much traffic at all." She's pink faced because it's humid today, and because she isn't good with the heat.
She seems to know what I'm thinking. "I'm not good with the heat at all, never have been to be quite truthful."
The short-term memory is shot. The long-term memory is failing, but parts of it are still intact. Fewer of these memories—records stored up on the higher ground, the flood waters lapping against their green hillock—present themselves as autobiographical lately, though random instances of likes and dislikes remain, and rise casually to the surface at unexpected moments.
"You're hot. Maybe you should take some of your cardigans off, then," I tell her.
She looks down, holds her arms out from her body. "Oh. Yes. I didn't think of that."
I help her to take the three extra ones off. But when I see her a few minutes later she's got them all on again, and is just in the act of buttoning the top one, badly and askew.
WITHIN A FEW short weeks we fall into a sort of pattern that we should probably call a life. It's my in-laws' life, at least. The challenge for us is not to let it be all of ours. For now, there is optimism. The new life is full of structure. Structure and comforting sameness, that's what The Book says Nancy needs, and I am, at least for now, keen to do things by The Book. We get Nancy and Morris up once the children are off to school, put them back to bed at night when Morris is ready, and in between the days unfold almost identically. Only meals remind them of the time—meals and the television, their lives parceled out in programming. The day begins with the delivery of breakfast to their sitting room. Morris announces his arrival and his readiness for the teapot by means of several penetrating coughs. He'll stay there all day, politely accepting lunch and supper like a passenger on a long-haul flight. They've elected to eat all their meals there, on lap trays with padded bases.
Nancy will sit down for short periods, but the rest of the day she follows me around. She comes with me to walk the dogs in the morning, dawdling along and bending to look at things like a young child does.
"Don't pick that up, Nancy. It's dirty."
"What are you talking about? It's perfectly clean."
She brings sticks home, the tops of ineptly picked flowers, sprigs of dried grass, a stone, a leaf she liked the look of, and puts them on the table. Within five minutes, she'll be complaining about them—"Who left these horrible things here?"—and ferrying them individually to the wastepaper basket.
After the dog walk, we come in and have a cup of tea and deliver one to Morris with cake. Cake's become a big part of the day. The dishwasher's kept busy with teacups, and I am learning to measure out my life in coffee spoons. The chirpy drone of a home-improvement show burbles through the kitchen door. Nancy helps load the dishwasher, handing me things one by one, wiping the jam from knife to hands to trousers. We put some washing on, and then, because Nancy loves housework, we zip round the ground floor vacuuming and dusting. Nancy is delegated little jobs. She gets to tidy the newspapers and magazines and does this conscientiously, glancing at me to ensure I'm happy. She is given a duster and some spray polish and sings as she polishes: "When Irish Eyes Are Smiling." This is her song. Since we got here she's sung nothing else. The trouble is, she doesn't know the words and fills her own into the stanzas, experimentally.
"When all the things are lovely, dee dee, dee deeee de dee, And I am a milkmaid and I have a car, de dee, de deee, de deeeeeee."
OUR MEMORIES FOR music are stored in a different part of the brain from the ordinary language memory, and tunes survive longer in Alzheimer's than words do. Capitalizing on this, dementia singing groups are springing up around the world. One reports great success with the Beatles songbook. What's interesting is that the music memory appears to bring the words along with it, unlocking the language block. These groups have reported success with quite advanced dementia, citing cases of people with very little residual language who find, after a few sessions, that they can recall and sing lyrics without trouble.
Music professionals with dementia make for interesting case histories. The American composer Aaron Copland (1900 _–_ 1990), who died of Alzheimer's, seems to have had a slow fade, having first developed symptoms in the early 1970s. He didn't compose much after 1973 other than for reworking a couple of old pieces, but was still conducting his best-known work, _Appalachian Spring_ , almost to the end—though critics complained that he lost the thread in the very last performances. Conducting is done from a different part of the brain again—squirreled away in the cerebellum, where our highly practiced, automatic gestures are delegated and stored. The cerebellum is one of the last places reached by Alzheimer's disease.
Copland seems to have had a lonely end. He was dropped by old friends as dementia took hold; two such who ventured to his home on his ninetieth birthday, three weeks before his death, expecting there to be a gathering, a party and a cake, found there were no other visitors. People are afraid of this disease. I know of people who find that when their parent becomes demented, the rest of the family and all the old friends cut them off. One of the people I have "met" on Internet Alzheimer's forums, an American who's returned from her city life to live with her ill and widowed mother, tells me that not only have people stopped calling or visiting, but when challenged about it they grow hostile, pointing out that it's her mother's "bad behavior" and "madness" that are to blame for their absence. People act as if dementia were contagious, she says, and the social stigma is as strong as ever. It can't help if you're a gay man, like Copland, without even the grudgingly given support of family. When things get difficult for old colleagues and fans, it's easier for them to turn away, untroubled by duty.
Nancy begins to sing variants of "When Irish Eyes Are Smiling" all day, on and off, for days at a time. I know what it's like to get a song stuck in your head, one that seems to be there in the background of thought, unbidden, like Muzak playing in a shopping mall. But it's possible that Nancy is suffering musical hallucinations. This isn't just music that's imagined but music that's heard, as if by the ears. PET scans have shown that in hallucinators all the same areas of the brain light up as they do when people listen and pay attention to external music, other than for the principal auditory cortex that does the listening, and in this respect the inner music is exactly like a visual hallucination. The brain is "hearing" (again and again) music that isn't coming in at the ears but is in every other way perfectly replicated. It doesn't seem able to turn it off. Nancy, I suspect, is stuck in a hallucinatory loop, and is singing along to hers. I try her on some other songs I think she might remember, with no luck, though she can hum some of them, in snatches. _Three blind mice, see how they run_.... We start together, and sing the same line again, and then we both come to a halt, look at each other, break out laughing. Neither of us can recall what comes next. Farmer's wife, carving knife, but how does it go exactly?
# Chapter 3
_A memory is what is left when something happens and does not completely unhappen_.
—EDWARD DE BONO
WE'RE A MONTH IN TO THE EXPERIMENT AND I DECIDE that for now at least, I'll give up the struggle to work. It's the school holidays and for now at least, the pressure can justify itself in being off. I kick to the back of my mind the persistent question: What will Nancy do all day when you're busy? That isn't really the question, of course. The question, the real question, is: How will you get any work done with Nancy in the house?
I go out into the hallway and see Nancy rubbing away at the same table I left her at ten minutes earlier, sweeping her cloth over the table legs and round the rim. The spray polish isn't consulted. It's too difficult to use the push button. The squirting stuff comes out at unpredictable angles. She places the tall can at the other end of the room, under another table, so it can be overlooked. I rescue it. I hover with it, my finger on the trigger.
"Shall I spray a little for you? It makes the furniture shinier."
"No, no, don't bother. Don't bother yourself. I don't like it. I don't like any of it."
"Oh dear. What's the matter? Are you tired? Would you like to go and sit down?"
"Certainly not. I'm fine. I'm in the prime of life. I'm not going to let a little thing like the woman get me down."
"Woman?"
"The woman. She comes here. She tells me what I'm to do and I've to do it if there isn't to be trouble."
She's talking about me. Doubt surges in. I thought she liked to clean. I thought it made her happy. That's the only reason I bought the spray.
"Stop then," I tell her, taking hold of her hand. "It doesn't matter. It doesn't really need doing again. You did it yesterday. It's fine."
She throws the duster across the hall. "This is not my job. I wasn't brought here to do all the work and I'm not doing it."
"Come on. Let's go and find Morris."
"He's a lazy bugger, that one. Old buggerlugs. All he does is sit there."
"Well, he has bad legs, Nancy."
"I know. I know that. You don't need to tell me that, thank you very much." She stalks off, disappearing into her sitting room and banging the door. There's dark muttering from within, Morris saying, "What? You're not making any sense at all!"
Nancy helps me get the lunch ready. She butters the table, puts the bread together at odd angles, picks ham off the pile with grimy fingers and eats it, hungrily and a slice at a time.
"I thought you were making Morris a roll."
"No, no. Not for him. He doesn't deserve it."
"Here. Let me." I take charge of the buttering knife. "Let's take him a chair picnic and go into the garden. We can hang the washing and have a walk."
"Ooh lovely. I like a walk."
AFTER LUNCH WE make soup for tomorrow. Nancy wants to help and spends ten minutes scraping a carrot. What's left of it when it's scraped collapses into three weedy bits. I have sympathy for the carrot. The carrot and I have a special spiritual connection. I need to get away from Nancy and her wittering for a bit. I can't stand any more. I am missing my laptop and silence and words that come biddably out of finger ends. I am beginning to feel, quite suddenly, rather desperate. I jump up.
"Well," I say, "that was great, but I have to do some work now, on my computer."
"Oh dear," she says without feeling. "You'd better not be late." Her face has a customary betrayed look. Betrayed and stoical. I know which way this mood will lead and have a silly tactic at hand. It's called Can You Walk Like This?
"Can you walk like this?" I cry, putting my hands over my ears, sticking my elbows out and my knees, losing a foot or more in height and walking like a robot, making nerp-nerp robot noises.
"Oh no! No! I can't do it!" she screams with laughter. I glance round and find her wiggling her fingers and moving her head from side to side, and laugh, too. Then I take her firmly by the shoulders from the back and maneuver her into her sitting room and put her in her armchair, still in robot character, Nancy giggling, Morris looking puzzled.
I go to the drawing room and swoop on the laptop, sinking into my customary chair, my customary spot by the window, with a happy sigh. This is where I should be. This is home. But then I find I can't work. I'm listening for Nancy. And sure enough, there she is. I hear her wandering the house, trying each door in turn.
"Oh look, look at this one. This is a nice one. A bit dark. Very big. They don't know how big it's got lately. Somebody should tell them, to be quite truthful. I think they might come and see it and be surprised. Hmmm, hmmm, hmmm." The singing starts up. "When all the place is ready, and the place is fine and free, and the man isn't there and the man isn't there, and that is all for me."
She can still rhyme. She butters the table but she can still rhyme, can still come up with lyrics that scan, can, in effect, still write poetry. This is a very peculiar disease.
I hear her rattling the conservatory door. "Nancy?" I call out, expectantly.
She doesn't answer. More rattling. The rattling grows louder, as does the muttering. I jump up to go see. "Nancy?"
"What?" Now she's rattled, too.
"Don't let the dogs out, remember."
Paddy is a reddish gold golden retriever, dim and soppy and mildly bowlegged. Left to himself, he'd be happy to carry paired socks round the house all day and wag things off tables. Unfortunately he's easily led astray by his little white friend, Sparky, a Parson Jack Russell, who's long legged and well muscled, has adorable cupped ears, is formidably cunning and a merciless killer. Sparky is quick to learn and Paddy quick to follow. Paddy is his friend's dopey apprentice. They hang around the outer doors waiting for Nancy to let them out.
"Oh doggies. Hello, doggies. Nice doggies. You want to go out? Here, off you go." Whoooosh: a flash of streamlined white and an ungainly portly ginger thing cantering gamely behind are seen disappearing into the horizon line. So this is how it is. About the dogs, we are dogmatic. They can't ever go out on their own. Our neighbor has eight cats. Beyond the lure of feline sniffs, the mesmerizing cat trails leading through the wood, there's an open gate, miles of quiet roads and open fields. There's trouble out there. Farm dogs to get into fights with. Unsuspecting pet rabbits. Plus, and here's the clincher, they've proven not completely trustworthy with sheep.
IT'S A BEAUTIFUL day and Nancy wants to be out in the garden. Nobody has the time (at least Chris and I don't) or the inclination (Morris doesn't) to be in the garden with her in the way that she wants: attentive, lazy, gossiping, devoted to her amusement. But it's a really beautiful day and I have to face the fact that I'm not going to get any writing done. I go out to do some weeding and take Nancy with me. Every inner piece of wall has a herbaceous border, and every border is overgrown. It's the myth of Sisyphus, horticultural version: I push the wheelbarrow up the hill and it just rolls down again. Dandelions see extermination as a challenge: lop their heads off and they'll grow twice as many new ones overnight. But it's a beautiful day and the heart wants to be outdoors. The heart wants it and Nancy is determined. Nancy and I go out singing, each our own versions of "When Irish Eyes...". Hers, I note, involves a pudding and a gate. I give her the two gardening cushions to hold and a rug to sit on, and I fetch the trowel and the shears and a scarf for my head. We go to the border beyond the north lawn, which is wide and full of white and yellow flowers of uncertain parentage, and the whole dishearteningly grassy. I kneel and have the trowel poised and sense Nancy standing behind me. She's right behind, leaning forward to touch my shoulder.
"Do you want to help?" I ask her.
"I'd love to help. Nobody ever asks me, though. They're not nice people here. They don't let me out."
"Well, you're out now."
"It's the first time in about five years, I can tell you that."
"Here. Kneel down next to me." She can't, of course. She's almost eighty and her weeding days are over. But she's already bending, apparently effortlessly, to pull dead flowers from last year's _Potentilla_ and secreting them in her palm. She makes a heap on the rug, and then she pulls a tulip out, bulb and all, holding it aloft like a sword and looking vaguely triumphant.
"Let me tell you what needs doing," I say pleasantly.
"You should because I'll make a terrible mess of it otherwise. I can't remember things anymore."
Is it memory that's returned to her, in this gobbet of self-awareness, transitorily, almost freakishly, or is it language, allowing her to express thoughts she can't ordinarily articulate? Has the singing done its work?
"I haven't been very well lately," she tells me. "But I can't quite put my finger on it. What it is that's wrong. Something's wrong, I know that. But I can't... I can't seem to find it."
She knocks the heel of her palm against her forehead and her eyes fill with tears. "My memory doesn't work. I can't remember things. Even quite little things. And they tell me it doesn't matter. At the hospital, they tell me that. But it does matter. It matters to me. They say I'll get better. The doctors, they do say that. I have to be patient. But it's difficult, you know. And I know that something is very badly wrong."
There's a shout from behind and Jack's there. "Hey, Mum. Hi, Gran."
"Hello, hello, hello," she says to him. "And how are you today? How's my little man? You look very smart in your jacket."
"I'm fine, thanks. Well... I'm just going in. Starving," he says, turning and loping away.
"You must be, yes, you must get something nice to eat," Nancy sympathizes. She runs her fingers through her hair. She looks at her old hands and shakes her head.
"Nancy, you have an illness," I blurt. "In your brain. It's the illness that makes you lose your memory." Why do I feel so urgently that I want to make her aware of her predicament? There's no point in it, truly none, when I know that the waters are about to close over her again.
"They told me in the hospital that I would get better, though," she says. "The doctors said to me when they sent me out, 'It will take a while but you'll get better slowly.' 'Slowly, but you'll be fine.'"
"You won't get better," I say. What am I doing?
"Well, that's what they told me, that's what the men said."
"You'll get worse. You won't be able to remember anything. And eventually it will make you ill and you'll die of it." I hate myself. I do.
"We all have to go sometime," Nancy says. She sits down on the rug beside me and we look at each other. I find myself taking her hand, which is gnarled and blue veined and ruched with age. She's wearing eight rings. There's a large wart on her knuckle.
"Tell me what you remember," I say.
"Not very much at all and that's the trouble," she says. Her voice is warbly with emotion. She has a look of great concentration. "My father. My father worked in the garden when I was born. We lived in a big house, you know, huge. It was absolutely beautiful. So many flowers and trees. Then I got sent away in the war. He was so clever, such a good gardener, you know. My brothers were always busy. They got whatever they wanted but I didn't. I minded that and I've minded all my life."
She keeps talking about the "brothers." But she only has one—doesn't she?
"What's your name?" I ask her.
"Nancy. But I've always hated it—ugh! It's a horrible name."
"Do you know your married name?"
"Oh, that. No. Old buggerlugs just sits there."
"He has bad legs, Nancy."
"We all have bad legs sometimes. There's people a lot worse off than him. But he won't get up. He needs to walk about a bit. It doesn't do him any good just sitting. I tell him and I tell him and it doesn't do the first bit of good. He won't listen to me. He doesn't hear a word I say."
"Do you remember living in Edinburgh? Meeting Morris?"
"Oh yes, of course."
"And working at the company? You were the office dynamo, you know. You ran the place more or less single-handedly. Morris and the children never saw you. Worrying over your computer system...."
"Oh, those were such good years. Such good years."
"Do you remember getting the babies?"
"Of course I remember, course I remember."
"What are their names?"
"Oh, it was such a long time ago. I've forgotten all about it."
"You adopted them, didn't you? A boy and then a girl. I'm married to the boy baby."
"Are you? That's wonderful. Oh, they were lovely. What wonderful days."
IT'S THE BOOK—one of the many books that tell the caregiver how to care—that instructs me that I should be straightforward with Nancy about her illness. The Book says that acceptance is the key. It makes two assertions: (1) those who accept the truth handle the situation better and (2) dementia sufferers don't have to be violent and unmanageable. No explicit connection is made between these two remarks but I sense there may be one. Caregivers, in this psychotherapeutic view, are like parents of potentially unruly children. The same scary nanny woman who haunts British TV, who diagnoses faulty parenting in almost every case of trouble, might have a parallel, a scary caregiving expert, sent around to three-generation households to watch, diagnose, prognosticate. She, no doubt, will pronounce _bad caregiving_ as a self-fulfilling road to trouble. Like being a parent, being a caregiver is fraught with expectations, duties, and blame.
The Book says that dementia sufferers should be told what's wrong with them. Trust is vital, The Book explains; it is of paramount importance between caregiver and care-receiver, and there can be no trust without disclosure. If a person has Alzheimer's, she should be informed it's Alzheimer's. She has a right to know and to plan accordingly (to kill herself while still able, I think they mean). A doctor on the Internet writes quite candidly that he would plan for suicide if he were diagnosed with dementia. Suddenly, this approach is commonly talked of in Web land. It's boosting the social respectability of self-inflicted death in the United States, one of my forum contacts tells me. There's quite a bit of it about, she says: the stockpiling of pills, and unemotional discussions about the best, most reliable, and least painful methods. But Alzheimer's has a different profile in the United States—an additional profile to the one we know in the United Kingdom, I'd say, from reading around American forum culture. The sheer numbers of dementia sufferers and much earlier diagnosis: put these together and what happens is that a lobby group grows up. That's what has happened in North America. Articulate, professional, early-diagnosed dementia sufferers, as yet showing few signs of the disease, able to talk and write and head up campaigns, are lobbying for better drugs, for widespread scanning, for more and better research. There is a good deal of humor in the mix: black humor and self-satire. If these new stars of the dementia lobby seem impressively to "go gentle into that good night," it's only because they've been diagnosed years earlier than used to be the case. Their good night is likely to be a ten-, fifteen-, twenty-year event, and they are engaged in a long journey through a gradually encroaching twilight.
The Book says planning is crucial. They make it sound like investing in stocks and bonds. Get organized, they implore. Do the research early, and get the nursing home of your choice sorted out in readiness for the inevitable day. (How do you do that, then? Join a waiting list, presumably. If the home of choice doesn't operate one, it's difficult to plan anything. And how do you coordinate the timing of your relative's need and his or her getting to the top of the list?) In their world, that of the writers of The Book, whom I'd guess to be city-medical or psychiatry-southern-suburban in type and origin, nursing homes are as plentiful and local as boarding schools; the client is king and cost isn't an issue.
The Book expects us to be saints. Make Alzheimer's fun, they exhort. Give it your all as a caregiver. Stay upbeat. Get help. Make tasks with the sufferer as lively as you can. Learn coping strategies. Manage the behavior in imaginative ways. Keep score so that bad days can be analyzed and caregiver behavior adjusted. Assess your approach and change it. Keep notes. Make sure you're not inadvertently making things worse. Don't punish the demented, ever, or reproach or scold; these are all forms of abuse. Remember, the demented are no longer responsible for their actions. Keep calm. Step back. Ignore bad behavior. Don't try to reason with them. Distraction is more effective. Reward good behavior. Above all, create tranquillity. Keep reassuring them. Look at things from their perspective and adjust your pace and attitude. Create routines they enjoy and can relate to. Keep them busy, involving them in home life where you can. Create an environment that makes them feel safe. Give them space, light, warmth, activity, companionship, love. Find somewhere to take exercise safely indoors and out. Adopt child-proofing techniques around the home. Don't allow change—it will upset them. Minimize new things, new situations, new people, noise. _Basically, caregiver, your life is over_. (My italics. My conclusion.)
In general this caregiver manifesto holds true across the media, other than for the first point, about telling the sufferer what's wrong with them, which it turns out is an out-of-date approach. Nobody any longer seems to advise the telling of hard truths. The buzz phrase in dementia now is _person-centered care_. Person-centered care takes its cue from the misapprehensions of the ill person and plays along with her, joining in with the delusions that dementia unfolds. In the United States, a process related to this way of thinking has been called "validation."
I try this with Nancy, try living in her reality unquestioningly, but the present keeps intruding, rearing inconsistently into view. I have a go at providing some props for her as Nancy the office worker. I set up the computer for her and some pens, files, a stapler, and she fiddles with these, opening drawers and making piles and pressing buttons, for all the world like a preschooler at a play group. But the absorption is short-lived and very soon I'm asked who the hell I am and what I'm doing there, as this isn't my office. Then, instead of evicting the interloper, it's Nancy who storms off. I go after her and try to steer her back with some guff about a meeting, and our being late. I take her to her bedroom and help her pick a jacket, but she bursts into tears and demands Morris, and shuts the door on me, and is inconsolable. It's as if the two worlds, mine and hers, are ocean liner and iceberg and can't come together happily for very long. She goes for her nap unhappy and wakes full of confusion, and I redirect her to her lunch, but she wonders where her work colleagues are and I stumble over my responses, watched by a baffled Morris. I hold a warning finger up at him as he begins to say, "You're not in the..." and take Nancy into the hall. She wonders when the children will be home and I ask about them and she tells me they're lost. I suggest we go and make biscuits ready for their coming back from school, and she sits and watches me make them, looking forlorn, not wanting to join in. I give her a bowl of her own and some flour and butter and a cutter, and then because she's lost and anxious, I mix the sugar in for her and roll the dough, and my cheeks burn with shame. She dips her hands into the mixture and is happy for a few minutes but then she isn't sure what to do next and nothing I say and nothing I do can stop the tears from coming back, her head bent low over the bowl, her fingernails full of mixture and her nose streaming. When I say that it might be time for her to go back to the office, my own voice is cracking.
"They don't want me there," she tells me.
"Course they do. Let's go now or you'll be late."
We go, hesitantly, back to the office in the house and she pauses at the door. "This isn't it. Where are you taking me? Why are you lying to me?"
After three or four days of effort, Nancy more unhappy than ever and her humiliation complete, I determine to take the days as they come. I'll try to engage her in real-time activities. I will join in with her if she insists on being in the past, though I won't try to lead the action or develop it. That seems to take us to a place where we both feel patronized and unhappy.
# Chapter 4
_Lulled in the countless chambers of the brain,
Our thoughts are linked by many a hidden chain_.
—ALEXANDER POPE
UNLIKE THE HEART AND LUNGS, THE HUMAN BRAIN doesn't look from the outside as if it's _for_ anything in particular. It doesn't appear to have moving parts. It looks very like an inert lump, silent and motionless. Ancient peoples tended to underestimate its importance. From our brain-savvy, brain-centric world, it's easy to scoff at this, and at why it was that the heart, so obviously merely a blood pump, should once have been thought the seat of the moral self. Plato was considered radical in his idea that the brain might be its real location, as was Hippocrates, who declared, in the manner of a man who expected to be contradicted, that "from the brain and the brain alone comes pleasure, happiness, laughter, as well as sorrow and pain." The ancient Egyptians discarded the brains of those they were mummifying, removing them in bits through the nose with long hooks and binning them, though they preserved the hearts of their kings for the kings' later use.
The brain is a big thing and heavy. It can weigh three pounds. The outer layer, the cortex (from the Latin for _tree bark)_ , is wrinkled so as to cram more surface area into the limited space. The cauliflower shape is actually a twinned half-cauliflower pair, two halves, left and right, with functions allocated, divided, and shared. People who've had the connection between the two halves severed medically experience a troubling dual consciousness in which one hand really doesn't know what the other one's doing.
Brain use is tiring work. Whether purposely trying to think or not, conscious or unconscious, the system takes a lot of energy. Up to a fifth of food energy is dedicated to fueling brain functions. Glucose is the brain's gasoline, and brain glucose levels plummet in Alzheimer's; one of the newer diagnostic tests measures these levels in living subjects. Aside from the 100,000 million neurons, there are ten times as many _neuroglia_ (from the Greek for _glue)_ , cells that form the support network, feeding and repairing the lead actors. This support network also suffers devastating cell loss in the Alzheimer's forest fire.
They're gray, these neurons, Hercule Poirot's _leetle gray cells_ —gray with white axons. And so many of them: 100,000 million is a big number. There are fewer than 7,000 million people in the world. I read somewhere that the phone system covering the whole planet, with all its connections and interconnections, parts of it at rest and parts of it firing with calls, is nowhere nearly as complicated as the interior of one human brain.
While neurons are different shapes according to function, those illustrated in neurology texts are generally star shaped. Neurons are microscopic, but an axon can be as much as half an inch long, directing its communications network in particular sequences, though most stay very local, passing information along like firemen passing water in buckets (albeit huge numbers of firemen and buckets), passing water too fast for the naked eye to see.
The neurons are packed tight in the cortex, and the cortex divides into four main areas, or lobes. The _frontal lobe_ , the front third of the brain, in and behind the forehead, is where we think in the most obvious, self-conscious sense, plan, imagine, debate, decide. It's the area that develops last in the growing child. It's the area that best distinguishes us from the rest of the animal kingdom. It's our executive center, the seat of the executive I. It has vital secondary roles in all kinds of brain function and is crucial in the retrieval of memory.
The _temporal lobes_ , worn like earmuffs at the sides of the head, are memory banks and instrumental in language and the comprehension of language. They analyze sensory input and, with the auditory cortex, interpret sound. The temporal lobes work in emotion as well as memory. The so-called God spot is here, the mysterious brain area that may give us our sense of the divine. In an experiment done with nuns, it was the same small location in the right temporal lobe that lit up within each, shown on a scanner, when they were asked to focus on communication with the Almighty. Richard Dawkins, the biological theorist who wrote _The God Delusion_ , thinks that this God spot, in evolutionary terms, has to do with belonging to a tribe and the socially unifying effects of tribal genuflection. The bishop of Oxford thinks it's provided by the Lord as an interface.
The _parietal lobe_ , at the upper part of the back of the head, helps orient us, giving us spatial awareness, our three-dimensional sense of the world, our own detailed body map, and our orientation to left and right. Number recognition, and the ability to manipulate numbers, is worked on here also.
The _occipital lobe_ , at the lower rear of the head, is responsible for vision. Vision takes up a lot of space and energy. Other centers in the brain collaborate to process visual information. Among its visual tasks, the occipital lobe helps interpret writing.
Across the top of the head like a stretchy headband runs the _motor cortex_ , and behind it lies a second headband-type strip, known as the _somatosensory cortex_ , where messages from the nerve endings in the body arrive for processing and analysis from the spine.
Deep beneath the cortex, the _limbic system_ , folded away in its own compartment, includes the hippocampus and amygdala and our sense of smell. A dulled sense of smell (like Nancy's) may be a predictor of Alzheimer's and contributes to problems with appetite. The amygdala has been described as the fear zone, the seat of primitive emotions, instinctive, fearful, and aggressive. The egglike thalamus, at the center of the system, acts as mediator between the limbic system and the cortex, between instinct and abstract thought, and may be the brain area that most specifically corresponds to the experience of consciousness. The hippocampus processes short-term memory, which may or may not then be laid down into long-term memory. It's called hippocampus because it's supposed to look like a sea horse.
The _brain stem_ is in evolutionary terms the original organ and resembles the whole brain of simpler animals like lizards. It handles all the basic regulatory functions, the heart rate, hormones, sleep, breathing, blinking, blood pressure. It's a bulbous small area at the top of the spine.
The _cerebellum_ , at the base and back of the skull, is an onion-shaped organ that's thought to be a minibrain in itself, a minicomputer, and may be a sort of backup generator for the rest. Traditionally, its main responsibilities are thought to be for movement, coordination, posture, balance. It's also the seat of our most secure, most deeply embedded memories. How to walk, for instance. Automatic actions, the kind we don't need to think about anymore—cleaning our teeth, riding a bike—are handled from here. The cortex learns things and then delegates, once we have the thing mastered. Forty million fibers connect the cerebellum to the cortex.
The romantic view of the brain as an interior landscape predated the Romantic movement by over two thousand years, in its using cave and weather and smoke metaphors. "Caverns there were in my mind," Wordsworth writes, "which sun could never penetrate." Coleridge's "intellectual breeze, / At once the soul of each and god of all." Erasistratus, born three hundred years before Christ, talked about "vital spirit," the _pneuma_ , a liquid life force flowing around our bodies like blood. The second-century doctor-scientist Galen thought the cerebrospinal fluid to be the pneuma and discounted the hard-boiled-egg-consistency, gray-and-white matter that surrounded it as merely protective.
J. K. Rowling uses this antique idea of selfhood as something vaporous, silvery, swirling through the caverns of the mind like mist. In the Harry Potter books, memories can be decanted, studied, held in a _pensieve_. The dying Snape's memories emanate with his final breath and are caught by Harry in a flask, to be reviewed later. Nurses on intensive care wards open windows to let the souls of the just-deceased escape the walls of the hospital. Absurd though the idea of memory as a silver mist might be, it's in truth far closer to our own idea of the workings of our thoughts than the actual mechanism is. The actual mechanism has all to do with electricity made by the body. How can a body make an electrical impulse? Chemistry provides the answer, down at the cellular level—the fact that chemical molecules carry electrical charges that react with others. Your body may be a temple but it's also, far more intriguingly, a laboratory within which chemical reactions are ongoing.
The _resting potential_ of a cell is created by potassium leaking out of it. There's a high concentration of potassium inside cells, and a weak solution of it outside, where there's a high concentration of sodium. The potassium flowing out of a cell creates a negative charge (–70 millivolts). That's the resting potential of a cell. Along comes electrical information—from a pain in your leg, say, or something seen, or something learned, or a memory—and astoundingly, it seems that the information in every case is of the same order; it's just the question of where it comes from and where it's directed in the brain that translates it into pain, vision, knowledge, recall. What happens is that the sodium outside the cell flows in through a hinged gate, creating a wave of positive charge, which happens to be 110 millivolts, so that the balance from the original ‒70 is +40 millivolts. Sodium flows in, potassium flows out: It takes about a thousandth of a second. The electrical charge passes to the next excited cell, and onward in waves, at fantastic speeds. After the sodium/potassium exchange has occurred, a protein inside the cell is responsible for ushering out the excess sodium, chaperoning back the potassium, so that the cell is reduced to its usual state, ready for the next impulse (which is called an _action potential)_.
All this was confirmed, incidentally, by Alan Hodgkin and Andrew Huxley's research using the squid, which has a giant axon, a millimeter thick and visible to the naked eye. They were awarded the Nobel Prize for their work in 1963. Hodgkin commented that the prize should have gone to the squid.
# Chapter 5
_Here reign the simplicity and purity of a primitive age, and a health and hope far remote from towns and cities_.
—HENRY DAVID THOREAU
OUR FIRST BED-AND-BREAKFAST GUESTS ARRIVE, A foursome of young friends, three Scots and an Australian. It is a perfect golden day, windless and warm, and I serve their afternoon tea in the garden. As I'm handing out cups and setting down a plate of warm scones, I'm wondering what to say about Nancy. I may be guilty of being defensive about her. Should I warn people who come to stay about the potential for encounters with Alzheimer's? That's what is going through my mind as I pour tea. Nancy may well want to meet them. She's at an insistently sociable stage. I put the teapot down, take a steadying breath, point out that I made the rhubarb jam, and leave them to it, Nancy's name unspoken. It's too difficult to pitch it. I need to give some thought to this.
Later, I find the visitors gathered in the hall, looking very much as if they want to waylay somebody with a query. A question about eating: How do I rate the pubs in the village? Before I can answer, a hesitant voice says, "Hello?"
"It's all right, Nancy," I say, "just houseguests." And then, to our visitors, "This is Nancy, my mother-in-law."
"Hello, Nancy," they chorus.
"And how are you today?" the Australian asks her. "You have a beautiful house here. We're just admiring the plaster-work."
Nancy, beaming, shuffles forward. Some days she has an old-lady gait, uncertain of her footing. She has her arms outstretched as she comes, grinning. "Oh, it's you! Hello!" she cries to the Australian, and for a horrified moment I think she's going to kiss him.
"Let's go into the morning room; this is where you'll be having breakfast," I say, taking Nancy's hand and yanking her forward.
"Oh, you'll be comfortable in here," Nancy assures the visitors. "This is my house, you know. I was born here. I've always lived here. My father is here, too. He's in the garden. He doesn't mean to be rude." She smacks her lips together. "Well, I don't know, actually. Perhaps he does." She's giggling now.
The visitors are looking at Nancy in a new way. Wondering. I'm wondering, too. She's determined that her father's here, and perhaps he is.
"So is there anything you'd like or don't like for breakfast?" I ask them. "Eggs, bacon, baked flat mushrooms, baked tomatoes, black pudding?"
"Oh, I don't like black pudding," Nancy says gravely. "I hope that won't put you to too much trouble."
"I'm not sure I know what black pudding is," the Australian pipes up.
"It's horrible," Nancy tells him. "But I like tomatoes."
Everybody laughs. It's going to be fine.
AT BREAKFAST TIME, though, Nancy is having one of her restless mornings. This is difficult when I'm trying to cook the Full Hot Scottish. I come down at seven thirty and find her in the hall, stock-still in her pink nightdress, with a pensive expression and one shoe. I put her back to bed—she's cold and has probably been wandering for a while—and go and start the cooking. Then I go and check on Nancy. She's lying in bed flat on her back with her arms crossed over her chest like a medieval marble tomb effigy, eyes open and unblinking. I return to the kitchen. Apples are fried up in butter for the black pudding eaters, and then I go and check on Nancy again. She seems to be asleep. Chris swings into action, cooking up pancakes and collecting the eggs. I take dishes of grapefruit salad through and encounter the guests, who are just coming down. "Good morning," they bellow, and I wince, because they'll wake her. I go to make the coffee and when I come back, I see Nancy, naked other than for a large pair of lilac-colored underpants, coming out of the dining room, followed by a suppressed burst of laughter.
"She's quite a character, isn't she?" the Australian says.
BY THE END of the first summer we have the measure of the house. It is only late in August, as the light begins to fail and the first cold snaps descend, that measuring announces itself and the charming dark green of every inner wall is revealed as algal, the invoker of Stygian gloom. When it rains, water comes into the library. The chimneys are blocked by decades of nests. There are three kinds of heating, all expensive and inadequate. When the wind blows in a certain direction, the rain is driven under the roof and the children are called to bucket duty. When the wind blows in a certain direction, the kitchen stove and central heating are snuffed out and all the fireplaces puff out choking smoke. Needless to say, the wind blows in that certain direction quite a lot. And before you wonder aloud what all this has got to do with September, believe me, even August can be cold. Not cold as in chilly out of the sun. Cold as in _hailing_. Summer happens on about May 20 (with the occasional parting of clouds, an increase in the temperature of the wind) and if you're lucky will stutter through, notwithstanding the occasional storm, a surprise snow shower or two, the odd monsoon, until the final week of July. August marks the start of autumn. On the twelfth, when ice cream's dripping down the hands of children at the southern English resorts, the north of Scotland's putting on its tweed for the beginning of the field sports year.
Despite this hard-won knowledge, the peninsula agricultural show is held in August. It doesn't always rain. It isn't always freezing. The show's sprawling and immensely competitive, whether you're entering a bullock, a Labrador, or a bouquet of onions. Also, Boots the chemist runs out of hair dryers. I know this because I went to buy two for the B and B rooms. "You'll not get one in show week," the assistant told me. I must have looked blank. "They wash the beasts, you see, and then they need to dry them." And indeed, when idling round the stalls on Saturday afternoon, the penned-up, fed-up-looking sheep are unnaturally white and fluffy, fat clouds anchored on legs.
Among the promotional goodies on offer there's free local steak and also rather good beer, in unlimited, plastic pint-glass quantity. I join the steak queue and Chris gets in line for the beers. We notice that some people are eating steak and drinking beer while queuing for another. I, not driving that day (a nondriver, in fact), get back in the beer queue three or five times. After this, I am fearless about breaking the ice. And this is how we come to order the chickens.
We've never kept hens before and I expect to be able to take them home immediately, have already wondered aloud whether they'll stay put on the backseat or, like a Jack Russell, indicate a preference for driving, but alas, the chickens on view are representative only of style and color: _display_ chickens, not for sale. We order six: two Dutch Blacks (good layers, we're told), two Marans (delicious brown eggs), and a pair of French Bluebells (pretty), plus henhouse and accoutrements.
Two weeks later they arrive, dealt out from a vanload of identical hole-punctured cartons. There inside are murmuring little chicken bodies, fluffed up and warm and faintly disgusting smelling. Poor Audrey, a Dutch Black, dies on day three, found slumped on the henhouse floor, and is buried with full state honors in the wood. As in Hollywood, so in the henhouse: Ava and Bette are sworn enemies, plotting behind each other's backs. Doris is chirpy, Lauren sardonic, Marilyn the one that can't find the door (ouch). Nancy enjoys the chickens. Ava turns out the tamest and will crouch low to be stroked, or consent to be tucked under an arm and petted. Nancy bends to touch, her gnarled hand strikingly discolored as it moves very softly and with concentration along the dark feathers of Ava's back. For a few minutes after this triumphant rural experience, I feel vindicated in bringing Nancy here and confident in everything that semiremoteness can offer her.
I DECIDE I need to crack on with the garden, which now, as summer's turning, is gloriously profuse, though the weeds are as profuse as the flowers. There are thistles, docks, a firm infiltration of giant dandelion, and even—horrors—abundant patches of nettle lurking at the back. Morris is supposed to be helping with the garden, if only in an advisory capacity. He kept an allotment for thirty years. But Morris seems to have lost interest. The August wind is shockingly cold, but at least in the separate pair of formal gardens, high walled and south facing, it's still possible to work without a coat. I help him outside, and take chairs out, and make a point of asking his advice about the care and pruning of various shrubs. I'm hoping that sitting in the flower garden will engage him a little. There's a lot to do here, its former grandeur overgrown and tatty, but it could be wonderful.
"It could be wonderful, couldn't it?" I say.
"Uh-huh," he agrees.
Poor Morris seems more depressed than ever. Nothing we do to try to cheer him up—outings, lunches, tea parties, lavish amounts of grandchild attention—seems to make any difference. Several times I almost embark on a conversation designed to let him talk about his sadness—about how he feels now, deprived of his last remnants of independence, brought into our lives and his own effectively over, the two of them smothered by our inept attempts at kindness, unable to salvage anything meaningful from it all. Nancy, her condition, her deterioration, must be overwhelming for him, and it's inescapable that his role now as husband is to accompany her with as much strength as he can muster to the end. He must feel like the oarsman directing the rowboat across the Styx. What would you wish for in his situation, that the current would pull faster or that it would linger longer? It's an impossible question. The impossibility of things getting better, ever, of this being the final slow descent to the end of time—that must haunt him every waking minute, mustn't it? Though I don't know if that's how he feels. We don't have the conversation. I'm not sure why. I can only suppose it is out of some instinct that he wouldn't be glad of it, would find it intrusive and final. Once darkness is admitted, self-consciously, to our situation, then all hope of lightness will be lost. I can't bear to introduce this intensity to all our lives. And so I talk to him about the garden.
Morris must see that the garden could be lovely. It's just that he's found himself on holiday in the valley of the shadow of death and so, understandably, can't speak enthusiastically of such temporal things. Loveliness may offend him, and our happy, irreverent family silliness offend him also. He does, increasingly, seem to offer an unspoken opinion that our continuing, apparently undaunted and unaffected, to be a cheery child-parental group is somehow a failure of tact. Pushing this to the back of my mind, I press on with discussing the planting. The bare bones of loveliness are all here. The walls are of the most beautiful old stone, patina-rich in grayish cream, luxuriant in places with ivy. The trees along the west wall are twisted and furry with lichen. In the center of the garden there's a horseshoe-shaped pond, green with weed, and four borders bracketing it in an interrupted circle, each of them matted tightly with grass. There's a lot to do. The Victorian greenhouse has broken and jagged glass, its roof open to the weather. The grapevine inside has tiny pips of grapes, and its interior raised bed is dense with shell-pink poppies. I am trying to talk to Morris about the plants, but it isn't easy. He is gray, sullen, chain-smoking, and much more interested in what Nancy's doing. It reminds me of conversations with a girlfriend in my then-kitchen, way back, when our children were small and she couldn't focus on anything but toddler discipline.
"What do you think about this _Pulmonaria?"_ I say to him.
Nancy offers him six blades of grass in her hand. "What do I do with this?"
MORRIS: It's just a bit of grass, you silly woman! Put it down! Put it down, Nancy!
She turns to me, holding the six blades at arm's length as if they will bite. "What will I do with it?"
ME: Just put it in the wheelbarrow. Look. Over there. See? Wheelbarrow. With all the grass in it.
She looks around helplessly and I take her to the wheelbarrow. She starts to rearrange the weeds, talking to them. "Now you're a nice yellow one. And, oh look, you have a friend."
I return to the digging. After a while I see that Nancy has gone to the paved area by the greenhouse and is moving stones about.
MORRIS: Nancy, will you stop doing that! Leave the gravel alone!
NANCY: I'm just tidying it; it's my day to sort it out.
MORRIS: What are you talking about, your _day_?
ME: Morris, do you think these primroses will survive if I move them over there?
NANCY: The people who live here want me to do it so I'd better do it.
MORRIS: What are you talking about? You live here. We _are_ the people who live here.
NANCY: Don't be silly. I never heard such nonsense. You say that to all the people but they know who you really are. (She stalks off, through the archway and out.)
ME: I'll fetch her. Shall I get you some tea? A hat? The sun's quite warm.
MORRIS: That would be lovely, dear. I really don't understand what Nancy was on about.
ME: She has Alzheimer's, Morris. You do know that, don't you? You know what Alzheimer's is?
MORRIS: Oh yes, yes. But she talks such rubbish now. I worry she might really be ill.
I return with Nancy and with a tray of tea and biscuits. Tea and biscuits are consoling for Morris and work also as a sort of Nancy-sedative. Then I resume the gardening. I start at one of the borders that frame the pond, digging out deep-rooted mats of trespassing lawn. Jack comes in his Wellies and shorts to help, and gets into the water, a foot or so deep in its concrete mold, pulling out great green ropes of smelly weed and shrieking. The dogs rush about chasing birds into bushes.
But I'm aware that an argument is brewing over in the corner.
"All I'm asking you for..." His voice.
Then hers, shrill with irritation. "Well, I'm not doing it. I don't know where it is or what it is and I can't do it."
"Of course you can do it. I told you, it's on the chair in our sitting room. Our sitting room where the telly is."
"I am not going anywhere and you have no right! No right at all! To ask me anything at all!"
Out of confusion, anger springs.
I lay down my tools. "What is it, Morris? Something I can do?"
MORRIS: I'm feeling a bit cold and Nancy won't go and get my cardigan.
ME: Why didn't you say so? I can go.
When I get back, Nancy is over in the far corner, pulling hollyhocks out by the roots, and Morris is chiding her.
"I was told to do it," Nancy says, "by that man there."
"Here is Morris's cardigan, Nancy," I say to her. "Why don't you take it to him?"
Morris clears his throat. "Actually, do you think we could go in now, dear? My legs are bothering me."
Morris is supposed to be in charge of the kitchen garden, and the seed catalogs sit on the coffee table. The newspaper gets put on top, and the mail, and a packet of biscuits. I give him a notebook and pen and ask for ideas about how to lay the vegetables out, but these are put aside and forgotten about. There's a run of warm days, moist and fuggy. Deep white fogs roll in from the sea and engulf us. Chris gets the acres of grass cut, tractoring up and down on the ride-on mower in the fog, audible from the windows but invisible. The waves crash onto the beach with rhythmic suddenness, sounding bizarrely close, the distance distorted by the bowl of the sea fret, the bay licked by humid and milky mist. Directly overhead it thins a little like a balding head and the sky beyond is a rich and brilliant blue.
Morris and Nancy are invited to go to the Thursday Club in the village. Morris doesn't want to go, but neither is he able to withstand the entreaties of the two women from the club who storm the house to persuade him otherwise. Other than for this weekly outing, Morris sits in front of the television almost all of the time. I've given him a little silver bell to ring if he needs tea, or the fire fed, or a sweater, or help with Nancy, and he uses it with enthusiasm. I have spiritless phone conversations with my mother about the _invalid role_. On sunnier days, I have been encouraging the children to ask him if he'd like a walk in the garden, to be pushed in the chair. He hates the chair and doesn't appear much to enjoy his excursions. In any case, the program comes to an abrupt halt one afternoon when Jack is taking him across the lawn to show him the greenhouse and the chair trips on a tussocky bit of grass, sending Morris flying out forward. Luckily the grass is soft and thick and only masculine pride is dented.
"Do something about it! Do something!" Nancy shouts as I rush to Morris's aid. "I can't do it. I can't do anything. It's only my first day here!"
As summer cools and the days shorten, the true nature of the life we have landed in begins to sharpen and clarify. Caregiving permeates everything and nothing is spared. If Chris and I leave Nancy and Morris alone, something occurs, some small but pertinent crisis. Teapots are dropped and people near scalded. Nancy is found with black hands, black handprints on her trousers and chair, having put coal in the fire without using the hearth tools. Outer doors have been opened and dogs let out. While we are gone retrieving them, there are other crises at home. Nancy has wandered off, leaving Morris panicky, unable to keep up with her. Nancy trips and injures herself. She puts herself to bed and promptly falls out, bashing her head on the bedside table and giving herself an impressive black eye. She is sent by Morris into the kitchen to get him a drink or a snack, and returns with the wrong thing, prompting an argument and Nancy leaving home (again). Morris answers the phone and is stern with B and B guests, demanding to know what they want.
It has, in short, reached a point, _a point_ , of constant supervision. If we go out we have to take Nancy and Morris with us, levering Morris into the high front seat of the Land Rover, belting Nancy into the back, taking them into town, round the stores with us, sitting in tea shops, dealing with Nancy's car-sickness on the way home. Anything that is done without Morris and Nancy in attendance is done at risk, and risk assessment becomes a part of life. We don't go for walks anymore. We don't go out as a family anymore—just the five of us—unless we can go as seven. We go as seven to the cinema, out to dinner, to visit new friends. We're not often invited back.
I begin waking in the night in a panic, heart thumping, clammy. What is this future I appear to have solicited? I never imagined that the in-laws would become so immediately passive, and it didn't occur to me that it'd become so particularly my job to look after them, but that's how it falls, when men have proper jobs and women don't. The conscience is sated, plump and shiny, but the appetite for the day is shrunken, alarms going off all over the internal city. What kind of person is it that can give of herself this much, I think, jadedly, lying awake and waiting for dawn. People who feel guilty about happiness and freedom, I think (shamingly cynical). People who crave dependents, perhaps. People who never really had a life of their own and relish the absolute vocation of this role. This isn't me. This really isn't. I came looking for the Sublime. The hunt for the Sublime, however, has become a grimly private joke. It isn't out there, is stubbornly absent. And actually it's worse. Something else is out there, when I go off running toward the beach, sit on the dunes chewing on the pale inner bits of grass, wander entirely aimlessly along the neck of the headland toward the sandstone plunge of the cliffs. The anti-Sublime. The wilderness will only give me back what I yield up to it and all I have to offer is disheartedness. It lends me its own, magnified and in multiple. I'm not just uninspired, but positively oppressed by outdoors.
_Uncontradicting solitude
Supports me on its giant palm
And like a sea-anemone
or simple snail, there cautiously
Unfolds, emerges, what I am_.
Philip Larkin has the idea of the Sublime in his sights. I want to unfold and emerge. But I'm having the opposite experience. It occurs to me, during one of these walks, that Nancy and I are engaged on parallel journeys, hers into death and mine into depression, though this is grandiose and probably also offensive. My problems are contingent, after all. Life will shift, the sun will come out (the plain fact that her death may be the engine of this improvement is something I prefer not to think about). I'd never claim her metaphorically, poor Nancy, who's twice my age and terminal, when all I can complain of is that I'm demoralized and low. But the beginnings of unhappiness are here, poised at the end of summer, for Nancy and also for me. And unhappiness distorts perspective. Thus it is that when I read on in my Larkin edition and come across one of his many poems about death, I see Nancy there and then myself.
_This is what we fear—no sight, no sound
No touch or taste or smell, nothing to think with
Nothing to love or link with
The anaesthetic from which none come round_.
Caring produces a kind of anesthetic in this narrow sense, in its full immersion into near-intolerable practicality. Ludwig Wittgenstein wrote that "philosophy is a battle against the bewitchment of our intelligence by means of language," and I'm aware that my feelings are being distorted by the anti-Sublime, the terrible useless self-pity I find down on the beach and transpose onto the pitying sea, the pitying sky, the pitying cliffs. Latching onto poetic sentiment has become a sort of literary defeatism. I am falling for the Romantic idea of myself as a victim. Caring is taking me somewhere new, somewhere poetry can't follow without hindering my settling into it. This is a life-and-death struggle I'm engaged in now, someone else's life-and-death struggle. It seems to blot the point of fiction out. I find I can't read novels anymore and turn to biography. Biography and nonfiction. Read Wittgenstein, not Keats, I tell myself. "Beauty is truth, truth beauty"—phooey. Read Wittgenstein. "The world is everything that is the case." That's all. Get used to it.
# Chapter 6
_Life does not consist mainly, or even largely, of facts and happenings. It consists mainly of the storm of thought that is forever flowing through one's head_.
—MARK TWAIN
ALZHEIMER'S DISEASE CAUSES ONLY AROUND 65 PERCENT of the dementia cases recorded, but people tend to use the terms _Alzheimer's_ and _dementia_ interchangeably. A friend of mine does this, referring to her mother's Alzheimer's, when it's fairly clear, meeting the mother in question, that some other kind of dementia is to blame for her illness.
_Senile dementia_ as a term was coined in 1838 by one Jean-Étienne Esquirol, a doctor who noted a progressive loss of memory and initiative and creeping emotional instability in people over the age of sixty-five. The phrase may have been new, but the syndrome was already as old as the hills. Plato was involved in a discussion about dementia in the fifth century B.C. Lucullus, a Roman general, died of dementia, and his decline, as written about by Plutarch, is persuasively of the Alzheimer's kind. Marcus Aurelius, the Roman philosopher-emperor, writes in the second century A.D. that "even if a man lives a long time, it's doubtful his mind will survive him." He goes on to say that "the coming of senility may not be accompanied by respiratory or digestive disorders, no loss of the sensory life or of one's desires, but even so, the power of the faculties, of knowing and doing your duty, dealing with crises, sensing that the time has come to die—all of the decisions, in short, that demand proper thinking about, all of these will nonetheless already be fading away.
"We must get on and live life," he says, "not just because life is brief, but because our understanding may be briefer." It's an issue that has taxed individuals, their families, and the workings of society in all the years since. They worried about it in the fourteenth century: A test was discovered in the 1970s, an equivalent to the question-and-answer diagnostic test (MMSE) used today, that dated from 1383 and had been used to assess the competence of a woman in Cambridgeshire to run her own affairs. The 1970s were an important decade for dementia. It wasn't until the end of the 1960s that it was realized just how prevalent Alzheimer's is. Before that it was thought to be a rare disease, one small exotic branch of senility. Most cases were assumed to be of the vascular type, a furring up of brain arteries, which was still considered a normal part of aging. Alzheimer's was listed in the textbooks as uncommon. It was only when autopsies began to be done on brains in huge numbers, and retrospective microscopic examination of stored brains was undertaken, that it became startlingly clear that Alzheimer's was the main cause of dementia.
Dementia has been important in our history, then, and perhaps more important than is generally recognized. The tradition, or at least tendency, to elect men and women of mature years into power, and to allow people of over seventy to hang on to power, increases the risk that we will have leading statesmen and stateswomen—governors, presidents, and prime ministers—suffering from some form of dementia. It's only fairly recently been discovered that Harold Wilson stepped down as prime minister in 1976 because he'd become aware of his own mild cognitive impairment (MCI), and foresaw accurately that dementia was on its way. Not all politicians have the insight to abdicate so early in the disease. It's alleged that Woodrow Wilson had dementia in office, and that the resulting capriciousness of his decision making culminated in his failure to get Congress to approve the Versailles Treaty that ended World War I. It is also suggested that Stalin was a dementia sufferer, his failing intellect combining suggestively with increasing levels of aggression and paranoia. Roosevelt was evidently quite ill and possibly suffering symptoms of dementia when he had to negotiate with Stalin at Yalta in 1945 (he died two months later of a cerebral hemorrhage). The Labour prime minister Ramsay MacDonald is said to have struggled with dementia in office. Lenin died of dementia, which, as in the case of Stalin, was most likely brought on by syphilis. There seems little doubt that Urho Kekkonen, the president of Finland from 1956 to 1981, had Alzheimer's while in office, a fact actively covered up from about 1978 onward. Ronald Reagan showed early signs of the disease during his presidency.
Dementia is fast becoming the condition that's cited by the young and healthy as the disease that is most feared. It's not curable, unlike cancer. It's not able to be tackled with drastic measures, unlike heart disease and its bypasses and transplants. It's more fundamental than that. We don't _have_ brains; we _are_ our brains. You can lose a leg or an arm, or accept the gift of another person's heart and lungs, and still be yourself. The brain is where the self lives. Lose the use of your brain by degrees and the self is stripped away, layer by layer. In the early stages, the middle stages, even in the early part of the late stage this may well be something you are conscious of, the lights going out one by one.
The dementia numbers are ascribed to our soaring life expectancy rates. It's only an epidemic, so the orthodoxy goes, because we are living long enough to develop it. In 1910, when very little dementia was recorded, only 15 percent of people lived longer than the age of fifty. Life expectancy then was around forty-eight for men and fifty-two for women. We live, on average, around thirty years longer than we did a hundred years ago. Add to this another salient statistic: namely, the number of people over sixty-five worldwide is expected to double in the next twenty years. There's the engine of the epidemic on a plate.
Vascular dementia, the artery-furring sort, is the second biggest dementia disease group by numbers of sufferers. Around 20 percent of dementia victims have this one, and another 20 percent may have a vascular/Alzheimer's combination. It's the dementia that's most equivalent to heart disease. Vein damage prevents blood from getting to parts of the brain; neurons are starved and die. Vascular dementia can be caused by stroke: single-infarct dementia, if it's a single serious stroke; multi-infarct dementia, if it's lots of little strokes, some so tiny as barely to register symptoms, and this is the most common sort. A rare variant called Binswanger's disease begins in blood vessels deep in the brain and may start to show itself with walking problems.
King Lear has been diagnosed, from the verbal evidence of the play, to have suffered from vascular dementia. There's no doubt he suffered from one kind or another of dementing illness. "Methinks I should know you, and know this man," he says in act 4. "Yet I am doubtful; for I am mainly ignorant / What place this is; and all the skill I have / Remembers not these garments; nor I know not / Where I did lodge last night. Do not laugh at me."
The third most common sort is dementia with Lewy bodies (DLB); in fact, some studies claim it's the second most common. Notoriously difficult to diagnose, it overlaps with other dementias. At least 20 percent of Americans with dementia are thought to have DLB, and among the elderly demented the percentage is much higher. Dr. Frederich Lewy identified this variant in 1912, having spotted tiny foreign bodies (proteins again) in the neurons in the brain. Parkinson's disease also has these bodies, though in the case of Parkinson's they're confined to one brain area, the substantia nigra. Symptoms can mimic Alzheimer's, though DLB victims may have more specific problems, with near-normal memory and language skills but trouble with abstract thinking. Hallucinations are so common as to be diagnostic, much more so than in Alzheimer's. Sufferers may also have Parkinson's-like symptoms, trouble with movement and tremors. It's a very up-and-down disease with good days and bad days, good hours and bad hours.
The best known of the frontotemporal dementias is Pick's disease, named after a Czech neurologist, Arnold Pick (1851 _–_ 1924). Pick's was isolated and named in 1892. Specks known as Pick's bodies are found in the frontal and temporal lobes, to which this variant is confined. Pick's can be nasty: It comes on early, can instigate massive personality change, and sufferers exhibit an unfortunate tendency toward lechery.
Frontotemporal dementia (FTD) also includes particular niche dementias, like aphasia dementia (loss of language) and semantic dementia, in which the connections between words and meanings are lost. Frontotemporal dementia sufferers have specific problems with language, behavior, and emotional response. In research results, FTD has been more strongly linked with tau proteins than with plaques. No drugs are available: Alzheimer's medications seem only to make things worse.
Other conditions can lead to dementia. Parkinson's has already been mentioned. Variant CJD is another. AIDS can lead to dementia. People with Down syndrome or Huntington's disease are at risk. Damage caused by long-term alcoholism can mimic dementia (Korsakoff's syndrome), as can B vitamin deficiency, diabetes, kidney failure, thyroid problems, liver dysfunction, anemia, or electrolyte imbalance, though these are only apparent dementias from which people can recover.
# Chapter 7
_Our business in this world is not to succeed, but to continue to fail in good spirits_.
—ROBERT LOUIS STEVENSON
IT'S AUTUMN ON THE PENINSULA, AND MORRIS IS RUSHED into hospital. He gets up out of his chair to go to bed, puts his foot forward awkwardly, and goes down hard, breaking his leg at the top by the hip. Nancy doesn't know this, although she was with him when he fell, and held his hand until the ambulance came, and has visited him every afternoon. We keep it from her—or so she accuses, when it occurs to her to ask where he has gone, two or three dozen times a day. She doesn't take the news very well. Her face puckers up pinkly.
"Why didn't anyone tell me he was in hospital? That's just ridiculous." She sweeps out of the room in a huff, or tries to, her shuffling waddle a little faster than usual, her hands outstretched to grasp the door handle, like a great outsize wrinkled toddler.
There are days when the toddler similarity is persuasive and bizarre. Days when I feel like a babysitter, a new and inexperienced one, given care of a reluctant and stroppy child, having to make it up as I go along, trying more and more desperately to mollify and distract, and feeling that dark needle of fear when nothing I do makes any difference. She shakes her head and stamps her foot and has tantrums. She asks for her father and gets agitated when he can't be produced, looks horrified when I confess that he and her mother are dead. Then, shocked by my own bluntness, I add, "But that's because you are an old lady now." She looks baffled. "You see, you're almost eighty. You don't have parents any longer, but you have children, and grandchildren. Six grandchildren. Three of them are in Canada, and three of them live here with you."
Nancy's face acquires a stony set look. "My. Parents. Are. Coming for me."
When the forest fire of Alzheimer's causes havoc in the frontal lobe, it attacks the site that most approximates our adult selves. Frontal lobe damage can return dementia sufferers to childlikeness, and also childishness. Childishness is the worst because it's coated in a veneer of adult power, assumed authority, and physical strength. Sufferers can become unpredictably emotional, and this is likely to worsen until—probably late in stage 6—it burns itself out, the sufferer too ill to feel anything much. This is a fact I take comfort in, and the idea, leading on from this, that consciousness itself is eroded, so that by stage 7 there's too little left of the self to experience anything much of what's happening. In dementia, emotions can become dislocated from feelings. Emotions are bodily reactions, and feelings intellectual ones. The emotions are produced but the feelings—emotional impulses translated by the thinking mind—are lost or locked off. Nancy is emotional, now that Morris is in hospital, but she doesn't understand it. She cries and is grumpy and cries again and apologizes to us all. "I don't know what's the matter with me," she tells us, and that's literally true. It takes a facility for remembering in order to know what it is you are feeling and why.
Poor Morris is likely to be in hospital for a while. He's been to the county hospital for an operation to reset the bone, and is now back in the town, in the cottage hospital there, in his own room, with a television and a lifetime supply of toffee. The toffee is a way of dealing with the forcible giving up of nicotine. He's become a chain toffee eater. We speak to the doctors about Nancy's urgent need to have him home, in his usual chair. No dice; Morris won't be released until he's a bit more mobile. Nancy sits holding his hand and looks utterly blank. Having no memory of the accident, and unable to remember the hospital from one day to the next, she's having trouble with the context of his being there sufficient to undermine her ideas about who Morris is, exactly. She's no longer entirely sure.
We have horses now, two cobs: a chestnut one and a gray. Mine, the gray, is huge, like a medieval war horse with a long wavy mane. In the evenings, when the children are in place and happy to Nancy-sit, Chris and I ride out onto the headland. Curious bullocks come to the fences and snort, or dash across the pasture kicking their muddy heels, fizzing our horses into a froth. Sheep take off in a sinuous swarm, sticking together but running scared. Blown shreds of feed and fertilizer bags flap against barbed wire. I'm reminded that there's another way of being out in wild places, something that supersedes introspection. Staying on board, the physical harmony of it, negotiating hazards and the intermittent thrill of speed: I may be beginning to see the point of sport.
October stretches out mild and sunny, and the horses sit together, legs tucked sweetly under their tummies, in the long meadow grasses of the lower paddock, fed to satiation and drugged on sunshine. I take Nancy with me, under the white tape of the electric fence, presenting unexpected carrots from coat pockets, scratching under chins and into furry ears.
"Nice doggies," she says. And then, "Listen to me, saying nice doggies! What a fool I am sometimes. They're not doggies, of course. I can't think precisely of the word, though."
"Horses. They're horses."
"Course they are!"
She puts a tentative hand out to a velvety nose. "Nice doggies."
* * *
NOW THAT WE have her to ourselves, Nancy comes everywhere with us. She sits with us to have breakfast, belching and apologizing, a faraway look in her eyes. (Something's wrong, something's missing, but what?) She's becoming vague, losing track of where she is and what for. It was Morris who anchored her days. We need to prompt her to go to the bathroom now, and prompt her to come out again, or she'd sit there most of the morning. Hands are washed with transfixing care, each finger done scrupulously, like a surgeon scrubbing up, and then a lengthy towel-drying ritual begins. I have to remove the towel after a few minutes, when this threatens to go on and on and fingers are beginning to be rubbed red and raw.
After breakfast, if we're not going to town for shopping, we go to the village. Nancy likes to go into the shop and look at all the packets, the piles and rows, the colors. She picks up biscuits and cake and bars of chocolate: things she knows she likes. It's another Alzheimer's way-marker, this childish craving for sugar. She likes to talk to the shop assistant. "Look at that, you're very clever the way you do that," she says as the assistant rings up the prices. "I'm sure I could never do anything remotely like that, to be quite truthful."
We drive home the two miles along the winding seaside road, Nancy holding onto the carrier bag in the backseat and keeping up a steady monologue.
"Look at those things there. Look how far apart they are."
"You mean the sheep?" I crane my neck to look round at her.
"Sheep, is that what it is? They're animals of some kind." She sounds as if she's reminding herself. In Alzheimer's, the learned subtleties of categories of objects become less and less refined over time: A butterfly becomes an insect, and then an animal, and then a thing.
"Look, Nancy, there are cows, on the right; no, the right."
"They're amazing, aren't they? They're so big. You don't think they're going to be so big but they are. They probably always have been. Just me being daft again."
"You're not daft, you've just lost your memory," I tell her.
"That looks like the sea," she says, sounding surprised.
"It is the sea. You live by the sea."
"I didn't used to. I never saw it from one year to the next, to be quite truthful."
WE TAKE THE shopping into the kitchen. "Can I help you with all of this?" she asks, eager to be useful.
"Okay, then. You pass me the milk and I'll put it in the fridge."
"Is this the right thing?" She hands me the newspaper. The matching up of word and object is seriously adrift.
"No, the milk. Big tall carton, white. Cold. There. In the bag."
She hesitates. The old hands, mauve and white and heavy with their burden of rings, hover over the bags. "I can't see it."
It's not happening for her now. It's disquieting. The words _milk, big, tall, carton, white:_ They don't add up anymore to the object right in front of her.
"That's fine, just pass me anything."
She hands me the newspaper again. I put it aside. "Thank you. Now, something else?" She's holding the new jar of coffee.
"Can you put that in the coffee cupboard for me?"
"Yes, I'd be delighted, if you tell me where it is."
"Go toward your room a bit. No. The other way. That's right. Along there and... that's it, stop. You're there. Right there. The cupboard."
She bends to the floor and runs her hand along it.
Why does she do this? Are words and objects jumbling themselves, so that _cupboard_ mismatches itself with _floor?_ Or is it just that she sees I expect something of her, and the floor's the first thing that comes to mind?
"No, not down there. The cupboard. The door. Open the door."
She pulls one of the kitchen chairs out from the table.
I go to rescue her. "Look, here it is. Turn around a bit. There you go. See the cupboard?" She goes to pick up the kettle off its stand. I put my hand over hers and lift it slightly to the handle of the wall unit. "There you go. Cupboard. Remember? Where we keep the coffee and tea. Open it and look inside."
"You're quite right. All the things are in there."
I give her the coffee jar. "Can you put this in for me?"
She holds the jar up and pushes it a little further until it hits a can of chocolate powder. Then she brings it down again. "It won't do it for me."
She has enough latent knowledge to understand what's needed to put something somewhere, lifting the jar into place, but doesn't seem to recognize any longer that there needs to be a gap, a coffee-jar-shaped vacancy on the shelf. Nor can she coordinate the movements to place an object down and let go of it.
Nancy enjoys going into town. It's an ideal size and shape for her, our little town. There's just one main road, snaking from the harbor round into the high street, where for a brief stretch it's been pedestrianized before opening out again by the church. There are side streets leading off, but most of the shopping is here, along the high street, which is crowded with eighteenth-and nineteenth-century buildings and twentieth-century shop fronts, some of them early twentieth century by the look of them, with prewar sign writing. There's a stationer, a music shop, a delicatessen, a ladies' wear shop with just the kind of skirts and shirts and cardigans that appeal to Nancy (but not to teenage girls), two small boutiques (that do appeal to teenage girls, but sell £80-a-pop jeans, so not to their mothers), old-fashioned drapers, Boots the chemist and Woolworths, plus jewelry and gift shops that rely on the tourists. Socially Nancy's become entirely liberated from convention. Meeting people for the first time, she'll likely as not embrace them and, tears welling, say how glad she is to see them again. "I knew you once when I was very small," she'll say emotionally when introduced. Random strangers are hailed in the street.
Nancy likes to go round the ladies' wear shop commenting on things. If I want to make her laugh I suggest something for her in green. She can't abide green: It's one of the few things I can count on for her to remember.
One of the assistants might approach. "Need any help?"
"These were very popular, I was just saying to this lady here"—Nancy gestures toward me—"and everybody wore them when I was young."
The assistant looks at the cardigans dubiously.
"They're everywhere the ones like this in the place, all around us," Nancy says. "And it's really nice to see them again, you know, all together again because they like that. What am I saying? Listen to me blathering on. 'They like that,' she says. Honest to god, I'm losing my mind, I think. But, you know, they're nice, these things, aren't they?"
"Oh yes."
"It's awfully pretty! I was just saying to this lady here when you came in"—she gestures toward me again—"but she thought it wasn't. So. We all have different tastes and it would be a dull world if we didn't."
"That's very true," the assistant says, giving me a special look and retreating. Nancy grins after her and her teeth are appalling, yellow and coated and every crevice jammed with food. She no longer wants to take her false teeth out and clean them and won't countenance their cleaning in situ. Attempts have been made and abandoned. There were tears, fisticuffs, and biting. We're used to the state of them but strangers recoil.
Nancy looks lingeringly after the assistant.
"What a lovely person she is," Nancy says.
Babies in strollers are followed round the drugstore, waved at, sung to, engaged in one-sided conversations, which their mothers consent to warily.
"Look at you, you're gorgeous, and you know it, don't you? You're much more beautiful than the others." She leans forward, tries to pinch a rosy cheek. "The others are nasty about you, but don't you listen."
She's transfixed by the sight of so many racks of nail polish.
"You used to wear this all the time when you were younger, when you were working, do you remember?" I say to her. She'd spend part of every evening laboriously doing her nails, taking off the day's keyboard-chipped varnish, filing into soft points, applying various unguents and then color and topcoat from a vast collection on her dressing table. Morris filled her Christmas stocking with polishes.
She smiles delightedly. "How do you know that?" she asks. "You've been talking to someone, haven't you? But it wasn't this sort. It was the other kind, not in the same one, I mean. Jings crivens, I'm having trouble expressing myself today."
I take her to the shampoos. These now take up a whole aisle. Every time I come in here there are more brands, more daring claims.
"Which one shall we get?" I ask Nancy.
"Oh, don't ask me. I wouldn't know where to start."
"Between you and me I think they're all the same, except that some of the expensive ones are terrible, too busy trying to do other stuff to your hair to clean it."
"That's very true," Nancy says.
"So why don't you just pick the prettiest?" I say to her.
She starts to laugh. "No, no, I couldn't."
"Go on. The prettiest bottle."
She stands with her hand raised, looking embarrassed. She seems to find it impossible to make a choice, or understand what choosing is. Perhaps memory is essential for selecting. How else do we know what we like?
"Just choose one. Anything. Whatever appeals."
She begins to mutter to herself. Her blush deepens.
"What about this pale blue one?" I say. "You like blue."
"Oh yes! That's wonderful."
We pay for our toiletries and Nancy tells the checkout girl that she has lovely hair. She reaches out to touch it but I intercept her arm and hold on to it.
In Woolworths, while I'm buying magazines and having a look at the latest films on offer, Nancy wants to talk to small children. Preschool children, little girls in pink anoraks, small square-jawed boys with buzz cuts and suspicious eyes, hang on tighter to their mothers as Nancy stalks the aisles looking for somebody of three or four to talk to.
"Look at you, you totey wee thing," she says, bending to smile her toxic-toothed smile. A little girl with fair curls smiles back, twisting her Barbie in her hands. Nancy reaches out. "Can I see your dolly?" The Barbie is handed over. "Oh, look at this, she's absolutely beautiful, look at her gorgeous dress." Nancy beams.
"I'm buying her a new dress," the poppet squeaks.
"Kelly! Kelly!" The mother approaches, looking alarmed. "Come on, I said." Poppet is dragged off unwillingly and we hear her mother saying, "I told you not to talk to strangers. How many times have I told you?"
I leave Nancy choosing a chocolate bar and go to get a magazine for Jack, and when I come back, I find her standing by the pick-and-mix with a fistful of truffles, mouth working furiously, three gold sweet papers at her feet.
Obviously, this is a very minor kind of criminal behavior (though I'm glad that the staff here know us, nonetheless), but the principle that makes Nancy feel entitled to the chocolates is one that's dangerous to apply to life in general. The loss of frontal lobe wisdom, moral sense, any kind of brakes on her impulses: It might just as easily apply to a soft-top car, a diamond bracelet, somebody's baby in a buggy. For Nancy, everything is available. It's fair game. If she wants something she takes it. And she believes that everybody in the world operates that way. The idea of ownership is gone, which isn't to say that she doesn't assert that things are hers and hers only; what's gone, specifically, is the idea of other people's ownership of things she might want for herself. This is becoming a problem in the United States, where the huge number of Alzheimer's sufferers means that the legal system is having to grapple with issues of culpability surrounding dementia-sufferer crime. It's a thorny problem. If repeatedly you steal things because you no longer understand what stealing is, what is the state to do with you? What can the mechanics of civilian control do with otherwise fully functioning and peaceable adults who can no longer be reasoned with?
Nancy's hesitant, out on the street. She doesn't like the paving stones, carefully avoiding the cracks, looking down and adjusting her feet as she goes, first with small steps and then at a stride—an inconsistency that is tricky when your arms are linked together. She no longer deals well with changes of level, either, hesitating before going up or down from road to curb. At home, she has developed a thing about the black-painted slate floor in the back corridor, pausing as she comes off the carpet and dipping a toe in the "water" first. She thinks it's going to be wet. Sometimes she thinks it's a hole and I have to go first.
If it's blowy on the high street she hangs on to her hat, laughing near hysterically. We go into the council-run coffee shop, the Victorian Gothic ex–council headquarters, and eat a subsidized bit of apple pie with scalding weak coffee. Nancy likes it in here. She eats her pie with relish and licks the plastic container.
She's happy in town. It's when we get back that the trouble starts. She has no memory of being here, but emotional associations with things remain, subconscious associations, and Nancy's begun to associate the house with incarceration. Her spirits wilt visibly as she trundles back into the kitchen, and is steered toward her sitting room. She doesn't want to take her coat off or her shoes. She retreats to her bedroom, putting her bathrobe on over her coat. I go and crouch at her knee and take her hands and look up at her. She looks angry.
"Don't speak to me. Don't say a word," she growls.
"What's the matter?"
"Nobody talks to me. Nobody wants anything to do with me. They invite me here but then they ignore me. I'm going to take my things and go."
"But Nancy. We're your family. We look after you."
"You DO NOT."
"But Nancy. You don't know where you are."
She laughs mirthlessly. "That's what you say."
"Okay, then." My dander is up. The apple pie will go unrewarded. No kindness will go unpunished.
"Tell me where we are, then. Go on. Tell me."
She looks out the window. "We're here, of course."
"But where are we?"
"Edinburgh."
"We're not in Edinburgh."
"Well, you aren't." She puts great emphasis on the _you_. Is this a metaphysical point? I wonder. "I'm going home," she adds.
"How are you going to get home?"
"I'll be fine."
"It's a two-mile walk to the village bus, and even then..."
"Well, that's me, then. I'll say good-bye to you."
I follow her out. She's standing in the yard, looking in astonishment at the great open garden, the wall, the sea, the sky. What must it be like, to be as sure as you can be sure of anything that you're awake and in your own city, and open the door and find the ocean there?
"Come on, let's go in, it's starting to rain," I say. She follows me meekly indoors.
NANCY'S MOODS TAKE a decisive downturn. Daily she tells us that none of us love her, that none of us like her, that none of us want her here. We spend a lot of our weekends trying to convince her that she's wrong. Trying to convince somebody that you love them is exhausting work. Particularly when you need to reiterate it all, almost word for word, twenty minutes later. Chris takes on the job of trying to distract her at the weekends so that I can have a break. But, of course, I feel terrible, sitting by the fire with the Saturday papers and hearing it all going on. I go into the kitchen to make coffee and find the two of them, Chris and Nancy, sitting at the table making soup. Nancy has her own chopping board, her own knife, and is busy mangling a potato. Chris is saying, "The thing is, Mother, that we don't ignore you on purpose. The thing is that we all have busy lives, and things we want to do. We both have to work, we're working people who have to make a living in order to pay the mortgage on this big house and look after you. And when we're not working, we have other things we want to do sometimes. We want to go out, and spend time with our children, and paint pictures, and read books. We can't be sitting with you talking every minute of the day and you have to understand that."
Nancy says nothing, her knife jabbing at the potato.
Later on, when everybody is called for supper, Nancy refuses to get out of her armchair.
"I'm not coming."
"Supper time. Soup and lovely homemade bread and apple pie. You helped make it, remember?"
"I'm. Not. Coming."
"Aren't you hungry?"
She looks at me, takes a breath. Thinks better of it. Then takes another.
"I know very well what you're trying to do. You're trying to get me out of here."
"That's right. It's supper time. It's just through there. That door. The kitchen. Your food is waiting for you."
"I know very well what you're doing. I'm not allowed out of here. I'm to stay here. I've been told that I'm to stay here and not move. I'm not to move a muscle. I'm not allowed in there, oh no, that's what she said, she said I was to stay put and not move. The people who own this place told me that I'm to stay right here and _I'm_ not allowed anything to eat at all."
Scraps of information in Nancy's paranoia are traceable. I do tell her to "stay there" when I go off to get her supper. I do tell her she can't have a whole tin of shortbread to herself, that she's not allowed it (especially not five minutes before supper). These oddments get mangled, garbled, by the disease. Jigsaw pieces that don't fit are forced together to make a whole new picture.
I take her supper to her on a tray. The rest of us eat our dinner in silence, subdued by the outburst, hearing Nancy telling her soupspoon her troubles.
When I take her coffee in, she is crying. I crouch by her.
"What is it? What on earth's the matter?"
"It's nothing," she sobs. "Nothing at all. Just people being nasty to me. It's always happened so I shouldn't be surprised."
"What people?"
"Not you. I'm not talking about you. You're very nice. You're the only one that's nice. The rest of them are nasty to me. And they laugh at me, those nasty children. They laugh at me behind my back."
I can feel my hackles rising. "Don't you dare talk about my children that way," I say hotly. "Those are your grandchildren, and one of them, your grandson, tried to help you today and you called him an arsehole."
The idea of Nancy calling anybody anything remotely this rude is pretty funny, in retrospect, but it isn't amusing at the time.
"Granny's gone to the dark side," Jack warns his friends when they come to tea. Jack and his friends can handle it, raising their eyebrows at each other and making themselves scarce, but I am a lot less sanguine.
I've wondered, since this period, whether Nancy was bullied as a child. Whether long-term memory is creating long shadows in her dealing with children, now that she's ill. Her grandchildren are indescribably sweet and tolerant, rushing to her aid whenever she's troubled, trying to anticipate her needs. So either the brain is inventing maliciously—can misfiring neurons be said to be malicious?—or there's something from the past that's got mixed in, released by the subconscious and bobbing to the surface.
Nancy's wandering at night, presumably looking for Morris. I am dimly aware of this, surfacing from sleep half a dozen times, aware of noises below. Doors opening and closing. Someone talking. But I can't wake myself up enough to go and do anything about it. When we come down in the mornings we find all the doors open, things rearranged, piles of Nancy's clothes on the pool table.
# Chapter 8
_Trust the instinct to the end, though you can render no reason_.
—RALPH WALDO EMERSON
NANCY AND MORRIS HAVE HAD FIVE PROPERTY STAGES of life, which have mirrored what you might call the Five Ages of Nancy. These have been, in brief, apartment-house-apartment-bungalow-us. Property number one, the original apartment, was part of a period-house conversion in the city. They bought it before they were married and worked on it on weekends leading up to the big day. They graduated to property two, a stone-built row house at the end of a row, with its family bedrooms, little garden, handiness for the park and the primary school, as a prelude to the adoption of two children, first Chris and then his sister, who lives in Canada with her own young family and is rarely in touch. Downsizing to number three, a ground-floor retirement apartment, came at the point at which there were young granddaughters, and Morris could walk only with sticks. They were intending to stay at property three for the rest of their lives, but there were two other unexpected moves to come, as a response to encroaching dementia. Three, if you count the nursing home.
Alzheimer's crept up slowly, like Granny's footsteps in the game "What Time Is It, Mr. Wolf?" (Mr. Wolf isn't sure. Mr. Wolf has forgotten how to read a clock.) The first obvious clue was the classic one: forgetting things, being _absentminded_. "Where are the keys? Where? _What_ clay pot on the dresser?" But then, over three, four years a bad memory became something else. Nancy needed reminding about money and its rudimentary mathematics, how the door opened and was locked, and about things we'd just talked about. Eventually, under pressure, she saw her doctor and ended up at the memory clinic at the hospital. We'd had no encounters with dementia before, and stupidly we failed to realize that _memory clinic_ is a euphemism. We'd ask, on the phone and on our visits, how things were, what the clinic had to say, but Nancy was vague and Morris studiedly vaguer. "I think she's getting on fine, though her memory's terrible. They've put her on some pills."
The pill, it turns out, was galantamine. Alzheimer's wasn't mentioned. It was on her file, though. A later doctor would mention it almost casually, would see our faces, would be shocked by our not knowing. "It helps with memory loss" was all that was said at the time, about the prescription; a fudge enacted by the clinic for an anxious spouse's sake, perhaps, or, more disingenuously, by Morris for ours. There's no cure for dementia. There's no partial cure. All that's available is a slowing down of the symptoms of fire damage. Sufferers' experience of the drugs available is patchy and inconsistent. They don't work for everyone. They're hit and miss, and usually only of short- to medium-term use.
The current drugs work by targeting the synapses, the gaps between neurons through which all information must pass if a brain is to function. To cross the synapse the electrical impulse becomes briefly a chemical entity, a neurotransmitter. It's as if lorries bring goods to the port and unload them briefly onto ships to cross a river to the other side, where other lorries wait to be loaded for the next road journey.
Of the four Alzheimer drugs currently prescribed, three of them—Aricept (donepezil hydrochloride); Reminyl, known as Razadyne in the United States (galantamine); and Exelon (rivastigmine)—work as acetylcholinesterase (AChE) inhibitors. (The first of this class of drug to be launched in the United States, Cognex [tacrine], is now used only rarely because of the severity of side effects.) AChE is the enzyme that breaks down the neurotransmitter acetylcholine, getting rid of it once it's been used so that fresh neurotransmitter can be produced. In Alzheimer's fresh neurotransmitter isn't being produced as much, so the drugs work by keeping the old stuff going for longer and preventing it from breaking down. A fourth drug, a newer one called Ebixa, also known as Namenda (memantine), which may herald a new generation of drugs called NMDA receptor antagonists, blocks the overproduction of another neurotransmitter called glutamate, a glut of which causes cell damage. U.S. studies have found that Ebixa can be taken with one of the other sort of drugs to beneficial effect, but this, of course, is an expensive approach, and the UK National Health Service is unlikely to go for it. (More important, I'm told that its early promise has come to naught and it isn't much good anyway.) Additionally, it seems that these Alzheimer's drugs are proving in clinical practice to be quite good with other forms of dementia, specifically with suppression of bizarre behaviors, and are beginning to be used in place of antipsychotics. Antipsychotics are widely overprescribed. There's no doubt that in many cases dementia patients in homes are given them in order to make the staff's lives easier. It's estimated that around 100,000 people in nursing homes in the United Kingdom are misdosed or overdosed in this way, and many more in the United States, despite regular outcries in media such as the _New York Times:_ people rendered doped-up and compliant. There must be a great many more on the drugs cared for at home, thanks to doctors' prescriptions. The legitimate use of antipsychotics may be helpful in moderate doses for particular problems in dementia, but there's no doubt they impair thinking and speaking abilities. In addition, it's estimated that 25 percent of those taking them will die prematurely as a direct result. Stroke is a particular risk.
By 2012 there should be another name on the market, to add to the Big Four. The benefits of Rember (methylene blue) have been the cause of excitement among those involved in the trials. Methylene blue isn't, it seems, a new drug at all, but one new to the treatment of Alzheimer's. It appears to be effective as a tau protein inhibitor, attacking the tangles. Unfortunately, it's likely that political questions, having to do with who will get it and how much it will cost, will shadow its UK launch publicity. Plus, it remains to be seen how effective the wonder drug will be in real time and on actual humans. As someone working in the field told me, "Brains are so blooming complicated that the potential benefits from drug effects are often not really well understood for years after the drug hits the general market and doctors in ordinary places on earth (as opposed to research clinics) get to use them routinely."
Until then, only the usual pharmaceutical suspects are available. It's no wonder that people look for alternative approaches, ginkgo biloba, fish oil, folic acid, turmeric (curcumin), and HRT (hormone replacement therapy) among them. The medical establishment is trying to think laterally. Blood pressure medications and cholesterol-reducing statins have shown promise. Great things are claimed for vitamin E, which in vast doses has been shown to slow the disease, though other studies pinpoint vast doses of vitamin E as a killer. Anti-inflammatories like ibuprofen may slow or even prevent Alzheimer's; inflammation in the brain is a busy area of research. More cheeringly, it seems that one generous alcoholic drink a day may be protective. Some little-publicized research has shown that smoking might, also. Of course, at this level of selection, it's what you die of that becomes the issue.
I wonder now how many years Nancy was having problems before memory loss became obvious. Was there an even longer, slower fade than we thought? She'd not wanted to deal with meals and cooking for years, had been strongly averse to supermarket shopping for many years before that. Nancy's known for hating supermarkets. Is that where it started, that first tickle of dementia—in not being able to deal with the navigational demands of Safeway? Recently researchers have pinpointed the entorhinal cortex, which feeds into the hippocampus, as the more exact starting point of Alzheimer's in the brain. There isn't merely memory consolidation at stake in the entorhinal zone, but also mapping, mapping the location of objects in relation to the self. Alzheimer's may announce itself with navigational problems rather than memory loss as such.
It has the subtlest of beginnings, fuzzy and subjective. There's no absolute starting point. It doesn't start with pain, a suspicious lump, blue spots, an _attack_. It's easy for dementia to be self-diagnosed, or even medically diagnosed, as something else entirely. Natural aging. Middle-aged confusion. _Senior moments_. The doctor confesses he forgets just as easily and is just as erratic—him, at only fifty-six—and not to worry, and it's hard to argue. It's a creeping illness. It creeps up on people. People get used to their own very slow changes and make allowances for themselves. It's possible to muddle through life with conviction and survive perfectly well as a muddler, and among the elderly it's considered reasonably normal. We use words like _dotty_ , and smile about it; being in a muddle is often endearing. MCI, mild cognitive impairment, is after all only _mild_ , and only an _impairment:_ It doesn't sound too threatening.
In his account of the seven stages of Alzheimer's, the behavior-based guide to severity devised by Dr. Barry Reisberg in the United States twenty-five years ago and still widely consulted, other people notice the changes only when they've reached stage 3. Stage 1 has no impairment evident. That's the slow burn. Stage 2 is the dotty stage, with intermittent memory lapses (and intermittent _anything_ can be explained away), forgetting where you put things, forgetting words and names. Described like that, most of us post-forty-five could, with the aid of hypochondria, believe ourselves at stage 2. Stage 3 has taken us only as far as MCI, when others begin to notice something amiss, and less endearing kinds of muddle arise—trouble retaining information, with short-term memory, with reading and organizing. We're still on the fringes of what we'd consider normal for the elderly in stage 4: trouble talking about world issues, trouble with mental arithmetic, trouble with trivial activities like making supper—these in themselves may not ring alarm bells. And yet it could have been ten years, the road from stages 1 to 4. Someone's had Alzheimer's for ten years and nobody's yet recognized it.
Stage 5 is considered the first of the three obvious dementia stages. At stage 5 help is needed. The sufferer might forget his own address, phone number, the names of family members, and be hazy on major autobiographical events. He begins to lose his grip on what day, month, year it is. He may need help with dressing. At stage 6, there are noticeable personality changes in play. The sufferer loses a coherent sense of his environment, is cut off from his own history, may not recognize his spouse. All of this brings fear, and in its wake, aggression. The person needs help with all domestic tasks, including toileting. There may be some incontinence. Wandering becomes an issue. Delusions and hallucinations may begin or worsen. He may talk to himself in the mirror and adopt compulsive behaviors.
Stage 7 casts the darkest shadow. The sufferer loses the ability to talk, walk, eat. Gradually he becomes bedridden, immobile, and helpless. The horrific thing, should we dwell on it too long, is that stage 7 has been known to last for six or more years. But not everybody gets to stage 7. These markers are, remember, just averaged out, just the rule from which exceptions spring.
Ralph Waldo Emerson (1803 _–_ 1882) was an example of the long, slow fade. The American poet and essayist died of dementia (probably Alzheimer's, which is the only backdated diagnosis possible), a fate he'd already recognized by 1866, the date his son gives to "Terminus," which begins:
_It is time to be old,
To take in sail:—
The god of bounds,
Who sets to seas a shore,
Came to me in his fatal rounds,
And said: "No more!
No farther shoot
Thy broad ambitious branches, and thy root.
Fancy departs: no more invent;
Contract thy firmament
To compass of a tent_.
Emerson would live for another sixteen years after writing this. The epigrammatic and quirky nature of his transcendentalism helped disguise his affliction until very late. Mark Twain famously made fun of him in a speech at a dinner five years before Emerson's death, parodying one of his poems, not realizing he was ill (though, in fact, Emerson was by then far too ill to take offense, as his daughter explained in a letter). Robert Graves (1895 _–_ 1985), poet and author of _I, Claudius_ , had an even slower fade. It's said by his friends and biographers that his dementia was apparent even in the 1960s, though he masked it well with outrageous eccentricity. Among elderly artists and writers, there is sometimes a difficulty in knowing where art stops and Alzheimer's starts. Graves became passionate about astrology and bowed to the moon. He became convinced that he was the mortal mouthpiece for the White Goddess. He was ill for a long time and spent the last decade of his life virtually in silence. Dementia took its time.
It's an oddity of the disease that the first phase can present itself quasi-positively. Sufferers may not be seen to be suffering. They might seem to be happier. Among the great ocean of Alzheimer's writing online and all its many miseries, there's the occasional early-diagnosed commentator, reporting on experiences in a way that seems near euphoric. People with early Alzheimer's have reported a heightened sensory experience of life: Perhaps the senses sharpen as the thinking and remembering self begins to dwindle.
THIS GIVES ME an understanding of how it seemed with Nancy in the mildly muddled early years. Nancy seemed a happier person and more relaxed. She made more of a fuss over her grandchildren, especially her toddler grandson, apple of her eye. The children were greeted with outstretched arms and Jack was invited onto her knee shortly after. The granddaughters were allowed to take Granny off to the bedroom and give her a full makeover. She'd return merrily in odd combinations of evening clothes, her hair stiff with gel, wearing stripes of blue eye shadow and heavily rouged.
The truth of things came out piecemeal. Things had reached a stage beyond forgetfulness. Morris and Nancy were beginning to fight: him accusing her of not listening; her defending herself energetically. Things had taken a downturn domestically—though even in her prime Nancy was never much of a cook. They both worked long hours and Nancy became a dab hand at short-cut cuisine; macaroni made with Campbell's tomato soup sauce stands out in the memory. But everybody worshipped Nancy's raspberry jam. There were never jokes or winks about the jam. A jar had such currency that my father hid his from the rest of us at home, on a high shelf in the kitchen. One of my earliest memories of Nancy sees her presiding over great bucket-sized preserving pans, sterilized jars set out ready. Ironically, one of my first conversations with her—while helping with the raspberries, and I can't remember how the subject came up—was about euthanasia. It would turn out to be a pet subject. "There's no point keeping people alive who are useless," she'd say, in addition to the usual line about "dogs in that state not left to suffer by the vet." I was afraid to inquire what "useless" meant. Looking back on it, I'm certain that elderly dementia sufferers would have been counted among the condemned.
After she and Morris moved from their house to their apartment, Nancy began sifting. It's typical Alzheimer's behavior to embark on pointless projects like turning out all the old linens, or books from the bookcase, and abandoning them half sorted on the floor. The power to organize simple tasks, like the few sequential steps that make up emptying out a drawer and restoring it to tidiness (life is all about sequential steps), is hindered and then obliterated by the advance of Alzheimer's into the frontal lobe. The impact of frontal lobe damage is enormous. When lobotomies were commonplace, its victims became passive and easy to handle. Unfortunately, they also lost the ability to make a sandwich or tie their shoes.
Then, on one visit to Edinburgh, we found the kitchen cupboards bare. Eventually the truth was admitted to. Morris couldn't any longer walk as far as the shop. Nancy could, but she couldn't find it. She'd go out to get milk and bread, a tin of luncheon meat, and come back without them. Some days it took her a long while to come back empty-handed, as if she'd been lost. Nancy, it transpired, hadn't been to the corner grocer for six months.
"Six months! But why didn't you tell us?"
"It's all right, though, because our neighbor does it for us, when we ask her," Morris said.
A home help was appointed for three mornings and things were stable for a while. I took charge of their food shopping. Morris dictated his list on the phone and I organized the supermarket to deliver.
They began to be prey to unscrupulous salesmen. Morris amassed a suitcase full of jewelry and watches, bought from a "friend" nearby who _knew somebody_ and appeared regularly at the door with bargains. Four televisions were purchased, two hi-fi systems. Boxes of wine began appearing, stacked behind armchairs: Morris spent thousands ordering a massage chair, then canceled it and was only partially reimbursed. Brochures for new kitchens, double glazing, time shares, and computer systems littered the house, with local agents' names attached stapled on business cards. Rug hawkers called in unmarked vans.
Nancy continued to be busy. The bath and bed sheets were gray with grime, but the kitchen surfaces were cleaned over and over. The sink was scrubbed silver, and the dishes, greasy or not, were washed under cold running water.
Visiting with the children was becoming difficult. In the old days, Chris and I would have been able to take up the offer of an evening's babysitting and go to the cinema. Now that had to be abandoned. We'd get back and find wakeful children crying in the sofa bed the three of them shared in the study.
"What on earth did you do?" I'd ask.
"Nothing, really. We might have been a bit noisy. We were just excited. But Granny got annoyed and Granddad completely lost his temper." In the old days, there'd be a child-grandparental conspiracy to get us out of the apartment. It was a game we played and that we all enjoyed. Chris and I would lay down the rules—an early night, no adult TV, no sweets or fizzy drinks—knowing that flouting them was entirely the point. Sure enough, we'd get home at midnight to see children scurrying to bed, chocolate round their mouths, and hear happy giggling. But those days seemed to be over.
Nancy and Morris became isolated, cut off from old friendships. Did they jump or were they pushed? It's hard to say. In all likelihood there was both jumping and pushing. Nancy had become a social liability and Morris, perhaps, thought himself protective of her in cutting the ropes that had so long linked them to their Edinburgh circle. For whatever reason, the Saturday night out with the gang came to a dead stop. Nancy's oldest friend, Carol, was deterred from visiting. People were fended off and kept at a distance. Morris dug the moat and took up the drawbridge. And gradually, his relationship with Nancy changed from husband to keeper.
# Chapter 9
_There seems something more speakingly incomprehensible in the powers, the failures, the inequalities of memory, than in any other of our intelligences_.
—JANE AUSTEN
THE MOVE UP FROM EDINBURGH WAS PROMPTED BY health incidents. Morris had fallen several times, and on one of these occasions broken his shoulder. Nancy had been admitted to hospital twice with a blockage in the bowel, thanks to poor nutrition and dehydration. Chris found his life punctuated by panicked phone calls. The rest of us found our lives punctuated by long absences, the Land Rover roaring off late at night on another rescue mission.
There were high expectations of the bungalow. Theoretically the bungalow was perfect: two miles from us in a small development just off the center of a charming village. We were confident that this, a supervised life, would work, so the fact of its being so spectacularly disastrous was doubly surprising.
Health professionals have said to us since that it was a mistake to move Nancy to the peninsula, but they are missing the point. Moving her from the city apartment was the error. She'd moved into the apartment ten years earlier healthy, mildly impaired at worst. When she left she was ill with Alzheimer's, but she had a residual identity there, residual mapping of her location, and this provided her with some bearings. It doesn't really matter how many times you move an Alzheimer's patient once functionally the hippocampus is gone. Everything is new, every day, and, living in the present tense, the best that can be hoped for is present happiness.
By the time she moved north, Nancy had lost the ability to lay down new memories, literally lost it. The hippocampus was wiped, deleted, shot through with holes. It wasn't that she _had trouble remembering things_ , a phrase easy to use but that hints at a patchy up-and-down kind of unreliability, books misfiled in the personal mental library. This wasn't about memory retrieval, but about memory formation. It wasn't a Romantic matter, of locked-off bookshelves and mental caverns without sunshine, but physiological and technical. No new memories could be made. Nancy couldn't learn her new surroundings. She couldn't map the village, nor learn the layout of the bungalow. She'd ask Morris, twenty times a day, how to get to the bathroom and where the kettle was. She patrolled the house, wearing a groove in the new carpets, wringing her hands and weeping. Every day she asked if they could go home. Morris wasn't sympathetic.
"When will you get it into your head? We live here. This is home now."
But that was exactly and entirely the problem. She couldn't get it into her head. Morris spent long periods sitting in the courtyard at our house smoking, Nancy dancing attendance, asking what she should do for him and failing to carry out his instructions.
"I said my stick! My stick! I need my stick to get back into the house, not the newspaper!"
"Ashtray, Nancy, ashtray. You know what an ashtray is. Go and find Andrea. What do you mean, 'who?' Look, that's her there, in the kitchen. Kitchen! Through the door. Where are you going now? The door, door! For Christ's sake!"
"Why are Granny and Granddad always arguing?" Jack asked. "And why doesn't Granny know what a fork is?"
When Nancy had a minor stroke and spent three days in hospital, Morris moved in with us. He sat crumpled by the fire, lamenting. "Nancy, my Nancy! I can't lose her. She's all I have. What will I do? What will become of me? I don't think she's going to make it. She's going to die and leave me all alone." Nancy, meanwhile, was fast becoming the ward's most challenging resident. She heckled the nursing staff. All the doors had to be locked and windows secured because she was determined to escape. She did a remarkable impression of somebody not remotely at death's door.
ZOOM FORWARD A year. It's the end of October, our first autumn in the north, and we have friends to stay for Halloween. Nancy is a lamb. Nancy is a trouper. Nancy is conversational and light of heart. Then the health visitor drops in to see us and chats to Nancy, and Nancy handles this magnificently—seems actively, intently to be _handling_ it.
"How are you, Nancy?" the health visitor asks.
"Can't complain," Nancy says. "Well, I could complain, of course, but I won't." (Laughter.) Wit will always win the argument. "And how are you, yourself?" she goes on, all frowning and earnest solicitude. "Are you keeping well? You're obviously very busy. But keeping well, I hope." The health visitor is charmed. The health visitor, on her departure, is heard to use the word _sweet_. Instinctively, provoked by the institutional air and antiseptic smell of the visitor, Nancy knew she had to perform well and Nancy dug deep. The health visitor was managed. She was wrangled. This seemingly contrived approach to social situations is a new feature in Nancy's decline but also, it turns out, classic Alzheimer's. The American dementia blogs, particularly, are full of astounded remarks about severely ill and abusive relatives being winning, engaging, almost like their old selves when doctors come to call.
Dramatic news from the hospital: Morris needs a second operation, having dislocated his hip while recuperating, probably through inadvisable crossing of legs. He's sent back to the city, the rest of us into despondency.
November 5 is the windiest Guy Fawkes Night we've ever seen. We abandon the idea of the traditional bonfire (though we've never made and burned a puppet Guy Fawkes to go on the top, so it wasn't ever that traditional), and then even the packet of sparklers is judged too dangerous to use. The gale blows and howls round the eaves and the heating's snuffed out. It's 7:00 P.M. Up the back stairs, in the guest apartment, the last B and B guests of the year have retreated to bed and are watching a DVD from under the duvet. The wind roars down the chimneys and the drawing room is full of smoke. It appears to be windy _inside_ the house. The next morning, our visitors attempt to walk the beach in a sandstorm, though summer's lovely strand is kelp and refuse scattered. They attempt (inadvisably) a walk on the cliffs. They go out in the car for lunch and come back soaked to the bone. They are relieved to go home.
The north of Scotland needs a broader vocabulary for weather, such as the Inuit are said to have. We need thirty words for wind in all its variations. Its principal variant is aggressive. It slams. It blows dog ears flat, knocks children over, gets into a coat and sends it soaring. It's best not to open an umbrella unless the full Mary Poppins experience is desired. It forces entry into the esophagus, making breathing feel like work. Hats become offerings. They're whisked away, dropped into far fields, into the sea, off cliffs, deposited muddily on roads awaiting their next victim. An old scarf, very long and broad in a brown check, has become my constant companion. The scarf is applied to the head, wrapped securely and tied at the neck. Scarf, long waxed coat, Wellies: I don't leave the house without these three items. It's my new silhouette. A Barbour bag lady.
In late November, we wake after a stormy night to find that an enormous chunk of the flower garden wall is missing, some twenty feet across and eight feet high. Sheep were sheltering behind the wall and two of them were killed, crushed under falling stone. The farmer comes round to see us. His reputation has preceded him so we expect to be presented with a bill, or with some cunning and unusual revenge. Instead he's philosophical. These things happen. He won't hear of payment. He helps stack a delivery of hay bales into the barn, and goes off whistling.
I start writing in bed, late at night and very early in the morning, the rest of the household asleep. In daylight hours it's proving impossible, that is, until 4:00 P.M. when the cavalry arrives, disheveled and hungry with schoolbags on shoulders. I leave toast-eating granny-sitters in charge and sneak off to get on with things. It doesn't work. Nancy comes shuffling in after me.
"Excuse me, lady. I'm sorry to bother you, but I need to ask you something. Have you seen my husband?"
I close the laptop lid and we have the conversation, the full-length one that reprises her life so far, her marriage, her children, her retirement from work, her move here, and Morris's accident.
"Oh," she says. "I wasn't told any of that. Would be nice if people told me things."
She shuffles out. Then she shuffles back. "Excuse me, lady, I need to ask you something...." It all begins again, an almost word-perfect repetition. The only way out is to hide, and we take turns spending prolonged periods secreted upstairs, where Nancy doesn't go; she has developed a fear of stairways.
My transformation into her mother is complete. She wants to be where I am. If we're apart for more than a few minutes, she begins to fret. When she finds me in the drawing room her voice is full of relief.
"Ahh, there you are, I was worried."
I explain that I'm trying to get a few minutes' peace, reading a book by the fire. Would she like to sit with me and look at a book?
"Yes, I'd love it. I've been all on my own, nobody speaking to me at all." She then proceeds to free-associate. I put the book down.
"Okay, but as I say, I'm trying to read just now, so... Listen, do you think you could mind the children for me for a while?"
"Oh yes, I'd be happy to. Where are they?"
I take her back to her sitting room and introduce her to her grandchildren, whom she greets as if for the first time, introducing herself conscientiously. Then I rush back to the fireside and the novel, knowing I have seven or eight minutes until she shuffles in again.
"Ahh, there you are, I was worried."
EMOTIONAL, SUBCONSCIOUS ASSOCIATIONS are in the ascendant. She can spend all day bearing a grudge and render the consolations of the afternoon film useless by monologuing over the top of it. And the delusions are beginning. Nancy's chief delusion is that she's in charge.
"Nobody ever talks to me here. Nobody pays any attention to me at all. I may as well be dead. People are always telling me what to do _—me!_ And I own this house. They work for me. They all work for me." Jack comes into the room and then retreats again. "Those horrible children laugh at me and call me names behind my back...."
They go on and on for hours, these monologues. Luckily a response isn't expected. It's more in the way of a performance. I am learning to tune her out. I do the ironing standing in her sitting room and train my mind elsewhere, while Nancy sits and narrates her way through a series of grisly daytime TV programs.
Eventually even television fails. Jack, trying to tell her not to go outside because it's raining, is called a bastard and then a bitch for good measure. It's difficult to convey just how sinister these verbal attacks are, so out of the blue and so quietly passionate, her expression so malevolent. They wrong-foot all of us emotionally, but the children especially, who are shocked by the suddenness with which the mood turns. I understand their tears and hurt. It's like having your face slapped, is very like it, by someone you thought was on your team—slapped hard and unexpectedly.
There's an upset almost every afternoon, and because it's the children under attack and nothing else is as provoking as a bully, I find myself yelling at her, Nancy yelling back. I contrive to do this when there are other people in the house, friends of the children or plumbers or electricians, the tradesmen pausing to listen, shocked rigid by the shouting. "Listen to that! She's screaming at her poor mother-in-law, that sweet, gray-haired old lady!" They're rooted to the spot. Visiting children turn wide-eyed and silent as I rip out of the kitchen and tear a strip off Granny. Nancy denies everything, always. "I did not. Did not. Did not. That boy is a liar. A liar, always a liar, a nasty little liar, telling lies. It's all lies and rubbish."
I can't help myself from insisting that she is wrong and that she's behaved badly. I can't seem to stop myself from insisting she change her ways. Why do I waste my breath? Morality, misplaced and useless, is at the heart of it. Families are constructed from a shared sense of justice, and sail out on its complicated hidden currents. The children have been indoctrinated in the ways of fairness. Granny's dementia smashes right through their early training and leaves a trail of moral wreckage that constantly needs to be accounted for.
Help is close at hand. The Charity. The Charity employs people to provide short home-based bursts of respite care. The dictionary defines _respite_ as an interval of rest and relief, but we define it as time to act, to move and act unhindered, to resume life, if only for the short period in which someone else is in charge. If it wasn't for The Charity, many families would have no home-visit respite at all. We are offered two sessions a week, on Monday and Wednesday mornings. The Monday caregiver, Sian, a Rubenesque Essex blonde, deals with Angry-Nancy by imposing routine on her. The routine is a brisk itinerary encompassing hot shower, hairstyling, car trip, shopping, coffeehouse, and home. The Wednesday caregiver, Harriet, on the point of retirement herself, is a lithe, warmhearted northerner with kind blue eyes. If Sian's approach is dogged, unemotive, unflappable persistence, then Harriet's is more in the way of love-bombing. "Come on now, lovey, you can put your shoes on yourself, a big girl like you.... Yes! That's right! That's brilliant. Clever girl." The only thing Nancy really doesn't like is the packet of felt tips Harriet brings with her, and the bumper book of coloring, featuring girls with lambs and flower baskets, and elephants in lederhosen driving cars. Nancy can't color inside the fat black lines and doesn't see why she should try. Once Harriet is waved off on Wednesday lunchtimes, the pages left behind for Nancy to finish are verbally abused, ripped into pieces, and thrown in the fire.
THE CHARITY MANAGER is on the phone, asking how things are going with the caregivers, and I am properly grateful but frank about Nancy's bitter mood swings. "You'll need to think about residential care at some point," she says.
"At some point," I agree, "but we're hoping that Morris's return will restore the equilibrium. We should have a few years in the arrangement yet, I hope."
"You should talk to the social work department," she says brightly. "You could get some more help. You could get a care manager to coordinate everything."
"I'm sure," I say, "but we'd really rather not institutionalize things or get too many experts wading in. We manage fine. We struggle along."
The following week, we admit our first expert to the house. He talks briefly to Nancy and at length to the two of us and convinces us that joining in is possible as a nonjoiner, in situations like this. Chris and I are reluctant. We don't want to be part of the care machinery and enter the social work filing system. We don't think Morris would want it, either. How will he react, in any case, on his return, to the new Nancy that Sian and Harriet have created, the one that's talked to as if she is five and behaves accordingly? How will he deal with their taking charge of Nancy and redefining her, leaving him a helpless witness?
"We'll leave things as they are for now," the expert says. "But what I can do for you is organize a program of nursing home respite for next year. You're entitled to six weeks. I'll book Nancy in for six separate weeks, and I'll be in touch with dates."
# Chapter 10
_Ich hab mich verloren_.
—AUGUSTE DETER, THE FIRST CONFIRMED CASE OF ALZHEIMER'S DISEASE
ALZHEIMER'S DISEASE IS NAMED AFTER A GERMAN psychiatrist and neuropathologist. In a lecture in 1906, Alois Alzheimer reported seeing the characteristic plaques and tangles in the brain of fifty-five-year-old Auguste Deter, a sufferer of early-onset disease who'd died earlier that year. _"Ich hab mich verloren_ [I have lost myself]," she had said to Alzheimer, when first she was admitted to the Frankfurt asylum at age fifty-one.
Alzheimer didn't discover the disease as such. It had been observed and written about before, notably by an early brain researcher named Beljahow, who reported brain plaques in dementia in 1887. Likewise, the presence of tangles had been announced by other neuroscientists before Alzheimer gave his lecture, but nothing formal had got into the textbooks. What was remarkable and newsworthy about Auguste Deter's case was her disease's very early onset. This helped make a splash. It was a sensation. Publicity was the key. Alzheimer's boss, friend, and coresearcher, Emil Kraepelin, happened also to be a powerful figure in neurology and in science publishing. Today regarded as the father of mental disorder classification, Kraepelin named the disease Alzheimer's in 1910, in the course of a description in his new textbook. He did so almost casually, in a paragraph of notorious vagueness, referring to "this Alzheimer's Disease," a reference that surprised his readers and colleagues. To quibble, the label _Alzheimer's_ ought in all justice to have been confined to the early-onset variant only, since that's what the Auguste Deter research was concerned with. It's also interesting to note that Alzheimer himself didn't think the late-onset condition ought to be classified as a disease as such. It was his contention that Alzheimer's happens to all brains in the end—they wear out, like hips and knees—it's just the speed and volume of plaque growth, he argued, that marks the syndrome out. The speed is remarkable. Neuron loss in Alzheimer's has been calculated as ten times the speed of that entailed in normal aging.
A sense of theater, a gift for PR, the pressing financial need to impress funding bodies: It's suggested that all these led Kraepelin to make Alzheimer the eponym. Kraepelin's battle with his great rival Sigmund Freud, as to whether such disorders were organic or psychiatric in origin, was another factor; Alzheimer had been supportive in putting Freud in his place. It helped that Alzheimer's time was one of diagnostic breakthrough. Modern Zeiss microscopes (the original slides have been discovered and preserved) and the advent of silver nitrate tissue-staining process (the innovation of Alzheimer's colleague Franz Nissl), which illuminated slices of the brain as never before, meant Alzheimer's work had the benefit of technological advance over that of his rivals.
Alzheimer had left Frankfurt and was working for Kraepelin in Munich by the time Auguste Deter died in 1906. Her brain and spinal cord were sent to him by train in a box.
Her case was unusual and even today would be regarded so, not only because she was so young—fifty-one is very early onset and fifty-five a very early dementia death—but also because it came on so aggressively and fast. The polite confusion she showed on admission very quickly declined into raving and wailing and wordless wounded animal misery. A stark black-and-white photograph of her while in the asylum shows a woman who appears to be in her seventies, her brow furrowed into deep ridges, her face ravaged and baggy, a bewildered look in her eyes.
ALZHEIMER'S OWN RECORD of their first conversation survives. He writes that she looks helpless. He asks her name.
"Auguste," she tells him.
"And what is your husband's name?"
"Auguste."
"Your husband?"
"Ah, my husband."
She doesn't appear to understand that it's a question.
"Are you married?" he persists.
"To Auguste," she says.
"How long have you been here?" he asks her.
"Three weeks," she says with confidence (though in fact she was admitted the day before).
She can still identify a pen, bag, key, diary, and cigar.
She is given pork and cauliflower for lunch, but when asked what she's having, answers "Spinach." When asked again she says, "Potatoes and horseradish."
He notes that objects shown to her are forgotten about almost immediately. In between she seems to have an obsessive interest in twins.
He asks her to write her name. She starts to write "Frau" and then gives up. Several attempts are made to write Auguste. First Augh. Then Auguse D, leaving out the _t_. That evening, Alzheimer writes, her conversation is full of non sequiturs and obsessive elements like perseverations, in which sufferers return to a subject, an idea, a phrase, again and again without making their meaning clear. (It's not so much that they persevere with a subject, but that the subject perseveres with them.)
* * *
THERE'S AN IRONIC footnote to the story. Auguste Deter's cause of death appears principally to have been arteriosclerosis of the brain: what's now classified as vascular dementia, rather than Alzheimer's disease. There's also a poignant footnote. Auguste's husband, on delivering her to the Frankfurt asylum, complained about her unreasonable jealousy. She was convinced, he said, that he was having an affair with a neighbor and had become irrational about it. Apparently he married this neighbor the year after Auguste died. Just because you're paranoid, it doesn't mean they're not out to get you.
On his way to a new job in Breslau in 1912, Alzheimer became ill on the train with a sore throat, which led in turn to rheumatic fever. He died in 1915 of heart failure, in effect from complications of tonsillitis. He was fifty-one.
# Chapter 11
_The real voyage of discovery consists not in seeking new landscapes but in having new eyes_.
—MARCEL PROUST
DECEMBER, AND WE ARE PLUNGED INTO DARKNESS. IT'S dark when the children go off to school, dark long before they come home. The daylight window shrinks to seven hours of pearly gray, the morning half opening its eye; the nights are profoundly black and long. But into this darkness comes a great light. A twinkle light. Christmas. Thank God for Christmas.
Morris is still in the city hospital but, as a result of Chris's ongoing appeals, is returned to the local hospital a week before the day. He's been put on Prozac for his depression. The hospital consultant wonders aloud if Morris is succumbing to dementia. Depression, particularly in the elderly, is often misdiagnosed as Alzheimer's, and vice versa.
I sit upstairs in the bay window of the upper sitting room and try to work. It's difficult to concentrate. I'm becoming a connoisseur of weather systems, watching fronts invade and recede, break, roll, gather, disperse. There are weird lighting effects, lurid green clouds, and the winter sea is mercury silver, thick and lazy, lapping viscous onto the beach. Harriet's singsong voice echoes up from downstairs. Harriet's adopted approach is Not to Stop Talking. She blocks Nancy's monologues by providing her own.
"Now, dear, shall we get a nice cup of tea? A cup of tea would be lovely, wouldn't it? A lovely cup of tea. Right then, here we go, into the kitchen. Oopsadaisy and along to the kettle. Here's the kettle and we lift it. Now what do we do with it? Fill it with water, that's right. Here we go then, darling, here we go, along to the tap. Can you turn the tap on? That's right, what a girl you are, what a clever girl. And off again. Good. That's very good. Now we need cups...." She must be exhausted to the point of annihilation when she leaves here. She doesn't stop narrating for three hours straight.
Nancy's become fixated on the kitchen. Perhaps it's to do with Christmas busyness, the spice fruit pie making, the aromas, or perhaps it's to do with her own definition of womanliness, this not being able to stay away, haunting the stove and the sink. I can tell it's her who's coming into the kitchen, even if the children are home, because of the struggle with the handle, and the way the door opens ever so slowly. Then she's there, saying the same thing every time. "Oh. Sorry. Sorry, didn't know anybody else was here. Sorry."
"Come in, Nancy," I say. "It's fine. It's just me."
"Oh, it's yourself. Well, that's fine." She walks the length of the room, turns, and goes back, wringing her hands and muttering. Goes along the length of the hall, wringing her hands. "I'm so useless, I'm not any good for anything, I'm no use to anybody, I may as well be dead."
She's become compulsive about going through doors. Later, I come across a paper about frontal lobe damage and visual prompts, in connection with wandering, which advises covering doorknobs with handkerchiefs or something similar. Alzheimer's can damage the decision-making process so badly that a prompt like a door handle becomes, in the sufferer's mind, an instruction. There's no particular content to the trying of door handles, the neurologist author says. It isn't necessarily anything to do with escape. Just a recognition that handles need turning, translating itself into the burden of having to turn them. It might actually be a relief, he says, to have the stimulus removed.
The busier the household gets, cooking and decorating and preparing for Christmas, the meaner Nancy's mood. There's protracted complaining about The Woman, the one who makes her do everything and does none of the work herself. I find myself wondering if it would be better for female Alzheimer's sufferers to have male caregivers. Nancy would be happier with a man in my place. She'd get to flirt with him. She'd probably be more respectful—women of her generation have a natural deference, the kind that feeds men first and takes the leavings. She'd not be preoccupied with this jostling for matriarchal precedence. Though sexual confusion might become an issue. Dementia patients can suffer an embarrassing loss of inhibition. Lately, Nancy has made suggestions that make Chris squirm. "Come to bed with me, come on," she says to him. "I'll warm you up all right. Come and cuddle up," patting the mattress suggestively.
Among the confusion and rage are flashes of good humor. I can make her good humor flash. If I worked harder to make her happy, it would flash oftener and longer. I'm aware of this. It's a fertile source of self-reproach. I can take her by the hands and cheer her up, with smiles and tone of voice and suggestions of things we might do together. They have to be things we do together. Things she could do alone don't work. But if I begin a sentence with "Why don't you and I..." her face lights up.
"Will you come and look at the drawing room with me?"
"Yes, of course. What will we do there?"
"Come and tell me what you think about the decor." I lead her into the room. We hold hands now when we're together. "Look. Green walls. Horrible. Pond green. A pond-that's-gone-off-and-smells kind of green."
Nancy's giggling. "You're right, you're right! It's horrible. I've always hated green."
I know that, Nancy.
"I'm going to get a repro wallpaper," I say. "I'm thinking of one of the chinoiserie style ones with birds and trees. What do you think?"
"Oh, but look at that," she says, pointing in admiration at one of the twin sofas by the fire. "That's gorgeous. That's gorgeous, isn't it?"
"Do you like it?"
"I love it."
"Even though it's green?"
"It isn't green, is it?"
"It's a sort of greeny-blue. Turquoise."
"Oh no. It's green. I hate green. It's the only color I really hate, to be quite truthful. I don't like it at all."
I persuade her to sit on one of the green sofas, and go and fetch a pot of tea.
"Ooh, cake. I like cake."
I know that, Nancy.
She eats and drinks and rubs her hands together, looking into the fire, logs crackling and spitting.
"Did they say when they'd be coming back?" she asks suddenly.
"Who?"
"The boys."
"Do you mean..."
"I mean the boys. My brothers."
"You only have one brother, don't you? I thought you only had one. Angus. In Australia."
"He's in Australia? Why didn't anybody say so?"
"He's been there over forty years, Nancy. Have some more cake."
She eats it, looking thoughtful. Then she says, "When are my brothers coming?"
HER FAMILY IS on her mind, it seems. Later, in the early evening, as I sit with her watching a nature documentary, she says, "When did he say he was coming back?"
"Who?" I'm reading the paper, not really listening. That's the kind of bad mother I am.
"My father."
The newspaper dips. Our eyes meet.
"Your father died a long time ago, Nancy, when your children were small."
(As you can see, I'm inconsistent with the validation habit. Often it's simply forgetfulness. I'm too used to telling the truth, bad at remembering to play along with the delusions.)
"I don't have children," Nancy says, mildly outraged. "I never married."
_Don't bite. Don't_. I obey, raising the paper again.
"When did he say he'd be back?"
"Who, Nancy?"
"My father. He isn't usually this late."
Nancy often refers to her childhood, but always in this same impersonal way. It's on her mind, but floats free of content. I've never heard her refer to the far past, to the 1930s, for instance, when she was a child, in any other way than this. I'd expected that the loss of short-term memory in Alzheimer's would bring the long-term memory sharply into focus. I've been reading Patrick Leigh Fermor's books about his epic walk across Europe just prior to the Second World War, which he wrote largely from memory decades later, the earliest of his notebooks having been lost during the trek. He said that writing the account was like trying to reconstruct a dinosaur from a miscellaneous bag of bones. I didn't expect the whole dinosaur skeleton from Nancy, but I expected to be shown the bag of bones at least, and nothing remotely like that has happened. She has vague persistent cravings for her parents—her father especially—and talks a lot about the "brothers," but that's as far as it goes. No telling details emerge, have ever emerged, and didn't, not even when I first knew her. I don't know anything about her. Chris doesn't know much more. Morris isn't expansive on the question.
I sit with Nancy in front of the television and escape down my own wormhole, the one provided by the Internet, laptop balanced on lap. Somebody out in the odd, dislocated world of anonymous, typed-and-not-spoken conversation makes a lighthearted remark about the spiritual advantages of Alzheimer's. I don't bother shouting him down, as I know from previous experience that hundreds will be racing to do just that. Dementia caregivers are everywhere and fatuity isn't tolerated. The person (no gender ascribed, even) makes the point that living in the moment, only in The Now, is surely the target state of Buddhist teachings; that Nirvana is a state of perfection attained by being cut off from past and future and their attendant states of wanting and anxiety. I can see what they're driving at and it's an interesting starting point for a discussion, but it's a debate that will never take place, as the self-appointed moral guardians who cluster at all such sites zoom in for the kill, hungry for the acclaim that will follow. Unfortunately, a state of bliss isn't the end point of Alzheimer's. Quite the reverse. The reality of having no past or future is that it isn't a state of perfection but of absence. The brain can't handle the absence and a chaotic, scrabbling sort of panic for order and meaning ensues. The Buddhist idea of living in The Now is, surely, something achieved through dealing with past and future, and not their absence—of quenching their demands and silencing their voices. These are sleeping dogs, not missing dogs. In a state of Nirvana we'd have total control over them, reconciled, having triumphed. An end to wanting and anxiety isn't ever going to be achieved through amnesia.
WILD WEATHER. IT'S too cold to do anything much, other than huddle by the stove and drink tea and eat muffins. I could do this with Nancy, of course, and do. But doing it alone seems like a treat. It amuses visitors that all of us in the family have adopted a great thick blanket each and wear these like cloaks in the house. I sit by the stove with my blanket tightly wrapped, just a hand and a paperback sticking out. Bitter, scouring north winds blow, gale force 8 or 9, not the strongest we've had but impressively gusty, and sufficient, earlier, to blow my new striped hat off my head, right over the top of the coach house and into the sheep field next door. It's my own fault. I know about hats. In any case, it's irretrievable: There's a foot of mud at least to navigate and the damn thing has disappeared under a stampede of ewes, charging toward me, hungry and hoping for food. The chickens huddle together under the laburnum tree. I take hay to the horse paddock and am almost blown over. The poor horses chase wispy escaped sections around the field as it blows along like tumbleweed.
We have a Christmas party at home and invite everybody we know or have come across since we arrived. Neighbors, other school parents, villagers, people we've met through gallery visits, the organic farmers from over the hill. Over a hundred people turn up and I worry that I'll have to introduce everybody but, of course, it's only us who are strangers. Everyone else is connected in the firm, elastic web of rural relationships and they greet each other like the old friends that they are. We have hired musicians for the evening, a fiddle player and a guitarist. I place them in a corner of the drawing room, they start to play, and immediately, as if given the nod, the fire begins billowing smoke. Nancy circulates, asking everybody if this is their house, thanking bemused guests for inviting her, reprimanding six-year-olds for running, and getting sloshed on whisky. She retreats to her armchair and half a dozen kind souls gather round her to chat. I am visited by a surge of well-being, rightness, and good fortune. This is a good community, and it's Christmas. The hall table is piled with presents. The night is crisp and snowy. Cars are parked all along the drive, half on the lawns and paddocks, all along the verges of the road. People arrive with cold feet and go home tipsy, squeaking along, leaving trails in the white, shouting "Happy Christmas" into the dark.
I've been reading about Marcel Proust. His hero in _In Search of Lost Time (À la recherche du temps perdu_ , also known as _Remembrance of Things Past)_ has an epiphany, a moment of physical and psychic joy, synthesizing childhood recall and a sense of the unity of life, while dipping a madeleine into a cup of lime blossom tea. This much is well known. Proust's experience of memory was multisensory, engaging taste and smell, and was auditory. Other people have told me they hear music in their dreams and in recollecting, though I never hear music in mine, which tend to be faded celluloid moments with the sound turned down low. Proust insists that it isn't necessary to go out in search of a connection with the Sublime. He finds meaning in small domestic details. He says that the ordinary world of eating and drinking and being alive is full of wonder and beauty and charm, and ours is the ordinary collective failure of failing to look at and relish it properly. (I do sometimes wonder what else was in the lime blossom tea.) In any case, Proust's life can hardly be described as dreary, or burdened by necessary altruism. He was blessed by untroubled self-absorption. Ill, fragile, hypochondriac, he still managed a lively Parisian social life, had servants, lived at home, and was fussed over throughout by his mother until she died when he was thirty-four. He inherited the equivalent of £3 million, and spent as much of his life as he wished in writing absurdly long novels in bed. "Proust on the Peninsula" is a book that will never be written.
We have neighbors for Christmas lunch, neighbors whose old kitchen's been ripped asunder, but whose new kitchen didn't quite arrive in time, and who were contemplating microwaved turkey for five. We go into town to see Morris late on Christmas afternoon. The starkly cube-based bungalow interior of the hospital smells mildly of figgy pudding, a new aroma to add to its usual bleach-and-plastics, sweat-and-gravy smell. The nurses are giddy, teasing, prompting jollity like summer-camp counselors. Sad swaths of thin green and red tinsel droop from ceilings. A little plastic tree is gaudy with baubles. Despite all this, Morris is not going to play the Christmas game. Presents don't cheer him up. He opens them all with an air of great monotony, says little, doesn't wish the children a happy Christmas. Nancy sits by him with her eyes fixed on me, the question in them unmistakable: What on earth am I doing here? The difference is that she means it literally.
# Chapter 12
_When a lot of remedies are suggested for a disease, that means it can't be cured_.
—ANTON CHEKHOV
BECAUSE THERE'S NO SINGLE, UNIFIED THEORY OF ALZHEIMER'S, teams across the world are working on different bits of it, hoping that their approach will prove to be the key that unlocks the disease. Rather like the Indian fable about the six blind men trying to identify an elephant, some research programs deal with the trunk, some with the tail, some with the leg and others with the ears, some with the torso and others with the tusks, and all have different ideas about what it is they're identifying. It's a race, not just because cracking Alzheimer's this year could save hundreds of millions of people in years to come (it's a prestigious race) but also because the saving of hundreds of millions of people will involve billions of packets of drugs (it's also a profitable race). There are fortunes to be made here, beyond the dreams of avarice. The latest estimate online put the dementia drugs market at $37 billion, and that was several years out of date, and is, of course, the pre-breakthrough figure, should the breakthrough ever come. Though it's possible—some scientists say probable—that Alzheimer's will never be cured as such, but will be managed with better drugs that keep it at bay, rather like HIV is managed today.
As far as causes of Alzheimer's go, there are two main camps. One of them, what you might call the plaque orthodoxy, is much bigger and better funded than the other. The plaque orthodoxy says that the formation of plaques is the key process. The other camp, the tau heresy, begs to differ. It suggests that plaque is a red herring and that the answer lies in the tangles, the crumpling and snagging of the ladder rungs in the microtubules, the neuron's internal communication lines. Work by the tau camp has focused on the adherence of surplus phosphorous molecules onto the tau protein, arguing this is the key process of Alzheimer's, most probably caused by a genetic switch that could be turned off.
The plaque orthodoxy looks for ways to defuse the beta-amyloid crisis that seems to take place in the brain of sufferers, creating characteristic aggregations between neurons, clogging the neural glue with what look like suspect moles, crusty rounded lumps of something unevenly brownish. Bits of the cells that produce the neurotransmitter acetylcholine break down and become part of the plaque blobs. The plaque orthodoxy, which has been in place now for over twenty years, sees tangles as mysteriously secondary. Some tau heretics have a stiff rebuttal to this: Plaques, they say, are not only a red herring, but might be heroes rather than villains. Plaque, they say, is a sign of the brain trying to protect itself from something (the something that causes the tau to mangle). Plaque, they point out, has been seen in quantity in other instances of brain damage. Alzheimer himself was a tau man.
Don't go away with the impression that this is a gentlemanly dispute (though it's true that gentlemanly disputes have been taken into the woods at dawn with pistols). Pistols at dawn is more like it. Pro-tau scientists have had enormous trouble getting research grants or getting their results published in medical journals. The plaque orthodoxy has become quite medieval in its tolerance of heretics. So it's heartening to see that the latest wonder drug, Rember, is a tau-directed one. Perhaps the orthodoxy is being challenged now. I hope so. I hope the tau-research-blocking drug companies have the good grace to look a little sheepish.
The question is, what's really behind the mechanism, whether beta-amyloid or tau? What's making beta-amyloid run wildly out of hand; what's going on with the tau-tangling phosphorous? Presently, there are two answers: (a) genetic and (b) environmental. It's conventional to ascribe the root causes of Alzheimer's to a combination of both. A genetic predisposition and an environmental trigger: that's mainstream science's current best guess.
There are those inclined to lay the blame entirely on environment. Those who insist that Alzheimer's, though not a new condition, is essentially a modern condition, a zeitgeist condition, the defining twenty-first-century disease, also tend to point the finger at multiple pollution. And once pollution is mentioned, a whole underworld of subgroups comes to light. In the United States, there's been a widespread scare about amalgam fillings in teeth, and the resultant potential for mercury poisoning. Another points the finger at aluminum and other metals in the drinking water; yet another claims it can show evidence that pesticide poisoning in food is to blame. Manganese is spoken of suspiciously. These lines of inquiry may get somewhere or nowhere.
Two other theories doing the rounds:
1. That Alzheimer's is a prion disease. Prions are peculiar things, infectious proteins whose molecules nudge up to healthy cells and corrupt them. The best known prion disease is called variant CJD, or the human form of mad cow disease. In research, transgenic (genetically modified) mice were given Alzheimer's by being injected with autopsied brain material. Thus Alzheimer's was shown to be infectious, though only in this rather specialized and perverse manner.
2. That Alzheimer's is caused by a virus. Hepatitis C has been mentioned. There has been more convincing talk of herpes simplex, which infects a good many of the world's population in a dormant state and can be activated by illness, stress, or inflammation.
IN TERMS OF treatment, immunization may be the way forward, certainly in the plaque camp. This idea was scorned when first suggested, but has been piloted with some success. The idea is that it works rather like the polio vaccine: The body attacks the injected beta-amyloid and with any luck also sweeps up the plaques in the brain (though whether this is fundamentally a good idea remains moot). Tiny amounts of vaccine get past the fine mesh of the blood-brain barrier, which protects the narrow blood vessels in the head from clogging. Early results claimed a 50 percent plaque clearance rate in some brain areas in transgenic mice. Unfortunately, when it was trialed on humans, the immune reaction was so pronounced that brains swelled dangerously. The latest research in Tokyo, going on as I write this, is focusing on injecting DNA that "codes for" beta-amyloid, provoking a gentler immune response. There are also trials going on for a pill that binds the amyloid and stops it from accumulating.
If Alzheimer's proves to be a genetic condition, then genetic work is the way forward, and in terms of gene therapy we live in an age of wonders, with new diseases being identified and marked with a highlighter pen in the DNA all the time. There's a rare inheritable form of Alzheimer's, familial Alzheimer's disease, that strikes people young—perhaps even in their thirties—and will go on to affect 50 percent of their children on average. It's one very small subset of early-onset Alzheimer's, which is classified as beginning at under the age of sixty-five. Familial Alzheimer's has been found lurking at chromosomes 14 and 21 (and also at chromosome 1 in the American population). Regular Alzheimer's—for want of a better description—the noninheritable kind, is thought to be indicated genetically by the gene _Apolipoprotein E (APOE). APOE4_ , one of its four variants, is associated with high risk. If both parents carry _APOE4_ , their children will have ten times the average risk of developing Alzheimer's. It's not all bad news. _APOE2_ appears to indicate a much lower risk than is average of developing the disease.
Genetic diagnosis and manipulation is at a stage roughly similar to that of the New World three hundred years ago. It's still in the maps, marvels, and flag-planting era, but once domination of the genome begins in earnest, and the railroad is built to traverse it, capitulation will surely follow in a rush. For now, it's at an early, exploratory stage, a Wild West full of frontier towns and gunslingers. The encouraging thing is that it's begun.
# Chapter 13
_I have said nothing. I leave nothing. I have not said what
I wanted to say. I have so much more to say._
—MAURICE RAVEL
WINTER ARRIVES WITH A VENGEANCE AND THE PENINSULA feels intensely vulnerable, low-lying in an angry sea. It's as if, were we to go up another notch in the storm conditions, it might all be swept away, scouring the place of buildings, washing the cattle barns and cottages into the ocean. I imagine us all, humankind and livestock together, found floating in the bay, distended, by a passing ship. It's a sailing ship and it seems to be the early eighteenth century. _What disaster hath happened here?_ I'm having peculiar daydreams. I sit at the drawing room window watching the hedges and shrubs blown flat and gray in mouse-colored rain, the sky low and woolly, looking at the distant dark peak of the mountain across the bay. A cloud sits over this hilltop, raining on it.
Morris is still in hospital. Progress, officially, is "slow." The idea of mobility has been abandoned, though nobody's explicit. Instead, the _W_ word begins to be mentioned: _wheelchair_. Nancy sits in her chair for long periods, rubbing her hands. It's too cold in the hallway for wandering, even with all the bottle-gas heating chugging away, the old electric storage heaters pumping out on max, every coal fire in action. No one hangs about between internal destinations. Washing and dressing are done at a gallop. Nancy is feeling the cold acutely, despite the customary five cardigans. She has taken to wearing a hat indoors, a dark blue felt number with a jaunty feather. Her hands are cold and so I retrieve a cashmere blanket from her cupboard, one in fine-woven gray that we gave her for Christmas once, tucking it around her lap and legs. Thus immobilized, Nancy's a picture of a woman lost in a dream. Her hand rubbing has become systematic, ritualized. First, palms are placed together at right angles making a cross shape, rubbed as if rolling dough into a sausage. Then the back of each hand is rubbed briskly by the other palm in turn. After this, she interlocks her fingers, jamming them tight together and releasing, jamming and releasing, before the sequence begins again. Later, I read an identical account online of somebody's mother's behavior, and begin to see Nancy as pulled helplessly along bizarre, well-trodden tracks of disease, along Dr. Reisberg's railway.
At intervals in the hand rituals, she twiddles her hair. Her gaze is averted from the television pictures flashing in front of her, off toward the window, toward the fire, or dipped carpet-ward, as she captures and twirls sections in turn. Her hair begins to develop twiddled kinks. These, the hand rubbing and hair twiddling, must be signs of distress. But what's to be done about them?
We realize that Morris is a skilled Alzheimer's companion, in his way. His way consists of the sharing of all-day television. He watches everything, anything, is a habitual channel flicker, and keeps up a steady dialogue with the programming that is really a way of talking to his wife. The sharing is crucial.
"Just look at that suit! What does she look like?"
"Three hundred fifty pounds? I wouldn't give him ten bob for that vase, would you?"
"Look at the colors in the trees! I'd no idea Bulgaria was hilly, did you, dear?"
He uses television as an intermediary. He talks, and then from time to time, when it's clear a response is required, Nancy joins in. "Oh yes, you're quite right...." She might almost pull it off, the illusion that she understands the question, but then likely as not will blow it: "And I have always thought so, but not many people, not many other people I should say, can see it that way, you know, and it's been like that my whole life."
Morris, in Alzheimer's denial, can't let such wittering pass uncontested.
"What are you talking about, you silly woman? It's bloody Bulgaria!"
But Morris is still in hospital, and Nancy finds me lacking as a television companion. I'm not given to being surprised by television, or to arguing with it, or to commenting on people's appearance. I don't really watch it. I sit with her in front of it but my mind is elsewhere. Her mind is elsewhere, too. Or nowhere. It's hard to know what or where her mind is now or whether she has, in any meaningful sense, a mind at all. Her brain functions well in instructing her body as to movement, forward propulsion, the signals required to bend and pick up a crumb from the carpet and eat it. If you ask a question, she'll answer it, in a fashion: _Are you hungry?_ I could eat something nice. _Do you like this color?_ I think it's very nice. _Do you want to go to bed?_ Bed might be nice. But I'm not sure this is evidence of a mind. Are these her answers, the answers she gives me, or are they any old answers found in a box, retrieved from those that have survived the fire? Alzheimer's robs the brain of time travel, of its customary and constant roaming forward and back, the past stretching behind and the future ahead. That's how people operate. We put everything into context. But Nancy's marooned in the present. I'm only just beginning to see how fundamental this is. As Milton writes in _Paradise Lost_ , "that must end us; that must be our cure— / To be no more. Sad cure! For who would lose / Though full of pain, this intellectual being / Those thoughts that wander through eternity." That's what I want from my experience of outdoors, from my walks along the cliffs: thoughts that wander through eternity, escaping the tyrannies of the lists, and the intimate tedium of the caregiver's day.
I'm not entertaining enough a television companion, though I try. If Nancy's especially restless or upset, I'll try to talk her through what's happening on-screen. But it isn't the same as having Morris here. He's belligerent, and that amuses her. He's a fund of trivia about various actors and their lives, who they married and divorced, their war service, other things they've been in. He makes cynical remarks about the soap plotlines, predicts what's going to happen, whoops with pleasure when he's right. Nancy loves all this and whoops along. Not much whooping is going on now. None. My heart isn't in it. I remember, guiltily, that I had much the same trouble once with children's television, and it occurs to me, not for the first time, that it might be me that has the problem.
"Do you like that dress, Nancy?" I'll ask, rather desperately, as Nancy gets up to leave (leave home). "What do you think of that pink sari?"
"It's very nice," she says, but she isn't engaged. It may be that my voice isn't the right trigger. There's nothing left that's equipped to recognize me anymore. I'm a stranger, a pleasant enough stranger staying at the same hotel. Those are today's assigned roles, it seems, to judge from the things Nancy tells me. She complains about the service, the temperature of the tea, the quality of the lunch. "They're really not very good here at all," she confides, leaning in toward my ear. "It's gone downhill, this establishment, since the old days." I sit with her after lunch and find "a nice film," but eleven minutes later she's up and out of the room. Eleven minutes has become her attention span for television pictures.
"I'm just going for a wee walk."
Shortly after, I find her in Chris's office, interrupting a business call.
"Perhaps this gentleman can help me," she says, gesturing toward her son, her voice full of emotion. "Can you tell me where I am?" Tears course down her cheeks.
A NEW CAMPAIGN of making Nancy happy commences. We do housework together after breakfast—five times more housework than I would normally do—Nancy poignantly grateful at being allowed to share in the tasks. We clean windows, Nancy rubbing away at one pane of glass with a piece of kitchen towel and singing. "When all my eyes are finished, and the world is bright and free, then I will be there and I know I can come, and that's the one for me." She can still rhyme. Though she's unhappy about having to do a different pane.
"There's nothing wrong with this one and it said it liked it."
"Yes, but look, the other ones are dirty."
"Nonsense."
"Its friends, you see them, look, here and here. Its friends are dirty and they will be embarrassed."
"Oh dear. Oh deary, deary me, that will never do." She changes panes. I'm aware that a lot of what I do with Nancy plays on the subject of shame. For this generation it's powerfully embedded in, the worry about what people will think, and I exploit it thoroughly.
We vacuum rooms, Nancy going over the same piece of carpet back and forth. I leave her vacuuming and go to put some washing in the machine, and as I bend to slam the porthole door, I hear shrieking of a familiar kind. I go back to her sitting room at a run and find Nancy standing with her hands in her mouth and the Dyson lying on its back roaring. "It doesn't work! It's falling down!"
I get her into a rhythm again, but when I leave the room she begins to howl. My head comes round the door. "What is it now?"
"Could you do it? Could you take it? I can't do it." She withdraws to the safety of the wall but can't take her eyes off the thing. It's as if she's forgotten how vacuum cleaners behave. This one might do something spontaneous and dangerous.
I vacuum the room and Nancy watches. "You're so clever," she says. "You make it look like nothing at all." She begins to sing her "Irish Eyes" variant.
I'VE BEEN READING about music and dementia. The composer Maurice Ravel (1875–1937) had (probable) early-onset disease from about the age of fifty-two and died ten years later following unsuccessful exploratory neurosurgery. His most famous work, _Boléro_ , whose sweeping repetitions are now firmly associated with ice-skating, has been cited as an example of dementia composing. He wrote _Boléro_ in the year after becoming ill. The question that's unanswerable is: Would a healthy Ravel have written the same score, or is it one of the best-known examples of perseveration in art? His own judgment seems clear; he referred to _Boléro_ as "orchestral fabric without music." Ravel's dementia first presented itself as confusion about his touring schedule. He lost luggage, lost his tickets, and traveled with hoarded letters in his pockets. He forgot how to swim—the procedural memory having failed him—and almost drowned. In 1933, four years before his death, he told friends that he wouldn't after all be able to write his planned opera _Jeanne d'Arc_ , saying he could hear the music in his head but couldn't access it. "It's over," he said.
We deal with laundry, Nancy and I. Laundry is a big part of the day. But now, it's too cold to hang it outside and make laundry into a journey. The truth is, there aren't many days when washing dries outside hereabouts: The growing season is short and the drying season shorter. It's too windy much of the time, the clothes disappearing over the wall, or cast into bushes, muddy and torn. Laundry goes out warm and wet and comes back in cold and wet so there's little point. Today it would return stiff and white with frost. Nancy stands at one end of the old-fashioned pulley in the utility room and I stand at the other. I pass her clothes from the basket and she pauses to pass comment on them.
"Those are underpants. They are not my underpants. They are horrible, actually."
"They are your underpants."
"Are they? Are they? Who said that was a very bad person because it's entirely the other way."
"Hang them over the pole. Like this, look."
She has the elastic waist of a pair of pink underpants clutched tight in her fingers. She moves her whole hand forward but it doesn't connect with the pole. Watching her, it occurs to me that the trouble she's having is like using a mirror to try and fasten an earring or snip at a stray bit of hair. Fingers and scissors don't move in quite the expected way. The mind plays spatial tricks of distance and direction.
"It doesn't want to go," she says. She jerks her hand forward further and overshoots. The underpants disappear over the other side and flutter to the ground. "There," she says. "Give me another one."
I give her a pair of socks. I put my hands over hers and guide them to the pole. She puts one on top of the other and fusses with them, trying to get them straight. I get on with the rest of the hanging up. "It's quite something watching you do that," she says admiringly. "You know just how it works. I think I used to know but I don't know now, that's for sure."
I stop and look at her, a child's school sweatshirt in my hand. "Do you remember being young and having washing to do?" I ask.
"It's all somewhere else. My father was there."
"You're always talking about your father but you never mention your mother," I say.
"Well, she was there, I expect. But I didn't know her, really."
"Why's that? Didn't you get along well?"
Nancy's eyes are watering. It's hard to know if this is weeping or whether the smell of detergent has triggered her allergic rhinitis. The rhinitis distracts her.
"I used to go along the street, you know, and I'd be crying. I have these terrible watery eyes and people would stop me and say, 'What on earth's the matter?' and I'd say it's just my watery eyes but you could see they didn't believe it."
"That's a shame, poor you."
"They'd stop me in the street and say, 'What's the matter with you?' and I'd tell them straight, I have this condition but you could see they didn't believe a word of it."
"Didn't they? That must have been annoying."
"I'd be walking along the road, and people would see me crying and it was amazing really—"
"Shall we stop now and have some coffee?" I interject.
What would we do without coffee breaks? They stitch the caregiver's day together.
We sit at the table and eat cake. Nancy is buzzing with energy but I am exhausted. The only real benefit is to the state of the house. I am that most conflicted of creatures, a house-proud sloth, so all this effort pays off in terms of the secret pleasure I take in order. Like nature, I can't abide a vacuum (cleaner), but on the other hand I can't settle to writing if the carpets are filthy, a curse that may be peculiar to womankind. So this is how I live now. I make an effort to be Mrs. Tiggy-Winkle in the mornings, and to include the elderly, confused hedgehog of my acquaintance in housewifely activities. Then, after lunch, counting on the aftereffects of physical effort and the sedating power of a cheese toastie, I chum my mother-in-law in her TV room. Nancy mutters and twiddles and hand-rubs away, and then dozes in front of the television and I—though sitting with her, in Morris's electric armchair—disappear into my preferred world of words.
THERE IS CHANGE afoot and changes come as steps and not as slopes. There are sudden downward movements and this is the latest one. It seems quite suddenly true that Nancy doesn't know her grandchildren. This seems to be another instance of parietal lobe damage. Alzheimer's patients rarely have trouble with vision as such: The occipital lobe isn't usually affected, but family-member recognition is a subtler, deeper-buried form of word-object connection. The truth is that she hasn't known the children for quite a while. If you asked her about them, in the abstract, while cleaning windows, she'd deny having any or say they were all grown up and worked at Kmart, or some such random answer. She hasn't known their names for two or three years. But now she doesn't respond to them visually, either. The visual prompt of a little boy face appearing and grinning at her might elicit a happy response, but only because it's a little boy and (usually) she loves small children.
"Hello," she'll greet him with exaggerated surprise. "Look, a wee boy. Come in, come in, I won't bite. Let me look at you. You're a fine fellow. What's your name?"
Jack falls for this sometimes, despite knowing that Granny will turn on him in the end. "You're a little bastard, aren't you? Get out of here."
The granddaughters, being self-possessed young women, are ignored or dealt with in tones of wilting sarcasm. She mutters into her hand when they talk to her, as if her palm is an improvised gossiping friend.
Children following her into her bedroom—"You all right, Gran? You looking for something?"—are rounded on.
"Why do you keep following me everywhere? Why can't you just leave me alone?"
Then she'll come and find me, complaining bitterly. "I have to say I'm absolutely pig sick of all these young people that seem to live here."
ONE AFTERNOON THERE'S an unseemly and pointless row, utterly counterproductive. It starts when Jack is called a series of unfortunate names: Alzheimer's dished up with a side order of Tourette's. Jack's so upset that Chris is drawn into the row. Nancy's told the blunt facts: that we can only put up with so much. The _H_ word is mentioned: the _home_. The one she'll be shipped off to if she doesn't mind her mouth. Net result: two days of hand-wringing.
"I didn't do it! I didn't do anything bad!" Over and over and over. It's difficult to distract her from these ongoing, all-day denials.
"Would you like a cup of tea?"
"But I didn't do it! I didn't! I didn't do anything bad!"
"No. Listen. Tea. Do you want some? A biscuit?"
"But I didn't do anything! It's all a load of rubbish!"
And then, on the third day, calmer but no less angrily: "I'm afraid I have to tell you that unfortunately your children are liars. They're all bitches."
All of which begs the question: How did she remember the incident so long? Or did she? Perhaps it was another example of a contentless verbal loop—something that bypasses memory—and rage is just very sustaining. Emotional events have their own kind of longevity. I look up swearing and Alzheimer's, and it seems that it's to do with damage deep in the limbic system, in the amygdala. Amygdala damage has been linked to bad language, undressing in public, lechery, unprovoked hostility. Amygdala atrophy has been seen in Alzheimer's autopsy.
Morris comes home for a day visit, with an occupational therapist and various mobility aids. Nancy stays out of the way.
"Come on, Nancy," I chivy, putting my arm round her shoulders. "Morris is here. He's home for the day."
"Oh. Is he. Is he. Right," she says, pretending to watch the television.
"Morris! Your husband! He's here. Come on. Let's go and say hello."
"Oh, all right, then. If you say so."
We make our way with exaggerated slowness through the kitchen.
"You do know who Morris is," I venture.
"No."
"Your husband."
"Oh."
"Come on, then." I open the bedroom door and there's Morris, looking absolutely spent.
"Hello, dear," Morris says.
Nancy is blushing. "Hello," she says timidly.
"Amn't I going to get a kiss?"
She goes over and kisses him and then returns to my side.
"I'm just home for the day," Morris tells her. "But I should be back soon."
"Oh," Nancy says.
HARRIET DECLARES HERSELF available for granny-sitting, and we're invited to eat with neighbors who have a tree-growing business. Jane has a mother with Alzheimer's in a nursing home in England. Whenever Jane and I begin to talk, we fall down the same conversational black hole. I hate it that I seem able to talk only about Nancy these days. I have become very boring—not least to myself—a judgment confirmed by another supper party, where I fall into the black hole again, monologuing on the Nancy subject, even though nobody else present has caregiving problems. I must do something. I must do something about this. I must restrain myself from downloading. I see the tedium cross people's faces, the light go out of their eyes. I am beginning to repel people. Dementia caregiving is isolating in more subtle ways than I'd imagined. Though the community here is a friendly one, real friendships are slow to take shape. I don't go out of the walled kingdom of the house often on my own, and when I do I'm very dull company, and people don't visit much. Who can blame them? Nancy's likely to want to join in, seating herself close by and chuntering. And when she doesn't, she and Morris are all I seem able to discuss.
# Chapter 14
_It is useless to attempt to reason a man out of a thing he was never reasoned into_.
—JONATHAN SWIFT
MY BIRTHDAY FALLS IN FEBRUARY AND IS A BLIZZARDY day, with snow brought by a north wind that puts the heating out and makes the chimneys unusable, so I spend most of the occasion wrapped in a duvet in the drawing room, alone and reading, Chris having taken on the Nancy-minding in my honor. In the evening, determined not to stay at home with pizza and a movie, we go into town with Nancy and end up at the American diner, eating pizza, and going to see a movie. The pizza is disgusting. Then the movie's delayed. After a half-hour wait in the cinema, house lights up, joking that they can't find the right movie reels, the manager comes in to explain that the bulb has blown in the projector and they can't find a spare.
I am not feeling well. This not-feeling-well feeling is persistent and low-key. I am not going to the doctor. I don't want to have a conversation about stress and embarrass myself. _Stress_ would be the word used at the consultation. It's an easy word, protectively imprecise, a useful box to tip your feelings into. But a better word would be _incompatibility_. It's the shock of daily, ongoing proximity to this "vegetable universe" of my in-laws: their lives pared back to the bone, to the medical, physiological, placed squarely in the raw, reduced to material struggle, an easy decline the most that can be hoped for. It's William Blake, the source of the phrase: "Imagination is the real and eternal world of which this vegetable universe is but a faint shadow." There are plenty of people who'd phrase it directly in reverse, insisting that the vegetable universe is what's real and the world of the mind the shadow, Morris and Nancy among them. There can be no communication. There never has been on any real level, but since moving here we've both of us, both camps, been pushed further to the extreme of our position and the gulf has been rendered unbridgeable. This, I have self-diagnosed as the root of the trouble.
I am having funny turns, two or three a week, in which I get attacks of something like vertigo. The room spins, my head swirls, and I am intensely nauseous. Sometimes it happens in the car and that's the worst. I have to hold on to my head because it feels too heavy to hold up. The scenery rushes past the window at odd angles and small hills feel like a fairground ride. It goes beyond nausea; it's more like my whole body seeking to turn itself inside out.
When I can I confine myself to bed, propped up with pillows, and read, and surf the Internet for celebrity dementia sufferers, a strange new hobby. Jonathan Swift (1667–1745), author of _Gulliver's Travels_ , spent his healthy adult years fearing dementia in just the same preoccupied way that Philip Larkin brooded on death. In 1717, on a country walk, he remarked of a tree whose upper canopy was shriveled, "I shall be like that tree, I shall die at the top." His forebodings were spot-on, though ironically what he believed to be early signs of brain disease was probably Ménière's, a vertigo-causing disorder of the inner ear, a condition it's possible I'm also suffering from. Swift started losing his memory in 1735 and was declared incompetent to handle his own affairs in 1742. His obsession with senility, while still in his prime, led him to invent the Struldbruggs, immortals born randomly among a mortal race, in book 3 of the _Travels_ (written in 1724). The Struldbruggs, far from satisfying Gulliver's excited anticipation of wisdom, have forgotten the common names for things, can no longer read, and are emotionally incontinent, nasty and feared. They are in rapid decline at about the age of eighty and are then demented for eternity, unable to die, "peevish, covetous, morose... uncapable of friendship, and dead to all natural affection." Swift's probable Alzheimer's was misunderstood as ill temper and as lunacy by contemporaries and by critics. Samuel Johnson, writing in 1779, blamed Swift's dementia on his refusal to wear glasses. "His ideas, therefore being neither renovated by discourse nor increased by reading, wore gradually away, and left his mind vacant to the vexations of the hour, till at last his anger was heightened into madness."
Morris is still in hospital.
Nancy doesn't mention him anymore. I ask her about her family sometimes (her family are elsewhere; her family, she suspects, are all dead), reminding her that she has a brother who lives in Australia.
"He's in Australia? Has he been there long?"
I mention Morris's name. Does she know who that is? I show her a photograph.
"He's my brother."
"No, he's your husband."
"Oh. Is he. Oh." Indifferently.
"He's in hospital."
"Is he? Oh dear, oh dear. Oh dear me. Well. I'd better just go a wee walk."
Some days, though, the story is different.
"I don't like to pry," one of the day center volunteers says to me, "but she's been saying that she hasn't been allowed to see her husband...."
That'll be the husband that she sits with in the afternoons, then. She sits at his bedside and holds his hand and becomes tearful, as does he. We leave them to it and go off to do errands and when we get back find Morris looking perplexed.
"She's been talking absolute gibberish."
"Morris. You do understand what Alzheimer's disease is, don't you?"
"Of course, but even so, absolute gibberish."
Nancy begins going to the day center in the town. A minibus comes and picks her up at the conservatory door on Tuesday mornings. This means that all her weekdays now have something going on in them other than Friday. Friday becomes a black day. We dread Fridays.
Her old love of housework has quite abruptly evaporated.
"I wasn't brought here to do this!" she cries, throwing the duster down.
"Would you like to vacuum your sitting room for me? Look at all the dog hair."
"I don't think that's my job."
"Help me hang the washing, then. You like that. We'll get the pulley down."
"I'm not at all interested." Eyelids nervously fluttering. "And you should know that I'm going home in a minute."
I put the television on for her, provide magazines to look at, cake and tea, her favorite soft toys, and put her beloved blue handbag onto her lap, the one with the assortment of things in it that she likes to take out and put in again. But none of this interests her on Fridays.
She turns the television off—or rather gets me to—and sits looking forlorn. This is niggling. It isn't possible to work. I crouch at her knee looking up into her face, taking her cold hands in mine, trying to coax her into wanting something out of the day.
"Just leave her to it," Chris says. His patience is worn very thin.
"I can't just leave her sitting there."
"That's what she wants. If she wants to brood, then let her."
"But..."
"You've spent most of the last five months with her. It's all right. Look. Just go. Go and do something somewhere else. I'll sit with her for a bit."
Chris installs himself with his work project in Morris's chair, and makes business calls there from his mobile. His mother sits a few feet away staring straight ahead and not saying anything. But the expression on her face is pretty noisy and her hands are in continual motion.
I sit in the drawing room at the window, wrapped in my blanket, watching foul weather hurtle itself toward us from the Atlantic, great low barrels of cloud rolling in, gray rain turning sleety. The wind is oddly comforting, a soothing white noise, baffing at the windows like a heartbeat. _There there. There. There. There there now_. But despite this I find it difficult to work. My concentration skims the surface of the job at hand. I need to write about the people in the novel that I'm supposed to be writing as if they were real. I need to build walls around them, conjure up the house that they live in, which is as important a character in the book as anyone; feel their breath, read their minds, give their situation depth. The vividness of the impression needs to come off the page at you. I know all this. But it isn't happening. The problem is that I'm not really present in the story. I read and reread paragraphs of the draft hoping and failing to catch the tide. But I'm half listening for trouble, and for the door to open, which it does, Nancy shuffling in and saying, "Ah, there you are, I thought I'd lost you," and Chris coming in after her, "No, Mother, she's working just now; come on, come back with me."
I go back through to make coffee a half hour later and find Chris watching a western and Nancy fast asleep.
Next, Nancy turns against her galantamine. She secretes her pills up her sleeve and is cunning about it. She takes them apparently happily, and pretends to glug them down with water, but when I check later, they're tucked up her cardigan arm, saliva-softened and encased in tissue paper.
"They're not mine!"
"Yes, they are. It's important you take them."
"What's it got to do with you?"
"I'm the one that looks after you."
"Oh, are you. Are you indeed."
"Here's one, look, take it from my hand."
She clamps her jaw shut and looks like she can't hear me, and only intense negotiation gets the pills properly swallowed.
SHE BEGINS TO suffer quite dramatically from _sundowning_. There's a noticeable deterioration of mood, shortening of temper, worsening of speech, a sharp downturn in reasoning capacity, at around four to five o'clock in the afternoons. There are sundowning incidents. She begins hitting. Caitlin's slapped hard, and Nancy is completely and utterly unrepentant. This is worrying. I look it up and it seems to be connected with frontal lobe damage, again. The "prefrontal" lobe (the very front, behind the forehead) appears to be the site of our moral selves, and is crucial in triggering appropriate emotional responses. It bestows on us the very human qualities of guilt, embarrassment, and self-reproach, things Nancy no longer feels. Everything I read at this time seems to have some link into dementia. Even, unexpectedly, _Middlemarch_ , in which George Eliot writes: "With memory set smarting like a reopened wound, a man's past is not simply a dead history, an outworn preparation of the present... it is a still quivering part of himself, bringing shudders and bitter flavors and the tinglings of a merited shame." Nancy's tingling days seem to be over.
She gets a wild look in her eye when anyone asks anything of her.
"I'm going to report you to the manager," she says to me one day as I'm handing her a glass of her favorite lemon squash.
"What?"
"I think it's only fair to forewarn you."
"Nancy. Who do you think I am?"
"I know very well who you are."
"Who? Who am I?"
"What insolence," she says.
Chris comes into the room. "Trouble?"
"You have no idea who I am. To whom I am," Nancy says in a more formal voice than is usual.
"Oh? Who are you, then?" Chris asks her.
"I. Am. The. King of Scotland," she says.
"You're what?"
"Yes. Nobody knows yet. The king of Scotland and this is all mine."
One evening, when Chris is in London, Nancy is mislaid for a while. Eventually I find her in the tractor shed in the yard. She's closed the door behind her and is standing expressionless by the ride-on mower.
"Nancy. Thank god. What are you doing in here?" It's dark in there and oily smelling.
"I'm looking for my father."
"Come inside, it's cold, you'll catch your death out here."
"No! I won't! I'm going home!"
She looks meaningfully at the mower.
"Nancy, you're in the yard in February wearing a cotton nightdress and a bath towel. You need to come inside."
I take her by the arm and she snaps it away violently. I take it again more firmly. She resists and a tussle develops, Nancy shouting at the top of her voice.
"Go away! Leave [shrug] me [yank] alone [push]."
I manhandle her back into the boot room. She's icy to the touch and bluish. She stands in front of the open glass door into the house, barring the way, hands spanning the width, holding on tight to the door frame, bracing herself and white-knuckled.
"You're not going in," she says in a low voice.
"What do you mean? Those are my children in there."
"No. I was born here. And you are not welcome."
"Nancy. Let me in."
"You are not welcome." It's said with genuine menace and I feel a shiver of fear. I'm reminded of something Nancy's friend Carol said, the last time she phoned. Carol's late mother had Alzheimer's and thought Carol an intruder. She'd stalk the house with a carving knife, intending to do poor Carol harm. What if Nancy had a knife right now? By the look of her, she'd not be shy in brandishing it.
I try to pry her fingers but she's supernaturally strong. She lets go all of a sudden, lurches forward, and pushes me backward. I fall on some shoes in an undignified heap. It must be hilarious from her viewpoint: my surprised face as I land legs akimbo in a muddy pile of Wellies. I'd be laughing if it were me. But Nancy has a very serious look on her face, and glinting malice in her eye. She returns to the barricade.
"Nancy. Listen. You are my mother-in-law. I am married to your son."
She snorts. "That's ridiculous."
"Nancy. Just let me into the house, please."
"No."
"I need to go in now. It's cold. And I need to get your supper ready."
She relaxes her grip. "I'm hungry right enough."
"Let me make you something tasty to eat. And we'll have a pot of milky coffee, shall we?"
Now she's all smiles, clutching her hands under her chin like a Victorian heroine.
"Oh, that would be lovely! You're very kind to me."
She's already had some pasta with us, but I make her scrambled eggs on toast and she wolfs it down.
Then it starts again.
"My house. They say it's not but it is and they'll be sorry. They'll find out forever. They say what they like but they don't know anything." She sees a pound coin sitting on the carpet and picks it up. For the next half hour she turns it over in her hands, examining it closely.
Eventually she speaks.
"This is all there is," she says to me, holding the coin in front of her face.
"There was only one," I admit.
"What have you done with the rest of it?" she demands.
"Sorry?"
"My money. This is all I have left. This is all I have now. The rest has been stolen. You took it. You took it, you stole my money."
Distraction is the only way out.
"Would you like some more toast?" I ask her brightly. "Because I'm jolly well going to have some. Toast with cherry jam and tea. What do you say?"
"Oooh, lovely. I like jam."
She follows me into the kitchen. "How nice you are and how beautiful you are," she says, running her hand over my neck.
When I put her to bed she stands with her arms outstretched, consenting to be undressed, accepts the flannel nightdress, gets into bed with her blue teddy, then bursts into tears.
"What on earth's the matter now?"
"It's just that you're all, all so k-kind t-to m-me," she blubbers, nose and eyes streaming. "You're all so very kind and I don't know what to d-do about it."
"You don't need to do anything. I hope you have a good night's sleep."
I put a gathered-up tissue over her nose and instruct, "Blow, please." She blows.
"I have to reward you somehow," she gushes. "I need to give you something. Some money. Would you like some money?" She reaches for the pound coin on the bedside table.
"No, thank you. Just go to sleep." She's sleeping within ten minutes.
Half an hour later she's up again. I hear her coming, the slamming of doors, the raised old-lady inflection. A voice declaring, "I never heard anything so stupid in my life!"
I go out into the corridor. "Nancy?"
She's holding a toilet roll. I take her to the bathroom. When I try to get her back into bed she takes a swing at me with her arm, her fist balled up. "Don't you touch me. Don't ever touch me again!" she screams.
That's Saturday. Sunday is worse. Caitlin is slapped again, and Jack's yelled at, and then I'm denounced for interfering. There is trouble from breakfast to bedtime, one thing after another. Caitlin spends most of the afternoon sobbing and Jack joins in. "W-why is G-granny so m-m-mean to us?" We have another (one in a series) of our conversations about Alzheimer's.
"That's not really your granny any longer," I find myself saying. "Your granny is gone." Harsh words, but they pour oil on the waters.
On Monday we make a telephone call to the health visitor. She turns up that same afternoon, with a colleague and a file full of paperwork, to talk about The Future. The great steam engine of the National Health Service is seen lurching from standstill to a creaking, groaning forward crawl. There are things nobody tells you unless you know the right questions to ask. I stare at the paperwork, the forms, the pen poised over them. At least drugs are free here, doctors and hospitals are free, day centers are free—aside from a small charge for tea and cake. I feel sorry for friends in the United States in a similar position, caught up in long tangles of red tape: the details of Medicare benefits, rights to Medicaid, drug bureaucracy, health insurance companies and their cunning opt-outs, and almost everything coming back to money, money, money. At least the health visitor (another free service in the United Kingdom) isn't advising us to see a lawyer specializing in dementia in order to have our rights and our options explained at the outset. We will need to see the doctor, see the memory team, be assessed. We shrink from institutionalizing. Just three months ago we'd been dismissive about getting help, but now we're greedy for information. We say that we've read somewhere that it was a good idea to get names down as soon as possible for residential care, a long way in advance. The health visitor doesn't laugh, but her eyes register our naïveté. That's not going to be available until things are much, much worse. And it isn't going to be free. Instead, it's going to become clear that care of the elderly is a business, and a highly profitable one at that.
Millie's had good exam news, so we decide at the last minute to go to the Indian restaurant in the town to celebrate. There isn't time to get Nancy in the bath and we're aware, in the restaurant, that she's rather grimy, her cardigan food spattered and the smell unmistakable. She doesn't say anything all evening, just rubs her hands in her usual three-stage way, twiddles her hair, sings a bit. But then, for the first time ever, she has trouble with the cutlery. We put her fork in her hand but she can't seem to coordinate it from plate to mouth, and abandons it for her knife. We cut her food up and take her knife away and give her the fork again. She puts it down and refuses to use it and eats curry with her fingers. We've ordered her a chicken korma, and it's difficult to eat with the hands. We give her plenty of bread but she doesn't seem able to scoop. She picks up rice a few grains at a time in her fingertips and stuffs it in; then pieces of saucy chicken, licking her fingers with pleasure. People stare.
THE GLOOM OF February is eased by decorating, every room de-greened, and the weather is briefly in harmony, giving us a week of crisp and sunny weather, windless and almost balmy, days that end in stupendous sunsets, orange and red over a lilac-colored sea. Thousands and thousands of bulbs are sprouting in the wood, bouquets of white and purple crocuses under the trees, a carpet of snowdrops stretching between; the taller, stouter foliage surrounding them promising a field of daffodils preparing to show itself, and bluebells for later on.
Nancy has her first respite week at the nursing home, and the rest of us go to London. It's our final weekend before Morris's return, and we need to gird our loins. We are all in frisky, happy form, cavorting like horses let out of winter stabling onto spring grass. We can't stop chattering. Morose tourists with heavy cameras look at us thoughtfully on the train.
Morris is brought home in an ambulance on the last day of the month. Preceding his arrival, the home care manager visits. She's heard already that Morris is severely disabled and will need home support. He is weak and can't walk at all, has trouble even taking a step, we confirm, and is unlikely to improve, though it's hard to say the words. The home care manager takes us in hand. Over the few days before Morris returns, a steady stream of assessors, builders, physical therapists, and equipment specialists tramp in and out measuring, advising, talking gadgets. We're advised that he'll need a wet room and given consultation on where and how to construct it. Morris is equipped with an inflating and lifting bed, humming along on its electrical supply, a ramp for wheelchair access from bedroom to sitting room, toilet frames and urine bottles. I'm reminded of his and Nancy's old family home, the row house, which had a handrail fixed onto the wall of the downstairs toilet, a relic of Nancy's mother's long-term stay in what had been their dining room. A Zimmer frame sits by his electric chair, for getting in and out, which I'm learning to call "transferring." He can't transfer alone.
There are gales, and heavy rain. But at least there are snowdrops in the wood. Our decorator, an ex–creel fisherman, goes diving in the sound and finds a treasure trove, the seabed littered with scallops. He brings me a fertilizer bag full, a gift. Chris spends a whole day shucking. We eat some of them cooked in garlic butter with pasta. We have some for breakfast. Then we freeze the rest for another day, a day when we'll have visitors and can share our good fortune.
# Chapter 15
_Alas, must it ever be so?
Do we stand in our own light, wherever we go,
And fight our own shadows forever?_
—EDWARD BULWER-LYTTON
IT'S DOUBTFUL THAT MORRIS WILL EVER WALK ANY BETTER than this. He comes home with exercises the physical therapist gave him, a cardboard tube he's supposed to manipulate with his legs, but he isn't doing it and has no intention of bothering. Reminding him that he's supposed to exercise constitutes nagging. "Don't nag me," he says. "I have a wife for that." I look at Nancy, sitting looking blank in her chair, and raise an eyebrow.
"It's not nagging," I tell him. "It's what you need to do to improve. Don't you want to improve?"
"All right! All right, all right!" he snaps. I withdraw. But I can't leave it at that. Somebody has to chivy him. So every day after breakfast I place the cardboard tube and the instruction sheet on his side table and pat it. "Here's the thing for your exercises," I say. "Yes, boss," he says, not looking away from the television.
Boss is my new name. Nancy likes it. It makes her giggle when Morris calls me boss. It's plain to her, you see, that I have a junior position in the household and that Morris is in charge. Men are always in charge. I get called boss every day. I only get called boss when I exceed my authority. Thus is it made clear that I exceed my authority a lot. I do it when I suggest that he eat more fruit and fewer biscuits. I do it when I suggest that he try to walk a few steps, or recommend that he call me before the fire has gone out or the tea's soaked into the chair, or before his need for the bathroom is urgent. It seems that our relationship, mine and Morris's, has entered a new phase. Perhaps it's to do with having been nursed and unhappy in hospital for so many months. He has begun to see my role as caregiver as an attempt at power over the two of them, and has begun to resist.
We'd been looking forward to him coming home. We'd been counting on it stabilizing Nancy. But with Alzheimer's there's rarely any hope of going back a step. And actually it's worse. Morris's return has the opposite effect. She finds his sudden appearance in the chair next to hers, the bed next to hers, alarming. An enormous downward step is in the process of taking place. Something momentous. Not recognizing Morris.
"Who is that man?" she'll ask me in a low and confidential tone, as we clear the breakfast things away.
"Man?"
"That one in there. The one sitting in there." And then, as I go to look in the direction that she's gesturing in, "Don't look! He'll see you."
"That's your husband."
"Is it?"
"Yes."
"Oh."
It's "oh" for now because she trusts me. She won't trust me much longer and then her reaction will be different.
We have reached the apotheosis of forgetting. Morris the man, Morris the husband, and all the years and memories that Morris encapsulates. In 2004, the anterior cingulate was identified as the missing link in memory production, controlling the storage and retrieval of long-term memory. The area lights up in brain scans once memories the subject is engaging with are several months old. When the anterior cingulate, which is part of the wrap of functions around the limbic system, is damaged by dementia, its malfunction means that only parts of the long-term memory are accessible, and these bits can't come together in a meaningful way. Perhaps, in terms of emotional impact, it's worse to have parts than none at all. If the anterior cingulate is disabled entirely, the access to the past is completely blocked off.
It isn't in Nancy's case. Not yet. Instead it fluctuates like bad radio reception, going on and off station. It's worse for Morris to have her memory of him come and go so seemingly arbitrarily. From minute to minute, he doesn't know whether she knows him or not. He's exhausted by the inconsistency. When she does know him, she's loving, fussing over him, and getting onto his lap (getting onto his lap and crying). It's as if she has missed him. When she doesn't know him, the manner of her rejection chills the blood, and it's Morris's turn to weep.
Now that Morris is home, the television is on all day. He's getting a bit deaf and the volume's turned up high. Thus we all get to share in his televisual life, which kicks off with the confessional shows first thing in the morning, the accusations flying, and follows an itinerary that soaks up as many soaps and police and hospital dramas as can be managed. Their sitting room becomes a little self-contained universe of televisual strife: assault, adultery, divorce, neglect, crime, rip-roaring arguments: all these keep Nancy smiling and the rest of us depressed. (I'm with Groucho Marx when he said, or is said to have said, "I find television very educational—every time someone switches it on, I go into another room and read a book.") A constant melodramatic state of crisis thrums under the door, seeping out and spreading its antilove around. Nancy likes it in there when the soaps are blaring. She sits and twiddles away at her hair with one finger and smiles. The rest of the house is too quiet, too harmonious for her. We bore her. She wants talking, talking, all the time talking; can no longer tolerate anyone reading, has to interrupt; comes and looms over people until they pay attention to her. I'm reminded of a particularly needy cat we had once that attacked opened newspapers. Though he didn't go in much for existential angst.
Nancy's care plan arrives in the post from the Tuesday day center. Two copies. We're to read it, sign it, add any comments, send a copy back. "Nancy may try to put used toilet tissue paper up her sleeve," it says. "In the interests of her dignity she has begun eating lunch with a caregiver in a separate room." Nothing is said specifically about mood swings, nor about pinning the lunchtime helper's hand to the table with a fork, so we are quietly encouraged.
Morris decides against resuming at the Thursday club and no longer leaves the house, but the trappings of a sort of a social life are the unpredicted benefit of having the home care team in. A lady comes in at 9:30 A.M. to get him up, another at 8:30 P.M. to get him to bed. The lead home care aide, the one usually on duty, is between generations, Morris's and ours, and in the early days proves a useful bridge between Chris and his father. She shares Morris's cynicism, his knee-jerk conservatism, enjoys his anecdotes, laughs at his jokes. And slowly, she becomes his confidante. This is the point at which things change. She and Morris have long and private conversations, while she's getting him up and getting him to bed again. Getting him up and getting him to bed begin to take longer. She stops telling us what Morris thinks, though useful snippets are cast our way. We hear that he despairs of Nancy getting better. We hear that he doesn't ever want to go to a day center again. We hear whether or not he's enjoying the meals we cook for him. It's the things we don't hear about that intrigue us. And in time, this new relationship in the house becomes both a bridge and a wedge.
The phone starts ringing, now that Morris is home. Professionals on the line. We're one of the plates the care profession has to keep spinning. People checking up on Morris, and also on me, who's been flagged, perhaps, as most likely to crack. Something I have said in a low moment has been talked about, has been the subject of an e-mail, or has become a sentence in a file. There may be a blue Biro question mark by a note about my own mental health. The only outcome of this intuition, however, is a renewed determination on my part to clam up. I feel bad about this, about my concerted noncompliance. It's their job, their jobs collectively, after all, to follow up on Morris and Nancy, to make sure they're okay and that their caregivers are caring and coping. I feel bad about resenting their concern. I ought to be grateful. It's churlish to be otherwise. But it's difficult to convey just how little I like to talk about my problems with concerned professionals on the phone. I hate the phone, anyway, always have. I hate it ringing when I'm doing something else and people blithely expecting me to be ready to give considered answers to their questions. I'm not, in any case, a talker. The words in my head go from brain to fingertip, not from brain to tongue. I can't do justice to myself, and especially not in these circumstances, when it's _off the cuff_ but also _on the record_ , which is how these phone calls seem. Being caught on the hop makes it infinitely worse. I say the first thing that comes into my head, and this might very easily be untrue, unfair, something true or fair today but not tomorrow (caregiving is a mood-swinging business), irrelevant, repetitive, self-pitying. Quite often, increasingly often, it's all of these things crammed triumphantly into one conversation. People ask me how things are. That's generally the first sentence spoken. "It's _X_ here; how are things?" I can't tell them I have no idea. I have to produce something. What I produce is entirely random. Quite often it's blacker than is accurate, because I'm irritated by being interrupted and by being expected continually to give an account of the state of things. I don't know how I am, how things are. I'm just doing this. I do it, over and over, every day. There isn't any alternative, is there? Other than a nursing home. "So, have you talked to Morris about residential care lately?" the professionals ask. As if it's like ordering coal.
There's no room in the nursing homes, anyway. There are waiting lists. And it will cost a fortune, eroding their life savings away until the money's all gone. How can we be responsible for the loss of their life savings? It's not something anybody could enter into lightly. There's only one private home in our county and it doesn't have an Alzheimer's unit. Nancy isn't severe enough a case, nothing like severe enough, to be taken into council-sponsored care. That has been made very plain. She'll have to be completely doolally, physically frail, incontinent, not eating, and have sawn off one of Jack's limbs with a bread knife, before urgent action will be taken of that kind. The state of things is that I am lumbered, and am resentful about it, but resigned. That is how I am. But I don't say that.
I say, "How am I? I'm okay. Functioning. But it's hard work and getting harder." That's on a good day. On a good day, I'll follow this up with the latest Nancy anecdotes and will tell them in such a way that we both end up laughing, my interlocutor and I. On a bad day, I'm impatient with the phone person and incapable of being funny about the anecdotes. I sound terrible. I'm aware of this, and I mind. I wish they'd stop phoning to sympathize because in this situation, pity isn't any use, and an emotional phone conversation throws long shadows on the day. The worst of it is that nobody ever phones to sympathize in brief. The conversations go on for twenty, thirty minutes, forty. We go through it all again. Her behavior. Her condition. Her decline. Morris's state of health and prognosis. His behavior. His condition. His decline. His interaction with Nancy, his desperate unhappy bullying. The effect that all the above are having on the family.
I become markedly less grateful for official concern as time goes on. I hit the ball back over the net with topspin. "Hello there, bad timing, unfortunately, just going out." They ask if they can ring back. "Tell me what it is you want, if you like, but can you make it quick? I'm really busy."
I interrupt them midstream. "Thanks for this, for phoning, but now I need to get on." I invent things. "Can't talk now, Nancy is in the bath." "Nancy and I were just about to play badminton." "Sorry to interrupt, but I see Nancy making off down the drive." I start making my answers short, pointedly so. Eventually it's open rebellion. "Look, can we not have any more meetings? Can you not call quite so often? I already have very little time to myself and having meetings isn't what I need. Sympathy isn't what I need. If you can suggest anything we can do to make Nancy less angry and Morris less depressed and me less tired, that'd be great. Otherwise, perhaps you could e-mail? E-mail's better for me."
PEOPLE ARE OFFENDED left, right, and center. They're not accustomed to being headed off at the pass. Perhaps they categorize my unwillingness as indicative of neglect? Perhaps, in their training, their trained way of seeing the world, my being flippant is considered a marker of something. I don't know. They must be accustomed to other sorts of people being caregivers, I decide. Perhaps they're used to people who like this endless going over things. It must be seen as therapeutic, cathartic. It's supposed to be bad to bottle things up, isn't it, unhealthy? I, however, am of the opposite view. I don't relate to the lanced-boil metaphor. I tend to think that a problem shared is a problem doubled. Or quadrupled. A problem shared, hereabouts, is generally a problem that's gossiped about.
March brings its usual last blast of winter to blow away the spring. We find ourselves snowed in for almost a fortnight, transport canceled, schools closed, the world quiet in that uniquely quiet, snow-muffled way. We don't get that much snow, but the little that falls is blown impressively into drifts sufficient to immobilize a bus, and subzero temperatures freeze the slush on the roads, and that brings the whole region to a halt. We can get no farther than the village shop. Potatoes, cabbage, frozen fish, bacon: these become, temporarily, a big part of the diet.
In late March I have a new project. I organize myself a whole other life. A house in Turkey. We've sold our ruin in Normandy that we bought to renovate and never will, and want spring and autumn sun. The project starts as a house in Bulgaria, as I have seen houses on eBay going for £8,000, though Bulgarian trawling ends in a cul-de-sac. It's Greece we want but we can't afford Greece. I go online across the water into Turkey, and find that a tiny cement house by the sea is possible at the £25,000 level. And so I spend two weeks doing nothing but chasing leads, drawing up a short list, pestering agents. The in-laws, Chris, and the children are kept at arm's length. Dealing with the inevitable, inescapable day-to-day slog, unpaid and thankless—running the household, dealing with Morris's mild but needling hostility, coping with Nancy's bizarre and darkening world—all of that, I can deal with because round the corner, in laptop-land, life-changing things are afoot.
# Chapter 16
_One need not be a chamber to be haunted;
One need not be a house;
The brain has corridors surpassing
Material place_.
—EMILY DICKINSON
WHAT HAPPENS WHEN THE ABYSS OF AMNESIA IS opening constantly at your feet, as it appears to be with Nancy? Some days it appears that her brain is compensating by creating and supplying its own answers, its own improvisations: fictions that keep her afloat. It isn't that nothing is going on in there, in her brain. She improvises her reality from minute to minute. This is on my mind today, a stormy April morning, and as it happens, the first anniversary of our agreeing to buy the house, because I dreamed last night that Nancy was a sales representative, working in a postapocalyptic landscape. There was a war-blackened ruinous backdrop of burnt-out skyscrapers; it didn't much resemble Edinburgh. The company she worked for had gone, as had all the other personnel, leaving her alone in the city, but she kept going, working out of her car and flying by the seat of her pants. My dreams lately have tended to the metaphorical.
It isn't possible to have identity without a history. Pascal was wrong when he wrote, "If somebody loves me for my judgment, or my memory, do they love me? Me myself? No, because I could lose these qualities without losing my self." The more I think about this statement of his, the odder it seems. It's normal for selves continually to be evolving, and in that sense Nancy's improvisations are reassuring. Something is happening. It hasn't all come to a halt. I am writing this in bed, early in the morning, the rest of the family asleep. Since waking I have thought, apparently randomly, fleetingly, about a whole array of insignificant things and in making decisions, reflecting on them, I have become a new person, albeit in a trivial sense. As Heraclitus had it, you can never step into the same river twice.
Which rivers does Nancy put a toe in? Which river is she wading in, thigh deep, in those periods of sitting in her chair hand rubbing and looking deep in thought? It's clear from her eye movements, her mouthings, her shifting expressions that something is happening. Are they words, pictures? Is she thinking in the first person, or does her voice come at her like dictation? Where is her mind taking her right now, lying awake in her twin bed, the light vivid at the edges of the curtains, Morris snoring lightly and curled in a fetal circle? She's lying facing the door, facing her upright wood-framed chair, the clothes laid on it from yesterday; she can see Morris and the entrance to the bathroom. What occurs to her about these things that she's looking at? What content and format do the improvisations take? The little output that does reach us— _this is my house; you work for me; I was born here; my father is in the garden; I must get to the office; the friends are on their way_ —doesn't hint at much in the way of a coherent alter ego, nor the creation of whole new identities, if you discount one-offs like the king of Scotland episode. She doesn't claim to be anybody else, other than, on occasion, her younger self, unmarried, unburdened, childless, her whole life ahead of her (and don't we all imagine our immortal souls, our essential selves, to be fixed at around the age of twenty-eight?). Nancy's fictions are more to do with her brain coming up with scenarios that explain her life now. For half-hour periods, she is the owner of a big house in Edinburgh with staff (the rest of us), and/or somebody who has lived here her whole life, confusing it with the estate where she was born, and/or must get the house ready for a party because the friends (everybody from her past that she can remember, I assume) are on their way. They're called the friends, collectively. She no longer has a handle on any particular name or face, is just hopeful that they're out there and on their way to rescue her.
Waking this morning at six and listening to the wind rattling at the windows, I tried to fake being a person without a memory but it was impossible. Everything we are is the sum of our history, augmented by every new experience, each stone added to the cairn and modified by our thoughts about that stone, and about the shape the cairn is taking. Our selves are fed by our narrative, the story of our past and our imagined futures. Ask me who I am and I turn immediately to memory. It isn't possible to answer the question "Could you tell me something about yourself?" without recourse to biography. Even aside from replies that start, "Well, I was born in..." (which are the most obviously memory driven), other kinds of responses, ones that try to avoid the straight biographical—"I am intelligent, curious, anxious, and usually hungry"—also rely entirely on memory. You only know yourself because of your memory. If you ask Nancy who she is, she can quote her name, but that's all that's likely to arise from her unprompted. If you ask her, "What are you like?" or "What kind of person are you?" she isn't able to answer. She'll appear to think about it. The eyes dart from side to side. But then she says, "I don't know, really," or "I couldn't exactly say" or laughs defensively. At a fundamental level there has been a disconnection and Nancy's self is locked in a room with no windows.
Who I am is what I've done and experienced, and what I think about it all; how other people make me feel about it all, how the books I've read and films I've seen have made me think and feel about it all, creating a unique and labyrinthine web of connections that is my self. I have a library of self at hand. I can wander the halls of this library and choose whichever bit I like, and read from it and enjoy the indulgence of having new ideas about the past. I find in the last few years that I am dipping into it more and more and finding surprising new connections between things. This, I suppose, is what people mean when they talk about personal growth and one of the few compensations of being post-forty.
The only (inadequate) way I can relate to what Nancy experiences when she wakes is in recalling moments when I haven't been sure where I was. Waking from an anesthetic. Or waking up in a strange hotel room, with the wrong furniture, the wrong shadows, the wrong smell, the door in the wrong place, and that first mildly alarming recognition that this isn't home. The alarm is barely formulated before it's redundant. The brain steps in hurriedly with information, clears its throat: the efficient personal secretary. _Ahem. I think you'll find you're in the Travelodge. Half-term trip_. Ah. Yes. An instantaneous connection is made between this room—the Travelodge, the half term, the life I have—and the library of the past, which is always with me, wherever I am. The Travelodge becomes another pebble on the cairn.
This morning when I opened my eyes the room I put together was there, the anticipated objects, Chris sleeping, everything familiar and as it should be. I'm looking around it now. It's cold and I'm wearing a sweater in bed. The clean laundry is piled on the chair awaiting sorting. My new handbag is hanging from the wardrobe door. I recognize the handbag. I remember buying it. I don't bother to have the memory, in full, of the shopping and acquisition; it's more like, in computer terms, a shortcut on the desktop that I am confident leads to the memory. I don't open the file, though I did, the morning after buying it, reviewing the choice that was available and reassuring myself that I didn't want the red one with the too-short handles that I was drawn to initially. That's all I needed to do. Now when I see the handbag all I see—and I don't even see it, I don't need to—is the shortcut on the mental desktop that connects me to the object. It's so brief as to be a shortcut to the shortcut. Recognition. It fits into my narrative and that's all that's needed. If I wake and see something that doesn't fit—a book I haven't seen before that's appeared on the bedside table, say—then my first instinct is to try and make it fit. There's a book there I don't recognize. How has it appeared on the bedside? I didn't put it there, did I? I do a brief file search. Oh yes, it was purchased in a rush in the city—I see the bookstore, I see the face of the assistant—bought for a birthday, and last night Chris was emptying the bags. He must have put it there when he came to bed, thinking it was for me. It's good to _see_ the bookstore and the face of the assistant again. It reassures me that my narrative is intact.
But Nancy doesn't have any of this anymore. I don't know what she does have. Her mind, unable to deal with not being able to make sense of things, makes its own sense, delivering explanations up from fragments, inventing new scenarios that make things seem coherent. Whatever the case, it's clear that her self has been pared back to the minimal. She is operating on the level of the _core self_ , which Antonio Damasio, a neurologist-neuroscientist at the University of Southern California, describes as "a transient entity, re-created for each and every object with which the brain interacts."
Nancy says to me almost every morning, "I'm sorry, I don't know where I am," and in the circumstances that seems a remarkably gracious response. It's her face that betrays her fear. The reason I'm not afraid on waking is that, stirring and stretching in bed, everything I see around me is explicable; it was _put_. Personal history isn't just about the CV, executive or social. We have history with everything surrounding us. The house is one we bought having sold the previous one. Our possessions carry with them their own stories, of how they were acquired and where, and their thing biographies, things that have happened since we got them. A chair used for reading is a highly evocative thing, or a sofa owned since the children were small. Look hard at that sofa and you'll see them, little pink and white people, fresh out the bath in clean pajamas, waiting for a story. An old pair of jeans carries history with it, that's why they're hard to part with. This isn't just sentimentality, but context. Imagine waking in the morning and finding everything around you is new: the building, the garden outside the windows, the people who talk to you as if you know each other, the shirt the stranger hands you, the chair they take you to, the man sitting in the other chair. If your brain were still intact enough to want to make a history out of things, it might get around the novelty of all this by explaining your real life as somewhere else. You are somewhere new and your life is somewhere else. All you're going to want to do is get back there.
I'm getting up now to make breakfast. The house layout is known to me. Rooms are subsequent in the expected way. The kitchen cupboards hold the things I put in them. I know where the frying pan is, the olive oil, the glazed bowl, and the whisk. There are leftover potatoes, garlic, some tomatoes for the omelet. As I rise and dress and go down to make the breakfast, I'm running through visual anticipations of how it will be, barely consciously if at all; each next step conjured and satisfied in turn. In a way, I'm remembering things before they happen.
Six months ago, Nancy may have appeared in the kitchen, hearing me up and about, and once I'd reassured her by appearing to know her and offering her tea, she'd ask if she could help. I miss helpful Nancy, wanting to do things. Although, asked to put eggs in a bowl, she couldn't even then have grasped what was being asked. She might, with encouragement, have put the eggs in the bowl, entire, shell on, and stared at them as if expecting them to act. She remembered "egg" then, though "bowl" was trickier. Always, with the progress of Alzheimer's, life is bound up with lists and ranks of objects, and tiny gradations of loss.
# Chapter 17
_Time changes everything except something within us which is always surprised by change_.
—THOMAS HARDY
SPRING BRINGS SUICIDE TALK. WE NO LONGER LEAVE Nancy alone with Jack, since she started seeking counseling from him about her urge to go and jump in the canal. She means the canal that passed by their Edinburgh apartment, I presume; a canal she's known twelve years, hanging grimly on in episodic memory. Heaven knows why she picks on Jack, but she does, singling him out and pinning him to the wall with conversational monologuing, from which there is no easy escape; Jack signaling for help wild-eyed, like a desperate guest at a cocktail party cornered by a bore. I look up dementia and suicide and find there's no consensus on their interplay. The orthodoxy is that suicidal feelings are burnt out early because of the self-awareness that's required, though there are heretics who say otherwise. Giving up eating and drinking, voluntarily and abruptly, in a way that seems to have been considered and decided upon, is known in nursing homes as Alzheimer's suicide. It's a not-uncommon way to go. Whether it is suicide is debatable. If people live long enough, the disease will reach the brain area that sponsors sensations of hunger and thirst.
Alzheimer's is set to become a hot potato when, eventually, assisted suicide is legalized. Dignitas, the Swiss suicide organization, which offers a legal framework to the suicidal—providing a house and a lethal dose of barbiturate, and leaving it to the client's discretion whether they take the drug or not—was questioned a few years ago about the death of an Alzheimer's patient, and whether he could be guaranteed to have been of sound enough mind to make the contract between them legal. Dignitas responded that even someone with advanced dementia may have moments of sufficient lucidity to want to die. This idea has been shouted down as ludicrous but I'm not so sure. Nancy keeps talking about the canal.
She talks to her address book about it. She's carrying her address book everywhere with her now, and won't put it down. Not on the toilet. Not in the bath. Not in bed. She tells me she's heard "the people" plotting to take it. She sits in her armchair and talks to it. Sometimes to herself, about it, flicking the pages and narrating; sometimes to the book directly, asking it questions. She seems to recognize that it has something to do with her past. An address book is, after all, as personal as a photograph album. She's had this one for thirty years or more. Addresses in it are written in various inks, the handwriting varying according to mood and circumstances, health or tiredness, and whether the writing was done on lap or table or against the wall, the telephone receiver held tight under her chin. Numbers and names have been scratched out and replaced. Old friends, some of them long dead. People she worked with once. Relatives, former neighbors, and fellow school parents. The addresses Chris and I have lived at in our many nomadic wanderings, and those that track the life path of her daughter. It's a fairly small address book, small enough to fit in an ordinary business envelope, cream bound, tatty, its cover dotted with seventies-looking flowers and butterflies. The hinge of the binding is beginning to wear through, and pages are coming loose from the stitching.
Every time I go into the in-laws' sitting room, there's Nancy with the address book, head bent in concentration, flicking through and talking to herself in a low and urgent tone.
"And that's the one I want. That was always the one. Yes." She taps her finger on the page decisively. Her nails have grown long again and mysteriously they appear well shaped, with smooth semicircular ends. She must be using an emery board at night; there's one in her underwear drawer. Nail care, so much a part of her life once, must be embedded deep in long-term memory, one of those automatic activities that don't need thinking about. (Thank god for the cerebellum. Imagine having to learn to walk afresh every morning.) "And there it is," she says, confidentially to an entry in the address book, a pointed finger hovering. "There's the one. That's the one." Flick flick. "Ah, now look. That's the one I was meaning. I meant that one. That one there. That's it."
"You're not having it!" she shrieks at me when I try to put it on the coffee table.
"I'm not taking it, just putting it down. Right there. So you can eat your supper."
"It's not any of your business!"
"Here's your knife and fork, look. You're going to need two hands to eat this piece of chicken."
"I will do it as I do it and I thank you to keep your nose out."
"Okay, then."
She holds on, her grip white-fingered, to the book, and uses the fork a little. Then she puts the fork down and eats with her hand. The book is not relinquished.
She starts taking her teeth out in bed. This is a new development. It's been impossible to get her to take her false teeth out for years. But now she wakes with a jaw-caved-in look and speaks in a different way and it's obvious that they are out. We play hunt the falsies. They get put away in unpredictable places in the night: in drawers, under things, on windowsills. One morning we find them in the toilet, sitting unflushable under the water. On another, the Tuesday bus is kept waiting, engine running, at the conservatory door while we do a last frantic search.
"She'll just have to go without them," Chris says.
I pick up her wallet, her latest pet, which unaccountably she's left sitting by the bed, and find it is strangely lumpy. Sure enough the dentures are within, stuffed in tight and straining at the leather.
Next, the invasion begins. The in-laws are getting lots of visitors. Since I said to one of the health team that they should come straight in and not bother to knock, everybody now does this. Which is fine. We might not hear the doorbell, we might be on the phone. We don't, in any case, want to have to go to the door repeatedly and make stilted conversation with health visitors. Far better if they just come straight in. They know where to find Morris. Morris no longer moves. He keeps the urine bottle by his armchair and uses it in situ.
The house no longer feels the same. It no longer feels entirely like a house. The emphasis, once firmly on the children and on child raising, has shifted. We seem to have reached a tipping point, and tipped. Now it's all about Morris and Nancy. They sit at the center of it, two fat spiders in a web (not fat as it happens but you get the gist). I can't help thinking of _Charlie and the Chocolate Factory_ and the four bedridden grandparents, living in the giant bed in the sitting room being fed cabbage soup. That is how home is becoming. The rest of us are satellite creatures with satellite lives.
An ancillary to this is the feeling that we are constantly on show. The house has to be kept tidy, approvably clean and swept of personality. Overdue bills can't be left sitting on tables, nor open books, letters, sales catalogs that might provoke comment, any signs of extravagance, unusual foodstuffs, alcohol of any kind, drawings half drawn and half-done crosswords nor anything to do with business. Sometimes when visitors come looking for me, wanting consultation about some Morris-and-Nancy matter, they find me on the sofa reading. Apparently doing nothing.
"How do you get the time to read so much?" one of the care ladies asks me, blushing pink, "because I _never_ get the time," her eyes flickering toward the ironing pile. Our home aides aren't the kind of people who sit down much, and nor were Morris and Nancy in their prime. Working hours were long, for them, at home as well as the office. None of them associate sitting down with earning a living. Sitting down is something that's available when all possible chores have been done, late at night or not at all—and there's considerable kudos attached to never getting the chance. Nancy, when first we moved in, was heard to complain about the state of the housekeeping. "They're terrible here," she'd say, looking in horror at the dishes by the sink. "Look at this! It's terrible." Sitting down during the day might be construed as immorality. The owning of too many books, to some ways of thinking, is an admission that one misuses one's time.
Out here, far from everything, a village is a village and also a world. People talk. News and rumor are the lifeblood of isolation. Talking is a major activity and we're all too conscious of people's curiosity, their assumptions and misconceptions. Private conversations can't be had until late at night, phone calls can't be enjoyed: not in daylight hours when there are likely to be outsiders present. I can't be seen in the kitchen in my bathrobe (not after 8:00 A.M.! "Have you had a nice lie-in today? Lucky for some"), nor am I happy to have the children seen in theirs. Judgment, both real and imaginary, hangs heavy over us. Downstairs has become a public zone. Dogs must be locked away upstairs in case they escape. Chris and I start having daytime conversations by e-mail. Doing otherwise risks being overheard, at least before 9:00 P.M., when the back door closes the final time and the home care lady is gone. At nine o'clock we all relax. But that's also Jack's bedtime. The window of ordinary family interaction has shrunk alarmingly.
The phone calls continue and are on occasion deeply aggravating. The physical therapist rings to ask why we haven't had the wet room installed that she recommended. From health and social workers, recommendations are usually orders. Someone else rings to report that Morris, at the day hospital, has complained that Nancy's being unsettled at night is keeping him awake and that he worries we won't hear him if he calls for help. She is insistent that we install a baby monitor, so we can listen to Nancy and Morris, reassure ourselves that they are sleeping, and be alert to anything we ought to go and sort out. This phone call has a peculiarly depressing effect. The house has become an institution and we are its night staff. And we ought to be aware that a part of our duties is lying awake listening to Nancy monologuing away in the early hours, and Morris shrieking at her to shut the fuck up and go to sleep.
Then one of the nurses at the day hospital telephones to say that Morris is complaining of being lonely. She mentions the name of the day center, the Thursday day center he point-blank refuses to attend.
"But he doesn't like the day center!" I retort, perhaps too vociferously. "He used to go to it. He canceled it. He hated it."
"Well, I'm just ringing to let you know that I have booked him in to recommence. He'll start on Thursday. Okay?"
Morris makes a face when I pass the message on. "I didn't really have any choice," he grimaces, though when Thursday comes he goes off on the bus cheerfully enough. What does he really think about the day center? What does he really think about anything?
As if in punishment for our not agreeing to the baby monitor, we have a series of late-evening and early-morning crises. Nancy begins getting up and getting dressed at two or three in the morning, and trying to get to Somewhere Else. I no longer believe the "doorknob prompts" theory. These are breakouts. I find her downstairs rattling at the door that leads from the main hall into the porch. It's a half-glass-, half-wood-paneled door, a Victorian door, and heavy. When it's rattled it swings in its housing and echoes through the house like thunder.
"Nancy. It's you. Couldn't figure out what the noise was. You gave me a fright."
"I need to go now. I'm late."
"Come back to bed. It's the middle of the night."
"That's all right for you to say but I'm not supposed to be here!"
We meet Nancy almost every evening, on one of her moonlight sojourns. The drawing room door opens in spooky slow motion and she shuffles in, waddling from side to side, shoes on the wrong feet, holding some combination of possessions: her handbag, clothes, a pair of shoes, her address book, her teeth in a handkerchief. Quite often she's singing, to the usual tune.
"When I am young and busy, and the world will have to be, and the thing that comes down is the thing I brought here, and that's the same to me."
She will be in one of two moods, black and white. Either very glad to see us and intent on joining in our late-night whisky, or misanthropic and full of gloom. And she can still rhyme.
I feel bad about putting her to bed so early, but this is how it is. Morris has no choice but to be put to bed at eight-thirty; that's the only slot he could get in the home care schedule, and quite often he's glad of it, his legs bothering him, bed wanted. Nancy must go with Morris. There has to be some granny-free time and this is it. Nine P.M. to 11:00 P.M. is sacred. I've gone the other way on occasion: taken pity on her restlessness, sat with her in front of the television till after midnight, till she began at last to flag, remade the fire, made her toast and hot milk and been tolerant about the ranting. But I can't do it anymore. Besides which, if Nancy is absent, Morris can't sleep. He stays awake waiting for her return. He grows agitated, wondering what she's up to.
Late one weekend evening, while the rest of us are upstairs in the family room, there is a sudden hullabaloo from downstairs. A frantic impassioned yelling. It takes a few moments to register that it's somebody calling out Chris's name and sounding desperate about it. Chris and I jump up and go down, insisting the children stay put.
"Oh god, it's Granddad!" Millie cries out. Jack bursts into tears and Caitlin follows. Millie joins in and the three of them stand on the top landing, snuffling and clutching each other.
When we get to Morris's bedroom the door's open and he's by the threshold, on the chair at the end of Nancy's bed. He's managed to stagger to the door and open it in order to shout for help, but has not been able to get further.
He looks ghostly, yellow, terrified.
"Oh, son," he says, emotionally. "It's Nancy. My Nancy. I think she's dead." She's lying on her back, utterly still with her arms by her sides. Chris listens to her chest and puts his ear to her mouth. Then he listens to her chest again.
"Her pulse is regular," he says.
"Oh thank god, thank god. I've been trying to wake her up for a good half hour. Normally she's awake at this time and chatting to me. I couldn't get her to answer."
Chris helps his father back into bed. Meanwhile I sit with Nancy and talk to her. "Nancy. Na-an-cy. Nancy! Nancy! Wakey wakey. Hello-o. Are you there?" I squeeze her hands. She doesn't respond. Ten minutes pass like this: me tickling and squeezing and shouting and demanding that she wake up, get up, right now; Nancy unresponsive and apparently unconscious. I keep going. Finally, when I squeeze her big toe, she kicks out at me with a sleepy growl.
"Nancy," I say, very firmly, my head dipped close to hers. "You need to get up now. Come on. I need to speak to you. Open your eyes."
"No," a small voice says.
"Come on." I pull at her arms and she rises, eyes still closed, and puts her legs out of the side of the bed. I pull gently on her arms and she glides to a standing position. I take her, slowly, eyes still closed, to the bathroom, where she sits and has a pee. Then I put her, eyes still closed, back to bed. She's grumbling under her breath.
"I just need to speak to you for a moment," I say, knowing Morris is still in a panic. "I need to ask you a question. Would you like a drink of water?"
"Bugger off," the voice says, from between near-closed lips.
"Thank god," Morris says. "She's fine."
AT THE END of the month, as the school Easter holidays begin, Chris goes off with his good mate Michael for a week's sailing course. I've known about this booking for a long while and am in favor, despite dreading it. An annual week off from family life to do something independently is official marital policy; it's just that I never take mine. It isn't a happy week for the nonsailors. The weather is cold and rainy. The children and I have to be on hand, on duty, in case of upset or crisis, so we hang out in the drawing room most of the day, where we can keep an ear and eye on things. Nancy comes in at regular intervals to ask for help with _the chap through there_.
"I can't help you with that, Granny, sorry," the children say, as coached.
"No. No. You don't understand," Nancy tells them. "You have to come and talk to him. He seems to think we know each other and he's being annoying."
Eventually, weary of interceding, the four of us take to hiding upstairs, reading by the fire, Jack playing his self-absorbed role-playing games with guns, coded messages, cloaks, and light sabers. We work through a stockpile of films, magazines, and chocolate. Chris and Michael have a freezing cold, challenging week on the water, with bad weather and plenty of chucking-up and have the time of their lives. It isn't such a vintage week at home: I'm up early, mucking out horses in a gale force 7 and horizontal hail showers.
Each of the seven days Chris is away there are tears. Sometimes mine. Almost daily Nancy's. "This man is NOT MY HUSBAND," she insists. "He's NOT, he's NOT, I've never seen him before in my LIFE." She means it. She's frantic, and she can't understand why I'm not equally exercised by the stranger in the sitting room. It's particularly bad timing, as poor Morris is beginning to have embarrassing "toileting issues," to use the social care argot. He is soiling himself in bed. As is the way of things, this begins in spectacular style while Chris is away. The morning home care lady arrives to find her charge awake and mortified. She deals with the worst of it (getting him up and clean is her remit) and then hands me the marigolds (getting the bed clean isn't). I text Chris, who is night sailing in the Cromarty Firth. _Yr parents hell. You owe me big time fr this chum_. It's decided that we'll look for a privately employed home help to work twenty hours a week.
The council announces a residential care crisis (another in a series), and all respite at the town nursing home is revoked for the year, all six weeks that were so painstakingly negotiated. It's a financial crisis, one being experienced simultaneously all over the United Kingdom. The council doesn't have enough in the budget to run their homes, to hire their staff (nor can they keep them: We hear constantly how low morale is among the workers at the town home), and so, in this situation, cancellation of the bookings of respite clients—who will only be there for a few days, or a fortnight at a time—is the immediate money-saving mechanism. They know, even so, at the council that there will have to be some negotiating and exceptions will have to be made. Several emotional phone calls later, Nancy's April week is rearranged. She's to go to the new home, the countryside home, a swish bungalow-based home with an Alzheimer's unit, half of which is locked off unused because there isn't the funding to run it. So Nancy goes there, and Morris goes to the seaside home: separate destinations because it's judged that he needs a break also. Respite requires a formal social work assessment on each occasion, which entails a pot of tea with our care manager in the conservatory, answering detailed questions about the in-laws' abilities or lack thereof, and the exactitude of need. Questions, I can't help thinking, that could have been asked on the phone. It can only be that they want to run an expert eye over the household, sniff the air, and gauge the mood.
It's a busy week, full of incident. My horse has become ungovernable with the coming of spring, and first Chris and then I am bucked off, in his case flat on his back onto tarmac, and in mine upside down against a wall, denting a drainpipe and slamming into a water tank. I hobble round the house doing the holiday packing, limp in and out of care meetings, shuffle bruisedly onto the plane. We go to Turkey and buy a teeny house, not much bigger than a beach hut, but with drainage and a veranda, on a holiday site on the Aegean coast, spending the bulk of the French _ruine_ money.
The plan to get extra paid help doesn't go quite to plan. The person I had in mind has taken a job at the town nursing home. I consider putting an advertisement postcard in the post office but am dissuaded by a friend. Mightn't that be awkward—interviewing people you know from the village and then not hiring them? She has a point. Still, there doesn't seem to be any alternative. I ask the lead home aide what she thinks, and she offers herself for two hours each weekday, to keep an eye on the in-laws and help with the housework. I accept with gratitude.
The B and B takes off with a whoosh, with a rush of inquiries by e-mail and others by phone. My irritation with the telephone means that e-mail queries are far more likely to be successful. I develop a highly unscientific method of weeding people out. I attempt to instigate an e-mail conversation. If people are irritable, blunt, or evasive, if they can't spell or want a discount, they're turned away. My approach is fearlessly partisan. But the days when there are guests departing and arriving are always, whether by coincidence or not, the days when Nancy's most troublesome.
# Chapter 18
_Nothing is at last sacred but the integrity of your own mind_.
—RALPH WALDO EMERSON
THE RELATIONSHIP BETWEEN MIND AND BRAIN APPEARS, at first sight, to be a relatively easy one to grasp, even for the amateur neurologist. Brain is the machine, mind its creation. Brain is the cinema equipment, mind the feature film. Brain is the cluster of tiny lasers on the podium, and mind the holographic image of the Fabergé egg. Brain is the instrument, and mind the consciousness that arises out of it, orchestrated by millions of neurons working in concert. It's your brain, not your mind, that the surgeon sticks the scalpel into. It's your mind, not your brain, that feels nervous at the prospect. Simplistic, but so far so good. It's when you get into the relationship between brain, mind, self, and soul that things become more speculative and more prone to prejudice, not least of the religious kind.
Aristotle set the agenda in the fourth century B.C. as a materialist, arguing that the soul (mind) can't exist without the body, which sounds impressively modern until you take into account his insistence that the heart was the location of the thinking self, and the brain some kind of body-cooling device. In general the more modern the thinker, the more integrated brain and self are assumed to be. So it's mildly shocking to read something as recent as Carl Gustav Jung's _The Interpretation of Nature and the Psyche_ (1955) and find him asserting that "we must completely give up the idea of the psyche being somehow connected to the brain." Mary Baker Eddy, founder of the Christian Science movement, agreed. "Give up the belief that mind is, even temporarily, compressed within the skull, and you will quickly become more manly or womanly," she wrote. "You will understand yourself and your Maker better than before."
In contradiction of this, the most recent crop of popular science writing is at pains to point out that in every way that really matters, we are our minds, and that our minds and our brains are wholly interdependent. In his idea that psyche is something separate, Jung isn't far from the mind-set of René Descartes (1596–1650) and his firm division of body and self, the self (soul) merely residing in the (mortal, transient) body until such time as immortality can be earned and achieved. It's assumed that this philosophy is biblical, but in fact you'll struggle to find supportive evidence there: the idea of dualism is essentially Greek, and man in the Bible is a holistic, whole creature, body and soul together, anticipating bodily resurrection. The Greek idea is that immortality is a fundamental human attribute; in Christianity it's a gift from God. Plato was Descartes's model, in his belief that an immortal self enters the body somehow, and departs it intact after death. (Descartes struggled with his faith. Having coined _cogito, ergo sum_ —I think, therefore I am—he worried that perhaps his thinking self was all that he was, and no more.)
It looks like a two-horse race. Either the brain is all there is to us, personalized through genetic inheritance and through the individuality of experience into a mind, creating the illusion of soul through its clever holographic tricks, and we die with our neurons, _or_ the brain is simply the machinery the self/soul employs for its brief stay on earth and in time, and the self/soul, the _ghost in the machine_ , survives us. Any mortal creature would wish Descartes fervently to be right. Added to which, the idea that there is some higher order of personal reality beyond the body, the state of the brain, the workings of the mind—this has a special resonance for dementia sufferers. It introduces the hope that their essential self survives the apparent disintegration dementia brings, locked away safe from the banality of disease.
Descartes thought the soul entered the body through the pineal gland, choosing this entry point because there wasn't then any other obvious use for it, and it was thought to be specifically a human piece of kit. He was, for obvious reasons, an established church favorite, despite his doubts. The establishment was less keen on Franz Joseph Gall (1758–1828), inventor of phrenology (head bump reading), who having surveyed the head shapes of the criminal class, placed subtleties of personality in specific brain regions, which seem to us now entirely random: self-esteem in the parietal lobe, for instance, secretiveness in the temporal lobes, and friendship in the occipital. Less eccentrically, this led him, and the population at large, to the conclusion that self is biology. This was enough to get him expelled from Austria by the emperor Francis I.
If brain is mind, and mind's thought equivalent to self, self equivalent to soul, theological problems are going to arise. There are neurologists writing now who are confident that consciousness itself will before long be "located" and explained as utterly physiological, a line of thought that Francis Crick, the DNA Nobel winner, popularized in his book _The Astonishing Hypothesis_ (1994). In his last paper (2004), Crick suggested the claustrum, a "sheet" located beneath the inner surface of the neocortex, which receives information from all areas of the cortex and returns information back into it, might be the seat of consciousness. The truth is that science doesn't yet have the answer to the mystery: how it is that a subjective self comes about at all (known as the Easy Problem) and achieves self-awareness (the Hard Problem).
The phrase _ghost in the machine_ , incidentally, was coined by a British philosopher, Gilbert Ryle, in 1949, in mockery of Descartes's dualism. Arthur Koestler's book of the same name (1967) was interested in a different kind of ghost, one associated with the amygdala, deep in the limbic system, creator of impulses concerned with gut instinct, fear, aggression. He suggested that our social evolution has far outstripped our brain evolution, and that we are held back by the primitive emotions and functions of obsolete but still-powerful remnants of our prehistoric selves, which can be held accountable for our being warlike, suspicious, and bigoted.
When the frontal lobe is damaged by Alzheimer's and the self is fractured by the forest fire of neuron death, maybe other parts of the brain rise up to compensate. When rationality is damaged or lost, it is perhaps more primitive parts of the brain and the great hidden sea of the unconscious that prompt facets to rise unexpectedly into view, redirecting the personality of the dementia sufferer into something the caregiver doesn't recognize, with new preoccupations, hostilities, and weirdness. As Freud wrote, though we are more sure of ourselves than of anything, confident that a self is something autonomous and self-contained, the truth is that "the ego extends inwards with no clear boundary into an unconscious psychical entity." As social philosophers of the seventeenth century might have put it, Nancy has lost her Natural Government, and is in danger of relapsing into a state of nature.
It isn't necessarily a two-horse race. The German philosopher Arthur Schopenhauer seems to have been an adherent of a third way, the idea that though there is no immortality of the individual earthly self, we are more than our brains, and return after death to the same state of existence we enjoyed before birth, giving up (with relief, he claimed) the painful and limited animal consciousness of being human and existing in time. "Consciousness is destroyed in death," he wrote, "but that which created it is by no means destroyed." He wasn't the first to see things this way. Anaxagoras, in the fifth century B.C., is thought to have introduced the idea of mind _(nous)_ as something infinite and immortal, emanating from The One, the collective human entity that organizes matter and survives it.
Others take a more Platonic route. As the Scottish psychiatrist R. D. Laing wrote, "If my physical frame dissolves, I can't live in this world any more, because this world is a transform: the brain is the transformer and is itself a transform." (A _transform_ is reality as delivered up by our perceptions.) It's perhaps Laing who puts the problem of brain and self most succinctly when he goes on to say, "[T]his collection of cells has the impression that it is I. This is a proposition I do not necessarily agree with."
# Chapter 19
_The trouble with troubleshooting is that trouble shoots back_.
—ANONYMOUS
HERE'S SOMETHING NOT COVERED BY THE BOOK, other than in a veiled and decorous way. _Sufferers will need help using the bathroom_. Unfortunately, Nancy won't accept help using the bathroom. She won't _go_ if anyone else is present, and is outraged by the suggestion that she use the paper hanging by the toilet. We change her underwear at least once a day. Bedding is also affected, and added to Morris's problems, this means that quite suddenly we have masses of washing to do. But that's not the worst of it. That's just laundry. The much worse thing that's new is that Nancy has started squatting on the floor. Sometimes she tries to hide it. Excreta are found behind the toilet, lurking behind curtains on windowsills, and on one memorable occasion, hidden behind paperbacks in the library—the kind of discovery that's unexpected late at night when you're looking for something to read. If she can be persuaded to use paper it is rare that it's flushed away. Unpleasant sections of toilet roll emerge from cardigan sleeves. She's taken to cleaning the toilet bowl with her hands, so we are careful about not letting her touch food. The long fingernails have to go.
Stuck at home a lot of the time, companion to a man she believes a stranger, Nancy becomes despondent. The fact of her no longer recognizing Morris has a huge emotional cost. She's alone in the world now, and unhappy, and it's as if her unhappiness is beginning to leach out. It fills the air. It coats the walls and furnishings. This isn't just metaphorical. Despite twice weekly baths, my mother-in-law has acquired a smell, a sweet and sweaty smell with a dark undercurrent: feces, armpit, old organs, fear.
Morris is ill and the doctor comes to visit. They have a long conversation together. Later she rings and tells me that Morris is convinced that Chris and I won't stay on the peninsula long, that we'll up and move to the south of France and leave the two of them here all alone.
"The south of France?" I echo. "What on earth gave him that idea?"
It's embarrassing. Is that what he's been telling his home care confidante, or perhaps, what she has been telling him, having misinterpreted something overheard—and is that why people in the village have been asking how long we'll be here?
Morris has grown dangerously unsteady on his feet. His aides have doubled in number so that twosomes can cooperate on the heavy lifting. The consequences of this prove far-reaching. He can't manage his bathroom visits alone any longer. He's too unstable to manage the pulling up and down of trousers. He has to keep his hands on the Zimmer frame or he will fall. This puts the poor man in a very tricky situation. What makes it especially tricky is that Nancy stops cooperating. The manner of his demanding that she help, and her refusing, becomes an explosive part of every day. Chris helps Morris when he can, but often Chris is on the phone at the crucial moment. Morris would, in any case, far prefer that Nancy help him. He certainly doesn't want me in there, a scruple for which I'm grateful.
"Nancy! Nancy!" we hear, urgently from the sitting room. "No! Not that! I need the Zimmer. The Zimmer frame, there. The silver thing with the... the frame, Nancy, the frame! The bars, the rack, the frame thing, there. _There!"_ His conversations with her have become thesaurus-like. "The Zimmer. Right there, right in front of you. The thing right in front of you, the big silver thing, the Zimmer. No! Not the biscuit tin!" He takes a sharp breath inward and bellows, "THE ZIMMER!" Nothing I say to Morris can make him understand that _Zimmer_ is part of the English that's become a foreign language to Nancy. She relies on cues now, cues and context, a "now you're hot, now you're cold again" kind of verbal directing, an impersonal in-car Sat Nav approach. He may as well use the word _zangle_.
Unfortunately, by the time I get Morris into the bathroom, Nancy has had enough of being yelled at, and is marching off in the other direction. I leave him standing, balancing precariously while I run after her, taking her hands in mine and imploring.
Her reaction is predictable.
"Why on earth should I help anybody? It's got nothing to do with me."
This is the crux of it. Nothing has anything to do with Nancy anymore. She floats free of connections to the world. Alzheimer's has invaded her empathy and placed its flag.
"That's your husband, though, Nancy," I tell her.
"No, it is not. It most certainly isn't."
"Yes, it is. Yes. Yes. You have to come and give him a hand."
"I never heard anything so ridiculous."
"He needs help with the bathroom. And you are his wife. You need to go and help."
I'm wasting my breath giving her the backstory. All she is listening to, responding to, is my authority over her, my determination that she should act. That alone will save the day. Her respect for my presumption of power is all that drives her acceptance of my orders.
Afterward she retreats to her bedroom, and that's where I find her, sitting on the side of her single bed, hands in her lap, staring downward and utterly dejected. The conversation that we have now, the daily conversation about her living here with us, her family, and her forty-seven years of marriage, is becoming grindingly repetitive. As the saying (often attributed to Einstein) goes, the definition of madness is doing the same thing over and over again and expecting different results, and from that point of view my explaining things is a mad enterprise.
Except on one occasion. On that day, instead of staring at the floor, she turns to look at me as I come in.
"I'm so glad it's you," she says with feeling.
"What's up, Nancy? What's up, dear?" I ask, sitting by her. "It was only poor Morris having to go to the toilet. He has to go to the toilet every day in the afternoon and every day you help him. It isn't worth getting this upset about."
She fumbles with a paper handkerchief, twisting and untwisting it.
"I don't understand it," she says, looking into my face, her eyes wet with tears.
"What don't you understand?"
"I don't understand at all what is going on here."
I put my arm round her shoulders. She rests her head on my shoulder and cries, abjectly and with abandon.
"None of it makes sense," she says when she's able to talk. "What's happening here? How did I get here? What's happening? Please. Please."
What will validation do for us now? There's no cozy pretend world to slip into, averting our eyes from the present, taking refuge in dementia-fantasy, stepping through into dementia-time. What else is there to tell her but the truth? If it were me, that's what I'd want.
"Here's the thing," I tell her. "You have a condition. Your memory doesn't work properly. You don't remember things. That might seem like something quite trivial but, actually, it undermines your whole life. It means that you don't really know who you are."
She snuffles through the handkerchief and tells me her maiden name.
"That used to be your name, but when you were about thirty, a little over thirty, you got engaged to a nice chap called Morris. You got married to him and you adopted two babies."
"But I don't remember that. I don't remember any of it," she says.
"That's because you have this illness and..."
"I am not ill. I am absolutely fine."
"You have a condition that lots of old people get. You're in great health otherwise, fit as a fiddle, but your memory is almost gone. Lots of old people get it."
"Am I old?"
"You're seventy-nine. Nearly eighty."
"Am I? Am I? I'm not. I'm not eighty. Am I?"
"Yes. Nearly."
"Oh my god. That's right, is it. I'm eighty." She laughs nervously. I consider taking her to the mirror but then think better of it.
"Yes. Do you remember living in Edinburgh?"
"That's where I live. Edinburgh."
"Do you remember the apartment by the canal? Feeding the moorhens and the swans with bags of bread?"
"No. I don't remember that."
"Do you remember going to work, your little car, your rack of navy blue suits and shirts? Doing your nails every night?"
"No. Not really."
"Do you remember the children? Getting the children, the babies?"
"No. No."
She looks at me as if hopeful that the two of us might be in this mess together.
"That was forty years ago, more. You're retired now and you live with us. With your son and his family. Morris is here, too."
"Is he?"
"Yes. He's the man sitting through there."
"Can I see him? I've been wondering where he was." She starts to cry again.
"Come on. Quickly. Come and see him and give him a kiss."
We go through the kitchen and up the step, through the second door. When she sees him, Nancy goes into reverse, trampling my feet.
"No! No, no. You're wrong. That isn't Morris."
"Of course I'm Morris," Morris says.
Nancy turns to me. "That. Is not. Morris."
"Yes, it is. That's Morris. He's got old, just like you. And he's disabled. His legs don't work, remember?"
"You're all mad. You're all mistaken. That isn't Morris. You think I don't know who Morris is? Well, I do and that isn't him."
All this is horribly upsetting for Morris, and his confidante has been counseling him at length in the mornings. The rejection is hard for him to bear. Nancy seems so hard-hearted, so impermeable suddenly. Morris tries to talk her round, to jostle her memory, insisting on the truth.
"But darlin'! It's me! You must know me! You must!"
"I most certainly do not. You're all liars."
"Nancy, please. Please don't say you don't know me. Please."
She goes and sits on the edge of her bed for hours and hours, refusing to eat or drink, sitting looking hopeless in her cold bedroom; me going in from time to time and trying to coax her to come back to the fire. Sometimes I find her in bed fully clothed, the duvet pulled up over her mouth, her shoes sticking out the other end, flat on her back and deeply asleep. Sleep is good. Sleep is her friend. The event is wiped clean away. As long as I don't use the words _Morris_ or _husband_ when I reintroduce her to her sitting room, things are fine for a while. If only Morris could resist asserting himself and conjuring up their shared history, things would stay calmer for longer, but he can't.
The rages spill out of the sitting room. There begin to be rumblings from day centers. They cope well at the Tuesday one, where they are trained to deal with Alzheimer's old folk and have others with dementia attending. Dementia's meat and drink to them. But the Thursday club, held in our own village, is a different matter altogether. It's a social club for over-sixties. They are kind, good-hearted people, the people who run the Thursday club, and they try diversionary tactics first, before they call. If Nancy's stroppy, somebody takes her out for a walk. They might go to the shop and get her an ice cream and go look at the boats in the harbor. If diversions don't work she's brought home early, delivered to the door, she and Morris alone in the bus, Morris embarrassed. "She's not had too good a day today," the helpers say. "Not too happy today." Sometimes she comes home wearing borrowed underwear and I am full of admiration for the volunteers who deal so stoically with that.
In May, we experience the first of the major Thursday club upsets. It's heralded by a phone call after lunch, from one of the helpers, who happens to be the mother of one of our doctors.
"I'm afraid Nancy's in a terrible state," she says. "Do I have your permission to take her to the surgery?"
"What on earth's going on?"
"She's... well, she's just in meltdown, really. We can't do anything with her and I, um, think she might need medical intervention."
"You mean she's having a tantrum, she's upset? Is she being rude to people?"
"You might say."
Everyone's far too nice to be explicit. I have visions of Nancy going at the other members with a hail of china-saucer-fire, a volley of cutlery artillery, kicking old men in the groin and felling them with karate chops. Nancy is whisked off and given emergency sedation. They keep her there until it kicks in and then she is driven home, arriving monosyllabic and irritable. The doctor rings me later. "That was quite something," she says. "I've never seen her like that before." A frank conversation follows. Nancy is prescribed a mood-improving drug to add to the galantamine, the blood pressure drug, and the aspirin in the dosette box. Though we don't give it to her beyond forty-eight hours, as its principal effect seems to be to tranquilize her in the daytime and make her more restless and agitated than usual at night.
Can I cope with this? Should I be trying? These are the questions that whirr in the brain at five in the morning when the long, gray summer days of the far north dawn early. We are already way out of our depth and we've been here for less than a year. How much longer will this go on? How much longer can I stand it?
Internally, I'm fervently apologetic to all those unknown, anonymous people I ever maligned for _dumping_. Dumping their parents in nursing homes, when they should have been clasping them to their familial bosoms, for better or worse. Movie grannies, with their crumpled-and-smoothed tissue-paper faces and gray plaits worn Heidi-like across their heads, and tea dresses and crochet cardigans—the kind of grannies who are pliant and hygienic, who dispense old-world wisdom to the children of the house, and are amusingly direct—they have a lot to answer for. Movie grannies don't refuse point-blank to clean their teeth. They don't yell obscenities at their grandchildren or accuse their daughter-in-law of stealing all their money or tell outsiders they're being kept a prisoner. They never pull down their trousers and touch their toes and ask you if their bottoms are clean, or get sent home early from the Thursday club for disruption.
I am out in the garden tackling the weedy borders and planting new shrubs quite a bit of the day, having abandoned the novel, again. It isn't something I'm happy about. It's not been a matter of choice, but writing isn't possible with Nancy on the rampage and the constant interruptions. Nor is writing possible when so much unhappiness is at large. Nancy's. Morris's. My own. So instead of struggling on, there will be reading and gardening and strategies for psychic survival. Weeding is good for impotent rage. Old neglected borders full of grass and dandelion are gone at with energy. I take Nancy into the garden with me for part of the day and try to filter out her wittering, and try to be calm about her standing in the middle of the flower beds, trampling new plants. Her white skin is sun sensitive and burns in a trice. She is dressed in her customary elastic-waist slacks, a long-sleeve cotton shirt, and a wide-brimmed hat. She smiles, when she's outdoors. There's Nancy in the photographs, flushed pink and grinning. It's indoors that she hates, the terrible boredom of indoors. Outside, she gets all my attention. And maybe it's more than that. Perhaps it's Nancy who's found a relationship with the Sublime.
Clinging to the idea that we ought to make her days as fulfilling as possible, and having given up on work for now, I try to include her, like mothers of preschoolers do, in the daily domestic tasks. It's exhausting work: "Now here's your peg, here's the sock—see, you open the peg like this and put it on the line and when you let go it grips it. See? No, don't put them all on top of one another, they won't dry. Ah no, see, you've got the peg upside down, that won't work." On Mondays, when Morris is out at the day hospital, I make a special effort. Then, having spent five unbroken hours together, I put her in her chair and go to make her a coffee and when I get back she is purple-faced with rage.
"I've been left here all day on my own! People have been going by and not speaking to me! If I'm not wanted here you had better just say so!"
I've been reading a book that claims to explain consciousness and its neurological mechanisms as something entirely animal. The writer pulls off the disarming logical trick of spending the entire first chapter disparaging Descartes, and the rest of the book coming face-to-face with a series of pro-Descartes (Cartesian) scenarios—suggesting in their various ways that there _is_ a ghost in the machine, operating the machinery—and dealing with them, one by one, by pointing out that since Descartes must be wrong, because his ideas are preposterous (a favorite authorial word), there must be some other explanation. It is, weirdly, a book that seems unconsciously to be prey to subconscious tides pulling its conclusions in opposite directions to those intended.
I also come across some rather startling research to do with the electrical impulses that carry information between neurons. Apparently, studies of the _action potentials_ have found that they fire up _before_ we decide they should be doing whatever it is that we've asked of them: for instance, to turn a page or flip a fried egg or pick up a stone on the beach. Experiments showing this to be true were begun by the research scientist Benjamin Libet in the 1970s, and continued in 1985 in a scientific trial done with people who flexed their wrists at will and signaled the moment of deciding by marking the position of a rotating disk. Extraordinarily, it was discovered that the appropriate neurons fired up a full half second before the moment the subjects "decided." The interval is known as Libet's delay. In terms of the speed of the electrical impulse, a half second is a very long time. What seems to be happening is that something below or aside from consciousness is making decisions before we think we are making decisions. Something else in us, backstage of our deciding, appears to be deciding before we decide. It reminds me of a British TV series called _Yes Minister_ , in which civil servants manipulate a member of the government, convincing him that he's in charge when the truth is that the real decision making is going on elsewhere. In April 2008, an experiment using fMRI scanning not only confirmed that Libet's delay exists, but went further, showing decisions can be predicted up to _ten seconds_ before deciders "decide." (Of course, it's possible to argue that these are ten seconds in which the subject is observed in readiness, preparing to do something as instructed by the experimenter.)
I DON'T KNOW what all this has to do with Alzheimer's. Probably nothing. But it increases the sense of there being some other self beyond the one we're confident of living within, that feels contained and definite—an alternative self that in our more exhausted moments, in my mother-in-law's case, we've taken to calling Nancy's evil twin. The interesting question, for me at least, is whether this new Nancy's simply a part of Nancy that's always been there, long suppressed and now unleashed by loss of inhibition, or is it something properly new? The validation thesis suggests that it's the former: that Nancy's self is still intact; it's just that we're seeing a different part of it now, one kept at bay previously by the frontal lobe but given liberty by dementia—the rise of aspects from her subconscious, perhaps. It's quite a Cartesian idea when you think about it, the idea that Nancy's self is still intact but trapped within failing machinery, and it's just her superficial way of dealing with the world in the old way that's been lost (if thinking can be said to be superficial). But my own suspicion is that it's something new—that the amygdala and more primitive parts of the brain, dedicated to survival, selfish and aggressive, are being allowed to come forward and create a new self; one that, in the circumstances, we can only continue to call Nancy. I'd always hoped, until recently, that I had a soul that would survive me, but I see now that I will have to locate it somewhere hidden from consciousness, unknown by what I think of as my self, if what I know now about the consequences of brain damage isn't to have the effect of extinguishing that hope. Reading about caregivers' experiences of looking after loved ones—husbands and wives, but particularly husbands—who have suffered catastrophic head injuries in accidents or assaults and have become _different people_ isn't reassuring.
The weather's quite outstandingly foul for May. A hailstorm in May seems like the end of the world. Though the gloom is mitigated by being offered more paid help, by one of the other aides. After a brief crossover period the first aide bows out, Morris's confidante, citing tiredness and illness; I've no doubt that Nancy's sniping was the cause of both. It's difficult (that's putting it mildly) to find people who can work with Alzheimer's sufferers and not become short-tempered, bewildered, bored, exhausted, or demoralized. Our second aide is one such, someone who doesn't and isn't. She's cheerful, assertive, robust, and unoffendable. But she also has young children, a farm to run, is the school cook, and caters for weddings. We squeeze extra hours out of her when we can but that's the most she can offer.
There are seasonal signs of hope in the garden. The wood is teeming with bluebells, thousands of them in a purple haze, and a melancholy bluebell scent drifts up the garden. People have been along in their cars to visit them. Nobody thinks to ask us if it's okay to wander in there, but we don't mind too much, at least not until we see, one late afternoon, somebody standing on the drive with his hands on his hips, looking up toward the house, standing guard it turns out, while his accomplice is at work. They see me at the window and retreat. But when I go down there, the earth has been disturbed in several areas under the trees. It was the whole plant, bulb and all, that they were after, and they've made off with armfuls.
# Chapter 20
_It's a poor sort of memory that only works backwards_.
—LEWIS CARROLL
ELSEWHERE I DESCRIBE MEMORY BANKS AS A LIBRARY that we can visit in our heads. That's the traditional way of seeing it, but it isn't remotely accurate. Memory is an activity and not a vault. The brain stores different aspects of any one memory in different parts of the brain. What was seen, what was heard, the smell, touch, taste, the emotional input—all are contributed by their specialist areas. Visual memory's called up from the occipital lobe, auditory memory from the temporal, working together in a synchronized way. It's not a place, but a process, and a process not unlike music made by an orchestra. In short, it works in just the same way that consciousness does.
Why do some people have good memories and others bad? My sister has an extraordinary memory for our shared childhoods, which puts me at a disadvantage, when I'm quoted at age eight in a fight over an ice-cream scoop. Partly, the reason some people retain the "film" of the past in such vivid detail is that they use their memories more. To keep a memory you have to keep having the memory, revisiting the memory, using it, so as to keep that collection of neurons imprinted and those synaptic connections in place. If they're not used, then they wither. To remember things you have to go through the process of remembering them again. You make a new memory each time you remember, revisiting the route from neuron to neuron. Researchers have discovered that there is an actual anatomical change in the laying down of long-term memories. The axons grow new synapses and new proteins are made in the nucleus of the neuron. There's a change at the cellular level, something that doesn't occur in the making of short-term memories. In his book _In Search of Memory_ , Eric Kandel, who was awarded the 2000 Nobel Prize for medicine, elaborates on this idea that in order to convert a short-term memory into a long-term one, we need to care about it enough, whether for happy or unhappy reasons, and that our caring has physiological effects. One hit of neurotransmitter and the synapse is improved. Five hits and the cell is alerted to this (whatever it is) being something important. It sends the information, via a protein, to the nucleus that triggers the genetic switch for the growth of the new synaptic port. There are two ways in, it seems, via quantity or quality: either via repetition, thinking about something over and over, or by means of the intensity of a shock or equivalent emotional event.
The things that stick aren't always the obvious things. Odd, oblique, incidental, tangential things stick. As the writer Elizabeth Bowen said once in an interview, "The charm, one might say the genius, of memory is that it is choosy, chancy, and temperamental: it rejects the edifying cathedral and indelibly photographs the small boy outside, chewing a hunk of melon in the dust." Montaigne, in the sixteenth century, was more succinct but less alluring, as is his way, in writing that "[t]he memory represents to us not what we choose but what it pleases." What you care about isn't necessarily what you think you care about.
When you remember, it's a memory of the memory that you're having. You don't go into the library of your memory and pick up the book and read your past. In a sense, you write the book all over again. And research shows that if you don't take the trouble to rewrite the books, the books disappear. It's rather like those wardrobe nannies who insist that anything not worn for twelve months ought to be put in a bin bag. _You haven't thought about this for years so I'm chucking it out_. It sends the nanny in and chucks, and it's only when you open the wardrobe that you discover your fake fur jacket/caravan holiday memory is missing. Or, to use another analogy, we need to keep digging out paths in the snow. If we don't, snow eliminates them. Get out there and dig those paths. Maintain them and you can keep walking on them. Don't maintain them and they are gone. How does the brain do this? The nanny in question's an enzyme called PP1 that removes the phosphate from the target protein and deactivates it, in effect wiping a particular memory from the slate.
There are four levels of memory. The first, sensory memory, isn't really memory at all. It's stuff that the eyes see, that the brain may know (far more goes in than is retrievable), but the conscious self doesn't notice. Take the scene in front of my eyes just to the side of the laptop, right now, for instance. The books and papers, used coffee cups, the tin of salted almonds, the box of old photographs waiting to be put into albums, the postcards, pens, mobile phone, plus the jewelry and homework the children left there—everything that's spread on the coffee table beside me as I write this—made a brief sensory imprint in my mind, but hadn't been processed any further until I turned my attention tableward. Perhaps a probe could find it in my head, if probes and scanners grew that sophisticated. Perhaps I might be an unwitting witness to a crime that my eyes saw but I didn't register, while looking out of the window in the city at the cherry blossom on a busy street, where among the traffic and pedestrians, somebody was quietly and efficiently killed with a knife. I saw it but I didn't register it. It was among the things my eyes were seeing while I was concentrating on something else. That's the first level of memory.
The second level is the working memory. This is the material we _hold in mind_ , temporarily, like part of a mathematical calculation we put aside while doing the second part, ready to add the two numbers together, or a phone number we need to remember that was given to us when we didn't have a pen. It's recited in the head and retained for as long as we need it. Then we forget it. Nancy is beginning to forget things that have just happened, things that have just been said to her, and how to finish a sentence she's only halfway through speaking. She's losing her working memory and is unable to hold things in mind. The man known only by the initials H.M., a neurologically much-quoted epilepsy victim—run over by a bicycle at age nine, and in his twenties at the time of being a research subject in the 1950s—with his temporal lobe function diminished and hippocampus removed, could still remember new things done or said for a few minutes. His working memory survived although his short-term memory, ordinarily the next phase in the process, no longer functioned.
Scientists classify short-term memory differently according to length. Neurologists tend to talk about it as short-to-medium term. The things we did yesterday, last weekend, even the wedding we danced at the weekend before that, can be described as held in short-term memory. The process of converting a select few of these into long-term memory, forming strong memories that survive, can take weeks, and it's thought most of the work's done while we sleep.
Memory making is a single-track road. To get from the sensory memory stage through working memory into short-term and thus into long-term memory is like going along one of those winding narrow routes that stretch out into the fingers of the coast of Argyll in remote western Scotland. There's only one road from the village of Sensory to Long Term. To get to Long Term you need to go through the other three villages first. In other words, if there is a break in the road, a flash flood, say, and then a road slip, a section of the road sagging and tipping down the hill, and the road becomes impassable, nothing can get to Long Term. That's what happens in Alzheimer's. The short-term memory fails, is gone for good, and so nothing new can be processed into long-term memory. The poor old village of Short Term is obliterated entirely. The brain has an alternative route _out_ of Long Term, though not in. Eventually it, too, will be obliterated.
Once you get into long-term memory, the road branches. Down one road there's implicit memory, and down the other, explicit. Explicit further branches, into episodic and semantic. Implicit is another way of saying procedural memory, the one that deals with the things that we do as if automatically. Riding a bike, driving a car, knowing a dance, playing the flute: these skill memories are taken care of by the cerebellum in league with the basal ganglia, four clusters of neurons at the base of the brain that help initiate and control movement. Serotonin is the neurotransmitter of choice in the making of implicit memories, and dopamine in the creating of explicit memory. Researchers think that implicit memories are laid down while we're in REM (rapid eye movement) sleep, in which our dreams are most vivid, and that explicit memories are made during non-REM sleep.
Explicit memory is the sort we need actively to call up, "thinking" in the familiar conscious sense. Episodic is autobiographical, and locates things in time and sequence: "I ate eggs for breakfast, went to the life drawing class in the village, and after lunch Nancy and I took the dogs to the beach." That's episodic memory. Semantic memory is encyclopedic, intellectual, for facts.
Alzheimer's damages the episodic (autobiographical) memory first and worst. The semantic survives longer. Sufferers might know very little about themselves, nothing whatever about what happened ten minutes ago, and yet might be able to talk at length about the history, the battles, and the princes associated with a ruined castle visited on a Sunday afternoon outing, using long-term semantic memory. Alzheimer's sufferers of a certain generation, taught screeds of poetry by rote at school, find they can still recite their twenty verses of Longfellow with perfect accuracy, until quite late in the disease.
Because memory is a process, relying on neurons to fire up in the same sequence each time we remember, memory can be wrong. Memory, indeed, is notoriously unreliable. Why should it be, though, when we rely on it for survival? Perhaps that's the point. Perhaps our brains are more dedicated to our psychic health than to the truth. What we see, the way that we see it, and the way we remember it are essentially subjective. The process of making memories and then remembering them is both technical and personal. The synapses may not reproduce their original pattern. It's like the old fable of the bad carpenter's table, in which leg number two is drawn from leg number one (and is a bit out), and then leg number three is drawn from two and is even more wrong, and number four, drawn from number three, isn't anything like the same length or shape as number one. Something we thought, imagined, doubted, added on one occasion of remembering distorts the memory for next time it's called up. How then can I be sure of what I have done and experienced in my life? There are some slices of time, moments, collections of moments, from the deep past that are unlikely, eccentric, unaccountably preserved, and which I treasure. But are they accurate, or are they a story I tell myself for my own reasons? There's no way of knowing for sure. Not only do you, the reader, perhaps suspect that not all of what I write about life with Nancy is exactly as it happened, but strictly speaking, knowing the mechanism to be emotional, I ought to suspect the same. Our memories of things are never objective. We interact with them and add meaning; highlight certain aspects and throw others into shadow.
The brain is selective about memory. Not only about the details, but about the quantity. This selection and editing is important in life having a shape. The truth of this is illustrated by the problems encountered by people who have too much memory. There have been neurological cases of people who can't forget things. Their brains can't filter out or edit and everything is retained. They can tell you in detail exactly what was said or done on this day last year. What happens to them is that they lose the big picture, a sense of perspective, and are overwhelmed by detail. No choices can be made, no judgments. Everything is of equal importance. Because of this, they don't always function well as humans. So it appears that in principle and in moderation, forgetting is important. As Nietzsche wrote, "There could be no happiness, cheerfulness, hope, pride, immediacy, without forgetfulness."
# Chapter 21
_Between the acting of a dreadful thing
And the first motion, all the interim is
Like a phantasma or a hideous dream_.
—WILLIAM SHAKESPEARE
I FIND, ON A ROUTINE SELF-AUDIT, THAT I HAVE BECOME very low. Nancy is succeeding in sucking the optimism out of me, a strange new place to find myself in the middle of summer: flat in the heart, empty in the head, craving solitude and sleep. Peninsula sunshine is blinding, uninterrupted by geography, reflected and magnified by the sea, but I am peculiarly unmoved by the sun at the edges of the curtains in the morning. It's worse, in a way, when the weather is good. On top of my own lowness is overlaid recrimination. Here it is, a summer day at last, and I don't want to have anything to do with it. Worse, I don't seem to be _able_ to have anything to do with it. I can't step into it and be warmed. The heat on my skin is an irritant, the warmth on my head provoking. It's best if I keep my distance.
What I want is a sofa to myself. I want to be left alone to read, reassured by the sun pouring in at the windows, which is a novelty in itself, but Nancy's reaction to the arrival of something like summer is an inability to sit still. Daily she wanders the hall. Up and down to the conservatory and back. Through one kitchen door and out the other, twenty, thirty times a day. Up and down from bedroom to sitting room. In and out of the bathroom with an absent expression, her mouth drooping, her eyes blank and hooded. She's taken on what's called the _lion face_ of Alzheimer's. But if sunshine makes her twitchy, it's worse, far worse, on the bad weather days that follow, which come as we know they will, a wind and rain corrective. Seasons arrive and depart in self-contained daily chapters that seem to have little to do with one another or with the conventions of the calendar.
"Hey, you. Hey. What you doing?" I say cheerily. "Wandering again? Can't we find something for you to do?"
The voice of doom speaks in monotone. "There is nothing to do. There is nothing at all to do. It's just all meaningless. It happens again and again and it doesn't mean anything." She goes to the window, gestures out at the bay, the headland, the sea roaring and the wind howling in another midsummer gale. "Look. There's nothing there."
She's right, I find myself thinking. We must get back to the city! Though it's my mother-in-law's thought, this Edinburgh-craving thought I'm having. I'm channeling my mother-in-law. She has possessed me and I am diminished.
After the gales recede there are days of gray mugginess, the midges gathering in clouds on the road beside the wood. Then there's a cold snap, which coincides with the B and B booking of two women from New Mexico engaged on a European tour. They live in perpetual drought at home with a cactus garden, and though they say that they expected Scotland to be cool, their idea of cool is 68 degrees, a temperature regarded as sweltering hereabouts. The house isn't warm enough for them, even with all of our inadequate heaters blasting on max. They huddle by their coal fire in three layers of thermal and fleece. One of them wears a bobble hat for breakfast. They go out on a cool June morning wearing all the sweaters and coats and earmuffs and scarves and gloves they could find in an emergency dash to the knitwear shop in town. Nancy meets them in the conservatory and makes a fuss of their itchy Fair Isle acquisitions. "Oh, that's just brilliant!" I hear her exclaim. "Where on earth did you get it? Because I want one just the same! Can I try it? Can I? Can I if I'm extra nice?"
Written feedback arrives from Nancy's April respite week. It seems that she was a happy bunny at the residential home. "She taught some of the other residents to dance." She "only became distressed when the time came for her to leave." Next, feedback arrives from the care manager, verbally over tea and biscuits. There's bad news from the bed allocation committee. Our attempt to get Nancy onto the list bounced back with little consideration. "More help could be offered at home." Seven words. Nancy's not anywhere near eligible even for the waiting list yet. Meanwhile, following more day center shenanigans, and spurred on by this bounce back, one of our doctors writes a letter saying that in her opinion Nancy has already reached the point of requiring twenty-four-hour medical care. This cuts no ice whatever.
It's getting more difficult to persuade Nancy to take her clothes off at bedtime. Underwear is a particular bone of contention. She's physically as strong as ever and holds on tight to her underpants with both hands as they descend.
"Come on, you know you don't wear underpants in bed," I say.
"I've never heard anything so ridiculous in my life," she thunders, Miss Jean Brodie to a T.
SHE BEGINS A concerted series of kitchen raids. She seems to be hungry all the time. This is explained when I go back into the in-laws' sitting room one morning and find Morris eating Nancy's cereal. He has a rabbit-stunned-by-headlights look when he sees me coming.
"Morris, you really shouldn't eat Nancy's breakfast," I tell him. "I can get you another bowl if you're hungry."
"It's not that. It's just that she won't eat it and she's anxious about it, so I'm helping her out."
"The thing is, we need to know how much she's eating."
"Okay, boss, I hear you."
"It's important that she eats properly. I can find her something else to eat if she doesn't want Weetabix."
"You're the boss."
He's also helpful when she decides that she isn't going to take her pills any longer. More than once, I approach their door and hear Morris saying, "Quick, she's coming, give them to me." I find them cupped in his palm, the five little pills that Nancy doesn't want to take.
What does he think, that this helps Nancy? What does he think, that I'm the nasty matron who will insist on cod liver oil, and ought to be outwitted? That's how it seems. His attitude produces a twin. Mine. Cast in the role of nagger.
"You have to be in charge of Nancy eating enough, and Nancy getting her medication," I tell him.
"Uh-huh," he says, not taking his eyes off the television.
"Morris! This is serious stuff. This is important."
"All right, all right! I hear you!" he snaps.
When I leave the room, they mutter together. The word _she_ begins to be heard a lot. I am she. She is me. The nasty matron.
NANCY'S DIET BECOMES difficult to manage. She is rejecting most of the things I present her with. Morris gets given them when I leave the room. Her breakfast cereal, her toast. Her lunch. She'll eat the potato chips and the yogurt but not the sandwich. Morris gets first dibs on that, and if he declines, it's offered to a passing dog, and if the dog turns his nose up, it goes into the fire. I find the evidence in the grate among the cinders—a jumble of foodstuffs, tipped and scraped. Sandwiches, grapes, baked potato, small heaps of rejected salad. I fill up the biscuit barrel on Morris's side table every day, and every day it empties again. Every day, hungry, Nancy comes into the kitchen when I'm not there and carries out a snack raid. Things begin disappearing that nobody can account for. Packets of biscuits, packets of nuts, half-pound bricks of cheese, a bowl of strawberries, tomatoes brought in from the greenhouse and left in the colander. Things put out on the worktop ready to cook suffer random losses. An aubergine with give-away bite marks appears in the wastepaper basket. I come into the kitchen one afternoon and find Nancy standing in front of the stove, stuffing buns into her mouth. They'd been left out to cool on a rack ready for the children coming home. She has one in each hand and her mouth is full, working hard at another. There are four others missing from the tray. I'd been gone from the kitchen for ten minutes, gone to bring washing in. I'm beginning to understand what lies behind her constant opening of the kitchen door, and her retreat when she sees I am in there.
"Oh. Oh, there is somebody. Well, it doesn't matter. I'll come back."
"Something you want, Nancy?"
"No, no. Just wondering if there was anybody here."
"I'm here. Just reading the paper."
"Yes, yes. Well, then. I'll come back."
"Sure I can't get you anything?"
"No, no. It's fine."
Occasionally she's caught red-handed. She's defiant about the buns.
"Oh, Nancy—no. They were for later."
Nancy (through a mouthful of cake, spitting liberal crumbs): "What's it got to do with you?"
"I made them. For later."
"Well, then. Well, then. Enough of your nonsense."
She reaches for another.
"Nancy. Leave the buns just now. Go and see Morris. Here, take some peanuts with you. How about some cheese and crackers?"
Nancy ignores these offers. Her beady eye is fixed birdlike on the cakes.
"Leave the buns till later, Nancy."
"I will not. They're just as much mine as yours."
"Actually, no, because I made them."
"Oh. Oh, sorry."
But when she leaves the kitchen she takes one with her.
She picks things up and inspects them even if she doesn't eat them, which is worse in a way. Bathroom hygiene has been abandoned, so nobody else in the family wants to risk eating food that Nancy has touched. The fruit bowl becomes unpopular. Jack will only eat bananas and oranges, things that have to be peeled.
There are several alarming toilet incidents. The kind that make a person gag when called upon to deal with them. By "a person" I mean me. Chris is made of stronger stuff and this is chastening. I didn't have any problem dealing with the children's bottoms when they were little, after all. I try to think of Nancy as a big stroppy baby—one shouting, "I didn't do it! I didn't do anything! It's nothing to do with me!" But it doesn't help. One morning we come down to find—and I'm sorry to be so graphic, but this account is only of any use if it's honest—what can only be described as a trail, leading from the day bathroom out into the hall. The beige-colored carpet is smeared. Turds have been deposited at intervals and then trodden in. Opening the bathroom door, the floor is awash. Chris steps in and rolls up his sleeves and deals with it. I offer to go and have a look at the perpetrator. She's fast asleep. The feet are easily sorted out, courtesy of a series of wet wipes and a supermarket bag, breath held and eyes averted. The carpet will never be the same.
It's not a good month for bottom issues. Morris has been given a toilet aid, a mobile lightweight frame with a higher seat. One day when I'm cleaning, I move it away from the toilet in the day bathroom and forget to put it back again. Nancy, not able to understand the significance of its being moved, goes in and sits on the frame rather than the toilet, and pees gallons on the carpet. She's getting enough to drink. That's obvious. The kitchen door is often open in the morning and I suspect she's sleep-snacking. Sleep-snacking and filing her nails. Having refused tea, coffee, juice, squash, or water all day, she's possibly waking thirsty and going to the kitchen tap. But how? She can't any longer manage a tap. Nor find a glass in the crockery cupboard. Nor find the cupboard, come to that. Not when she's awake, at any rate.
The worst day with Nancy, the worst ever, comes unexpectedly as disasters tend to. One morning, when everybody is out of the house except Nancy and me—Morris has gone to the day hospital, the children are at school, and Chris is away—she summons me to her bedside with shouts. I can't be sure of this, but it might be _"Service!"_ that she's shouting. When I open the door to her room, she's red faced, her eyes blazing. It's 9:00 A.M. and twenty minutes ago she was snoring. The twenty minutes have been spent profitably. She's lying in bed wearing miscellaneous layers of clothes: a shirt and a pair of trousers next to her skin, followed by a bra (backward, clips across the bosom, cups flapping behind) and another pair of trousers, then underpants, then a cardigan worn round her waist like a skirt, another two cardigans worn properly, her bathrobe on top.
"Where have you been? What time do you call this, then? About bloody time, too," she says to me.
It takes almost an hour to get her undressed and showered and redressed—she's soiled and peed herself in bed—by which time she is in pugnacious form.
"Is that all you've got for me? It's not very much, is it?" she says when I light the fire. "Is that all you can do?" when I put her television on. I ask if she'd like a cup of tea.
"Not likely, not if you've anything to do with it," she says. "Get me somebody else. Go on! Go on! Fetch me the manager because I want to complain."
"Look," I say. "I do not work here. For your information. I live here. This is my house. You are my mother-in-law and that's the only reason that I put up with you."
She looks taken aback.
"Well," she says with great theatricality, "I've never been so insulted—"
"Really? It's early yet," I shout, leaving the room and slamming the door.
I sit at the table, heart thumping. What am I doing? Why am I so angry? My hands are shaking.
I get up and open the door. Smiling like a maniac.
"Nancy! How lovely to see you!" I trill. Ordinarily, this tactic would work. Short-term memory loss can work to a caregiver's advantage. But Nancy isn't going to be bought off anything like that easily. This is a seriously bad day, heroic in the pantheon of Bad Days with Nancy. Ordinarily an immediate about-turn of mood on my part has an almost magical effect. Ordinarily she would grin at me and greet me back. "I've not seen you for so long! My friend! Come and see me, come and sit by me." Patting her lap as if I were six. This would be a normal about-turn and all would be well. But today is different.
"I'll just get your breakfast, back in a tick," I say.
I take her a bowl of two Weetabix with milk and sugar and a piece of jammy toast for her tray, a glass of orange juice and cup of tea for her side table. She stares forward, rubbing her hands together. She takes her pills without comment, without resistance, still staring ahead and rubbing. I stand in front of her for a moment. She doesn't seem to see me. I crouch down. "Nancy," I say. "Here's your cereal." She is rubbing more urgently now and her eyes are wide. I put the spoon in and offer a little to her lips. I offer up the toast. She takes no notice.
Righto, then, I think. Just leave her be. Get your own breakfast. I eat some porridge and drink some coffee and read yesterday's _Times_. Then I go back into her sitting room.
She's sitting in her armchair with that same unseeing stare, hands rubbing rhythmically. Her lap tray is on the table by the window. Her orange juice has been drunk, and her tea, and her bowl is almost clean.
"That's great, you managed it all this morning," I say. Then my eye is caught by the fireplace. The Weetabix is on the hearth in a scraped milky heap. As I approach I see that the orange juice and tea have been poured onto the grate among the cinders, over a wigwam of toast. Nancy sits rubbing her hands quite frantically together, the pace increasing as I approach, her face streaked with crimson and blue, her expression defiant.
"What on earth have you done?" I say. "Why on earth did you do that?"
"What?"
"That. Look. Cereal all over the hearth." I poke it. It's setting into quite a useful Weetabix-based cement.
"I didn't do that. It was that other woman. That bitch, the other woman."
"Come over here," I command.
"I told you, I didn't do it," a high wavery voice insists. "It's nothing to do with me."
"Come over here. Come here. Come here," I say, attempting authority.
She gets up, sighing.
"Look," I say. "You did that. You made that mess and I want you to clean it up." I give her the coal shovel and brush.
"I'm not doing anything of the kind," she says.
"Yes. You. Are," I say.
"NO."
"YES. YES YES YES." I'm shrieking now. I'm losing it. Months and months of holding back and being reasonable have their price and here is their invoice.
"I am sick of you!" I yell. "I am so sick of you and looking after you and the endless bloody drudgery!"
Nancy roars. That's the only word for it. She roars like a lion, like an old skinny lion with a mangy coat, left behind by the pride to starve. She's dangerous. She's a cranky old lion and still has teeth. She swings the shovel toward my head, and I make a reflex movement and it misses. She throws the shovel against the wall, where it breaks into two pieces and falls. She brandishes the brush and jabs forward with it at my chest. I grab it and we're struggling. She lets go of the brush and has me by the upper arms with tight white fingers. She pushes suddenly and I fall backward. She's shouting incoherently; I can't make out the words. I barely hit the carpet before I'm up again and grabbing her. I have her by the upper arms now. Now it's her turn to topple backward.
"Don't you ever, EVER, get aggressive with me, you vicious old cow, or you will be in a nursing home before you can say _tea bag!"_ I screech. I can't catch my breath and it occurs to me that I'm going to have a heart attack and die. All I can think is, What if it had been one of the children? What if she'd taken a swing at Jack and the sharp edge had hit home, across his cheek, his ear, his eye?
By the time she falls, in graceful slow motion, onto her bottom on the carpet, still holding on to my arms, rolls onto her back, and is pushing herself upright again, one knee and one foot braced, I am screaming. I can't remember what I say. I remember that I step backward and am holding on tight to the door handle. I don't trust myself not to hit her. I'm yelling my head off. She's standing up by the fireplace now—thank god, thank god that neither of us fell toward it: I have a flash, a vision of Nancy cracking her head on the hearth and going limp. She's rubbing at her upper arms with both hands and saying, "Oh Christ, oh Christ."
I slump onto the floor, my back against the door. I'm shaking violently. I can't believe that I threatened the nursing home as a punishment. Elder abuse. Elder abuse is all I can think of. Strictly speaking it was self-defense—or retaliatory, at least, as she pushed me first—but even so. She has Alzheimer's. What in hell am I doing?
The rest of the day is spent making ostentatious amends, singing songs, taking her on walks and garden visits, and brushing her hair and making her laugh, all upset forgotten. But when I put her to bed I see that she has bruises. Small, faint fingerprints encircle both arms. Mine are sore to the touch but are unmarked. She bruises easily, dramatically easily these days, but even so. These I inflicted. Aristotle in the _Poetics_ describes how _hamartia_ , a serious lapse of judgment, can all too easily lead to _peripeteia_ , a calamitous reversal of fortune. The dark shadow of _peripeteia_ is hanging over me. Guilt, in other words.
I pour myself a (large) glass of malt, noting idly that daytime drinking is becoming the norm, and administer a self-directed pep talk. There's no point in rising, in engaging, in any of the negative energy because it's only me who suffers. I give away my power, and I'm not going to do it again. Tomorrow, if the same happens, I will scrape the breakfast from the hearth and leave the room and go find other things to do. I'll do it quietly and without comment. I will find a way of not minding. It's not caregiving that's exhausting, but minding. It's minding that will make me ill.
I get it all out of my system in the classic modern way. I write e-mails. How could Weetabix lead to violence? In the calm of the aftermath, in cold words on the screen, it's hard to say. Pauline, a good friend, replies almost immediately. "Course it's not the breakfast cereal. It's the incessant drip drip drip, the relentlessness of it. Not surprising that you should crack. She's okay, I take it, and blessed with goldfish memory, so that'll be that as far as she is concerned—but dear god, this is real lulu for you. You still sound pretty shaken up. I wish you better things for tomorrow. Poor you. Poor battered soul."
It occurs to me to worry that the aides will see those faint amulets of bruises and will imagine that they know the truth. It isn't the first time I've worried about what people might think. Nancy is constantly in the wars, walking into doors and tables, tripping over steps, falling out of bed and blackening her eye; do the paid caregivers ever mention the bruising that ensues to the social work department? I launch an Internet hunt on the question of caregiver abuse, and the words in the search box bring up other, unexpected results. Abuse _of_ caregivers, and not just by them. Here's a dementia victim, a husband who killed his caregiver wife with a hammer. Another woman, assaulted by her Alzheimer's husband, who declares herself afraid of him, describing him as "cunning, nasty, aggressive, menacing." On the Web pages I glance over, though, these stories are outnumbered by attacks by caregivers upon the demented. Most of them detail a "snap" moment. Most appear to be about male caregivers attacking wives. Some of these have been dubbed mercy killings: a wife strangled by a husband who said he couldn't any longer bear her to suffer, another with her throat cut. More often, though, the attacks are attributed to rage. Faced with an impossible situation, people can fail in a dramatic manner to cope. Here's a man who smothered his wife because she wouldn't stop taunting him, in her demented, perseverating way. Another who tied up and gagged his wife after she'd kept him awake for days and nights on end roaring and shouting; she died. All elicit a mixture of horror and sympathy, the two vying feebly for precedence.
I read wider and come across the Greek myth of Eos and Tithonus. Eos, goddess of the dawn, falls for Tithonus, a hunky Trojan prince. She asks Zeus to grant him immortality, and he does. But she neglects to ask for immortal youth. Tithonus gets old at the usual rate and then keeps on getting older. He becomes senile, and stays that way for eternity. Eos, driven to distraction, locks him up in a room. In some accounts she turns him into a grasshopper.
Ralph Waldo Emerson wrote, "Finish each day and be done with it. You have done what you could. Some blunders and absurdities no doubt crept in; forget them as soon as you can. Tomorrow is a new day; begin it well and serenely." This is good advice, though he also wrote, more pertinently, "Sometimes a scream is better than a thesis."
# Chapter 22
_Between the essence
And the descent
Falls the Shadow_
—T. S. ELIOT
HER MIND IS OPAQUE NOW, HER MOODS IMPOSSIBLE to read. Does she know that, as Iris Murdoch put it, she's "sailing into the dark"? Does she spare us this knowledge as a kindness, by speaking in metaphors?
"I want to go home," she says over and over, and she doesn't mean to the bungalow we rescued them from, that lonely suburban isolation, the washing piling up, tea made from the hot tap and packets of biscuits for lunch. She means home to her old self. She's aware that she's lost her somehow, the woman who was a company secretary, with long painted nails and a wardrobe full of blue jackets, who made raspberry jam every summer, who knitted exquisite baby clothes for each of the children. She knows, but she can't quite put her finger on it. There are days when the delusions are full throttle. In the throes of the hospital one, she has begged me for medicine for the dying patient in the next bed, all the time standing in her nightdress in the hall.
THE SUBLIME SEEMS of no use to me now, in this dark time of bright sunlight. I go out into it desperate for something good, for a taste of the Epic, for help: hoping to encounter something even mildly similar to Wordsworth's "spots of time / That with distinct pre-eminence retain / A renovating virtue," but come back feeling far worse. I can relate, at the least, to his pre-visionary moments.
_O'er my thoughts
There hung a darkness, call it solitude
Or blank desertion. No familiar shapes
Remained, no pleasant images of trees,
Of sea or sky, no colours of green fields;
But huge and mighty forms, that do not live
Like living men, moved slowly through the mind
By day, and were a trouble to my dreams_.
On bad Nancy days, the really bad days, the beauty out there seems tainted, all of it, by her animosity, which begins to seem like a misconceived fight against disease, against the lights going out, like a misdirected energy in her struggle to emerge from the dark. The anti-Sublime, purposelessly and destructively ruminative, reveals a landscape full of death. Death we think trivial. A broken cat in a ditch. A seagull neatly bisected on the side of the road. A baby seal dead on the beach, and then a dolphin, part eaten before it was washed ashore. I begin to feel an overwhelming, disproportionate pity for the sheep and the bullocks that watch me from their pasture as I pass. It's all suffering and cruelty out there, I think, stomping along the beach in a summer dress and raincoat and Wellies; it's cruelty disguised by landscape, by our fetish for views. I blame Wordsworth for that. I come across the archconservative Joseph de Maistre—a man named by philosopher Isaiah Berlin as one of the six Enlightenment enemies of liberty. "In the whole vast dome of living nature," de Maistre wrote, "there reigns an open violence, a kind of prescriptive fury, which arms all the creatures in their common doom; as soon as you leave the inanimate kingdom you find the decree of violent death inscribed on the very frontiers of life." Unsurprisingly, this doesn't help much.
I read that Charles Baudelaire said that de Maistre taught him how to think, and find this quotation from the poet: "We are weighed down, every moment, by the conception and the sensation of Time. And there are but two means of escaping and forgetting this nightmare: pleasure and work. Pleasure consumes us. Work strengthens us. Let us choose." Yes, I think, that's so true; work is all that will keep me afloat. I must work and make everything else secondary. Though it's slightly disheartening to discover that Baudelaire died at forty-six.
I dip in and out of books, following dementia trails like snail tracks across the paper, but seem unable to settle on anything new. I revisit John Bayley's magnificent account of life with his wife, Iris Murdoch. Her demise seems to have illustrated the fact that though highly educated people are somewhat less likely to get Alzheimer's, when they do succumb it tends to be an aggressive, fast-acting, fast-forward disease. Iris Murdoch's Alzheimer's first presented itself as trouble with finding words. It manifested itself in her last book, _Jackson's Dilemma_ (1995). Neurological study has found her vocabulary much reduced in it. She died in 1999, three years after diagnosis.
More care meetings, more assessments. The professionals come and settle themselves in chairs; they drink tea, they talk to us and make occasional notes. Conversations with Nancy are brief. Nancy is produced, and as is usual when faced with a social situation is utterly charming. Some emergency facility, buried deep, is mined and polished: that of the social bluffer.
"So. Can you tell me your date of birth?"
Nancy's grinning. "I've absolutely no idea. But I can tell you it was a very long time ago; my memory's terrible."
The professionals are reassured. They give us quizzical looks. She's really not at all bad. Crisis, what crisis?
Rita Hayworth (1918–1987), the Hollywood film star, developed Alzheimer's early. She was diagnosed formally in 1980 but had been ill five years, since the age of fifty-seven, and her daughter says in interviews that she had shown the first symptoms twenty years earlier. Alcoholism confused the issue, but reported agitation, hand rubbing, paranoia, mood swings, vacancy of the gaze, obsessive reorganizing of cupboards—this all sounds like Alzheimer's. Her daughter reports that even quite late in the disease, her mother continued to turn on the charm with doctors. Asked a direct question like "Who's the president?" she'd switch into flirtation and change the subject. The performance was remarkable, it's said; it was as if she was constantly auditioning. That sounds familiar. Nancy's winning smile and stock of phrases dredged from the past are a form of doctor repellent.
JULY BRINGS A series of bed-and-breakfast food sensitives. They bring their own tea bags and ask for hot water at breakfast. They want to discuss what's in the bread, the provenance of the bacon, and they have a raft of food aversions. _Rhubarb? Sorry. Wrong sort of acids. Oops, no, I don't do fungus. Is the coffee fair trade? Are the mushrooms organic? I'm a celiac, didn't I mention? Do you have any gluten-free bread?_ And then, as plate goes down on table, _Ahhh, sorry, but I can't actually eat tomatoes, or anything that's been in proximity to a tomato_.
It's warm some days, but the wind blows. The wind blows most days. It's so much our default weather that the days when it stops are puzzling. The silence takes its time to penetrate the senses. Being here in quietness is a different experience. The landscape is different. The air hangs heavy over it and its shapes settle into quadrilaterals—broad stripes of sky and sea, thinner strips of garden and wall, layered, irregular blocks of headland—all of this replacing its customary dynamic wildness, the sea in ferment, clouds scudding, the sharp diagonals of trees blowing. An elderly neighbor I meet one chilly morning down on the road while dog walking (in the Barbour, in midsummer, with head-wrapped scarf in place and Wellies) tells me that she has decided she can't face another peninsula winter and is moving down to Somerset. "Make sure you get off frequently, regularly; quarterly if you can manage it," she told me. "You have to get off sometimes."
Off?
"Away south. Four times a year. It's important. Otherwise you get unhappy. Cabin fever."
I've lost all sense of where my feelings about caring for Nancy and Morris end and where those about living out here begin. The longer we're here, the more the two things, the social isolation entailed in caregiving and the physical remoteness of the house, seem bound up in one another. Wordsworth's exalted observation, "How exquisitely the individual Mind /... to the external World / Is fitted:—and how exquisitely, too— /... The external World is fitted to the Mind," seems unintentionally close to taunting.
I go out into the landscape and see it all externalized: Nancy's panic and my own resentment. The view is soaked in both varieties of unhappiness. I am beginning to think that I will love the peninsula only after I've left it. Flaubert had a similar relationship with Egypt, which bored and depressed him when he was there, at twenty-seven, but grew in his mind in all the years afterward, some forty years of ripening. The Egypt of memory, his idea of Egypt, followed him to the end. He was thinking of it, longing for it, just before his death. That's probably what will happen here and it's all down to elementals. The sea and sky, beach and cliff, meadow and wall and wood. These will prove irresistible, once memory has charge of them.
* * *
SCHOOL BREAKS UP and the summer holiday stretches lengthily ahead. On days when it's grotty weather, bucketing rain, autumnally cold, we struggle to find things to do indoors that Nancy will tolerate. I invite her to come and do some art with me and the children at the kitchen table one afternoon.
"I can't really do it. I'm terrible at drawing," she tells me.
"Doesn't matter. But would you like to have a go? Play with some color on the paper?"
"I'd love it. Just show me how."
We sit down with a still life between us, a jug with flowers, a cup, a glass. Quietness descends with concentration. Nancy sits with an oil pastel in her fingers, looking uncertain.
"What do I do?"
"Just draw what you see. Or part of it. Or whatever you'd like to draw. Anything at all," I say. "Or just make colors on the page." She looks more uncertain. I've given her too many instructions. "Just use the crayon," I tell her.
The rest of us have settled into a slow breathing rhythm, glancing from objects to paper and sketching in the shapes. Nancy looks at the jug of flowers, the glass, the cup, and at her paper, which is a large white sheet. The oil pastel makes contact at the left-hand side and she begins. What emerges is very like her signature, what her signature's become, repeated with variations: a long unbroken series of what look like _n_ 's and _v_ 's interrupted by the softer contours of what might be _m_ 's and the occasional punctuating _y_. It goes across the paper and up, forming a ribbon of letters about half an inch broad. All the time that she's writing, she's looking, as the rest of us do, toward the still life grouping and back at the paper, as if what she's doing reflects what she sees. She talks to herself as she draws, murmuring along. "There, that's up and away and here's the next part, and there it goes, and it's like that, and it's like that again, that's right." The children maintain a tactful silence. "There," Nancy says, putting her oil pastel down. "That's it, I think."
The American abstract artist Willem de Kooning (1904 _–_ 1997) died of Alzheimer's. His late work, his dementia art, is very different from the hectic, intricate colors and anxiety of the art of his prime. It's much simplified, has curving open lines (lines dominate), allows generously for white space, and fills in the shapes created by intersections with color. It's as if Jackson Pollock turned into Mondrian. In a series of huge canvases the white field is interrupted by curling tendrils, "ribbon paintings" that tangle their ribbons together, and more than one critic has said, rather tactlessly, that they resemble the tangling of tau protein in the Alzheimer's brain. There had been symptoms as early as 1980, when the output is visibly taking on its dementia-period look. In that same year a friend reported that de Kooning, formerly a voracious eater-up of books, had given up reading entirely. Instead he painted and drew all day long obsessively. He gave up preparatory drawing in 1983 and used old drawings as the basis for compositions. He was extraordinarily productive: 340 paintings were produced in the 1980s alone.
There's been much debate about whether de Kooning's 1980s work is properly art or not. Some critics have raised issues of intention. Is the demented de Kooning still essentially himself, still giving of the same self that made his pre-eighties work so valuable? (Be in no confusion. This is a question about value. About money. His canvases sell for many millions of dollars.) There needs demonstrably to be an artist in control of his material, especially in the modern art market. In terms of value (monetary), the artist's brand is more important than the material. The self that is implied by the work, that is sold in the selling of the work, has outstripped the power of the work itself. In the age of Damien Hirst and Tracey Emin it's the artist that's the artifact. The artwork is secondary.
De Kooning wasn't diagnosed until 1989. In that same year, another friend reported that though he remained physically robust, affable, appeared to recognize people, and was still painting, when asked questions he'd give answers that were utterly unintelligible. Images remained with him but word and meaning were lost. He gave up painting in 1990. What would the art world have thought about that final decade if de Kooning had died undiagnosed? Mightn't there have been a rather different tone to the discussion, talk of an old man's retrospection, his sense of peace and optimism? Funny you should mention that, for lo and behold, reevaluation can take us there. The late works are beginning to be exhibited and written about as complex marvels, and in this spirit of reassessment de Kooning's dementia is given quotation marks and spoken of as irrelevant. There is plenty of evidence that he was a happier man in these brief years, having recovered from the alcoholism that precipitated a breakdown in 1980 and seemed to bring Alzheimer's on.
Nancy begins to pick on the dogs. Perhaps it's a pecking order thing, and this is her only remaining outlet for the exercise of authority. Asked if she'd like the dogs in her sitting room, Nancy will always claim that she does, warmly and with apparent sincerity. The dogs have spent a lot of time in the in-laws' room this year. Nancy asks them in, blocks off their exits and bosses them about. The terrier prefers to be by the fire but Paddy goes and sits by Nancy's feet, enjoying having his ears fiddled with, tolerant about the ranting. They get biscuits, they get sandwiches, are bemusedly offered orange peel, and they get told what Nancy calls _stories_. That was the case until recently at least. Now, though, they're unwelcome. Doors open and dogs are evicted.
"Out! Out! You're a bad dog. You don't deserve to be in here. Off you go you little bastard. Away with you, you bad dog."
Paddy and Sparky are untroubled by "bastard" but they know what "bad dog" means and are suffering intermittent crises of self-esteem, slinking out with their ears held flat and tails pulled in tight.
A fortnight of summer respite looms and the care manager is preparing a dossier, again, on Morris and Nancy. Chris, in consultant mode, gives her the full consultation, though I'm not convinced that the comprehensiveness of his reports is appreciated. I join in with anecdotal examples and occasionally quite daring moments of honest noncopingness, though I don't like myself much for doing this. The care manager doesn't respond emotionally to anything we have to say. She is a highly trained professional, unflappable, has no doubt heard it all before, but we are new to the business of dealing with social workers and find her stoicism unsettling. Well, I do. And I find I might be exaggerating the problems slightly, in order to get a rise out of her. (It doesn't work. You could say to her, "Nancy is eating dog poop now," and she'd say, "Is she? And how's that impacting on the rest of you?" She is unshockable.) We're listened to with sympathy but very little is written down. She's a conduit of the system, and the system is only after one thing: the facts of the matter, and translating these into a score on a sort of geriatric Richter scale. The assessment is printed up and returned to us for checking. It amounts to ticked boxes, mapping what's possible and not in terms of their physical and mental status.
The assessment has a dual purpose: It will go to the bed allocation committee (we're still trying to get onto the waiting list for a nursing home) and to the home where residential respite is being offered. Because it's also for the respite home, the report strives to be positive where possible. The temptation to minimize Nancy's problems so that a home without dementia facilities will take her is irresistible. We've been warned that the residential care waiting list is long, two years long or more, and the situation is worsening. I know I won't last another two years without cracking. So it's important that this assessment is frank. Can this assessment, the one assessment, do both jobs? No. It's a bit like being self-employed and preparing one set of annual accounts. These are rounded down as much as possible for the tax man (small income, please don't bill me). But they also need to be bigged up for the mortgage company (sizable income, we can afford the house). Which way to jump? A happy medium is best in most things, but possibly not in the case of social care assessments. In the case of social care assessments the report is left dangling, nowhere, compromised, full of euphemisms. Reading over the report is a disheartening business because it bears very little relation to the reality of all our lives.
Morris is to be in the seaside home again, and Nancy is to go into the town home. Then Nancy loses her place in the town home, thanks to bed blocking, which happens when there are no new beds available because patients who have nowhere else to go continue to occupy a spot for days or weeks (or in some cases years). The care manager is frank about the problems. Sometimes people don't go home again. Sometimes families refuse to take them back. We've been asked to sign a statement pledging to collect Nancy on the agreed date. The care manager says she's hoping she's got Nancy into the same home as Morris. _Hoping_. The stumbling block is that Morris's home has no Alzheimer's unit. Luckily the compromised report does the trick. At the time, a week before our holiday is very nearly canceled, this seems like the most important thing. We can go. And we do, though expecting, even at the gate to the aircraft, an announcement over the loudspeaker calling us back to duty. _Would the woman who's abandoned her mother-in-law please come to the information desk_....
Reading up on neurology on holiday, I find that the buzzword in brain talk is _plasticity_. The brain can be molded, reshaped, even in adult life; it responds to demands made on it. An experiment done with adults taught intensively to play the piano, from scratch, who were then brain-scanned, showed that their motor cortex had expanded significantly to cope with the workload. Not only that, it was commandeering neurons in neighboring zones to help with the job of learning and playing the piece in question. Cabdrivers in England doing "the Knowledge," which involves learning the entire driving map of London, have been shown to have enlarged hippocampi.
Brains that are damaged try to compensate for their losses. They set up connections elsewhere, get adjacent areas to set up lost functions, march in to other bits and clear the desk and lay down the law. It's a situation rather like that of soldiers interrupted in their task by a higher ranking officer: "Yes, I know you're supposed to be in North Africa but, actually, we need to move you into Crete. No arguments!" (Groans in background.) "There's a war on and we need to pull together, be a bit adaptable." A damaged brain is a war zone and its efforts to keep pushing on, delegating jobs to other areas, opening up and staffing new fronts, appear nothing short of heroic.
Size isn't everything. It's wiring that matters, connections, pathways. The wiring process is called myelinization. Myelin is the "white matter" that forms the insulating sheaths that coat the stems of the axons and ensure the signals are full strength when they're emitted from the end. Without insulation there would be seepage and slowness and incompleteness. It's a slow and steady process, and has its own sequence. Section by section, parts of the brain are brought online by being myelinized and connected. Motor functions are first in the infant human (we hold our heads up), then touch, better vision and hearing, and language skills. The frontal lobes (the executive self) and the memory-forming hippocampus are the last to be brought onboard, beginning at about three, when most people's earliest memories commence. Frontal lobe evolution is still taking place in teenagers. By the end of the teen years, the pruned neuronal forest has its myelinization completed.
One hot potato—hotly fought over—is the question of when we come to consciousness. The answer seems to be that consciousness grows as we grow, and that the experience of being alive is what makes us more and more conscious. Thus is consciousness linked inextricably with memory, not just in the process of knowing and doing, but _knowing_ that we know and have done. Once we're on the move and encountering the world, our consciousness grows rapidly, until at about the age of three we're fully conscious beings. As Nietzsche wrote, "Only then, through the power of using the past for living and making history out of what has happened, does a person first become a person."
The reason I'm telling you all this is that Alzheimer's reverses the processes of turning from baby to toddler to child to adult in a way that's almost pointed. Almost uncanny. In Alzheimer's there's a gradual loss of intellectual ability, stage by stage in grotesque mimicry. There's a last-in-first-out kind of logic to it. The two "adult-making" brain zones, last to develop in children, the hippocampus and frontal lobe, are decimated first, the basic motor functions last. The memory goes, the memories that form the context for all our adult judgments, our own hard-won experience of what's right and what's good, what works and what doesn't, what we like and don't, what's safe and dangerous. The self that debates these things, that uses memory as an intellectual tool, as a consequence is pared away. We're returned to a second childhood, one jammed in reverse gear. The intellectual capacity of the teen is lost, and then that of the primary school child, and we are returned to toddler-dom. Toddler milestones go last. The power to govern oneself, to dress, go to the bathroom, manage our own eating, the things all learned and perfected then, begin to falter and disappear. Finally the Alzheimer's sufferer, should she live long enough, is returned to a state of infancy and to incontinence. Language and recognition of language, then the infant powers of walking, bending, grasping; the ability to sit up, to lift the head and to smile, all these are lost. Advanced Alzheimer's cases resemble newborns in their total dependence on others.
# Chapter 23
_We live in the mind, in ideas, fragments. We no longer drink in the wild outer music of the streets_.
—HENRY MILLER
NOTHING IS HARDER THAN COMING HOME FROM HOLIDAY. Having undergone the serene psychic rebooting that comes about, mysteriously, by means of daily immersion in jade green water, the return to the house and its duties is shockingly difficult. The sense of dread starts at the airport. By the time we get in the car I am properly nervous. So much of coping is contained in its being got on with, prompted by adrenaline, habitual and unconsidered. It isn't always helpful to be allowed to step away and see the big picture. The hawk's-eye view, hovering above the landscape, is likely to induce an abrupt case of vertigo. Now, with Aegean salt still crystallizing on my skin, driving north through a cold gray summer, the return to the battlefield seems too difficult to contemplate. We have a day before the in-laws' return from the nursing home and it's spent feeling ill at ease. We're all subdued and I'm barely talking. The unspoken questions come round in a loop. Can I do it, keep doing it? What if I can't switch back? What if, perversely, relaxation and refreshment and that renewed sense of self I had on a Turkish veranda have been destructive after all? What if that transforming alchemy, the one brought on by aimless fiction reading and grit between my toes, has made me intransigent? Am I prepared to give it all away again? It may not even be a question of that, of choosing. I fear I may have lost the knack. When Morris and Nancy's bus trundles up the drive, it's hard to assume a welcoming face. I smile but I'm close to revolt. Morris looks glad to be back, but doesn't ask about our holiday. Nancy and I have one impulse in common, at least, one that's instinctive but lacking in specifics. Within a half hour she's telling Morris that she's leaving.
"You can stay here if you want but I'm not staying here. I'm going home."
"What are you talking about, you daft woman? This is your home."
"You think I don't know where my home is?"
"Yes! That's what I think!"
AUGUST IS ALL about doors. All the doors must be locked as Nancy is intent on escape. Every morning she is asking, even before breakfast, when she can go home, if she can go out, if it's time for her to leave yet; have they come for her, is her father coming, is it time to take the train, are the friends here yet? Morris shouts her down in his customary manner. That's usually the spur to action.
Until the school holidays arrived, the outer doors were locked all day and the keys were left in them. When we arrived here, Nancy couldn't manipulate a key. But now, apparently illogically, she can. She seems to have learned how, and retains the ability, day on day. She gets better and quicker at it, opens doors and is off. How, in someone lacking a functioning memory, is this possible? The answer may lie in the history of H.M., of whom we heard earlier, the epileptic research subject run down by a bike. He had his hippocampus removed and other bits of his brain modified, as a typically gung ho 1950s approach to defusing the extreme severity of his fits, and couldn't remember things. But when scientists asked him to draw, to copy the image of a star, following its outline, and then do the exercise again and again and again in the days following, H. M.'s performance improved. He couldn't ever remember, from one day to the next, that he'd seen the star before, but he got significantly better, progressively better, at drawing it. The answer lies in the cerebellum. The procedural memory appears to be able to learn, bypassing the conscious mind and the ordinary routes of memory. And this, I suppose, is what's happening to Nancy.
So now, official policy is that the key is taken out of the lock and hung on a hook. She can't manage to get the key off the hook and into the lock; she doesn't identify it, hanging on its hook, as the key to the door. It's the school holidays, however, and the children are constantly in and out of the house. Neighboring children are in and out of the house. There are bikes piled on the lawn, alien sneakers and sweaters in the boot room, strange strident voices on the stairs. And the doors are hardly ever locked.
Nancy's escaping becomes a big part of our lives. It wouldn't, of course, be an escape, if it weren't imperative that somebody always be with her. How lovely it would be for us all if she could be left to saunter round the garden picking the tops off flowers and chewing random stalks of found rhubarb, shutting the hens in their house because they've been naughty, telling her troubles to horses. Alas, this isn't possible. Thirty seconds after getting out of the door and into the world, anxiety descends. Some days we find her walking round the laundry green in ever diminishing circles, wailing; on other days, standing tightly, closely in to the elbow of a wall, chewing on her cardigan and paralyzed with fear. But then she starts leaving the garden. She begins to make a beeline for the road, charging down the driveway with whatever possessions she has judged vital to take home (cardigans, the address book, a singing toy kestrel I brought from the airport, the blue handbag) and out onto the lane, without pausing to look out for traffic. She is found one afternoon in our nearest neighbor's garden, talking to workmen building his extension, asking them if they know where she lives. She is discovered on another afternoon lying on the road.
On this occasion she is found before we realize that she's lost. Her habit of going to her room to be alone, two, three times a day, and the taking of lengthy afternoon naps: these are to blame for our not noticing. Morris doesn't raise the alarm if she disappears anymore, assuming, like the rest of us, that that's where she's gone. A stranger ringing at the doorbell is the first alert in this instance. Are we missing an old lady, she wonders. There's a white-haired old lady in a red cardigan lying on the road by the farm, and they've already called 999. We rush down there, two hundred yards down the lane toward the beach, and sure enough, there's an ambulance, parked; two ambulance personnel crouched by a seated figure, freshly cloaked in a tartan rug; and three cars pulled into the verge, with concerned (nosy) locals, waiting to see what happens next. They assume she's been run over, although statistics suggest not: Their three cars amount to 60 percent of the local traffic, from houses on the headland beyond us, and they're intimately related to the drivers of the other two cars. If somebody had bopped Nancy, word would be out by now. As Douglas Adams, the author of _The Hitchhiker's Guide to the Galaxy_ , observed, "Nothing travels faster than the speed of light, with the possible exception of bad news, which obeys its own special laws." The ambulance people are just about to whisk Nancy off to hospital, when we come running up. We have to explain that dementia makes her do this kind of thing. She has no apparent injuries and is determined about not going anywhere. We bring her home and keep an eye on her as promised. The first hint of illness, and she's to be whisked off to the emergency room.
I bring her back into Morris's orbit, the TV room, his occupied chair and her empty one, the afternoon movie in full flow, and tell Morris where we found her. Nancy interrupts me.
"That's just a pack of lies and you know it."
"Nancy. You were down on the road. Don't you remember?"
"No."
"You were lying on the road. The ambulance came along. They sat you up and put a blanket round you."
"Lies. It's all lies."
"Look. You've got a scrape on the side of your hand. I'll clean that up. I think you must have toppled off the sidewalk."
"It's a load of nonsense. I haven't been anywhere. I've been here the whole time."
The weather's peculiar and the hens aren't laying consistently. Some days there are five eggs, other days none. The trouble is that when both guest rooms are occupied, we need four eggs a day. On one particularly grim and windy morning, there are four Canadians to feed and no eggs at all: none in the fridge, all backups depleted. It's too early for the shop. Before the guests appear downstairs, we nip off in the car to a neighbor's roadside honesty box and are saved by the half dozen eggs sitting in it. It's only later that afternoon, following Chris's hunch, that we find a pile of fourteen eggs in the herbaceous border, and another pile of twenty-two among some brushwood on the lawn. I ask Nancy to help sort them, inventing a task I think she'll enjoy. She crouches down in front of the hoard of twenty-two, sitting on her haunches, and talks her way through it. "Two piles," I tell her. "Brown and white." I start her off. "Brown in this pile, white in this. See?" She murmurs her assent. I clean the chickens out and watch, with horrified fascination, as Nancy fails to be able to do the task. She can't seem to distinguish brown and white, can't make two piles. The browns and whites are mixed up again. But she enjoys handling the eggs, so delicately and gently, in cupped hands, and seems absorbed in her task.
AUGUST BRINGS NANCY'S eightieth birthday. I speak to Morris about this great event a week before the day. Would he like me to do some shopping on his behalf?
"I don't think so, dear, no."
"But Morris, her eightieth. Surely you want to get her a gift. Look at all these mail-order catalogs. Wouldn't you like to choose something? A new cardigan, a bracelet, some of these Velcro slippers?"
"Nup," he says, not tearing his gaze from his television quiz show.
I buy him a card to write and stand over him. His handwriting is affected by age, though only insofar as greater concentration on signing has exaggerated its sweeping ascenders and descenders. Nancy's attempts to sign her name—on bank things, for instance, since she and Morris still have a joint account—have become stressful and also hilarious. She's the one who's amused, laughing till she cries as she tries and fails to write her name. I give her a practice few goes on a blank piece of paper, and these aren't too bad. They don't look like a name but at least they are done with brio. It's when the form is produced that the trouble starts. No matter how many times I explain that the name has to go in the box, resting my fingertip as a guide, Nancy can't get it in there. She signs above, below, or on the wrong part of the sheet entirely. Nor can she sign in a straight line. Morris doctors her signature afterward, adding vowels.
"What shall I write?" Morris asks, his pen hovering over the card.
"Oh, I don't know. Something about her birthday." What does he mean?
_Happy Birthday Nancy, love Morris, x_. That's all she's getting. Not that it really matters.
Nancy enjoys opening her cards, but insists on putting them back in the envelopes afterward, and needs help getting them in. She carries the stack of mail around with her all day. She admires her Fair Isle cardigan, her new necklace, but won't try them on.
"These aren't my things."
"Yes. They're new. We bought them for you. And the chocolates we ate earlier, remember? And the bath bubbles. For your birthday."
"Is it my birthday?"
"Yes. You're eighty."
"Am I? I'm not. You're joking. You're funny. Eighty, she says!"
When we bring her cake in, crowded with candles, and sing to her, she claps her hands and her eyes fill with tears. "Oh! Look at that! It's so beautiful. I haven't seen anything that beautiful for many a long day."
She joins in with the singing, eats three pieces of lemon sponge, sips at her champagne, and sleeps most of the afternoon.
BUT THE DAY after, she's noticeably tired. And that evening, when I've taken supper through to Nancy and Morris, and have settled down to our own meal, there's a sudden explosive ruckus from next door.
"Just leave them to it," Chris says, mid-potato.
I eat a leaf of salad and hear the door from their sitting room out into the hall closing with a slam, Morris remonstrating, "Nancy! Please!"
I rise from the chair. Chris puts his hand over mine. "Just leave it. Eat your supper," he says.
"Na-an-cy! Nancy!" Morris shouts. I put my head round the door. "What's up? Nancy stomped off?"
"Yes, and she's taken her plate."
This is a first.
I check the bedroom first. No Nancy. Nor is she in the bathroom. I go round the ground floor calling. No response. I can't see her. She must have escaped, I think: Is there an eighty-year-old woman inching down the driveway with a plate? Then I hear a noise. Scraping. Coming from the library. Sure enough, there's Nancy, standing in the dark tipping her supper into the bookcase. Sausages, potatoes, radicchio, off the plate and onto the paper tops of a row of novels. I'm only grateful there isn't any gravy, though vinaigrette has already bled its way into Richard Ford and Edith Wharton.
I take Nancy back into her sitting room by the elbow and the moment she sees Morris, before I can speak, she turns to me pointing her finger and says, "This woman's a liar." She sits and fumes. But when I take her to bed she's all smiles.
"You're wonderful. You're my friend," she says.
"Well, thanks, Nancy. That's nice."
"I can't tell you how grateful I am that you look after me so well. You're a lovely person. No, I mean it. A really lovely person and kind."
"You're very welcome," I say, smiling at her.
"Not like those other ones. Those other people here. I don't like them."
# Chapter 24
_When ideas fail, words come in very handy_.
—GOETHE
MORRIS IS CONFIDING IN CAREGIVERS BUT NOT IN US, and now the contrast has become explicit. His favorite stays on past her duties to talk and I overhear things I wish I hadn't. Standing in the kitchen making a shopping list, I hear the aide's voice from next door, raised slightly, arguing a point. "But it's your home, too, Morris." I'm beginning to wonder if factions are forming.
He's ill, with ongoing kidney problems, and has disappeared somewhere within himself that Prozac can't reach. He no longer makes an effort to speak to the children. Any of the grandchildren who dare run the gauntlet of Nancy's heckling and threats and go into their grandparents' sitting room find that not even Granddad seems happy to see them. He has nothing to say to them and has lost all curiosity. On Millie's birthday he is unforgivably morose. Nancy's quite chipper: Presented with a slab of cake and a dog on the next sofa cushion, begging for a bit, she seems perfectly content. She has a lengthy conversation with the terrier about whether he's a good boy and deserves pudding. Morris sits in his chair and looks at the carpet. He eats a bit of cake and leaves the rest. He doesn't talk. He doesn't wish Millie a happy birthday until prompted. He pointedly doesn't take an interest in her gifts. This makes the whole day seem heavy. It's hard to rise above the heaviness set by such unwarranted indifference.
Nancy's nighttime restlessness has a new flavor to it, one of strident noncompliance. Nancy's ranting into the wee small hours and Morris is at his wits' end. We check up on them every half hour, standing outside their door and listening.
_9:30 P.M._ Nancy's voice, chivying Morris, trying to galvanize him to get out of bed.
"We've got to get out of here, come on, we've got to get home. We need to go. We've got to get up and get dressed and go now. But you aren't listening. No. You're not listening to me. You never listen. You just lie there, a useless lump. All the people here hate you but you don't know that. They hate you. They hate me, too, but that's beside the point. I'm used to it. I don't say anything. They tell me what to do all day. All day I have to do the work. They should do the work but I have to do it. You just sit there. And they—they have got you all wrapped up. All wrapped up. Yes, miss, whatever you say. She thinks she's in charge but she isn't. She's going to get a shock. I'm going to surprise her one of these days. I'll sort her out once and for all. She doesn't know the first thing about it, not a thing."
_10:00 P.M._ Nancy is moving around. We can hear her dressing. She is pulling clothes out of the cupboard. The wire coat hangers tinkle and clank as they fall. We hear her walking, her heavy breathing, and her continuing monologue.
"This is what we need. We need these things to go home. We need to go home now. I have the things, the things and the other things and the rest of the things. You have to get the other things now. But you won't do that, will you? No. You just lie there, doing nothing. Doing nothing as usual. I have to do everything."
The door opens. I'm standing in the corridor. Nancy looks at me and closes it again.
"She's there. She's standing there," she says to Morris, who doesn't respond. "I need more things to get the things. She is going to take them. She will take them away and sell them. She doesn't want anything or anything but the money. She will take the money. That's right. I know that. I have always said that but you won't listen. You don't say anything to her. It's all left to me as usual."
The door opens cautiously and Nancy peers out.
"Hello, Nancy," I say. "Time for bed now, isn't it?"
"No," she says. "It isn't time for bed. I'm just going out for a walk."
I go in and open the curtain a little. "But look out there," I say. "Look how dark it is."
"Oh. Oh dear."
"It's very late and everybody's going to bed."
"Oh."
"Time for you to go to bed now."
She gets into bed, muttering under her breath. I close the bedroom door and put an ear to it.
"She's such a bitch. A bitch, and you don't say so. You don't do anything about it ever."
_10:30 P.M._ Nancy is still talking, though more quietly. I can't make out individual words but they pour out of her in a stream. She sounds as if she's sitting up in bed.
_11:00 P.M._ She's out of bed again. I go in and take her to the bathroom—usually her bedtime cup of tea has caught up with her by now—and put her back to bed. "Good night, then, have a lovely sleep, see you in the morning," I say, tucking her in firmer.
She glares at me.
"Don't even talk to me. Don't you dare even say a word."
I should walk away. Usually I do but sometimes I don't. Sometimes I mind having my evening fragmented. On those days I have to have the last word.
"Well, that's charming," I chide. "That's very good manners, isn't it?" As I leave the room, the low monotone starts up again, the words "she" and "bitch" just audible.
_11:30 P.M._ Nancy's wandering the halls. She looms in her white nightdress out of unexpected directions in the dark like a phantom. She's not keen on returning to her room. Morris speaks up. "For god's sake, for pity's sake, Nancy, shut up and get into bed. I'm not sleeping and I'm getting really fed up with you now."
"Oh right then. Oh fine," she says, getting into bed and pulling the duvet over her nose. Two affronted rheumy blue eyes stare at me from the covers.
Midnight is the fighting zone.
MORRIS: Nancy, I've told you, I need to go to sleep now. Will you just shut up and go to sleep?
NANCY: I certainly will not. Who the hell do you think you are?
MORRIS: I'm your husband and what I say goes.
NANCY: Oh are you. Are you indeed. Well, we'll see about that.
MORRIS: Be QUIET.
NANCY: I'm going home. I can't stand another minute of this place.
MORRIS: Get back into bed RIGHT NOW. I mean it.
NANCY: You're a fine one to talk. You just do everything she says. She tells you what you can do and what you can't do. She has everything she wants and you have nothing. You don't have any of it anymore. You just lie there. You are never helpful to me. You never do what I say. But she. Oh she. She is ever so that way, and you know it. She has you wrapped round her finger.
MORRIS: What are you talking about now? Who does?
NANCY: You know very well.
MORRIS: You're talking rubbish. Shut up and go to sleep.
NANCY: I won't shut up. Not until you tell her. You must tell her, this is my house.
MORRIS: Who? Who are you talking about?
I know who she's talking about. And so does he. But when he's exasperated he can't resist reminding her of her failing memory.
NANCY: I don't have to tell you her name.
MORRIS: You don't know it, do you? You don't know anybody's name.
NANCY: Don't be ridiculous.
MORRIS: Go on then. What's mine?
NANCY: You know your name very well.
MORRIS: I do. But you don't, do you?
_12:30 A.M._ Morris is quiet again, probably asleep. Nancy rants on undauntedly. At this point Chris might give her a spoonful of something prescribed to help her sleep, though we try to minimize its use because of the hangover that will follow. If she gets sleeping syrup, she'll doze most of the following day and be awake all that night angry, which demands another dose, another day of dozing and another wakeful night, leading to more and larger dosages. So is it that a care facility syndrome is born.
In any case the sleeping syrup doesn't always work. What she needs, I tell Chris, is rhino tranquilizer. We stand outside her door and whisper and stifle our giggles at the idea of a rifle with a tranquilizer dart. We're not quite ourselves. Later, I look up sleep disturbance in dementia and find that melatonin levels in the pineal gland fall in Alzheimer's sufferers, who as a result no longer take darkness as a cue. Melatonin can be given as drops, apparently, but you can't buy it over the counter in Britain; caregivers on message boards get theirs from the United States. As Alzheimer's progresses into the final dark phase there will be a complete turnaround, and sleeping will be the norm, as the disease goes further and deeper and cell damage is such that wakefulness can't be supported. The brain is then so damaged that it demands unconsciousness in order to muster all its forces of repair.
_1:00 A.M._ Nancy's monologues are sleepier, with more pauses. We go to bed, hopeful of rest. Most nights now there is another breakout, or a succession of them, typically at 2:00 A.M., 3:30, 5:15. Chris jumps out of bed and goes blurrily off, insisting I go back to sleep. I hear their two voices echoing up the stairway.
"No, Mother. You're not going anywhere."
"I'll do as I please."
"Be quiet, you're waking the children."
"I will not."
But when he's away working I have to get up and do the night shift. This is difficult. There's a reason, other than for filial duty and husbandly kindness, that otherwise it's always Chris who goes off to sort his mother out at night. Nancy has taken against me, me specifically. She's dramatically less cooperative with me than with anyone. She sees me coming and bristles.
There's been an abrupt switch around. I've gone from most favored to least. "And who the hell do you think you are?" she asks when I take her elbow and try to steer her bedward. "This has always been my house, do you hear, and I want you to leave RIGHT NOW." She turns, her face set with hatred, cheeks reddened, mouth turned decisively down. "You, you are not worth anything, do you hear me? Nothing. You think you're somebody, don't you? You really think you're somebody. Well, you're not. You're NOTHING. NOTHING. You're not worth the shit on my shoe."
The home aides are getting some of this treatment, too. Nancy wakes in a foul temper most days, and the appearance of the home care team, cheerily wishing her a good morning as they go to get Morris out of bed, is the trigger for ranting verbal abuse. "She's a bit upset this morning," the ladies say to me, looking rather shaken. "Not a happy bunny." "Nancy's been crying and upset and she's set Morris off." Morris is wheeled through with red eyes, a wet hankie.
"You know it's just the Alzheimer's talking, don't you?" I say to him. "It isn't you she's having a go at. It's just the disease speaking."
"I know, dear," he says, his voice cracking, "but that doesn't make it easier to take."
How hideous this is for him. How intolerable and cruel, to spend your seventies in this state (he's three years Nancy's junior), witnessing the death of your wife by slow degrees and having to deal with this protracted and ongoing grief, a pre-death bereavement spread over a decade. Not for nothing is Alzheimer's known as _the long good-bye_. Other people in their seventies go traveling, have adventures. What's worse, having Alzheimer's or being handcuffed to it and forced to watch? If it _is_ just the Alzheimer's talking and Nancy is already gone, then Morris's seems to me much the worse of the two fates.
Some mornings, Nancy gets in the way of his routine and one of the aides brings her out of the bedroom and chums her while Morris is dressing. I hear Nancy's voice, monologuing away as if it were still 1:00 A.M., as if she'd not paused for breath. "And I say so but he doesn't listen to me. He's useless. He just sits there and won't come home. And he won't stand up to her. Oh no. The bitch has it all her own way. Yes. She says what goes and what doesn't go. He won't say a word. Not a word." The aides don't comment but I can see from their reactions that they know the bitch is me. They shut the kitchen door so I can't hear. What they don't know is that when they leave the house, even before their fingers have left the door handle, Nancy's delivering her parting shot toward their receding footsteps. "She's a terrible bitch, that one. Don't let her in the house again. I'll not have her in the house. Coming into my house and talking to my husband like he was hers. She should get her own husband."
_Husband_ is a word she uses only when cornered, these days. _Husband_ is a word available only by means of the emergency generator, the same one that powers conversation with the health visitor. _Husband_ is a concept dredged up under pressure. We're all a threat, as she perceives us, all the other women who populate the house, to her marriage and her matriarchal rule. I come to realize that the days in which Morris has more protracted contact with the home caregivers, the days they hang about and are animated and make Morris laugh—those are Nancy's worst days. She sits in her chair watching, and rubbing her hands, saying nothing until they are gone. Then she'll be foul and ungovernable all morning. She'll likely as not be foul all day.
THE QUARTERLY ASSESSMENT is due. We sit in the conservatory and take tea, as is habitual.
"I'm afraid I have bad news," our care manager tells us.
"Oh?"
"We failed to get Nancy and Morris onto the waiting list again."
"Oh, what? How come?"
"It's bounced back from the bed allocation committee. I'm sorry. I was sure they'd get onto the list this time, but no."
"But why?"
"It's just the way it goes. There is a lot of competition. There aren't enough beds."
"But we're not asking for a bed," I say, more emotional than I would like. "We're just asking to go on the list."
"Can you tell us what criteria are applied in the decision-making process?" Chris asks.
"What happens is that they look at the assessment, and at other reports, and we talk to them, and they look at all the evidence, and then they make a decision. Once they've made it, there isn't a lot we can do. Just put the application in again. Do a new assessment. Have another go. I realize this must be very disappointing, but we'll keep plugging away for you."
"What other reports?" Chris asks.
It transpires that our care manager had visited Nancy and Morris while they were last in respite. A conversation was had with Morris about the future. The care manager confirms this much but won't comment. Can't comment. Disclosure would be against the law. Nonetheless it seems pretty clear to Chris and me that Morris, whether intentionally or not, has ambushed Nancy's route onto the residential waiting list this year. It's our guess that he's done this by taking advantage of our absence to insist that she's no trouble at home.
"I thought this may be of interest," the care manager says, holding out a large brown envelope. Inside there's a glossy brochure for a swanky residential community, one that's geared to active retirement, Florida style. Everybody owns their own property within its walls and pays annual premiums to fund the social life, the golf, the recreational facilities. Chris and I look at this prospectus when the social workers have gone and are rendered speechless. They can't really think, can they, that this clinically depressed, poorly wheelchair-bound old man and his demented, aggressive wife would fit in here, into this tea-dancing, bowls-and-bridge-playing culture? Have they been here, really been present, the social workers, at these quarterly meetings? Have they been listening to us at all?
Nancy, meanwhile, has taken to carrying her turds around the house. She no longer recognizes them. She carries them in her hands and brings them to us. "I found this and I don't know what to do with it" or "Somebody put this thing in my underpants and I don't know who it was but when I find them there'll be words, I can tell you that." She's begun using her bedside chair as a toilet in the night.
All of which has a curious effect. I am becoming squeamish about dealing with the B and B guests, about cleaning their bathrooms and de-hairing the plughole and changing their linens. I find, in my demoralized state, that I am beginning to resent the poor holidaymakers and their ordinary holidaymaking habits. Their leaving jammy knives on the tablecloth, coffee rings on the bedside table, and sticky kitchen refuse in the raffia wastebasket. It's too like dementia, this behavior, this not knowing what's appropriate. I'm not related to them, these paying guests. I have no burden, no duty, I don't have to be tolerant. I fume at their flushed, toilet-blocking sanitary towels, at novels that have had their spines bent; rage about books that go missing because somebody "inadvertently" packed something plucked from our library, a £20 full-color guidebook the same weight as a brick. Muddy footprints on a pale carpet provoke immoderate tutting. I bridle, not always silently, at luggage dragged along corridors, against newly painted walls and off the dressing table, leaving characteristic black smears.
The promised Indian summer doesn't materialize and the weather is gray and sullen with a stiff breeze. We wrap up and go out on the boat, leaving Morris and Nancy locked in the house. It's the only way we can go out on our own at the weekend, just the five of us, since our private care hours have shrunk to the minimum. It's also a complete no-no, leaving them alone and locked in, something I'd guessed even before the phone call, the caller tipped off by one of our aides. Locking in is unacceptable. What about the fire risk? It's a mustn't in the lexicon of don'ts that surround geriatric care.
"But what if I can't find a sitter?" I ask.
"Well, it's obvious, isn't it?" the voice on the phone says. "You'll have to stay at home."
We decide that we'll fish for mackerel but have no clear idea how to do it. We make fishing lines out of string hanging from twigs plucked from the wood. We can't buy hooks in the village, so we have to settle for safety pins, which are difficult to fix. The bacon won't stay on the pins. We go out for two hours and catch nothing. Bacon doesn't seem to be very desirable to ocean life. Or perhaps there's just nothing there, the sea a vast empty bowl. That has become my suspicion.
Our assigned October respite week is canceled, and this time our begging falls on deaf ears: There's no money available, not enough staff, and that's that. It means there'll be no half-term Turkish trip. I put my bucking bronco up for sale and buy a horse unseen on the Internet. He arrives in a borrowed lorry late at night, huge and brown and quivering with alarm.
# Chapter 25
_World is crazier and more of it than we think,
Incorrigibly plural. I peel and portion
A tangerine and spit the pips and feel
The drunkenness of things being various_.
—LOUIS MACNEICE
I AM LOSING IT. LOSING MY GRIP. THE THINGS THAT ARE various are spinning out of control. One afternoon in October, as I'm sitting in the drawing room in my pajamas, working hopelessly but with energy on the fiction project, surrounded by dogs and dog hair, toast rinds, watermarked coffee cups and old newspapers, the doorbell rings out. I decide to ignore it. Then it rings again. I fling the laptop aside on the sofa and go to the conservatory door, muttering loudly about bloody visitors.
Two people stand there, unmistakably American. A shiny rented Peugeot sits parked in the B and B spot. Light dawns. And there seems no other option but to respond with a sharp expletive.
"I'm sorry. I'm so sorry," I say, opening the door. "And sorry about swearing. I completely forgot you were coming."
Chris keeps them talking in the hall until I can tidy the drawing room. Then he keeps them talking in the drawing room while I hurtle round the apartment.
We spend that weekend moving a woodpile out of the old stable. The woodpile fills it floor to ceiling. People here are hoarders, particularly of anything that will burn, as trees are so few and stunted, and this pile amounts to the accretions of an era: wormy limbs of furniture, old sash windows, planks, logs, twigs picked up for kindling that are brittle and silvered with age, rotted-out joists, wartime ships' boxes with faded stenciled labels.
The respite booking system seems to have turned into a form of roulette. We're awarded six days in October at the council-owned home. Then these days are canceled again because of staff shortages. Then, out of nowhere and at short notice, we're awarded two and a half weeks at the privately owned Victorian home in the town. Though it's a private home, the social work department (under pressure, no doubt, from relatives who have had their respites canceled) has made the decision to buy in extra placements. We hadn't inquired about permanent care at the private home because the truth is we'd not wanted to use it. It doesn't have Alzheimer's provision, which is crucial, but in any case we'd heard unflattering reports. The private home is glad of the business and a representative begins chatting us up about permanent places even before the in-laws' stay.
Two and a half weeks later there is unexpected news. It comes four hours after Morris and Nancy were due back and, weary of hovering at windows, we phone the home to see what's happening. The reason for the delay, it transpires, is that Morris doesn't want to leave. Chris goes off to see him. When Chris appears, Morris backtracks. He wants to go home, please. No, he doesn't want to stay here. It's hard to know whether this is properly considered decision making, or whether he feels in some misplaced way embarrassed for electing to be in residential care, as if he is rejecting us and our hospitality. These are the only two possibilities that occur to us at the time. Chris reminds his father that last night he had cried and begged to be allowed to stay (according to the home). All Morris will say now is that he feels quite the opposite. He wants his chair and his fire. He doesn't want to stay here where everybody is old. It's fixed that Morris and Nancy will come back the following day.
The chap from the home rings again, to reiterate that Morris has had a lovely stay and (until Chris turned up) was heartbroken at the prospect of leaving. He says he'll talk to Morris again. Then he rings back and we have almost the identical conversation, word for word. Morris is adamant he's coming home and does, with Nancy in tow looking baffled. The chap rings to see if they're happy to be back. Then he rings the following day to advise me that the twin room is still available, but that they can't hold it for long. I tell him not to hold it. I don't think Morris will change his mind. He rings the day after, and the day after that. Finally, I'm short with him and he stops calling.
"He's a persistent character, isn't he?" Chris says.
"Odd, how he keeps on calling," I say.
"Makes you wonder just how keen Dad was on staying on permanently," Chris says. "And whether the guy was trying to cover himself, in his insisting that it was all Morris's idea."
This is a shocking idea but rings true. It occurs to me that it was just business.
AT HOME IT'S business as usual. Morris tells one of the aides, on Nancy's day out in town, that he doesn't know how much longer he can go on; that he thinks he made a mistake, preventing her going into care. He doesn't say so to me, though, even when I've been tipped off and prompt him directly. Chris is barely communicating with his father by this point, so there's no prospect of confidences arriving via that route.
When Nancy comes back from town she's all smiles, but the peace is short-lived. Caitlin finds her heading out the door to the garden, asks her to come in, and is slapped hard. She suffers a volley of verbal abuse from her grandmother, which is overheard by Chris. He takes Nancy into the sitting room and sits her down and tells Morris how angry he is with her. Morris isn't altogether impressed with this. He reacts rather like a mother at the door, when another mother comes to complain about her child being beaten up. He's not quite sure how to react or whose side he's on. But later, he seems surer. I'm in the kitchen preparing supper and overhear him reassuring Nancy.
"It's just you and me, you know," he says to her. "All we have is each other. It's just you and me against the world. Do you believe me? Because there's no one else. When push comes to shove, there's no one else that matters to me but you."
MY HEART IS hardening. I can feel it hardening and contracting. I begin handling Nancy's kitchen incursions differently. I turn the radio up louder and mouth, "Sorry, can't hear!" If she comes into the kitchen in a rage I don't say a word, just turn her round and open the door and eject her. If she comes back in, I have taken to shouting, "No!" just as the door opens and her angry red face appears. This is usually enough to prevent another annunciation for a while. I can hear her ranting about me next door, but she is in rant mode most of the time now anyway, so it doesn't matter. None of it matters in the least, I say to myself, turning the radio up louder. The radio is on in the kitchen all day now, radio or the CD player. Hendrix turns out to be an excellent granny-repellent. Mozart brings Nancy in asking questions and Sinatra sparks something that has the tone of reminiscence, but is a random putting together of words and ideas, presented as urgently true.
This may sound harsh and uncaring. Maybe it is. But it comes after a long, long campaign. Take battle weariness into account. The only way of continuing with this is to disengage emotionally. That is what has happened here. Self-protective distance has kicked in. I no longer intercede unless it's necessary—and even then only briefly, to call for a truce and move on. I no longer feel I have to wade in and referee. I no longer have that old burning impulse to convince Nancy of anything. I've spent a lot of time this year in trying to help Nancy to orient herself. However well intentioned, the information that she is ill was a mistake, misguided. She didn't believe it; she saw it as a form of aggression, as another lie from a hornet's nest of liars.
What I do now is keep my distance. I service their physical needs. I make sure they are warm enough, not too warm, that they have everything they want—newspapers, books, possessions, food. They are passive and need prompting. The television might need retuning. Phone calls may need to be made and shopping commissioned. Morris would never ask, but is grateful to be second-guessed.
What's upsetting and awkward, what begins to be seriously upsetting and awkward, is that Nancy's ranting about me to helpers and caregivers has become ubiquitous and vicious. _She's only after the money; that's the only reason she's here. She steals things and has to be watched. She hits me. She shouts. She smells. She hates everyone and nobody's her friend, nobody loves her. She's thinks only ever of herself. She's greedy. She eats all the food. She doesn't feed me. She's lazy. She makes me do the work. She's nasty and cruel but nobody knows that, nobody guesses. Don't be fooled, it's all pretend. Somebody should tell the manager. Somebody should tell the police_.
They ignore her, let her rant on, which is considered the correct response, but that doesn't seem the most important thing suddenly. Why is she the only one with rights? What about mine, not to be slandered and bullied? I want the aides so badly to intervene and tell her off. It would do my soul good to hear someone just once saying, "Don't talk about your daughter-in-law like that; she looks after you and works hard, and it isn't fair to call her names." But nobody ever does. They don't mention it to me, either. Which makes it more embarrassing. Chris gets told what she's been saying about me, but never me, and it's a distinction that seems close to pointed. Why wouldn't they mention it to me directly? Is it because they don't want to upset me? Or is it because it's their judgment that there's rarely smoke without fire? I get to wondering if Nancy's allegations are mentioned in a file somewhere. My continuing and deepening depression, my evident detachment, these might lend a little weight and credence.
Old friends call them from time to time. Nancy's friend Carol is on the phone again, and Nancy chats away to her as if it were fifteen years ago, her voice spookily youthful. Only the content is lacking. She has nothing to say for herself and what emerges is a string of clichés strung together. "She couldn't put a name to me," Carol says, "but she knows my voice and knows enough to treat me as a friend." Carol's an excellent antidote to the correctness of professionals, quite ebulliently partial. She says she's come across "this bitch thing" before. Her mother was the same. "It's not you, you know," she says. "It's hatred of your being young and able-bodied and running the household."
Nancy's brother Angus rings from Australia once every couple of months. Nancy takes the phone and starts off well. "I'm not too bad, thanks, plodding along, you know how it is, life is never simple, you better look before you leap, but you know, wasn't that always the way?" Then she runs out of steam abruptly and hands the phone over to Morris midsentence. "Here, you take it. You do it."
Our little sailing boat's parked out in the bay in front of the house for a few days, halfway on its journey to its winter berth in the town marina. While it's here, the weather turns wild and the boat is smashed against the pontoon. Her rudder is broken and Chris says he'll have to organize a lifting crane. Before he can do this, the storm winds worsen and we wake to find, looking out the bedroom window one morning toward the mooring, that only the top of the mast is showing above the surface. The rest of her is wedged down tight, badly holed and her fin keel ripped off. Poor Chris is submerged in gloom. Mahler's Fifth is on the CD player.
The weather turns apocalyptic, with floods, hail, schools closed, the horses cantering in hysterical circles in the road as we try to bring them up from pasture. It abates for a few days and we think it's over, though the truth is that we're at the eye of the storm. By Halloween wild gales are battering us, windows banging in their housings, the house creaking, sea in uproar. The house itself appears to undulate, rising and falling in queasy ocean-liner rhythm. Chris is out wrestling with the horse fencing—the electrified white tape is blown out of its posts and flies about. The wheelbarrow and henhouse are both blown across the garden, the hens roosting cluckily in the _Hebe_ bush.
Nancy changes into her outdoor shoes and rattles the doors. She sits on her bed in two coats and a hat. She's bolshie at the village day center. She gets brought home early one Thursday afternoon, having verbally abused the other members, having denounced Morris, having taken off her jewelry and thrown her full complement of rings across the floor. But that's not all. It seems there has been an incident. The day center manager rings. "When Ruth tried to take her to the bathroom, she went for her," she tells me.
"She tried to hit her?"
"Afraid so."
"I think the time may have come to expel her from the center," I say. "I know it's harsh, but I don't think you should be expected to put up with this. It must be spoiling everybody else's day."
"We're okay, we manage fine most of the time," she says. "We'll continue as we are for now."
# Chapter 26
_Any idiot can face a crisis; it is this day to day living that wears you out_.
—ANTON CHEKHOV
WINTER CREEPS IN MUNDANELY, UNSPECTACULAR with a gray wind. Wallace Stevens wrote that "the mind is the great poem of winter." Nancy's mind, perhaps. It's somewhere frosty winds moan, earth stands hard as iron, water like a stone. All human warmth has gone.
She's passed beyond some unnamed, unmarked point. She's unhappy and angry nearly all the time, and when that's true, the power of unhappiness and anger is lessened. They begin to seem inauthentic. As caregivers we become desensitized. Unhappiness and anger are Nancy's defining characteristics, and the few residual things that made her herself are fast fading. Her knowing her Christian name, and her hanging grimly on to the idea that she lives in Edinburgh: these are really the only two things she any longer knows for sure.
I've done a lot of research on dementia, but this doesn't make me any better at excusing her behavior. I do excuse it, intellectually. I know what it means and what it doesn't. But emotionally it's much more difficult, and I find that I'm behaving accordingly. Her being angry all the time, her seeming to single me out for particular contempt, has switched off something in myself that used to feel responsible for her being entertained, for her life having _value_. I might have observed something causal in the chain that's got us here. She becomes more difficult to handle and thus harder to like, and I react to this by being less friendly and less keen to spend time with her, which (possibly) makes her more difficult to handle.
It's hard to love somebody who hates you. It's hard to care for her, in either sense. The Book says it isn't Nancy to blame—not really, not in any meaningful sense—it's the plaques and tangles. Validation theory disagrees. It says that people with dementia become angry because they are trying, before dying, to express feelings and ideas they have long suppressed, which validation workers help them to release. It seems that Nancy might hate me, after all, might always have hated me. I've asked this question before and don't feel any closer to an answer: Is the person that Nancy has turned into as a result of the disease someone new, or someone who appears new because previously hidden? Validation has it that dementia exaggerates aspects mined deep from the buried self. I'm not sure how much credence should be attached to this idea. After all, how could anyone know what was or wasn't buried deep? And what does this say about human nature? Almost all of the Alzheimer's sufferers I've heard of have become abusive and contrary. Is it really possible to make a case for the greater part of humankind suppressing feelings of rage all their lives, feelings only let out of the box by the loss of inhibition that dementia brings?
Looking for more on this, I read a book about dementia caregiving that takes the playing-along-with-delusions approach, and simplifies it into three golden rules. It insists that all that's wrong with Alzheimer's sufferers is the loss of short-term memory, and all that's needed to make them happy is to stop asking questions, to stop contradicting anything they say, and to live in the past with them unquestioningly, using sympathetic conversational techniques, because the long-term memory is still intact and working perfectly logically in its way. It claims that there cease to be behavioral problems and upset once the caregiver acts accordingly—for instance, in going along with a loved one's impression that it's 1970 and they're both teachers at a school—and advises that the caregiver brief everyone else the ill person comes into contact with, so that others can join in with the dementia drama. Caregiver, family, friends—all should agree to leave the world of real time and engage with an individual's "dementia reality," another dimension in the present where time is more bendy than even theoretical physicists imagine. Could the ballad of Morris and Nancy be rewritten as an illustration of the truth of this? I doubt it. Not only because my experience of Nancy's dementia convinces me that memory loss was just the start of her story, and not the story itself, but because Nancy wants only very rarely to engage with her past self. She's much more concerned with grappling with the present, caught at the meeting point of the two worlds, a place at which neither past nor present makes sense, and that seems insoluble.
In terms of keeping her occupied, I'm a burnt-out case. Nancy doesn't remember making pastry/collecting eggs/listening to Beethoven; not a trace of it survives the subsequent five minutes. Is there a point to having experiences we don't remember? Yes, probably, in terms of mood. But Nancy's mood is no longer improved by my giving of myself. Quite the reverse. That being the case, I have justified my withdrawal. There are other demands on my time. Children and work, preeminently, both of which have been neglected in the last eighteen months. But it's more than that. I'm bored with Alzheimer's. I'm bored with her decline. I'm bored with being yelled at. I'm bored with dealing with it all.
Unfortunately for Nancy, this is the last thing she needs. What she needs is twenty-four-hour, one-to-one love, and new ideas about what she might do in the great and engulfing boredom that comes from not remembering. The problem of keeping her happy is desperate and paradoxical. Her presence in the moment, where she now lives, is one of constant questing for something: something to do, the _something_ that she should be doing. The last thing I tried to do with Nancy was over a month ago. I took her into the dining room and asked for help with setting the table. She couldn't put the plates out. She couldn't put the cutlery at the sides of the plates. She ended up throwing the forks on the carpet. I think that might have been the moment at which something in me gave up.
DOWN. MID-NOVEMBER.
I've woken in the morning lately feeling heavy. It's hard to get out of bed. I can't face the day. I surprised Chris by saying to him, "Life is just work, isn't it, hard work, unrelenting and hard. I'm beginning to wonder if it's worth it."
I feel as if I might be vanishing. My physical weight might just be imaginary.
"All I do is service other people's needs," I say to him. "I don't think I really exist. Not as something independently. Just a function."
I start thinking about being dead. Not thinking about it, but imagining, fleetingly, what it would be like not to be here. I have transient longings for life to stop, but what I mean is This Life. It's just a way of despairing. The idea of life really stopping—a terminal illness diagnosed and definite—would be quite a different matter. There'd be constant pleading with a hoped-for deity, then.
I am beginning to find it difficult to make housework happen. Physical objects are becoming more powerful than I am. It's a huge effort to rearrange them in the usual everyday way. Cups and books and laundry have become uncooperative. I am not any longer sure that I can bend the world to my will.
I am no longer a person who speaks freely. There's a lot that can't be said and plenty else that can't be said in the public spaces of home. I have secrets, though I'm not sure what they are. The framework of secret keeping is present but not the content. No one who lives here is frank any longer. Unsaid things become a kind of pollutant. The things that some people won't say. The things that other people daren't.
I begin to be convinced that I have early-onset Alzheimer's. I'm losing my memory, groping for words. I forget easily, can't seem to hold things in mind, need to make lists, have become mathematically illiterate. I begin to understand what a train of thought is, the sequential synaptic journey from one carriage to another. I lose my train so easily now, grasping for a foothold; if I could only get back onto one carriage along the sequence, I'd be able, surely, to revisit the whole train, but it's not happening. I think about things and think about them and develop them and tell myself I'll remember and on the way to find paper and a pen they're gone. There's a name for this syndrome but I only discover it later. Caregiver's dementia.
I write long e-mails. Some of them are sent, though most are severely edited. I reread before pressing Send and think, What _tedious_ self-pitying drivel; you can't burden your friends like this. Some days I do burden them and their kindness in responding is almost unbearable. It can't be done every day. It's boring for people to hear it and keep hearing it. Not when there isn't any resolution possible other than death of the aged dependents. Nobody could be so crass as to hope for that.
And I mind people knowing. I don't want pity. I think of my state of mind as failure: Looking after aging parents is a normal fact of life for millions of people, after all. And it's the right thing to do. It's impossible to argue otherwise. Life has a circular shape. First we are helpless and mothered. Later, the mothers are helpless and mothered in turn. That's how it works. It's ungrateful, selfish to abstain from obligation. It smacks of the worst kind of individualism, of duty overthrown by the will. Me me me.
Gandhi said, "The best way to find yourself is to lose yourself in the service of others." Which is ironic, because I begin to feel an indefinable loss of substance. (It isn't, of course, ironic in the least. I refuse staunchly to lose myself and thus will never experience Gandhi's revelatory self-dissolved identity.) I'm becoming intolerant. I feel bad about my tolerance levels dropping. Protest is a source of shame. I should put up with verbal abuse because the abuser can't be held accountable. If I don't tolerate it, if I shout back, then I know that I have put myself into the wrong. I put myself in the wrong more and more. I'm bad-tempered and demanding. I'm sarcastic with Nancy, impatient with Morris. I become sour with the children, with Chris, with people on the phone.
All this has physical repercussions. I eat the wrong things, drink to excess, put on weight. Exercise feels like it will take too much out of me. There isn't enough of me left to play with the children or ride a bike. Taking the dogs to the beach feels like a major undertaking. Coming back is exhausting, my heart racing and thumping up the casual hill. I collapse on the sofa and sleep.
"It feels like a life sentence, this task we have taken on," I say to Chris, and then, in a silly accent, "We should not have made this bargain," a _Star Wars_ joke. It's important to go on making light of things.
How to bring up the subject of permanent residential care? It's not easy when your relationship has dwindled to the surface dwelling and pragmatic; even less so when your relationship has _always_ been surface dwelling and pragmatic. Morris doesn't want to face the inevitable. Big subjects are shrugged off. Attempts to talk about the future are repelled: The future's the future; it will come and then we'll see, won't we?
"Can we cross that bridge when we come to it?" he says, exercising his most boyish smile.
The thing is, he'll never come to that bridge—not in his own mind, anyway, even though I can see that we're standing on it right now, and have been for some time.
I read and reread all I can find about dementia. The grim books are piled by the bed. What I'm looking for, though I don't admit this even to myself, is reassurance that this will all soon be over, that we are entering end game. I want to see it in print, that we are approaching the beginning of the final phase, and that somebody soon will see that Nancy needs to be elsewhere. "Once the turds are in the cardigan pockets and she will only eat biscuits...." is how this craved-for paragraph will start. But I wouldn't wish the final stage on anyone, not least my mother-in-law. What I want for her, really, is to die peacefully at home in bed before getting to that point. That would be the best thing for her. And for those who love her. Her life seems to have become a prolonged form of suffering. "You wouldn't leave a dog in that state," Nancy's voice says from the past. Among her papers in the desk, we find a yellowing membership pack, at least twenty years old, from the Euthanasia Society.
IT'S OFTEN SAID that people don't die from Alzheimer's, but from complications arising. It's true that many dementia sufferers die of the conditions that overwhelm the old: blood clots, stroke, water infection, pneumonia, or blood poisoning from infected bed sores (Auguste Deter died of this). But people do die of Alzheimer's. If patients live long enough, brain atrophy will get them in the end, the brain stem under attack, the body-maintenance circuits going down like city zones in a power failure. It makes the hairs prickle on the back of my neck, watching the film _2001: A Space Odyssey_ and hearing the quiet panic of HAL, the onboard computer, talking as he's being dismantled and saying, over and over and in a determinedly quiet and rational way, that he can feel his mind going, and then admitting that he's afraid.
The Internet is thoroughly trawled, in search of camaraderie. Camaraderie at arm's length. I'm looking for others who wish their loved ones would die and feel degraded by their hope. Instead, I find the opposite. I come across the blog of a woman—let's call her Marigold—who has decided against institutional care for her Alzheimer's-stricken husband, and has kept him at home throughout a terrible final stage. She reports going through a phase of profound doubt about her decision. But then her story takes an unexpected turn, about which she's evangelical. She's found a way through, by embracing her role as caregiver, and, unlikely as it seems, finding joy in it. She sees her role as a privilege, almost a sacrament. It's become a spiritual awakening for her. I read all this avidly but fail to be convinced. I find myself agreeing with her friends, the ones who plead with her to find residential care and not let two lives be spoiled when only one need be. That's how I am now. Hardnosed about numbers. A utilitarian.
People ask me how I am. "Oh fine, fine, though Nancy's hard work," I tell them. This is the person I've constructed: the cheerful coper. A forgery.
Sydney Smith, the essayist, farmer, and founder of the _Edinburgh Review_ , wrote a letter to his friend Georgiana Morpeth in 1820, advising on a twenty-point plan for dealing with depression. "Always take a short view of life—not further than dinner or tea," he advocates, which is good advice, at least if somebody else is cooking. He also suggests that she live as well as she dares, take tepid baths, get as much exercise as possible, and see people who amuse her. "Avoid poetry, dramatic representations (except comedy), music, serious novels, melancholy, and sentimental people," he writes. He tells her to confide in her friends. "Low spirits," he says, "are always worse for dignified concealment." He adds, "Don't expect too much from human life, a sorry business at the best." Nothing is said about vodka, but I suspect he may have disapproved.
I disappear whenever I can into a book, taking solace in other lives and others' eloquence. I am hungry for proxies. I become particularly keen on people in trouble. Biographies of the besieged, bankrupted, and maritally abandoned are particularly welcome. I have repetitive, variant dreams about being trapped in buildings. I try to negotiate broken stairways, stairs that turn into steep ramps or ladders with rungs missing. I need to escape out of windows onto ledges, down onto lower roofs, walls, slipping unseen into dark gardens. I'm chased by faceless, unknown enemies, from whom I must hide.
I am supposed to be working when the caregivers are here. I try. I give every impression of working. But it's all done without breaking the surface of imagination. I'm one of those water boatmen whose long feet straddle the top of the pond, indenting it like a skin, deep water stretching away beneath. I can't seem to go beneath the surface of the novel anymore. It occurs to me that perhaps this is how I am now, this is what I am, and what I will be when caregiving is over with. I've changed for good. I'm no longer a writer. Marigold's transformation has taken place, a darker version of that, inverted, its subject ungrateful and in revolt, like one of Milton's rebel angels.
I go out onto the headland, getting as close as I dare to the cliffs, which are bronze red and steeply raked, the sea crawling up them with agitated gray fingers. I go down onto the beach, enduring the wind's ranting and roaring, showering me with stinging sand, and sift through stones on the shoreline, looking for something perfect and lovely. Tennyson comes into my head.
_Strange, that the mind when fraught
With a passion so intense
One would think that it well
Might drown all life in the eye,—
That it should, by being so overwrought,
Suddenly strike on a sharper sense
For a shell, or a flower, little things
Which else would have been passed by_.
Time moves very slowly with Nancy, unendurably so. Empathy takes me into her world and I don't want to be there. Wherever she goes, fear goes with her. How will it end, this hideous ticking-away day? There is no relief. She has begun to be severely carsick and throws up even on a trip to the village. The caregivers are housebound with her. The weather's increasingly stormy and she paces like a caged cat, growling at the world outside her bars.
# Chapter 27
_One's real life is so often the life that one does not lead_.
—OSCAR WILDE
I'M BETTER THAN IN NOVEMBER, BUT I CAN'T SEEM TO stay in a good mood, or in any mood. The alarm clock, set to local radio, switches on at 7:30 every morning, and the room fills with dread at the prospect of the day. Chris and I lie in bed listening to the presenters in the dark. Cattle prices, sheep sales, council controversies, travel news, sporting and artistic triumphs, lengthy descriptions of lost and found cats. The wind howls round the house. It's black dark when the girls go to the end of the drive to meet the high-school transport.
I'm beginning to feel afraid, though it isn't clear what there is to fear. That I won't be able to do it anymore, perhaps. That Nancy will hurt one of the children and that I will hurt her. That she sees through me, my plastic attempts at love. That this is a test of character that I'm failing, D minus. That I will say something to Morris I will always regret. I'm irritable with him and his apparent not caring. I'm having thoughts and feelings of which I'm ashamed. Dislike. Resentment. Regret. Things a caregiver isn't allowed to feel; our moral relationship isn't individual, but universal, cultural, social. Morris is so far in denial now that he no longer registers Nancy's behavior as anything unusual. Oblivious, he affects puzzlement if I bring the subject up of his taking more of a role in entertaining and watching over her. He sees a magazine piece I wrote about our lives with Nancy and is shocked by it—not because I wrote it, but because, as he tells me, he'd no idea that things had got so serious and gone so far.
December contrives to be both vile and uplifting. The weather's atrocious but Christmas with children is a guaranteed solace. I am busy and the calendar fills up and I find, on some days, some half days or half hours, that I feel almost normal. The downward spiral is also an upward one. That's how spirals are.
I like the peninsula Christmas, the modesty and gusto of its series of concerts and events, its precommercialized spirit. The village lights are unshowy in primary colors. The tree in the square is tall and twinkly, and held down by guy ropes so that the wind can't take it. The official village tree lighting is preceded by the Salvation Army, at length. There was no music last year (we gathered, the lights were switched on, we funneled into the hall for mulled wine), but this time the band swings into action with a program of carols to get through, out in the square, and we're all caught out, in thin coats and hatless, shivering as we sing along.
Nancy has a new friend. She finds her, unexpectedly, standing at the dogleg from rear passage to hall. An expanse of wall there has been enlivened by an Arts and Crafts mirror, a large rectangle with a carved oak frame. Nancy finds her new friend here, three-dimensional and in color, backlit by the glass outer door. It's shocking to discover that she no longer recognizes herself. A year ago, compulsively washing her hands (this urge has passed), she would have a good look at her reflection while doing so, adjusting her hair and muttering her displeasure at being so dilapidated. The year before that, she could still be funny about it. "God, but you're ugly," she'd say, laughing. Self-recognition is a major hallmark of consciousness. Chimps, dolphins, and apparently also elephants recognize their own reflections. Elephants are the newest additions to the list: an experiment in 2007 at a New York zoo found that once they got used to the mirror, they'd use it to have a good look inside their own mouths. Nancy's loss of self-recognition is, it seems, to do with severe right hemisphere damage, right frontal lobe damage.
Not that talking to herself in the mirror is alarming in itself. People talk to themselves all the time, with or without a reflection on hand. When you think about it, this is rather odd behavior. Who is it that's talking and who listening? Perhaps it's simply and unmysteriously true that we're all two people, two in one. One of us, the actor, is out there in the world, interacting and reporting back, doing things and saying things, out on a limb, a free agent. This self might behave badly, be easily led, go astray, come back with ludicrous notions or shameful confessions. The other one of us is deeply embedded, the sum of everything we know, and thus is infinitely wiser and more cautious. That's the editor. While the actor's out shopping, the editor stays at home in the mind, and makes judgments.
"You were so stupid to buy that jacket."
"Shut up, it was half price."
It's undeniable that there's a dialogue going on.
People refer to this internal double act all the time. "I'm not myself today," they say. "What was I going to do?" "Why did I do that, why do I do these things?" and my personal philosophical favorite: "What do I think I'm doing?"
Sometimes, we're hard on ourselves. We speak of ourselves as dual creatures: self-knowledge, after all, requires a self and a knower of self, which seems to leave the field wide open for Team Descartes. The Cartesian view isn't needed, though, if you accept that consciousness and self aren't strictly equivalent; that self extends beyond and below what we know of it. Aristotle said, "We are not able to see what we are from ourselves." We do what we can. Polonius, in _Hamlet_ , tells his son, "[T]o thine own self be true, / And it must follow as the night the day, / Thou canst not then be false to any man." _I know you better than youknow yourself_ is probably the most irritating thing anyone can tell you (other than stuff about Aristotle).
It's universally agreed that having a dialogue is better than having a war and perhaps that's the point of the inner conversation. Different parts of our selves, instinctive and rational, conservative and liberal, get to debate things. That's how the mind seems to work. That's how information is presented and assessed, teased out, opinions formulated and actions decided upon. The editor doesn't always triumph. Sometimes other parts of the brain win the argument. The gut instinct, for example, which is delivered via a red phone from the limbic system.
"This man's trouble."
"Nonsense. Just because he has tattoos."
"This man's trouble, shut the door on him."
"He showed me his card, don't be so paranoid."
"He's trouble. Look, I'm shutting the door if you don't." Slam. Locks are shunted into place.
"Well, I hope you're happy now because you look like an idiot. He was from the electricity board."
"He was a fraud. Couldn't you sense it?"
"Ridiculous."
It seems sometimes that Nancy is traveling through what survives of her life asleep. Life is so odd, so unaccountable, so disengaged from reality to her brain-damaged perceptions, that it might be like being in a dream. The once Amazonian-sized forest of nerve cells and axons and synapses, its millions, trillions of connections, seems now to have reduced to just a few well-trodden tracks through a wood, a few broad footpaths that have been worn into deep ruts. The rest looms dark and unknowable. Things seem out of control, bizarre, to her. People around her look familiar but unfamiliar—I think sometimes that it must be like a constant process of déjà vu. They make statements that cannot be true. She can't convince the people in the dream that her life is elsewhere and that they are all, all of them, engaged in the joint hallucination that takes place through the looking glass. The dream goes on and on, for months and years, and there's no waking up. Jung wrote that it's likely we continually dream, but that consciousness makes so much noise that we're not aware of it. What if Nancy's consciousness has stopped making much of a noise? Is it a kind of waking dream that takes its place?
NOW THAT SHE'S found the woman in the mirror, Nancy's talking to herself for much of the day and her mood is miraculously lifted. I find her there one afternoon as I come in from outside. I come up behind her.
"Hello!" I say. "Looking in the mirror again? How are you looking today?"
"Here's my friend!" Nancy says, gesturing toward herself and looking absolutely delighted.
"Hello, there," I say to the grinning reflection. "And who are you?"
"She won't tell you that," Nancy says. "I keep asking her to come in, but she won't."
"How's she going to come in?"
"Through the door," Nancy says. "Through here." The mirror is a doorway. She takes a step backward and her arms are raised, beckoning. "Come on then. Come on. Come in for a little while. Won't you come in? Because I'd love it if you would."
"This is a mirror, isn't it, Nancy?" I say. "Look." I knock on the glass. "It's a mirror, and that's you."
Nancy looks at me as if I'm really idiotic. "I know that. I know that. Do you think I'm stupid?" She laughs at her reflection and it laughs back. "Look! Look! She's laughing at you," she says.
"But that's you," I say. Why can't I let it go? I don't really know. Perhaps it's to do with being a mother, this habit of correcting people's misapprehensions. Or perhaps it's something worse.
"That's you, isn't it?" I say, waving at her in the mirror. "Look. I'm waving at you. See, here's my hand waving"—she looks briefly at my hand—"and here's my hand again, waving in the mirror."
"That's my friend," Nancy says.
"Why don't you wave, too?" I say.
The woman in the mirror is frowning.
Nancy's face falls. "She doesn't like it," she says.
"Okay, then." I am conceding defeat. "I'll leave you two to chat."
When I pass by again an hour later, seeing to laundry, Nancy's still there, chatting away to the mirror. And she's smiling, laughing, giggling with her friend. I go into the kitchen and make a pot of coffee and sit staring at the cup. Nancy isn't always unhappy. Nancy still has her moments of fun. She's getting something out of life. She doesn't always hate you. She doesn't hate you at all. What on earth made you think that? How could you be so deranged? And how—oh dear god, this is appalling—how on earth could you wish her dead?
WE DECIDE NOT to do any more bed-and-breakfast, a decision sparked by discovering that one of our guests this year has stolen quite a number of DVDs. We're not sure how many. We only cotton on because Chris asks if anybody knows where the Humphrey Bogart films have gone—all of them are gone, it transpires—and then we discover more films are missing, another twenty or so classics. The idea that somebody friendly, somebody who wrote admiringly in the visitors' book, put thirty or more DVDs in his suitcase, shook our hands and thanked us very much again for the fantastic weekend, and drove off with our stuff, is fatally dissuasive.
By mid-December, I have a permanent sharp pain in my head and neck and the doctor diagnoses tension, bad posture, a trapped nerve. He prescribes a muscle relaxant (tranquilizer), which I daren't take. Jack is ill with one of his epic bouts of tonsillitis, and in mid-December he's admitted to the men's ward at the local hospital—there's no children's ward—to put him on a stronger regimen of drugs and monitor him. I don't sleep. I sit downstairs in the in-laws' sitting room, warmest in the house, at 5:00 A.M. with herbal tea. Nancy is up and wandering the halls. She comes in the sitting room door, then goes out through the kitchen saying, "Well, you're no use, are you? Typical, typical," and puts herself back to bed. We go to the hospital in the morning and find Jack up, dressed, playing PlayStation. They've found a heart murmur and will want to see him again. He's discharged and comes home. Purple and white and gray.
Jack, the trapped nerve, the stresses of the night shift, anxiety about making the children's Christmas happy—evidently it's all too much. December's recovery turns out to be veneer. One day when Nancy has been making the perfectly routine complaint that I'm the only person that's nice to her, but that the children who live here are nasty and call her names, my facade suffers a small additional crack. Small but structural. I leave her sitting on her bed and go to find Morris, landing in Nancy's chair with a thud.
"I'm at the end of my tether. I can't stand any more. I can't. Stand. Much. More. I mean it," I say. Morris looks appropriately alarmed, which is to say that he looks just the same but his eyes are wider. I seem to be having a breakdown, right there in the middle of _Cash in the Attic_. "Why don't you respond? Say something!" I tell him. "Are you listening? I'm telling you something important, Morris. I can't go on with this. I'm at the end of the road. Do you understand me?"
"Yes. I understand you," he says, looking at me as if he's the bank teller and I'm the madman with the gun.
I'M ON THE Internet a lot, finding refuge from the too-specific gravity of life in the weightless world of e-mail, which floats free of consequence. But Alzheimer's has its teeth in me and before long I find myself trawling dementia forums. There's a lot of guilt out there: blame, self-blame, and confused thinking. A physician in the _New York Times_ remarks that, though American citizens believe that the modern generation of elderly is being dumped in vast numbers in care facilities, the reality is rather the opposite, with a huge percentage being cared for at home. I think about that word _dumped_ all day. All across the World Wide Web there is praise for those who keep caregiving in the family, and the widespread assumption that opting for residential care is a kind of failure, only mitigated by personal circumstances. Rita Hayworth's daughter, interviewed about the care of her mother and whether she'd considered an institution for her says no, never, not even at her worst, and the interviewer hands out the appropriate admiration. Even I, the battered soul and incipient alcoholic, whose blood could be used to inoculate others against the taking in of parents, can see that hers is the right answer, the honorable answer. Recognizing this, people agonize on the dementia forums about whether they can go on, looking for permission from their peers to capitulate. Among caregivers who have capitulated, who've gone for the nursing home option after years of keeping loved ones with them, there's almost unanimous self-loathing for giving up. The words _giving up_ are used a lot. That and _dumping_.
Dumping. Do we _dump_ people in hospitals when they're ill? Is that the language used? I worry that we're all confusing a physical disease with natural aging, believing that we ought to be able to contain aging and death within the family, recognizing the failure and stigma of doing otherwise. We confuse dementia with old age, and it's a moral given that old age oughtn't to be punished by exclusion; put in those terms, there's no argument.
I sit close to two girls in a coffee shop. One says to the other, "Is it true your gran has gone into a home?"
The other one nods. "My mum said she couldn't cope with her anymore."
"God, your poor granny, those places are terrible."
"I know. I'm so angry with my mum for putting her in there. We went to see her and she was crying." She's blushing a deep red.
There's a strong whiff of shame about parents going into a nursing home. I worry that the medical profession colludes in this. I see them colluding all over the Internet. One American woman reports that her Alzheimer's-suffering mother, at about Nancy's stage of the disease by the sound of things, has four doctors and a therapist, all of whom have agreed that putting the mother into residential care would be "like killing her."
The weekend before Christmas we have our party. Nancy and Morris go into respite at the private home for the weekend and appear to enjoy it. Nothing is said about staying on, this time, and there are no sales calls. We have two hundred guests, a magician, a movie-and-pizza splinter group upstairs for those under four foot six, and tipsy teenagers gathered round the pool table.
"It looks pretty likely that Nancy and Morris will go into residential care next year," I find myself telling people when they ask (and everybody asks). I don't seem to have any other form of conversation. I seem to feel the need to brief everybody there, individually, about the situation. I'm properly defensive about the reasons.
"And if they do go into care, will you be moving away?"
That's what everybody wants to know. Will we be staying or not. There's a strong chance we won't be, but I hesitate to admit this.
We're an odd assortment of souls, gathered here together tonight. Most of us are _incomers_ (less charmingly, _blow-ins)_ , who have come to live on the peninsula from the outside—outsiders and not locals. The sheer intrepidity of incomers is impressive, like the organic farmer over the hill, who persists in trying to grow vegetables on an economic scale in these weather conditions, despite constant setbacks. Some people I've met came because they're artists and because it's relatively cheap to buy an artist's house on the shore; many are here because of house prices. There's a lot of sea view, a lot of fresh air, a lot of unspoiled wilderness on offer per pound spent. And very little crime. People don't lock their cars in town. People don't lock their houses. But once people are installed, many of them become possessive of the place. They don't like to hear it criticized, and leaving is seen as rejection. They're openly perplexed by people not staying on.
"Don't tell people you might not be here more than two or three years," somebody says to me at the party. "They won't bother to get to know you."
"It's irritating when you make friends and then they leave," someone else tells me. "You invest all this time in them and then they're gone."
The occasional local die-hard grows donnishly disapproving. "Why come here if you don't like extreme weather, though?" they ask, in a tutorial manner. "But why come here at all if you didn't intend to stay?"
Living here can be a trial of strength. That's one way of looking at it. The challenges of meteorology, of isolation, of making a living are looked upon by long-established residents as a test of true grit. "Naah, they left after eighteen months, they couldn't hack it," they say, at one of the many village socials, dismissively of some poor soul. It's important to show that you can hack it, that you relish hacking it, that you're man or woman enough.
"It's only a force nine, what are you talking about; that's just for drying your washing." "Call that winter? That was only a shower. You just wait." As Sydney Smith observed: "No nation has so large a stock of benevolence of heart as the Scotch. Their temper stands anything but an attack on their climate." But the joke's on Sydney Smith. Almost everybody I know who feels this (on occasion quite savage) defensiveness of the superiority of _here_ and the inexorable decline of _there_ is English. Adopted Scots.
NANCY AND MORRIS come back glumly from their weekend, and remain glum over the holiday. On Christmas morning we gather in the drawing room for the children's present-opening ritual, something my in-laws would once have enjoyed, despite carping about the overgenerosity, the waste of wrapping paper. Nancy is kept busy with a tin of gaudily wrapped chocolates and Morris is silent. At Christmas lunch Morris picks at his food in silence and Nancy is occupied trying to eat gravy with her fingers. Afterward, I give Morris the option of an afternoon by the TV with a box of chocolates, and this is gratefully taken up. The two of them sit by a roaring fire in paper hats, eating truffles and drinking from the various bottles I put on the tray, flicking between Christmas Day programs, and seem almost jolly.
After Christmas, friends come to stay for New Year's Eve, another family of five. One evening while the ten of us, crammed round the kitchen table, are having supper, and Nancy's supposed to be eating hers with Morris, she appears at the door, hands on hips, nodding slowly at us all, her face bright red, veins pulsing in her neck.
"Nancy! What can we do for you? Have you finished eating?"
She stares at me and then says, "So, you're all still here, then?"
"Er, what?"
"I said. I said so you're all still here, then. You're all still here."
"What are you talking about? We live here. And these are our friends, visiting for New Year."
"I'm not having it. You all have to go now. Now. I mean it. Out. I said now."
I bundle her back into her sitting room and close the door. I am angry.
"Now listen to me. Don't you dare embarrass me in front of our friends. I've had quite enough of your mouth lately. Stay in here just now. Stay here. Stay put," I snap at her, and then I return to the kitchen, aware of a subtle shift in mood. I'm embarrassed by my own reaction as much as by Nancy's rudeness. Everybody heard me shouting at her. Various explanatory sentences are born and die in my head, particular, and then at last general: None of it makes any sense if you haven't been here and lived through it, though saying as much sounds trite. I shrug it off, as I have learned to. Mr. Bennet in _Pride and Prejudice_ flashes into my mind, chastened by his part in Lydia's disgrace, and feeling, despite Elizabeth's soothings, that morally he ought to suffer. "I am not afraid of being overpowered by the impression. It will pass away soon enough." And even while I'm thinking this and feeling bad, I'm also feeling grateful that I have this store of associations on hand, and the healthy brain that delivers them up.
On New Year's Eve, Morris expresses a positive disinclination toward staying up late. They'd rather go to bed at the usual time, thanks, he tells me. We drink a lot, have champagne at midnight with a dozen or so neighbors, watch the fireworks going off out on the headland, and wish each other, with greater sincerity than is ordinary, a very happy New Year.
# Chapter 28
_This long last childhood
Nothing provides for.
What can it do each day
But hunt that imminent door
Through which all that understood
Has hidden away?_
—PHILIP LARKIN
THE NEW YEAR BRINGS NEW DEVELOPMENTS. THE FIRST of these is that Nancy begins to declaim. She's a mobile declaimer, addressing herself to each of the rooms she walks through. As she walks she makes three statements. She hasn't spoken to me directly since Christmas. Instead, she has fixed on three repeated lines:
"And I will never be.
"And I will never know.
"And I will never be again."
If I'm in the kitchen when she passes by, she doesn't seem to notice me. She looks straight ahead. "And I will never be."
"Hello, Nancy," I say. Even when I address her she doesn't look toward me.
"And I will never know."
"Just having a walk? Morris's through that door there, straight ahead, if you're looking for him."
She goes up the step and rattles the handle of her sitting room door.
"And I will never be again."
When she gets tired, she goes and sits by Morris and needles him.
"I've been waiting for you for twenty-five years!"
"It's forty-seven years, actually, that we've been married," Morris corrects her.
"I've been waiting for you and you haven't said anything to me."
"What do you mean? We spend all day together. We talk to each other all day."
"You haven't said a word. Not a word. A real word, I mean, and not one of the other ones."
"What are you talking about?" he bellows. "I'm always here and you're always here and we're always talking."
Much later, passing by their door, I find the same conversation's still going on.
"You never talk to me. I sit here and I talk and you don't answer," Nancy's saying.
"That's rubbish," Morris says emphatically. "That's total rubbish. Think before you speak. Think what you're saying because it's rubbish and you know it's rubbish."
"I talk and I talk and you don't listen."
"Shut up! Just shut up, will you," he cries.
"Don't you dare to tell me to shut up."
"Well, be quiet then. I want to watch this TV program."
"I want to watch it but you won't let me."
"What do you mean? You're sitting right in front of it."
"You won't let me do anything."
"Can you just be quiet so I can watch it?"
"I didn't say a word."
SHE'S DISCOVERED THAT the mirror in her bathroom also has a friend in it. She goes there in the evening and talks to her reflection in the moonlight.
"Oh yes, and I always said so. I said that about you but nobody believed me. That's what happens, though. To me, I mean." She pauses as if the reflection is speaking. Perhaps it is. "Oh my goodness, yes. You're quite right and no mistake."
"Hello, Nancy, what are you up to in here in the dark?" Chris asks amiably, putting the light on.
He doesn't often call her Mother anymore. She won't answer to it, might query it, might want to make declarations that are best avoided.
"That's my friend," Nancy says, smiling at herself. "I only have one friend and that's her."
BLACK SUNDAY. NANCY'S in a state of perpetual rage. Jack is threatened. He has the temerity to touch the dog in her presence.
"What are you doing that for? Get out of here. Do as you're told."
He leaves the dog and goes to stand by the fire.
"Get out of here, you little bugger."
"What? Why, Gran?"
"Come closer to me and I'll get you. I'll kill you. I will. I'll kill you. I mean it. You filthy little bastard."
We don't talk to her about these outbursts anymore. There's no point and everybody gets pointlessly upset. Morris pretends he hasn't heard and Nancy's determined she didn't do anything wrong. She'll be difficult for the rest of the day, if she's told off. The reason for the telling off doesn't register. Nothing is learned by it and nothing is gained. If she isn't told off, chances are her rage will subside pretty quickly. So, strictly in terms of the balance sheet, it's better to ditch the moralizing. Though this is difficult to explain to Jack.
Like an anorexic girl finding power over her mother in not eating, Nancy begins to decline food, any food, whether left out for her to forage in the kitchen or offered on a plate. A fish pie with a mashed potato top, served to her in a bowl with a dessert spoon, is rejected untasted. I go and kneel by her chair and try to spoon some of it into her.
"I don't like it! I'm not going to eat anything if I don't like it!"
"You need food, though, Nancy. Usually you love fish pie. It's got lots of cream in it, and prawns. Just try it. Just have a bit."
She takes a spoonful from me then talks with her mouth full, spitting haddock. "You've given me far too much! Ask them and they'll tell you straight. There's too much in my mouth."
"Just stop talking and eat it."
She chews and chews, looking pained.
I offer another spoonful. A protective, shielding hand goes up, her fingernails an ominous dark brown.
"You've got to eat something or you'll get ill."
"Don't make me laugh."
"No, I mean it. You can't live on biscuits. You need some protein and some vitamins."
Nancy's head goes back disdainfully. "No no no. No, they don't. That's stupid. You don't know what you're talking about. You really have no idea about anything or any education."
"I mean it. You need some real food or you'll get poorly."
Her hand is slammed on the dinner tray.
"Well that's _not_ what they do in Edinburgh."
"Perhaps you should go back to Edinburgh, then, where you could eat biscuits all day."
"Yes. Yes. I'm going back tonight."
Morris mutters something that I half hear.
"Is that what you want?" I say to him. "I can arrange an Edinburgh residential home if that's what you'd like."
"I'd go tomorrow if I knew where to go," he says.
* * *
NEXT, THE WHISPERING starts. It's curiously disconcerting, this whispering. Nancy talks to herself under her breath all day and for much of the night, rehearsing imagined wrongs. Almost all of what she has to say begins with "she." The whispered undertone follows her, precedes her, announcing her arrival at the half-opened doors of other rooms. It's difficult to make out what's being said unless you're up close to her face. I find her early one morning inserted tight behind the wide-opened door of the day bathroom, pressed hard between the wall and the door, a length of toilet paper held up to her chin, and only know she's there because of the whispering. "She can't and she won't, it won't be like that, I'll find it again, I'll take it there, and there will be the end of it, and then they will come, and I will tell them, and they will be glad, and I will be there again, and then I will come home, come here, or not here, where is here, I don't know, and then we will know, we will all know, and I will be right, and she will be wrong."
Then Nancy stops washing or wanting to wash. The caregiver arrives for the Monday morning session and finds that she can't get Nancy in the bath. The bath is run but Nancy won't get in it. Nancy gets her way. The caregivers feel that they can't pressure clients into being clean if they don't want to be. I step in. I pressure without a qualm.
"Come on, Nancy, time for your bath."
"I'm not having a bath. I don't need one."
"You are. You smell."
"I do not. Don't be ridiculous. I never smell."
"I hear what you're saying, Nancy, but unfortunately you're going in the bath anyway."
"No, I'm not." A little scream. A foot stamped hard.
"Yes, you are. I'm not taking any nonsense from you about this, you have to have a bath every now and then, and you are beginning to smell bad."
"It doesn't bother me so why should it bother you?"
This is actually a really good question and surprisingly sophisticated in the current scheme of things.
"It bothers me because you smell and I have to look after you," I tell her. "It bothers me because you are making the house smell. And you will get ill if you stay dirty. So come on. None of your nonsense [historically, a favorite child-chiding phrase of her own]."
"You're NOT LISTENING." She's shouting now. "I'm NOT GETTING IN."
"Yes, you are. Get your clothes off. Get in the bath. You're filthy. Your underwear is filthy." Inspiration strikes. "Everyone can smell you. They will talk about you and say how dirty you are."
As ever, alluding to what the neighbors might think does the trick. She starts to take her sweater off, kicks off her shoes.
"Well, all right then, but I'm not happy."
Once she's in the bath she loves it. She starts to sing, war-blingly.
"When all the men are dead now, and the world has come to me, and the way I bring home and the sort I do then, and it's the same for me...."
She plays with the bubbles, purrs when her hair is washed, and is reluctant to get out. And she can still rhyme.
Eating problems escalate. Like a choosy toddler in a high chair, she clamps her lips shut and then her eyes and turns her head away from the spoon. The Battle Royal of the Baked Beans is typical. When she's refused meals for more than twenty-four hours, beans usually break the fast. But not any longer.
"I'm not having it! I'm not," she cries, jumping to her feet, throwing her tray across the room and exiting. I find her in her usual retreat, talking to her bathroom mirror, a stray thread of moonlight reflecting off one eye.
"And she says the same; always the same bloody lies...."
Then she sees me. "And what do you want?" Her most imperious tone.
Later, after she has consented to toast and jam, eaten a quarter of a slice and passed the rest to Morris, I find her in the corridor.
"Hello, Nancy," I say cheerily. "How nice to see you. How are you?"
The Book insists that a caregiver's tone is paramount.
She stares. "I'm not. Speaking to. You."
Chris appears and takes her by the hand: "Come and find Morris, come on," steering her through the kitchen. I go into the hall and _bouf_ , there's a small explosion. Chris, renowned for not losing his temper, has lost it and is yelling. "Don't you dare, don't you ever, ever call my wife a bitch again!" I go into the kitchen and make a vodka tonic and hear them at it through the door. I'm thinking that I'll go in and change the subject, offer whisky, get Chris out of there. But Chris is in full flow. He is talking, and then Morris, and then Nancy, and all of them calmly, taking their turn. A most bizarre half hour ensues in which Chris and his father talk Nancy through her recent behavior.
I hear Nancy responding in her shrill defensive voice. "What have I done to anybody? Nothing, nothing at all."
MORRIS: You've been very rude to people and you're upsetting them.
NANCY: When have I been rude to anybody? I wasn't rude. Who told you that?
CHRIS: Nobody told me, Mother. I was there, standing right next to you. You called my wife a bitch and it isn't the first time and it has to stop.
NANCY: I've never done anything of the kind. I've not used that word my whole life.
"Why did you bother?" I ask him when he emerges, having enraged Nancy into sulking and silence.
"No point at all, not for her, but it was good for my father. He got to air some recent grievances."
This is true.
"You've been very rude to me, too, and sneering; you sneer at me and I don't like it," Morris told his wife.
THE DAY AFTER this, I wake feeling certain that I'm at the end of the road. I have to do something. I can't go on, can't physically. My legs are leaden, my heart heavy. I can't face another day. I ring the surgery, and the doctor on duty says he'll call by. He'll reassess Nancy and perhaps prescribe something else. Her drugs may need adjustment.
The drug regimen of Alzheimer's patients is one of the chief bugbears of their and their caregivers' lives. The neurotransmitter breakdown inhibitor that boosts communications in surviving brain cells and at best slows the sufferer's decline, the one that has four manifestations, four brand names: that's the only drug available. Everything else an Alzheimer's sufferer is prescribed is tried out from a menu of drugs developed for other conditions, tackling individual symptoms. That's the best that can be done. Antipsychotics, benzodiazepines like Valium, epilepsy drugs, mood stabilizers, antihistamines, antidepressants, sleeping pills, Parkinson's disease drugs, in rare cases even Ritalin: all might be dipped into, on a suck-it-and-see basis, and every Alzheimer's patient has her own cocktail and combination. Every individual is an individual drug trial. Things are tried, don't work, are adjusted. That's how it is.
"You sound like you're at the end of your tether," the doctor says.
"Not quite," I tell him, "I'm not quite there. But I can see it now, the end of it." It's in my mind's eye, the end of a fat sailing rope, looming frayed up ahead.
The doctor has been in touch with the social work department, and so have we, and a care meeting has been fixed for tomorrow in town.
When the doctor arrives, I take him into the drawing room—respectably tidy, coal fire lit—and go and fetch Nancy. She is civil when she shakes his hand but begins to look suspicious when he sits by her on the sofa.
"I'm just going to ask you a few questions, Nancy."
"If you must you must. But be quick about it." Her disdain is penetrating.
The doctor has the laminated sheet out of his bag, the standard Alzheimer's memory test known as the MMSE (mini mental state examination). Points are given out of thirty. There aren't thirty questions. Ten marks are given for orientation to time and place, three marks for registering three words, five marks for attention and calculation, three marks for remembering three words, eight marks for language, and one for visual construction.
"Right then. Do you know what year it is?"
"No idea."
"What is the month?"
She thinks a moment, shakes her head.
"What's the date today?"
"Haven't a clue."
"What day is it?"
"No idea at all."
"Right. Do you have any idea what the season is? What season are we in?"
She looks blank.
"Do you know what a season is? What's a season?"
She purses her lips and looks straight ahead.
"I do, of course. It's one of those things that's over there, which is to say it's one and two and three, that kind of thing."
"Right. Next I'm going to give you three words to remember, and in a minute I want you to remember them and tell me what they are, okay?"
"What would I want to do that for? I don't want anything to do with your things, it isn't anything to do with me."
He gives her the words, three short common nouns. Ball, car, man. "Can you say them for me? Ball, car, man."
"I'm not remotely interested in that," Nancy tells him.
"Can you repeat this phrase for me? No ifs, ands, or buts."
She stares at him.
"Say this: No ifs, ands, or buts."
She keeps staring.
"Do you know the name of this house?" the doctor asks her.
"No." Annoyed. "And don't ask me that again."
"Where do you live, where is this house?"
"Edinburgh!" exasperatedly. "It's Edinburgh! That's where I live."
"What's the area called? This area we live in?"
"Edinburgh! Are you stupid? Edinburgh! Edinburgh's where I live."
"What floor of the house are we on? Are we on the ground floor, or upstairs?"
"Not a clue." With some satisfaction, folding her arms.
"Can you spell this word— _world_. World. Like the world we live in. World."
"What?"
"World. Can you spell it?"
"No, and I don't want to. What would I want to do that for? All stupid questions! You and your wode."
Nor can she spell it backward, or remember the three words he gave her to remember. The math test is skirted over quickly. It's pointless, really, asking Nancy to subtract seven from one hundred.
"Right. Here's a piece of paper. I'd like you to hold it in your right hand."
Surprisingly, she can do this one and is happy to oblige. "This is my right." She extends her right hand and picks the paper up.
"Can you fold the paper in half?"
She can do this, too. Two points have been earned. She puts the short sides carefully together and smooths the fold crisply and precisely in place.
"Now, put the paper on your knee."
"Where?"
"Put it on your knee."
She leans down to put it on the floor.
"No, put it on your knee."
"This is my knee." She lifts her left knee up and looks puzzled. The paper drops to the ground.
"Right, Nancy. Can you tell me what this is?" (It's a pen.)
"Yes, of course, it's one of those things that's for you, and that's yours, and it's for holding and it goes along there. It's yours, just take it yourself. Why are you asking me?"
"And can you tell me what this is?" (It's a watch.)
Very irritably and shrill. "I've told you already, it's yours, just take it, if you want something you just take it, don't you, you don't ask stupid questions about it, just take it!"
The doctor is sounding properly nervous now. His mouth is dry when he speaks. Nancy's intimidating. Wild-eyed, spittle flying.
"Can you write a sentence for me—a short one, absolutely anything?"
"Like what?"
"Anything you like. A short sentence of your choice. Just a few words. Whatever you want."
"But what do you want me to do that for?"
"It's a test. I just want to see you write something down. Just write one word if you like."
He gives her the pen. She holds it, looks at the paper. Her fingers work their way round the Biro. She pauses and considers. Then she hands the pen back.
"I've got absolutely no need to do that and no interest in doing it, either." High dudgeon. "Why should I do these things for you when you do nothing for me?"
The doctor clears his throat. He holds out the laminated sheet. "Okay then. Can you read this?"
"Where?"
"Just here. These three words. Can you read them?"
She looks at the sheet, at where his finger indicates, for a moment.
"That's a _c_. And another _c_. And that's an _o_. There's a _c_ and an o."
"Right. Do you still have that block of paper? Here's a pen. Can you copy these shapes onto the paper?"
"What shapes?"
"These shapes here, on the sheet." They're intersecting pentagons.
She looks at the pen and at the sheet of paper and her hand hovers. She looks at the sheet and at the pen and at her hand, frowning.
"Why would I want to write that down? I don't want to," she says eventually. "I don't see the point and the point and not that at all."
"It's just a short test. It's over now."
"I know who did this," she says, putting her hair behind her ears. "I know his name. I know why he did it but I'm not going to say, oh no. I'm not telling anything to any of you at any time."
Less than twenty-four out of thirty indicates substantial cognitive impairment, the Internet tells me. Healthy people over the age of eighty should be able to score twenty-five. Nancy scored two. Just two out of thirty.
The Alzheimer's Society says in its MMSE fact sheet that a patient should score twelve or more for there to be any point in taking the dementia-specific drugs. Nancy's galantamine is to be phased out with immediate effect and a new drug given in its place, one recommended by the psychogeriatrician at the city hospital for mood swings and aggression (one we give her for less than a week as it makes her ill at night, comatose in daylight).
The final thing the doctor wants to do is to check Nancy's blood pressure.
"No, I don't think I want to do that."
"It will only take a moment. Just want to check your blood pressure."
"You're not taking any blood from me, I can tell you that."
"I'm not taking blood, just checking it," he reassures her.
"Well, if you say so, but I'm not happy about it."
She consents to her sleeve being rolled up. The plastic is wrapped round her upper arm. He begins to inflate it. Puff puff. Puff puff.
"Christ, that's cold."
"Sorry. It's been in the car."
Puff puff. Her face droops, her eyes close.
"You do that one more time," Nancy says in her low warning voice, "and it will be the worse for you. And I'm not joking. I'm not kidding around. You will regret it."
He puffs another puff and her other arm comes up with the fist balled tight. The doctor ducks.
# Chapter 29
_The very tones in which we spake
Had something strange, I could but mark;
The leaves of memory seemed to make
A mournful rustling in the dark_.
—HENRY WADSWORTH LONGFELLOW
THE EMERGENCY CARE MEETING IS HELD IN THE TOWN, at the swanky new social work offices that look, appropriately enough, like a cross between a medical practice and a solicitors' group and smell pungently of carpet. There are four of us present. Me, Chris, our care manager, and the care manager's boss, whom we're meeting for the first time. The first thing the boss has to say is that we can't see the file because Morris would have to give consent. (Why would we want to see the file? Is there something in it that's material? We'll never know.)
"So how are you, and how are things?" we're asked.
"Desperate, and desperate," I say.
"We need to establish why Nancy isn't on the residential waiting list," Chris says, "and how we can get her onto it."
"We can do another assessment," the boss says. "Things seem to have deteriorated badly since the summer."
"All that's changed lately," I tell her, "is that she's no longer so charming with outsiders, and doesn't mask her condition so well."
We talk about Nancy's MMSE result, which the doctor has been in touch about. "That really is quite a marked deterioration," the manager comments. I point out that she hasn't had the test before.
"We're confused about what the criteria are for getting onto the waiting list," Chris says again.
The sea change is coming. It's seconds away. And it happens by accident.
"The thing is, we can no longer cope," Chris says. "We can't do it anymore."
"Every day is a struggle," I concur.
"Are you saying you can't go on?" the boss asks.
At once, I begin backtracking, feeling as if I'm about to be judged inadequate.
"We're at the end of the road," Chris tells her. "We can't any longer care for them, unfortunately."
"So. You're saying that you can no longer look after them. Is that what you're saying?"
"Yes. That's what we're saying."
"It's much more difficult now," I add, "to manage Nancy at home, because she's threatening the children, and hitting them."
The professionals look at each other.
"Well, in that case, emergency respite will have to be arranged with immediate effect," the boss says. "We'll have to consider the long term. And the Family Division will have to be informed."
Family Division? I have visions of cars arriving at dawn, the children hauled off, and protest energetically. There's an instinctive fear of social workers, the extravagance of their powers, buried just below the surface of all my dealings with them. The social workers say, with regretful finality, that they have no choice, now that the abuse has been reported; they have a legal obligation to report violence upon children.
No care places are available locally, so respite will involve a journey into the next county. We mention that Morris is keen to return to Edinburgh, and they tell us they can get names onto the Edinburgh waiting list. In the United Kingdom, it's worth campaigning to have the list placement done by the social work department, rather than just going ahead and finding somewhere yourself. If it's their referral, the nursing portion of the fee is paid by the sponsoring council. This isn't to be sniffed at, being £150 or so a week _each_ , saving £15,000 a year. Even so, it will be around £35,000 a year for the two of them (by national standards, this isn't expensive). Discreet inquiries are made about means.
"They have the money to pay their own way for the first few years," Chris says. "They have their life savings. The rest of it's invested and will have to be de-invested." It's our house we're talking about; that portion of it that Nancy and Morris own. The house will have to go on the market. All assets count. Their savings, investments, all of it will be liquidated into a pot from which the state will drink hungrily. It will leach away, month by month, until there's £20,000 left. Only at that point will the state begin to contribute. At the time of writing, the rules dictate that only the final £12,250 will be left intact, untouchable by the state. This information hits Morris hard. It's a disheartening thing to face, for those who've always been frugal. The people who held on to their ancient washing machine until it gave out, who were content with the old linoleum in the kitchen, who put money by for a rainy day—their rainy day has come. The make-do-and-mend philosophy was all about providing an inheritance for Chris and his sister. But getting old and ill will take almost everything. It's raining hard now.
Once you have it clear in your mind that Alzheimer's is a disease, whose sufferers are ill, and that what's needed for it is treatment, the idea that a nursing home is optional, a luxury, and will be invoiced on that basis, is deeply offensive and wrong.
Next, the care twosome pays a house call. It's Morris they want to talk to and they go into closed session. He's anecdotalizing and hoots of appreciative laughter boom out under the closed doors. On this occasion laughter isn't a good sign. It means, almost certainly, that Morris is illustrating that the problems, such as they are, are blackly comic at worst. Later, Chris tries to talk to his father about the meeting but is met by the usual studied vagueness.
Three days later. The phone rings. Our care manager. Can she come out today to talk to Morris again? She'd want us in on the conversation this time. "Will you sit in? Because he doesn't seem to believe that there's a problem." She asks me if I will be frank with Morris about being at the end of the road, at the end of my tether, and needing the two of them out. No, I can't do that, I tell her. I'm not going to be a part of any staged resolution, no emotional pleas, no histrionics. I'm certainly not going to confirm anybody's dark suspicions that this has all been about me.
Nancy's up all that night, wandering, rattling doors, and ranting. She won't take the sleeping syrup; she clamps her lips together and flat refuses to comply. Her nocturnal narratives begin to remind me of somebody with a head injury trying to keep herself from losing consciousness. One foot is put in front of another, literally but also verbally, without there being any really strong thread at work. Just keeping going is the thing, keeping walking and keeping awake, with whatever words come to hand.
She has extraordinary stamina. Nighttime sleep is intermittent and daytime naps have been given up but she keeps on going nonetheless. She badgers Morris all the next day and is still badgering when I deliver the afternoon tea.
"I've told you already, I want to go for a walk."
"You can't go for a walk, it's dark outside," Morris tells her.
"I have asked you a hundred times," Nancy says.
"What are you talking about? I can't walk. I have a wheelchair. You'll have to push me. We'll do it tomorrow. It's too late now."
"It was the same with my father. He was standing at the door and he said something to me and you closed it."
"Your father?"
"You closed the door on him when he was here. He was talking to me."
"Your father's been dead for thirty years."
"I know the truth and you don't. He died last night."
She's weeping now.
"Thirty years!" Morris roars. "He's been dead for thirty years!"
"It all comes to money," Nancy tells him. "They want my money."
_"Who_ does?"
"They know who they are and where they went and you don't."
"I haven't the faintest idea what you're talking about."
"Well, that's what I'm saying. You're my father."
Morris (apoplectic): "I am _not_ your father. I am your husband."
Nancy's whimpering. "I've told you a thousand times but you don't listen. We could go home if it wasn't for you."
Morris is yelling now at the top of his voice. "I've told you! I can't walk! You'll need to push me in the chair." And then, calmly, "It's dark now; we'll do it tomorrow."
ON THE MORNING of the assessment I have a brief conversation with Morris.
"You know that it's coming, don't you? We can no longer cope with Nancy in the family. You know that, don't you?"
"Yes."
"And you will want to go with her, yes?"
He pauses. "I think so."
He's decided against Edinburgh, though. He'd rather stay up here. He doesn't think the old Edinburgh friends would visit them, he says. Not once the novelty had worn off.
The care manager and the boss arrive, and ask, ominously, if they can talk to Chris and me first. We repair to the drawing room. The boss appears to have a speech prepared. She tells us that it's going to make this whole process a lot more difficult if Chris and I won't speak to Morris directly about our feelings and won't agree to go on the record as having done so.
She's talking about the phone call, the one in which I was asked to tell my father-in-law that I need him to leave, in circumstances (social workers present) that might look engineered. But this isn't about feelings. It isn't about Morris. It's about Nancy, and Nancy's unhappiness. Nancy's health. I don't think the social workers see that. I don't think they understand Alzheimer's. I think they look at us, Chris and me, and see people giving up, capitulating, _dumping_.
Morris needs to give permission for Chris and me to sit in on the meeting. He says he wants a confidential talk with the two ladies first. I say to him that I feel at this stage of things that he ought not to have anything to say to the social workers that he couldn't say to us. This angers him. The social workers are hovering so I leave the room, embarrassed. Foolishly, I pause at the door, and hear Morris berating me for wanting to know everything, wanting always to be consulted, for wanting to be in charge, for being interfering and bossy.
Fifteen minutes later, we are admitted to the room. Nothing's said about the confidential talk. Long explanations follow about how the waiting list works—not as a queue, it turns out, but strictly according to need. Every time a place comes up, the whole list is consulted for the best match. And it's possible that a double room will become vacant this spring.
Morris is emotional, his eyes brimming, his voice quavery, when the boss asks if he is happy to go into the nursing home. I feel like I might cry myself. Pity and relief are fighting for top billing.
"Not really," he says, "but what's the alternative?"
"You could stay here. We could offer more help. But I think you're aware that your family are having difficulty coping."
"Well, if that's the case, then there isn't any choice," he says.
The boss says something about confidentiality. Morris's reply is surprising.
"There's nothing you say to me that you shouldn't say in front of my son and daughter-in-law. They have looked after us magnificently...." The phone rings out and I have to excuse myself to answer it, so I miss the rest of this tribute.
The day before they're to go off to respite, Nancy is spoiling for a fight. By lunchtime she and Morris are in open warfare.
NANCY: You are getting on my nerves, I wish you'd clear off out of here.
MORRIS: Oh that's very nice. We'll both shut up, then.
NANCY: You talk to me as if about I was a child.
MORRIS: No. I talk to you as if you _were_ a child. There's no "about" in the sentence. That's what you should have said. Why don't you rub your fingers together?
NANCY: Why should I do that?
MORRIS: You do it all day. Rub rub rub. I'd like to know why myself. Why don't you twiddle your hair?
NANCY: That's just ridiculous. I do nothing of the sort.
MORRIS: Why don't you stick them up your nose?
NANCY: That's completely ridiculous.
MORRIS: That's right. I'm being ridiculous again. Which is another way of saying that I'm totally fed up.
Cases to pack. Morris doesn't want to be involved in this. He watches television and I take suggested clothes and books through for approval.
When I go into their sitting room later that evening to take Nancy to bed, she has that look on her face: the warning look, pinkish violet, with lips in a tight purse. She goes to the bathroom and I sit on her bed to wait for her, feeling uneasy. Things can get out of hand at this point in the day. She sits on the toilet and free-associates. At least I think that's what she's doing. But it occurs to me after a while that she might be talking to her urine.
"You're going to go there and do the right thing and go down, and that's right. And I will have to talk to her about you and where you need to go next."
Now she is having trouble with her underwear.
"Are you all right in there?" I call through.
"Yes. No. I'm coming. I'm coming, I tell you. I'm telling you straight." Her head bends to deal with errant clothing and her voice is muffled accordingly. "You don't go on right anymore. You used to know where to be but now you don't. And she will have something to say about that."
A few minutes later she trundles through.
"Right then," I say. "Let's get you changed for bed."
"I'm not getting changed, I'm bloody freezing."
"It's warm in here, Nancy, with the heater on. I'm really warm. Aren't you warm?"
I start to unbutton her cardigan. An old veined hand clamps itself over mine.
"Now come on," I say. She just stares at me. She is as ever astoundingly strong. "We just need to get your nightie on."
"But I've spent all my life trying to get it and it isn't there. You don't know the first thing about it."
I have her cardigan off now and she consents to her sweater coming over her head.
"Now that'll do fine. I'll just go like this," she says, holding on to her blouse hem tight, her knuckles white.
"Look. Your nightie and your fleecy bed jacket. You'll be warm as toast."
"Hum huum, hum huum, hum hm-hm." Her favorite tune. She hums it now. She can no longer rhyme.
I whip the vest off and get the nightie on, the zip jacket. "You're going on your holidays tomorrow," I say, grinning at her.
"I am not."
"Yes, you're going away on your holidays, with Morris."
"Who's Morris?"
"That's your husband."
"Oh. That's what you say. That's my husband, is it. You don't look like my husband."
"No. I'm married to your son. Yours and Morris's."
"I don't have a son. I never had children and I'm glad because they just disappoint you."
Here comes the crunch. The trouser and underpants removal from under the long skirt of the nightdress, pulled quick and altogether like the magician's tablecloth. "Oh Christ! What the hell do you think you're doing?"
"Good night, Nancy! Sweet dreams!"
# Chapter 30
_Life is the continuous adjustment of internal relations to external relations_.
—HERBERT SPENCER
MID-FEBRUARY. THICK SNOW, BLUE SKIES. THE RETURN from respite is delayed by transport problems, and Chris has a series of phone conversations with the nursing home manager, in the course of which she says, "Can I ask, are you thinking about permanent care for your parents?"
Chris launches into the abridged story of our lives in the last two years and then gets to the point and confirms that he is.
"Because they could stay on, you know," the manager says. "They seem pretty happy here."
"Well, that would be ideal for everyone, if that were the case," he tells her. "Let's take it a step at a time. Let's wait and see what Morris has to say to the social worker when she calls by."
Next, we hear that Morris has booked to stay on in the nursing home for another fortnight. The extra fortnight morphs, within twenty-four hours, into the real possibility of permanence. Morris likes the nursing home. Nancy has been moved into a single room in the Alzheimer's unit and he sees her at mealtimes. He doesn't tell us about his plans directly; a message is conveyed through our care manager, who's also instructed to tell us that he hopes we aren't offended.
There's a phone call from The Charity manager, Mary, who's a good egg and empathic. She mentions that she had a run-in with someone in authority at the council about Nancy's not being on the waiting list. The someone had maintained the until recently unanimous stance that Nancy wasn't severe enough a case to be on the residential care list.
"Not severe enough?" Mary had echoed, incredulously. "She's a lot more severe than plenty of the people you do have in nursing homes. You just try spending a couple of days looking after her. You'd see what she's really like."
Then she makes an illuminating point about the fundamentals of how these things work. "When you get to the point that you can no longer go on as a caregiver, and state that you can no longer go on, that you can't do it anymore, then the state has to step in and take charge and find alternative arrangements. You say baldly that you can no longer cope, they have to take them in, it's that simple."
And that's the reason they're on the waiting list now. That's the reason that the care manager has been on the phone to Chris this morning, saying that permission will almost certainly be granted for a council-sponsored permanent placement at the respite home, though they have to go through the motions of soliciting a local place first.
Chris calls his father and is relieved to find him remarkably cheery.
"I hear you are thinking of staying on, Dad."
"Yes, thinking about it, yes."
"I think that's great news for you, really good news, if the two of you are happy and settled there. I think you should go for it, Dad. Stay on. We'll make all the arrangements."
"Right, son."
"So you'll be staying on?"
"I'm coming round to that way of thinking, yes."
# Chapter 31
_Only the paradox comes anywhere near to comprehending the fullness of life_.
—C. G. JUNG
THE BOOK—AND PARTICULARLY, THE AMERICAN VERSION of The Book—says that taking up the nursing home place is just the beginning of a new life for both of you, the dementia sufferer and also the caregiver. The Book doesn't seem to want caregivers to have any sense of relief or liberation when caregiving comes to an end. That might be unseemly, mightn't it? Being glad. Being glad is taboo. The Book envisages that the caregiver will live close enough to the home to visit every day—though this might involve moving. The nursing home staff will be glad to see you popping in and out, The Book says. The nursing home staff will be glad to share tasks and to devolve some of them. You (the caregiver) can help with dressing, and feeding, and bathing, and care, quite apart from providing a face that still might be familiar and all its resonances of love and family. Quite apart from providing outings.
Interestingly, career nursing home managers with expertise in dementia often feel differently. Those I've come across, at least. Some of them, at least, are prepared to speak up for the wisdom of letting go. Having spent a good many months immersed in the online Alzheimer's community, I find that one common experience shines out: that at some point along the dementia journey, the fact that the caregiver is close to the person stops helping them and starts hindering. So often, people find that it's their own interactions with the dementia sufferer that are triggering their senile anxiety, however unwittingly. Sometimes people see this and remark on it. Sometimes they don't and it's only the outsider who sees.
The caregiver hates leaving their demented relative at the home, which is, after all, an institution, and may jar aesthetically on the nerves of the healthy, with its hospital look, its easy-clean surfaces and perceived lack of comforts. But then they discover that the demented one's health and morale improve at the home. They find this hard to believe, because when they visit, all they see is unhappiness. The caregiver speaks to the staff about this—about their loved one's unhappiness—and may be told tactfully that actually the outbursts only happen when the caregiver visits, and possibly it might be better for the dementia sufferer to be left to get on with this new phase of life undisturbed. That's a hard thing to be told, and even harder to accept. After all, we think of ourselves as essential. We have been essential to the person for months and years, day in and day out without pause. Part of what disturbs us about leaving them at the nursing home is that we are missing from that environment; family and personal history are missing, and no one, surely, could be better off without those. The truth is that sometimes they are. Sometimes family and history exert too much pressure, provoking a chaotic mental state that can't deal with the presence of pieces of the old life, things sparked in a damaged brain that dementia can't make sense of.
TWO THINGS APPEAR to be true.
1. When Alzheimer's sufferers get to the point of constant unhappiness at home, they are ready to leave. Nothing you do or try will make any difference to this.
2. All that matters, at this advanced stage of the disease, is that they are as happy as they can be, even if that means your having restricted access to them. If they can be got through the day with minimal fear, anxiety, rage, then that's a good day.
It's time to stop reading the dementia books. There's a shelf full, the books placed there in acquisition order. They start at the left with the how-to caregiver books, proceed into the more medical tomes, the more specialized, soften abruptly into memoir, then take a swerve into American publishers, books of alien sizes and typefaces, titles dredged for and blundered into on Amazon. Some of these are rather wacky. Some I'm not sure what to make of. I meet a retired neuroscientist on a bus, a stranger, who sees what I'm reading and strikes up a conversation. He agrees with the basic premise of the book in my hands, one I'd taken to be eccentric, that Alzheimer's disease is a myth. "There isn't really any such thing, you know," he says, eyes twinkling. "It's just that some people age faster than others." This brings me up short. But do the two camps really have to be at odds, normal aging versus abnormal event? Isn't it possible that a condition that presents itself as the acceleration of brain aging is itself a disease? In any case, what about the 115-year-old woman, in the news recently, whose brain showed no signs of dementia at autopsy and was said to be as healthy as that of someone fifty years her junior? If she was merely an exception, how many exceptions are there? Are we two kinds of human, those who do and those who don't experience brain failure, the key to the mystery mundanely genetic?
Members of the Alzheimer's backlash, the dissenters like the writers of this "myth" book, have to admit that early-onset dementia is a disease. But like the neuroscientist on the bus, they regard late-onset dementia as a human-condition condition, one that will never find a cure. They cite the work of scientists at Harvard University in pinpointing the disintegration of myelin (our white matter) in the brain, which makes electrical signals between neurons weak and diffuse, as just as likely to explain _senility_. They point out that plaques may be beneficial—something, of course, that the pro-tau research body would also agree with. They point out that some dementia autopsy brains show neither plaques nor tangles, and that some autopsy brains of people who didn't suffer from dementia are plaque riddled. Alzheimer's as a disease, they say, is a myth fostered by pharmaceutical companies. I find others sympathetic to these ideas in the online community: those who think that labeling people as having dementia is a form of bone pointing (an Aboriginal idea, in which the subject toward whom the bone is pointed convinces himself he is doomed, and dies accordingly); others who claim there's no actual loss of self entailed in the progress of dementia, other than for that imposed by society, which conditions the ill culturally and socially to behave in a demented manner.
If some or all of this is bonkers, there's no doubt that the way healthy humans regard those with brain illnesses, brain damage, brain disability is in general shockingly uncivilized. I find myself averting my eyes in the supermarket as a woman with dementia of some kind rants and accosts passersby, all of whom avert their gaze in turn, and begin to create a margin as they pass, a collusive semicircle of safety. We're embarrassed by dementia. I'm embarrassed by dementia. The unpredictability of how somebody may act, what they may say to you—these are factors. I smile at an old lady with dementia on the bus and she shouts at me for the rest of the journey home. It's safer to keep your distance. In addition, there's the shame of being old, your body failing. We treat the old with contempt, their weakness provoking bullying by the healthy. Is it because they rub our noses in our own mortality? Because we see mortality as a failing, after all? Perhaps subconsciously we feel the bone's been pointed, and that we should keep ourselves clear. Mortality is contagious; we catch it from our parents. It's hard to treat somebody with failing thinking and language skills as a person as fully human as yourself. You may not think so. Every instinct in you might insist otherwise. In the supermarket, though—how are you then, when the old lady thanks you for finding the jar of mayonnaise, then smashes it on the ground, hits you with her basket, calls you her daughter, tells the checkout staff and everyone on the street outside that you're taking her home? It's so much worse if the person is haphazardly dressed, dirty, smells; it's so much harder to treat them like an equal.
These may sound like accusations but in fact I'm trying to excuse the behavior of people toward the demented, which tends to be awkward at best, unkind, and thereafter progresses on a sliding scale into savagery. It's untrained. It's untrained, certainly, though should the behavior of humans toward others need training to be fair? If it does, it can only be because some unfortunate _ghost in the machine_ has survived, subconsciously, in our own minds, that marks out brain afflictions as dehumanizing and dehumanizes accordingly. This is the only way I can account for the behavior of the many doctors, specialists, and consultants written about in dementia forums who are on the record as having treated those with Alzheimer's so badly and so dismissively. And how else can we explain the treatment of the elderly in nursing homes (some elderly, in some nursing homes), who are talked to like bad children, neglected, ill fed, abused, or even—as in a recent case in the newspapers—tied onto their chairs?
The old woman on the bus who shouted—she was real. But the woman in the supermarket—I made her up. Or rather my brain did, while I was sleeping. The thing is, she was me, the woman in the dream, the Barbour bag lady in thirty years' time. She wasn't like me, she was me. It was me. All that separated us was time, and the dementia roll of the dice that will determine whether in 2040 I'll be one of the 90 million, or not.
It's time to stop reading the dementia books.
# Chapter 32
_Things that were hard to bear are sweet to remember_.
—SENECA
THIS IS WHAT MIGHT HAPPEN TO OLD PEOPLE WHO GO into nursing homes. They have secure, dull lives looked after by gentle Eastern Europeans in easy-care aprons. Their families ring when they get around to it. Their tidy, institutional bedrooms with the matching floral duvet and curtains show little sign of life, other than for the book and glasses by the bed, and the two propped cards that have arrived from old friends.
It's two hours' journey from the house to the home, which is a ranch-style bungalow, purpose-built, U-shaped with deep wings and a steeply pitched hipped roof. Inside it's warm, draft-proof, with tight modern windows, wide carpeted corridors, jolly pine furnishings, and chain-store chandeliers.
We go to visit at Easter; our holiday route almost passes the door. There has been excited anticipation of our visit. Nancy and Morris are washed, pressed, have had their hair cut that very morning. Because we are en route and in a rush, we anticipate we'll be there for less than an hour, but that seems to have been forgotten. Nancy and Morris's lunch has been delayed indefinitely in our honor. As it turns out, we're running late and spend twenty-seven minutes in the building. Chris speaks to the manager and asks where we can have our meeting. He doesn't use the word _meeting_ , of course, but that's what this feels like, a hurried business conversation in a motel conference facility. We use an alcove designed for the purpose just by the front doors, a cubbyhole furnished with three chairs.
Morris isn't around. He's wheeled himself off to his room to get a birthday card for Caitlin, one that we bought and mailed to him in readiness. I go into the dayroom and see an old lady there. She's sitting alone in a winged armchair, rubbing her hands energetically together and muttering.
"Oh, and she said, she said it was all right, so I suppose it is. It must be if she says so. She knows everything."
Nancy. In some bizarre unexpected way I have been missing her. Her eyes are fixed unblinking on the table twenty feet away, where six residents and two staff are sitting having a conversation, desultorily engaged in nursing home activities. Playing cards. Sticking down a magazine cutout collage. I stop at her chair and stoop to touch her arm and her face shifts from its blankness briefly to a look of alarm and then into a great beaming smile.
"Oh! Hello! What on earth are you doing here?"
She seems genuinely to recognize me. Chris had this experience last month, when he came to see that they'd settled in, and his mother greeted him with "So how is the family?" Her improvement since she's been here has been quite remarkable, say the staff. That's why the social workers have already let it be known that they wouldn't let her out to live in the civilian world again. Morris, a couple of weeks ago, was on the phone to Chris to say that he was thinking about returning to live with us, and Chris had had to tell him so.
"I've come to visit you," I say, kissing her and offering an armful of supermarket tulips. "Look, flowers for your room. And a box of Dairy Milk, your favorites."
"Oooh, thank you," she says. "Gimme gimme!" Making playful grabbing motions.
"Your grandchildren are just round the corner, in the visitor chairs," I say to her. "Do you want to come and see them?"
"Oh no. No, thank you. I don't think so. Not today."
"Come and see them. Just for a minute." I hold my hand out and she takes it, getting up and toddling after me. "Well, all right, then. Seems like I don't have any choice. As usual."
Nineteen minutes left. I look covertly at my watch. Morris arrives in his chair with Chris, all smiles, looking well.
"Here he is," Nancy says. She seems to know who Morris is. She tells him off when he starts talking about their new life: "Oh, hold your wheesht, you," slapping him flirtily on the knee. We talk about our trip south, and ask about the home. They have a pretty good social life, it seems, though Morris has been reprimanded for preferring to sit with a book than join in, he tells me. He's reading about Hitler's last days in the bunker. He urges me to visit his room and see his certificates and there they are, thumbtacked to the wall. One is for winning at the Beetle drive, the other for being placed third in carpet bowls. The children don't say much and neither does Nancy, but she smiles at everybody and her teeth are clean. She is altogether immensely clean, her fingernails white and her silver bob immaculate. Morris has no complaints, he says, other than about being kept waiting for meals once they're seated in the dining hall. Their timekeeping here isn't that impressive and it's annoying having to wait. The food's pretty good, though. And he has television in his room.
"How's she been?" I whisper to Morris, while Chris is chatting to his mother.
"Pretty good," he says. "She's in this part of the home with me some of the time... but the rest of the time she's in the Alzheimer's wing, locked up." He looks embarrassed.
"What's it like through there?" I ask, looking toward the security doors.
"Just like this side, a bit smaller, but otherwise a mirror image of here," he tells me.
Nancy has gone through the looking glass.
"She wanders into other people's rooms, apparently," Chris tells me later. "Ignores them, goes and looks out of the window and then leaves without saying anything."
It's time to go.
Nancy tries to follow us out of the building.
"Will you be back soon?" she asks, anxious, trying to grasp at my hands.
"Yes," I say. "Very soon," nudging at her arms so I can close the door.
Morris is barking instructions at her. She turns to him and she says, "Just wait till I get you home."
ALMOST EXACTLY A year later, Morris died at the home from kidney failure. During his last illness he spent a lot of time anxious about Nancy, unwilling to stay in bed. She might be up to something. He ought to be there. He couldn't rest.
Nancy didn't understand or mark his passing and his funeral service was held in Edinburgh as he wished. Nancy was judged too ill to make the journey, which would have entailed spending most of a day on the train south, a night in the city, and another day on the train back. In some ways she's very ill, and in others amazingly robust. Physically, she continues to be good for her age. The nursing home staff speak about her with touching protectiveness. She remains somebody who wants to be doing things and busy and finds it hard to sit still. If she's awake, she's usually on the move. She's prone to aggression; there are long gaps between haircuts as they have to pick the right day to embark. Haloperidol (Haldol), an antipsychotic, is administered in small doses, a half teaspoon two or three times a week, on an ad hoc basis of need, and doesn't appear to slow her down. She's unaware who anybody is, but keen to be included in the group, despite her communication difficulties. Shrunken, but insistent on wearing favorite old trousers, she gathers spare fabric bunched up at the waistband with one hand as she shuffles about. She looks like a very old lady now. Having rejected, finally, the wearing of false teeth, her food's mashed up for her; she eats only with her fingers, and can be insistent about keeping moving, eating a little on each circuit of the home. Often she hasn't the patience for dealing with food at all, and steals a neighbor's tea biscuits later. She gets dietary supplement drinks. On a good day she'll react positively to the photographs in her room, though she can't put a face to a name. She talks to staff about her parents sometimes. Occasionally she mentions a man she used to know, a man in a wheelchair.
As for me, I've arrived, already, at a state of self-protective forgetting. People are good at that, at moving on, dwindling the past into a story we tell ourselves, into parables, and choosing the future over the past. It's true that every now and then mistakes that I made rear up in memory, like splinters surfacing out of a finger. Memories of bad days, revisited synaptically in sound and vision, far outweigh the good. But it's also true that, as Oscar Wilde put it, "The great events of life often leave one unmoved; they pass out of consciousness, and, when one thinks of them, become unreal. Even the scarlet flowers of passion seem to grow in the same meadow as the poppies of oblivion."
# Additional Reading
## **Online Resources**
Useful resources on the Internet divide into two main types: those that offer information, and those that offer support. I can't urge you strongly enough, if you are a caregiver of someone with dementia, to join a forum community and share your day-to-day struggles, concerns, and questions. "Talking" online to others who have just the same kind of issues and crises is invaluable. There is a lot of genuine companionship and good advice out there. Just to have an ongoing conversation with others in a similar situation to yourself, and to forge friendships, is immensely helpful and can vastly improve caregiver morale.
**www.alz.org**
The Alzheimer's Association. Education, advice, publications, and support. A guide to your rights and options. A very popular forum, where you can "talk" via message to other caregivers and sufferers. Twenty-four-hour toll-free advice on the phone.
**www.alzfdn.org**
Alzheimer's Foundation of America. Practical advice about the technicalities of caring for and supporting someone with dementia. A not-for-profit organization that acts as an umbrella for 1,200 other organizations. A useful list of government organizations to contact, under caregivers' tips/government/federal resources.
**forum.alzheimers.org.uk**
"Talking Point," a useful, busy caregivers' forum hosted by the Alzheimer's Society in the United Kingdom. Though the drug names and care procedures and legislation may be different, you'll find British caregivers have many of the same kinds of problems and solutions.
**www.alzheimersreadingroom.com**
A useful digest of Alzheimer's stories in the news, and articles on aspects of dementia and dementia care, edited by a caregiver.
**www.alzinfo.org/forum**
A forum for those affected by Alzheimer's, hosted by the Fisher Center for Alzheimer's Research Foundation (www.alzinfo.org).
**www.caps4caregivers.org**
CAPS—Children of Aging Parents. A nonprofit charitable organization that aims to offer support and information for caregivers to the elderly. There is also an online support group.
**www.caregiver.org**
Family Caregiver Alliance. Campaigning on behalf of caregivers, providing education and support. State-by-state navigator for care options. You can sign up to "talk" to other caregivers online.
**www.caregiving.com**
Information, personal stories, resources.
**www.caregiving.org**
National Alliance for Caregiving. A nonprofit coalition of national organizations focusing on issues surrounding family caregiving. An advocacy group.
**www.ehealthforum.com**
Hosts forums for a whole list of illnesses. Choose "Alzheimer's" from the menu. Post problems and get feedback. This Web site says that medical personnel can give feedback also; you need to sign up for this.
**www.eldercare.gov**
A search facility that puts Americans in touch with finding the organizations in their own area that can help and advise about home-based and community care.
**www.thefamilycaregiver.org**
The National Family Caregivers Association. Information and advocacy: giving a voice to the vast silent army of caregivers in the United States. The Family Caregiver Forum has message boards where you can post your problems and solutions and share with others.
**www.healthboards.com**
Forums for all sorts of illnesses: Go to the message board index and choose Alzheimer's and Dementia, or put "Alzheimer's" into the search box.
**www.helpguide.org**
Navigate via the Seniors & Aging link, to Alzheimer's/Dementia. The "Support for Caregivers" section has good advice.
**www.mayoclinic.com**
The Mayo Clinic's own information pages about Alzheimer's (follow the links) offer good basic guidance. The blog strives to be positive and can-do. There is a chance to comment; the comments from caregivers present an interesting contrast.
**www.nia.nih.gov**
National Institute on Aging Web site. Click on Alzheimer's Disease Information. Clear and simply written information on diagnosis, treatment, and caregiving.
**www.tangledneuron.info**
"A layperson reports on memory loss, Alzheimer's and dementia." A useful digest of dementia news.
## **Blogs**
More and more caregivers are beginning to use the Internet to post diaries of their thoughts and experiences. You may find some of the following blogs useful, though be warned that sometimes their sadness and courage can prove overwhelming.
**www.alzheimersdad.blogspot.com**
**www.alzheimersspeaks.wordpress.com**
**www.acaregiversjournal.com**
**www.eldercarecafe.blogspot.com**
**blog.seattlepi.com/witnessingalzheimers**
**www.mindingoureldersblogs.com**
**www.ajourneywithalzheimers.blogspot.com**
**www.knowitAlz.com**
**www.thelastofhismind.com**
**www.killingmyfather.com**
## Popular Books About Dementia and Alzheimer's
_The Alzheimer's Action Plan_ , by P. Murali Doraiswamy and Lisa P. Gwyther—A guide that looks at what can be done about MCI and early stages of dementia.
_Alzheimer's from the Inside Out_ , by Richard Taylor—Essays written by a psychologist and early-onset-dementia sufferer, observing his own deterioration.
_The Brain That Changes Itself: Stories of Personal Triumph from the Frontiers of Brain Science_ , by Norman Doidge—An interesting book about brain plasticity, its ability to rewire itself.
_A Caregiver's Guide to Alzheimer's Disease: 300 Tips for Making Life Easier_ , edited by Patricia R. Callone—A very practical guide popular with caregivers.
_Creating the Good Will_ , by Elizabeth Arnold—A comprehensive guide to financial end-of-life planning.
_Dementia Diary: A Care Giver's Journal_ , by Robert Tell—An account by a son of his mother's slow descent into disease.
_Elder Rage, or Take My Father... Please!_ by Jacqueline Marcell—A popular and cathartic howl of frustration by a child looking after a difficult aged parent.
_Learning to Speak Alzheimer's_ , by Joanne Koenig Coste—Personal experience and person-centered home-based-care advice from the wife of an Alzheimer's sufferer.
_Losing My Mind_ , by Thomas DeBaggio—An account of his own decline, by an early-onset-Alzheimer's sufferer.
_Mothering Mother: A Daughter's Humorous and Heartbreaking Memoir_ , by Carol O'Dell—Another account of a mother with dementia being taken into her child's home and life and the struggles that ensued.
_Still Alice_ , by Lisa Genova—A bestselling novel about a psychology professor with early-onset Alzheimer's.
_The 36-Hour Day_ , by Nancy L. Mace and Peter V. Rabins—The best-known of the how-to-care guides for family and home-based caregivers.
# Acknowledgments
**Introduction**
Luis Buñuel, _My Last Sigh_ , translated by Abigail Israel, published by Alfred A. Knopf, a division of Random House, Inc. A paperback edition is published by University of Minnesota Press.
**Chapter 2**
Aaron Copland quotation reprinted by permission of the publisher from _Music and Imagination: The Charles Eliot Norton Lectures 1951–52_ by Aaron Copland. Harvard University Press, Cambridge, Mass. Copyright © 1952 by the President and Fellows of Harvard College, Copyright © renewed 1980 by Aaron Copland.
**Chapter 5**
Philip Larkin, "Best Society," from _Collected Poems by Philip Larkin_. Copyright © 1988, 2003 by the Estate of Philip Larkin. Reprinted by permission of Farrar, Straus, and Giroux, LLC.
Philip Larkin, "Aubade," from _Collected Poems by Philip Larkin_. Copyright © 1988, 2003 by the Estate of Philip Larkin. Reprinted by permission of Farrar, Straus, and Giroux, LLC.
Ludwig Wittgenstein, aphorism 109, from _Philosophical Investigations_ , 1953. New edition edited by Hacker and Schulter, and published by Wiley-Blackwell.
Ludwig Wittgenstein, _Tractatus Logico-Philosophicus_ , 1922. New edition translated by Pears and McGuinness, and published by Routledge Classics.
**Chapter 16**
Antonio Damasio, _The Feeling of What Happens_ , 2000, published by Harvest Books.
**Chapter 18**
C. G. Jung and W. Pauli, _The Interpretation of Nature and the Psyche_ , 1955. Translated by R. F. C. Hull and published by Routledge.
Sigmund Freud, _Civilisation and Its Discontents_ , 1930, translated by David McLintock, published by Penguin Modern Classics.
R. D. Laing, _The Facts of Life_ , 1976, published by Allen Lane. Reproduced with kind permission of the R. D. Laing Estate.
**Chapter 20**
Elizabeth Bowen, _Vogue_ magazine interview, September 1955. Reproduced with permission of Curtis Brown Group Ltd, London, on behalf of the Estate of Elizabeth Bowen. © Elizabeth Bowen 1955.
**Chapter 22**
T. S. Eliot, "The Hollow Men," from _Collected Poems, 1909_ – _1962_ , © 1936 by Harcourt, Inc. and renewed 1964 by T. S. Eliot. Reprinted by permission of Houghton Mifflin Harcourt Publishing Company.
John Bayley, _Iris_ , 1998, published by Duckworth.
**Chapter 23**
Henry Miller, _Black Spring_. Reproduced with permission of Curtis Brown Group Ltd, London, on behalf of the Estate of Henry Miller. ©
Henry Miller 1936.
Douglas Adams, _The Hitchhiker's Guide to the Galaxy_ , 1979, published by Macmillan.
**Chapter 25**
Louis MacNeice, "Snow," from _Collected Poems_ , 2007, published by Faber & Faber.
**Chapter 26**
Wallace Stevens, "Man and Bottle," from _The Collected Poems of Wallace Stevens_ , 1990, published by Vintage Books, a division of Random House, Inc.
**Chapter 27**
C. G. Jung, _Children's Dreams, Notes from the Seminars Given in 1936–40_ , translated by Ernst Falzeder and Tony Woolfson, published 2007 by Princeton University Press.
**Chapter 28**
Philip Larkin, "Long Last," from _Collected Poems by Philip Larkin_. Copyright © 1988, 2003 by the Estate of Philip Larkin. Reprinted by permission of Farrar, Straus, and Giroux, LLC.
**Chapter 31**
C. G. Jung, _Psychology and Alchemy_ , translated R. F. C. Hull, and published by Routledge.
# About the Author
ANDREA GILLIES is a writer and journalist. _Keeper_ , her first book, won the 2009 Wellcome Trust Book Prize for the best book on a medical topic published in the United Kingdom. She lives with her family in St. Andrews, Scotland, and has just completed her first novel.
**Keeper**
Reader's Guide
# About This Book
_Keeper_ is an account of a crisis in a family's life: a long drawn-out crisis without simple boundaries or solutions. When Nancy comes to live with her son and daughter-in-law and their children, along with her husband, Morris, she brings her Alzheimer's disease with her. They move together to a romantic mansion out on a headland in remote northern Scotland, expecting life to be easier for Nancy and manageable for the rest of them. But Alzheimer's is a ticking clock, building to a sustained emotional explosion. When Nancy no longer knows who she is, when her autobiography has been whittled away by brain disease, the learning curve is steeper than anyone could have imagined. _Keeper_ takes the form of interweaving a vivid diary of Nancy's day-to-day unraveling with the author's own research into what Alzheimer's is and what it means for us as human souls.
# About This Guide
Dear Reader:
Here are some questions that seem obvious to me, as the author of the book. They're the questions I ask myself, now that I read it again, given the benefit of some distance from the events described.
My best to you,
Andrea Gillies
# Discussion Questions
1. Did Andrea and Chris have any real choice when they offered to take Nancy and Morris into their home? Where would another choice have led the family as a whole?
2. On what basis does the author rationalize the choice she's made to care for Nancy herself? Is her reasoning sound, or based on idealism and ignorance?
3. To what extent does the romantic setting—the landscape surrounding the house and the beauty of the house itself—become metaphorical and emblematic of the progress of the story?
4. How do Nancy's and the author's "journeys" (in terms of mood and state of mind) come to mirror each other?
5. Why do the other characters, particularly the author's husband and children, play so small a part in the book?
6. What's Morris's role in the progression of Nancy's dementia?
7. What's the turning point in the story, and why? When does it become clear that the experiment isn't going to work?
8. How successful is the author in explaining how a disease can affect a personality? Were you convinced that, as the author comes to believe, memory is the same thing as identity and that "self" is a biological entity?
9. Can the author's getting aggressive and indifferent with Nancy be justified? And if so, how?
10. At the end of the book, when change comes, do Andrea and Chris make the right decision for Morris and Nancy? Should they have taken this course of action at the beginning?
Copyright © 2009 by Andrea Gillies
Reader's Guide copyright © 2011 by Andrea Gillies
All rights reserved.
Published in the United States by Broadway Paperbacks, an imprint of the Crown Publishing Group, a division of Random House, Inc., New York.
www.crownpublishing.com
Broadway Paperbacks and its logo, a letter B bisected on the diagonal, are trademarks of Random House, Inc.
Originally published in slightly different form in Great Britain by Short Books, London, in 2009, and in hardcover in slightly different form in the United States by Broadway Books, an imprint of the Crown Publishing Group, a division of Random House, Inc., New York, in 2009.
Library of Congress Cataloging-in-Publication Data
Gillies, Andrea.
Keeper: one house, three generations, and a journey into
Alzheimer's / by Andrea Gillies.
1. Alzheimer's disease—Patients—Care—
Scotland. 2. Gillies, Andrea.
3. Caregivers—Scotland—Biography. 4. Alzheimer's disease—Patients—Scotland—Biography. I. Title.
RC523.G476 2010
616.8′31—dc22 2010006659
eISBN: 978-0-307-71913-3
_Cover design by Laura Duffy_
_Cover photograph:_ © _Claire Morgan/Trevillion_
v3.1
| {
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{"url":"https:\/\/www.varsitytutors.com\/hotmath\/hotmath_help\/topics\/developing-a-probability-distribution-from-empirical-data.html","text":"# Developing a Probability Distribution from Empirical Data\n\nIn real-world situations, statisticians obtain data by means of observation and experimental methods. Data obtained in this manner is called empirical data.\n\nA probability distribution obtained by means of observation and experimental methods is referred to as an empirical probability distribution , or a relative frequency distribution based on observation.\n\nExample:\n\nLet $X$ be the number of movies a high school student watches in a given month.\n\nA survey conducted at one particular high school in the month of December gives by the following table:\n\n Number of Movies Watched $1$ $2$ $3$ $4$ $5$ $%$ of students $17$ $28$ $34$ $15$ $6$\n\nIf we assume that the students at this high school are typical, and that December is a typical month, then\n\nThe probability that a high school student will watch $1$ movie per month is $17%$ ;\n\nThe probability that a high school student will watch $2$ movies per month is $28%$ ;\n\netc.\n\nUse this empirical probability distribution to find the expected value for the number of movies a high school student will watch in a month.\n\nConvert the percentages to decimals.\n\n$17%=0.17$\n\nUse the weighted average formula.\n\n$E\\left(x\\right)=\\sum {x}_{i}\\cdot P\\left({x}_{i}\\right)$\n\n$\\begin{array}{l}=\\left(1\\right)\\cdot \\left(0.17\\right)+\\left(2\\right)\\cdot \\left(0.28\\right)+\\left(3\\right)\\cdot \\left(0.34\\right)+\\left(4\\right)\\cdot \\left(0.15\\right)+\\left(5\\right)\\cdot \\left(0.06\\right)\\\\ =0.17+0.56+1.02+0.90+0.30\\\\ =2.95\\end{array}$\n\nSo, we can expect the average high school student to watch $2.95$ movies per month.","date":"2018-02-20 07:49:41","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 20, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.47855132818222046, \"perplexity\": 454.0896508510842}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-09\/segments\/1518891812913.37\/warc\/CC-MAIN-20180220070423-20180220090423-00023.warc.gz\"}"} | null | null |
Special Pubs need Protecting
Special Pubs that are historical, have strong social or cultural qualities and benefits need to be given protection and saved. There is nothing more rewarding than giving an old Pub building a new lease of life with the enthusiasm of a new owner and with community behind it too.
29 Pubs a week are closing according to statistics from The Campaign for Real Ale (CAMRA)
Pubs need Protection
Whether a Pub is in such poor condition and is going to be demolished or plans are to convert the Pub for other uses such as residential or for another business Pubs need saving. Pubs are a centre for a community, a hub and a place where people can meet – many are recognised as an Asset of Community Value (ACV).
I come from a culture where the Pub is the centre of the community. The Pub is the Internet. It's where information is gathered, collated and addressed.
Rhys Ifans
Pub Building Surveys
Save You Time & Money
Free Phone 0800 298 5424
Why are Pubs Closing?
Many Pubs across the UK have closed, there has been a decline since 1980 and in the last 6 years, according to the Cask Ale Report, 5000 pubs have closed their doors for the last time.
There are several factors that have influenced the decline of the Pub including rising taxation, changes in regulations as well as an economy where the public's disposable income has declined. On 1st July 2007 smoking was banned in public places which many believe had an effect on the Pub trade and in 2008 the Government introduced the Alcohol Duty Escalator, which automatically increased duties at 2% above inflation.
Other factors that have aided the decline in Pubs include cultural changes and a reduction in alcohol drinking and in particular beer consumption decline. Many have preferred at home drinking rather than a trip to their local pub which has also impacted the Pub trade's decline.
How has Pub trade changed?
Almost half of our British Pubs are run by Pub companies with the two largest being Enterprise Inns and Punch Taverns. If you are a tenant running a Pub and paying rent on the Pub building you are bound via contract to purchase drinks from the Pub company at a price that is higher than the wholesale market price. Rather than buying a Pub freehold taking on a lease is beneficial to the licensee as they find this a lower cost method of entering the Pub trade however it also leaves them open to increases in rent and tied into buying drinks at set prices.
The Monopolies and Mergers Commission
The Monopolies and Mergers Commission in 1989 changed the Pub trade deeming it unfair that a small number of brewers dominated the Pub trade. The legislation limited the number of Pubs a brewery could own to 2000. The other important change brought in with this legislation was that Pub tenants could sell a guest beer from another supplier thereby not being totally tied into the brewery.
It is reported that today that even after the efforts of The Monopolies and Mergers Commission 75% of beer drunk in Britain is from four breweries – as in many sectors there are a few companies that dominate the trade.
Don't take on a Problem Pub Building
without a Pub Survey
Get peace of mind
Pubs closing
Recession hit Britain and with the property crash the Pub trade declined this had a big effect on the share prices dropping to a fraction of their previous rate for the likes of Punch Taverns and Enterprise Inns. This brought about closure of many Pubs although some that were closed by the large breweries were taken on by independents.
Read More about Buying a Pub CLICK HERE
Save Your Pub
"Holes in the current planning system allow pubs to be sold off, demolished or converted to many other uses without planning permission or the involvement of the local community. However when a pub is nominated as an Asset of Community Value it automatically receives planning protection meaning it is no longer a soft target to would-be developers looking to quickly purchase and convert or demolish the pub – which in some instances has literally happened overnight."
Tom Stainer, CAMRA's Head of Communications.
Communities have come together to buy their Pub such as the Craufurd Arms in Maidenhead, Berkshire, which was identified as an Asset of Community Value and was purchased by a local community group. The campaign to save the Craufurd Arms and bringing together of the community to take on and run their local Pub is encouraging to anyone wishing to keep the spirit of their local Pub as a centre of a community alive.
Are You taking on a Pub lease?
A Schedule of Condition
will protect you and save you money
Useful information for saving your Pub
Camra.org.uk/save-our-pubs-forum
MyCommunity.org.uk/resources/community-owned-pubs-a-quick-guide-to-saving-your-local/
Urban75.org/info/save-your-pub
Gov.uk/government/get-involved/take-part/take-over-a-local-pub-shop-or-green-space-for-the-community
Do you need to make a quick informed decision on a pub? A Pub Survey will enable you to make the right decision and save money too
Don't wait just give us a call our Surveyors are here to help You Free Phone 0800 298 5424 | {
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Articles from Johns Hopkins
Home > News and Publications > For the Media > News Release Archive
Children With Cystic Fibrosis Not Well Covered By Guidelines For Vitamin D Needs - 10/08/2008
Children With Cystic Fibrosis Not Well Covered By Guidelines For Vitamin D Needs
Hopkins Children's experts call for higher doses to address deficiencies
Existing recommendations for treating vitamin D deficiency in children with cystic fibrosis (CF) are too low to cover the serious need, leaving most at high risk for bone loss and rickets, according to researchers at Johns Hopkins Children's Center.
For more information; http://www.hopkinschildrens.org/newsDetail.aspx?id=5448
Contact JHM Media Team
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Contact a Media Representative | {
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%% Nicholas Czarnek
% SSPACISS/Mazurowski Laboratories, Duke University
% Modified 3 August 2015
%
% balance_data_v2(X_train,Y_train,numObsPerClass,classLabels)
% This function outputs a subset of the input X_train and Y_train based on
% the inputs.
% Example:
% labeledSubset = balance_data(X_train,Y_train,5000*[10 1 1 1 1],[1 2 3 4 5])
function [xSubset,ySubset] = balance_data_v2(X_train,Y_train,numObsPerClass,classLabels)
xSubset = zeros(sum(numObsPerClass),size(X_train,2));
ySubset = zeros(sum(numObsPerClass),1);
dataIdxStart = 1;
dataIdxEnd = numObsPerClass(1);
for cInc = 1:numel(classLabels)
classLogicals = Y_train == classLabels(cInc);
classSubset = X_train(classLogicals,:);
subsetLabels = Y_train(classLogicals);
if numel(subsetLabels)>numObsPerClass(cInc)
keepIdx = randsample(numel(subsetLabels),numObsPerClass(cInc));
else
keepIdx = 1:numel(subsetLabels);
dataIdxEnd = dataIdxStart + numel(subsetLabels) - 1;
end
xSubset(dataIdxStart:dataIdxEnd,:) = classSubset(keepIdx,:);
ySubset(dataIdxStart:dataIdxEnd) = subsetLabels(keepIdx);
dataIdxStart = dataIdxEnd + 1;
if cInc<numel(classLabels)
dataIdxEnd = dataIdxStart + numObsPerClass(cInc + 1) - 1;
end
end
%% If a class did not have enough observations, remove these.
xSubset = xSubset(ySubset ~= 0,:);
ySubset = ySubset(ySubset ~= 0);
end | {
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\section{#1} \setcounter{equation}{0}}
\newenvironment{proof}
{\noindent {\bf Proof.\;}}{$\qed$ \vspace{-0.2 in}\\}
\newenvironment{proofof}[1]{\mbox{} \\
\noindent {\bf Proof of #1. \;}}{$\qed$ \vspace{-0.2 in}\\}
\newenvironment{HW}[1]{\noindent #1.\\ }{$\qed$\vspace{.1in}}
\newenvironment{itm}{\vspace{-1ex}\begin{itemize}}{\end{itemize}}
\def\begin{itm}{\begin{itm}}
\def\end{itm}{\end{itm}}
\newenvironment{oride}[1]{\hspace{-6pt}\tiny \begin{array}{l} #1
\\[-.8ex]}{\end{array} \hspace{-6pt}}
\newenvironment{Hypo}[1]{\par \vspace{-7.3ex} \no $
\begin{array}{cc}
&\hspace{\textwidth} \hspace{-4ex}\\
\makebox[0pt][l]{#1}& \displaystyle } {\end{array} $ \\}
\def\equ_ind{\arabic{section}.\arabic{equation}}
\def\sec_ind{\arabic{section}}
\font\nrm=cmr10 at 9pt \font\nit=cmti10 at 9pt
\font\nsl=cmsl10 at 9pt
\begin{document}
\title{\bf On Quadratic $g$-Evaluations/Expectations and Related Analysis}
\author{
Jin Ma,\thanks{ \noindent Department of Mathematics, Purdue
University, West Lafayette, IN 47907-1395; Department of
Mathematics, University of Southern California, 3620 S. Vermont
Ave., KAP 108, Los Angeles, CA 90089. Email: jinma@usc.edu. This
author is supported in part by NSF grant \#0505427. }\quad Song
Yao\thanks{ \noindent Department of Mathematics, University of Michigan,
Ann Arbor, MI 48109; email: songyao@umich.edu. } }
\date{}
\maketitle
\centerline{\bf Abstract }
\bigskip
In this paper we extend the notion of $g$-evaluation, in particular
$g$-expectation, of Peng \cite{Peng-97, Pln} to the case where the
generator $g$ is allowed to have a quadratic growth (in the variable
``$z$"). We show that
some important properties of the $g$-expectations, including a
representation theorem between the generator and the corresponding
$g$-expectation,
and consequently the reverse comparison theorem of quadratic BSDEs
as well as the Jensen inequality, remain true in the quadratic case.
Our main results also include a Doob-Meyer type decomposition, the
optional sampling theorem, and the up-crossing inequality.
The results of this paper are important in
the further development of the general quadratic nonlinear
expectations (cf. \cite{HMPY-07}).
\vfill \vspace{.5cm} \noindent {\bf Keywords: }\:
Quadratic $g$-evaluations, quadratic $g$-expectations, BMO, reverse
comparison theorem, Jensen's inequality, Doob-Meyer Decomposition,
optional sampling, upcrossing inequality.
\eject
\section{Introduction}
\setcounter{equation}{0}
In this paper we extend the notion of {\it $g$-evaluations},
introduced by Peng \cite{Pln}, to the case when the generator $g$
is allowed to have quadratic growth in the variable $z$. This will
include the so-called quadratic $g$-expectation as a special case,
as was in the linear growth case initiated in \cite{Peng-97}. The
notion of $g$-expectation, as a nonlinear extension of the
well-known Girsanov transformations and originally motivated by
theory of expected utility, has been found to have direct relations
with a fairly large class of risk measures in finance. When the
nonlinear expectation is allowed to have possible quadratic growth,
it is expected that it will lead to the representation theorem that
characterizes the general convex, but not necessarily ``coherent"
risk measures in terms of a class of quadratic BSDEs. The most
notable example of such risk measure is the entropic risk measure
(see, e.g., Barrieu and El Karoui \cite{BarKar}), which is known to
have a representation as the solution to a quadratic BSDE, but falls
outside the existing theory of the ``filtration-consistent nonlinear
expectations" \cite{CHMP}, which requires that the generator be only
of linear growth. We refer the readers to \cite{Peng-97},
\cite{BCHMP}, \cite{CHMP}, and the expository paper \cite{Pln} for
more detailed account for basic properties of $g$-evaluations and
$g$-expectations, as well as the relationship between the risk
measures and $g$-expectations. A brief review of the basic
properties of $g$-evaluations and $g$-expectations will be given in
\S2 for ready references.
The main purpose of this paper is to introduce the notion of
quadratic $g$-evaluation and $g$-expectation, and prove some of the
important properties that are deemed as essential. In an
accompanying paper \cite{HMPY-07} we shall further extend the notion
of filtration consistent nonlinear expectation to the quadratic
case, and establish the ultimate relations between a convex risk
measure and a BSDE. The main results in this paper include the
Doob-Meyer decomposition theorem, optional sampling theorem,
upcrossing inequality, and Jensen's inequality. We also prove that
the quadratic generator can be represented as the limit of
the difference quotients of the corresponding $g$-evaluation,
extending the result in linear growth case \cite{BCHMP}. With the
help of this result, we can then prove the so-called {\it reversed
comparison theorem}, as in the linear case.
Although most of the results presented in this paper look similar to
those in the linear case, the techniques involved in the proofs are
quite different. We combine the techniques used in the study for
quadratic BSDEs, initiated by Kobylanski \cite{Ko} and the by now
well-known properties of the BMO martingales. Since many of these
results are interesting in their own right, we often present full
details of proofs for future references.
\medskip
This paper is organized as follows. In section 2 we give the
preliminaries, and review the existing theory of
$g$-evaluation/expectations and BMO martingales. In section 3 we
define the quadratic $g$-evaluation and discuss its basic
properties. Some fine properties of $g$-evaluations/expectations are
presented in Section 4. These include a representation of quadratic
generator via quadratic $g$-evaluations, a reverse comparison
theorem of quadratic BSDE, and the Jensen's inequality. In section 5
we prove the main results of this paper regarding the quadratic
$g$-martingales: a Doob-Meyer type decomposition, the Optional
Sampling theorem, and the Upcrossing Inequality.
\section{Preliminaries}
\setcounter{equation}{0}
Throughout this paper we consider a filtered, complete probability
space $(\Omega,{\cal F}, P, {\bf F})$ on which is defined a $d$-dimensional
Brownian motion $B$. We assume that the filtration ${\bf F}\stackrel{\triangle}{=}
\{{\cal F}_t\}_{t\ge0}$ is generated by the Brownian motion $B$,
augmented by all $P$-null sets in ${\cal F}$, so that it satisfies the
{\it usual hypotheses} (cf. \cite{Pr-90}). We denote $\mathscr{P}$ to be
the progressively measurable $\sigma$-field on $\Omega\times [0,T]$; and
${\cal M}_{0,T}$ to be the set of all ${\bf F}$-stopping times $\tau$ such
that $0\leq\tau\leq T$, $P$-a.s., where $T>0$ is some fixed
time horizon.
In what follows we fix a finite time horizon $T>0$, and denote $\mathbb{E}$
to be a generic Euclidean space, whose inner product and norm will
be denoted by $\langle\cdot,\cdot\rangle$ and $|\cdot|$,
respectively; and denote $\mathbb{B}$ to be a generic Banach space with
norm $\|\cdot\|$. Moreover,
the following spaces of functions will be frequently used in the
sequel. Let ${\cal G}$ be a generic sub-$\sigma$-field of ${\cal F}$, we denote
\vspace{-5pt}
\begin{itemize}
\item for $0\le p\le\infty$, $L^p({\cal G};\mathbb{E})$ to be all $\mathbb{E}$-valued,
${\cal G}$-measurable random variables $\xi$, with $E(|\xi|^p)<\infty$.
In particular, if $p=0$,
then $L^0({\cal G};\mathbb{E})$ denotes the space of all $\mathbb{E}$-valued,
${\cal G}$-measurable random variables; and if $p=\infty$, then
$L^\infty({\cal G};\mathbb{E})$ denotes the space of all $\mathbb{E}$-valued,
${\cal G}$-measurable random variables $\xi$ such that $\|\xi\|_\infty
\stackrel{\triangle}{=} \underset{\omega \in \Omega}{\mathop{\rm esssup}}|\xi(\omega)|<\infty$;
\item $0 \le p\le\infty$, $L^p_{\bf F}([0,T];\mathbb{B})$ to be all
$\mathbb{B}$-valued, ${\bf F}$-adapted processes $\psi$, such that
$E\int_0^T\|\psi_t\|^pdt<\infty$. In particular, $p=0$ stands for
all $\mathbb{B}$-valued, ${\bf F}$-adapted processes; and $p=\infty$ denotes
all processes $X\in L^0_{\bf F}([0,T];\mathbb{B})$
such that $\|X\|_\infty \stackrel{\triangle}{=} \underset{t,\omega}
{\mathop{\rm esssup}} |X(t,\omega)|<\infty$;
\item $\mathbb{D}^\infty_{\bf F}([0,T];\mathbb{B})=\{X\in L^\infty_{\bf F}([0,T];\mathbb{B}): \hbox{ $X$ has c\`adl\`ag paths}\}$;
\item $\mathbb{C}^\infty_{\bf F}([0,T];\mathbb{B})=\{X\in \mathbb{D}^\infty_{\bf F}([0,T];\mathbb{B}): \hbox{ $X$ has continuous
paths}\}$;
\item ${\cal H}^2_{\bf F}([0,T];\mathbb{B})=\{X \in L^2_{\bf F}([0,T];\mathbb{B}): \hbox{$X$ is predictably
measurable}\}$.
\end{itemize}
Finally, if $d=1$, we shall drop $\mathbb{E}=\mathbb{R}$ from the notation (e.g.,
$L^p_{\bf F}([0,T])=L^p_{\bf F}([0,T];\mathbb{R})$,
$L^\infty({{\cal F}_T})=L^\infty({{\cal F}_T};\mathbb{R})$, and so on).
\bigskip
\noindent {\bf $g$-Evaluations and $g$-Expectations}
\medskip
We first recall the notion of $g$-evaluation introduced in Peng
\cite{Pln}. Given a time duration $[0,T]$, and a ``generator"
$g=g(t,\omega,y,z): [0,T] \times \Omega \times \mathbb{R} \times \mathbb{R}^d \mapsto \mathbb{R}$
satisfying the standard conditions (e.g., it is Lipschitz in all
spatial variables, and is of linear growth, etc.), consider the
following BSDE on $[0,t]$, $t\in[0,T]$:
\begin{eqnarray}
\label{BSDE}
Y_s=\xi+\int_s^t g(r,Y_r,Z_r)dr - \int_s^t Z_r dB_r, \qquad s \in
[0,t],
\end{eqnarray}
where $\xi \in L^2({\cal F}_t)$. Denote the unique solution by
$(Y^{t,\xi},Z^{t,\xi})$.
The {\it $g$-evaluation} is defined as the family of operators
$\big\{{\cal E}^g_{s,t}: L^2({\cal F}_t) \mapsto L^2({\cal F}_s) \big\}_{0 \le s
\le t \le T}$ such that for any $t \in [0,T]$,
$ {\cal E}^g_{s,t}[\xi]\stackrel{\triangle}{=} Y^{t,\xi}_s$, $s \in [0, t]$.
In particular, for any $\xi \in L^2({\cal F}_T)$, its {\it
$g$-expectation} is defined by ${\cal E}^g(\xi)\stackrel{\triangle}{=} Y^{T,\xi}_0$, and
its {\it conditional $g$-expectation} is defined by
${\cal E}^g[\xi|{\cal F}_t] \stackrel{\triangle}{=} {\cal E}^g_{t,T}[\xi]$, for any $t \in [0,T]$. We
shall denote (\ref{BSDE}) by BSDE$(t, \xi,g)$ in the sequel for
notational convenience.
\medskip
\begin{rem} {\rm An important ingredient in the definition of
$g$-evaluation is its ``domain", namely the subset in $L^0({\cal F}_T)$
on which the operator is defined (in the current case being
naturally taken as $L^2({\cal F}_T)$). The domain of a
$g$-evaluation/expectation may vary as the conditions on the
coefficients change, due to the restrictions on the well-posedness
of the BSDE (\ref{BSDE}). For example, owing to the nature of
quadratic BSDEs, in the rest of this paper we shall choose
$L^\infty({\cal F}_T)$ as the domain for quadratic $g$-evaluations. We
refer to our accompanying paper \cite{HMPY-07} for a more detailed
discussion on the issue of domains for general nonlinear
expectations. \qed
}
\end{rem}
By virtue of the uniqueness of the solution $(Y^{t,\xi},Z^{t,\xi})$,
one can show that the $g$-evaluation ${\cal E}^g_{s,t}$ has the following
properties:
\begin{itm}
\item[\(1\)] ({\it Monotonicity}) For any $\xi, \eta \in L^2({\cal F}_t)$ with $\xi \ge \eta $, {\hbox{$P$-a.s.}}, $ {\cal E}^g_{r,t}[\xi] \ge {\cal E}^g_{r,t}[\eta]$, ${\hbox{$P$-a.s.}}$;
\item[\(2\)] ({\it Time-Consistency})~~${\cal E}^g_{r,s}\big[{\cal E}^g_{s,t}[\xi]\big]
={\cal E}^g_{r,t}[\xi]$, ${\hbox{$P$-a.s.}}$, $\xi \in L^2({\cal F}_t)$, $0 \le r \le s \le
t \le T$;
\if{0}
\item[\(3\)] ({\it Constant-Preserving})~~${\cal E}^g_{s,t}[\xi]=\xi$, ${\hbox{$P$-a.s.}}$, $\xi \in
L^({\cal F}_s)$, if it holds ${\hbox{$dt \times dP$-a.s.}}$ that
\begin{eqnarray}\label{g0}
g(t,\omega,y,0)=0, \quad~~~y \in \mathbb{R};
\end{eqnarray}
\fi
\item[\(3\)] ({\it Constant-Preserving})~~${\cal E}^g_{s,t}[\xi]=\xi$, ${\hbox{$P$-a.s.}}$, $\xi \in L^2({\cal F}_s)$,
if it holds $dt\times dP$-a.s. that
\begin{eqnarray}\label{g0}
g(t,y,0)=0, \qquad y \in \mathbb{R};
\end{eqnarray}
\if{0}
\item[\(4\)] (``{\it Zero-one Law}")~~$ {\bf 1}_A{\cal E}^g_{s,t}[{\bf 1}_A \xi]={\bf 1}_A
{\cal E}^g_{s,t}[\xi]$, ${\hbox{$P$-a.s.}}$, $\,\forall \, A \in {\cal F}_s$, $\,\forall \, \xi \in \Lambda_t$, if
\begin{eqnarray} \label{ga1}
{\bf 1}_A \xi \in \Lambda_r, \quad~~ \,\forall \, A \in {\cal F}_r,~ \,\forall \, \xi
\in \Lambda_r, ~ \,\forall \, r \in [0,T];
\end{eqnarray}
\fi
\item[\(4\)] (``{\it Zero-one Law}")~~For any $\xi \in
L^2({\cal F}_t)$ and any $A \in {\cal F}_s$, $s \in [0,t]$, it holds that
$$ {\bf 1}_A {\cal E}^g_{s,t}[\xi]={\bf 1}_A{\cal E}^g_{s,t}[{\bf 1}_A \xi], \quad{\hbox{$P$-a.s.}}
$$
Moreover, if $g(t,0,0)=0$, $dt\times dP$-a.s., then ${\bf 1}_A
{\cal E}^g_{s,t}[\xi]={\cal E}^g_{s,t}[{\bf 1}_A \xi]$, ${\hbox{$P$-a.s.}}$;
\if{0}
\item[\(5\)] (Translation Invariance)~~${\cal E}^g_{s,t}[\xi+\eta]={\cal E}^g_{s,t}[\xi]+\eta$,
${\hbox{$P$-a.s.}}$, $\xi \in L^2({\cal F}_t)$, $\eta \in L^2({\cal F}_s)$, if $g$ is
independent of $y$ and $\Lambda$ is closed under addition.
\fi
\item[\(5\)] ({\it Translation Invariance})~~Assume that $g$ is independent of $y$, then for
any $\xi \in L^2({\cal F}_t)$ and $\eta \in L^2({\cal F}_s)$,
it holds that $ {\cal E}^g_{s, t}[\xi+\eta]={\cal E}^g_{s,t}[\xi]+\eta$,
${\hbox{$P$-a.s.}}$
\end{itm}
Clearly, if $g$ satisfies (\ref{g0}), then one can deduce from (2)
and (3) above
that
\begin{eqnarray}\label{coin}
{\cal E}^g[\xi|{\cal F}_s]={\cal E}^g_{s,T}[\xi]={\cal E}^g_{s,t}\big[{\cal E}^g_{t,T}[\xi]\big]={\cal E}^g_{s,t}[\xi],\quad
{\hbox{$P$-a.s.}}, ~~~ \xi \in L^2({\cal F}_t), ~~~ 0 \le s \le t \le T;
\end{eqnarray}
and the conditional $g$-expectation ${\cal E}^g\{\cdot|{\cal F}_t\}$ possesses
the following properties that more or less justify its name
(assuming (\ref{g0}) for (2a) and (3a) below):
\begin{itm}
\item[\(1a\)] ({\it Monotonicity}) For any $\xi, \eta \in L^2({\cal F}_T)$ with $\xi \ge \eta $, {\hbox{$P$-a.s.}}, ${\cal E}^g[\xi|{\cal F}_t] \ge {\cal E}^g[\eta |{\cal F}_t] $, ${\hbox{$P$-a.s.}}$;
\item[\(2a\)] (Time-Consistency)~~${\cal E}^g\big[{\cal E}^g[\xi|{\cal F}_t]\big|{\cal F}_s\big]={\cal E}^g[\xi|{\cal F}_s]$,
${\hbox{$P$-a.s.}}$, $\xi \in L^2({\cal F}_T)$, $s \in [0,t]$;
\item[\(3a\)] (Constant-Preserving)~~${\cal E}^g[\xi|{\cal F}_t]=\xi$, ${\hbox{$P$-a.s.}}$, $\xi \in
L^2({\cal F}_t)$;
\item[\(4a\)] (Zero-one Law)~~For any $\xi \in
L^2({\cal F}_T)$ and $A \in {\cal F}_t$, it holds that
${\bf 1}_A {\cal E}^g[{\bf 1}_A \xi |{\cal F}_t]={\bf 1}_A {\cal E}^g[ \xi |
{\cal F}_t]$, ${\hbox{$P$-a.s.}}$; Moreover, if $g(t,0,0)=0$, ${\hbox{$dt \times dP$-a.s.}}$, then ${\bf 1}_A
{\cal E}^g[\xi|{\cal F}_t]={\cal E}^g[{\bf 1}_A \xi|{\cal F}_t]$, ${\hbox{$P$-a.s.}}$;
\item[\(5a\)] (Translation Invariance)~~Assume that $g$ is independent of $y$, then
for any $\xi \in L^2({\cal F}_T)$ and $\eta \in L^2({\cal F}_t)$ it holds that
$ {\cal E}^g[\xi+\eta|{\cal F}_t]={\cal E}^g[\xi|{\cal F}_t]+\eta$, ${\hbox{$P$-a.s.}}$
\end{itm}
\medskip
\noindent {\bf BMO Martingales and BMO Processes}
\medskip
An important tool for studying the quadratic BSDEs, whence the
quadratic $g$-expectations, is the so-called ``BMO martingales"
and the related stochastic exponentials (see, e.g., \cite{HIM}). We
refer to the monograph of Kazamaki \cite{Ka} for a complete
exposition of the theory of continuous BMO and exponential
martingales. In what follows we shall list some of the important
facts that are useful in our future discussions for ready
references.
To begin with, we recall that a uniformly integrable martingale $M$
null at zero is called a ``BMO martingale" on $[0,T]$ if for some
$1\le p<\infty$, it holds that
\begin{eqnarray}
\label{BMOp}
\|M\|_{BMO_p}\stackrel{\triangle}{=}\sup_{\tau\in{\cal M}_{0,T}}\Big\|E\{|M_T-M_{\tau-}|^p\big|
{\cal F}_\tau\}^{1/p}\Big\|_\infty<\infty.
\end{eqnarray}
In such a case we denote $M\in$BMO$(p)$. It is important to note
that $M\in$BMO$(p)$ if and only if $M\in$BMO$(1)$, and all the
BMO$(p)$ norms are equivalent (cf. \cite{Ka}). Therefore in what
follows we shall say that a martingale $M$ is BMO without specifying
the index $p$; and we shall use only the BMO$(2)$ norm and denote it
simply by $\|\cdot\|_{BMO}$. Note also that for a {\it continuous}
martingale $M$ one has
\begin{eqnarray*}
\|M\|_{BMO}=\|M\|_{BMO_2}=\sup_{\tau\in{\cal M}_{0,T}}
\Big\|E\{\langle M\rangle_T-\langle M\rangle_\tau\big|{\cal F}_\tau\}^{1/2}\Big\|_\infty.
\end{eqnarray*}
For a given Brownian motion $B$, we say that a process $Z\in
L^2_{\bf F}([0,T];\mathbb{R}^d)$ is a BMO process, denoted by $Z \in$ BMO by a
slight abuse of notations, if the stochastic integral $M\stackrel{\triangle}{=} Z\cdot
B=\int Z_tdB_t$ is a BMO martingale.
Next,
for a continuous martingale $M$, the Dol\'eans-Dade stochastic
exponential of $M$, denoted customarily by $\mathscr{E}(M)$, is defined as
$\mathscr{E}(M)_t\stackrel{\triangle}{=} \exp\{M_t-\frac12\langle M\rangle_t\}$, $t\ge 0$. If $M$ is
further a BMO martingale, then the stochastic exponential $\mathscr{E}(M)$
is itself a uniformly integrable martingale (see \cite[Theorem
2.3]{Ka}).
The theory of BMO was brought into the study of quadratic BSDEs for
the following reason. Consider, for example, the
BSDE$(T,\xi,g)$ (see (\ref{BSDE})) where the generator $g$ has a
quadratic growth. Assume that
there is some $k>0$ (we may assume without loss of generality that
$k \ge \frac{1}{2}$) such that for $dt \times dP$-a.s. $(t,\omega) \in
[0,T] \times \Omega$,
\begin{eqnarray}
\label{gquad}
|g(t,\omega,y,z)| \leq k(1+|z|^2), \qquad (y,z) \in \mathbb{R} \times \mathbb{R}^d,
\end{eqnarray}
and denote $(Y,Z)\in \mathbb{C}^\infty_{\bf F}([0,T]) \times
{\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$ be a solution of the BSDE$(T,\xi,g)$.
For any $\tau\in{\cal M}_{0,T}$, applying It\^o's formula to $\displaystyle e^{4kY_t}$ from $\tau$ to $T$ one has
\begin{eqnarray*}
e^{4kY_{\tau}}+8k^2\int_{\tau}^T e^{4kY_s}|Z_s|^2ds &=& e^{4kY_T} +
4k\int_{\tau}^T e^{4kY_s}g(s,Y_s,Z_s)ds-4k\int_{\tau}^T e^{4kY_s}Z_sdB_s
\nonumber \\
&\leq& e^{4kY_T}+4k^2\int_{\tau}^Te^{4kY_s}\big(1+|Z_s|^2\big)ds
-4k\int_{\tau}^T e^{4kY_s}Z_sdB_s.
\end{eqnarray*}
It is then not hard to derive, using
some standard arguments, the following estimate:
\begin{eqnarray}\label{bmo1}
E\big[\int_{\tau}^T |Z_s|^2 ds|{\cal F}_{\tau}\big] \leq e^{4k \|Y \|_\infty}
E\big[e^{4k\xi}-e^{4kY_{\tau}}|{\cal F}_{\tau}\big]+e^{8k \| Y \|_\infty}(T-{\tau}).
\end{eqnarray}
In other words, we conclude that
$Z\in$ BMO, and that
\begin{eqnarray}
\label{BMO1}
\|Z \|^2_{BMO} \leq (1+T)e^{8k \| Y \|_\infty}.
\end{eqnarray}
\section{Quadratic $g$-Evaluations on $L^\infty({\cal F}_T)$}
\setcounter{equation}{0}
Our study of the $g$-evaluation/expectation benefited greatly from
the techniques used to treat the quadratic BSDEs, initiated by
Kobylanski \cite{Ko}.
We first list some results regarding the existence, uniqueness, and
comparison theorems for the quadratic BSDEs.
Throughout the rest of the paper we assume that the generator $g$ in
BSDE($T,\xi,g$) (\ref{BSDE}) takes the form:
\begin{eqnarray*}
g(t,\omega,y,z)=g_1(t,\omega,y,z)y+g_2(t,\omega,y,z), \quad~~ \,\forall \, (t,\omega,y,z) \in
[0,T] \times \Omega \times \mathbb{R} \times \mathbb{R}^d,
\end{eqnarray*}
and satisfies the following {\it Standing Assumptions}:
\begin{itm}
\item[{\bf (H1)}] Both $g_1$ and $g_2$ are $\mathscr{P} \otimes \mathscr{B}(\mathbb{R}) \otimes \mathscr{B}(\mathbb{R}^d)$-measurable
and both $g_1(t,\omega,\cdot,\cdot)$ and $g_2(t,\omega,\cdot,\cdot)$ are continuous for any $(t,\omega) \in
[0,T] \times \Omega$;
\item[{\bf (H2)}] There exist a constant $k>0$ and an increasing function $\ell:
\mathbb{R}^+ \mapsto \mathbb{R}^+$, such that for $dt \times dP$-a.s. $(t,\omega) \in
[0,T] \times \Omega$, \begin{eqnarray*} |g_1(t,\omega,y,z)| \leq k \quad\mbox{and} \quad
|g_2(t,\omega,y,z)| \leq k+\ell(|y|)|z|^2,\quad~~ (y,z) \in \mathbb{R} \times
\mathbb{R}^d; \end{eqnarray*}
\item[{\bf (H3)}] With the same increasing function $\ell$, for $dt \times dP$-a.s.
$(t,\omega) \in [0,T] \times \Omega$,
\begin{eqnarray*}
\Big|\frac{\partial g}{\partial z}(t,\omega,y,z)\Big| \leq
\ell(|y|)(1+|z|),\qquad (y,z) \in \mathbb{R} \times \mathbb{R}^d;
\end{eqnarray*}
\item[{\bf (H4)}] For any $ \varepsilon > 0$, there exists a positive function $h_\varepsilon(t) \in L^1[0,T]$
such that for $dt \times dP$-a.s. $(t,\omega) \in [0,T] \times \Omega$,
\begin{eqnarray*} \frac{\partial g}{\partial y}(t,\omega,y,z) \leq h_\varepsilon(t)+\varepsilon|z|^2,\qquad (y,z)
\in \mathbb{R} \times \mathbb{R}^d. \end{eqnarray*}
\end{itm}
Under the assumptions (H1)-(H4), it is known (cf. \cite[Theorem 2.3
and 2.6]{Ko}) that for any $\xi \in L^\infty({\cal F}_T)$, the BSDE
(\ref{BSDE}) admits a unique solution
$(Y,Z) \in \mathbb{C}^\infty_{\bf F}([0,T]) \times {\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$.
In fact, this result can be extended to the following more general
form, which will be useful in our future discussion.
\begin{prop}\label{BSDEV}
Assume that $g$ satisfies (H1)-(H4). For any $\xi \in
L^\infty({\cal F}_T)$ and any $V \in \mathbb{D}^\infty_{\bf F}([0,T])$, the BSDE
\begin{eqnarray}\label{BSDE1} Y_t=\xi+\int_t^T g(s,Y_s,Z_s)ds +V_T-V_t-\int_t^T
Z_s dB_s, \qquad t \in [0,T], \end{eqnarray} admits a unique solution $(Y,Z) \in
\mathbb{D}^\infty_{\bf F}([0,T]) \times {\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$.
\end{prop}
{\it Proof.} We define a new generator $\tilde{g}$ by
$\tilde{g}(t,\omega,y,z) \stackrel{\triangle}{=} g(t,\omega,y-V_t(\omega),z),~ (t,\omega,y,z) \in
[0,T] \times \Omega \times \mathbb{R} \times \mathbb{R}^d$. Then it is easy to see
that for any $(t,\omega,y,z) \in [0,T] \times \Omega \times \mathbb{R} \times
\mathbb{R}^d$
\begin{eqnarray*}
\tilde{g}_1(t,\omega,y,z)&=& g_1(t,\omega,y-V_t(\omega),z),\\
\tilde{g}_2(t,\omega,y,z) &=& g_2(t,\omega,y-V_t(\omega),z)-g_1(t,\omega,y-V_t(\omega),z)
V_t(\omega).
\end{eqnarray*}
It can be easily verified that $\tilde{g}$ also satisfies (H1)-(H4).
We can then conclude (see, \cite{Ko}) that the
BSDE$(T,\xi+V_T,\tilde{g})$ admits a unique solution $(\tilde{Y},Z)
\in \mathbb{C}^\infty_{\bf F}([0,T]) \times {\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$. But this
amounts to saying that $(\tilde{Y}-V,Z)$
is the unique solution of (\ref{BSDE1}), proving the corollary.
\qed
\medskip
Proposition \ref{BSDEV} indicates that if $g$ satisfies (H1)-(H4),
then we can again define a $g$-evaluation ${\cal E}^g_{s,t}:
L^\infty({\cal F}_t)\mapsto L^\infty({\cal F}_s)$ for $0 \le s \le t \le T$,
as in the previous section. We shall name it as the ``{\it quadratic
$g$-evaluation/expectation}" for obvious reasons. More generally,
for any $\sigma$, $\tau \in {\cal M}_{0,T}$ such that $\sigma \le \tau$, ${\hbox{$P$-a.s.}}$, we
can define the quadratic $g$-evaluation ${\cal E}^g_{\sigma,\tau}:
L^\infty({\cal F}_\tau) \mapsto L^\infty({\cal F}_\sigma)$ by ${\cal E}^g_{\sigma,\tau}[\xi]
\stackrel{\triangle}{=} Y^\xi_\sigma$, where $\xi \in L^\infty({\cal F}_\tau)$, and $Y^\xi$
satisfies the BSDE:
\begin{eqnarray}\label{BSDEtau}
Y^\xi_t=\xi+\int_t^T {\bf 1}_{\{s < \tau
\}}g(s,Y^\xi_s,Z^\xi_s)ds-\int_t^T Z^\xi_sdB_s,\qquad t \in [0,T].
\end{eqnarray}
with $Z^\xi \in {\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$,
and
$Y^\xi_t=Y^\xi_{t \land \tau}$ and $Z^\xi_t={\bf 1}_{\{t < \tau \}}Z^\xi_t$,
${\hbox{$P$-a.s.}}$ In particular, if $\tau=T$, we
define the quadratic $g$-expectation of $\xi$ for any $\sigma \in
{\cal M}_{0,T}$ by ${\cal E}^g[\xi|{\cal F}_\sigma] \stackrel{\triangle}{=} {\cal E}^g_{\sigma,T}[\xi]$.
We note that, similar to the deterministic-time case,
${\cal E}^g_{\sigma,\tau} $ has the following properties: \begin{itm}
\item[\(1\)] {\it Time-Consistency:}~~For any $\rho$, $\sigma$,
$\tau \in {\cal M}_{0,T}$ with $\rho \le \sigma \le \tau$, ${\hbox{$P$-a.s.}}$, we have
\begin{eqnarray*}
{\cal E}^g_{\rho,\sigma}\big[{\cal E}^g_{\sigma,\tau}[\xi]\big]={\cal E}^g_{\rho,\tau}[\xi],
\quad~~ {\hbox{$P$-a.s.}} \quad \,\forall \, \xi \in L^\infty({\cal F}_\tau);
\end{eqnarray*}
\item[\(2\)] {\it Constant-Preserving:}~~Assume (\ref{g0}),
${\cal E}^g_{\sigma,\tau}[\xi]=\xi$, ${\hbox{$P$-a.s.}}$, $\,\forall \, \xi \in
L^\infty({\cal F}_\sigma)$;
\item[\(3\)] {\it ``Zero-one Law":}~~For any $\xi \in L^\infty({\cal F}_\tau)$
and $A \in {\cal F}_\sigma$, we have ${\bf 1}_A{\cal E}^g_{\sigma,\tau}[{\bf 1}_A \xi]={\bf 1}_A
{\cal E}^g_{\sigma,\tau}[\xi]$, ${\hbox{$P$-a.s.}}$; Moreover, if $g(t,0,0)=0$, ${\hbox{$dt \times dP$-a.s.}}$,
then ${\cal E}^g_{\sigma,\tau}[{\bf 1}_A \xi]={\bf 1}_A {\cal E}^g_{\sigma,\tau}[\xi]$, ${\hbox{$P$-a.s.}}$;
\item[\(4\)] {\it ``Translation Invariant":}~~If $g$ is independent of $y$, then
\begin{eqnarray*}
{\cal E}^g_{\sigma,\tau}[\xi+\eta]={\cal E}^g_{\sigma,\tau}[\xi]+\eta,\quad {\hbox{$P$-a.s.}} \quad
\,\forall \, \eta \in L^\infty({\cal F}_\sigma),\quad \xi \in L^\infty({\cal F}_\tau).
\end{eqnarray*}
\item[\(5\)] {\it Strict Monotonicity:}~~For any $\xi,\eta \in L^\infty({\cal F}_\tau)$ with $\xi \geq \eta$,
${\hbox{$P$-a.s.}}$, we have ${\cal E}^g_{\sigma,\tau}[\xi] \geq {\cal E}^g_{\sigma,\tau}[\eta]$,
${\hbox{$P$-a.s.}}$; Moreover, if ${\cal E}^g_{\sigma,\tau}[\xi] = {\cal E}^g_{\sigma,\tau}[\eta]$,
${\hbox{$P$-a.s.}}$, then $\xi=\eta$, ${\hbox{$P$-a.s.}}$
\end{itm}
We remark that the last property (5) above is not completely
obvious. In fact this will be a consequence of so-called ``strict
comparison theorem" for quadratic BSDEs, a strengthened version of
the usual comparison theorem (see, for example, \cite[Theorem
2.6]{Ko}). For completeness we shall present such a version, under
the following conditions that are similar to those in \cite{Ko}, but
slightly weaker than (H1)--(H4).
\begin{itm}
\item[{\bf (A1)}]~$g$ is $\mathscr{P} \otimes \mathscr{B}(\mathbb{R}) \otimes \mathscr{B}(\mathbb{R}^d)$-measurable
and $g(t,\omega,\cdot,\cdot)$ is continuous for any $(t,\omega) \in [0,T]
\times \Omega$;
\item[{\bf (A2)}]~For any $M>0$, there exist $\ell \in L^1 [0,T], k \in L^2 [0,T]$
and $C>0$ such that for $dt \times dP$-a.s. $(t,\omega) \in [0,T] \times
\Omega$ and any $(y,z) \in [-M,M]\times \mathbb{R}^d$,
\begin{eqnarray*}
\big|g(t,\omega,y,z)\big| \leq \ell(t)+C|z|^2 \quad\mbox{and}\quad
\Big|\frac{\partial g}{\partial z}(t,\omega,y,z)\Big| \leq k(t)+C|z|;
\end{eqnarray*}
\item[{\bf (A3)}]~For any $\varepsilon>0$, there exists a positive function $h_\varepsilon \in L^1[0,T]$
such that for $dt \times dP$-a.s. $(t,\omega) \in [0,T] \times \Omega$ and
any $(y,z) \in \mathbb{R} \times \mathbb{R}^d$,
\begin{eqnarray*} \frac{\partial g}{\partial y}(t,\omega,y,z)\leq h_\varepsilon(t)+\varepsilon|z|^2.
\end{eqnarray*} \end{itm}
\begin{thm}
Assume (A1)-(A3). Let $\xi^1, \xi^2 \in L^\infty({\cal F}_T)$ and $V^i$,
$i=1,2$ be two adapted, integrable, right-continuous processes null at $0$. Let $\big(Y_t^i,Z_t^i\big) \in
\mathbb{D}^\infty_{\bf F}([0,T]) \times {\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$, $i=1,2$
be solutions to the BSDEs:
\begin{eqnarray*}
Y^i_t=\xi^i+\int_t^T g(s,Y_s^i,Z_s^i)ds+\int_t^T d V_s^i-\int_t^T
Z_s^i d B_s, \quad~~ t \in [0,T],\quad i=1,2,
\end{eqnarray*}
respectively. If $\xi^1 \geq \xi^2$, ${\hbox{$P$-a.s.}}$ and $ V_t^1-V_t^2$ is
increasing, then it holds {\hbox{$P$-a.s.}}~that
\begin{eqnarray}\label{scomp1}
Y_t^1 \geq Y_t^2, \qquad t \in [0,T].
\end{eqnarray}
Moreover, if $Y_\tau^1 =
Y_\tau^2$ for some $\tau \in {\cal M}_{0,T}$, then it holds {\hbox{$P$-a.s.}} ~ that
\begin{eqnarray}\label{scomp2}
\xi^1 = \xi^2, \quad~~ \mbox{and}\quad~~ V^1_T- V^2_T= V^1_\tau - V^2_\tau .
\end{eqnarray}
\end{thm}
{\it Proof.} It is not hard to see that (\ref{scomp1}) is a mere generalization of \cite[Theorem 2.6]{Ko}, thus we only need to
prove (\ref{scomp2}). Let $M\stackrel{\triangle}{=}\|Y^1\|_\infty+\| Y^2 \|_\infty$,
and define $\Delta\eta=\eta^1-\eta^2$ for $\eta=Y$, $Z$, $V$,
respectively. Then $\Delta Y$ satisfies:
\begin{eqnarray}
\label{DeltaY}
d \Delta Y_t&=&
-\big(g(t,Y^1_t,Z^1_t)-g(t,Y^2_t,Z^2_t)\big)dt-d \Delta V_t+\Delta Z_t d B_t \nonumber\\
&=& -\int_0^1 \Big(\frac{\partial g}{\partial y}(\Xi_t^\lambda)\Delta Y_t + \frac{\partial
g}{\partial z}(\Xi_t^\lambda)\Delta Z_t\Big) d\lambda d t -d \Delta V_t+\Delta Z_t d B_t\\
&=& -a_t\Delta Y_t d t -d \Delta V_t+\Delta Z_t(-b_tdt+ d B_t),\nonumber
\end{eqnarray}
where
$\Xi_t^\lambda\stackrel{\triangle}{=} (t,\lambda \Delta Y_t + Y_t^2 ,\lambda\Delta Z_t + Z_t^2)$, and
\begin{eqnarray*}
a_t\stackrel{\triangle}{=} \int_0^1 \frac{\partial g}{\partial y}(\Xi_t^\lambda) d \lambda
\quad~\mbox{and}\quad~ b_t \stackrel{\triangle}{=} \int_0^1 \frac{\partial g}{\partial z}(\Xi_t^\lambda) d
\lambda, \qquad t \in [0,T].
\end{eqnarray*}
Note that $|\lambda \Delta Y_t + Y_t^2|\leq M,~\,\forall \, t \in [0,T]$, ${\hbox{$P$-a.s.}}$, by
using some standard arguments with the help of assumptions
(A1)--(A3) as well as
the Burkholder-Davis-Gundy inequality we deduce from (\ref{DeltaY})
that
\begin{eqnarray}
\label{bddab}
E \Big\{ \underset{t\in[0,T]}{\sup}\int_0^t a_sd s + \underset{t\in[0,T]}{\sup} \Big|
\int_0^t b_sd B_s \Big| \Big\} <
\infty.
\end{eqnarray}
Define $Q_t\stackrel{\triangle}{=} \exp \Big\{ \int_0^t a_sd s-\frac{1}{2}\int_0^t
|b_s|^2 d s +\int_0^t b_sd B_s\Big\}$, $t\ge 0$, and
\begin{eqnarray*}
\tau_n \stackrel{\triangle}{=} \inf\big\{ t \in [\tau,T]: Q_t>n
\big\}\land T, \quad~~ n \in \mathbb{N},
\end{eqnarray*}
we see that $\tau_n\mathop{\uparrow} T$, ${\hbox{$P$-a.s.}}$, and (\ref{bddab}) indicates that
there exists a null set $\mathscr{N}$ such that for each $\omega\in\mathscr{N}^c$,
$T=\tau_m (\omega)$ for some $m \in \mathbb{N}$. On the other hand, for any $n
\in \mathbb{N}$, integrating by parts on $[\tau,\tau_n]$ yields that
\begin{eqnarray*} Q_{\tau_n}\Delta Y_{\tau_n} &\neg \neg \neg=\negthinspace\neg \neg& Q_\tau \Delta Y_\tau -
\int_\tau^{\tau_n} Q_t \Delta Y_t a_t d t
- \int_\tau^{\tau_n} Q_t \Delta Z_t b_t dt - \int_\tau^{\tau_n} Q_t d \Delta V_t \\
& &+ \negthinspace \int_\tau^{\tau_n} \negthinspace Q_t \Delta Z_t d B_t
+ \negthinspace \int_\tau^{\tau_n}\negthinspace \Delta Y_t Q_t a_t d t +\negthinspace \int_\tau^{\tau_n}\negthinspace \Delta Y_t Q_t b_t d B_t
+\negthinspace \int_\tau^{\tau_n}\negthinspace Q_t \Delta Z_t b_t d t \\
&\neg \neg \neg=\neg \neg \neg& - \int_\tau^{\tau_n} Q_t d \Delta V_t + \int_\tau^{\tau_n}
Q_t \Delta Z_t d B_t + \int_\tau^{\tau_n} \Delta Y_t Q_t b_t
d B_t.
\end{eqnarray*}
Taking expectation on both sides gives:
$$ E\Big\{Q_{\tau_n}\Delta Y_{\tau_n} + \int_\tau^{\tau_n} Q_t d \Delta V_t \Big\}=0,
$$
which implies that there exists a null set $\mathscr{N}_n$ such that for any
$\omega \in {\mathscr{N}_n}^c$, it holds that $\Delta Y_{\tau_n(\omega)}(\omega)=0$ and $
\Delta V_{\tau_n(\omega)}(\omega) = \Delta V_{\tau(\omega)}(\omega) $. Therefore, for
any $\omega \in \Big\{ \mathscr{N} \cup \big( \underset{n \in \mathbb{N}}{\cup}\mathscr{N}_n
\big)\Big\}^c$, one has
\begin{eqnarray*}
\Delta Y_T(\omega)=0 \quad \mbox{and}\quad \Delta V_T(\omega) = \Delta V_{\tau(\omega)}(\omega) .
\end{eqnarray*}
This completes the proof. \qed
In most of the discussion below, we assume the generator $g$
satisfies (H1)-(H4) (hence (A1)-(A3)). We first extend a property of
$g$-expectations \cite[Proposition 3.1]{BCHMP} to the case of
quadratic $g$-evaluations.
\begin{prop}
Assume (H1)--(H4). Assume further that the generator $g$ is
deterministic. For any $t\in[0,T]$ and $\xi \in L^\infty({\cal F}_t)$, if $\xi $ is independent of ${\cal F}_s$ for some $s\in[0, t)$, then the random variable ${\cal E}^g_{s,t}[\xi]$
is deterministic.
\end{prop}
{\it Proof:} Let $0\le s<t\le T$ be such that $\xi\in
L^\infty({\cal F}_t)$ and that it is independent of ${\cal F}_s$. It suffices
to show that ${\cal E}^g_{s,t}[\xi]=c$, ${\hbox{$P$-a.s.}}$ for some constant $c$.
To see this, for any $r \in [0,t-s]$, we define $B'_r=B_{s+r}-B_s$,
${\cal F}'_r=\sigma\big(B'_u, u\in [0,r]\big)$, and ${\bf F}'=\{{\cal F}'_r\}_{r \in [0,t-s]}$.
Clearly, $B'$ is an ${\bf F}'$-Brownian motion on $[0,t-s]$. Since $\xi
\in {\cal F}_t$ is independent of ${\cal F}_s$, one can easily deduce that
$\xi \in {\cal F}'_{t-s}$. Now we denote by $\{(Y'_r,Z'_r)\}_{r \in
[0,t-s]}$ the unique solution to the BSDE:
\begin{eqnarray*}
Y'_r=\xi+\int_r^{t-s}g(s+u,Y'_u,Z'_u)du-\int_r^{t-s} Z'_u dB'_u,
\quad~~~ r \in [0,t-s].
\end{eqnarray*}
The simple change of variables $r=v-s$ and $w=s+u$ yields that
\begin{eqnarray*}
Y'_{v-s}&=&\xi+\int_v^t g(w, Y'_{w-s},Z'_{w-s})d w-\int_v^t
Z'_{w-s}d B'_{w-s} \\
&=& \xi+\int_v^t g(w, Y'_{w-s},Z'_{w-s})d w-\int_v^t
Z'_{w-s}d B_w, \quad~~~ v \in [s,t].
\end{eqnarray*}
In other words, $\{(Y'_{v-s},Z'_{v-s})\}_{v \in [s,t]}$ is a
solution to BSDE$(t,\xi,g)$ on $[s,t]$. The uniqueness of the
solution to BSDE then leads to that
$Y'_{v-s}={\cal E}^g_{v,t}[\xi],$ $ v \in [s,t]$. In particular, one has
${\cal E}^g_{s,t}[\xi]=Y'_0$, ${\hbox{$P$-a.s.}}$, which is a constant by the
definition of ${\bf F}'$ and the
Blumenthal $0$-1 law, completing the proof.
\qed
\medskip
As we can see from the discussion so far, so long as the
corresponding quadratic BSDE is well-posed, the resulting
$g$-evaluation/expectation should behave very similarly to those
with linear growth generators, with almost identical proofs using
the properties obtained so far. We therefore conclude this section
by listing some further properties of the $g$-evaluation/expection
in one proposition for ready references, and leave the proofs to the
interested reader.
\begin{prop}
\label{gevcomp} Let $g_i$, $i=1,2$, be two generators both satisfy
(H1)-(H4).
\begin{itemize}
\item[{1)}] Suppose that $g_i(t,0,0)=0$, $i=1,2$, and that
\begin{eqnarray} \label{gicomp}
{\cal E}^{g_1}_{0,t}[\xi]={\cal E}^{g_2}_{0,t}[\xi],\quad~~\,\forall \, t \in [0,T],~~~\,\forall \,
\xi \in L^\infty({\cal F}_t),
\end{eqnarray}
then for any $ \xi \in L^\infty({\cal F}_T)$, it holds ${\hbox{$P$-a.s.}}$ that
${\cal E}^{g_1}_{t,T}[\xi]={\cal E}^{g_2}_{t,T}[\xi]$, $\,\forall \, t \in [0,T]$.
\item[{2)}] Suppose further that $g_i$, $i=1,2$ are independent of $y$,
For any $t \in [0,T]$, if
${\cal E}^{g_1}_{0,t}[\xi]\le {\cal E}^{g_2}_{0,t}[\xi]$, $ \,\forall \, \xi \in L^\infty({\cal F}_t)$, then for any $\xi \in L^\infty({\cal F}_t)$, it holds
${\hbox{$P$-a.s.}}$ that ${\cal E}^{g_1}_{s,t}[\xi]\le {\cal E}^{g_2}_{s,t}[\xi]$, $\,\forall \, s
\in [0,t]$.
\end{itemize}
\end{prop}
\medskip
To end this section, we state a stability result of quadratic BSDEs
which is a slight generalization of Theorem 2.8 in \cite{Ko}. Since there is no
substantial difference in the proof, we omit it.
\begin{thm}
\label{stable} Let $\{g_n\}$ be a sequence of generators satisfying (H1) and (H2) with the
same constant $k>0$ and increasing function $\ell$. Denote, for each $n \in \mathbb{N}$, $(Y^n,Z^n) \in
\mathbb{C}^\infty_{\bf F}([0,T]) \times {\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$ to be a
solution of BSDE$(T,\xi_n,g_n)$ with $\xi_n \in L^\infty({\cal F}_T)$.
Suppose that $\{\xi_n\}$ is a bounded sequence in $L^\infty({\cal F}_T)$, and converges $P$-a.s. to some
$\xi \in L^\infty({\cal F}_T)$; and that
for $dt \times dP$-a.s. $(t,\omega) \in [0,T] \times \Omega$, $\{g_n(t,\omega,y,z)\}$ converges to $g(t,\omega,y,z)$ locally uniformly in
$(y,z) \in \mathbb{R}\times \mathbb{R}^d$ with $g$ satisfying (H1)-(H4). Then
BSDE$(T,\xi,g)$ admits a unique solution $(Y,Z) \in
\mathbb{C}^\infty_{\bf F}([0,T]) \times {\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$ such that
$P$-a.s. $Y^n_t$ converges to $Y_t$ uniformly in $t \in [0,T]$ and
that $Z^n$ converges to $Z$ in ${\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$.
\end{thm}
\section{Some Fine Properties of Quadratic $g$-Evaluations}
In this section we extend some fine properties of $g$-evaluation to
the quadratic case. These properties have been discovered for
different reasons in the linear growth cases, and they form an
integral part of the theory of nonlinear expectation. In the
quadratic case, however, the proofs need to be adjusted, sometimes
significantly. We collect some of them here for the distinguished
importance.
\smallskip We begin by a representation theorem for the generators via
quadratic $g$-expectation.
\begin{thm} \label{repre3}
Assume (H1)--(H4). Let $ ( t,y,z) \in [0,T) \times \mathbb{R} \times \mathbb{R}^d$. If $g$ satisfies
\smallskip \noindent \(g1\) $ \displaystyle \underset{ (s, \,y' ) \to
(t^+ \negthinspace ,\,y)}{\lim} \, g(s,y',z) = g(t,y,z)$, {\hbox{$P$-a.s.}}~ and
\medskip \noindent \(g2\) For some $\varepsilon_0 \in (0,T-t]$ and some $\delta>0$, there exists an
integrable process $\{\widetilde{h}_s\}_{s \in [t,t+\varepsilon_0]}$ such that
for $dt \times dP$-a.s. $(s,\omega) \in [t,t+\varepsilon_0] \times \Omega$,
\begin{eqnarray*}
\qquad \quad \frac{\partial g}{\partial
y'}(s, y',z) \ge \widetilde{h}_s,\quad \,\forall \, y' \in \mathbb{R} \hbox{ with } |y'-y| \le \delta ,
\end{eqnarray*}
then it holds ${\hbox{$P$-a.s.}}$ that
\begin{eqnarray*}
g(t,y,z)=\underset{\varepsilon \searrow 0}{\lim}\,\frac{1}{\varepsilon}\big({\cal E}^g_{t,(t+\varepsilon)\land \tau
}[y+z(B_{(t+\varepsilon)\land \tau}-B_t)]-y\big) ,
\end{eqnarray*}
where $\tau \stackrel{\triangle}{=}\inf\big\{s>t: |B_s-B_t|>\frac{\delta}{1+|z|} \big\}\land T$.
\end{thm}
{\it Proof:}
We set $M \stackrel{\triangle}{=} 1+ |y|+ \frac{\delta |z| }{1+|z|} $, and $\widetilde{M}\stackrel{\triangle}{=} k M + 2 \ell (4M ) |z|^2 $.
By reducing $\varepsilon_0$, we may assume that $\widetilde{M} \varepsilon_0 e^{k \varepsilon_0} \le \frac{\delta}{1+|z|} \land \frac{1}{4\ell (4M ) } $.
\medskip Fix $\varepsilon \in (0, \frac{\ln 2}{k } \land \varepsilon_0]$. Since $\|z(B_{(t+\varepsilon)\land
\tau}-B_t)\|_\infty \le \frac{\delta |z|}{1+|z|}$, there exists a unique solution $\big\{(Y^{\varepsilon}_s,Z^{\varepsilon}_s)\big\}_{s \in [t,t+\varepsilon]} \negthinspace
\in \negthinspace \mathbb{C}^\infty_{\bf F}([t,t+\varepsilon]) \negthinspace \times \negthinspace
{\cal H}^2_{\bf F}([t,t+\varepsilon]; \mathbb{R}^d)$ to the following BSDE:
\begin{eqnarray*}
~~~ Y^{\varepsilon}_s\negthinspace =y\negthinspace +\negthinspace z(B_{(t+\varepsilon)\land \tau}\negthinspace -B_t)\negthinspace
+\negthinspace \int_{s }^{t+\varepsilon } \negthinspace {\bf 1}_{\{r < \tau \}} g(r, Y^{\varepsilon}_r, Z^{\varepsilon}_r)dr\negthinspace -\negthinspace \int_{s }^{t+\varepsilon } \negthinspace Z^{\varepsilon}_r dB_r, ~~~ s \in [t,t+\varepsilon].
\end{eqnarray*}
We know from Corollary 2.2 of \cite{Ko} that $\|Y^{\varepsilon}\|_\infty \le (|y|+ \frac{\delta |z|}{1+|z|}+ k\varepsilon) e^{k\varepsilon}
\le 2 M $. Now let
\begin{eqnarray*}
\widetilde{Y}^{\varepsilon}_s \stackrel{\triangle}{=} Y^{\varepsilon}_s-y-z(B_{s\land \tau}-B_t), \quad
\widetilde{Z}^{\varepsilon}_s \stackrel{\triangle}{=} Z^{\varepsilon}_s-{\bf 1}_{\{s < \tau\}} z , \quad \,\forall \, s \in [t,t+\varepsilon] .
\end{eqnarray*}
It is easy to check that $\{(\widetilde{Y}^{\varepsilon}_s,\widetilde{Z}^{\varepsilon}_s)\}_{s \in
[t,t+\varepsilon]}$ is a solution of the BSDE:
\begin{eqnarray}
\widetilde{Y}^{\varepsilon}_s=\int_s^{t+\varepsilon} \tilde{g}(r,\widetilde{Y}^{\varepsilon}_r,\widetilde{Z}^{\varepsilon}_r)dr- \int_s^{t+\varepsilon} \widetilde{Z}^{\varepsilon}_r dB_r,
\quad~~~ s \in [t,t+\varepsilon] \label{BSDErev}
\end{eqnarray}
with $ \tilde{g}(s, \omega, y',z') \stackrel{\triangle}{=} \varphi(y') {\bf 1}_{\{s < \tau\}}
g\big(s, \omega, y'+y+z(B_{s\land \tau}(\omega)-B_t(\omega) ),z'+z\big)$, $
(s,\omega, y',z')\in [t,t+\varepsilon]\times \Omega \times \mathbb{R} \times \mathbb{R}^d$ where $\varphi: \mathbb{R}
\mapsto [0,1]$ is an arbitrary $C^1(\mathbb{R})$ function that
equals to $1$ inside $ [-3M, 3M]$, vanishes outside
$ (-3M-1 , 3M+1 )$ and satisfies $ \underset{ 3M <|x|<3M+1 }{\sup} | \varphi'(x) | \le 2 $.
For any $(s,\omega, y',z')\in [t,t+\varepsilon]\times \Omega \times \mathbb{R} \times \mathbb{R}^d$, we see that
\begin{eqnarray*}
\tilde{g}(s,\omega, y',z') = \tilde{g}_1(s,\omega, y',z') y' +\tilde{g}_2(s,\omega, y',z')
\end{eqnarray*}
with
\begin{eqnarray*}
\tilde{g}_1(s,\omega, y',z') &= & \varphi(y') {\bf 1}_{\{s < \tau\}}g_1\big(s, \omega, y'+y+z(B_{s\land \tau}(\omega)-B_t(\omega)),z'+z\big), \\
\tilde{g}_2(s,\omega, y',z') &= & \varphi(y') {\bf 1}_{\{s < \tau\}} g_1\big(s, \omega, y'+y+z(B_{s\land \tau}(\omega)-B_t(\omega)),z'+z\big) \\
&& \times \big(y+z(B_{s\land \tau}(\omega)-B_t(\omega))\big) \\
&& + \varphi(y') {\bf 1}_{\{s < \tau\}} g_2\big(s, \omega, y'+y+z(B_{s\land \tau}(\omega)-B_t(\omega)),z'+z\big).
\end{eqnarray*}
One can easily deduce from (H2) and (H3) that for $dt \times dP$-a.s. $(s,\omega )\in [t,t+\varepsilon]\times \Omega$, it holds for any $( y',z')\in \mathbb{R} \times \mathbb{R}^d$ that
\begin{eqnarray}
|\tilde{g}_1(s,\omega, y',z') | &\le & k \label{estimate1} \\
|\tilde{g}_2(s,\omega, y',z') |
& \le & k M + 2 \ell (4M )\big(|z|^2+ |z'|^2 \big)= \widetilde{M}+ 2 \ell (4M ) |z'|^2 \qquad \quad \\
\hbox{and } \quad \left| \frac{\partial \tilde{g} }{\partial z'}(s,\omega,y',z') \right|
&\le& \ell (4 M ) \big(1+ |z'|+|z|\big) . \label{estimate3}
\end{eqnarray}
Corollary 2.2 of \cite{Ko} once again shows that $\|\widetilde{Y}^{\varepsilon}\|_\infty \le \widetilde{M} \varepsilon e^{k\varepsilon} \le \widetilde{M} \varepsilon_0 e^{k \varepsilon_0}
\le \frac{\delta}{1+|z|} \land \frac{1}{4 \ell (4M ) }$. Applying It\^o's formula to $|\widetilde{Y}^{\varepsilon}_s|^2$ we obtain that
\begin{eqnarray}
|\widetilde{Y}^{\varepsilon}_s|^2\negthinspace +\negthinspace \int_s^{t+\varepsilon}\negthinspace |\widetilde{Z}^{\varepsilon}_r|^2 dr\negthinspace =
2\negthinspace \int_s^{t+\varepsilon}\neg \neg \widetilde{Y}^{\varepsilon}_r\, \tilde{g}(r,\widetilde{Y}^{\varepsilon}_r,\widetilde{Z}^{\varepsilon}_r)dr
\negthinspace- \negthinspace 2 \negthinspace \int_s^{t+\varepsilon} \neg \neg \widetilde{Y}^{\varepsilon}_r \widetilde{Z}^{\varepsilon}_r dB_r, ~~
s\in [t,t+\varepsilon]. \label{ito1}
\end{eqnarray}
Using \eqref{estimate1}-\eqref{estimate3} and some standard manipulations one derives easily
that
\begin{eqnarray*}
\qquad && \hspace{-2.5 cm} 2\negthinspace \int_s^{t+\varepsilon}\neg \neg \widetilde{Y}^{\varepsilon}_r\,
\tilde{g}(r,\widetilde{Y}^{\varepsilon}_r,\widetilde{Z}^{\varepsilon}_r)dr
= \negthinspace 2\negthinspace\int_s^{t+\varepsilon}\negthinspace
\widetilde{Y}^{\varepsilon}_r\,\tilde{g}(r,\widetilde{Y}^{\varepsilon}_r,0)dr\negthinspace+ \negthinspace 2\negthinspace \int_s^{t+\varepsilon}\negthinspace
\widetilde{Y}^{\varepsilon}_r \widetilde{Z}^{\varepsilon}_r \big(\neg \neg \int_0^1 \negthinspace \frac{\partial
\tilde{g}}{\partial z'}(r,\widetilde{Y}^{\varepsilon}_r,\lambda \widetilde{Z}^{\varepsilon}_r)d\lambda \big) dr\\
&\neg \neg \neg \le &\neg \neg \neg 2 \negthinspace \int_s^{t+\varepsilon}\negthinspace
|\widetilde{Y}^{\varepsilon}_r|\big( k |\widetilde{Y}^{\varepsilon}_r|+ \widetilde{M} \big)dr\negthinspace +\negthinspace 2\ell(4M)\negthinspace
\int_s^{t+\varepsilon}\negthinspace |\widetilde{Y}^{\varepsilon}_r|\,| \widetilde{Z}^{\varepsilon}_r| \big(1+|z|+ \frac12 |\widetilde{Z}^{\varepsilon}_r|\big)dr\\
&\neg \neg \neg \le &\neg \neg \neg \negthinspace \int_s^{t+\varepsilon}\negthinspace
|\widetilde{Y}^{\varepsilon}_r|\big( 2k |\widetilde{Y}^{\varepsilon}_r|+ 2\widetilde{M} + \ell(4M) (1+|z|)^2 \big)dr\negthinspace +\negthinspace 2 \ell(4M)\negthinspace
\int_s^{t+\varepsilon}\negthinspace |\widetilde{Y}^{\varepsilon}_r|\, |\widetilde{Z}^{\varepsilon}_r|^2 dr\\
&\neg \neg \neg \le &\neg \neg \neg C \varepsilon^2 + \frac12 \int_s^{t+\varepsilon}\negthinspace |\widetilde{Z}^{\varepsilon}_r|^2 dr, \quad~~~ s\in
[t,t+\varepsilon] ,
\end{eqnarray*}
where $C$ is a generic constant depending on $ |y|, |z|, \varepsilon_0, \delta, k$ and $\ell(4M)$,
which may vary from line to line. Taking the conditional
expectation $E[~|{\cal F}_s]$ on both sides of (\ref{ito1}) we have
\begin{eqnarray}
E\Big\{ \int_s^{t+\varepsilon}|\widetilde{Z}^{\varepsilon}_r|^2
dr\big|{\cal F}_s\Big\} \le C\varepsilon^2 , \qquad s \in [t,t+\varepsilon]. \label{zterm}
\end{eqnarray}
\noindent Now, taking the conditional expectation in the BSDE (\ref{BSDErev})
we have
\begin{eqnarray*}
\neg \neg \neg \frac{1}{\varepsilon} \widetilde{Y}^{\varepsilon}_t- \tilde{g}(t,0,0)&=& \frac{1}{\varepsilon} E\Big\{\negthinspace
\int_t^{t+\varepsilon}\neg \neg
\big(\tilde{g}(r,\widetilde{Y}^{\varepsilon}_r,\widetilde{Z}^{\varepsilon}_r)-\tilde{g}(t,0,0)\big)dr\Big|{\cal F}_t\Big\} \\
& \neg \neg \neg=& \frac{1}{\varepsilon} E \Big\{\int_t^{t+\varepsilon} \Big[\widetilde{Z}^{\varepsilon}_r
\int_0^1 \frac{\partial \tilde{g}}{\partial z'}(
r,\widetilde{Y}^{\varepsilon}_r,\lambda \widetilde{Z}^{\varepsilon}_r )d\lambda \\
&& + \widetilde{Y}^{\varepsilon}_r \int_0^1 \frac{\partial \tilde{g}}{\partial y'}(r,\lambda
\widetilde{Y}^{\varepsilon}_r,0)d\lambda + \tilde{g}(r,0,0)-\tilde{g}(t,0,0)\Big]dr\Big|{\cal F}_t\Big\}.
\end{eqnarray*}
\noindent We know from (g2) and (H4) that for $dt \times dP$-a.s. $(s,\omega) \in [t,t+\varepsilon] \times \Omega$,
\begin{eqnarray*}
\qquad \widetilde{h}_s \le \frac{\partial g }{\partial y'} (s, \omega, y'+y+z(B_{s\land \tau}(\omega)-B_t(\omega)), z) \le h_1(s)+|z|^2
\end{eqnarray*}
holds for any $ y' \in \mathbb{R} \hbox{ with } |y'| \le \frac{\delta}{1+|z|}$. It follows that for $dt \times dP$-a.s. $(s,\omega) \in [t,t+\varepsilon] \times \Omega$,
\begin{eqnarray*}
\left| \frac{\partial \tilde{g} }{\partial y'}(s,\omega,y',0)\right| &\neg \neg \neg=& \neg \neg \neg \left| \varphi'(y') {\bf 1}_{\{s < \tau\}}
g(s, \omega, y'+y+z(B_{s\land \tau}(\omega)-B_t(\omega)), z) \right| \\
&\neg \neg \neg & \neg \neg \neg + \left| \varphi(y') {\bf 1}_{\{s < \tau\}} \frac{\partial g }{\partial y'} (s, \omega, y'+y+z(B_{s\land \tau}(\omega)-B_t(\omega)), z) \right| \\
& & \hspace{-2.5 cm} \le 2k(1+4M)+\big(1+2\ell(4M) \big)|z|^2 +|\widetilde{h}_s| + h_1(s) \stackrel{\triangle}{=} h_s
\end{eqnarray*}
holds for any $ y' \in \mathbb{R} \hbox{ with } |y'| \le \frac{\delta}{1+|z|}$.
Clearly, $\{h_s\}_{ s \in [t,t+\varepsilon_0]}$ is an integrable process. Then applying \eqref{estimate3}, (\ref{zterm}) and the H\"older Inequality we have
\begin{eqnarray}
\big|\frac{1}{\varepsilon} \widetilde{Y}^{\varepsilon}_t- \tilde{g}(t,0,0)\big|\negthinspace
&\le & \frac{1}{\varepsilon} E \Big\{
\int_t^{t+\varepsilon}\Big[\ell(4M)\big((1 +|z|) |\widetilde{Z}^{\varepsilon}_r|+\frac12 |\widetilde{Z}^{\varepsilon}_r|^2\big)+
|\widetilde{Y}^{\varepsilon}_r| h_r \Big]dr\Big|{\cal F}_t\Big\} \nonumber \\
&& + E \Big\{\frac{1}{\varepsilon}\int_t^{t+\varepsilon} \big|\tilde{g}(r,0,0)- \tilde{g}(t,0,0)\big|
dr\Big|{\cal F}_t\Big\} \nonumber \\
&\le & C\big( \varepsilon+ \sqrt{ \varepsilon}\big)+ \widetilde{M} e^{k \varepsilon} E \Big[
\int_t^{t+\varepsilon}h_r dr\Big|{\cal F}_t\Big] \nonumber \\
&& + E \Big\{\frac{1}{\varepsilon}
\int_t^{t+\varepsilon} \big|\tilde{g}(r,0,0)- \tilde{g}(t,0,0)\big| dr\Big|{\cal F}_t\Big\}. \label{tilde-g}
\end{eqnarray}
As $\underset{ s \to t^+ }{\lim} {\bf 1}_{\{s < \tau\}}=1$ and $\underset{ s \to t^+ }{\lim} (B_{s\land \tau} -B_t ) =0 $, {\hbox{$P$-a.s.}},
one can deduce from (g1) that
\begin{eqnarray*}
\underset{ s \to t^+ }{\lim} \, \tilde{g}(s, 0, 0) =
\underset{ s \to t^+ }{\lim} \, g(s, y+z(B_{s\land \tau} -B_t ), z)
= g(t, y , z) =\tilde{g}(t,0,0) , \quad {\hbox{$P$-a.s.}},
\end{eqnarray*}
which implies that
\begin{eqnarray*}
\underset{\varepsilon \searrow 0}{\lim}\, \frac{1}{\varepsilon}
\int_t^{t+\varepsilon} \big|\tilde{g}(r,0,0)- \tilde{g}(t,0,0)\big| dr =0, \quad {\hbox{$P$-a.s.}}
\end{eqnarray*}
Since $ | \tilde{g}(s,\omega,0,0)| \le \widetilde{M}$ for $dt \times dP$-a.s. $(s,\omega) \in [t,t+\varepsilon] \times \Omega$,
Lebesgue Convergence
Theorem implies that the right hand side of \eqref{tilde-g} converges
$P$-a.s. to 0 as $\varepsilon \to 0^+$. Therefore,
\begin{eqnarray*}
g(t,y,z) &=& \tilde{g}(t,0,0)=\underset{\varepsilon \to
0^+}{\lim}\,\frac{1}{\varepsilon}\widetilde{Y}^{\varepsilon}_t=\underset{\varepsilon \to
0^+}{\lim}\,\frac{1}{\varepsilon}(Y^{\varepsilon}_t-y) \\
&=& \underset{\varepsilon \searrow 0}{\lim}\,\frac{1}{\varepsilon}\big({\cal E}^g_{t,(t+\varepsilon)\land \tau
}[y+z(B_{(t+\varepsilon)\land \tau}-B_t)]-y\big), \quad {\hbox{$P$-a.s.}},
\end{eqnarray*}
where we used \eqref{BSDEtau} in the last equality. The proof is now complete. \qed
\if{0}
\begin{cor} Assume that $g$ is independent of $y$.
For any $t \in [0,T)$, if the process $\{g(s,0)\}_{s \in [0,T]}$ is
${\hbox{$P$-a.s.}}$ right-continuous at $t$, then for any $z \in \mathbb{R}^d$, it holds
${\hbox{$P$-a.s.}}$ that $ \displaystyle g(t,z)=\underset{\varepsilon \to
0^+}{\lim}\,\frac{1}{\varepsilon}\big({\cal E}^g[zB_{t+\varepsilon}|{\cal F}_t]-zB_t\big)$.
\end{cor}
{\it Proof:} It is easy to check that $Y_s\stackrel{\triangle}{=} {\cal E}^g[zB_{t+\varepsilon}
|{\cal F}_s]$, $s \in [t,t+\varepsilon]$ solves the BSDE (\ref{BSDErev}) with $
g^z(s,z') \stackrel{\triangle}{=} g(s,z'+z)$, $ \,\forall \, (s,z')\in[t,t+\varepsilon]\times \mathbb{R}^d$.
Then a simple application of Lemma \ref{repre} yields the
conclusion. \qed
\fi
A simple application of the Theorem above gives rise to a
reverse to the Comparison Theorem of quadratic BSDE:
\begin{thm}
Assume that $g_i$, $i=1,2$ satisfy (H1)-(H4) and (\ref{g0}). Let $\, t \in
[0,T)$. If ${\cal E}^{g_1}[\xi|{\cal F}_t] \le {\cal E}^{g_2}[\xi|{\cal F}_t]$, ${\hbox{$P$-a.s.}}$ for any $\xi
\in L^\infty({\cal F}_T)$, and if both $g_i$ satisfy (g1) and (g2) for any $(y,z) \in \mathbb{R} \times \mathbb{R}^d$, then it holds ${\hbox{$P$-a.s.}}$ that
\begin{eqnarray*}
g_1(t,y,z) \le g_2(t,y,z), \quad \,\forall \, (y,z) \in \mathbb{R} \times \mathbb{R}^d.
\end{eqnarray*}
\end{thm}
We also have the following corollary of Theorem \ref{repre3}.
\begin{prop}
Assume that $g$ satisfies (H1)-(H4) and (\ref{g0}). We also assume that ${\hbox{$P$-a.s.}}$, $g(\cdot, y,z)$
is continuous for any $(y,z) \in \mathbb{R} \times \mathbb{R}^d$. If
$g$ satisfies (g1) and (g2) for any $(t, y,z) \in [0,T) \times \mathbb{R} \times \mathbb{R}^d$, then $g$ is independent of $y$ if and only if
\begin{eqnarray*}
{\cal E}^g[\xi+c]={\cal E}^g[\xi]+c, \quad~~ \,\forall \, \xi \in
L ^\infty({\cal F}_T),~~~ \,\forall \, c \in \mathbb{R}.
\end{eqnarray*}
\end{prop}
{\it Proof:} ``$\mathop{\Rightarrow }$": A simply application of Translation
Invariance of quadratic $g$-expectations.
\noindent ``$\mathop{\Leftarrow }$": For any $c \in \mathbb{R}$, we define a new generator
$g^c(t,\omega,y,z)\stackrel{\triangle}{=} g(t,\omega,y-c,z)$, $\,\forall \, (t,\omega,y,z) \in [0,T]\times
\Omega \times \mathbb{R} \times \mathbb{R}^d$. It is easy to check that $g^c$
satisfies (H1)-(H4) as well as the other assumptions on $g$ in this
proposition. For any $\xi \in L^\infty({\cal F}_T)$, let $(Y,Z)$ denote
the unique solution to BSDE$(T,\xi,g)$. Setting $\tilde{Y}_t=Y_t+c$,
$t\in [0,T]$ one obtains that
\begin{eqnarray*}
\tilde{Y}_t=\xi+c+\int_t^T g^c(s,\tilde{Y}_s, Z_s)ds-\int_t^T Z_s
dB_s, \quad~~ \,\forall \, t \in [0,T].
\end{eqnarray*}
Thus, it holds ${\hbox{$P$-a.s.}}$ that
\begin{eqnarray*}
{\cal E}^{g^c}[\xi+c|{\cal F}_t]=\tilde{Y}_t =Y_t+c ={\cal E}^g[\xi|{\cal F}_t]+c, \quad~~\,\forall \, t \in [0,T].
\end{eqnarray*}
In particular, taking $t=0$ gives that ${\cal E}^{g^c}[\xi]={\cal E}^g[\xi]$
for any $\xi \in L^\infty({\cal F}_T)$. Since $g$ satisfies (\ref{g0}),
it easy to see that the condition (\ref{gicomp}) is satisfied for
$g^1\stackrel{\triangle}{=} g$ and $g^2 \stackrel{\triangle}{=} g^c$. Hence, Proposition \ref{gevcomp}
implies that for any $ \xi \in L^\infty({\cal F}_T)$, it holds ${\hbox{$P$-a.s.}}$
that ${\cal E}^g[\xi|{\cal F}_t]={\cal E}^{g^c}[\xi|{\cal F}_t]$, $\,\forall \, t \in [0,T]$.
Applying Theorem \ref{repre3} we see that for any $(t,z) \in
[0,T)\times \mathbb{R}^d$, it holds ${\hbox{$P$-a.s.}}$ that
$g(t,c,z)=g^c(t,c,z)=g(t,0,z)$. Then (H1) implies that for any $t\in
[0,T)$, it holds ${\hbox{$P$-a.s.}}$ that $g(t,y,z)=g(t,0,z)$, $\,\forall \, (y,z) \in \mathbb{R}
\times \mathbb{R}^d$. Eventually, by our assumption, it holds ${\hbox{$P$-a.s.}}$ that
$g(t,y,z)=g(t,0,z)$, $\,\forall \, (t,y,z) \in [0,T)\times\mathbb{R} \times \mathbb{R}^d$.
This proves the proposition. \qed
\medskip
To end this section we extend another important feature of the
$g$-expectation to the quadratic case: The Jensen's Inequality. We
begin by recalling some basic facts for convex functions, and we
refer to Rockafellar \cite{Rock} for all the notions to appear
below.
Recall that
if $F: \mathbb{R}^n \mapsto \mathbb{R}$ is a convex function, then by considering
the convex real function $f(\lambda) \stackrel{\triangle}{=} F(\lambda x)-\big(\lambda
F(x)+(1-\lambda)F(0)\big)$, $\lambda \in \mathbb{R}$, with $f(0)=f(1)=0$, it is easy
to check that for any $x \in \mathbb{R}^n$, it holds that
\begin{eqnarray}
\label{Convex}
\left\{\begin{array}{lll}
F(\lambda x) \le \lambda F(x)+(1-\lambda)F(0), \quad~~\mbox{if }\, \lambda \in [0,1],\\
F(\lambda x) \ge \lambda F(x)+(1-\lambda)F(0), \quad~~\mbox{if }\, \lambda \in (0,1)^c.
\end{array}\right.
\end{eqnarray}
Next, if $F:\mathbb{R} \mapsto \mathbb{R}$ is a convex (real) function, then we
denote by $\partial F$ the {\it subdifferential} of $F$ (see
\cite{Rock}). In particular, for any $x\in
\mathbb{R}$, $\partial F(x)$ is simply an interval $[F'_{-}(x),F'_{+}(x)]$,
where $F'_-$ and $F'_+$ are left-, and right-derivatives of $F$,
respectively.
The following result is an extension of the linear growth case (cf.
\cite[Proposition 5.2]{BCHMP}).
\begin{thm} Assume that $g$ is independent of $y$ and
satisfies (H1)-(H4) and (\ref{g0}). Let $\, t \in [0,T)$. If
$g(s, \omega, z)$ is convex in $z$ for $dt \times dP$-a.s. $(s,\omega) \in [t,T] \times \Omega$, then
\begin{eqnarray*}
F\big({\cal E}^g[\xi|{\cal F}_t]\big)\le
{\cal E}^g[F(\xi)|{\cal F}_t], \quad {\hbox{$P$-a.s.}}
\end{eqnarray*}
for any $\xi \in L^\infty({\cal F}_T)$
with $\partial F\big({\cal E}^g[\xi|{\cal F}_t]\big)\cap (0,1)^c
\ne \emptyset$, ${\hbox{$P$-a.s.}}$
\end{thm}
{\it Proof:} Since both $F'_{-}(x) $ and $F'_{+}(x) $ are non-decreasing functions, we can define another non-decreasing function:
\begin{eqnarray*}
\beta(x)\stackrel{\triangle}{=} {\bf 1}_{\{F'_{-}(x)\le
0\}}F'_{-}(x)+{\bf 1}_{\{F'_{-}(x)> 0\}}F'_{+}(x) , \quad x \in \mathbb{R}.
\end{eqnarray*}
Thus $ \beta_t \stackrel{\triangle}{=} \beta\big({\cal E}^g[\xi|{\cal F}_t]\big)$ is an ${\cal F}_t$-measurable random variable. Since $\beta(x) \in (0,1)^c$ for any $x\in \mathbb{R}$ with $\partial F(x) \cap (0,1)^c \ne \emptyset$, it follows that
\begin{eqnarray} \label{beta_01}
\beta_t \in (0,1)^c , \quad {\hbox{$P$-a.s.}}
\end{eqnarray}
One can deduce from the convexity of $F$ that
\begin{eqnarray} \label{jensen}
\beta_t \big(\xi-{\cal E}^g[\xi|{\cal F}_t]\big) \le
F(\xi)-F\big({\cal E}^g[\xi|{\cal F}_t]\big).
\end{eqnarray}
Since $\xi
\in L^\infty({\cal F}_T)$, it is clear that $F(\xi)$, ${\cal E}^g[\xi|{\cal F}_t]$,
$F\big({\cal E}^g[\xi|{\cal F}_t]\big)$ as well as $\beta_t
\big(\xi-{\cal E}^g[\xi|{\cal F}_t]\big)$ are all of $L^\infty({\cal F}_T)$. Taking ${\cal E}^g[~|{\cal F}_t]$ on both side of (\ref{jensen}), and using
Translation Invariance of quadratic $g$-expectation we have
\begin{eqnarray*}
&&{\cal E}^g[\beta_t \xi |{\cal F}_t]-\beta_t {\cal E}^g[\xi|{\cal F}_t] ={\cal E}^g\big[\beta_t
\big(\xi-{\cal E}^g[\xi|{\cal F}_t]\big)\big|{\cal F}_t\big] \\
&\le& {\cal E}^g\big[F(\xi)-F\big({\cal E}^g[\xi|{\cal F}_t]\big)\big|{\cal F}_t\big]=
{\cal E}^g[F(\xi)|{\cal F}_t]-F\big({\cal E}^g[\xi|{\cal F}_t]\big),\quad {\hbox{$P$-a.s.}}
\end{eqnarray*}
Hence, it suffices to show that $\beta_t {\cal E}^g[\xi|{\cal F}_t] \le
{\cal E}^g[\beta_t \xi |{\cal F}_t]$, ${\hbox{$P$-a.s.}}$ To see this, let $Y_t \stackrel{\triangle}{=}
{\cal E}^g[\xi|{\cal F}_t]$, $t \in [0,T]$. As $\beta_t \in {\cal F}_t$, one has
\begin{eqnarray*}
\beta_t Y_s =\beta_t \xi + \int_s^T \beta_t g(r,Z_r)dr- \int_s^T \beta_t
Z_r dB_r, \quad~~ \,\forall \, s \in [t,T].
\end{eqnarray*}
Since $g$ is convex and satisfies (\ref{g0}), using
(\ref{Convex}) and \eqref{beta_01} we obtain
\begin{eqnarray*}
\beta_t Y_s \le \beta_t \xi + \int_s^T g(r,\beta_t Z_r)dr- \int_s^T \beta_t
Z_r dB_r={\cal E}^g[\beta_t \xi|{\cal F}_s], \quad~~ \,\forall \, s \in [t,T].
\end{eqnarray*}
In particular, we have $ \beta_t
{\cal E}^g[\xi|{\cal F}_t] \le {\cal E}^g[\beta_t \xi |{\cal F}_t]$, ${\hbox{$P$-a.s.}}$, proving the
Theorem. \qed
\section{Main Results}
\setcounter{equation}{0}
\medskip
In this section we prove the main results of this paper regarding
the {\it quadratic $g$-martingales}. To begin with, we give the
following definition. Recall that ${\cal E}^g_{s,t}[\cdot]$, $0\le s\le
t\le T$ denotes the $g$-evaluation.
\begin{defn}
An $X \in L^\infty_{\bf F}([0,T])$ is called a ``$g$-submartingale"
\(resp. $g$-supermartingale\) if for any $0 \le s \le t \le T$, it
holds that
\begin{eqnarray*}
{\cal E}^g_{s,t}[X_t] \ge (\mbox{resp. }\le) X_s, \qquad {\hbox{$P$-a.s.}}
\end{eqnarray*}
$X$ is called a $g$-martingale if it is both a $g$-submartingale and
a $g$-supermartingale.
\end{defn}
We should note here
that, in the above the martingale is defined in terms of
quadratic $g$-evaluation, instead of quadratic $g$-expectation as we
have usually seen. This slight relaxation is merely for convenience
in applications. It is clear, however, that if $g$ satisfies
(\ref{g0}), then the quadratic $g$-martingale defined above should
be the same as the one defined via quadratic $g$-expectations,
thanks to (\ref{coin}).
We shall extend three main results for $g$-expectation to the
quadratic case: the Doob-Meyer decomposition, the optional sampling
theorem, and the upcrossing theorem. Although the results look
similar to the existing one in the $g$-expectation literature, the
proofs are more involved, due to the special nature of the quadratic
BSDEs. We shall present these results separately.
\medskip
We begin by proving a Doob-Meyer type decomposition theorem for
$g$-martingales.
\begin{thm} \label{gmd}
{\rm (Doob-Meyer Decomposition Theorem)} Assume (H1)--(H4). Let $Y$
be any $g$-submartingale (resp. $g$-supermartingale) that has
right-continuous paths. Then there exist a c\`adl\`ag increasing
(decreasing) process $A$ null at $0$ and a process $Z \in
{\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$ such that \begin{eqnarray*} Y_t=Y_T+\int_t^T
g(s,Y_s,Z_s)ds-A_T+A_t-\int_t^T Z_s dB_s, \qquad t\in[0,T]. \end{eqnarray*}
\end{thm}
{\it Proof.} We first assume that $Y$ is a $g$-submartingale. Set $M
\stackrel{\triangle}{=} ( \|Y\|_\infty + kT) e^{kT}$ and $K \stackrel{\triangle}{=} \ell(M+1)$, we let $\phi:\mathbb{R}
\mapsto [0,1]$ be any $C^2(\mathbb{R})$ function that equals to $1$ inside
$\big[e^{-2KM},e^{2KM}\big]$ and vanishes outside
$\big(e^{-2K(M+1)},e^{2K(M+1)}\big)$. Let us construct a new
generator: For any $(t,\omega,y,z) \in [0,T] \times \Omega \times \mathbb{R} \times \mathbb{R}^d$,
\begin{eqnarray*}
\tilde{g}(t,\omega,y,z) \stackrel{\triangle}{=} \phi(y)\Big[2Ky \,g\Big(t,\omega,\frac{\ln{(y)}}{2K}, \frac{z}{2Ky}\Big)
-\frac{|z|^2}{2y}\Big].
\end{eqnarray*}
One can deduce from (H2) that for $dt \times dP$-a.s. $(t,\omega) \in
[0,T] \times \Omega$,
\begin{eqnarray*} \tilde{g}(t,y,z) \leq 2(M+2)kK\phi(y)y, \qquad
(y,z) \in \mathbb{R} \times \mathbb{R}^d.
\end{eqnarray*}
\if{0}
$$\phi(y) = \left\{
\begin{array}{ll}
1,\quad \mbox{if }y \in \big[e^{-2KM},e^{2KM}\big]\\
0,\quad \mbox{if }y \notin \big(e^{-2K(M+1)},e^{2K(M+1)}\big)
\end{array}
\right.$$
\fi
Since $2(M+2)kK\phi(y)y$ is Lipschitz continuous in $y$, we can
construct (cf. \cite{Ko}) a decreasing sequence $g_n(t,y,z)$ of
generators uniformly Lipsichitz in $(y,z)$ such that {\hbox{$P$-a.s.}}
\begin{eqnarray*} g_n(t,y,z) \searrow \tilde{g}(t,y,z), \qquad \,\forall \, (t,y,z) \in
[0,T] \times \mathbb{R} \times \mathbb{R}^d .
\end{eqnarray*}
Now fix $t \in [0,T]$, for any $\xi \in L^\infty({\cal F}_t)$ with
$\|\xi\|_\infty \leq \|Y\|_\infty$, we define $y_s \stackrel{\triangle}{=} {\cal E}^g_{s,t}[\xi]$, $s
\in [0,t]$. It follows from \cite[Corollary 2.2]{ Ko} that $\|y\|_\infty \le ( \|Y\|_\infty + kT) e^{kT}=M$. Applying It\^o's formula we see that $\tilde{y}_s \stackrel{\triangle}{=}
e^{2Ky_s}$, $s \in [0,t]$ together with a process $\tilde{z} \in
{\cal H}^2_{\bf F}([0,t];\mathbb{R}^d)$ is a solution of the following BSDE:
\begin{eqnarray*} \tilde{y}_s=e^{2K\xi}+\int_s^t
\tilde{g}(r,\tilde{y}_r,\tilde{z}_r)dr-\int_s^t \tilde{z}_r dB_r,
\qquad \,\forall \, s \in [0,t].
\end{eqnarray*}
Since $g_n$ is Lipschitz, a standard comparison theorem implies that
\begin{eqnarray*} e^{2K {\cal E}^g_{s,t}[\xi]}=\tilde{y}_s \leq
{\cal E}^{g_n}_{s,t}[e^{2K\xi}], \quad~~~ s \in [0,t], \quad~ {\hbox{$P$-a.s.}}
\end{eqnarray*}
In particular, taking $\xi =Y_t$ shows that
\begin{eqnarray*}
e^{2K Y_s} \le e^{2K {\cal E}^g_{s,t}[Y_t]} \le {\cal E}^{g_n}_{s,t}[e^{2K Y_t}],\quad s
\in [0,t], \quad~ {\hbox{$P$-a.s.}}
\end{eqnarray*}
Namely, $\tilde{Y} = e^{2KY}$ is a right-continuous
$g_n$-submartingale in the sense of $g^n$-evaluation for any $n \in
\mathbb{N}$. Applying the known $g$-submartingale decomposition theorem for
the Lipschitz case (see \cite[Theorem 3.9]{Pln}), we can find a
c\`adl\`ag increasing process $A^n$ null at $0$ and a process $Z^n
\in {\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$ such that
\begin{eqnarray}
\label{gnrep}
\tilde{Y}_t=\tilde{Y}_T+\int_t^T g_n(s,\tilde{Y}_s,Z^n_s)ds
-A^n_T+A^n_t-\int_t^T Z^n_s dB_s, \qquad t\in[0,T],
\end{eqnarray}
from which we see that $\tilde{Y}$, whence $Y$ is c\`adl\`ag. Note
that, in the representation (\ref{gnrep}), the martingale parts must
coincide for any $m$ and $n$. In other words, one must have
$Z^m=Z^n$ as the elements in ${\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$. Thus, for any
$n \in \mathbb{N}$, (\ref{gnrep}) can be rewritten as
\begin{eqnarray*}
\tilde{Y}_t=\tilde{Y}_T+\int_t^T g_n(s,\tilde{Y}_s,\tilde{Z}_s)ds
-A^n_T+A^n_t-\int_t^T \tilde{Z}_s dB_s, \qquad
t\in[0,T].
\end{eqnarray*}
Since $g_n\searrow \tilde{g}$, the Lebesgue Convergence Theorem
implies that
\begin{eqnarray*}
\int_0^T \big[g_n(s,\tilde{Y}_s,\tilde{Z}_s)-\tilde{g}(s,\tilde{Y}_s,
\tilde{Z}_s)\big]ds \to 0, \quad~~ {\hbox{$P$-a.s.}}
\end{eqnarray*}
Consequently, it holds ${\hbox{$P$-a.s.}}$ that
\begin{eqnarray*}
A^n_t \to \tilde{A}_t \stackrel{\triangle}{=} \tilde{Y}_t-\tilde{Y}_0 +\int_0^t
\tilde{g}(s,\tilde{Y}_s,\tilde{Z}_s)ds-\int_0^t
\tilde{Z}_s dB_s, \quad~~ \,\forall \, t \in [0,T].
\end{eqnarray*}
It is easy to check that $\tilde{A}$ is also a c\`adl\`ag increasing
process null at $0$.
Now let us define a new $C^2(\mathbb{R})$
function $\psi$ by $ \displaystyle \psi(y) \stackrel{\triangle}{=} \frac{\phi(y)\ln{(y)}}{2K}$, $y \in \mathbb{R}$.
Applying It\^o's formula to $\psi(\tilde Y_t)$ from $t$ to $T$ one
has
\begin{eqnarray*}
Y_t& = & Y_T+ \int_{t+}^T\frac{1}{2K\tilde{Y}_{s-}}\big[\tilde{g}(s,\tilde{Y}_s,\tilde{Z}_s)
ds-d\tilde{A}_s-\negthinspace \tilde{Z}_sdB_s\big] +\frac{1}{2}\int_{t+}^T
\frac{|\tilde{Z}_s|^2}{2K\tilde{Y}^2_{s-}}ds\\
&& - \underset{s\in (t,T]}{\sum} \big\{\Delta Y_s - \frac{\Delta
\tilde{Y}_s}{2K\tilde{Y}_{s-}}\big\} \\
&= &Y_T+\negthinspace \int_t^T
\frac{1}{2K\tilde{Y}_s}\big[\tilde{g}(s,\tilde{Y}_s,\tilde{Z}_s)ds-d\tilde{A}^c_s-\tilde{Z}_sdB_s\big]
+\frac{1}{2}\int_t^T \frac{|\tilde{Z}_s|^2}{2K\tilde{Y}^2_s}ds-\neg \neg\underset{s \in (t,T]}{\sum}\Delta Y_s \\
& = &Y_T+ \negthinspace \int_t^T g(s, Y_s,
\frac{\tilde{Z}_s}{2K\tilde{Y}_s})ds-\int_t^T\frac{1}{2K\tilde{Y}_s}d\tilde{A}^c_s
-\int_t^T\frac{\tilde{Z}_s}{2K\tilde{Y}_s}dB_s- \underset{s \in (t,T]}{\sum}\Delta Y_s,
\end{eqnarray*}
where the second equality is due to the fact that $\Delta
\tilde{Y}_s= \Delta \tilde{A}_s>0$ and $\tilde{A}^c$ denotes the
continuous part of $\tilde{A}$. Clearly, $ A_t \stackrel{\triangle}{=}
\int_0^t\frac{1}{2K\tilde{Y}_s}d\tilde{A}^c_s + \neg \neg \neg \underset{s
\in (0,t]}{\sum}\Delta Y_s $ is a c\`adl\`ag increasing process null
at $0$, finally we get \begin{eqnarray*} Y_t=Y_T+\int_t^T g(s, Y_s,
Z_s)ds-A_T+A_t-\int_t^T Z_s dB_s, \qquad t\in[0,T]. \end{eqnarray*}
\medskip
On the other hand, if $Y$ is a $g$-supermartingale, then one can
easily check that $-Y$ is correspondingly a $g^{-}$-submartingale
with \begin{eqnarray}\label{gneg} g^{-}(t,\omega,y,z)\stackrel{\triangle}{=} -g(t,\omega,-y,-z),\qquad \,\forall \,
(t,\omega,y,z) \in [0,T] \times \Omega \times \mathbb{R} \times \mathbb{R}^d. \end{eqnarray}
Clearly, $g^{-}$ also satisfies (H1)-(H4), thus there exist a
c\`adl\`ag increasing process $A$ null at $0$ and a process $Z \in
{\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$ such that \begin{eqnarray*}
-Y_t=-Y_T+\int_t^T g^{-}(s,-Y_s,Z_s)ds-A_T+A_t-\int_t^T Z_s dB_s, \qquad t\in[0,T].
\end{eqnarray*} We can rewrite this BSDE as: \begin{eqnarray*} Y_t=Y_T+\int_t^T
g(s,Y_s,-Z_s)ds-(-A_T)+(-A_t)-\int_t^T (-Z_s) dB_s, \qquad t\in[0,T].
\end{eqnarray*} The proof is now complete. \qed
\if{0}
\medskip The following corollary is significant.
\begin{cor}
Assume that $g$ satisfies (\ref{g0}) and satisfies
(H2) with $\ell$ being constant in (H2). For any right-continuous $Y
\in L^\infty_{\bf F}([0,T])$ and any $\zeta \in L^\infty_{\bf F}([0,T];\mathbb{R}^d)$,
if the process $Y_t+\int_0^t \zeta_s dB_s$, $t \in [0,T]$ is a
$g$-submartingale \(resp. $g$-supermartingale\) in the sense of
$g$-expectation, then there exist a c\`adl\`ag increasing \(resp.
decreasing\) process $A$ null at $0$ and a process $Z \in
{\cal H}^2_{\bf F}([0,T];\mathbb{R}^d)$ such that \begin{eqnarray*} Y_t=Y_T+\int_t^T
g(s,Y_s,Z_s+\zeta_s)ds-A_T+A_t-\int_t^T Z_s dB_s, \qquad t\in[0,T]. \end{eqnarray*}
\end{cor}
{\it Proof.} Since $g$ satisfies (\ref{g0}), $Y_\cdot+\int_0^\cdot
\zeta_s dB_s$, is also a $g$-submartingale \(resp.
$g$-supermartingale\) in the sense of quadratic $g$-evaluation. Then
a BSDE transformation shows that $Y$ is correspondingly a
$\tilde{g}$-submartingale \(resp. $\tilde{g}$-supermartingale\) in
the sense of $\tilde{g}$-evaluation with \begin{eqnarray*}
\tilde{g}(t,\omega,y,z)\stackrel{\triangle}{=} g(t,\omega,y,z+\zeta_t(\omega)),\qquad \,\forall \, (t,\omega,y,z) \in
[0,T] \times \Omega \times \mathbb{R} \times \mathbb{R}^d. \end{eqnarray*} It is easy to check
that $\tilde{g}$ satisfies (H1)-(H4). Hence Theorem \ref{gmd}
implies the conclusion. \qed
\fi
We now turn our attention to the {\it Optional Sampling Theorem}. We
begin by presenting a lemma that will play an important role in the
proof of the Optional Sampling Theorem.
\begin{lem}
Let $\tau \in {\cal M}_{0,T}$ be
finite valued in a set $0 = t_0 < t_1< \cdots <t_n= T$. If $t_i \le
s< t \le t_{i+1}$ for some $i \in \{0,1, \cdots n-1\}$, then for any $ \xi \in {\cal F}_{t \land \tau}$
\begin{eqnarray}\label{op0} {\cal E}^g_{s \land \tau , t \land \tau}[\xi]={\bf 1}_{\{\tau \le
t_i\}}\xi+{\bf 1}_{\{\tau \ge t_{i+1}\}}{\cal E}^g_{s, t}[\xi],\quad~~ {\hbox{$P$-a.s.}}
\end{eqnarray}
\end{lem}
{\it Proof.} For any $\xi \in {\cal F}_{t \land \tau}$, let $(Y,Z)$ be the unique solution
to the BSDE (\ref{BSDEtau}) with $\tau = t \land \tau$.
Then we have
\begin{eqnarray*}
{\cal E}^g_{r \land \tau, t \land \tau}[\xi]=Y_{r \land \tau} &=& \xi + \int_{r \land \tau}^T {\bf 1}_{\{u < t \land \tau \}}
g(u,Y_u,Z_u)du-\int_{r \land \tau}^T {\bf 1}_{\{u < t \land \tau \}} Z_udB_u\\
& =& \xi + \int_r^t {\bf 1}_{\{u < \tau \}}g(u,Y_{u \land
\tau},Z_u)du-\int_r^t {\bf 1}_{\{u < \tau \}}Z_udB_u, \qquad \,\forall \, r \in [0,t].
\end{eqnarray*}
For any $r \in [s,t]$, since $\{ \tau \le t_i \}=\{ \tau \ge t_{i+1}
\}^c \in {\cal F}_{t_i} \subset {\cal F}_r$,
one can deduce that
\begin{eqnarray}
\label{op1}
{\bf 1}_{\{ \tau \le t_i \}}Y_{r \land \tau}\negthinspace&\negthinspace =\negthinspace &\negthinspace
{\bf 1}_{\{ \tau \le t_i \}}\xi + \int_r^t {\bf 1}_{\{ \tau \le
t_i \}}{\bf 1}_{\{u < \tau \}}g(u,Y_{u \land \tau},
Z_u)du-\int_r^t {\bf 1}_{\{ \tau \le t_i \}} {\bf 1}_{\{u < \tau
\}}Z_udB_u \nonumber\\
\negthinspace&\negthinspace=\negthinspace&\negthinspace {\bf 1}_{\{ \tau \le t_i \}}\xi,
\end{eqnarray}
and that
\begin{eqnarray}
\label{part1}
{\bf 1}_{\{ \tau \ge t_{i+1} \}}Y_{r \land \tau}
&\neg \neg=\neg \neg&{\bf 1}_{\{ \tau \ge t_{i+1} \}}\xi + \int_r^t {\bf 1}_{\{ \tau
\ge t_{i+1} \}} {\bf 1}_{\{u < \tau \}}g(u,Y_{u \land \tau},Z_u)du-\int_r^t
{\bf 1}_{\{ \tau \ge t_{i+1} \}}
{\bf 1}_{\{u < \tau \}}Z_udB_u \nonumber \\
&\neg \neg=\neg \neg&{\bf 1}_{\{ \tau \ge t_{i+1} \}}\xi + \int_r^t {\bf 1}_{\{ \tau
\ge t_{i+1} \}}g(u,Y_{u \land \tau},Z_u)du-\int_r^t {\bf 1}_{\{ \tau \ge
t_{i+1} \}} Z_udB_u.
\end{eqnarray}
On the other hand, we let $Y'_r={\cal E}^g_{r,t}[\xi],~r \in [0,t]$. Then
for any $r \in [s,t]$, by the definition of quadratic
$g$-evaluation, one has
\begin{eqnarray}
\label{part2} {\bf 1}_{\{\tau \le t_i\}}Y'_r =
{\bf 1}_{\{\tau \le t_i\}}\xi + \int_r^t {\bf 1}_{\{\tau \le
t_i\}}g(u,Y'_u,Z'_u)du-\int_r^t {\bf 1}_{\{\tau \le t_i\}}Z'_udB_u.
\end{eqnarray}
Adding (\ref{part2}) to (\ref{part1}) shows that $\tilde{Y}_r \stackrel{\triangle}{=}
{\bf 1}_{\{\tau \ge t_{i+1}\}}Y_{r \land \tau}+{\bf 1}_{\{\tau \le t_i\}}Y'_r$ and
$\tilde{Z}_r \stackrel{\triangle}{=} {\bf 1}_{\{ \tau \ge t_{i+1} \}} Z_r+{\bf 1}_{\{\tau \le
t_i\}}Z'_r$ solve the following BSDE
\begin{eqnarray*} \tilde{Y}_r= \xi+
\int_r^t g(u,\tilde{Y}_u,\tilde{Z}_u)du-\int_r^t \tilde{Z}_u dB_u,
\qquad \,\forall \, r \in [s,t].
\end{eqnarray*}
Then it is not hard to check that $\hat{Y}_r={\bf 1}_{\{r\ge
s\}}\tilde{Y}_r+{\bf 1}_{\{r< s\}}{\cal E}^g_{r,s}[\tilde{Y}_s],~r \in [0,t]$
is the unique solution of BSDE$(t,\xi,g)$. Hence we can rewrite
$\hat{Y}_r={\cal E}^g_{r,t}[\xi],~ r \in [0,t]$. In particular, it holds
$P$-a.s. that
\begin{eqnarray}
\label{part3} {\bf 1}_{\{\tau \ge t_{i+1}\}}Y_{s
\land \tau}={\bf 1}_{\{\tau \ge t_{i+1}\}}\tilde{Y}_s={\bf 1}_{\{\tau \ge
t_{i+1}\}}\hat{Y}_s={\bf 1}_{\{\tau \ge t_{i+1}\}}{\cal E}^g_{s,t}[\xi].
\end{eqnarray}
Letting $r=s$ in (\ref{op1}) and then adding it to (\ref{part3}),
the lemma follows. \qed
We are now ready to prove the Optional Sampling Theorem.
\begin{thm}
Assume (H1)-(H4). For any $g$-submartingale $X$ \(resp.
$g$-supermartingale, $g$-martingale\) such that $\underset{\omega \in
\Omega} {\mathop{\rm esssup}}\underset{t \in [0,T]}{\sup} |X(t,\omega)|<\infty$, and for
any $\sigma$, $\tau \in {\cal M}_{0,T}$ with $\sigma \le \tau$, ${\hbox{$P$-a.s.}}$ Assume either that
$\sigma$ and $\tau$ are finitely valued or that $X$ is
right-continuous, then \begin{eqnarray*} {\cal E}^g_{\sigma,\tau}[X_\tau]\geq (\mbox{resp.
}\leq,\,= )\,X_\sigma, \quad~~~ {\hbox{$P$-a.s.}} \end{eqnarray*}
\end{thm}
{\it Proof.} We shall consider only the $g$-submartingale case, as
the other cases can be deduced easily by standard argument. To begin
with, we assume that $\tau$ takes values in a finite set $0 = t_0 <
t_1< \cdots <t_n = T$. Note that if $t \geq t_n$, then it is clear
that ${\cal E}^g_{t \land \tau,\tau}[X_\tau]={\cal E}^g_{\tau,\tau}[X_\tau]=X_\tau$, ${\hbox{$P$-a.s.}}$
We can then argue inductively that for any $t \in [0,T]$,
\begin{eqnarray}
\label{op2}
{\cal E}^g_{t \land \tau,\tau}[X_\tau] \ge X_{t \land \tau},
\quad~~~ {\hbox{$P$-a.s.}}
\end{eqnarray}
In fact, assume that for some $i \in \{1,\cdot \cdot\cdot n\}$,
(\ref{op2}) holds for any $t \geq t_i$. Then for any $t \in
[t_{i-1},t_i)$, the time-consistence and the monotonicity of
quadratic $g$-evaluations as well as (\ref{op0}) imply that
\begin{eqnarray*}
{\cal E}^g_{t \land \tau,\tau}[X_\tau]&=&
{\cal E}^g_{t \land \tau,t_i \land \tau}\big[{\cal E}^g_{t_i \land \tau,\tau}[X_\tau]\big]
\ge {\cal E}^g_{t \land \tau,t_i \land \tau}[X_{t_i \land \tau}]\\
&=& {\bf 1}_{\{\tau \le t_{i-1}\}}X_{t_i \land \tau}
+{\bf 1}_{\{\tau \ge t_i\}}{\cal E}^g_{t, t_i}[X_{t_i \land \tau}]\\
&=& {\bf 1}_{\{\tau \le t_{i-1}\}}X_{t \land \tau}+{\bf 1}_{\{\tau \ge
t_i\}}{\cal E}^g_{t, t_i}[X_{t_i \land \tau}], \qquad {\hbox{$P$-a.s.}}
\end{eqnarray*}
Since $\{\tau \ge t_i\}=\{\tau \le t_{i-1}\}^c \in {\cal F}_t$, the
``zero-one law" of quadratic $g$-evaluations shows that
\begin{eqnarray*}
{\bf 1}_{\{\tau \ge t_i\}}{\cal E}^g_{t, t_i}[X_{t_i \land \tau}] &=&{\bf 1}_{\{\tau
\ge t_i\}}{\cal E}^g_{t, t_i}[{\bf 1}_{\{\tau \ge t_i\}}X_{t_i \land \tau}]
={\bf 1}_{\{\tau \ge t_i\}}{\cal E}^g_{t, t_i}[{\bf 1}_{\{\tau \ge t_i\}}X_{t_i}]\\
&=&{\bf 1}_{\{\tau \ge t_i\}}{\cal E}^g_{t, t_i}[X_{t_i}]\ge {\bf 1}_{\{\tau \ge
t_i\}}X_t={\bf 1}_{\{\tau \ge t_i\}}X_{t \land \tau},\qquad {\hbox{$P$-a.s.}}
\end{eqnarray*}
Hence, (\ref{op2}) holds for any $t \geq t_{i-1}$, this completes
the inductive step. If $\sigma$ is also finitely valued, for example in
the set $0 = s_0 < s_1< \cdots <s_m = T$, then it holds $P$-a.s.
\begin{eqnarray}
\label{op4}
{\cal E}^g_{\sigma,\tau}[X_\tau]={\cal E}^g_{\sigma\land\tau,\tau}[X_\tau]=\sum^m_{j=0}{\bf 1}_{\{\sigma
= s_j\}}{\cal E}^g_{s_j\land \tau,\tau}[X_\tau]\ge \sum^m_{j=0}{\bf 1}_{\{\sigma = s_j\}}X_{s_j\land
\tau}=X_{\sigma \land \tau}=X_\sigma.
\end{eqnarray}
For a general $\tau \in {\cal M}_{0,T}$, we define two sequences
$\{\sigma_n\}$ and $\{\tau_n\}$ of finite valued stopping times such that
$P$-a.s.
\begin{eqnarray*}
\sigma_n \searrow \sigma, \quad~~ \tau_n \searrow \tau, \quad ~~ \mbox{and}
\quad~~ \sigma_n \le \tau_n, \quad \,\forall \, n \in \mathbb{N}.
\end{eqnarray*}
Fix $n \in \mathbb{N}$ and let $(Y^n,Z^n)$ be the unique solution
to the BSDE (\ref{BSDEtau}) with $\xi=X_{\tau_n} $ and $\tau = \tau_n$. We know from (\ref{op4}) that $P$-a.s.
\begin{eqnarray*}
Y^n_{\sigma_m}= {\cal E}^g_{\sigma_m,\tau_n}[X_{\tau_n}] \ge X_{\sigma_m}, \qquad \,\forall \, m \ge n.
\end{eqnarray*}
In light of the right-continuity of $X$ and $Y^n$, letting $m \to \infty$ gives that
\begin{eqnarray*}
Y^n_\sigma \ge X_\sigma, \qquad {\hbox{$P$-a.s.}}
\end{eqnarray*}
Now let $(Y,Z)$ be the unique solution
to the BSDE (\ref{BSDEtau}) with $\xi=X_{\tau} $. It is easy to see that for $dt \times dP$-a.s. $(t,\omega) \in [0,T]
\times \Omega,~{\bf 1}_{\{t \le \tau_n\}}g(t,\omega,y,z)$ converges to ${\bf 1}_{\{t
\le \tau\}}g(t,\omega,y,z)$ uniformly in $(y,z) \in \mathbb{R}\times
\mathbb{R}^d$. Theorem \ref{stable} then implies that $P$-a.s.
$Y^n_t $ converges to $ Y_t $ uniformly in $t
\in [0,T]$. Thus we have
\begin{eqnarray*}
{\cal E}^g_{\sigma,\tau}[X_\tau] = Y_\sigma = \underset{n \to \infty}{\lim} Y^n_\sigma \ge X_\sigma, \qquad {\hbox{$P$-a.s.}},
\end{eqnarray*}
proving the theorem.
\qed
Finally, we study the so-called {\it Upcrossing Inequality} for
quadratic $g$-submartingales, which would be essential
for the study of path regularity of $g$-submartingales.
\begin{thm}\label{ucism}
Given a $g$-submartingale $X$,
we set $J \stackrel{\triangle}{=} \big(\|X\|_\infty+kT\big)e^{kT}$ and denote $\widetilde{X}_t
= X_t+k(J+1)t,~t \in [0,T]$. As usual, for any finite
set ${\cal D}=\{0 \le t_0<t_1<...<t_n \le T\}$, we let
$U^b_a(\widetilde{X},{\cal D})$ denote the number of upcrossings of the interval
$[a,b]$ by $\widetilde{X}$ over ${\cal D}$. Then there is a BMO
process $\big\{\beta_{\cal D}(t) \big\}_{t \in [0,t_n]} $ such that
\begin{eqnarray*}
E\Big[U^b_a(\widetilde{X},{\cal D})\exp{\big(\int_0^{t_n} \beta_{\cal D}(s) dB_s -
\frac{1}{2} \int_0^{t_n} |\beta_{\cal D}(s)|^2 ds \big)} \Big] \le
\frac{\|X\|_\infty+k(J+1)T+|a|}{b-a},
\end{eqnarray*}
and that
$E\int_0^{t_n} |\beta_{\cal D}(s)|^2 ds \le C$, a constant independent of the choice of ${\cal D}$.
\end{thm}
{\it Proof.} For any $j \in \{1,\cdots n\}$ we consider the
following BSDE:
\begin{eqnarray*}
Y^j_t=X_{t_j}+\int_t^{t_j}
g(s,Y^j_s,Z^j_s)ds-\int_t^{t_j} Z^j_sdB_s, \qquad \,\forall \, t \in [t_{j-1},t_j].
\end{eqnarray*}
Applying Corollary 2.2 of \cite{Ko} one has
\begin{eqnarray}
\label{usis1}
\|Y^j\|_\infty \le
\big(\|X_{t_j}\|_\infty+k(t_j-t_{j-1})\big)e^{k(t_j-t_{j-1})}\le J.
\end{eqnarray}
Now let us define a $d$-dimensional process
$\beta_{\cal D}(t)=(\beta^1_t,\cdots \beta^d_t)$, $t \in [0,t_n]$ by
\begin{eqnarray*}
\beta^l_t \stackrel{\triangle}{=} \sum^n_{j=1}{\bf 1}_{t \in (t_{j-1},t_j]} \int_0^1
\frac{\partial g}{\partial z_l} \big(t,Y^j_t,(Z^{j,1}_t, \cdots
\lambda Z^{j,l}_t ,0,\cdots0) \big) d\lambda,\qquad l \in \{1,\cdots,d\}.
\end{eqnarray*}
It is easy to see from Mean Value Theorem that for any $t \in
(t_{j-1},t_j]$,
\begin{eqnarray}
\label{uci2}
&& g(t,Y^j_t,Z^j_t)-g(t,Y^j_t,0)\nonumber\\
&=&\sum^d_{l=1}\Big\{g\big(t,Y^j_t,(Z^{j,1}_t, \cdots
Z^{j,l}_t,0,\cdots, 0)\big)\negthinspace
-\negthinspace g\big(t,Y^j_t,(Z^{j,1}_t, \cdots Z^{j,l-1}_t,0,\cdots,0)\big) \Big\} \nonumber\\
&=&\sum^d_{l=1} Z^{j,l}_t \beta^l_t =\langle Z^j_t, \beta_{\cal D}(t)\rangle.
\end{eqnarray}
Moreover, (H3) implies that
\begin{eqnarray}
\label{uci4}
\big|\beta^l_t\big|\le \ell(J)\sum^n_{j=1}{\bf 1}_{t \in (t_{j-1},t_j]}(1+ |Z^j_t|),\quad~~
t \in [0,t_n],\quad l \in \{1,\cdots,d\}.
\end{eqnarray}
We see from \eqref{BMO1} that each $Z^j$ is a BMO
process, thus so is $\beta_{\cal D}$. In fact, for any $\tau \in
{\cal M}_{0,t_n}$, one can deduce from (\ref{uci4}) that \if{0}
\begin{eqnarray}\label{uci3} E\big[
\int_\tau^{t_n} |\beta_{\cal D}(s)|^2 ds \big| {\cal F}_\tau \big]
&\neg \neg \neg \le \neg \neg \neg& 2d \ell(J)^2t_n +2d \ell(J)^2\sum_{j=1}^n
E\big[\int_{(\tau \vee t_{j-1})\land t_j }^{t_j} \negthinspace \negthinspace |Z^j_s|^2 ds \big| {\cal F}_\tau \big]\\
&\neg \neg \neg \leq \neg \neg \neg& 2d \ell(J)^2T+2d \ell(J)^2\sum_{j=1}^n \Big\{{\bf 1}_{\{\tau \le t_{j-1}\}}
E\big[\int_{t_{j-1}}^{t_j} \negthinspace \negthinspace |Z^j_s|^2 ds \big| {\cal F}_{\tau \land t_{j-1}}\big]\nonumber\\
& &+{\bf 1}_{\{t_{j-1} < \tau \leq t_j\}}E\big[\int_{(\tau \vee t_{j-1})\land t_j}^{t_j}
\negthinspace \negthinspace |Z^j_s|^2 ds \big| {\cal F}_{(\tau \vee t_{j-1})\land t_j}\big]\Big\} \nonumber\\
&\neg \neg \neg \leq \neg \neg \neg& 2d \ell(J)^2 T+2d \ell(J)^2\sum_{j=1}^n \Big\{{\bf 1}_{\{\tau \le t_{j-1}\}}
E\big[E[\int_{t_{j-1}}^{t_j} \negthinspace \negthinspace |Z^j_s|^2 ds|{\cal F}_{t_{j-1}}]
\big| {\cal F}_{\tau \land t_{j-1}}\big]\nonumber\\
&&+{\bf 1}_{\{t_{j-1} < \tau \le t_j \}}|\negthinspace| Z^j_s |\negthinspace|^2_{BMO_2} \Big\}\nonumber\\
&\neg \neg \neg \leq \neg \neg \neg& 2d \ell(J)^2T +2d \ell(J)^2\sum_{j=1}^n |\negthinspace|
Z^j_s |\negthinspace|^2_{BMO_2} \nonumber
\end{eqnarray}
\fi
\begin{eqnarray}
\label{uci3} && E\big[
\int_\tau^{t_n} |\beta_{\cal D}(s)|^2 ds \big| {\cal F}_\tau \big]
\le Ct_n +C\sum_{j=1}^n E\big[\int_{(\tau \vee t_{j-1})\land t_j }^{t_j}
\negthinspace \negthinspace |Z^j_s|^2 ds \big| {\cal F}_\tau \big]\\
&\neg \neg \neg \leq \neg \neg \neg& CT\negthinspace+\negthinspace C\sum_{j=1}^n \Big\{{\bf 1}_{\{\tau \le t_{j-1}\}}
E\big[\negthinspace \int_{t_{j-1}}^{t_j} \negthinspace \negthinspace \negthinspace |Z^j_s|^2 ds \big| {\cal F}_{\tau \land t_{j-1}}\big]
\negthinspace+\negthinspace {\bf 1}_{\{t_{j-1} < \tau \leq t_j\}}
E\big[\negthinspace \int_{(\tau \vee t_{j-1})\land t_j}^{t_j} \negthinspace \negthinspace |Z^j_s|^2 ds
\big| {\cal F}_{(\tau \vee t_{j-1})\land t_j}\big]\Big\} \nonumber\\
&\neg \neg \neg \leq \neg \neg \neg& CT\negthinspace+\negthinspace C\sum_{j=1}^n \Big\{{\bf 1}_{\{\tau \le t_{j-1}\}}
E\big[E[\int_{t_{j-1}}^{t_j} \negthinspace \negthinspace |Z^j_s|^2 ds|{\cal F}_{t_{j-1}}] \big|
{\cal F}_{\tau \land t_{j-1}}\big]\negthinspace+\negthinspace {\bf 1}_{\{t_{j-1} < \tau \le t_j \}}
|\negthinspace| Z^j_s |\negthinspace|^2_{BMO_2} \Big\}\nonumber\\
&\neg \neg \neg \leq \neg \neg \neg& CT \negthinspace+\negthinspace C\sum_{j=1}^n |\negthinspace| Z^j_s
|\negthinspace|^2_{BMO_2}, \nonumber
\end{eqnarray}
where $C\stackrel{\triangle}{=} 2d \, \ell(J)^2$. Thus $\big\{\mathscr{E}\big( \beta_{\cal D} \bullet
B \big)_t\big\}_{t \in [0,t_n]}$ is a uniformly integrable martingale.
By Girsanov's theorem we can find an equivalent probability $Q^{\cal D}$
such that
\if{0}
\begin{eqnarray*} \frac{dQ^{\cal D}}{dP}=\exp{\Big\{
\int_0^{t_n}\beta_{\cal D}(s)dB_s-\frac{1}{2}
\int_0^{t_n}|\beta_{\cal D}(s)|^2ds \Big\}} \end{eqnarray*}
\fi
$dQ^{\cal D}/dP=\mathscr{E}\big(\beta_{\cal D} \bullet B \big)_{t_n}$.
\if{0}
Moreover, it is well-known that $\displaystyle B^{\cal D}_t \stackrel{\triangle}{=} B_t-\int_0^t
\beta_{\cal D}(s) ds~ t \in [0,t_n]$ is the Brownian Motion under
$Q^{\cal D}$.
\fi
Then (\ref{uci2}) and (H2) show that for any $j \in\{1,...,n\}$ and
any $t \in [t_{j-1},t_j]$,
\begin{eqnarray*}
Y^j_t &=&X_{t_j} + \int_t^{t_j}\big[g(s,Y^j_s,0)+\langle Z^j_s, \beta_{\cal D}(s)\rangle\big]ds-\int_t^{t_j} Z^j_sdB_s\\
&=& X_{t_j}+ \int_t^{t_j} g(s,Y^j_s,0)ds-\int_t^{t_j} Z^j_s dB^{\cal D}_s \\
&\le& X_{t_j}+ k(J+1)(t_j-t)-\int_t^{t_j} Z^j_s dB^{\cal D}_s,
\end{eqnarray*}
where $B^{\cal D}$ denotes the Brownian Motion under $Q^{\cal D}$. Taking the
conditional expectation $E_{Q^{\cal D}}[\cdot|{\cal F}_t]$ on both sides of
the above inequality one can obtain that
\begin{eqnarray*}
{\cal E}^g_{t,t_j}[X_{t_j}]=Y^j_t \le
E_{Q^{\cal D}}[X_{t_j}|{\cal F}_t]+k(J+1)(t_j-t), \quad~~{\hbox{$P$-a.s.}} \quad \,\forall \, t \in
[t_{j-1},t_j].
\end{eqnarray*}
In particularly, taking $t=t_{j-1}$ we have
\begin{eqnarray*}
X_{t_{j-1}} \le {\cal E}^g_{t_{j-1},t_j}[X_{t_j}] \le
E_{Q^{\cal D}}[X_{t_j}|{\cal F}_{t_{j-1}}]+k(J+1)(t_j-t_{j-1}),\quad~~ {\hbox{$P$-a.s.}}
\end{eqnarray*}
Hence $\{\widetilde{X}_{t_j}\}^n_{j=0}$ is a ${Q^{\cal D}}$-submartingale.
Applying the classical upcrossing theorem one has
\begin{eqnarray*}
E_{Q^{\cal D}}\big[U^b_a(\widetilde{X},{\cal D})\big] \leq
\frac{E_{Q^{\cal D}}\big[(\widetilde{X}_{t_n}-a)^+\big]}{b-a} \le
\frac{\|X\|_\infty+k(J+1)T+|a|}{b-a}
\end{eqnarray*}
Furthermore, we denote $C>0$ to be a generic constant depending only
on $d,T, J, k, \|X\|_\infty$, and is allowed to vary from line to line.
Letting $\tau=0$ in (\ref{uci3}) one can deduce that
\begin{eqnarray*}
&& E\int_0^{t_n} |\beta_{\cal D}(s)|^2 ds \le C +C\sum_{j=1}^n E\int_{t_{j-1}}^{t_j} |Z^j_s|^2 ds \\
&\le& C +C\sum_{j=1}^n \Big\{ e^{4\tilde{K}J} E\big[e^{4\tilde{K}Y^j_{t_j}}
-e^{4\tilde{K}Y^j_{t_{j-1}}}\big]+e^{8\tilde{K}J}(t_j -t_{j-1})\Big\} \\
&\le &C+C \sum_{j=1}^n
E\big[e^{4\tilde{K}X_{t_j}}-e^{4\tilde{K}X_{t_{j-1}}}\big]= C+ C
E\big[e^{4\tilde{K}X_{t_n}}-e^{4\tilde{K}X_{t_0}}\big] \le C,
\end{eqnarray*}
where we applied (\ref{bmo1}) and (\ref{usis1}) with
$\tilde{K} \stackrel{\triangle}{=} \frac{1}{2}\vee k(J+1) \vee \ell(J)$ to derive the
second inequality and the third inequality is due to the fact that
$Y^j_{t_{j-1}} = {\cal E}^g_{t_{j-1},t_j}[X_{t_j}] \ge X_{t_{j-1}}$. The
proof is now complete. \qed
\medskip
With the above upcrossing inequality, we can discuss the continuity
of the quadratic $g$-sub(super)martingales.
\begin{cor}
\label{cgsm}
If $X$ is a $g$-submartingale\(resp. $g$-supermartingale\),
then for any denumerable dense subset ${\cal D}$ of $[0,T]$, it holds
$P$-a.s. that \begin{eqnarray*} \underset{r \nearrow t,\, r \in {\cal D}}{\lim}
X_r\mbox{ exists for any }t \in (0,T]\mbox{ and }\underset{r
\searrow t, \, r \in {\cal D}}{\lim} X_r\mbox{ exists for any }t \in [0,T).
\end{eqnarray*}
\end{cor}
{\it Proof.} If $X$ is a $g$-supermartingale, then $-X$ is
correspondingly a $g^{-}$-submartingale with $g^{-}$ defined in
(\ref{gneg}).
Hence, it suffices to assume that $X$ is a
$g$-submartingale. Let $\{{\cal D}_n\}_{n \in \mathbb{N}}$ be an increasing
sequence of finite subsets of ${\cal D}$ such that $\underset{n}
{\cup}{\cal D}_n={\cal D}$. For any two real numbers $a<b$, Theorem
\ref{ucism} and Jensen's Inequality imply that:
\begin{eqnarray*}
\widetilde{C}&\stackrel{\triangle}{=}&1+\frac{\|X\|_\infty+k(J+1)T+|a|}{b-a}\\
&\geq & 1+E\bigg[U^b_a(\widetilde{X},{\cal D}_n)\exp\Big\{\int_0^{t_n} \beta_{\cal D}(s) dB_s
- \frac{1}{2} \int_0^{t_n} |\beta_{\cal D}(s)|^2 ds \Big\}\bigg] \\
&=& E\bigg[\big(1+U^b_a(\widetilde{X},{\cal D}_n)\big)\exp\Big\{\int_0^{t_n} \beta_{\cal D}(s) dB_s
- \frac{1}{2} \int_0^{t_n} |\beta_{\cal D}(s)|^2 ds \Big\}\bigg] \\
&\geq& \exp
\bigg\{E\Big[\ln{\big(1+U^b_a(\widetilde{X},{\cal D}_n)\big)}+\int_0^{t_n}
\beta_{\cal D}(s) dB_s - \frac{1}{2} \int_0^{t_n} |\beta_{\cal D}(s)|^2 ds
\Big] \bigg\},
\end{eqnarray*}
from which one can deduce that
\begin{eqnarray*}
E\Big[\ln{\big(1+U^b_a(\widetilde{X},{\cal D}_n)\big)}\Big] \le \ln\widetilde{C}+\frac12
+ \|\beta_{\cal D} \|^2_{L^2_{\bf F}([0,t_n]; \mathbb{R}^d)} \le C',
\end{eqnarray*}
where $C'$ is a constant independent of the choice of ${\cal D}$. Since $
U^b_a(\widetilde{X},{\cal D}_n)\nearrow U^b_a\big(\widetilde{X},{\cal D} \big)$ as ${\cal D}_n
\nearrow {\cal D}$, Monotone Convergence Theorem implies that
$\ln{\big(1+U^b_a\big(\widetilde{X},{\cal D} \big)\big)}$ is integrable, thus
$U^b_a\big(\widetilde{X},{\cal D} \big)< \infty$, ${\hbox{$P$-a.s.}}$ Then a classical
argument yields the conclusion for $\widetilde{X}$, thus for $X$. The proof
is now complete. \qed
\noindent{\bf Acknowledgment.} We would like to express our sincere gratitude to the anonymous referee for
his/her careful reading of the original manuscript and many valuable suggestions, which helped us to
improve the quality of the paper significantly.
\bibliographystyle{plain}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,600 |
Apple's Phil Schiller introduces the A6 processor at last week's event.
How did Apple arrive at the A6 chip in the iPhone 5? A longtime chip analyst documents the long and winding road.
While endorsing a report that the A6 is a unique Apple design, Linley Gwennap, who heads The Linley Group, a chip consultancy, posted a brief history of the A6's chip development in a research note on Saturday.
Some of the history has been documented before, but other parts are not as well known. | {
"redpajama_set_name": "RedPajamaC4"
} | 8,343 |
Q: How Backbone.js with rest based service using Asp.NET MVC 4 I implemented some code using backbone.js in Asp.NEt MVC3 and found backbone.js very helpful. Actually I am developing the data warehouse application where user at view side can run/save his data analysis. And after relogin can re-run the saved analysis. On clicking the save button at toolbar system persist the analysis in DB.
My question is that can someone point links where backbone.js interacts with rest based service using Asp.NET MVC 4 so to save/retrieve the data in DB .
Please do advice me the better way as well.
A: I've built several very large systems on top of ASP.NET MVC4 and WebAPI, with Backbone, recently. I highly recommend WebAPI. It's very easy to use, and works very well with Backbone.
http://www.asp.net/web-api
As one example of an app that I've built with it:
https://ravenhq.com/
The registration, login, management, and account settings are all Backbone on top of WebAPI.
| {
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} | 1,631 |
"You're the One" is a song recorded by R&B group the Emotions released as a single in 1984 on Red Label Records. The single reached No. 34 on the Billboard Hot Soul Songs chart.
Background
You're the One was composed and produced by Billy Osborne and Zane Giles. The song came from the Emotions 1984 studio album Sincerely.
Critical reception
Hugh Wyatt of the New York Daily News wrote "The Emotions are a threesome, but their sound at times resembles a full choir, particularly on such cuts as the title tune and You're the One."
References
1984 songs
1984 singles
The Emotions songs | {
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} | 7,888 |
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org.apache.xmlrpc</FONT>
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Interface XmlRpcRequest</H2>
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<DT><PRE>public interface <B>XmlRpcRequest</B></DL>
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Interface to an XML-RPC request made by a client. Replaces the
class <code>org.apache.xmlrpc.XmlRpcClientRequest</code> from
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<TD><CODE><B><A HREF="../../../org/apache/xmlrpc/XmlRpcRequest.html#getConfig()">getConfig</A></B>()</CODE>
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getConfig</H3>
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<A HREF="../../../org/apache/xmlrpc/XmlRpcRequestConfig.html" title="interface in org.apache.xmlrpc">XmlRpcRequestConfig</A> <B>getConfig</B>()</PRE>
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java.lang.Object <B>getParameter</B>(int pIndex)</PRE>
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11/13/2020, 10:33am
By Becca Benner
The virtual concert featured a meet and greet .
The University Program Council's Major Entertainment Committee held a virtual concert of Auburn's The Brook and the Bluff on Oct. 27, 2020.
Leading up to the concert, the UPC Instagram page also announced a giveaway in which one lucky winner would have the opportunity to meet with the band virtually before the show.
Dalton Odom, junior in public relations, was the winner of this giveaway. He said he wasn't sure what to expect with a virtual meet and greet.
"I thought it kind of defeated the whole purpose of a 'meet and greet,' but I had so much fun," Odom said. "It was cool just to chat with a band that I've listened to for a long time. You still don't get that experience every day."
Director of the Major Entertainment committee, Julia Kern, coordinated the virtual concert.
"Major Entertainment's virtual concert featuring TBATB – The Brook and the Bluff – was overall a success for our very first virtual concert," Kern said.
She said there were some benefits and challenges that came with doing a virtual concert. One thing the committee had to do was book the band early.
Get The Plainsman straight to your inbox.
"We booked the band in early September, and we were so excited to get to work with Alabama natives and Auburn alums," Kern said. "Of course, we would have loved to have this concert in person, but UPC as a whole has had to be very creative in its event planning this semester."
Kern said working with the artist through virtual mediums was a different experience, but she was happy with the concert in the end. It also was a great way for students to release from the day-to-day school work, she said.
"We are continuing to create unique experiences for Auburn students to help get their mind off of how COVID-19 has affected this semester," Kern said. "I believe this virtual concert did just that and also provided some joy and entertainment along the way."
Kern said she believes the virtual concert was a success.
"The Brook and the Bluff has a cult-like following in Auburn, and they definitely showed up to support the band, even if it was just through a live stream," Kern said.
Elaina Eichorn, sophomore in marketing, was one of the virtual audience members.
"The virtual Brook and the Bluff concert really brightened my day," Eichorn said. "I watched it off my laptop, and it was so good to hear them play live music again. I'm a huge fan of their songs."
She has been to three other concerts of theirs and she said while virtual was a different experience, she enjoyed it all the same.
However, Eichorn said there were some aspects of live performances she missed.
"I missed being in the crowd with other fans; it was always really fun when everyone did the little step-touch dance to their songs," she said.
Eichorn said this concert did have its bonuses though.
"They played two new songs, which was very special to hear before they released it to the public," she said.
The concert did provide a music outlet Eichorn had been missing, she said.
"Overall, I'm so glad that UPC was able to get The Brook and the Bluff to Auburn and give the students something that we've all been missing," she said.
Do you like this story? The Plainsman doesn't accept money from tuition or student fees, and we don't charge a subscription fee. But you can donate to support The Plainsman.
Support The Plainsman
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The Auburn Plainsman welcomes thoughtful discussion on all of our stories, but please keep comments civil and on-topic. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
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Q: Можно ли запустить python скрипт так чтобы в pstree он отображался не как python а как я хочу? Использую Ubuntu, хотелось бы знать как сделать чтобы при просмотре активных процессов командой pstree, запущенный python скрипт писался не как python а, например, как MyProgram. То есть как изменить имя процесса при старте.
Либо хотел бы узнать можно ли изменить имя процесса при старте не вручную командой из консоли, а из другого скрипта. Например я использую Subprocess.
| {
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Q: Excel Macro pause, resume, or end based on clicking of OK and Cancel in a input box I am building a data validation between two databases workbook and need the macro to pause when a data mismatch is detected. Id like to make the macro pause when the mismatch is detected and pop up a dialog box with an "OK" button and a "cancel" button. I can't figure out how to make the macro resume when i press okay and stop when i press cancel. Can anyone help? btw im fairly new at VBA :)
A: If MsgBox("Your message here", vbOkCancel) = vbCancel Then Exit Sub
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,577 |
Ladislav Kovács, ps. "GuardiaN" (ur. 9 lipca 1991 w Komárnie) – słowacki profesjonalny gracz e-sportowy w serii gier Counter-Strike, aktualnie będący nieaktywnym graczem organizacji Natus Vincere. Były reprezentant takich formacji jak 3DMAX, TCM Gaming, Virtus.pro czy FaZe Clan. 2 najlepszy gracz CS:GO 2015 roku według rankingu HLTV oraz najlepszy słowacki gracz w historii Counter-Strike. Dotychczas w swojej karierze zarobił ok. 805 tysięcy dolarów.
Życiorys
Przed wydaniem Counter-Strike: Global Offensive w 2012, Słowak grał profesjonalnie zarówno w Counter-Strike 1.6, jak i Counter-Strike: Source. Grę pokazał mu jego brat, który kupił płytę CD z nieznaną wówczas grą FPS. Ladislav Kovács tak polubił produkcję, że grał w nią codziennie. Niedługo później znalazł drużynę i wziął udział w swoich pierwszych turniejach LAN. Ze względu na małą liczbę turniejów rozgrywanych w Counter-Strike 1.6, postanowił przejść na Counter-Strike: Source. Ladislav przeniósł się na CS:GO w połowie 2012 roku. 2 czerwca 2013 roku dołączył do Virtus.pro, która to drużyna była pierwszą profesjonalną organizacją Słowaka w CS:GO. 9 grudnia 2013 roku GuardiaN dołączył do organizacji Natus Vincere. To właśnie z tą organizacją, Ladislav zaczął osiągać największe sukcesy. W kwietniu 2016 roku, GuardiaN wraz z kolegami z Natus Vincere uplasował się na 1 miejscu w rankingu najlepszych drużyn CS:GO, tworzonych przez serwis HLTV. 3 sierpnia 2017 Ladislav opuścił Natus Vincere i dołączył do FaZe Clan. W tej organizacji wygrał m.in. ELEAGUE CS:GO Premier 2017, ESL One: New York 2017 czy EPICENTER 2018. 20 września 2019 GuardiaN opuścił FaZe Clan i powrócił do Natus Vincere, gdzie obecnie został przeniesiony na ławkę rezerwowych.
Wyróżnienia indywidualne
Został uznany najlepszym graczem turnieju Game Show League Season 1.
Został uznany najlepszym graczem turnieju ESL Pro League I.
Został uznany najlepszym graczem turnieju StarLadder StarSeries XIII.
Został uznany najlepszym graczem turnieju Electronic Sports World Cup 2015.
Został uznany najlepszym graczem turnieju Intel Extreme Masters X San Jose.
Został uznany najlepszym graczem turnieju GAMEKIT Counter Pit League Season 2
Został uznany najlepszym graczem turnieju Intel Extreme Masters XIII Sydney.
Został uznany najlepszym graczem turnieju ELEAGUE CS:GO Invitational 2019.
Został wybrany 10 najlepszym graczem 2013 roku według serwisu HLTV.
Został wybrany 11 najlepszym graczem 2014 roku według serwisu HLTV.
Został wybrany 2 najlepszym graczem 2015 roku według serwisu HLTV.
Został wybrany 17 najlepszym graczem 2016 roku według serwisu HLTV.
Został wybrany 9 najlepszym graczem 2017 roku według serwisu HLTV.
Został wybrany 11 najlepszym graczem 2018 roku według serwisu HLTV.
Został wybrany 10 najlepszym graczem 2018 roku według serwisu Thorin's Top.
Osiągnięcia
2. miejsce – Intel Extreme Masters VII – Katowice
3. miejsce – StarLadder StarSeries VI
3/4. miejsce – DreamHack Summer 2013
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4. miejsce – ESWC 2014
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1. miejsce – ESL Pro League I
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1. miejsce – Electronic Sports World Cup 2015
2. miejsce – CEVO Season 7 Professional
1. miejsce – CS:GO Champions League Season 1
2. miejsce – Gaming Paradise 2015
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2. miejsce – ESL ESEA Pro League Season 2 – Finals
2. miejsce – StarLadder i-League StarSeries XIV
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2. miejsce – MLG Major Championship: Columbus 2016
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3/4. miejsce – ELEAGUE Season 1
1. miejsce – ESL One: New York 2016
3/4. miejsce – EPICENTER 2016
2. miejsce – Adrenaline Cyber League 2017
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Przypisy
Słowaccy zawodowi gracze komputerowi
Urodzeni w 1991
Ludzie urodzeni w Komárnie | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,812 |
'use strict';
const Audit = require('../audit');
const URL = require('../../lib/url-shim');
const BLOCKLIST = new Set([
'click here',
'click this',
'go',
'here',
'this',
'start',
'right here',
'more',
'learn more',
]);
class LinkText extends Audit {
/**
* @return {!AuditMeta}
*/
static get meta() {
return {
category: 'Content Best Practices',
name: 'link-text',
description: 'Links have descriptive text.',
failureDescription: 'Links do not have descriptive text',
helpText: 'Descriptive link text helps search engines understand your content. ' +
'[Learn more](https://webmasters.googleblog.com/2008/10/importance-of-link-architecture.html)',
requiredArtifacts: ['URL', 'CrawlableLinks'],
};
}
/**
* @param {!Artifacts} artifacts
* @return {!AuditResult}
*/
static audit(artifacts) {
const failingLinks = artifacts.CrawlableLinks
.filter(link => {
if (
link.href.toLowerCase().startsWith('javascript:') ||
URL.equalWithExcludedFragments(link.href, artifacts.URL.finalUrl)
) {
return false;
}
return BLOCKLIST.has(link.text.trim().toLowerCase());
});
const headings = [
{key: 'href', itemType: 'url', text: 'Link destination'},
{key: 'text', itemType: 'text', text: 'Link Text'},
];
const details = Audit.makeTableDetails(headings, failingLinks);
let displayValue;
if (failingLinks.length) {
displayValue = failingLinks.length > 1 ?
`${failingLinks.length} links found` : '1 link found';
}
return {
rawValue: failingLinks.length === 0,
details,
displayValue,
};
}
}
module.exports = LinkText;
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,732 |
\section{Introduction}
\label{intro}
Conventional multiple-layer neural networks are entirely defined through \emph{explicit} expressions of its entering layers and loss functions. These expressions are typically provided in the form of a function that maps the input $y^{(k)}$
of the $k^{th}$ layer to its output $y^{(k+1)}$, as
\begin{equation}
y^{(k+1)} = f(y^{(k)}).
\end{equation}
Here $y^{(k)}$ may contain the output of the previous layer as well as the trainable parameters (commonly denoted $\theta$).
This explicit approach has the advantage that training through back-propagation, a method that operates on the partial derivatives associated with each layer, is straightforward to implement. However, this approach has also proven to be rather restrictive in a sense that
there are limited types of the layers that can be included in an end-to-end trainable network.
This work investigates the implicit approach. The term \emph{implicit layer} refers to a neural network layer that is defined implicitly by an implicit equation:
\begin{equation}
F(y^{(k)},y^{(k+1)}) = 0.
\end{equation}
An intuition into why this reformulation is beneficial can perhaps be found in multi-variable calculus, where the notion of \emph{functions given by a formula} has invariably been seen as too limited for many purposes. There are countless examples of functions that cannot be expressed explicitly, for instance, the locus of the expression
\begin{equation}
y^5 + 16y- 32x^3 + 32x = 0
\end{equation}
defines a precise and sketchable subset of $\mathbb{R}^2$, yet no formula for it exists.
As a matter of fact, the set of implicit functions is a proper superset of the set of explicit functions. This follows trivially as any explicit function in the form $y^{k+1}=f(y^k)$ can be defined implicitly as $y^{k+1}-f(y^k)=0$. As feedforward layers are conceptually functions mapping the input onto the output, a similar conclusion can also be made here. That is, not all implicit layers can be expressed explicitly. The reverse, however, is indeed true, that all explicit layers can be expressed implicitly. Figure \ref{fig:intro_fig} illustrates the premise.
This work presents a general framework of implicitly defined layers. Section~\ref{sec:implicit} addresses much of the theoretical analysis of implicit layers through the implicit function theorem. Section~\ref{sec:prop_through_imp_layers} describes the treatment of backpropagation of implicit layers; specifically, it is demonstrated how our framework is directly applicable to current automatic differentiation techniques for use in backpropagation based training. In section~\ref{sec:experiments} a number of diverse showcases demonstrate the versatility and practical benefit of the proposed approach.
\section{Related Works}
\label{sec:review}
Optimization plays a key role in a wide array of machine learning applications as a tool to perform inference in learning. Differentiation through optimization problems, e.g., $\argmin2$
operators, has seen a number of advances in recent years, among which, there are techniques that come up in bi-level optimization~\cite{gould2016differentiating,kunisch2013bilevel} and sensitivity analysis~\cite{bonnans2013perturbation,johnson2016composing,mairal2011task}.
More specifically, ~\cite{kunisch2013bilevel} proposed semi-smooth Newton algorithms that could efficiently find optimal regularization parameters, leading to efficient learning algorithms. The proposed bi-level learning framework could be applied to variational models, including the non-smooth functions, but not including data fidelity terms that are different from quadratic ones. The authors of ~\cite{gould2016differentiating} presented results for differentiating parameterised $\arg\!\min$ and $\arg\!\max$ optimization problems through equality constraints, but did not consider inequality constraints, and thus could only be applied to a limited class of problems with smooth functions within the $\arg\!\min$ and $\arg\!\max$ domain. The work of~\cite{mairal2011task} considered $\arg\!\min$ differentiation for a dictionary learning problem, and presents an efficient algorithm to solve it.
Moreover, \cite{amos2017input} considered $\arg\!\min$ differentiation within the context of the bundle method, and learned the inference step along with the network itself, without building structured prediction architectures explicitly. ~\cite{johnson2016composing} used implicit differentiation on convex objectives with coordinate subspace constraints but was unable to cope with general linear equality constraints and inequality constraints.
All the aforementioned approaches have limited applications in a sense that they either consider equality constraints in their implicit differentiation~\cite{gould2016differentiating,kunisch2013bilevel} rather than both equality and inequality constraints, or they can only insert the optimisation problem in the final layer of the network~\cite{johnson2016composing}.
Most closely related to our work is the recent method of~\cite{amos2017optnet}, in which, the implicit differentiation can be performed through both inequality and equality constraints, and the optimisation problems can be inserted anywhere in the network. To derive the gradients from the KKT matrix of the optimisation problem, OptNet~\cite{amos2017optnet} makes use of techniques from matrix differential calculus. However, this work is restricted to convex quadratic problems only.
This work differs from the existing works in that we are proposing a more general framework applicable to any layer expressible as an implicit function. Furthermore, most of the above mentioned work requires manual derivations and implementation of analytic gradient expressions, which is not needed in our framework.
\section{The Implicit Layer}
\label{sec:implicit}
In this work, we present a principled treatment of \emph{implicitly} defined layers in
feedforward neural networks. We formally define this concept as follows.
\begin{definition}[Implicit Layer]
A neural network layer is implicitly defined if its output $y^{(k+1)}\in\mathbb{R}^m$ is given
as the unique solution of the system of equations $F: \mathbb{R}^n \times \mathbb{R}^m \mapsto \mathbb{R}^m$,
\begin{align}
F(y^{(k)},y^{(k+1)})=0,
\label{implicit_def}
\end{align}
for some input $y^{(k)}\in\mathbb{R}^n$.
\end{definition}
\noindent
We distinguish this from the usual explicitly defined feedforward layers
where the relationship between input and output is given as $y^{(k+1)}=f(y^{(k)})$.
As before, $y^{(k)}$ does not only denote the output of the previous layer but also
the trainable parameters of the current layer.
\subsection{The Implicit Function Theorem}
To overcome the limitations of the naive definition of functions as explicit expressions, functions are instead commonly defined in a set-theoretic sense \cite{hamilton_1983}.
\begin{definition}[Function]\label{def:fun}
Here a function $f$ from a set $X$ to a set $Y$ is formally defined as a set of ordered pairs
$(x,y)$, $x\in X$ and $y\in Y$ with the properties that
\begin{itemize}
\item for each $x\in X$ there exist a pair $(x,y)\in f$;
\item if both $(x,y_1)\in f$ and $(x,y_2)\in f$, then $y_1=y_2$.
\end{itemize}
\end{definition}
\noindent
With this definition, each $x\in X$ defines a unique $y\in Y$ for which $(x,y)\in f$.
That is, it describes the process of associating each element of $X$ with a single
unique element in $Y$. It is common to use the more convenient notation of letting $y=f(x)$
denote $(x,y)\in f$.
The system of equations in \eqref{implicit_def} define an arbitrary closed (if $F$ is continuous) subset of $\mathbb{R}^m$. Although no explicit expression might exist, it can be shown that, under certain conditions, such implicit expressions can be \emph{locally} expressed as functions (with Definition~\ref{def:fun}). The details of sufficient conditions for this to hold is provided by the \emph{Implicit Function Theorem}~\cite{krantz2012implicit}, see theorem~\ref{implicit_function_theorem}.
\begin{theorem}[Implicit Function Theorem]
\label{implicit_function_theorem}
Given three open sets $X \subseteq \mathbb{R}^n, Y \subseteq \mathbb{R}^m$, and $Z \subseteq \mathbb{R}^m $, if function $F : X \times Y \mapsto Z$ is continuously differentiable, and $(\hat{x}, \hat{y}) \in \mathbb{R}^{n} \times \mathbb{R}^m$ is a point for which
\begin{align}
F(\hat{x}, \hat{y}) = \hat{z},
\end{align}
and the Jacobian of $F$ with respect to $y \subseteq Y$
\begin{align}
J_{F, y}\Big|_{i,j}=\left[{\frac {\partial F_{i}}{\partial y_{j}}}\right]
\end{align}
is invertible at $(\hat{x}, \hat{y})$, then there exists an open set $W \subset \mathbb{R}^{n}$ with $x \in W$ and a unique continuously differentiable function $\phi : W \mapsto Y$ such that $y = \phi(x)$ and
\begin{align}
F(x, y) = \hat{z}
\end{align}
holds for $x \in W$.
\end{theorem}
In addition, it can be shown that the partial derivatives of $\phi$ in $W$ are given by
\begin{align}\label{eq:thm1_J}
J_{y, x}=-\left[J_{F,y}\right]^{-1}\left[J_{F,x}\right].
\end{align}
This theorem states that under certain mild conditions on the partial derivatives, the solution to a system such as \eqref{implicit_def} is locally the graph of a function. Note that these functions might also only be available implicitly. However, according to this theorem, if such functions exist, they must be continuously differentiable and their derivatives can have a simple analytical expression~\eqref{eq:thm1_J}. Most of the results in this paper will build on this latter consequence of the implicit function theorem.
\section{Propagating through Implicit Layers}
\label{sec:prop_through_imp_layers}
\subsection{The Forward Pass}
As in conventional neural networks pipelines, in our proposed approach, the forward path and the backward path
of an implicit layer are independent.
The forward pass in an implicit layer is directly realised through the solution of \eqref{implicit_def}.
The most appropriate choice of solver is highly task specific, hence we will assume that a method of performing the forward pass is given along with the an implicit definition of a layer.
Our proposed framework is entirely agnostic to the choice of forward pass solvers, we can therefore make this
assumption without loss of generality.
Examples of different forward pass solvers are given in section \ref{sec:experiments}.
\subsection{The Backward Pass}
\label{sec:backprop}
The remaining question then relates to the backward pass through the implicit layer. To form a backward pass of a neural network layer we require the partial derivatives of its output with respect to its input, including the previous layer's output and all the trainable parameters of this layer. Hence, we need an expression for all these partial derivatives and an efficient way to calculate them. We will show that the former is provided by the implicit function theorem and the latter can be obtained by utilising existing automatic differentiation techniques.
The backward pass of an implicit layer is obtained as follows. Let the current state of the layer be given by $(\hat{y}^{(k)}, \hat{y}^{(k+1)})$ such that $F(\hat{y}^{(k)}, \hat{y}^{(k+1)})=c$, where $c\subseteq \mathbb{R}^n$ a vector of constants. Our premise is that there then exists, in the set-theoretic sense, a function $\phi: \mathbb{R}^m \mapsto \mathbb{R}^n$ such that $y^{(k+1)}=\phi(y^{(k)})$ and that $\phi$ is differentiable in some neighbourhood of $(\hat{y}^{(k)}, \hat{y}^{(k+1)})$.
Let the partial Jacobian of $F$ with respect to the output $y^{(k+1)}$ be denoted by
\begin{align}
J_{F, y^{(k+1)}}\Big|_{i,j}=\left[{\frac {\partial F_{i}}{\partial y^{(k+1)}_{j}}}\right],
\end{align}
then from the the implicit function theorem, theorem~\ref{implicit_function_theorem}, and the
differentiability assumption on $\phi$,
$\left[J_{F,y^{(k+1)}}\right]$ will have full rank at $(\hat{y}^{(k)}, \hat{y}^{(k+1)})$
and the sought partial derivatives of $\phi$ are given by
\begin{align}\label{backward_pass}
\begin{aligned}
J_{y^{(k+1)},y^{(k)}} =~& -\left[ J_{F,y^{(k+1)}}\right]^{-1}\left[ J_{F,y^{(k)}}\right],
\end{aligned}
\end{align}
evaluated in some neighbourhood of $(\hat{y}^{(k)}, \hat{y}^{(k+1)})$.
The Jacobian of the output with respect to the input is the key to applying the chain rule to compute the derivatives of the loss $L$ with respect to the input $\hat{y}^{(k)}$ and so to propagate gradients backwards:
\begin{align}\label{eq:chain_rule}
\begin{aligned}
\left(\frac{\partial{L}}{\partial{\hat{y}^{(k)}}}\right)^T &= \left(\frac{\partial{L}}{\partial{\hat{y}^{(k+1)}}}\right)^T \left[J_{y^{(k+1)},y^{(k)}}\right]\\
&= - \left(\frac{\partial{L}}{\partial{\hat{y}^{(k+1)}}}\right)^T \left[ J_{F,y^{(k+1)}}\right]^{-1}\left[ J_{F,y^{(k)}}\right].
\end{aligned}
\end{align}
It is the above expression that underpins our formal treatment of implicitly defined layers and is what
permits their inclusion into standard backpropagation training techniques.
As previously discussed any explicit layer can also be defined implicitly.
This then implies that the resulting backward pass for such a layer should be
invariant to the manner in which it is defined, i.e. explicitly or implicitly.
To verify that Equation~\eqref{eq:chain_rule} conforms to the standard treatment for explicit layers, consider an explicit layer that is defined by function $y^{(k+1)} = f(y^{(k)})$ and is currently at the state $(\hat{y}^{(k)}, \hat{y}^{(k+1)})$. The
implicit form of this layer is thus
\begin{align}\label{eq:explicit_fun}
F(y^{(k)}, y^{(k+1)}) = f(y^{(k)}) - y^{(k+1)} = 0.
\end{align}
It follows from~\eqref{eq:explicit_fun} that
\begin{align}
\begin{aligned}
\left[ J_{F,y^{(k+1)}}\right] = -I~~\text{and}~ \left[ J_{F,y^{(k)}}\right] = \left[ J_{f,y^{(k)}}\right],
\end{aligned}
\end{align}
which, substituted into~\eqref{eq:chain_rule}, leads to the familiar equation for explicit layer backpropagation
\begin{align}\label{eq:explicit_chain}
\begin{aligned}
\left(\frac{\partial{L}}{\partial{\hat{y}^{(k)}}}\right)^T = \left(\frac{\partial{L}}{\partial{\hat{y}^{(k+1)}}}\right)^T \left[ J_{f,y^{(k)}}\right].
\end{aligned}
\end{align}
Lastly, as stated in \eqref{eq:chain_rule}, calculating $\left[J_{y^{(k+1)},y^{(k)}}\right]$ requires the explicit construction of
the Jacobian $\left[J_{F,y^{(k)}}\right]$. As this matrix is typically
large this operation can be very costly.
Modern deep learning packages seldom, if at all, explicitly construct
the entering matrices in the backward pass, but instead derive
Vector-Jacobian Products for the right hand side of~\eqref{eq:explicit_chain}.
This technique results in greatly reduced memory requirements and computational
costs.
This approach is applicable to backpropagating implicit layers as well.
Note that the right hand side of~\eqref{eq:chain_rule} is also a Vector-Jacobian Product of the vector
$-\left(\partial{L}/\partial{\hat{y}^{(k+1)}}\right)^T \left[ J_{F,y^{(k+1)}}\right]^{-1}$
and Jacobian matrix $\left[ J_{F,y^{(k)}}\right]$, thus it is feasible to improve computational efficiency by carefully analysing the expression of and the structure of $\left[J_{F,y^{(k)}}\right]$; see Section~\ref{sec:showcase_mnist} for an example of efficiently backpropagating an implicit layer doing quadratic programming.
\subsection{Automatic Differentiation}\label{sec:auto_diff}
Deriving the analytical expression for the implicit backward pass
can be prohibitively time consuming and error-prone, particularly in situations where the models or network architecture is expected to change frequently. On the other hand, techniques for \emph{automatic differentiation} in deep learning packages provide accurate, efficient, and reliable computation of partial derivatives in a fully automated manner, thus eliminate the need for manual derivation and implementation of analytical gradient formulae. In light of this, we propose a method by which the partial gradients of implicit layers could be automatically calculated.
In this section we show how existing implementations of automatic differentiation can be used to provide backward passes through implicitly defined layers with little modification.
Let our implicit layer be defined as in \eqref{implicit_def}. Now consider the related explicit layer defined by
\begin{align}\label{eq:autodiff0}
\begin{aligned}
&z^{(k+1)} = F(z^{(k)}), ~~\text{where}\\
&z^{(k)} = [y^{(k)},y^{(k+1)}]\in \mathbb{R}^{n+m}~~\text{and}~~z^{(k+1)}\in 0^m.
\end{aligned}
\end{align}
As this layer is defined explicitly we can apply existing automatic differentiation methods directly to yield the partial derivatives
\begin{align}
\frac{\partial z_i^{(k+1)}}{\partial z_j^{(k)}}= \frac{\partial F_i}{\partial z_j^{(k)} }, \hspace{4mm} i\in[1,m],\ j\in[1,m+n],
\label{eq:autodiff1}
\end{align}
or more compactly
\begin{align}
\frac{\partial z^{(k+1)}}{\partial z^{(k)}}=
\Matris{
J_{F, y^{(k+1)}}
\ \Big| \
J_{F, y^{(k)}}
}.
\label{eq:autodiff2}
\end{align}
Comparing \eqref{eq:autodiff2} with \eqref{backward_pass}
we note that the elements $J_{F, y^{(k)}}$ and $J_{F, y^{(k+1)}}$ for calculating the backward pass in \eqref{backward_pass}
are provided by automatic differentiation of an explicit layer as of \eqref{eq:autodiff1}.
Consequently, the automatic differentiation of implicit layers
can be realised as a matrix inversion and multiplication, with no need
for manual derivation or model specific implementations.
\noindent
Let us consider the following simple example:
\begin{exmp}
\begin{align}
\begin{aligned}
&F_1(x,y)=x^2 + y_1^2 + y_2^2 - 4 , \\
&F_2(x,y)= xy_1 - 1.
\end{aligned}
\end{align}
A solution to $F(x,y)=0$ is given by $(\hat{x}=1, ~\hat{y}=[1, \sqrt{2}])$.
To calculate the backward pass, (i.e. the partial derivatives of the output $y$ with respect to the input $x$) through such a layer we instead look at the related explicit layer defined by
\begin{align}
z^{(k+1)} = F(z^{(k)}),~~~ z^{(k)}\in \mathbb{R}^3, \ z^{(k+1)}\in \mathbb{R}^2,
\end{align}
and specifically in this example
\begin{align}
\begin{aligned}
&z^{(k)} = (x, y) = (\hat{x}, \hat{y}_1, \hat{y}_2), \\
&z^{(k+1)} = (0,0).
\end{aligned}
\end{align}
As the layer is now defined explicitly, we can apply automatic differentiation directly to provide
\begin{align}
\frac{\partial z_i^{(k+1)}}{\partial z_j^{(k)}}(z)= \frac{\partial F_i}{\partial z_j^{(k)}}(z), \hspace{4mm} i=1,2,\ j=1,2,3,
\label{eq:autodiff_1}
\end{align}
for some $z\in \mathbb{R}^3$.
In compact form, this equation reads
\begin{align}
\left[J_{z^{(k+1)}, z^{(k)}}(z)\right] = \left[J_{F, z^{(k)}}(z)\right].
\end{align}
The term on the right hand side of the above equation is instantiated as
\begin{align}\label{eq:dev}
\begin{aligned}
\matris{
J_{F, z^{(k)}} (z^{(k)})
}
&=
\left[ \arraycolsep=1.8pt\def\arraystretch{2.2}
\begin{array}{cccc}
\frac{\partial F_1}{\partial x} & \frac{\partial F_1}{\partial y_1} & \frac{\partial F_1}{\partial y_2} \\
\frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial y_1} & \frac{\partial F_2}{\partial y_2}
\end{array} \right]
= \matris{ J_{F, x}(z^{(k)})~ & \Big| & ~J_{F, y}(z^{(k)})}\\
&=
\left[ \arraycolsep=1.8pt\def\arraystretch{1.8}
\begin{array}{cccc}
2x & 2y_1 & 2y_2 \\
y_1 & x & 0
\end{array} \right]
=
\left[ \arraycolsep=1.8pt\def\arraystretch{1.8}
\begin{array}{cccc}
2 & 2 &2\sqrt{2}\\
1 & 1 & 0
\end{array} \right].
\end{aligned}
\end{align}
And finally, the partial derivative of the output $y$ of the implicit layer with respect to the input $x$ is calculated by~\eqref{backward_pass}:
\begin{align}
\begin{aligned}
\left[J_{y,x}([\hat{x},\hat{y}])\right] &= -\big[J_{F,y}(\hat{x},\hat{y})\big]^{-1} \big[J_{F,x}(\hat{x},\hat{y})\big] \\
&=-\left[ \arraycolsep=1.8pt\def\arraystretch{2.2}
\begin{array}{cc}
2 &2\sqrt{2}\\
1 & 0
\end{array} \right]^{-1}
\left[ \arraycolsep=1.8pt\def\arraystretch{2.2}
\begin{array}{c}
2\\
1
\end{array} \right]
\end{aligned}
\end{align}
Recall that as ~\eqref{eq:dev} is supported by the automatic differentiation implementations of over-the-shelf machine learning libraries, we eventually get the partial derivative $\left[J_{y,x}\right]$ without manually deriving algebraic expressions of the derivatives.
\pushQED{\qed}
\qedhere
\popQED
\end{exmp}
\section{Example Applications}\label{sec:experiments}
In this section we present a number of example applications to demonstrate the usage and features of the proposed framework for implicit layers.
In particular, we show:
how to model a real problem as implicit functions;
how implicit layers enables end-to-end training;
how the backward path is independent to the forward solver; and
how the automatic differentiation feature copes with complex functions.
Note, our intention here is not to propose new methods for solving specific tasks nor attempting to improve state-of-the-art algorithms but rather to highlight the versatility and accessibility of the proposed framework.\footnote{Demo codes are available on \url{github.com/qgzhang/Imp_layers_demo}.}
\subsection{Quadratic Programming Layers }
\label{sec:showcase_mnist}
We begin with an introductory example showing how the proposed framework
models a Quadratic Programming (QP) problem as an implicit layer and
then determines the necessary expressions for backpropagation
for use in an end-to-end trainable neural network.
Including QP in such a manner was also proposed
in the notable work~\cite{amos2017optnet} as a way to encode
constraints and dependencies that conventional explicit layers are unable to capture.
The efficiency of this approach was demonstrated on a number of problems
including, signal denoising and handwritten digit recognition. However, this work
is restricted to convex quadratic problems only, in addition it also requires the
backward step to be explicitly implemented. These are limitations our proposed
framework do not possess.
To better illustrate our example, we follow the existing work of
OptNet~\cite{amos2017optnet} closely but elaborate relevant
concepts from the point of view of implicit layers.
We define a convex QP as
\begin{align}\label{eq:QP}
\begin{aligned}
& \underset{y}{\arg\!\min}~~\frac{1}{2}y^TQy + q^Ty\\
& s.t.~ Ay = b,~ Gy <= h,
\end{aligned}
\end{align}
where $y$ is the optimisation variable; and $Q\succeq 0, q, A, b, G$, and $h$ are parameters
of the QP problem.
$Q, q, A, b, G$, and $h$ collectively represent both the input and the trainable
parameters of the layer. They may or may not dependent on the input of the layer
but, conceptually, they can all be classified as explicitly differentiable functions
of the input. Therefore, the key to backpropagate this layer is to determine the derivatives of the loss $L$ with respect to $Q, q, A, b, G$, and $h$.
The remaining task is then simply direct application of the standard chain rule.
Using the KKT conditions we can write~\eqref{eq:QP} into a system of implicit
functions.
The Lagrangian function of~\eqref{eq:QP} is given by
\begin{align}
\begin{aligned}
L(y, \lambda, \nu) = \frac{1}{2}y^TQy + q^Ty + \lambda^T(Ay-b) + \nu^T(Gy-h),
\end{aligned}
\end{align}
with dual variables $\lambda$ and $\nu$.
The corresponding KKT conditions then become
\begin{align}\label{eq:QP_KKT}
F:
\left\{
\begin{aligned}
Qy + q + A^T\lambda + G^T\nu&=0, &(stationarity)\\
Ay-b&=0, &(feasibility) \\
D(\nu)(Gy-h)&=0, &(compl.\ slackness)
\end{aligned}
\right.
\end{align}
where $D(\nu)$ denotes a matrix with $\nu$ in the diagonal and $0$ everywhere else.
As \eqref{eq:QP_KKT} forms the necessary conditions for the solution of
\eqref{eq:QP} it also forms a system of equations that we can be used to define
an implicit representation of the QP layer, i.e., $F$ as in \eqref{implicit_def}.
Note that under the above definition, the primal $y$ and the dual $\lambda$ and
$\nu$ correspond to the output of the layer, i.e., $y^{(k+1)}$ as in~\eqref{eq:chain_rule}; and $Q, q, A, b, G$ and $h$ correspond to the input of the layer, i.e., $y^{(k)}$
in~\eqref{eq:chain_rule}:
\begin{align}
\begin{aligned}
&y^{(k+1)} = (y, \lambda, \nu)\\
&y^{(k)} = (Q, q, A, b, G, h).
\end{aligned}
\end{align}
Now, in order to apply~\eqref{eq:chain_rule} to complete the backpropagation process, we need $J_{F, y^{(k)}}$ and $J_{F, y^{(k+1)}}$, which are obtainable by differentiating~\eqref{eq:QP_KKT}:
\begin{align}\label{eq:JFyk1}
\begin{aligned}
\left[J_{F, y^{(k+1)}}\right] &= \left[ J_{F, y} \Big| J_{F, \lambda} \Big| J_{F, \nu} \right]\\
&= \left[
\begin{array}{c}
Q \\
A \\
D(\nu)G
\end{array}
\Bigg|
\begin{array}{c}
A^T \\
0 \\
0
\end{array}
\Bigg|
\begin{array}{c}
G^T \\
0 \\
D(Gy-h)
\end{array}
\right]
\end{aligned}
\end{align}
and
\begin{align}\label{eq:JFyk}
\begin{aligned}
\left[J_{F, y^{(k)}}\right] &= \left[ J_{F, Q} \Big| J_{F, q} \Big| J_{F, A} \Big| J_{F, b} \Big| J_{F, G} \Big| J_{F, h}\right]\\
&= \left[
\begin{array}{c}
I\otimes y^T\\
0 \\
0
\end{array}
\Bigg|
\begin{array}{c}
I \\
0 \\
0
\end{array}
\Bigg|
\begin{array}{c}
\lambda^T\otimes I \\
I\otimes y^T \\
0
\end{array}
\Bigg|
\begin{array}{c}
0 \\
-I \\
0
\end{array}
\Bigg|
\begin{array}{c}
\nu^T\otimes I \\
0 \\
D(\nu)I\otimes y^T
\end{array}
\Bigg|
\begin{array}{c}
0 \\
0 \\
-I
\end{array}
\right],
\end{aligned}
\end{align}
where $\otimes$ denotes the Kronecker product.
Furthermore, we show that, as in the case of backpropagating explicit layers, explicit construction of $\left[ J_{F, y^{(k)}} \right]$ is avoidable. Taking the partial derivatives of the loss with respect to $Q$ for example, observe in the following expression that the multiplication operation on $J_{F, Q}$ is replaced with a Kronecker product on vector $y^T$:
\begin{align}\label{eq:simplification}
\begin{aligned}
\frac{\partial{L}}{\partial{Q}} &= \frac{\partial{L}}{\partial{y^{(k+1)}}} \left[ J_{F, y^{(k+1)}} \right]^{-1} \left[ J_{F, Q} \right] \\
&= \frac{\partial{L}}{\partial{y}} \left[ J_{F, y^{(k+1)}} \right]^{-1} \left[ I\otimes y^T \right] \\
&= \frac{\partial{L}}{\partial{y}} \left[ J_{F, y^{(k+1)}} \right]^{-1} \otimes y^T.
\end{aligned}
\end{align}
Other components of $\left[ J_{F, y^{(k)}} \right]$ can be simplified in a similar fashion.
\subsubsection{Automatic Differentiation of the QP Layer}
With the techniques introduced in~\ref{sec:auto_diff} we are even able to avoid to derive~\eqref{eq:JFyk1} and~\eqref{eq:JFyk} (as well as any attempt for simplification, e.g.,~\eqref{eq:simplification}) at all. We introduce the following auxiliary variables:
\begin{align}
\begin{aligned}
&z^{(k)} = \left[ y^{(k+1)} \Big| y^{(k)} \right] = [y, \lambda, \nu, Q, q, A, b, G, h]\\
&z^{(k+1)} = \left[ z_1 \Big| z_2 \Big| z_3 \right] = 0,
\end{aligned}
\end{align}
and thus obtain an explicit representation $F(z^{(k)})=z^{(k+1)}$ of the QP layer by rewriting~\eqref{eq:QP_KKT} as
\begin{align}\label{eq:QP_KKT2}
F:
\left\{
\begin{aligned}
Qy + q + A^T\lambda + G^T\nu &= z_1\\
Ay - b &= z_2\\
D(\nu)(Gy - h) &= z_3.
\end{aligned}
\right.
\end{align}
As $z_1$, $z_2$ and $z_3$ are effectively explicit functions of $z^{(k)}$, we are able to use existing automatic differentiation implementations to compute $\frac{\partial{z_1}}{\partial{z^{(k)}}}$, $\frac{\partial{z_2}}{\partial{z^{(k)}}}$ and $\frac{\partial{z_3}}{\partial{z^{(k)}}}$ without any manual derivation, which constitute
\begin{align}
\frac{\partial{z^{(k+1)}}}{\partial{z^{(k)}}} = \frac{\partial{F}}{\partial{z^{(k)}}} = \left[ J_{F, y^{(k+1)}} \Big| J_{F, y^{(k)}} \right].
\end{align}
The automatic differentiation feature of the implicit layer comes extremely handy when the expression of the functions is complex, as is shown in the example in Section~\ref{sec:showcase_graphmatch}.
\subsubsection{Experiments - QP Layers}
We verify the formulation of the QP layer on the task of hand digit recognition on MNIST. The upper branch of figure~\ref{fig:mnist_pipeline} shows the pipeline. Two fully connected layers take input from the vectorised $28\times28$ image, followed by a layer that solves a QP problem as defined by~\eqref{eq:QP}; the solution of the QP problem goes through softmax and yields the negative log likelihood loss. We model this QP solving layer as an implicit layer.
\begin{figure}[ht]
\centering
\begin{subfigure}{0.48\textwidth}
\centering
\includegraphics[width=\linewidth]{figs/mnist_pipeline.pdf}
\end{subfigure}%
\caption{The hand written digits recognition pipeline. The solid branch represents using an implicit layer to solve the QP problem and the dashed branch represents using OptNet.}
\label{fig:mnist_pipeline}
\end{figure}
The lower branch shows of figure~\ref{fig:mnist_pipeline} the pipeline designed by OptNet. The only difference between the two pipelines is the component that solves the QP problem. As for the choice of the solver of the forward pass, we adopt the same primal-dual interior point method. This provides an equal comparison of the backward pass between the two pipelines.
As expected, these two pipelines produce very similar convergence curves on
both training and testing data, see figure~\ref{fig:mnist_loss_acc}, thus validating the
efficacy of our proposed framework.
\begin{figure}[ht]
\centering
\begin{subfigure}[ht]{0.40\textwidth}
\centering
\includegraphics[width=\textwidth]{figs/Opt_all_train_loss.pdf}
\end{subfigure}
\vspace{1em}
\begin{subfigure}[ht]{0.40\textwidth}
\centering
\includegraphics[width=\textwidth]{figs/Opt_all_train_acc.pdf}
\end{subfigure}
\caption{Hand written digits recognition learning curves. Both pipelines achieves test accuracy above $98\%$ upon termination at 15 epochs.}
\label{fig:mnist_loss_acc}
\end{figure}
The main purpose of this initial example application was to introduce the basic usage of
implicitly defined layers and to demonstrate that its performance is comparable to
existing state-of-the-art method .
However, the key difference between these two methods is that
\cite{amos2017optnet} is restricted to convex QPs and requires a manual implementation
of the resulting backward pass, whereas our implicit framework can easily be adapted
to a much broader class of layers. It also does not rely on dedicated backward pass
implementations but can instead take advantage of existing automatic differentiation
techniques to calculate the necessary partial derivatives.
\subsection{Normalised Cuts Layers}
In this example we show how an implicit layer can be used to include a
Normalised Cuts (NCut) framework~\cite{shi2000normalized} into an end-to-end
trainable architecture. Normalised Cuts is a well established method for solving
the perceptual grouping problem in computer vision. This is done by treating images as graphs and the image segmentation task as finding the cuts in such graphs that minimise the Normalized Cuts criterion.
Given an image $I$ an undirected graph $G$, with vertices $V$ and edges $E$ where each vertex in $V$ corresponds to a single pixel in the image and the weights on the edges $E$ encodes the similarity between two pixels. The non-negative weights of each such edge are represented by an affinity matrix $W$, with only non-negative entries and of full rank. A Normalised Cut is then defined as the non-trivial partitioning of the graph $G$
into disjoint subsets $A$ and $B$ such that the criterion
\begin{eqnarray}
\label{eq_ncut_def}
N_{cut}=\frac{cut(A,V)}{assoc(A,V)} +\frac{cut(B,V)}{assoc(B,V)}
\end{eqnarray}
is minimised. Here $A\cup B=V$, $A\cap B=\emptyset$ and the normalizing term is defined as $assoc(A,V)=\sum_{i\in A, j\in V} w_{ij}$.
It is shown in \cite{shi2000normalized} that
a continuous underestimator of the (minimal) Normalized Cut can be efficiently computed
as the second smallest eigenvalue of the generalised eigensystem
\begin{align}
\label{eq:ncut}
L v = \lambda D v,
\end{align}
where $L$ denotes the discrete Laplacian of the adjacency matrix $W$ of the image
and $D$ is a diagonal matrix of the weighted graph order.
The implicit form of the NCut layer, defining the relationship between the
resulting cut and the graph affinity, is obtained directly by rewriting~\eqref{eq:ncut}
as
\begin{align}
F:
\left\{
\begin{aligned}
(L - \lambda_2 D)v_2 = 0.\\
v_2^Tv_2 -1=0,
\end{aligned}
\right.
\end{align}
Here $\lambda_2$ and $v_2$ denotes the second smallest generalised eigenvalue
and corresponding eigenvector of~\eqref{eq:ncut}.
\subsubsection{Experiments - NCut Layers}
The Normalised Cuts method is a graph theoretic formulation that aims to partition
an image based on some measure of similarity between pixels (vertices) such that
similar pixels are grouped together and dissimilar ones may be separated.
This similarity measure is typically handcrafted,
\cite{malik2001contourand,yu2003multiclass}. To evaluate the implicit Ncut layer
we instead aim to learn this similarity metric from training data.
The proposed implicit framework permits us to do so in an end-to-end fashion.
We design a prototypical neural network with the NCut layer and compare it with the
classical, non-learning based Normalised Cut method~\cite{shi2000normalized}, as shown
in figure~\ref{fig:ncut_pipeline}.
\begin{figure}[ht]
\centering
\begin{subfigure}{0.48\textwidth}
\includegraphics[width=\linewidth]{figs/ncut_pipeline.pdf}
\end{subfigure}
\caption{Image segmentation pipeline. The solid branch represents an end-to-end trainable pipeline. It has an implicit NCut layer to decompose the Laplacian $L$, which is constructed based on learned features. In contrast, the standard non-learning NCut method (the dashed branch) construct the Laplacian based on handcrafted features.}
\label{fig:ncut_pipeline}
\end{figure}
We evaluate this pipeline
on the HazySky dataset~\cite{Song2018Sky} which contains 500 natural images with ground truth sky/non-sky segmentation mask. Training data consists of 400 randomly sampled images with the rest for testing data. Measured by the Intersection over Union (IOU) score, the accuracy of the learning based pipeline converges to just over$85\%$. By comparison, the non-learning based NCut method only achieves an IOU score of about $70\%$. See figure~\ref{fig:hazysky_acc}.
\begin{figure}[ht]
\centering
\begin{subfigure}{0.40\textwidth}
\centering
\includegraphics[width=\linewidth]{figs/seg_train_only_loss.pdf}
\end{subfigure}
\vspace{1em}
\begin{subfigure}{0.40\textwidth}
\centering
\includegraphics[width=\linewidth]{figs/train_iou.pdf}
\end{subfigure}
\caption{Convergence curve of the image segmentation pipeline. The left panel plots the convergence of loss as training for the learning based branch. The right panel compares the IOU scores of the learning based and standard method.}
\label{fig:hazysky_acc}
\end{figure}
\subsection{Level Set Layers for Shape Inference}
Next we demonstrate how Level Sets, an established tool for
numerical analysis of surfaces and shapes can be
incorporated into current deep learning frameworks through an
implicit formulation. We evaluate this approach on the task of inferring object 3D shapes from a
single image.
Representing shapes in end-to-end trainable neural networks has proven to be
challenging task. A majority of existing learning-based approaches involving shapes or
structures relying on either voxel occupancy
\cite{girdhar2016learning,choy20163d,rezende2016unsupervised,richter2018matryoshka},
sparse point clouds \cite{qi2017pointnet,fan2017point} or explicit shape
parameterisation \cite{liao2018deep}.
Each of these representations comes with its own advantages and disadvantages, in
particular for the application of shape inference in a learning framework.
Recent work \cite{Park_2019_CVPR,Michalkiewicz_2019_ICCV} has instead argued that \emph{Level Sets} constitute a more appropriate choice for the task of learned shape inference.
The Level Set method for representing moving interfaces was proposed independently
by \cite{osher1988fronts} and \cite{dervieux1980finite}.
This method defines a time dependent orientable surface $\Gamma(t)$ implicitly as the zero
iso-contour, or level set, of a higher dimensional
auxiliary scalar function, called the \emph{level set function}
or \emph{embedding function}, $\phi(x,t): \Omega \times \mathbb{R} \mapsto \mathbb{R}$, as,
\begin{align}
\Gamma(t)= \left\{ x : \phi(x,t) = 0 \right\},
\label{eq:level_set}
\end{align}
with the convention that $\phi(x,t)$ is positive on the interior and
negative on the exterior of $\Gamma$.
The underlying idea of the level set method is to capture the motion of
the iso-surface through the manipulation of the level set function $\phi$.
However, owing to this implicit definition of shape, existing deep learning frameworks
can not incorporate this representation straightforwardly. Instead, the inference is
either
carried out on the embedding function $\phi$ \cite{Park_2019_CVPR} directly, rather than on the
shape itself, thus resulting in suboptimal reconstructions, or by approximating metrics
on the iso-surfaces of $\phi$ \cite{Michalkiewicz_2019_ICCV}.
In this section we will show how the level set representations can be included exactly
using implicit layers. Again, the aim here is not provide an exhaustive study of
implicit representations of shape but rather to demonstrate the applicability of our
proposed framework.
The definition of the implicit layer that realises \eqref{eq:level_set} is provided
directly from its definition as
$0=\phi(x)$.
However, here the input to this layer is a discrete representation of the embedding function $\phi$
and Theorem.~\ref{implicit_function_theorem} assumes a continuous function $F$.
This can be accomplished by constructing a continuous surrogate of $\phi$ and using this to represent the desired implicit layer. We define
\begin{align}
0= \phi_{tri}(y^{(k+1)};y^{(k)})
\end{align}
as our implicit layer representation of \eqref{eq:level_set}.
Here $\phi_{tri}{\cdot;y^{(k)}}$ denotes the tri-linear interpolation of $y^{(k)}$ at $y^{(k+1)}$.
The forward pass through this layer can be obtained by any iso-surface extraction algorithm,
see \cite{hansen2011visualization}. In this setup we use standard marching cubes \cite{lorensen1987marching}.
\subsubsection{Experiments - Level Set Layers}
To evaluate this formulation we followed the implementation details provided in
\cite{girdhar2016learning} and \cite{Michalkiewicz_2019_ICCV} as closely as possible with respect to preprocessing, image rendering and evaluation.
Our network was evaluated with the proposed formulation on 8,000 3D models from four different categories
({'cars'}, {'chairs'}, {'bottles'} and {'sofas'}) in the ShapeNet dataset
\cite{chang2015shapenet}. The results, using a $32^3$
resolution, are shown in figure \ref{fig:levelsets_cars} and table \ref{tab:table1}.
These results are comparable to, if not better than, those reported in
\cite{Michalkiewicz_2019_ICCV}.
Again the results are achieved with no dedicated implementation of the backward pass.
\begin{figure}[h!
\centering
\includegraphics[width=.27\linewidth]{figs/supp/b/GT2.png}
\includegraphics[width=.16\linewidth]{figs/supp/b/input1.png}
\includegraphics[width=.27\linewidth]{figs/supp/b/pred_ls_2.png}
\includegraphics[width=.27\linewidth]{figs/supp/b/pred_vox_2.png}
\includegraphics[width=.27\linewidth]{figs/supp/e/GT5.png}
\includegraphics[width=.16\linewidth]{figs/supp/e/input1.png}
\includegraphics[width=.27\linewidth]{figs/supp/e/pred_ls_5.png}
\includegraphics[width=.27\linewidth]{figs/supp/e/pred_vox_5.png}
\includegraphics[width=.27\linewidth]{figs/supp/g/GT7.png}
\includegraphics[width=.16\linewidth]{figs/supp/g/input1.png}
\includegraphics[width=.27\linewidth]{figs/supp/g/pred_ls_7.png}
\includegraphics[width=.27\linewidth]{figs/supp/g/pred_vox_7.png}
\def.24{.24}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_man_36.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_view_36.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_ls30_36.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_vox30_36.png}\\
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_man_37.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_view_37.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_ls30_37.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_vox30_37.png}\\
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_man_46.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_view_46.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_ls30_46.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_vox30_46.png}\\
\def.24{.24}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_man_1.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_view_1.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_ls30_1.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_vox30_1.png}\\
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_man_5.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_view_5.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_ls30_5.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_vox30_5.png}\\
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_man_11.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_view_11.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_ls30_11.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_vox30_11.png}\\
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_man_28.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_view_28.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_ls30_28.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_vox30_28.png}\\
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_man_32.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_view_32.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_ls30_32.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_vox30_32.png}\\
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_man_34.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_view_34.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_ls30_34.png}
\includegraphics[width=.24\linewidth]{./figs/s_figs/s_vox30_34.png}\\
\caption{Shape inference from a single image. Ground-truth (left),
input image (second column),
level sets (third column),
and voxels (right).
}
\label{fig:levelsets_cars}
\end{figure}
\begin{table}
\begin{tabular}{@{}lrrrcrrrcrrrr@{}}
&& \multicolumn{2}{c}{\textbf{IoU}} & \phantom{a}& \multicolumn{2}{c}{\textbf{Chamfer}} \\
\cmidrule{3-4} \cmidrule{6-7}
\textbf{Category}& \phantom{ab} & Voxels & Level Sets & & Voxels & Level Sets \\
\cmidrule{1-7}
Car && 0.814 & \textbf{0.868} & & 0.063 & \textbf{0.036} \\
Chair && 0.100 & \textbf{0.568} & & \textbf{0.083} & 0.089 \\
Bottle && 0.659 & \textbf{0.782} & & 0.067 & \textbf{0.050} \\
Sofa && 0.680 & \textbf{0.737} & & 0.071 & \textbf{0.056} \\
\end{tabular}
\caption{Average test errors using voxels and level sets representations.\label{tab:table1}}
\end{table}
\subsection{General argmin Layers - Graph Matching}
\label{sec:showcase_graphmatch}
As a final example we demonstrate how a general, non-linear, non-convex, constrained optimization
problem can be attained through an implicit layer representation. The task we substantiate here is that of
Graph Matching. This refers to the task of establishing correspondence between the nodes of two graphs based on similarity between nodes and edges.
Incorporating graph matching into an end-to-end trainable neural network was first proposed in
~\cite{zanfir2018deep}. The solution proposed therein relied on approximate, fixed iteration algorithms
for solving the forward and backward pass through its graph matching layer. We here show what we believe
is a more efficient and elegant procedure for the same task.
Graph matching is a fundamental combinatorial optimisation problems with a broad class of applications in vision. It is usually formulated as a Quadratic Assignment Problem (QAP)~\cite{bazaraa1982use}, in which exists an affinity matrix $M$ that encodes unary (node-to-node) similarity and pairwise (edge-to-edge) similarity between the two graphs.
We use notations similar to what appeared in~\cite{zanfir2018deep} to formulate graph matching as a QAP. Formally, given two graphs $\mathcal{G}_1 = (V_1, E_1)$ and $\mathcal{G}_2 = (V_2, E_2)$, with $|V_1|=n$ and $|V_2|=m$, let $y \in \{0,1\}^{nm}$ be an indicator vector such that $v_{ia} = 1$ if $i\in V_1$ is matched to $a \in V_2$ and $0$ otherwise. We build an affinity matrix $M \in R^{nm\times nm}$ such that $M_{ia;jb}$ measures the similarity between edge $(i,j)\in E_1$ and edge $(a,b)\in E_2$, and the diagonal entries of $M$ measures node-to-node similarity. The optimal assignment $y^*$ is obtained by solving the following QAP
\begin{align}\label{eq:QAP}
\begin{aligned}
&\underset{y}{\arg\!\max}~ y^TMy,\\
&s.t.~ Cy = 1,~~C \in \{0,1\},~~y \in \{0,1\},
\end{aligned}
\end{align}
where the binary matrix $C$ encodes one-to-one mapping constraints. General QAP problems are known to be NP-hard, so instead approximate solutions are typically considered for this class of problems.
In this section we will study two different relaxed versions of \eqref{eq:QAP}.
Firstly, the non-convex Quadratic Constrained Quadratic Programming (QCQP) problem obtained by dropping the constraint
$Cy = 1$ altogether and relaxing the binary constraint $y \in \{0,1\}$ to $||y||_2=1$:
\begin{align}\label{eq:rQAP_SM}
\begin{aligned}
&\underset{y}{\arg\!\max}~ y^TMy,\\
&s.t.~ \|y\|_2 = 1.
\end{aligned}
\end{align}
Even though \eqref{eq:rQAP_SM} is a nonconvex problem there exist efficient solvers for finding
the global minima of \eqref{eq:rQAP_SM}, and for this relaxation we use the Spectral Matching (SM)
algorithm~\cite{cour2007balanced}.
The implicit form of a graph matching layer is given by the KKT-conditions of \eqref{eq:rQAP_SM}
\begin{align}\label{eq:SM_KKT}
F:
\left\{
\begin{aligned}
& My + \lambda Iy=0\\
& y^{T}y-1=0.
\end{aligned}
\right.
\end{align}
Our proposed implicit layer solves \eqref{eq:rQAP_SM} directly using SM algorithm followed by a
bistochastic rounding to enforce the constraints $Cy = 1$ and $y \in \{0,1\}$.
This constitutes the forward pass of the QCQP relaxed graph matching layer.
The backward pass is then obtained by applying section~\ref{sec:prop_through_imp_layers} to
the above implicit form.
The method proposed in ~\cite{zanfir2018deep} was also based on the above relaxation. However,
the authors did not solve \eqref{eq:rQAP_SM} using standard SM but instead by carrying out a
fixed number of Power Iterations (PI). As this approach is not guaranteed to solve \eqref{eq:rQAP_SM}
globally and owing to the potentially known poor convergence rate of PI,
the overall performance of this algorithm is unclear.
A tighter relaxation of~\eqref{eq:QAP} is given by only relaxing the binary constraints on $y$ to
$||y||_2=1$, yielding the following non-convex, constrained problem
\begin{align}\label{eq:QAP_SMAC}
\begin{aligned}
&\underset{y}{\arg\!\max}~ \frac{y^TMy}{y^Ty},\\
&s.t.~ Cy = 1, ~y \geq 0.
\end{aligned}
\end{align}
As with the previous relaxation, despite not being a convex problem, the above problem can also
be solved efficiently with global optimality. Here we use the Spectral Matching with Affine Constraints
(SMAC) algorithm from~\cite{cour2007balanced}. The backward pass is obtained through the implicit
formulation of ~\eqref{eq:QAP_SMAC}, which is given by
\begin{align}\label{eq:SMAC_KKT}
F:
\left\{
\begin{aligned}
& 2\frac{My^*y^{*T}y^* - y^{*T}My^*Iy^*}{(y^{*T}y^*)^2} + C^T\lambda^* - I\nu^*=0,\\
& Cy^*-1=0,\\
& D(\nu^*)Iy^*=0
\end{aligned}
\right.
\end{align}
This aproach guarantees doubly-stochastic output and exhibits better robustness against noise and
outliers~\cite{zhou2012factorized}.
Note that the above relaxation cannot be solved using the approach of ~\cite{zanfir2018deep}.
We in addition point out that the automatic differentiation feature of implicit layers, as described in Section~\ref{sec:auto_diff}, obviates the necessity of deriving symbolic expressions of derivatives for such a complex system of functions as~\eqref{eq:SMAC_KKT}, thus greatly simplifies actually implementations.
\subsubsection{Evaluation on Graph Matching}
We end this showcase with an experiment on the CUB-200-2011 dataset~\cite{WahCUB_200_2011}. The dataset contains 11,788 images of 200 bird species, with a total of 15 semantic landmarks annotated by pixel location and visibility indicator. A neural network is introduced to perform graph matching in order to establish an assignment that matches landmarks of the source image to those of the target image, see figure~\ref{fig:qap_qualitative}. Since the purpose of this experiment is to verify the prototype of implicit layer rather than to show a comprehensive competition with other works, we simplify the task to operate on a subset of the CUB-200-2011 dataset that was built by~\cite{kanazawa2016warpnet}, which contains 5,000 images pairs with more than 50,000 ground truth matches. Training dataset and test dataset are set at a ratio of 9:1. We match fixed 8 randomly selected landmarks across images instead of matching up to 15 landmarks to avoid dealing with invisible landmarks.
We build on the network designed by~\cite{zanfir2018deep}. As shown in figure~\ref{fig:qap_pipeline}, the pipeline takes input a pair of images and for each image it constructs a graph (by e.g., Delaunay triangulation, or fully connecting the nodes) with the landmarks as the nodes of the graph. At the same time, a CNN backbone (e.g., VGG-16~\cite{simonyan2014very}) extracts for the landmarks high-level features, denoted as $F$ and $U$ in figure~\ref{fig:qap_pipeline}. $F$ and $U$ are respectively used to compute node-to-node affinities and edge-to-edge affinities, and subsequently construct an affinity matrix $M$. At this point the pipeline diverges into three branches. The upper dashed line branch represents the original pipeline in~\cite{zanfir2018deep}, in which the QCQP problem~\eqref{eq:rQAP_SM} is solved by the PI algorithm. The branch in the middle represents using an SM layer to solve~\eqref{eq:rQAP_SM}. The bottom branch represents using an SMAC layer to solve~\eqref{eq:rQAP_SM}. Note that the output of the SMAC layer is directly doubly-stochastic assignment vector.
The learning curves are plotted in figure~\ref{fig:cub_loss_acc}. The accuracy metric is defined as the Percentage of Correct Keypoints (PCK)~\cite{yang2011articulated}, by which a match is considered correct if the predicted location is within $\alpha\sqrt{w^2+h^2}$ from the ground truth ($w$ and $h$ are the width and height of the image, respectively, and we set $\alpha=0.1$ throughout the experiment). It is observed that the implicit layers converge at higher rates than the PI method.
Figure~\ref{fig:qap_qualitative} shows qualitative results of the pipeline that uses an SMAC layer.
\begin{figure*}[ht]
\centering
\begin{subfigure}{0.75\textwidth}
\centering
\includegraphics[width=\linewidth]{figs/qap_pipeline.pdf}
\end{subfigure}%
\caption{Graph matching network. The VGG-16 extracts high level landmark features ($F1, F2, U1, U2$) to compute the affinity matrix $M$. The upper, middle, and bottom branchs represent solving some relaxed QAP problem with the PI method, the SM layer, and the SMAC layer, respectively. The Bi-stochastic layer produces a doubly-stochastic confidence map ($C$ in~\eqref{eq:QAP_SMAC}). The voting layer converts a confidence map to predicted placement $d$ (refer to~\cite{zanfir2018deep} for more details).}
\label{fig:qap_pipeline}
\end{figure*}
\begin{figure}[htbp]
\centering
\begin{subfigure}{0.45\textwidth}
\centering
\includegraphics[width=\linewidth]{figs/qap_qualitative_4_con.pdf}
\end{subfigure}
\caption{Qualitative results of the pipeline with an SMAC layer. Landmarks are color coded. The source image is on the left, the target image with predicted landmarks is in the middle, and on the right is the target image with ground truth landmarks.}
\label{fig:qap_qualitative}
\end{figure}
\begin{figure}[htbp]
\centering
\begin{subfigure}{0.40\textwidth}
\includegraphics[width=\linewidth]{figs/GMN_train_pck_rmean_0-2000.pdf}
\end{subfigure}
\caption{PCK against training iterations. $PI=k$ represents the PI method with $k$ iterations.}
\label{fig:cub_loss_acc}
\end{figure}
\section{Conclusion}
In this paper we have presented a general treatment of implicitly defined layers
in feedforward neural networks. The proposed framework, which fits in seamlessly
with existing explicit formulations, provides a provably richer class of end-to-end trainable neural network architectures.
We also showed how this framework can be directly incorporated into
current automatic differentiation techniques for use in backpropagation based
training. This feature greatly improves the ease-of-use of implicit layers
by eliminating the need for any additional, problem-specific manual implementation of the backward pass.
The generality and applicability of implicitly defined layers was demonstrated
on a number of diverse example problems with very convincing results.
\bibliographystyle{spmpsci}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,203 |
Theophilus MacCartan (1700–1778) was an Irish Roman Catholic Prelate and Bishop of Down and Connor.
He served as Bishop of Down and Connor from 10/9/1760 until his death 16/12/1778.
MacCartan was a scion of a long established family McCartan in the wider Barony of Kinelarty, one of the seven baronies in County Down. It is the most central and contains the towns of Ballynahinch, Saintfield, Downpatrick, Clough, Dundrum, Newcastle and Castlewellan.
His early life and education took place against the backdrop of the Penal Laws and so firm details are scarce. But a generally respected source, writing at the end of the nineteenth century, asserts that the future bishop was born in Aughnagon near Mayobridge around 1700.
He was ordained priest at Ballykinlar by Bishop John Armstrong (1727-1739) and, as was often the case in the eighteenth century, was sent to study theology after ordination: in his case at the Sorbonne. Sometime around 1737 it is believed he was appointed parish priest of Loughinisland and erected the first church there.
Episcopal Ministry
There is a record of the then bishop and his kinsman Dominic McCartan from Clonvarghan House each subscribing £5 so that the popular devotional book Imitation of Christ by Thomas a Kempis could be translated into Irish. An early twentieth century account of the Bishops of Down and Connor suggests that Bishop MacCartan "always preached in Irish" and that it was during his episcopate that the first relaxation in the penal code was made. O'Laverty in his history of the diocese notes that MacCartan's sermons in Irish were remembered in the diocese for more than a generation after his death.
In 1773 he donated a chalice for use by any priest of the name McCartan, or the most senior priest of that name, in either the Roman Catholic Diocese of Dromore or his native Down and Connor. The chalice has been in continuous use since and there is a well-established historical record of the priests who have used this chalice.
His last will and testament has been preserved and much of the what is known about him and his episcopal ministry is drawn from that document drawn up days before his died. He is buried in Loughinisland, the parish which had been so central to his priestly and episcopal ministry.
References
18th-century Roman Catholic bishops in Ireland
1700 births
1778 deaths | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,035 |
Public Financing of Stadiums
Concussion Cover-up
Fixing U.S. Soccer
Sports Bettors' Bill of Rights
USSF Hearing
Fan Survey
Michael Rechan
Another Reason to Hate Stan Kroenke
I have been a St. Louis Rams my entire life. My family had season tickets to Rams games for nearly five years. Through all the consistently mediocre seasons, coaching changes, and poor decision- making, I stood by the Rams until the very end.
But when I first heard Stan Kroenke was trying to move the Rams to Los Angeles, I knew those days of cheering on my Rams in my hometown were numbered. Kroenke might have been born in Missouri, but at the end of the day, he and all of the other NFL owners just want to make money. And unfortunately for Rams fans like me, that meant bringing football back to the lucrative market of LA.
Moving the Rams to LA isn't what bothers me most. It's the fact that in taking our team away he slighted every single St. Louis Rams fan in the process, particularly the season ticket holders. A recent article by Deadspin tells of 40,000 St. Louis personal seat license holders (PSL's) losing the opportunity to buy season tickets in LA and even worse, failing to offer a refund to those directly affected by the relocation. The team is moving half way across the country, and they planned to not offer refunds?! The scam was clear: the Rams did not want PSL's to have their tickets transfer with the relocation, so that they could resell those tickets for exorbitant prices in LA.
Three groups have subsequently sued Kroenke and the Rams on behalf of the PSLs – lawsuits that will ultimately cost the Rams millions of dollars. The result? Some PSLs will be refunded, while others will be offered season tickets in LA depending on with whom the PSL signed their contract. So, Kroenke's profit-seeking plan was blocked, at least in part. But should it really have taken three different lawsuits to give ticket holders what they rightfully deserve?
This is just one of about a hundred different ways Kroenke has continually screwed St. Louis and their fans. Should I be surprised? Of course not. This is the guy who turned down hundreds of millions of public dollars to build a new stadium in St. Louis, the only city to offer public stadium funding. This is the guy who left St. Louis fans with millions in taxes to maintain a stadium no longer being used. This is the guy that charged fans the fifth highest beer prices in the NFL. What does St. Louis have to show for it? A useless stadium, millions still owed in taxes, and a city without football. Kroenke, thanks for nothing, we won't miss you.
Michael Rechan published this page in Blog 2016-10-10 18:14:11 -0400
All information you provide on this petition signing form will be public on the petition signatures page, except your email address, which will remain private. You may receive updates on this issue and other issues from Sports Fans Coalition though you're always welcome to unsubscribe anytime. Your email is always safe with us.
Follow @sportsfanorg on Twitter | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,711 |
{"url":"https:\/\/quant.stackexchange.com\/questions\/49089\/understanding-apr-via-programming","text":"# Understanding APR via programming [closed]\n\nI am trying to better understand different types of interest rates. However, I am having difficulties complete, consistent and pedagogically-efficient explanations online. Thus, I have decided to design and program a couple of scripts. I find that programming can be a powerful bridge to truly understand tricky concepts.\n\nThe first concept I am trying to understand is the so-called annualized percentage rate (APR). Common explanations basically mention this one as an interest rate that somehow accounts for the costs of the loan...\n\n1. The name is very misguiding... right?\n2. Does it have to be annualized if and only if the given rate is not quoted per annum?\n3. Also, is it computing using the real interest rate, rather than a nominal interest rate?\n4. How is the APR used in practice by loan providers? Do they establish a nominal rate that makes sense to them, add the fees and then compute the \"APR\" to then use it to compute interest payments?\n\nDoes the following piece of code make sense? In this piece of code, I use the term \"cost-of-borrowing interest rate\" to refer to the \"APR\":\n\n inflation_rate_per_year \/\/ Annual inflation rate.\nnominal_interest_rate_per_year \/\/ Nominal interest rate per year.\nreal_interest_rate_per_year \/\/ Inf-adj inf. rate per year.\ncost_borrow_interest_rate_per_year \/\/ Cost-of-borrowing int. rate.\nprincipal_money \/\/ Amount being loaned.\ntotal_fees_money \/\/ Total to be payed in fees.\ntotal_cost_loan_money_per_year \/\/ Total cost of loan per year.\ntotal_owned_money \/\/ Total to pay back.\n\n\/\/ Given: inflation_rate_per_year, nominal_interest_rate_per_year\n\/\/ Given: principal_money, total_fees_money\n\n\/\/ Compute real interest rate and percentage.\nreal_interest_rate_per_year = nominal_interest_rate_per_year - inflation_rate_per_year\n\n\/\/ Compute cost-of-borrowing interest rate and percentage.\ncost_borrow_interest_rate_per_year =\n(total_fees_money + principal_money)\/principal_money*\nreal_interest_rate_per_year\n\n\/\/ Compute appreciation on principal.\ntotal_cost_loan_money_per_year =\nprincipal_money*((1 + cost_borrow_interest_rate_per_year) - 1)\n\ntotal_owned_money = principal_money*(1 + cost_borrow_interest_rate_per_year)\n\n\nWhy do I mean by \"making sense\"? Well, is it OK, considering the given interest rates assume the borrowing period to be 1 year? If it were not a year... Should I have to add the annualization factor of 365\/n, with n equaling the number of days in the borrowing period.\n\nI understand the question is posed a little vaguely. I basically want to better understand the APR via this script :)\n\n\u2022 Are you asking about this from the statutory concept of an APR or what is often used in textbooks for students. They are not the same thing. Writing this in C++ isn't very helpful here either. Some of your definitions appear only to be valid in logarithms. What jurisdiction are you in? In the United States, there are differences of opinion as to what goes into an APR among the courts. Although usually minor differences, they do create locality based differences. \u2013\u00a0Dave Harris Oct 6 '19 at 22:39\n\u2022 @DaveHarris What is the statutory concept of an APR? \u2013\u00a0Eduardo Oct 7 '19 at 18:26\n\u2022 @DaveHarris: I have simplified the source code. \u2013\u00a0Eduardo Oct 7 '19 at 18:56\n\u2022 Essentially, the statutory concept is that if there is a cost you must bear only because you are getting a loan and it is financed into the contract, then it is a form of interest. For example, if you had \\$50 documentation fee on a \\$1,000 loan with a single payment of \\$1100 dollars, the law looks at it as a \\$950 loan with a final payment of \\\\$1100 if the fee is included in the credits. Where it becomes very complicated is that specific courts consider different things a finance charge. It is a highly specialized area of law. \u2013\u00a0Dave Harris Oct 8 '19 at 23:56\n\u2022 if you are concerned with the statutory meaning you are in the wrong forum. You should probably post it in the personal finance or the law forum. \u2013\u00a0Dave Harris Oct 9 '19 at 0:00","date":"2020-01-20 22:00:55","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.46635305881500244, \"perplexity\": 1608.628239481666}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-05\/segments\/1579250599789.45\/warc\/CC-MAIN-20200120195035-20200120224035-00086.warc.gz\"}"} | null | null |
{"url":"https:\/\/runescape.wiki\/w\/Small_dungeoneering_token_box","text":"# Small dungeoneering token box\n\nA small dungeoneering token box is an item that could be obtained from Treasure Hunter on 8 December 2015 and later from various other sources (e.g. during the Christmas Advent Calendar promotion or loot pi\u00f1atas).\n\nThe amount of tokens found in each box can be found with the equation ${\\displaystyle {\\frac {4}{9}}x^{2}+50}$ tokens, where ${\\displaystyle x}$ is your level in Dungeoneering. At level 120 Dungeoneering it awards 6,450 Dungeoneering tokens when opened.\n\nThe dungeoneering token box may be kept in the bank, and players can open them later for more tokens when they reach a higher Dungeonerring level.\n\n## Drop sources\n\nThis is an auto-generated list (update now). For all sources of this item, see here (include RDT). For help, see the FAQ.\n\n## Update history\n\nThis information has been compiled as part of the update history project. Some updates may not be included - see here for how to help out!\n\u2022 patch 15 January 2018\u00a0(Update):\n\u2022 Players can now auto-redeem Dungeoneering token boxes won from Treasure Hunter.","date":"2021-10-27 17:30:33","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 2, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.37593695521354675, \"perplexity\": 5310.102429803826}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-43\/segments\/1634323588216.48\/warc\/CC-MAIN-20211027150823-20211027180823-00531.warc.gz\"}"} | null | null |
Percy jackson grammar school reunion
Ken Cooke
History of the Percy Jackson Grammar School: Adwick-Le-Street, Doncaster, Yorkshire, 1939-1968: Recollections of Schooldays of the 1940s, 1950s & 1960 by Ken Cooke
The opening of the Percy Jackson Grammar School in 1939 coincided with the onset of the Second World War, an ominous start for the life of a new school. In this informal history, the author chronicles the lifetime of this coeducational school, and with a light touch he considers whether schooldays really are the best days of your life. Drawing upon abundant recollections from former pupils - many of whom, like him, were the children of mining families and the first generation to attend grammar school - Ken Cooke covers topics as diverse as war-time evacuees, the school visit to the Festival of Britain in 1951, reactions to the 1968 conversion to a comprehensive, and the achievements of some of the schools alumni. This history is faithful to the facts, yet true to the sentiments.
File Name: percy jackson grammar school reunion.zip
PJ healthedventure.org
Joyce Brooks obituary
Made With Serif WebPlus. It was so sad to see such a beautiful building go. It held so many happy memories f or me — my first teaching post where I laid the foundations of my career, the place where I met Ann Rowland, the junior secretary, who became my wife for 44 years before her early death in , and where I met so many wonderful people, many of whom became good, lifelong friends. Grammar Schools are often old, even ancient, foundations, but ours was very new, and needed to establish its own direction, ethos and traditions. Through this I was able to resume contact, after nearly 60 years, with my second-form English teacher, John Good, and, on his 90th birthday recently, to thank him properly at last for all his help and encouragement. I claim even earlier association; I was born just a hundred yards beyond the Adwick primary school, and as a toddler I used to play in the field that in became PJGS. It all holds a lot of memories.
Goodreads helps you keep track of books you want to read. Want to Read saving…. Want to Read Currently Reading Read. Other editions. Enlarge cover. Error rating book.
Preview Book. Review this Book. Author Website. He draws upon abundant recollections from former pupils, many of whom, like him, were the children of mining families and the first generation to attend grammar school. The first head, Mr Field, with the responsibility of establishing the new school, had to cope with the complications of a world war whilst his pupils decorated their gas marks and speculated on a romance between their teacher and the captain of a battleship.
Made With Serif WebPlus. Although the Registers were at one time available for viewing at Doncaster Archives, once The Data Protection Act of came in, this was no longer the case. Only personal information is now accessible to the individual and about any proven deceased former pupils. For this there would probably be a fee. In , a small group of PJGS alumni clubbed together to buy photocopies of the Admission Register from to This has taken a long time due to a confusing manner in which some of the years were recorded.
He later joined the Regimental Band of the Welsh Guards and pursued a highly successful career in the brass band movement as player, conductor and adjudicator. In David was awarded the MBE for services to music. Dr John Maxwell and his wife Sheila, both entrants, have tickets. By the planned capacity was reached, then in four prefab classrooms were added. Alumni include Fellows of the Royal Society, TV producers and writers, radiation and medical specialists, senior lawyers, a champion angler, a poetry publisher, a brass band adjudicator and a roller-skating judge, an RAF commanding officer and a brigadier in the Royal Army Dental Corps, a knighthood for services to education, a director of EMI Records, HM Principal Inspector of Engineering in Mines etc, not to mention many national honours and awards. Sign in Edit Account Sign Out. By By The Newsroom.
The style sheet is still retained on the Archive for the Yorkshire Viking Blog and on this memorial site. If you are a former Adwick School pupil and would like to use the WordPress theme for your own blog, do get in touch! My personal twin blogs have now dropped the Adwick Theme. The reasons for this are explained on the Archive for the Yorkshire Viking page. We are endebted to Gerald Sables who has lovingly and painstakingly documented our school, and its sad demise in
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Katie A. says:
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Covenronor says:
My mother, Joyce Brooks, who has died aged 86 after suffering from dementia, made a lasting impact on everyone who knew her, notably on countless schoolchildren who were her pupils at Percy Jackson grammar school, at Adwick-le-Street, near Doncaster. | {
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{"url":"http:\/\/www.neverendingbooks.org\/permutation-representations-of-monodromy-groups","text":"Today we will explain how curves defined over\n$\\overline{\\mathbb{Q}}$ determine permutation representations\nof the carthographic groups. We have seen that any smooth projective\ncurve $C$ (a Riemann surface) defined over the algebraic\nclosure $\\overline{\\mathbb{Q}}$ of the rationals, defines a\n_Belyi map_ $\\xymatrix{C \\ar[rr]^{\\pi} & & \\mathbb{P}^1}$ which is only ramified over the three points\n$\\\\{ 0,1,\\infty \\\\}$. By this we mean that there are\nexactly $d$ points of $C$ lying over any other point\nof $\\mathbb{P}^1$ (we call $d$ the degree of\n$\\pi$) and that the number of points over $~0,1~$ and\n$~\\infty$ is smaller than $~d$. To such a map we\nassociate a _dessin d\\\u2019enfant_, a drawing on $C$ linking the\npre-images of $~0$ and $~1$ with exactly $d$\nedges (the preimages of the open unit-interval). Next, we look at\nthe preimages of $~0$ and associate a permutation\n$\\tau_0$ of $~d$ letters to it by cycling\ncounter-clockwise around these preimages and recording the edges we\nmeet. We repeat this procedure for the preimages of $~1$ and\nget another permutation $~\\tau_1$. That is, we obtain a\nsubgroup of the symmetric group $\\langle \\tau_0,\\tau_1 \\rangle \\subset S_d$ which is called the monodromy\ngroup\nof the covering $\\pi$.\n\nFor example, the\ndessin on the right is\nassociated to a degree $8$ map $\\mathbb{P}^1 \\rightarrow \\mathbb{P}^1$ and if we let the black (resp. starred) vertices be\nthe preimages of $~0$ (respectively of $~1$), then the\ncorresponding partitions are $\\tau_0 = (2,3)(1,4,5,6)$\nand $\\tau_1 = (1,2,3)(5,7,8)$ and the monodromy group\nis the alternating group $A_8$ (use\nGAP ).\n\nBut wait! The map is also\nramified in $\\infty$ so why don\\\u2019t we record also a\npermutation $\\tau_{\\infty}$ and are able to compute it from\nthe dessin? (Note that all three partitions are needed if we want to\nreconstruct $C$ from the $~d$ sheets as they encode in\nwhich order the sheets fit together around the preimages). Well,\nthe monodromy group of a $\\mathbb{P}^1$ covering ramified only\nin three points is an epimorphic image of the fundamental\ngroup\nof the sphere\nminus three points $\\pi_1(\\mathbb{P}^1 \u2013 { 0,1,\\infty })$ That is, the group of all loops beginning and\nending in a basepoint upto homotopy (that is, two such loops are the\nsame if they can be transformed into each other in a continuous way\nwhile avoiding the three points).\n\nThis group is generated by loops\n$\\sigma_i$ running from the basepoint to nearby the i-th\npoint, doing a counter-clockwise walk around it and going back to be\nbasepoint $Q_0$ and the epimorphism to the monodromy group is given by sending\n\n$\\sigma_1 \\mapsto \\tau_0~\\quad~\\sigma_2 \\mapsto \\tau_1~\\quad~\\sigma_3 \\mapsto \\tau_{\\infty}$\n\nNow,\nthese three generators are not independent. In fact, this fundamental\ngroup is\n\n$\\pi_1(\\mathbb{P}^1 \u2013 \\\\{ 0,1,\\infty \\\\}) = \\langle \\sigma_1,\\sigma_2,\\sigma_3~\\mid~\\sigma_1 \\sigma_2 \\sigma_3 = 1 \\rangle$\n\nTo understand this, let us begin\nwith an easier case, that of the sphere minus one point. The fundamental group of the plane minus one point is\n$~\\mathbb{Z}$ as it encodes how many times we walk around the\npoint. However, on the sphere the situation is different as we can make\nour walk around the point longer and longer until the whole walk is done\nat the backside of the sphere and then we can just contract our walk to\nthe basepoint. So, there is just one type of walk on a sphere minus one\npoint (upto homotopy) whence this fundamental group is trivial. Next,\nlet us consider the sphere minus two points\n\nRepeat the foregoing to the walk $\\sigma_2$, that\nis, strech the upper part of the circular tour all over the backside of\nthe sphere and then we see that we can move it to fit with the walk\n$\\sigma_1$ BUT for the orientation of the walk! That is, if we do this\nmodified walk $\\sigma_1 \\sigma_2^{\\\u2019}$ we just made the\ntrivial walk. So, this fundamental group is $\\langle \\sigma_1,\\sigma_2~\\mid~\\sigma_1 \\sigma_2 = 1 \\rangle = \\mathbb{Z}$ This is also the proof of the above claim. For,\nwe can modify the third walk $\\sigma_3$ continuously so that\nit becomes the walk $\\sigma_1 \\sigma_2$ but\nwith the reversed orientation ! As $\\sigma_3 = (\\sigma_1 \\sigma_2)^{-1}$ this allows us to compute the\n\\\u2019missing\\\u2019 permutation $\\tau_{\\infty} = (\\tau_0 \\tau_1)^{-1}$ In the example above, we obtain\n$\\tau_{\\infty}= (1,2,6,5,8,7,4)(3)$ so it has two cycles\ncorresponding to the fact that the dessin has two regions (remember we\nshould draw ths on the sphere) : the head and the outer-region. Hence,\nthe pre-images of $\\infty$ correspond to the different regions of the\ndessin on the curve $C$. For another example,\nconsider the degree 168 map\n\n$K \\rightarrow \\mathbb{P}^1$\n\nwhich is the modified orbit map for the action of\n$PSL_2(\\mathbb{F}_7)$ on the Klein quartic.\nThe corresponding dessin is the heptagonal construction of the Klein\nquartic\n\nHere, the pre-images of 1 correspond to the midpoints of the\n84 edges of the polytope whereas the pre-images of 0 correspond to the\n56 vertices. We can label the 168 half-edges by numbers such that\n$\\tau_0$ and $\\tau_1$ are the standard generators b\nresp. a of the 168-dimensional regular representation (see the atlas\npage\n).\nCalculating with GAP the element $\\tau_{\\infty} = (\\tau_0 \\tau_1)^{-1} = (ba)^{-1}$ one finds that this permutation\nconsists of 24 cycles of length 7, so again, the pre-images of\n$\\infty$ lie one in each of the 24 heptagonal regions of the\nKlein quartic. Now, we are in a position to relate curves defined\nover $\\overline{Q}$ via their Belyi-maps and corresponding\ndessins to Grothendiecks carthographic groups $\\Gamma(2)$,\n$\\Gamma_0(2)$ and $SL_2(\\mathbb{Z})$. The\ndessin gives a permutation representation of the monodromy group and\nbecause the fundamental group of the sphere minus three\npoints $\\pi_1(\\mathbb{P}^1 \u2013 \\\\{ 0,1,\\infty \\\\}) = \\langle \\sigma_1,\\sigma_2,\\sigma_3~\\mid~\\sigma_1 \\sigma_2 \\sigma_3 = 1 \\rangle = \\langle \\sigma_1,\\sigma_2 \\rangle$ is the free group op two generators, we see that\nany dessin determines a permutation representation of the congruence\nsubgroup $\\Gamma(2)$ (see this\npost\nwhere we proved that this\ngroup is free). A clean dessin is one for which one type of\nvertex has all its valancies (the number of edges in the dessin meeting\nthe vertex) equal to one or two. (for example, the pre-images of 1 in\nthe Klein quartic-dessin or the pre-images of 1 in the monsieur Mathieu\nexample\n) The corresponding\npermutation $\\tau_1$ then consists of 2-cycles and hence the\nmonodromy group gives a permutation representation of the free\nproduct $C_{\\infty} \\ast C_2 = \\Gamma_0(2)$ Finally, a clean dessin is said to be a\nquilt dessin if also the other type of vertex has all its valancies\nequal to one or three (as in the Klein quartic or Mathieu examples).\nThen, the corresponding permutation has order 3 and for these\nquilt-dessins the monodromy group gives a permutation representation of\nthe free product $C_2 \\ast C_3 = PSL_2(\\mathbb{Z})$ Next time we will see how this lead\nGrothendieck to his anabelian geometric approach to the absolute Galois\ngroup.\n\n## One Comment\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.","date":"2020-02-27 18:41:01","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9222918748855591, \"perplexity\": 793.9309318980107}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-10\/segments\/1581875146744.74\/warc\/CC-MAIN-20200227160355-20200227190355-00005.warc.gz\"}"} | null | null |
Yahoo! Acquires Tumblr for $1.1 Billion
Published by BSIC on 25 May 2013 25 May 2013
Yahoo! (Market Capitalization as of 25/05/13: $28.5bn)
Tumblr (Market Capitalization: N/A)
On May 20th, Yahoo! continued its aggressive strategy since Marissa Mayer took over as CEO by announcing the takeover of New York-based blogging platform Tumblr for $1.1bn, all in cash. Tumblr was founded in 2007 and has more than 108mln blogs and nearly 117mln active monthly users. The transaction adds Yahoo! to the list of established Internet companies, including Google and Facebook, that have spent at least $1bn apiece to buy start-up companies in the hope of stimulating growth. Yahoo!'s stock, which has been on a +70% bull run in the year to date, was up 0.8% on the news.
The news comes about a year after Facebook snapped up the start-up photo sharing service Instagram for $1bn, the most closely comparable transaction to the Tumblr acquisition. In both cases, the target firms had little to no revenue compared to the size of the deal. However, it is worth noting that the Instagram deal was worth only 1% of Facebook's market cap, whereas Tumblr is valued at almost 3.9% of Yahoo!'s market cap. Yahoo! seems to have paid a chunky premium of $200mln over a precedent valuation of $800mln in a round of VC financing in 2011. The peculiarity of this deal is that it paves the way for a renaissance of the east coast, and New York City's Flatiron district in particular, as a technology start-up and venture capital hub. The first round of seed financing for Tumblr came from Union Square Ventures (NY), Spark Capital (Boston) and Betaworks (NY). These early backers have netted a 100-fold return on investment from the acquisition. The second major capital injection was carried out in 2011 and totalled $85mln from Greylock Partners, Sequoia Capital and Peter Chernin, among others. However, it is not clear that these later investors have obtained a satisfactory return over the implied valuation of $800mln at the time.
Tumblr's growth potential, particularly on smartphones, is the rationale being provided for the valuation, backed by the fact that the site has 300mln users. This means Yahoo! paid around $3 per user, while Facebook paid around $30 per user. The acquisition was carried out entirely in cash from Yahoo!'s $5.4bn reserves. This was also made possible by the fact that the firm has no debt on its balance sheet.
The acquisition is Yahoo!'s ninth this year, which highlights Ms. Mayer's strategic focus on returning the Sunnyvale-based firm to growth: one of the most advertised previous transactions is the purchase of Summly for $30mln, which made its 17 year old founder a millionaire. Nonetheless, the valuation has raised a few eyebrows in the industry given that Tumblr had revenues of only $13mln in 2012, which must be compared to estimated operating costs for the year of $25mln. Tumblr's management forecast revenues of $100mln in 2013, which would allow it to turn a positive profit but represent almost 800% growth YoY. The monetization of its customer activity is achieved through targeted advertising, even though the firm's track record is too limited to assess how it will perform in terms of revenue per user and clickthrough rates with respect to comparable social networks such as Twitter and Instagram.
With Yahoo!'s advertising revenue falling every quarter since 2011, the company will seek to utilize the untapped advertising revenue potential of Tumblr and Ms. Mayer has already hinted that she might create an ad exchange for Tumblr. Furthermore, Yahoo! can exploit the 100,000+ daily new users of Tumblr to stimulate its stagnating growth. Even though the double aim of Yahoo!'s acquisition is both to integrate Tumblr's user base into its own, and to offer a larger audience to its advertising partners, there are challenges to the smooth implementation of this plan. On the one hand Tumblr's young and independent core users do not show appetite for being "adopted" by Yahoo!: the stream of blog posts migrating to the competing WordPress platform reached a staggering 72,000 per hour on the night the deal was announced. On the other hand, Yahoo! might have trouble selling ad space to corporate image-sensitive advertisers on a website where one in six blogs contains adult material. Tumblr will keep operating independently with incumbent CEO and founder David Karp retaining his role to ensure "the same Tumblr irreverence, wit, and commitment to empower creators". His cash payoff from the acquisition is estimated at around $220mln from his 20% stake.
The deal illustrates that, despite a decline in its core customer base, a veteran like Yahoo still has the muscle – and cash – to take on its competitors and reposition itself in the current social networking-centred internet landscape. If Ms. Mayer can live up to her word of "We promise not to screw it up" one could see Yahoo! reaping the benefits of a larger and younger user base. On the other hand, the social networking landscape is intrinsically volatile in terms of customer preferences (as the fast migration has just shown) and feasibility of monetization schemes.
Ms. Mayer's bet may be an expensive and risky one, but it is one that in our opinion she has been right to make in the best interest of her firm and shareholders, in the search for a growth that has been missing for far too long at Yahoo!.
Yahoo! didn't hire external advisors for the transaction and relied on its internal M&A team, suggesting they have been accumulating expertise to be used in future deals. Tumblr was advised by Frank Quattrone and Jonathan Turner from specialised boutique Qatalyst Partners.
Categories: Corporate FinanceM&A Deals
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A snapshot into one of BSIC's activities: the Ge | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,233 |
\section{Introduction}
\label{sec:intro}
The peculiar double peaks in the $m_{\pi \pi}$ spectrum in $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ \cite{cusb}
have been lacking proper understanding. Although several suggestions have
been made, most of them reproduce the $\pi \pi$ spectrum only with limited
success \cite{prework}.
In the two recent works by us, we approached this problem in two different
ways. In the first work \cite{chakrako}, we assumed that the most general
form of the amplitude for $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ is
\begin{equation}
{\cal M} = A_{0} \left[ \left( q^{2} + B_{0} E_1 E_2 + C_{0} m_{\pi}^2
\right)~\epsilon \cdot \epsilon^{'} + D_{0} \left( p \cdot \epsilon
p^{'} \cdot \epsilon^{'} + p \cdot \epsilon^{'} ~p^{'} \cdot
\epsilon \right) \right],
\label{eq:amp1}
\end{equation}
in the lowest order in the pion momentum expansion.
Here, $p, p^{'}$ are the four--momenta of the final pions, $E_1$ and
$E_2$ are their energies, and $\epsilon$ and $\epsilon^{'}$ are the
polarization vectors of the initial and the final $\Upsilon$'s,
respectively. In Ref.~\cite{chakrako}, we found three sets of
parameters (P0, P1 and P2) by minimizing the $\chi^2$.
For these three sets of parameters, we predicted various angular
distributions which should be checked against experiments.
In the second work \cite{chakraetal}, we assumed that (i) the QCD
multipole expansion is applicable to $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ and (ii) $\Upsilon (3S)$ has
an admixture of a $D-$wave component :
\begin{equation}
| \Upsilon (3S) \rangle = \cos \phi |3S \rangle + \sin \phi | D \rangle.
\label{eq:mixing}
\end{equation}
The amplitude for $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ then
depends on three independent parameters, and we found two fits
which correspond to P1 and P2 of Ref.~\cite{chakrako}, respectively.
We then explored the consequences of our assumptions on other
hadronic and radiative transitions of $\Upsilon (3S)$ into the lower
level bottomonia.
In Ref.~\cite{chakraetal}, we again assumed that the lowest order
expansion in the pion momenta
\begin{equation}
\theta_{2 \pi}^{0} (q^2) \equiv \langle \pi \pi | \theta_{\mu}^{\mu}
| 0 \rangle = ( q^2 + m_{\pi}^2 )
\label{eq:theta1}
\end{equation}
be valid through the whole range of $m_{\pi \pi} = \sqrt{q^{2}}$.
Although these two approaches fit the $m_{\pi\pi}$ spectrum,
they still leave room for theoretical improvement in two aspects.
First of all, it is well known from the analysis of the $\pi \pi$ phase
shift that the dipion system in $I = L = 0$ experiences strong final
state interactions \cite{truong}. Since the dipion system
in $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ are in either $I = L = 0$ or $I = 0, L = 2$ state, one should
properly take into account of the $\pi\pi$ phase shift due to the
final state interactions in the $I = L = 0$ dipion system.
Secondly, the validity of (\ref{eq:amp1}) or (\ref{eq:theta1}) in $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$
is rather unclear, since the available $m_{\pi \pi}$ in $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ is large,
$ 2 m_{\pi} \leq m_{\pi \pi} \leq ( m_{i} - m_{f} ) = 895$ MeV.
The amplitude (\ref{eq:amp1}) with $B_0 = D_0 = 0$ gives a good description
for the $m_{\pi\pi}$ spectrum in $\Upsilon (2S) \rightarrow \Upsilon(1S)
\ \pi\pi$, where $m_{\pi\pi} \leq 563$ MeV. For higher value of
$m_{\pi\pi}$, we simply assume that it is valid and explore its consequences
on various spectra of decay products in $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$.
If any of the predicted angular distributions based on the amplitude
(\ref{eq:amp1}) including the $\pi\pi$ phase shift does not agree with
the experimental measurements, it would signal the importance of higher
order terms in the pion momenta which have been neglected in
(\ref{eq:amp1}).
In the present work, we follow the approach of Ref.~\cite{chakrako} and
include the phase shift of the $\pi\pi$ system in the $I=L=0$ channel
($\delta_{0} (q^{2})$) using the data
available in the literature \cite{pshift}.
The phase shift for the $I = 0$ $D-$wave $\pi\pi$ system ($\delta_{2}
(q^{2})$) is tiny enough to be neglected for the whole range of
$m_{\pi\pi}$ \cite{pshift}.
In Sec.~\ref{subsec:predict}, we decompose the amplitude (\ref{eq:amp1})
into the $\pi\pi$ $S-$ and $D-$waves, and incorporate the phase shift
$\delta_{0} (q^{2})$. It is found that
the importance of the phase shift due to the final state interactions is most
prominent in the $\cos \theta_{\pi}^*$ distributions, but the effect is
only moderate. Then, our results are compared with the recent data
from CLEO in Sec.~\ref{subsec:data}.
In Sec.~\ref{sec:three}, we generalize the amplitude (\ref{eq:amp1})
(in case of $D = 0$) to
include higher order terms in $q^2$, and discuss some general aspects of the
various angular distributions. It is also pointed out that one can extract
the difference between $\pi\pi$ phase shifts of the $S-$ and $D-$waves
in the $I=0$ channel, $\delta_{0} - \delta_{2}$,
by measuring the joint distribution ${\rm d}^{2}\Gamma / {\rm d}m_{\pi\pi}
{\rm d} \cos \theta_{\pi}^*$.
The results are summarized in Sec.~\ref{sec:four}.
\section{ effects of the $\pi\pi$ phase shift}
\label{sec:two}
\subsection{ Predictions on the $\cos \theta_{\pi^*}$ distributions}
\label{subsec:predict}
In this work, we use a modified form of the amplitude (\ref{eq:amp1}).
First of all, we ignore the recoil of the final $\Upsilon (1S)$ in the
amplitude, and make the $D-$term proportinal to the symmetric traceless
part :
\begin{equation}
{\cal M} = A~\left[ \left\{ q^{2} + B E_{1} E_{2} + C m_{\pi}^{2}
\right\} \hat{\epsilon} \cdot \hat{\epsilon}^{'} + D \left\{
\vec{p} \cdot \hat{\epsilon} \vec{p}^{'} \cdot \hat{\epsilon}^{'}
+ \vec{p}^{'} \cdot \hat{\epsilon} \vec{p} \cdot \hat{\epsilon}^{'}
- {2\over 3} \vec{p} \cdot \vec{p}^{'} \hat{\epsilon} \cdot
\hat{\epsilon}^{'} \right\} ~\right].
\label{eq:ampnew1}
\end{equation}
One can find relations between parameters in (\ref{eq:amp1}) and
(\ref{eq:ampnew1}) using
\[
\vec{p} \cdot \vec{p}^{'} = E_{1} E_{2} - {1\over 2}~(s - 2 m_{\pi}^{2}).
\]
It should be emphasized that our amplitudes (\ref{eq:amp1}) and
(\ref{eq:ampnew1}) satisfy Adler's condition by construction.
Now, we decompose the above amplitude into the $\pi\pi$ $S-$wave and
$D-$wave in order to take into account the phase shift of the $\pi\pi$
system in the $I = L = 0$ state. The $S-$ and $D-$waves have the
following tensor structures \cite{belanger} :
\begin{eqnarray}
{\cal S}_{ij} & = & f_{S}(q^2) \delta_{ij} + g_{S}(q^2) ( q_{i} q_{j} +
q^{2} \delta_{ij} ), \cr
{\cal D}_{ij} & = & f_{D}(q^2) \left( \cos^{2}\theta_{\pi}^{*} -
{1\over 3} \right)~
\delta_{ij} + g_{D}(q^2) \left[ r_{i} r_{j} -
{1\over 3} ( q_{i} q_{j} + q^{2} \delta_{ij} ) \beta_{\pi}^{*2} \right],
\label{sdij}
\end{eqnarray}
where $r_{\mu} = p_{\mu} - p_{\mu}^{'}$.
The amplitude (\ref{eq:ampnew1}) can be decomposed into the $f_{S,D},
g_{S,D}$ form factors as follows :
\begin{eqnarray}
f_{S} & = & q^{2} + C m_{\pi}^{2} + {B\over 4}~\left[ (E_{1} + E_{2})^{2}
- {1\over 3}~\vec{q}^{2} \beta_{\pi}^{*2} \right]
\nonumber \\
& & - {1\over 2}~D q^{2} -{1\over 6}~D~\left[ \vec{q}^{2} -
\beta_{\pi}^{*2}\left( q^{2}
+ {1\over 3} \vec{q}^{2} \right) \right],
\label{eq:fs}
\\
g_{S} & = & {1\over 2}~D~( 1 - {1\over 3} \beta_{\pi}^{*2} ),
\label{eq:gs}
\\
f_{D} & = & \left( {1\over 6} D - {1\over 4} B \right)~\beta_{\pi}^{*2}~
\vec{q}^{2},
\label{eq:fd}
\\
g_{D} & = & -{1\over 2}~D.
\label{eq:gd}
\end{eqnarray}
Multiplying the $S-$wave amplitude by the phase shift $\delta_{0}
(q^{2})$, we get
\begin{equation}
{\cal M} = \left[ {\cal S}_{ij} e^{i \delta_{0} (q^{2})} +
{\cal D}_{ij} \right]~\hat{\epsilon}_{i} \hat{\epsilon}_{j}^{'}.
\label{eq:ampnew}
\end{equation}
In case the final $\Upsilon (1S)$ is not reconstructed, summations
over polarizations of the initial and final $\Upsilon$'s are done with
\begin{eqnarray}
\sum \hat{\epsilon}_{i} \hat{\epsilon}_{j} & = & \left( \delta_{ij} -
\hat{z}_{i} \hat{z}_{j} \right), \label{eq:pol1} \\
\sum \hat{\epsilon}_{i}^{'} \hat{\epsilon}_{j}^{'} & = & \delta_{ij}.
\label{eq:pol2}
\end{eqnarray}
The fact that the initial $\Upsilon (3S)$ is transversely polarized
with respect to the beam directions (taken along the $z-$direction)
has been taken into account in (\ref{eq:pol1}).
This fact is very useful to test the existence of the $D-$term by
measuring the polar angle distributions of the final $\Upsilon (1S)$
(or, equivalently, the $\pi\pi$ system as a whole) and/or of a muon emerging
from the muonic decay of the final $\Upsilon (1S)$.
If one tags the muonic decay of the final $\Upsilon (1S)$, the polarization
sum over the final $\Upsilon (1S)$ (\ref{eq:pol2}) should be replaced by
\begin{equation}
\Sigma \hat{\epsilon}_{i}^{'} \hat{\epsilon}_{j}^{'} = ( \delta_{ij} -
\hat{l}_{i} \hat{l}_{j} ),
\end{equation}
where $\hat{l}$ is the three dimensional unit vector along the
direction of a muon in the rest frame of the initial $\Upsilon (3S)$.
For $D=0$, one gets
\[
dN/d\cos \theta_{l} \sim (1+\cos^{2} \theta_l),
\]
where $\cos\theta_{l} = \hat{l} \cdot \hat{z}$.
Using amplitude (\ref{eq:ampnew}), we fit the $m_{\pi\pi}$ spectrum
by minimizing $\chi^2$.
The best fit is given by three sets of solutions, P0, P1 and P2
(see Table~1), which are essentially the same as the ones given in Ref.~
\cite{chakrako}. Other angular distributions can be obtained by numerical
integrations as in \cite{chakrako}.
In Fig.~\ref{figone} (a) and (b), we show the $\cos \theta_{\pi}^*$
distributions of $\pi^+$ in the rest frame of the dipion system for P0
and P1 (P2), where $\theta_{\pi}^{*} = 0^{\circ}$
is along the direction of the dipion system as a whole in the rest
frame of initial $\Upsilon (3S)$.
For comparison, we show the corresponding plots with the phase shift
neglected ({\it i.e.} $\delta_{0} (q^{2}) = 0$) in Fig.~2 (a) and (b).
The phase shift moderately changes the $\cos \theta_{\pi}^*$ distributions,
but the overall effects may be hardly discernible in the experiment.
Before continuing on to the next section, we discuss how much the results
obtained in Ref.~\cite{chakraetal} will change when we incorporate the
$\pi\pi$ phase shift and possible corrections to (3) in higher orders
in the pion momentum expansion. In this case, it suffices to resort to
the QCD multipole expansion by our assumptions. The relevant matrix
element for $\langle \pi \pi | G_{\mu \nu}^{a} G^{a\mu\nu} | 0 \rangle$
has been obtained by Donoghue {\it et al.} using the dispersive approach in
conjunction with the chiral symmetry relations imposed in the chiral
limit \cite{gasser}. The result is that the above matrix element,
$\langle \pi \pi | G_{\mu \nu}^{a} G^{a\mu\nu} | 0 \rangle$, is
dominated by $\langle \pi\pi | \theta_{\mu}^{\mu} | 0 \rangle$, and that
(3) remains essentially unchanged up to $m_{\pi\pi} \sim 0.9$ GeV,
once it is regarded as the modulus of $\theta_{2\pi}$.
The phase shift is given by $\delta_{0} (q^{2})$. In short, one only
has to write (3) as
\begin{equation}
\langle \pi\pi | \theta_{\mu}^{\mu} | 0 \rangle = ( q^{2} + m_{\pi}^{2} ) ~
e^{i \delta_{0} (q^{2})}.
\end{equation}
Since the phase shift $\delta_{0} (q^{2})$ does not affect the $m_{\pi\pi}$
spectrum in $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$, our results derived in Ref.~\cite{chakraetal} do not
change at all. In particular, the predictions for $\Upsilon (3S) \rightarrow
\Upsilon (1S) + \eta$ remain the same, which excludes the fit P1
\cite{chakraetal}.
\subsection{Comparisons with the data}
\label{subsec:data}
Recently, the CLEO collaboration released a new set of data on hadronic
transitions in $\Upsilon (3S)$ decays \cite{newcleo}.
Their results on $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ can be summarized as follows :
\vspace{.2in}
(i) the $\cos \theta_{f}$ distribution is flat for the whole range of $
\cos\theta_f$.
(ii) the $\cos\theta_{\pi^*}$ distributions can be fitted by $(a + b \cos
\theta_{\pi^*}^{2})$, with $a = (1.24 \pm 0.06)$
{}~~~and $b = (-0.49 \pm 0.13)$.
(iii) the $\cos \theta_l$ distribution is consistent with $( 1 + \cos
\theta_{l}^{2} )$ for $ 0 < | \cos\theta_{l} | < 0.7$.
\vspace{.2in}
\noindent
Let us discuss the implication of each statement above to our fits, P0--P2.
Statement (i) excludes both P1 and P2, since these two lead to quadratic
functions of $\cos\theta_f$. Statement (iii) also partly supports this
conclusion, since (iii) implies $D=0$ in the amplitude (1).
Thus, (i) and (iii) select P0 as the final candidate.
However, the $\cos\theta_{\pi^*}$ distribution shown in Fig.~\ref{figone}
(a) does not agree with statement (ii) from CLEO.
We do not interpret this disagreement of the CLEO data with our
prediction on the $\cos \theta_{\pi}^*$ distribution
as a general failure of our approach based on the matrix element
satisfying the soft pion theorem. We rather regard
it as an indication that amplitude (1) needs to be modified to include
higher order terms in the pion momentum expansion.
This will be illustrated in the next section with a modified
amplitude for $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$.
\section{more on the amplitude with $D=0$}
\label{sec:three}
In this section, we consider the case $D=0$ in more detail, including
possible higher order terms in $q^2$ in the $S-$wave $\pi\pi$ amplitude,
$f_{S}$. It will be shown that the $\cos \theta_{\pi}^*$
distribution is sensitive to such higher order terms in $q^2$
contrary to other distributions.
Let us write the amplitude for $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ as
\begin{equation}
{\cal M} (\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi) = A~\left[~ f_{S} (q^{2}) e^{i \delta_{0} (q^{2})} +
f_{D} (q^2) ( \cos^{2} \theta_{\pi}^{*} - {1\over 3} ) ~\right]~\hat{
\epsilon}
\cdot \hat{\epsilon}^{'},
\label{eq:ampfin}
\end{equation}
where $f_S$ and $f_D$ satisfy the soft pion theorem.
The explicit forms of $f_{S,D}$ for the lowest order amplitude
(\ref{eq:ampnew1}) can be read off from (\ref{eq:fs})--(\ref{eq:gd})
with $D=0$. Therefore the differential cross section for $ e^+ e^-
\rightarrow \Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi \rightarrow \pi\pi\mu^{+}\mu^{-} $ is given by
\begin{eqnarray}
d^{3} \Gamma & \propto &
dm_{\pi\pi} d\cos\theta_{\pi}^{*} d\cos\theta_{l} ~~
m_{\pi\pi} | \vec{q} |~\beta_{\pi}^{*}~\left[ 1 + \cos^{2}
\theta_{l} \right] \nonumber \\
& \times &
\left[ ~f_{S}^{2} + f_{D}^{2} \left( \cos^{2} \theta_{\pi}^{*} - { 1 \over
3} \right)^{2} + 2 f_{S} f_{D} \cos\delta \left( \cos^{2} \theta_{\pi}^{*}
- { 1\over 3} \right) ~\right],
\label{eq:dcrosect}
\end{eqnarray}
where $\beta_{\pi}^*$ is the velocity
of a pion in the $\pi\pi$ rest frame and $\theta_{l}$ is the
angle between a muon and the $e^+ e^-$ beam in the rest frame of $\Upsilon
(3S)$.
Integrating the partial distribution (\ref{eq:dcrosect}) over
appropriate variables, one gets
\begin{eqnarray}
d\Gamma\over d\cos\theta_{f} & \propto & 1, ~~~~({\rm flat \ \ distribution}),
\label{eq:cosf} \\
d\Gamma\over d\cos\theta_{l} & \propto & ( 1 + \cos^{2} \theta_{l} ).
\label{eq:cosmu}
\end{eqnarray}
These two distributions are independent of the $q^2$ dependence
of the form factors, $f_{S}(q^{2})$ and $f_{D}(q^{2})$, as well as of
the $\pi\pi$ phase shift.
And, these results are consistent with the recent report from CLEO.
On the other hand, the $\cos\theta_{\pi}^*$ distribution is sensitive
to the actual forms of $f_{S}(q^{2})$ and $f_{D}(q^{2})$, and to the
$\pi\pi$ phase shift.
In principle, there are many possible terms to the next order in the
pion momentum expansion. Instead of writing down all possible
terms and fitting the $m_{\pi\pi}$ spectrum as in Ref.~\cite{chakrako},
we take the following amplitude for illustration :
\begin{equation}
f_{S} (q^{2}) = q^{2} \left[ 1 + C \left( {E_{1} + E_{2} \over
m_{\pi}} \right) \right]
+ {B\over 4}~\left[
( E_{1} + E_{2} )^{2} - {1\over 3} |\vec{p}_{f}|^{2} \beta_{\pi}^{* 2}
\right],
\label{eq:fsex}
\end{equation}
with the same $f_{D} (q^{2})$ as before. This amplitude has three
parameters, $A,B$ and $C$, and satisfies the soft pion theorem like
(\ref{eq:amp1}).
By $\chi^{2}$ fit to the $m_{\pi\pi}$ spectrum,
we found another fit (we will call it P3) with $\chi^{2}/d.o.f. = 11.0/7$ (see
Fig.~ \ref{figthree} (a)). The corresponding values of $A,B,C$ are given in
the last column of Table~1. This amplitude predicts the distributions,
(\ref{eq:cosf}) and (\ref{eq:cosmu}). The $\cos \theta_{\pi}^*$
distribution for P3 shown in Fig.~\ref{figthree} (b) ~differs a lot from
that for P0 in Fig.~\ref{figone} (a), and
gets much closer to the observed
data. The lesson from this example is that once we adopt
(\ref{eq:ampfin}) as the amplitude for $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$, we predict (i) the flat
$\cos\theta_f$ distribution, (ii) $( 1 + \cos^{2} \theta_{l} )$
distribution for
the polar angle of a muon, independent of actual forms of $f_{S}, f_{D}$
and $\delta_{0} (q^{2})$. This is not the case for $\cos \theta_{\pi}^*$
distribution and thus cannot be reliably calculated unless they are known.
The actual functional forms of $f_{S}, f_{D}$ and $\delta_{0} (q^{2})$ can
be extracted from the measurement of the joint distribution, $d^{2}\Gamma
/ dm_{\pi\pi} d\cos\theta_{\pi}^{*}$, as one can derive from
(\ref{eq:dcrosect}) :
\begin{eqnarray}
&&{d^{2} \Gamma \over dm_{\pi\pi}d\cos\theta_{\pi}^*} \nonumber \\
& \propto & m_{\pi\pi} | \vec{q} | \beta_{\pi}^{*}~\left[
f_{S}^{2} + f_{D}^{2} \left( \cos^{2} \theta_{\pi}^{*} - {1\over 3}
\right)^{2} + 2 f_{S} f_{D} \cos\delta_{0} \left( \cos^{2} \theta_{\pi}^{*}
- {1\over 3} \right) ~\right] \\
& = & \left[ C_{0} (q^{2}) + C_{2} (q^{2}) \cos^{2} \theta_{\pi}^{*}
+ C_{4} (q^{2}) \cos^{4} \theta_{\pi}^{*} \right].
\end{eqnarray}
For each $m_{\pi\pi}$ bin, one can measure the $\cos\theta_{\pi}^*$
distribution. This determines $C_{i} (q^{2})$'s, and in turn, three
unknowns, $f_{S}, f_{D}$ and $\delta_{0}(q^{2})$.
In particular, the decay $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ can be a
source of the $S-$wave $\pi\pi$ phase shift for the whole elastic region,
$2 m_{\pi} \leq m_{\pi\pi} \leq ( m_{i} - m_{f} ) = 895$ MeV \cite{fuchs}.
This may be important, since the existing data on the $\pi\pi$ phase shift
between $m_{K} \leq m_{\pi\pi} \leq 600$ MeV are rather poor in statistics
and one has to make some extrapolation \cite{fuchs1}.
\section{Conclusion}
\label{sec:four}
Concluding, we reanalyzed the $\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ decay using the most general
matrix element in the lowest order in the pion momentum expansion,
including the final state interactions of the $\pi \pi$
system in the $I = L = 0$ channel. The $\pi\pi$ phase shift changes
the $\cos\theta_{\pi}^*$ distributions moderately. (Compare Figs.~1 (a),
(b) with Figs.~2 (a), (b).)
Compared with the recent data from CLEO, P0 is selected, but
the $\cos\theta_{\pi}^*$ distribution does not agree.
In Sec. \ref{sec:three}, we argued that this distribution is
sensitive to possible higher order corrections in the pion momentum
expansion. As an illustration, we used a new ansatz for the $\pi\pi$
$S-$wave amplitude, (\ref{eq:fsex}), which is of higher order in the pion
momentum expansion, and satisfies Adler's condition. This amplitude
could fit the $m_{\pi\pi}$ spectrum (Fig.~3 (a)). The resulting
$\cos\theta_{\pi}^*$ distribution (Fig.~3 (b)) is
different from Fig,~1 (a), and gets closer qualitatively
to the measured distribution although not quantitatively.
It would be more proper to do the partial wave analysis
with the $S-$ and $D-$waves, and find out the form factors ($f_{S}
, f_{D}$) and the phase shift ($\delta_{0} (q^{2})$) in
(\ref{eq:ampfin}) from the
joint distributions in $m_{\pi\pi}$ and $\cos \theta_{\pi}^*$ using
(20) and (21).
Since our explanation involves both $S-$ and $D-$wave $\pi\pi$
systems, we may be able to find out the $S-$wave phase shift (or
$\delta_{0} - \delta_{2}$, more precisely) from the decay
$\Upsilon (3S) \rightarrow \Upsilon (1S) \ \pi \pi$ by measuring various joint distributions. Since the available
$m_{\pi\pi}$ is below the $K\bar{K}$ threshold, this decay may provide
information on the phase shift over the whole elastic
region, especially for $m_{K} \leq m_{\pi\pi} \leq 600$ MeV, where the
current data are rather poor in statistics.
\acknowledgements
We thank Y. Kubota, G. Moneti, R. Poling, S. Rudaz and M. Voloshin for
discussions on the subject.
We are grateful to J.L. Rosner for careful reading of the
manuscript and valuable suggestions.
P.K. thanks the Aspen Center for Physics for support and hospitality, where
part of this work was done.
This work was supported by Department of Energy
grant \# DOE--DE--AC02--83ER--40105.
S. C. thanks the Mechanical Engineering Department for financial support.
| {
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God's love is a pure love: unconditional, unilateral, unaffected by our merit or our sin. Jesus illustrates this love in the parable of the Prodigal Son. A wicked son, after having insulted his father and wasted his money, is accepted by his father again – with open arms! Why? Because it's his child. Because our own love is so conditional, we struggle – just like the older brother – to understand such love. We want to believe that God loves the good people, not the bad! They don't deserve love – they deserve condemnation, rejection, punishment! Wouldn't that be fair? Although God punishes bad people – or more correctly, their bad behaviour – he always loves them. Even the worst sinner is loved by God! When we sin, as we do, we often feel that we don't deserve God's love – and we don't, of course. Still, think of the Prodigal Son's father, think of how Jesus treated sinners – think of the cross. Did sin stop Jesus' love, God's love? No, sin cannot stop love!
Thank you, Lord, that my sin cannot and will not stop your love. I praise you for loving me always! Amen.
Look past your sin – see love! | {
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{"url":"https:\/\/plainmath.net\/55123\/prove-in-detail-the-inequality-of-this-formula-let-x-y","text":"# Prove in detail the inequality of this formula: Let x, y,\n\nProve in detail the inequality of this formula:\nLet x, y, z positive real number and $\\mathrm{\u25b3}ABS$ a triangle. $\\left[ABC\\right]$ denotes the triangle area and a, b, c the sides of the triangle. The inequality below is true:\n${a}^{2}x+{b}^{2}y+{c}^{2}z\\ge 4\\left[ABC\\right]\\sqrt{xy+xz+yz}$\nYou can still ask an expert for help\n\n\u2022 Questions are typically answered in as fast as 30 minutes\n\nSolve your problem for the price of one coffee\n\n\u2022 Math expert for every subject\n\u2022 Pay only if we can solve it\n\nstamptsk\n\nStep 1\nProof\nLet $\\alpha ,\\beta ,\\gamma$ denote the opposite angles to the sides $a,b,c$ respectively. R is the circumradius of $\\mathrm{\u25b3}ABC$. Observe that:\n${a}^{2}x+{b}^{2}y+{c}^{2}z\\ge 4\\left[ABC\\right]\\sqrt{xy+xz+yz}$\n${a}^{2}x+{b}^{2}y+{c}^{2}Z\\ge \\frac{abc}{R}\\sqrt{xy+xz+yz}$\n$\\frac{aRx}{bc}+\\frac{bRy}{ac}+\\frac{cRz}{ab}\\ge \\sqrt{xy+xz+yz}$\n$\\frac{1}{2}\\left(\\frac{4a{R}^{2}x}{2Rbc}+\\frac{4b{R}^{2}y}{2Rac}+\\frac{4c{R}^{2}z}{2Rab}\\right)\\ge \\sqrt{xy+xz+yz}$\n$x\\frac{\\mathrm{sin}\\alpha }{\\mathrm{sin}\\beta \\mathrm{sin}\\gamma }+y\\frac{\\mathrm{sin}\\beta }{\\mathrm{sin}\\alpha \\mathrm{sin}\\gamma }+z\\frac{\\mathrm{sin}\\gamma }{\\mathrm{sin}\\alpha \\mathrm{sin}\\beta }\\ge 2\\sqrt{xy+xz+yz}$\n$x\\frac{\\mathrm{sin}\\left(\\pi -\\alpha \\right)}{\\mathrm{sin}\\beta \\mathrm{sin}\\gamma }+y\\frac{\\mathrm{sin}\\left(\\pi -\\beta \\right)}{\\mathrm{sin}\\alpha \\mathrm{sin}\\gamma }+z\\frac{\\mathrm{sin}\\left(\\pi -\\gamma \\right)}{\\mathrm{sin}\\alpha \\mathrm{sin}\\beta }\\ge 2\\sqrt{xy+xz+yz}$\n$x\\frac{\\mathrm{sin}\\left(\\alpha +\\beta +\\gamma -\\alpha \\right)}{\\mathrm{sin}\\beta \\mathrm{sin}\\gamma }+y\\frac{\\mathrm{sin}\\left(\\alpha +\\beta +\\gamma -\\beta \\right)}{\\mathrm{sin}\\alpha \\mathrm{sin}\\gamma }+z\\frac{\\alpha +\\beta +\\gamma -\\gamma }{\\right\\}}\\left\\{\\mathrm{sin}\\alpha \\mathrm{sin}\\beta \\right\\}\\ge 2\\sqrt{xy+xz+yz}$\n\n###### Not exactly what you\u2019re looking for?\nEliza Norris\nStep 1\nHere is my algebraic proof.\nWe need to prove that:\n${\\left({a}^{2}x+{b}^{2}y+{c}^{2}z\\right)}^{2}\\ge \\sum _{cyc}\\left(2{a}^{2}{b}^{2}-{a}^{4}\\right)\\left(xy+xz+yz\\right)$\nor\n${c}^{4}{z}^{2}-\\left(\\left(\\sum _{cyc}\\left(2{a}^{2}{b}^{2}-{a}^{4}\\right)-2{a}^{2}{c}^{2}\\right)x+\\left(\\sum _{cyc}\\left(2{a}^{2}{b}^{2}{a}^{4}\\right)-2{b}^{2}{c}^{2}\\right)y\\right)z+$\n$+{a}^{4}{x}^{2}+{b}^{4}{y}^{2}-\\left(\\sum _{cyc}\\left(2{a}^{2}{b}^{2}-{a}^{4}\\right)-2{a}^{2}{b}^{2}\\right)xy\\ge 0$\nfor which its\n###### Not exactly what you\u2019re looking for?\nRizerMix\nStep 1 Writing the expression as $\\frac{{a}^{2}x+{b}^{2}y+{c}^{2}z}{\\sqrt{xy+yz+zx}}\\ge 4\\left[ABC\\right]$ the RHS is independent of x,y,z, therefore it is necessary and sufficient to prove the inequality in the worst possible case, i.e. when the LHS is minimized in x,y,z. In particular, by homogeneity we can fix a,b,c and consider the problem If two of x,y,z are zero the inequality is trivially proven. If just one of them is zero, say z, then the problem becomes by AM-GM, and clearly $2ab\\ge 2ab\\mathrm{sin}\\gamma =4\\left[ABC\\right]$ The last case is $x,y,z>0$. By the Lagrange method we obtain a critical point in the interior where that is $x={b}^{2}+{c}^{2}-{a}^{2}$ $y={c}^{2}+{a}^{2}-{b}^{2}$ $z={a}^{2}+{b}^{2}-{c}^{2}$ up to a multiplicative constant. Substituting above we obtain ${a}^{2}x+{b}^{2}y+{c}^{2}z=2\\left({a}^{2}{b}^{2}+{b}^{2}{c}^{2}+{c}^{2}{a}^{2}\\right)-\\left({a}^{4}+{b}^{4}+{c}^{4}\\right)=16\\left[ABC{\\right]}^{2}$ by Herons","date":"2022-05-18 12:57:22","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 27, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9614837169647217, \"perplexity\": 383.4448854143334}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-21\/segments\/1652662522270.37\/warc\/CC-MAIN-20220518115411-20220518145411-00352.warc.gz\"}"} | null | null |
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Nashik Directory offers easy to use information about various Transducers related resources around in Nashik (Nasik). There are plenty of Transducers sources available at the industrial sector in Nashik. A transducer is a device, usually electrical, electronic, electro-mechanical, electromagnetic, photonic, or photovoltaic that converts one type of energy or physical attribute to another for various purposes including measurement or information transfer (for example, pressure sensors). There are different types of the Transducers that are produced or supplied at the industries in the city. For example : Antenna, Fluorescent lamp, light bulb, Galvanometer, Vibration powered generator, Accelerometer, Loudspeaker, earphone, Thermocouple. If you are looking manufacturers, traders, importer and exporters of Transducers in Nasik than we can help you. Nashikdirectory.com is a leading B2B industrial online directory of Nashik (Nasik). | {
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{"url":"https:\/\/labs.thosgood.com\/translations\/PM-16-2002-47.html","text":"#### Translator\u2019s note\n\nFern\u00e1ndez S\u00e1nchez, P. \u201cAutomorfismo de foliaciones holomorfas sobre superficies racionales.\u201d Pro Mathematica 16 (2002), 47\u201359. (Available online.)\n\nThe translator (Tim Hosgood) takes full responsibility for any errors introduced, and claims no rights to any of the mathematical content herein.\n\nVersion: 7f60475\n\nIn this work we classify holomorphic foliations with infinite automorphism group on a rational surface. As a consequence of this result, we prove that the automorphism group of a foliation of general type with singularities on a rational surface is finite.\n\n# 1 Introduction\n\nSchwarz proved that the automorphism group of a Riemann surface of genus greater than 2 is finite. Andreotti, in [1?], generalises this result, proving that the bimeromorphism group of an algebraic variety of general type is finite, these being analogous to Riemann surfaces of genus greater than 2. In the case of algebraic surfaces, there is even a TO-DO for this number, cf. [14?].\n\nIn this work, we first classify holomorphic foliations with singularities on rational surfaces.\n\nLet {\\mathcal{F}} be a foliation on a rational surface M. If \\#\\operatorname{Aut}({\\mathcal{F}})=+\\infty, then {\\mathcal{F}} is bimeromorphic to a Riccati foliation or a rational fibration.\n\nWe then prove a result analogous to that of Andreotti for holomorphic foliations on rational surfaces (surfaces bimeromorphic to the projective plane).\n\nThe automorphism group of a foliation of general type with singularities on a rational surface is finite.\n\n# 2 Preliminaries\n\nFor the basic notions of holomorphic foliations, we recommend the books [4?,13?,2?]. A holomorphic foliation {\\mathcal{F}} with isolated singularities on an algebraic surface M can be defined as a family \\{X_i\\} of holomorphic vector fields defined on an open cover \\{U_i\\} of M that satisfies the cocycle condition X_i=g_{ij}X_j whenever U_i\\cap U_j\\neq\\varnothing, where \\{g_{ij}\\} are nowhere-zero holomorphic functions defined on U_i\\cap U_j. In this case, \\{g_{ij}\\} defines a line bundle T_{{\\mathcal{G}}}^* on M called the cotangent (or canonical) bundle. The set of singularities \\operatorname{Sing}({\\mathcal{G}}) of {\\mathcal{G}} is defined as \\operatorname{Sing}({\\mathcal{G}})_{\/U_i}=\\{X_i=0\\}, and it is always possible to suppose that \\dim(\\operatorname{Sing}({\\mathcal{G}}))=0.\n\nLet {\\mathcal{F}} be a holomorphic foliation on a complex surface M. Consider a compact curve C on M. Let p\\in C, and let \\{f=0\\} be a reduced local equation of C around p. Suppose that {\\mathcal{F}} is represented in a coordinate neighbourhood (U,(x,y)) of p=(0,0) by the holomorphic 1-form \\omega = a(x,y)\\mathrm{d}x + b(x,y)\\mathrm{d}y. If C is not F-invariant, then we define the tangency between {\\mathcal{F}} and C at p by \\operatorname{tang}_p({\\mathcal{F}},C)=\\dim_\\mathbb{C}\\mathscr{O}_p\/I, where I is the ideal generated by f and -b\\frac{\\partial f}{\\partial x}+a\\frac{\\partial f}{\\partial y}. We define \\operatorname{tang}({\\mathcal{F}},C)=\\sum_{p\\in C}\\operatorname{tang}_p({\\mathcal{F}},C). It is proven in [2?] that T_{\\mathcal{F}}^*C = \\operatorname{tang}({\\mathcal{F}},C) - C^2.\n\nIf M=\\mathbb{C}P(2) and C is a non-{\\mathcal{F}}-invariant line, then we have T_{\\mathcal{F}}^* = \\mathscr{O}_{\\mathbb{C}P(2)}(\\operatorname{tang}({\\mathcal{F}},C)-1).\n\nNow suppose that C is {\\mathcal{F}}-invariant, so that \\omega\\wedge\\mathrm{d}f=f\\Theta, where \\Theta is a holomorphic 2-form. Then there exist relatively prime holomorphic functions g and h defined on U, along with a holomorphic 1-form \\eta, such that g\\omega = h\\mathrm{d}f + f\\eta.\n\nWe define the Camacho\u2013Sad index by \\operatorname{CS}_p({\\mathcal{F}},C)=-\\frac{-1}{2\\pi i}\\int_\\gamma\\frac{\\eta}{h}, where \\gamma is a loop around p on \\{f=0\\}. We define \\operatorname{CS}({\\mathcal{F}},C)=\\sum_{p\\in\\operatorname{Sing}{\\mathcal{F}}\\cap C}\\operatorname{CS}_p({\\mathcal{F}},C) The Camacho\u2013Sad index theorem [5?] says that \\operatorname{CS}({\\mathcal{F}},C) = C^2. Recall that a reduced foliation {\\mathcal{F}} is a foliation such that every singularity p is reduced in the sense of Seidenberg, i.e., for every vector field X generating {\\mathcal{F}}, and for every singular point p of X, the eigenvalues of the linear part of X are not both zero, and their quotient, when defined, is not a positive rational number. If one eigenvalue is zero and the other is not, then the singularity is said to be a TO-DO: knot saddle?.\n\nLet {\\mathcal{F}} be a foliation on a complex surface S, and let {\\mathcal{G}} be an arbitrary reduced foliation that is bimeromorphically equivalent to {\\mathcal{F}}. The Kodaira dimension of {\\mathcal{F}} is given by \\operatorname{Kod}({\\mathcal{F}}) = \\limsup_{n\\to\\infty} \\frac{\\log h^0(S,K_{\\mathcal{G}}^{\\otimes=n})}{\\log n}.","date":"2021-09-18 21:53:44","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9252781867980957, \"perplexity\": 1376.4118114837347}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-39\/segments\/1631780056578.5\/warc\/CC-MAIN-20210918214805-20210919004805-00511.warc.gz\"}"} | null | null |
Twitter user tells PM his photo is becoming a meme, he says 'most welcome'
New Delhi [India], Dec 26 (ANI): Narendra Modi on Thursday pleasantly surprised his Twitter followers when he 'welcomed' a user who had tweeted that the photo of the Prime Minister viewing the solar eclipse is becoming a meme.
"This is becoming a meme," the Twitter user, Gappistan Radio said, alongside a picture of PM Modi where he is seen wearing dark glasses and is looking above in the sky.
The Prime Minister retweeted the tweet saying, "Most welcome....enjoy :)."
In less than half an hour, the Prime Minister's tweet garnered over 17,000 likes and more than 4,000 RTs.
Moments later, the Twitter user tweeted "Wait what".
Earlier, the Prime Minister had tweeted a few of his pictures and had said that like many Indians, he too was enthusiastic about the Solar Eclipse.
"Like many Indians, I was enthusiastic about #solareclipse2019. Unfortunately, I could not see the Sun due to cloud cover but I did catch glimpses of the eclipse in Kozhikode and other parts on live stream. Also enriched my knowledge on the subject by interacting with experts," the Prime Minister had tweeted.
Several parts of the country, including Odisha, Kerala, Gujarat, Tamil Nadu, Karnataka, Maharashtra and Delhi, are witnessing a solar eclipse on Thursday morning, which is said to the last one of the decade. (ANI) | {
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El género Phyllanthus, de la familia Phyllanthaceae, comprende árboles, arbusto y hierbas anuales o bieniales distribuidos en regiones tropicales y subtropicales.
Flor de hoja es el nombre común para todas las spp. Phyllanthus. La circunscripción del género ha sido tan confusa que en los 1990s, se ha llevado a cabo una reorganización de Phyllanthus.
Descripción
Son árboles, arbustos o hierbas sin látex; plantas monoicas o raramente dioicas (solo P. elsiae en Nicaragua). Con hojas alternas, simples, enteras, pinnatinervias, estipuladas. De flores en fascículos axilares o a veces caulifloras, apétalas; flores estaminadas con 4–6 sépalos, imbricados, disco generalmente segmentado o a veces ausente, estambres mayormente 2–6, libres o connados, pistilodio ausente; flores pistiladas mayormente pediceladas, sépalos 4–6, imbricados, disco entero, segmentado o a veces ausente, ovario 3 (4)-locular, 2 óvulos por lóculo, estilo libre o connado, generalmente bífido. El fruto es generalmente capsular o a veces drupáceo; las semillas son ecarunculadas.
Medicina
Las Phyllanthus han sido tradicionalmente usadas para curar un amplio espectro de enfermedades. En medicina ayurvédica, por ej., Phyllanthus es prescripta para ictericia (hepatitis A), gonorrea y diabetes (uso interno). Con cataplasma se tratan úlceras y otros problemas de la piel (uso externo). La infusión se hace de ramitas jóvenes para tratar la disentería crónica.
Las plantas básicamente tienen ligninas (e.g. fillantina e hipofillantina), alcaloides y flavonoides (e.g. quercetina).
Phyllanthus pareciera detener el desarrollo de la hepatitis B debido a su bloqueo del ADN polimerasa, la enzima necesaria para que el virus de la enfermedad se reproduzca. A este respecto, P. urinaria y P. niruri serían mejores.
Taxonomía
El género fue descrito por Carlos Linneo y publicado en Species Plantarum 2: 981–982. 1753.
Sinonimia
Anisonema, Aporosella, Arachnodes, Ardinghalia, Asterandra, Cathetus, Ceramanthus, Chorisandra, Cicca, Clambus, Conami, Dendrophyllanthus, Dicholactina, Dimorphocladium, Emblica, Epistylium, Eriococcus, Fluggeopsis, Genesiphylla, Hemicicca, Hemiglochidion, Kirganelia, Leichhardtia, Lomanthes, Maborea, Macraea, Menarda, Mirobalanus, Moeroris, Nellica, Niruri, Nymania, Nymphanthus, Orbicularia, Oxalistylis, Ramsdenia, Reidia, Rhopium, Roigia, Scepasma, Staurothylax, Synexemia, Tricarium, Uranthera, Urinaria, Williamia, Xylophylla
:en:Taxonomy of the Phyllanthaceae
Especies selectas
Phyllanthus abnormis
Phyllanthus acidus - grosella
Phyllanthus acuminatus - grosella de Jamaica
Phyllanthus brasiliensis (Aubl.) Poir. - barbascajo del Orinoco
Phyllanthus debilis
Phyllanthus diffusus
Phyllanthus emblica - grosella de la India
Phyllanthus epiphyllanthus
Phyllanthus fraternus
Phyllanthus juglandifolius - jobo jíbaro de Quito
Phyllanthus mirabilis - la única sp. suculenta del Gro.
Phyllanthus myrtifolius
Phyllanthus niruri - lepidio, rompepiedras (cálculos renales)
Phyllanthus nobilis
Phyllanthus reticulatus
Phyllanthus sellowianus sarandi blanco
Phyllanthus urinaria - chamberbitter
Para la lista completa: Lista de especies de Phyllanthus.
Galería
Referencias
Bibliografía
Burger, W. & M. Huft. 1995. Family 113 Euphorbiaceae. Fieldiana, Bot., n.s. 36: 1–169.
Forzza, R. C. & et al. 2010. 2010 Lista de espécies Flora do Brasil. https://web.archive.org/web/20150906080403/http://floradobrasil.jbrj.gov.br/2010/.
González Ramírez, J. 2010. Euphorbiaceae. En: Manual de Plantas de Costa Rica. Vol. 5. B.E. Hammel, M.H. Grayum, C. Herrera & N. Zamora (eds.). Monogr. Syst. Bot. Missouri Bot. Gard. 119: 290–394.
Idárraga-Piedrahíta, A., R. D. C. Ortiz, R. Callejas Posada & M. Merello. 2011. Flora de Antioquia. Catálogo de las Plantas Vasculares, vol. 2. Listado de las Plantas Vasculares del Departamento de Antioquia. Pp. 1-939.
Martínez Gordillo, M., J. J. Ramírez, R. C. Durán, E. J. Arriaga, R. García, A. Cervantes & R. M. Hernández. 2002. Los géneros de la familia Euphorbiaceae en México. Anales Inst. Biol. Univ. Nac. Autón. México, Bot. 73(2): 155–281.
Molina Rosito, A. 1975. Enumeración de las plantas de Honduras. Ceiba 19(1): 1–118.
Nasir, E. & S. I. Ali (eds). 1980-2005. Fl. Pakistan Univ. of Karachi, Karachi.
Standley, P. C. & J. A. Steyermark. 1949. Euphorbiaceae. In: P. C. Standley & J. A. Steyermark (eds.), Flora of Guatemala---Part VI. Fieldiana, Bot. 24(6): 25–170.
Stevens, W. D., C. Ulloa Ulloa, A. Pool & O. M. Montiel Jarquin. 2001. Flora de Nicaragua. Monogr. Syst. Bot. Missouri Bot. Gard. 85: i–xlii,.
Webster, G. L. & M. J. Huft. 1988. Revised synopsis of Panamanian Euphorbiaceae. Ann. Missouri Bot. Gard. 75(3): 1087–1144.
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\section{Introduction}
Much of the progress in our understanding of the capacity limits of wireless networks over the past decade has come from the pursuit of progressively refined capacity approximations. Generalized degrees of freedom (GDoF) characterizations represent a most significant step along this path because of their ability to capture arbitrary channel strength and channel uncertainty levels. The GDoF framework may seem counter-intuitive at first because it allows exponential scaling of signal strengths with various exponents. An intuitive justification for the GDoF framework is as follows. It is important to remember that the goal behind GDoF is to seek capacity approximations for a given wireless network with its arbitrary \emph{finite} signal strengths and channel uncertainty levels. Unlike the degrees of freedom (DoF) metric which linearly scales all signal strengths and loses the distinction of different channel strengths (every non-zero channel carries $1$ DoF), the GDoF formulation takes a more sophisticated approach. The key to GDoF is the intuition that if the capacity of every link in a network is scaled by the same factor, then the capacity region of the network should scale by approximately the same factor as well. Normalizing the capacity of the network by the scaling factor then yields a capacity approximation for the original network. Following this intuition, one allows the scaling factor to approach infinity, while guaranteeing that the capacity is always normalized by the scaling factor. The asymptotic behavior of normalized capacity is potentially easier to characterize than a direct approximation of the capacity of the original network. Let $\alpha_i$ represent the capacity of the $i^{th}$ link in the original network (in isolation from all other links), and let $\log(P)$ be the scaling factor applied to every link capacity. Then we obtain channels whose capacity scales as $\alpha_i\log(P)$, i.e., channels whose strength scales as $P^{\alpha_i}$, and, according to this intuitive reasoning, normalization of network capacity by $\log(P)$ in the limit $P\rightarrow\infty$ presents the approximation of the capacity of the original network. This approximation is what is known as the GDoF characterization, and along with its abstractions into deterministic channel models, over the past decade it has been the key to finding capacity approximations for many networks whose exact capacity remains intractable. Thus, the linear scaling of capacity naturally corresponds to an exponential scaling of signal strengths in the GDoF model.
GDoF studies started with settings where perfect CSIT is available \cite{Etkin_Tse_Wang, Jafar_Vishwanath_GDOF, Karmakar_Varanasi}. The opposite extreme of no CSIT was also explored under strong assumptions of statistical equivalence between users \cite{Huang_Jafar_Shamai_Vishwanath, Varanasi_noCSIT, Guo_noCSIT}. Lately, however, the focus has shifted to the broader assumption of finite precision CSIT \cite{Arash_Jafar_TC}, \cite{Arash_Bofeng_Jafar_BC}. Some of the more sophisticated concepts such as interference alignment \cite{Jafar_FnT} have turned out to be too fragile to be useful with finite precision CSIT, so that conventional achievable schemes are usually optimal. As such the main challenge for GDoF studies under finite precision CSIT tends to be the proof of optimality, i.e., the converse, or the GDoF outer bound. Finding tight GDoF outer bounds under finite precision CSIT is generally a hard problem, as exemplified by the conjecture of Lapidoth et al. \cite{Lapidoth_Shamai_Wigger_BC} which remained unresolved for nearly a decade. The main idea for these outer bounds is the Aligned Image Sets (AIS) argument that was introduced in \cite{Arash_Jafar_PN} in order to settle the conjecture of Lapidoth et al. Generalizations of the AIS approach have also helped settle the GDoF in other settings such as the X channel and the $2$ user MISO BC under finite precision CSIT in \cite{Arash_Jafar_TC}, and the 2 user MISO BC with arbitrary channel strengths and channel uncertainty levels in \cite{Arash_Bofeng_Jafar_BC}. Of particular relevance to this work is \cite{Arash_Jafar_IC} where the sum GDoF of $K$-user symmetric interference channel (IC) is characterized under finite precision CSIT (see Figure \ref{fig:Kusert}). This work is motivated by the goal of further broadening the scope of the AIS argument, so that the results of \cite{Arash_Jafar_IC} may be generalized to MIMO settings.
\begin{figure*}[h]
\center
\begin{minipage}[c]{0.54\textwidth}
\centerline{\includegraphics[width=3.7in]{p2+.pdf}}
\end{minipage}~~~~~
\begin{minipage}[c]{0.33\textwidth}
\begin{eqnarray*}
d(\alpha) = \left\{
\begin{array}{ll}
1-\alpha,&0\leq\alpha\leq \frac{1}{2}\vspace{0.03in}\\
\frac{K-2-(K-3)\alpha}{K-1}, &\frac{1}{2}<\alpha\leq\frac{K}{K+1}\vspace{0.03in}\\
1-\left(\frac{K-1}{K}\right)\alpha,&\frac{K}{K+1}<\alpha\leq 1\\
\frac{\alpha}{K},& 1<\alpha\leq K\\
1, &K<\alpha
\end{array}
\right.
\end{eqnarray*}
\end{minipage}
\caption{GDoF/user of the Symmetric $K$ User Interference Channel with Finite Precision CSIT \cite{Arash_Jafar_IC}.}\label{fig:Kusert}
\end{figure*}
In this paper, we characterize the GDoF for the symmetric $K$-user MIMO Interference Channel under the assumption that the CSIT is limited to finite precision. In this symmetric setting, each transmitter is equipped with $M$ antennas, each receiver is equipped with $N$ antennas, each desired channel (i.e., a channel between a transmit antenna and a receive antenna belonging to the same user) has strength $\sim P$, while each undesired channel has strength $\sim P^\alpha$, where $P$ is a nominal SNR parameter. GDoF per user take the form of a $W$-curve with respect to $\alpha$ for fixed values of $M$ and $N$. See Figure \ref{Fig1}. As usual for finite precision CSIT, achievability is fairly straightforward. While ostensibly the main result of this work is the GDoF characterization for the $K$-user symmetric MIMO IC, the deeper significance of this paper resides in a key generalization of the AIS approach that allows comparisons in the GDoF sense of the entropies of different numbers of linear combinations (finite precision versus perfectly known channels) of random variables under various power-level partitions. The generalization seems broadly useful for GDoF problems related to MIMO wireless networks.
{\it Notation:} The notation $|A|$ denotes the cardinality of the set $A$ and the notation $[n]$ is defined as $\{1,2,\cdots,n\}$ for any $n\in\mathbb{N}$ where $\mathbb{N}$ is the set of all positive integer numbers. The notations~ $X^{[T]}$ and $X_{i}^{[T]}$ also stand~ for $\{X(1), X(2), \cdots X(T)\}$ and $\{X_i(t): \forall t\in[T]\}$, respectively. Moreover, we use the Landau $o(\cdot)$ notation for the functions $f(x), g(x)$ from $\mathbb{R}$ to $\mathbb{R}$ as follows. $f(x)=o(g(x))$ denotes that $\limsup_{x\rightarrow\infty}\frac{|f(x)|}{|g(x)|}=0$. Finally, we define $\lfloor x\rfloor$ as the largest integer that is smaller than or equal to $x$ for any positive real number $x$ and the smallest integer that is larger than or equal to $x$ for any negative real number $x$. $A^\dagger$ is the transpose of matrix $A$. The support of a random variable $X$ is denoted as supp$(X)$.
\section{Definitions}
\begin{definition}\label{bd}[Bounded Density Channel Coefficients \cite{Arash_Jafar_PN}] Define a set of real valued random variables, $\mathcal{G}$ such that the magnitude of each random variable $g\in\mathcal{G}$ is bounded away from infinity, $ |g|\leq\Delta<\infty$, for some positive constant $\Delta$, and there exists a finite positive constant $f_{\max}$, such that for all finite cardinality disjoint subsets $\mathcal{G}_1, \mathcal{G}_2$ of $\mathcal{G}$, the joint probability density function of all random variables in $\mathcal{G}_1$, conditioned on all random variables in $\mathcal{G}_2$, exists and is bounded above by $f_{\max}^{|\mathcal{G}_1|}$. Without loss of generality we will assume that $f_{\max}\geq 1, \Delta\geq 1$.
\end{definition}
\begin{definition}[Power Levels] Consider integer valued random variables $X_i$ over alphabet $\mathcal{X}_{\lambda_i}$,
\begin{eqnarray}
\mathcal{X}_{\lambda_i}&\triangleq&\{0,1,2,\cdots,\bar{P}^{\lambda_i}-1\}
\end{eqnarray}
where $\bar{P}^{\lambda_i}$ is a compact notation for $\left\lfloor\sqrt{P^{\lambda_i}}\right\rfloor$ and the constant $\lambda_i$ is a positive real number denoting the \emph{power level} of $X_i$.
\end{definition}
\begin {definition}\label{powerlevel} For $X\in\mathcal{X}_\lambda$, and $0\leq \lambda_1\leq\lambda$, define the random variables $(X)_{\lambda_1}^{\lambda}$ as,
\begin{eqnarray}
(X)_{\lambda_1}^{\lambda}&\triangleq&\left \lfloor \frac{X}{\bar{P}^{\lambda_1}} \right \rfloor
\end{eqnarray}
\end {definition}
In words, $(X)_{\lambda_1}^{\lambda}$ retrieves the top $\lambda-\lambda_1$ power levels of $X$. Similarly, for the vector ${\bf V}=\begin{bmatrix}v_1&v_2&\cdots&v_k\end{bmatrix}^\dagger$, we define $({\bf V})^{\lambda}_{\lambda_1}$ as,
\begin{eqnarray}
({\bf V})^{\lambda}_{\lambda_1}&\triangleq&\begin{bmatrix}(v_1)^{\lambda}_{\lambda_1}&(v_2)^{\lambda}_{\lambda_1}&\cdots&(v_k)^{\lambda}_{\lambda_1}\end{bmatrix}^\dagger
\end{eqnarray}
\begin {definition}\label{deflc} For {real} numbers $x_1,x_2,\cdots,x_k{\in\mathcal{X}_{\eta}}$ define the notation $L_j^b(x_i,1\le i\le k)$ to represent,
\begin {eqnarray}
L^b_j(x_1,x_2,\cdots,x_k )&=&\sum_{1\le i\le k} \lfloor g_{j_i}x_i\rfloor
\end{eqnarray}
for distinct random variables $g_{j_i}\in\mathcal{G}$. The subscript $j$ is used to distinguish among multiple linear combinations, and may be dropped if there is no potential for ambiguity. For the vector $V=\begin{bmatrix}v_1&v_2&\cdots&v_k\end{bmatrix}^\dagger$ define the notation $L^b_j(V)$ to represent,
\begin {eqnarray}
L^b_j(V)=\sum_{1\le i\le k} \lfloor g_{j_i}v_i\rfloor
\end{eqnarray}
for distinct random variables $g_{j_i}\in\mathcal{G}$.
\end {definition}
\begin{definition}\label{defvec} For the two vectors $V=\begin{bmatrix}v_1&\cdots&v_{k_1}\end{bmatrix}^\dagger$ and $W=\begin{bmatrix}w_1&\cdots&w_{k_2}\end{bmatrix}^\dagger$ define the vector $V\bigtriangledown W$ as $\begin{bmatrix}v_1&\cdots&v_{k_1}&w_1&\cdots&w_{k_2}\end{bmatrix}^\dagger$.
\end{definition}
\section{System Model} {\label{sec-sys}}
In this work we consider only the setting where all variables take real values. Extensions to complex settings are cumbersome but conceptually straightforward as in \cite{Arash_Jafar_PN}.
\subsection{The Channel}
\noindent Define random variables $\mathbf{X}_{k}(t)$ and $\mathbf{Y}_{k}(t)$, $\forall k\in[K]$ as,
\begin{align}
\mathbf{X}_{k}(t)=&\begin{bmatrix}{X}_{k1}(t)&{X}_{k2}(t)&\cdots&{X}_{kM}(t)\end{bmatrix}^\dagger\\
\mathbf{Y}_{k}(t)=&\begin{bmatrix}{Y}_{k1}(t)&{Y}_{k2}(t)&\cdots&{Y}_{kN}(t)\end{bmatrix}^\dagger
\end{align}
where the channel uses are indexed by $t\in[T]$. $X_{km}(t), k\in[K],m\in[M],t\in[T]$ are the symbols sent from $m$-th transmit antenna of the $k$-th transmitter and are subject to unit power constraint, while $Y_{kn}(t), k\in[K],n\in[N],t\in[T]$ are the symbols observed by the $n$-th antenna of the $k$-th receiver. Under the GDoF framework, the channel model for the $K$-user MIMO IC is defined by the following input-output equations
\begin{align}
{\bf Y}_{k}(t)=&\sqrt{P}{\bf G}_{kk}(t)\mathbf{X}_k(t)+\sqrt{{P}^{\alpha}}\sum_{\hat{k}=1,\hat{k}\neq k}^K{\bf G}_{k\hat{k}}(t)\mathbf{X}_{\hat{k}}(t)+{\bf \Gamma}_{k}(t)
\end{align}
for all $k\in[K]$ and $t\in[T]$. The $N\times M$ matrix ${\bf G}_{k\hat{k}}(t)$ is the channel fading coefficient matrix between the $k$-th receiver and the $\hat{k}$-th transmitter for any $k,\hat{k}\in[K]$. The entry in the $n$-th row and $m$-th column of the matrix ${\bf G}_{k\hat{k}}(t)$ is ${G}_{k\hat{k}nm}(t)$. $\mathbf{\Gamma}_{k}(t)$ are $N\times 1$ matrices whose components are zero mean unit variance additive white Gaussian noise (AWGN) experienced by $k$-th receiver. Figure \ref{Fig1rr} illustrates a $3$-user $3\times 2$ MIMO IC. $P$ is a nominal SNR parameter that approaches infinity for GDoF characterizations. CSIR is assumed to be perfect. However, CSIT is limited to finite precision. Under finite precision CSIT we assume that $G_{k\hat{k}nm}(t)\in\mathcal{G}$ for any $k,\hat{k}\in[K],n\in[N],m\in[M]$ and $t\in[T]$, and since transmitters only know the probability density but not the realizations of channel coefficients, we assume that all ${\bf X}_k(t), t\in[T], k\in[K]$ are independent of $\mathcal{G}$.
\begin{figure}[tp]
\centering
\includegraphics[scale =0.28]{p8.pdf}
\caption{Three user $3 \times 2$ MIMO IC.}
\label{Fig1rr}
\end{figure}
\subsection{GDoF}
The definitions of achievable rates $R_i(P)$ and capacity region $\mathcal{C}(P)$ are standard. The GDoF region is defined as
\begin{eqnarray}
\mathcal{D}&=&\{(d_1,d_2,\cdots,d_K): \exists (R_1(P),R_2(P), \cdots,R_K(P))\nonumber\\
&&\in\mathcal{C}(P), \mbox{ s.t. } d_k=\lim_{P\rightarrow\infty}\frac{R_k(P)}{\frac{1}{2}\log(P)}, \forall k\in[K]\}
\end{eqnarray}
The maximum value of $d_1+d_2+\cdots+d_K$ over $\mathcal{D}$ is known as the sum GDoF value.
\section{Main Result}
\begin{theorem}\label{theorem:GDoF1} The sum GDoF value for the $K$-user symmetric MIMO IC for $M\le \frac{N}{K}$ is $KM$, and for $\frac{N}{K}\le M$ is
\begin{align}
{\sum_{k=1}^Kd_k}=&\left\{
\begin{array}{ll}
K\min(M,N)(1-\alpha)+\frac{K(N-M)^+\alpha}{K-1},&0\leq\alpha\leq \frac{1}{2}\vspace{0.03in}\\
\min\left(\frac{K}{K-1}\left((K-2)\min(M,N)(1-\alpha)+N(\alpha)\right)\right., &\vspace{0.03in}\\
N\alpha+K\min(M,N)\left(1-\alpha\right)\big),&\frac{1}{2}<\alpha\leq 1\\
\min\left(D(\alpha),K\min(M,N)\right), &1<\alpha
\end{array}
\right.\label {fg<}
\end{align}
where $N(\alpha)$ and $D(\alpha)$ are defined as,
\begin{eqnarray}
N(\alpha)&=&\min((K-1)M,N)\alpha+(N-(K-1)M)^+(1-\alpha)\\
D(\alpha)&=&(N-(K-1)M)^++\min(N,(K-1)M)\alpha
\end{eqnarray}
\end{theorem}
\begin{figure}[tp]
\centering
\includegraphics[scale =0.4]{p2c.pdf}
\caption{Sum GDoF of the three user $3 \times N$ MIMO IC.}
\label{Fig1}
\end{figure}
\begin{remark} The sum GDoF, i.e., \eqref{fg<} for $N< M$ yields,
\begin{align}
{\sum_{k=1}^Kd_k}=&{KN}\times \left\{
\begin{array}{ll}
(1-\alpha),&0\leq\alpha\leq \frac{1}{2}\vspace{0.03in}\\
\frac{K-2-(K-3)\alpha}{K-1}, &\frac{1}{2}<\alpha\leq\frac{K}{K+1}\vspace{0.03in}\\
1-(\frac{K-1}{K})\alpha,&\frac{K}{K+1}<\alpha\leq 1\\
\frac{\alpha}{K},& 1<\alpha\leq K\\
1, &K<\alpha
\end{array}
\right.\label {GDOFM>N}
\end{align}
\end{remark}
\section{Proof of Theorem \ref{theorem:GDoF1}: Converse}
The first step in the converse proof, identical to \cite{Arash_Jafar_IC},
is the transformation
into a deterministic setting such that a GDoF outer bound
on the deterministic setting is also a GDoF outer bound on
the original setting. We start directly from the deterministic
model.
\subsection{Deterministic Model}\label{DM_1}
\vspace{-1em}
\begin{align}
\bar{\mathbf{Y}}_{k}(t)&=[\bar{Y}_{k1}(t)\ \bar{Y}_{k2}(t)\ \cdots\ \bar{Y}_{kN}(t)]^\dagger\\
\bar{Y}_{kn}(t)&=L_{kn1}^b(t)\left({(\bar{\mathbf{X}}_{k}(t))}^{\max(1,\alpha)}_{\max(1,\alpha)-1}\right)+L_{kn2}^b(t)\left({(\bar{\mathbf{X}}_{j}(t))}^{\max(1,\alpha)}_{\max(1,\alpha)-\alpha},\forall j\in[K],j\neq k\right)\label{dm1}
\end{align}
for all $k\in[K],n\in[N],t\in[T]$. $\bar{\mathbf{X}}_{k}(t)$ are defined as,
\begin{align}
\bar{\mathbf{X}}_{k}(t)=&[\bar{X}_{k1}(t)\ \bar{X}_{k2}(t)\ \cdots\ \bar{X}_{kM}(t)]^\dagger\label{ggf1}
\end{align}
for any $k\in[K]$, $t\in[T]$ where $\bar{X}_{km}(t)\in\{0, 1, \cdots, {\bar{P}}^{\max(1,\alpha)}-1\}$, $\forall k\in[K],m\in[M],t\in[T]$.
\subsection{ Key Lemma}
The following lemma is the critical generalization of the AIS bound needed for Theorem \ref{theorem:GDoF1}.
\begin{lemma}\label{lemma} Define the two random variables $\bar{\bf U}_1$ and $\bar{\bf U}_2$ as,
\begin{eqnarray}
\bar{\bf U}_1&=&\left({U}_{11}^{[T]},{U}_{12}^{[T]},\cdots,{U}_{1N_1}^{[T]}\right)\label{lemmamimox1}\\
\bar{\bf U}_2&=&\left({U}_{21}^{[T]},{U}_{22}^{[T]},\cdots,{U}_{2N_2}^{[T]}\right)\label{lemmamimox2}
\end{eqnarray}
where for any $t\in[T]$, $U_{1n}(t)$ and $U_{2n}(t)$ are defined as,
\begin{eqnarray}
U_{1n}(t)&=&L_{1n}^b(t)\left((\bar{\mathbf{V}}_1(t))^{\eta}_{\eta-\lambda_{11}}\bigtriangledown(\bar{\mathbf{V}}_2(t))^{\eta}_{\eta-\lambda_{12}}\bigtriangledown\cdots\bigtriangledown(\bar{\mathbf{V}}_l(t))^{\eta}_{\eta-\lambda_{1l}}\right), \forall n\in[N_1]\label{lemmamimox3}\\
U_{2n}(t)&=&L_{2n}^b(t)\left((\bar{\mathbf{V}}_1(t))^{\eta}_{\eta-\lambda_{21}}\bigtriangledown(\bar{\mathbf{V}}_2(t))^{\eta}_{\eta-\lambda_{22}}\bigtriangledown\cdots\bigtriangledown(\bar{\mathbf{V}}_l(t))^{\eta}_{\eta-\lambda_{2l}}\right), \forall n\in[N_2]\label{lemmamimox4}
\end{eqnarray}
where $\bar{\mathbf{V}}_i(t)=\begin{bmatrix}\bar{V}_{i1}(t)&\cdots&\bar{V}_{iM_i}(t)\end{bmatrix}^\dagger$, $\bar{V}_{im}(t)\in\mathcal{X}_{\eta}$ are all independent of $\mathcal{G}$, and $0\le\lambda_{1i},\lambda_{2i}\le\eta$ for any $i\in[l]$. Without loss of generality, $(\lambda_{1i}-\lambda_{2i})^+$ are sorted in descending order, i.e., $(\lambda_{1i}-\lambda_{2i})^+\ge(\lambda_{1j}-\lambda_{2j})^+$ if $1\le i< j\le l$. Then, for any acceptable\footnote{Let $\mathcal{G}(Z)\subset\mathcal{G}$ denote the set of all bounded density channel coefficients that appear in $\bar{\bf U}_1,\bar{\bf U}_2$. $W$ is acceptable if conditioned on any $\mathcal{G}_o\subset (\mathcal{G}/\mathcal{G}(Z))\cup \{W\}$, the channel coefficients $\mathcal{G}(Z)$ satisfy the bounded density assumption. For instance, any random variable $W$ independent of $\mathcal{G}$ can be utilized in Lemma \ref{lemma}.} random variable ${W}$, if $N_1\le \min(N_2, \sum_{i=1}^lM_i)$ we have,
\begin{eqnarray}
&&H({\bar{\bf U}}_1\mid {W},\mathcal{G})-H({\bar{\bf U}}_2\mid {W},\mathcal{G})\nonumber\\
&\le&T\big((N_1-\sum_{i=1}^sM_i)(\lambda_{1,s+1}-\lambda_{2,s+1})^++\sum_{i=1}^sM_i(\lambda_{1i}-\lambda_{2i})^+\big)\log{\bar{P}}+T~o~(\log{\bar{P}})\label{lemmamimox5}
\end{eqnarray}
where $s$ must satisfy the condition $\sum_{i=1}^{s}M_i\le N_1< \sum_{i=1}^{s+1}M_i$.
\end{lemma}
Proof of Lemma \ref{lemma} is based on the AIS argument and is relegated to Appendix \ref{lemmap}.
\subsection{Some Insights For the Three User $2\times3$ MIMO IC}
\begin{figure}[!h]
\centerline{\includegraphics[width=6in]{p8d.pdf}}
\caption{ Three user $2\times3$ MIMO IC. The network is fully connected but only the channel strength parameters needed for the application of Lemma \ref{lemma} are shown in this figure.}\label{fig:intuit-}
\end{figure}
To gain some insights into the application of Lemma \ref{lemma}, consider the three user $2\times3$ MIMO IC illustrated in Figure \ref{fig:intuit-} for $\alpha\le1$. To apply Lemma \ref{lemma}, the random variables $\bar{\bf U}_1$, $\bar{\bf U}_2$, $\bar{\bf V}_1^{[T]}$, $\bar{\bf V}_2^{[T]}$ and $W$ are interpreted as $\bar{\bf Y}_2^{[T]}$, $\bar{\bf Y}_1^{[T]}$, $\bar{\bf X}_2^{[T]}$, $\bar{\bf X}_3^{[T]}$ and $\bar{\bf X}_1^{[T]}$, respectively. The first user receives the top $\alpha$ power levels of $\bar{\bf X}_2^{[T]}$ and $\bar{\bf X}_3^{[T]}$ while second reciever sees the top $1$ power levels of $\bar{\bf X}_2^{[T]}$ and the top $\alpha$ power levels of $\bar{\bf X}_3^{[T]}$. So we have $\eta=1, \lambda_{11}=1, \lambda_{21}=\alpha, \lambda_{12}=\alpha, \lambda_{22}=\alpha$. Therefore, $(\lambda_{11}-\lambda_{21})^+=1-\alpha$ and $(\lambda_{12}-\lambda_{22})^+=0$. From Lemma \ref{lemma} we conclude,
\begin{align}
&H(\bar{\bf Y}_2^{[T]}\mid \bar{\bf X}_1^{[T]},\mathcal{G})-H(\bar{\bf Y}_1^{[T]}\mid \bar{\bf X}_1^{[T]},\mathcal{G})\nonumber\\
\le&T\big(1\times 0+2\times(1-\alpha)\big)\log{\bar{P}}+T~o~(\log{\bar{P}})\label{jl0}
\end{align}
Let us also explain how intuitively we expect \eqref{jl0} to be true as well. Conditioned on $\bar{\bf X}_{1}^{[T]}$, $\bar{\bf Y}_{2}(t)$ is a linear combination of $\bar{\bf X}_{2}(t)$ and $(\bar{\bf X}_{3}(t))^{\alpha}$ while $\bar{\bf Y}_{1}(t)$ is a linear combination of $(\bar{\bf X}_{2}(t))^{\alpha}$ and $(\bar{\bf X}_{3}(t))^{\alpha}$. Consider the channel illustrated in Figure \ref{fig:intuit-}. First of all, observe that $\bar{\bf X}_{2}(t)$ appears in $\bar{\bf Y}_{2}(t)$ with the signal strength levels $1$ and appears in $\bar{\bf Y}_{1}(t)$ with the signal strength levels $\alpha$. Thus, due to the bounded density assumption the maximum difference of $2(1-\alpha)$ is possible in the GDoF sense between the two entropies. Note that, $\bar{\bf X}_{3}(t)$ appears in both the received signals $\bar{\bf Y}_{1}(t)$ and $\bar{\bf Y}_{2}(t)$ with the same signal strength levels of $\alpha$. Therefore, it cannot contribute positive difference of entropies as in the finite precision CSIT no interference alignment is possible.
Similarly, from Lemma \ref{lemma} we have,
\begin{align}
H(\bar{\bf Y}_3^{[T]}\mid \bar{\bf X}_1^{[T]},\bar{\bf X}_2^{[T]},\mathcal{G})-H(\bar{\bf Y}_2^{[T]}\mid \bar{\bf X}_1^{[T]},\bar{\bf X}_2^{[T]},\mathcal{G})\le&2T(1-\alpha)\log{\bar{P}}+T~o~(\log{\bar{P}})\label{jl00}
\end{align}
On the other hand, writing Fano's inequality for all the three users (and suppressing $o(T)$ terms for simplicity) we obtain the following bounds,
\begin{eqnarray}
TR_1&\le& H(\bar{\bf Y}_{1}^{[T]}\mid \mathcal{G})- H(\bar{\bf Y}_{1}^{[T]}\mid \bar{\bf X}_{1}^{[T]},\mathcal{G})\label{jl1}\\
TR_2&\le& H(\bar{\bf Y}_{2}^{[T]}\mid \bar{\bf X}_{1}^{[T]},\mathcal{G})- H(\bar{\bf Y}_{2}^{[T]}\mid \bar{\bf X}_{1}^{[T]},\bar{\bf X}_{2}^{[T]},\mathcal{G})\label{jl2}\\
TR_3&\le& H(\bar{\bf Y}_{3}^{[T]}\mid \bar{\bf X}_{1}^{[T]},\bar{\bf X}_{2}^{[T]},\mathcal{G})\label{jl3}
\end{eqnarray}
Therefore, for $\alpha\le1$, from \eqref{jl0}-\eqref{jl3} we have,
\begin{eqnarray}
TR_1+TR_2+TR_3&\le& H(\bar{\bf Y}_{1}^{[T]}\mid \mathcal{G})+4T(1-\alpha)\log{\bar{P}}+T~o~(\log{\bar{P}})\\
&\le&T(6-3\alpha)\log{\bar{P}}+T~o~(\log{\bar{P}})\label{oi}
\end{eqnarray}
\eqref{oi} is true as discrete entropy of any discrete random variable is bounded by logarithm of its cardinality.
\subsection{Equivalent Bounds}
Theorem \ref{theorem:GDoF1} is concluded from the following bounds,
\begin{enumerate}
\item If $\alpha\in\mathbb{R}^+,\alpha\le\frac{1}{2}$, then
\begin{eqnarray}
&&\sum_{k=1}^Kd_k\nonumber\\
&\le& \frac{K\left(\min(M,N)(1-\alpha)+(N-M)^+\alpha\right)+K(K-2)\min(M,N)(1-\alpha)}{K-1}\label{b1}
\end{eqnarray}
\item If $\alpha\in\mathbb{R}^+,\frac{1}{2}\le\alpha\le1$, then
\begin{eqnarray}
&&\sum_{k=1}^Kd_k\nonumber\\
&\le& \frac{K\left(\min((K-1)M,N)\alpha+(N-(K-1)M)^+(1-\alpha)\right)+K(K-2)\min(M,N)(1-\alpha)}{K-1}\nonumber\\
&&\label{b1+}
\end{eqnarray}
\item If $\alpha\in\mathbb{R}^+,\alpha\le1$, then
\begin{eqnarray}
\sum_{k=1}^Kd_k&\le& N\alpha+K\min(M,N)(1-\alpha)\label{b2}
\end{eqnarray}
\item If $\alpha\in\mathbb{R}^+,1\le\alpha$, then
\begin{eqnarray}
\sum_{k=1}^Kd_k&\le& (N-(K-1)M)^++\min(N,(K-1)M)\alpha\label{b2+}
\end{eqnarray}
\item For any $\alpha\in \mathbb{R}^+$,
\begin{eqnarray}
\sum_{k=1}^Kd_k&\le& K\min(M,N)\label{b3}
\end{eqnarray}
\end{enumerate}
Thus, in order to prove Theorem \ref{theorem:GDoF1}, the Bounds \eqref{b1}-\eqref{b3} should be proved.
\subsection{Proof of Bounds \eqref{b1}-\eqref{b3}}\label{mt}
The last bound, $\sum_{k=1}^Kd_k\le K\min(M,N)$ is the trivial combination of single user bounds. Let us prove the other four bounds, i.e., \eqref{b1}-\eqref{b2+}.
\begin{enumerate}
\item {\bf Proof of \eqref{b1} and \eqref{b1+}}\\
Writing Fano's Inequality for the first $K-1$ receivers we have,
\begin{eqnarray}
TR_1&\le& I(\bar{\bf Y}^{[T]}_{1};\bar{\bf X}^{[T]}_{1}\mid \mathcal{G})\label{c1}\\
TR_k&\le& I(\bar{\bf Y}^{[T]}_{k};\bar{\bf X}^{[T]}_{k}\mid \bar{\bf X}^{[T]}_{1},\cdots,\bar{\bf X}^{[T]}_{k-1},\mathcal{G}), \forall k\in[K-1],k\neq 1\label{c2}
\end{eqnarray}
{Summing} \eqref{c1} and \eqref{c2}, we have,
\begin{eqnarray}
&&T\sum_{k=1}^{K-1}R_k\nonumber\\
&\le& I(\bar{\bf Y}^{[T]}_{1};\bar{\bf X}^{[T]}_{1}\mid \mathcal{G})+\sum_{k=2}^{K-1} I(\bar{\bf Y}^{[T]}_{k};\bar{\bf X}^{[T]}_{k}\mid \bar{\bf X}^{[T]}_{1},\cdots,\bar{\bf X}^{[T]}_{k-1},\mathcal{G})\\
&=& H(\bar{\bf Y}^{[T]}_{1}\mid \mathcal{G})-H\left(\bar{\bf X'}^{[T]}_{K}\mid \mathcal{G}\right)\nonumber\\
&&+\sum_{k=2}^{K-1} \left(H(\bar{\bf Y}^{[T]}_{k}\mid \bar{\bf X}^{[T]}_{1},\cdots,\bar{\bf X}^{[T]}_{k-1},\mathcal{G})-H(\bar{\bf Y}^{[T]}_{k-1}\mid \bar{\bf X}^{[T]}_{1},\cdots,\bar{\bf X}^{[T]}_{k-1},\mathcal{G})\right)\\
&\le& H(\bar{\bf Y}^{[T]}_{1}\mid \mathcal{G})-H\left(\bar{\bf X'}^{[T]}_{K}\mid \mathcal{G}\right)+\sum_{k=2}^{K-1} T\min(M,N)(1-\alpha)\log{\bar{P}}+T~o(\log{\bar{P}})\label{c3}
\end{eqnarray}
where the new random variable, $\bar{\bf X'}_k(t)$ is defined as\\
\begin{eqnarray}
\bar{\bf X'}_k(t)&=&\begin{bmatrix}\bar{X'}_{k1}(t)&\bar{X'}_{k2}(t)&\cdots&\bar{X'}_{kN}(t)\end{bmatrix}^\dagger\\
\bar{X'}_{kn}(t)&=&{L}_{kn3}^b(t)\left((\bar{\bf X}_{k}(t))^{\alpha}\right), \forall n\in[N]
\end{eqnarray}
Let us explain how Lemma \ref{lemma} {yields} \eqref{c3}. Substitute the random variables $\bar{\bf U}_1$, $\bar{\bf U}_2$, $\bar{\bf V}_1^{[T]}$, $\bar{\bf V}_2^{[T]}$ and $W$ in Lemma \ref{lemma} with $\bar{\bf Y}^{[T]}_{k}$, $\bar{\bf Y}^{[T]}_{k-1}$, $\bar{\bf X}_k^{[T]}$, $\left(\bar{\bf X}_{j}^{[T]},j\in[K],j\notin[k]\right)$ and $\left(\bar{\bf X}_{j}^{[T]},j\in[k-1]\right)$, respectively. Next, we set $\eta=1, \lambda_{11}=1, \lambda_{21}=\alpha, \lambda_{12}=\alpha, \lambda_{22}=\alpha,M_1=M,M_2=(K-k)M,N_1=N_2=N$. Thus, we have $(\lambda_{11}-\lambda_{21})^+=1-\alpha$ and $(\lambda_{12}-\lambda_{22})^+=0$. Therefore, from Lemma \ref{lemma}, \eqref{c3} is concluded. Similar to \eqref{c3}, by symmetry we have,
\begin{eqnarray}
&&T\sum_{k\in[K],k\neq j}R_k\nonumber\\
&\le& H(\bar{\bf Y}^{[T]}_{j+1}\mid \mathcal{G})-H\left(\bar{\bf X'}^{[T]}_{j}\mid \mathcal{G}\right)\nonumber\\
&&+(K-2) T\min(M,N)(1-\alpha)\log{\bar{P}}+T~o(\log{\bar{P}})\label{c4}
\end{eqnarray}
for all $j\in[K]$. Summing \eqref{c4} for all $j\in[K]$ we have,
\begin{eqnarray}
&&T(K-1)\sum_{k=1}^KR_k\nonumber\\
&=&T\sum_{j=1}^K\sum_{k\in[K],k\neq j}R_k\nonumber\\
&\le& \sum_{k=1}^K\left(H(\bar{\bf Y}^{[T]}_{k}\mid \mathcal{G})-H\left(\bar{\bf X'}^{[T]}_{k}\mid \mathcal{G}\right)\right)+TK(K-2) \min(M,N)(1-\alpha)\log{\bar{P}}\nonumber\\
&&+T~o(\log{\bar{P}})\label{xx}
\end{eqnarray}
Now, let us consider the two cases of $\alpha\le\frac{1}{2}$ and $\frac{1}{2}\le\alpha\le1$ separately.
\begin{enumerate}
\item{$\alpha\le\frac{1}{2}$}
\begin{eqnarray}
&&H(\bar{\bf Y}^{[T]}_{k}\mid \mathcal{G})-H\left(\bar{\bf X'}^{[T]}_{k}\mid \mathcal{G}\right)\nonumber\\
&\le& T\left(\min(M,N)(1-\alpha)+(N-M)^+\alpha\right)\log{\bar{P}}+T~o(\log{\bar{P}})\label{c5}
\end{eqnarray}
Let us explain how \eqref{c5} follows from Lemma \ref{lemma}. Substitute the random variables $\bar{\bf U}_1$, $\bar{\bf U}_2$, $\bar{\bf V}_1^{[T]}$ and $\bar{\bf V}_2^{[T]}$ with $\bar{\bf Y}^{[T]}_{k}$, $\bar{\bf X'}^{[T]}_{k}$, $\bar{\bf X}_k^{[T]}$ and $\left(\bar{\bf X}_{j}^{[T]},j\in[K],j\neq k\right)$, respectively. Moreover, setting $\eta=1, \lambda_{11}=1, \lambda_{21}=\alpha, \lambda_{12}=\alpha, \lambda_{22}=0$, we have $(\lambda_{11}-\lambda_{21})^+=1-\alpha$ and $(\lambda_{12}-\lambda_{22})^+=\alpha$. Therefore, from Lemma \ref{lemma} we conclude \eqref{c5}. From \eqref{xx} and \eqref{c5}, \eqref{b1} is concluded.
\item{$\frac{1}{2}\le\alpha\le1$}
\begin{eqnarray}
&&H(\bar{\bf Y}^{[T]}_{k}\mid \mathcal{G})-H\left(\bar{\bf X'}^{[T]}_{k}\mid \mathcal{G}\right)\nonumber\\
&\le& T\left(\min((K-1)M,N)\alpha+(N-(K-1)M)^+(1-\alpha)\right)\log{\bar{P}}+T~o(\log{\bar{P}})\label{c5g}
\end{eqnarray}
\eqref{c5g} follows from Lemma \ref{lemma} similar to \eqref{c5}. Substitute the random variables $\bar{\bf U}_1$, $\bar{\bf U}_2$, $\bar{\bf V}_1^{[T]}$ and $\bar{\bf V}_2^{[T]}$ with $\bar{\bf Y}^{[T]}_{k}$, $\bar{\bf X'}^{[T]}_{k}$, $\left(\bar{\bf X}_{j}^{[T]},j\in[K],j\neq k\right)$ and $\bar{\bf X}_k^{[T]}$, respectively. The rest of the proof is concluded similar to \eqref{c5}. From \eqref{xx} and \eqref{c5g}, \eqref{b1+} is concluded.
\end{enumerate}
\item {\bf Proof of \eqref{b2}}\\
Summing \eqref{c1} and \eqref{c2}, we have,
\begin{eqnarray}
&&T\sum_{k=1}^{K}R_k\nonumber\\
&\le& I(\bar{\bf Y}^{[T]}_{1};\bar{\bf X}^{[T]}_{1}\mid \mathcal{G})+\sum_{k=2}^{K} I(\bar{\bf Y}^{[T]}_{k};\bar{\bf X}^{[T]}_{k}\mid \bar{\bf X}^{[T]}_{1},\cdots,\bar{\bf X}^{[T]}_{k-1},\mathcal{G})\label{c6+}\\
&=& H(\bar{\bf Y}^{[T]}_{1}\mid \mathcal{G})\nonumber\\
&&+\sum_{k=2}^{K} \left(H(\bar{\bf Y}^{[T]}_{k}\mid \bar{\bf X}^{[T]}_{1},\cdots,\bar{\bf X}^{[T]}_{k-1},\mathcal{G})-H(\bar{\bf Y}^{[T]}_{k-1}\mid \bar{\bf X}^{[T]}_{1},\cdots,\bar{\bf X}^{[T]}_{k-1},\mathcal{G})\right)\\
&\le& H(\bar{\bf Y}^{[T]}_{1}\mid \mathcal{G})+\sum_{k=2}^{K} T\min(M,N)(1-\alpha)\log{\bar{P}}+T~o(\log{\bar{P}})\label{c6}\\
&\le& T\left((N-M)^+\alpha+\min(M,N)\right)\log{\bar{P}}+\sum_{k=2}^{K} T\min(M,N)(1-\alpha)\log{\bar{P}}\nonumber\\
&&+T~o(\log{\bar{P}})\label{cv6}\\
&=& T(N\alpha+K\min(M,N)(1-\alpha))\log{\bar{P}}+T~o(\log{\bar{P}})\label{c7}
\end{eqnarray}
\eqref{c6} follows similar to \eqref{c3} and \eqref{cv6} is concluded as the entropy of a discrete random variable is bounded by logarithm of the cardinality of its support, i.e., \footnote{\eqref{cv6} follows from Lemma \ref{lemma} by substituting $\bar{\bf U}_1$ and $\bar{\bf U}_2$ with $\bar{\bf Y}^{[T]}_{1}$ and ${ C}^{[T]}$ where ${\bf C}^{[T]}$ is a $T$-letter constant variable. Then, substituting $\bar{\bf V}_1^{[T]}$ and $\bar{\bf V}_2^{[T]}$ with $\bar{\bf X}_1^{[T]}$ and $\left(\bar{\bf X}_{j}^{[T]},j\in[K],j\neq 1\right)$, \eqref{cv6} is concluded. Here, we assume $\eta=1, \lambda_{11}=1, \lambda_{21}=\alpha, \lambda_{12}=0, \lambda_{22}=0,,M_1=M,M_2=(K-1)M,N_1=N_2=N$.}
\begin{eqnarray}
H(\bar{\bf Y}^{[T]}_{1}\mid \mathcal{G})&\le& T\left((N-M)^+\alpha+\min(M,N)\right)\log{\bar{P}}
\end{eqnarray}
Dividing \eqref{c7} by $T\log{\bar{P}}$, \eqref{b2} is obtained.
\item {\bf Proof of \eqref{b2+}}\\
Similarly, from (\eqref{c6+}-\eqref{c6}) we have,
\begin{eqnarray}
&&T\sum_{k=1}^{K}R_k\nonumber\\
&\le& H(\bar{\bf Y}^{[T]}_{1}\mid \mathcal{G})+\sum_{k=2}^{K} T\min(M,N)(1-\alpha)^+\log{\bar{P}}+T~o(\log{\bar{P}})\\
&\le& H(\bar{\bf Y}^{[T]}_{1}\mid \mathcal{G})+T~o(\log{\bar{P}})\label{c6-}\\
&\le& (N-(K-1)M)^+T\log{\bar{P}}+\min(N,(K-1)M)\alpha T\log{\bar{P}}+T~o(\log{\bar{P}})\label{c6--}
\end{eqnarray}
\eqref{c6-} is true as $1\le\alpha$ and \eqref{c6--} follows similar\footnote{\eqref{c6--} follows similar to \eqref{cv6} from Lemma \ref{lemma}. Substitute $\bar{\bf U}_1$, $\bar{\bf U}_2$, $\bar{\bf V}_1^{[T]}$ and $\bar{\bf V}_2^{[T]}$ with $\bar{\bf Y}^{[T]}_{1}$, ${ C}^{[T]}$, $\left(\bar{\bf X}_{j}^{[T]},j\in[K],j\neq 1\right)$ and $\bar{\bf X}_1^{[T]}$ and assume $\eta=\alpha, \lambda_{11}=\alpha, \lambda_{21}=1, \lambda_{12}=0, \lambda_{22}=0,M_1=(K-1)M,M_2=M,N_1=N_2=N$.} to \eqref{cv6}. Dividing \eqref{c6--} by $T\log{\bar{P}}$, \eqref{b2+} is obtained.
\end{enumerate}
\section{Proof of Theorem \ref{theorem:GDoF1}: Achievability}
\subsection{A Useful Lemma}
Consider a $(M_1+M_2)$-user multiple access channel (MAC) where each transmitter is equipped with a single antenna, the receiver has $N$ antennas, $N< M_1+M_2$, and the $N\times 1$ received signal vector ${\bf Q}$ is represented as,
\begin{align}
{\bf Q}=&\sqrt{P}\sum_{k=1}^{M_1} {\bf H}_k{ T}_k+\sqrt{P^{\alpha}}\sum_{k=M_1+1}^{M_1+M_2} {\bf H}_k{ T}_k+\sum_{n=1}^{N} \sqrt{P^{\alpha_n}}{\bf G}_nZ_n\label{mac0}
\end{align}
where $T_1, T_2, \cdots, T_{M_1+M_2}$ are the transmitted signals, and $Z_1,Z_2,\cdots,Z_N$ are i.i.d. Gaussian zero mean unit variance noise terms. The ${\bf H}_k, {\bf G}_n$ are $N\times 1$ generic vectors, i.e., generated from continuous distributions with bounded density, so that any $N$ of them are linearly independent almost surely. The transmit power constraint is expressed as,
\begin {eqnarray}
\mbox{E}{|T_{k}|}^2&\leq&P^{-\eta_k},~\forall k\in[M_1+M_2]\label{mac1}
\end{eqnarray}
where for any $k\in[M_1+M_2]$, $\eta_k$ is a non-negative integer. Further, define $\gamma_k$ for $k\in[M_1+M_2]$ as,
\begin {eqnarray}
\gamma_k&=&\left\{
\begin{array}{ll}
{(1-\eta_k)}^+,&k\in[M_1]\\
{(\alpha-\eta_k)}^+, &\text{Otherwise}
\end{array}
\right.\label{gamma1}
\end{eqnarray}
Thus $\gamma_k$ is the received power level of user $k$ in the GDoF sense. The GDoF region $\mathcal{D}'$ is defined as
\begin{align}
\mathcal{D}'\triangleq&\{(d'_1,d'_2,\cdots,d'_{M_1+M_2}): \exists (R'_1(P),R'_2(P),\cdots,R'_{M_1+M_2}(P))\in\mathcal{C}'(P),\nonumber\\
& \mbox{ s.t. } d'_k=\lim_{P\rightarrow\infty}\frac{R'_k(P)}{\frac{1}{2}\log{(P)}}, \forall k\in[M_1+M_2]\} \label {region}
\end{align}
where $\mathcal{C}'(P)$ is the capacity region of MAC described in (\ref{mac0}).
\begin{lemma}\label{lemma:mac} The GDoF tuple $(d'_1, d'_2, \cdots, d'_{M_1+M_2})$ is achievable in the multiple access channel described above if $\forall k\in[M_1+M_2]$, and $\forall S\subset[M_1+M_2]$ where $|S|=k$,
\begin{align}
\sum_{i\in S} d'_i
&\le\max_{S_2\in S,|S_2|=\min(k,N)}\sum_{i\in S_2} \gamma_i -\min_{S_1\in[N],|S_1|=\min(k,N)}\sum_{i\in S_1} \alpha_i \label{mac3}
\end{align}
\end{lemma}
For proof of Lemma \ref{lemma:mac} see \cite{Arash_Jafar_MIMOsym_ArXiv}. It is sufficient to derive the achievability for Theorem \ref{theorem:GDoF1}, as Theorem \ref{theorem:GDoF1} is automatically concluded from it.
\subsection{Proof of Achievability in Theorem \ref{theorem:GDoF1}} \label{app3}
Now, let us achieve the bound (\ref{fg<}). We will suppress the time-index $t$ in this section to simplify the notation. For any $k\in[K]$ user $k$'s message $W_k$ is split into messages $(W_{kc},W_{kp})$, representing common message and private message, respectively. Let us consider the three cases of $\alpha\le\frac{1}{2}$, $\frac{1}{2}\le\alpha\le1$, and $1\le\alpha$ separately as,
\begin{enumerate}
\item {$\alpha\le\frac{1}{2}$.} Our goal here is to achieve $\min(M,N)(1-\alpha)+\frac{(N-M)^+\alpha}{K-1}$ GDoF per user where results in $K\min(M,N)(1-\alpha)+\frac{K(N-M)^+\alpha}{K-1}$ GDoF totally. In order to achieve $\min(M,N)(1-\alpha)+\frac{(N-M)^+\alpha}{K-1}$ GDoF per user, for any $k\in[K]$ the public message $W_{kc}$ is encoded into Gaussian codebooks $U_{k1c},U_{k2c},\cdots,U_{kMc}$ with powers $\mbox{E}{|U_k|}^2=1-P^{-\alpha}$ each carrying $\frac{(N-M)^+\alpha}{(K-1)M}$ GDoF. These codewords are transmitted through $M$ antennas along $M\times 1$ generic unit vectors ${\bf V}_{k1},{\bf V}_{k2},\cdots,{\bf V}_{k{M}}$. The private message $W_{kp}$ is encoded into Gaussian codebooks $U_{k1p},U_{k2p},\cdots,U_{k\min(M,N)p}$ with powers $\mbox{E}{|U_{kjp}|}^2=P^{-\alpha}$ for any $j\in[\min(M,N)]$ so that the total power per transmitter is unity. These codewords are transmitted through $\min(M,N)$ antennas along the $M\times 1$ generic unit vectors ${\bf V}_{k1},{\bf V}_{k2},\cdots,{\bf V}_{k{\min(M,N)}}$. Each of the private messages is carrying $1-\alpha$ GDoF. The transmitted and received signals are,
\begin{align}
\mathbf{X}_{k}=&\sum_{j=1}^{M}\mathbf{V}_{kj}{U}_{kjc}+\sum_{j=1}^{\min{M,N}}\mathbf{V}_{kj}{U}_{kjp}\label{TR21a-}\\
\mathbf{Y}_{k}=&\sqrt{P}{\bf G}_{kk}\mathbf{X}_{k}+\sum_{j=1,j\neq k}^{K}\sqrt{P^{\alpha}}{\bf {G}}_{kj}\mathbf{X}_{j}+\mathbf{\Gamma}_{k}\label{TR21a}
\end{align}
Using Lemma \ref{lemma:mac} we claim that each receiver, e.g., receiver $1$ can decode all the signals $U_{kjc}$ and $U_{1jp}$ for all $k\in[K]$ and $j\in[M]$ treating all the other signals as noise. Set the variables $M_1=M+\min(M,N)$, $M_2=(K-1)M$ and $\alpha_n=0$ for all $n\in[N]$. Moreover, define the codewords $T_{1},\cdots,T_{KM+\min(M,N)}$ as
\begin {align}
T_j=&\left\{
\begin{array}{ll}
U_{1jc},&1\le j\le {M}\\
U_{2(j-M)c},&M<j\le 2M\\
\vdots &\vdots\\
U_{K(j-(K-1)M)c},&(K-1)M<j\le KM\\
U_{1(j-KM)p},&KM<j\le KM+\min(M,N)
\end{array}
\right.
\end{align}
From \eqref{gamma1}, $\gamma_1,\cdots,\gamma_{KM+\min(M,N)}$ are derived as,
\begin {align}
\gamma_j=&\left\{
\begin{array}{ll}
1,&1\le j\le M\\
\alpha,&M< j\le KM\\
1-\alpha,&KM< j\le KM+\min(M,N)
\end{array}
\right.\label{ffd2}
\end{align}
Note that $N\le KM$ and $\alpha\le\frac{1}{2}$. Thus, from the received signal in (\ref{TR21a}), $T_1,\cdots,T_{KM+\min(M,N)}$ are decoded by first receiver as (\ref{mac3}) is satisfied for all $k\in[KM+\min(M,N)]$. For instance if we set $k=KM+\min(M,N)$, the condition (\ref{mac3}) is equivalent to,
\begin{align}
\sum_{i\in S} d'_i=&\frac{K(N-M)^+\alpha}{K-1}+\min(M,N)(1-\alpha)\nonumber\\
\le&\min(M,N)+(N-M)^+\alpha=\max_{S_2\in S,|S_2|=\min(k,N)}\sum_{i\in S_2} \gamma_i .
\end{align}
\item{$\frac{1}{2}<\alpha\leq1$.} Let us achieve $d$ GDoF where $d$ is equal to,
\begin{align}
d=\min\left(\frac{K}{K-1}\left((K-2)\min(M,N)(1-\alpha)+N(\alpha)\right),N\alpha+K\min(M,N)\left(1-\alpha\right)\right)
\end{align}
Similar to the case $\alpha\le\frac{1}{2}$, the public message $W_{kc}$ is encoded into Gaussian codebooks $U_{k1c},U_{k2c},\cdots,U_{kMc}$ with powers $\mbox{E}{|U_k|}^2=1-P^{-\alpha}$ each carrying $\frac{d-K\min(M,N)(1-\alpha)}{KM}$ GDoF. These codewords are transmitted through $M$ antennas along $M\times 1$ generic unit vectors ${\bf V}_{k1},{\bf V}_{k2},\cdots,{\bf V}_{k{M}}$. The private ~message $W_{kp}$ ~is encoded~ into Gaussian~ codebooks ~$U_{k1p},U_{k2p},\cdots,U_{k\min(M,N)p}$ with powers $\mbox{E}{|U_{kjp}|}^2=P^{-\alpha}$ for any $j\in[\min(M,N)]$. These codewords are transmitted through $\min(M,N)$ antennas along the $M\times 1$ generic unit vectors ${\bf V}_{k1},{\bf V}_{k2},\cdots,{\bf V}_{k{\min(M,N)}}$. Each of the private messages is carrying $1-\alpha$ GDoF. The transmitted and received signals follows the same as \eqref{TR21a-} and \eqref{TR21a}. From Lemma \ref{lemma:mac} each receiver, e.g., receiver $1$ can decode all the codewords $U_{kjc}$ and $U_{1jn}$ for all $k\in[K]$ and $j\in[M]$ treating all the other signals as noise. The details how receiver $1$ can decode all these codewords follows the same as the case $\alpha\le\frac{1}{2}$.
\item{$1\le\alpha$.} In this case, $\min\left(D(\alpha),K\min(M,N)\right)$ is achieved as follows. Recall that $D(\alpha)$ was defined as $(N-(K-1)M)^++\min(N,(K-1)M)\alpha$. All $\min(M,N)$ messages of each transmitter are
public in this case and are encoded into Gaussian codebooks
$U_{k1c},U_{k2c},\cdots,U_{k\min(M,N)c}$ with unit powers each carrying $\min\left(\frac{D(\alpha)}{K\min(M,N)},1\right)$ GDoF. These codewords are transmitted through $\min(M,N)$ antennas along the $M\times 1$ generic unit vectors ${\bf V}_{k1},{\bf V}_{k2},\cdots,{\bf V}_{k{\min(M,N)}}$. The transmitted and received signals are concluded similar to \eqref{TR21a-} and \eqref{TR21a} as,
\begin{align}
\mathbf{X}_{k}=&\sum_{j=1}^{M}\mathbf{V}_{kj}{U}_{kjc}\label{TR21a-1}\\
\mathbf{Y}_{k}=&\sqrt{P}{\bf G}_{kk}\mathbf{X}_{k}+\sum_{j=1,j\neq k}^{K}\sqrt{P^{\alpha}}{\bf {G}}_{kj}\mathbf{X}_{j}+\mathbf{\Gamma}_{k}\label{TR21a1}
\end{align}
Each receiver, e.g., receiver $1$ can decode all the signals $U_{kjc}$ for all $k\in[K]$ and $j\in[\min(M,N)]$ treating all the other signals as noise. Set the variables $M_1=\min(M,N)$, $M_2=(K-1)\min(M,N)$ and $\alpha_n=0$ for all $n\in[N]$ in Lemma \ref{lemma:mac} . Moreover, define the codewords $T_{1},\cdots,T_{K\min(M,N)}$ as
\begin {align}
T_j=&\left\{
\begin{array}{ll}
U_{1jc},&1\le j\le {\min(M,N)}\\
U_{2(j-\min(M,N))c},&\min(M,N)<j\le 2\min(M,N)\\
\vdots &\vdots\\
U_{K(j-(K-1)\min(M,N))c},&(K-1)\min(M,N)<j\le K\min(M,N)
\end{array}
\right.
\end{align}
From \eqref{gamma1}, $\gamma_1,\cdots,\gamma_{K\min(M,N)}$ are derived as,
\begin {align}
\gamma_j=&\left\{
\begin{array}{ll}
1,&1\le j\le \min(M,N)\\
\alpha,&\min(M,N)< j\le K\min(M,N)
\end{array}
\right.\label{ffd2}
\end{align}
Similar to the case $\alpha\le \frac{1}{2}$, $T_1,\cdots,T_{K\min(M,N)}$ are decoded by first receiver as (\ref{mac3}) is satisfied for all $k\in[K\min(M,N)]$. For instance if we set $k=K\min(M,N)$, the condition (\ref{mac3}) is equivalent to,
\begin{align}
\sum_{i\in S} d'_i=&K\min(M,N)\times\min\left(\frac{D(\alpha)}{K\min(M,N)},1\right)\nonumber\\
\le&\min\left((K-1)M,N\right)\alpha+\left(N-(K-1)M\right)^+=\max_{S_2\in S,|S_2|=\min(k,N)}\sum_{i\in S_2} \gamma_i
\end{align}
\end{enumerate}
\section{Conclusion}
Symmetric $K$-user MIMO IC with $M$ antennas at each transmitter and $N$ antennas at each receiver is considered. Sum GDoF of this channel is derived. The Sum GDoF is found to be a $W$ curve as a function of $\alpha$ for fixed $M$ and $N$ similar to the SISO case. Outer bound proof is obtained with the help of a key lemma that generalizes the AIS argument. The achievability follows from the achievability of the GDoF region of a MAC, combined with the `treating interference as noise' scheme.
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{"url":"http:\/\/www.askamathematician.com\/2011\/04\/q-how-do-you-calculate-6212-or-48293-whats-the-deal-with-this-orders-of-operation-business\/","text":"# Q: How do you calculate 6\/2(1+2) or 48\/2(9+3)? What\u2019s the deal with this orders of operation business?\n\nMathematician: Now that the Physicist and I have answered questions like these ones at least 9 times via email, I figure we should get this horrible topic out of the way once and for all with a short post.\n\nThe \u201corder of operations\u201d tells us the sequence in which we should carry out mathematical operations. It is merely a matter of convention (not a matter of right or wrong), giving us a standardized way to decide whether 6\/2*3 is (6\/2)*3 or 6\/(2*3). The convention has been accepted just so that when different people see an equation they can agree on its meaning, even if they hate each other\u2019s guts. You could come up with a different convention, but why bother? Let other people waste their time coming up with arbitrary conventions so that you can merely waste time reading posts about them.\n\nThe rules to follow are:\n\n1. Things in parenthesis get computed before things not in parenthesis (this is because of the (little known fact) that parentheses kick ass).\n\n2. Then, exponents get dealt with (including roots). Hence a^b*c = (a^b)*c.\n\n3. Next, multiplication and division get carried out from left to right. So a*b\/c*d is ((a*b)\/c)*d. Note that division can be thought of as still carrying out a multiplication since a\/b = a*(1\/b), so it makes a certain sense that neither division nor multiplication takes precedence over one another.\n\n4. Finally, additions and subtractions get carried out from left to right. Subtraction can be thought of as being like addition since a \u2013 b = a + (-b) so it makes sense that subtraction is treated on an equal footing with addition.\n\nPeople sometimes use the expressions PEMDAS, \u201cPlease Excuse My Dear Aunt Sally\u201d or \u201cPuke Exhaust Mud Dolls Autumn Sasquatch\u201d\u00a0to remember these rules. These all stand for \u201cParenthesis, Exponents, Multiplication and Division, Addition and Subtraction\u201d. It is easy to get confused by these\u00a0mnemonics\u00a0though because they don\u2019t capture the fact that multiplication is treated the same way as division (carried out left to right), and addition is treated the same way as subtraction (carried out left to right). That\u2019s why I prefer: \u201cPuke Exhaust Mud And Dolls Autumn And Sasquatch\u201d\n\nOne interesting thing to note about the order of operations is that there is nothing interesting about them whatsoever. But that implies there is something interesting about them. This is a contradiction, and hence the order of operations form a paradox. They are also about as fun as\u00a0getting to watch a fish crawl up the stairs as a reward for answering math questions.\n\nBut, that won\u2019t stop me from giving an example. Consider:\n\na^b*c\/d+e-f.\n\nWithout a convention, it has many possible interpretations, such as\n\na^(b*(c\/(d+e)))-f\n\nor\n\n(a^b)*((c\/(d+e))-f)\n\nwhich could be horribly confusing for people trying to communicate with each other. But, the correct interpretation according to the rules mentioned above, is:\n\n(((((a^b)*c)\/d)+e)-f).\n\nWhen in doubt about what the orders of operation are for an expression, there is a simple solution which works even better than tattooing\u00a0PEMDAS to your face. Just paste the expression into the google search bar and hit \u201csearch\u201d. For instance, if I paste in:\n\n4^2*3\/6+1-5\n\nit gives back\n\n(((4^2) * 3) \/ 6) + 1 \u2013 5 = 4.\n\nNot only is this the correct answer, but it shows you explicitely the order of operations that were used, so it\u2019s useful for learning.\n\nGoogle even gets this one right:\n\n4^2*3\/6 + 1 \u2013 5\/14*3 + 6\/10 \u2013 4*2\/14*6^3\/18 + 14\n\nwhich is the kind of question little sisters give you when they are old enough to finally sort of get what a mathematician is.\n\nIt\u2019s important to note that not all calculators handle the order of operations in the standard way. Why? Because people who make calculators hate children. The preceding sentence\u00a0was not intended to be a factual statement.\n\nAlso, a lot of programming languages use the standard order of operation rules, but then you have exceptions like Smalltalk.\n\nOh, and in case you were wondering:\n\na^b^c \u00a0= $a^{b^c}$.\n\nThis entry was posted in -- By the Mathematician, Conventions, Math. Bookmark the permalink.\n\n### 261 Responses to Q: How do you calculate 6\/2(1+2) or 48\/2(9+3)? What\u2019s the deal with this orders of operation business?\n\n1. Pedro says:\n\nAlso, according to this\nhttp:\/\/creativepro.com\/fine-tuning-your-type-setting-fractions\/\n\nis not the same 6\/2(2 + 1)=1 and 6\u20442(2 + 1) = 9\n\n2. Mathman says:\n\nThat\u2019s fine if you ONLY follow order of operations and ignore every other principle that was discussed, such as algebra, coefficients, product of factors, quantities, etc.\n\n3. Larry Scott says:\n\nA bit of non-science\nThere are numerous threads on numerous blogs re: 6 \u00f7 2(1+2) = ?\nI always distribute 2(1+2) which = 6, of course. And 6 \u00f7 (6) = 1. And have seen much concurrence on that.\nOn most of the blogs there is no shortage of good and bad math. Including, that with or without symbolic variables versus numbers it\u2019s reduced differently. And \u201cmy calculator returns \u2026\u201d as \u2018proof\u2019. And the bad math is too often supported by insistence, anger, and disrespect.\nOver the past several months I\u2019ve asked, in person, many people is it =1?, or =9?\nEveryone that I\u2019ve encountered in person supports =1.\nOnline is the only place I\u2019ve seen any insistence for =9. And also Excel, which requires verbose syntax.\n1\/2\u03c0 is one divided by the product, not one half pi.\n\n4. Larry Scott says:\n\nIn a physics textbook, UC Berkeley, Nobel laureate Dr Richard Feynman describes the inverse proportion to 2\u221aN. And writes that as 1\/2\u221aN. Expressly returning one over the product of two times sq root of N. And specifically NOT one half sq root N.\n\nFeynman physics lectures, UC Berkeley press\nSection 6-7. (Page 64 of the online version.)\n\nSix over two(one plus two).\n\n5. Pedro says:\n\nSoftware documentation\nhttps:\/\/qalculate.github.io\/manual\/qalculate-expressions.html\n\nImplicit Multiplication and Parsing Modes\nThe evaluation of short\/implicit multiplication, without any multiplication sign (ex. \u201c5x\u201d, \u201c5(2+3)\u201d), differs depending on the parsing mode. In the conventional mode implicit multiplication does not differ from explicit multiplication (\u201c12\/2(1+2) = 12\/2*3 = 18\u201d, \u201c5x\/5y = 5*x\/5*y = xy\u201d). In the\n\u201cparse implicit multiplication first\u201d mode, implicit multiplication is parsed before explicit multiplication (\u201c12\/2(1+2) = 12\/(2*3) = 2\u201d, \u201c5x\/5y = (5*x)\/(5*y) = x\/y\u201d). The default adaptive mode works as the \u201cparse implicit multiplication first\u201d mode, unless spaces are found (\u201c1\/5x = 1\/(5*x)\u201d, but \u201c1\/5 x = (1\/5)*x\u201d).\n\n6. matt grove says:\n\nThe original is poorly written because of how computers were originally programed with their limited capacities to work math problems.\nYou changed the equation.\nfrom\n6\/2(1+2)\nto\n6\/2*(1+2)\n\nPlease try and argue that there is some immutable law of mathematics that makes those two equal, because what made those two equal is how computers read math problems and not the rules of mathematics. In computer programing if that equation needed to be programed in, then it would have to be entered specifically as 6\/(2*(1+2)) so the computer would properly understand what was being asked.\n\nab\/cd = (ab)\/(cd) != ab\/c*d\nhttp:\/\/www.jstor.org\/stable\/2972726?seq=2#page_scan_tab_contents\n\nThe only thing that has changed is how computers read the equations, not the rules on how people should read equations.\n\nYou get 9 putting it into a computer because it is poorly formated to get the computer to answer the question being asked. Junk in = Junk out.\n\nHad a programing teacher predict this day, when people would forget the proper rule of mathematics because of how the computer solves equations like this one. Seems you are pretty poor \u201cexperts\u201d.\n\n7. John says:\n\n2(2+1)\/6= 1 , therefore 2(2+1)=6 , not 2\/3 . 6 over 2(2+1) = 6\/6 . 6 over 2\/3=9\n\n8. Beth Herman says:\n\nMath is a total lie. The truth of this equation is the correct answer. The fact that it requires a universal effort to see the problem the same way or you don\u2019t get the same answer is proof of the lie.\n\n9. Micheal Thomas says:\n\n@Beth Herman, that is just a cop-out answer. Everything we do in life is defined by a set grouping of rules. Language, biology, chemistry, driving, walking down the street, what you do at work. Same with mathematics. The arithmetic we do daily has a certain set of rules in place that define how it works. The truth with this problem, is that we are now calling out how computers are designed to answer problems versus how people have been intuitively (or being taught) to solve them in the past. For those that have been through the era of not having a calculator for everything, the 2(1+2) is telling me that it is a part of the same \u201citem\u201d and needs to be kept together (in the denominator of the system), thus I would get the (2+4) which makes the answer 1.\nPersonally, I read this equation as 6\/(2*(1+2)). If it was written as 6\/2*(1+2), I would read that as (6\/2)*(1+2), two seperate items being multiplied together, and I would get the 9 as an answer.\n\nSo yes, this is all semantics, but a sort of arguement that brings to light the differences between the different experiences people have had when learning how to process these equations, and what should be done about correcting these issues.\n\n10. Larry Scott says:\n\nYes, =1.\n\n2(a+b) is a denominator. But that\u2019s coming into conflict with the current Excel style math that is taught. On this site alone there have been obstinate dismissal of 2(1+2) as a quantity. And across the internet the argument is so uncivilized.\n\nApparently x(a+b) = (xa+xb) is no longer the zeitgeist. Although it is still the distribution identity. And distribution is called for.\n\n11. Larry Scott says:\n\nAnd, one over two pi is 1\/2\u03c0.\nBut that too has been distorted to:\n1\/2\u03c0 = one half pi.\n\nGo figure.","date":"2017-07-22 10:36:38","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 1, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7269884943962097, \"perplexity\": 1347.8507664388017}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-30\/segments\/1500549423992.48\/warc\/CC-MAIN-20170722102800-20170722122800-00364.warc.gz\"}"} | null | null |
Q: How to use message.properties in Spring Boot in code instead of hardcoded string I have a message.properties in my Spring Boot program and it works fine with Thymeleaf in HTML templates. However, I want to use key=value also in Java code. For example, I have:
public static final String PDF_PROGRAM_INFO_FIRST_LINE = "Some text"
In message.properties, I have:
text = Some text
How to put key into my string in code to use it value instead of typing string by hand?
I tried:
public static final String PDF_PROGRAM_INFO_FIRST_LINE =("${text}")
However, this doesn't work.
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"redpajama_set_name": "RedPajamaStackExchange"
} | 416 |
\subsection{Overview of Techniques}
\label{sec:informal-overview}
We will now give a high-level overview of the reduction. For simplicity of presentation, we will sometimes be informal; everything will be formalized in the subsequent subsections.
\paragraph{Previous Results.}
To explain the key new ideas behind our reduction,
it is important to understand high-level approaches taken in previous works
and why they fail to yield running time lower bounds
as in our Theorem~\ref{thm:running-time-lower-bound}.
Most of the known hardness results
for agnostic learning of halfspaces employ reductions
from Label Cover~\cite{AroraBSS97,FeldmanGKP06,GuruswamiR09,FeldmanGRW12,DiakonikolasKM19}\footnote{Some of these reductions are stated in terms of reductions from Set Cover or from constraint satisfaction problems (CSP). However, it is well-known that these can be formulated as Label Cover.}.
These reductions use gadgets which are ``local'' in nature.
As we will explain next, such ``local'' reductions \emph{cannot} work for our purpose.
To describe the reductions, it
is convenient to think of each sample $(\mathbf{x}, y)$
as a linear constraint $\left<\mathbf{w}, \mathbf{x}\right> \geq 0$ when $y = +1$
and $\left<\mathbf{w}, \mathbf{x}\right> < 0$ when $y = -1$,
where the variables are the coordinates $w_1, \dots, w_d$ of $\mathbf{w}$.
When we also consider a margin parameter $\gamma^* > 0$, then the constraints become
$\left<\mathbf{w}, \mathbf{x}\right> \geq \gamma^*$ and $\left<\mathbf{w}, \mathbf{x}\right> < -\gamma^*$, respectively. Notice here that, for our purpose, we want
(i) our halfspace $\mathbf{w}$ to be in $\mathbb{B}^d_{1}$, i.e., $|w_1| + \cdots + |w_d| \leq 1$,
and (ii) each of our samples $\mathbf{x}$ to lie in $\mathbb{B}^d_{\infty}$, i.e., $|x_1|, \dots, |x_d| \leq 1$.
Although the reductions in previous works vary in certain steps,
they do share an overall common framework. With some
simplification,
they typically let e.g. $d = |U| \cdot |\Sigma_U|$,
where each coordinate is associated with $U \times \Sigma_U$.
In the completeness case, i.e., when some labeling $\phi^c$ covers all vertices in $V$,
the intended solution $\mathbf{w}^c$ is
defined by $w^c_{(u, \sigma_u)} = \mathds{1}[\sigma_u = \phi(u)] / k$ for all $u \in U, \sigma_u \in \Sigma_U$. To ensure that this is essentially the best choice of halfspace,
these reductions often appeal to several types of linear constraints.
For concreteness, we state a simplified version of those from~\cite{AroraBSS97} below.
\begin{itemize}
\item For every $(u, \sigma_U) \in U \times \Sigma_U$,
create the constraint $w_{(u, \sigma_u)} \leq 0$.
(This corresponds to the labeled sample $(-\mathbf{e}_{(a, \sigma)}, +1)$.)
\item For each $u \in U$, create the constraint
$\sum_{\sigma \in \Sigma_U} w_{(u, \sigma)} \geq 1/k$.
\item For every $v \in V$, $\sigma_v \in \Sigma_V$ and $u_1, u_2 \in N(v)$,
add $\sum_{\sigma_{u_1} \in \pi_{(u_1, v)}^{-1}(\sigma_v)} w_{(u_1, \sigma_{u_1})} = \sum_{\sigma_{u_2} \in \pi_{(u_2, v)}^{-1}(\sigma_v)} w_{({u_2}, \sigma_{u_2})}$.
This equality ``checks'' the Label Cover constraints $\pi_{(u_1, v)}$ and $\pi_{(u_2, v)}$.
\end{itemize}
Clearly, in the completeness case $\mathbf{w}^c$ satisfies all constraints except the non-positivity constraints
for the $k$ non-zero coordinates.
(It was argued in~\cite{AroraBSS97} that any halfspace must violate many more constraints in the soundness case.)
Observe that this reduction does not yield any margin: $\mathbf{w}^c$ does \emph{not} classify any sample with a positive margin.
Nonetheless,~\cite{DiakonikolasKM19} adapts this reduction to work with a small margin $\gamma^* > 0$
by adding/subtracting appropriate ``slack'' from each constraint.
For example, the first type of constraint is changed to
$w_{(u, \sigma_u)} \leq \gamma^*$.
This gives the desired margin $\gamma^*$ in the completeness case.
However, for the soundness analysis to work, it is crucial that
$\gamma^* \leq O(1/k)$,
as otherwise the constraints can be trivially satisfied\footnote{Note that $\mathbf{w} = \mathbf{0}$ satisfies
the constraints with margin $\gamma^* - 1/k$, which is
$(1 - o(1))\gamma^*$ if $\gamma^* = \omega(1/k)$.} by $\mathbf{w} = \mathbf{0}$.
As such, the above reduction does \emph{not} work for us,
since we would like a margin $\gamma^* = \Omega(1/\sqrt{k})$.
In fact, this also holds for all known reductions, which are ``local'' in nature and possess
similar characteristics. Roughly speaking, each linear constraint of these reductions
involves only a constant number of terms that are intended to be set to $O(1/k)$,
which means that we cannot hope to get a margin more than $O(1/k)$.
\paragraph{Our Approach: Beyond Local Reductions.}
With the preceding discussion in mind, our reduction has to be ``non-local'', i.e.,
each linear constraint has to involve many of the non-zero coordinates.
Specifically, for each subset $V^j$, we will check all the Label Cover constraints
involving $v \in V^j$ at once. To formalize this goal, we will require
the following definition.
\begin{definition} \label{def:projection-matrix}
Let $\mathcal{L} = (U, V = V_1 \cup \cdots \cup V_t, E, \Sigma_U, \Sigma_V, \{\pi_e\}_{e \in E})$ be a decomposable Label Cover instance.
For any $j \in [t]$, let $\Pi^j \in \mathbb{R}^{(V \times \Sigma_V) \times (U \times \Sigma_U)}$ be defined as
\begin{align*}
\Pi^j_{(v, \sigma_v), (u, \sigma_u)} =
\begin{cases}
1 & \text{ if } v = v^j(u) \text{ and } \pi_{(u, v)}(\sigma_u) = \sigma_v, \\
0 & \text{ otherwise}.
\end{cases}
\end{align*}
\end{definition}
We set $d = |U| \cdot |\Sigma_U|$ and our intended solution $\mathbf{w}^c$ in the completeness case is the same as described in the previous reduction.
For simplicity, suppose that, in the soundness case,
we pick $\phi^s$ that does \emph{not} weakly cover any $v \in V$ and set
$w^s_{(u, \sigma_u)} = \mathds{1}[\sigma_u = \phi^s(u)]/k$.
Our simplified task then becomes:
\emph{Design $\mathcal{D}$ such that $\err_{\gamma}^{\mathcal{D}}(\mathbf{w}^c) \ll \err_{(1 - \nu)\gamma}^{\mathcal{D}}(\mathbf{w}^s)$,
where $\gamma = \Omega(1/\sqrt{k})$, $\nu > 0$ is a constant.}
Our choice of $\mathcal{D}$ is based on two observations.
The first is a structural difference between $\mathbf{w}^c (\Pi^j)^T$ and $\mathbf{w}^s (\Pi^j)^T$.
Suppose that the constraint graph has right degree $\Delta$.
Since $\phi^c$ covers all $v \in V$, $\Pi^j$ ``projects''
the non-zeros coordinates $w^c_{(u, \phi^c(u))}$ for all $u \in N(v)$
to the same coordinate $(v, \sigma_v)$, for some $\sigma_v \in \Sigma_V$,
resulting in the value of $\Delta / k$ in this coordinate. On the other hand,
since $\phi^s$ does not even weakly cover any right vertex,
all the non-zero coordinates get maps by $\Pi^j$ to different coordinates,
resulting in the vector $\mathbf{w}^s (\Pi^j)^T$ having $k$ non-zero coordinates,
each having value $1/k$.
To summarize, we have:
$\mathbf{w}^c (\Pi^j)^T$ has $k / \Delta$ non-zero coordinates, each of value $\Delta/k$.
On the other hand, $\mathbf{w}^s (\Pi^j)^T$ has $k$ non-zero coordinates, each of value $1/k$.
Our second observation is the following:
suppose that $\mathbf{u}$ is a vector with $T$ non-zero coordinates,
each of value $1/T$. If we take a random $\pm 1$ vector $\mathbf{s}$, then
$\left<\mathbf{u}, \mathbf{s}\right>$ is simply $1/T$ times a sum of $T$ i.i.d. Rademacher random variables.
Recall a well-known version of the central limit theorem (e.g.,~\cite{Berry41,Esseen42}):
as $T \to \infty$, $1/\sqrt{T}$ times a sum of $T$ i.i.d. Rademacher r.v.s converges in distribution to the normal distribution.
This implies that $\lim_{T \to \infty} \Pr[\left<\mathbf{u}, \mathbf{s}\right> \geq 1/\sqrt{T}] = \Phi(1).$
For simplicity, let us ignore the limit for the moment and assume that $\Pr[\left<\mathbf{u}, \mathbf{s}\right> \geq 1/\sqrt{T}] = \Phi(1)$. We can now specify
the desired distribution $\mathcal{D}$: Pick $\mathbf{s}$ uniformly at random from $\{\pm 1\}^{V \times \Sigma_V}$
and then let the sample be $\mathbf{s}\Pi^j$ with label $+1$. By the above two observations,
$\mathbf{w}^c$ will be correctly classified with margin
$\gamma^* = \sqrt{\Delta / k} = \Omega(1/\sqrt{k})$ with probability $\Phi(1)$.
Furthermore, in the soundness case, $\mathbf{w}^s$ can only get the same error
with margin (roughly) $\sqrt{1/k} = \gamma^* / \sqrt{\Delta}$.
Intuitively, for $\Delta > 1$, this means that we get a gap of $\Omega(1/\sqrt{k})$
in the margins between the two cases, as desired. This concludes our informal proof overview.
\subsection{The Reduction}
\label{subsec:reduction}
Having stated the rough main ideas above, we next formalize the reduction. To facilitate this, we define the following additional notations:
\begin{definition}
Let $\mathcal{L} = (U, V = V_1 \cup \cdots \cup V_t, E, \Sigma_U, \Sigma_V, \{\pi_e\}_{e \in E})$ be a decomposable Label Cover instance.
For any $j \in [t]$, let $\hat{\Pi}^j \in \mathbb{R}^{(U \times \Sigma_V) \times (U \times \Sigma_U)}$ be such that
\begin{align*}
\hat{\Pi}^j_{(u', \sigma_v), (u, \sigma_u)} =
\begin{cases}
1 & \text{ if } u' = u \text{ and } \pi_{(u, v^j(u))}(\sigma_u) = \sigma_v, \\
0 & \text{ otherwise}.
\end{cases}
\end{align*}
Moreover, let $\tilde{\Pi}^j \in \mathbb{R}^{(V \times \Sigma_V) \times (U \times \Sigma_V)}$ be such that
\begin{align*}
\tilde{\Pi}^j_{(v, \sigma'_v), (u, \sigma_v)} =
\begin{cases}
1 & \text{ if } v = v^j(u) \text{ and } \sigma'_v = \sigma_v \\
0 & \text{ otherwise}.
\end{cases}
\end{align*}
Observe that $\Pi^j = \tilde{\Pi}^j \cdot \hat{\Pi}^j$ (where $\Pi^j$ is as in Definition~\ref{def:projection-matrix}).
\end{definition}
Our full reduction is present in Figure~\ref{fig:reduction} below. Before we specify the choice of parameters, let us make a few remarks. First, we note that the distribution described in the previous section corresponds to Step~\ref{sample:label-cover} in the reduction. The other steps of the reductions are included to handle certain technical details we had glossed over previously. In particular, the following are the two main additional technical issues we have to deal with here.
\begin{itemize}
\item \emph{(Non-Uniformity of Weights)} In the intuitive argument above, we assume that,
in the soundness case, we only consider $\mathbf{w}^s$ such that
$\sum_{\sigma_u \in \Sigma_U} w_{(u, \sigma_u)}^s = 1/k$. However,
this need not be true in general, and we have to create new samples to (approximately) enforce
such a condition. Specifically, for every subset $T \subseteq U$, we add a constraint
that $\sum_{u \in T} \sum_{\sigma_u \in \Sigma_U} w_{(u, \sigma_u)} \geq |T|/k - \gamma^*$.
This corresponds to Step~\ref{sample:mass-lower} in Figure~\ref{fig:reduction}.
Note that the term $- \gamma^*$ on the right hand side above is necessary to ensure that,
in the completeness case, we still have a margin of $\gamma^*$.
Unfortunately, this also leaves the possibility of, e.g., some vertex $u \in U$ has
as much as $\gamma^*$ extra ``mass''. For technical reasons, it turns out that
we have to make sure that these extra ``masses'' do not contribute to too much of $\|\mathbf{w}(\Pi^j)^T\|_2^2$.
To do so, we add additional constraints on $\mathbf{w}(\hat{\Pi}^j)^T$ to bound its norm.
Such a constraint is of the form: If we pick a subset $S$ of at most $\ell$ coordinates,
then their sum must be at most $|S|/k + \gamma^*$ (and at least $-\gamma^*$).
These corresponds to Steps~\ref{sample:exceed-l2-norm} and~\ref{sample:neg-l2-norm}
in Figure~\ref{fig:reduction}.
\item \emph{(Constant Coordinate)} Finally, similar to previous works,
we cannot have ``constants'' in our linear constraints. Rather, we need to add
a coordinate $\star$ with the intention that $\mathbf{w}_{\star} = 1/2$,
and replace the constants in the previous step by $\mathbf{w}_{\star}$.
Note here that we need two additional constraints (Steps~\ref{sample:constant-lower}
and~\ref{sample:constant-upper} in Figure~\ref{fig:reduction}) to ensure that
$\mathbf{w}_{\star}$ has to be roughly $1/2$.
\end{itemize}
\begin{figure}[h!]
\begin{framed}
\textbf{Input:} Decomposable Label Cover instance $\mathcal{L} = (U, V = V_1 \cup \dots \cup V_t, E, \Sigma_U, \Sigma_V, \{\pi_e\}_{e \in E})$.
\textbf{Parameters:} $q, \gamma^* \in (0, 1), \ell \in \mathbb{N}$.
\textbf{Output:} Oracle $\mathcal{O}$ that draws a sample from a distribution $\mathcal{D}$ on $\mathbb{B}_{\infty}^{|U| \cdot |\Sigma_U| + 1} \times \{\pm 1\}$. \\
For notational convenience, we associate each coordinate of $(|U|\cdot|\Sigma_U| + 1)$-dimensional samples with an element from $(U \times \Sigma_U) \cup \{\star\}$. The oracle $\mathcal{O}$ draws a sample as follows:
\begin{enumerate}
\item With probability $0.25$, output the sample $2\gamma^* \cdot \mathbf{e}_{\star}$ with label +1. \label{sample:constant-lower}
\item With probability $0.25$, output the sample $2\gamma^* \cdot \mathbf{e}_{U \times \Sigma_U}$ with label +1. \label{sample:constant-upper}
\item With probability $0.25$, pick a random subset $T \subseteq U$ and output the sample $\mathbf{e}_{T \times \Sigma_U} - \left(\frac{|T|}{k} - 2\gamma^*\right) \mathbf{e}_{\star}$ with label +1. \label{sample:mass-lower}
\item With probability $0.25$, draw $j$ uniformly at random from $[t]$. Then, do the following:
\begin{enumerate}
\item With probability $0.5(1 - q)$, randomly pick a subset $S \subseteq U \times \Sigma_V$ of size at most $\ell$. Output the labeled sample $((\frac{|S|}{k} + 2\gamma^*) \mathbf{e}_{\star} - \mathbf{e}_S \hat{\Pi}^j, +1)$. \label{sample:exceed-l2-norm}
\item With probability $0.5(1 - q)$, randomly pick a subset $S \subseteq U \times \Sigma_V$ of size at most $\ell$. Then, output $(2\gamma^* \mathbf{e}_{\star} + \mathbf{e}_S \hat{\Pi}^j, +1)$. \label{sample:neg-l2-norm}
\item With probability $q$, sample $\mathbf{s}$ uniformly at random from $\{\pm 1\}^{V \times \Sigma_V}$ and, output $(\mathbf{s} \Pi^j, +1)$. \label{sample:label-cover}
\end{enumerate}
\end{enumerate}
\end{framed}
\caption{Hardness Reduction from Label Cover to $L_{\infty}$-margin Halfspace Learning.} \label{fig:reduction}
\end{figure}
The parameters of our reduction are set as follows:
\begin{itemize}
\item $C$ and $m_0$ are as in Lemma~\ref{lem:anti-concen},
\item $\Delta = \lceil 10^4 / C^2 \rceil$,
\item $\gamma^* = 0.5 C \sqrt{\Delta/k}$,
\item $k_0 = m_0 \Delta$,
\item $\delta = (0.1 / \Delta)^4$,
\item $\ell = \lceil\delta\sqrt{k}\rceil$,
\item $q = 0.001 / n^{\ell}$ (where $n$ is as defined is Theorem~\ref{thm:label-cover-hardness}),
\item $\epsilon^* = 0.6 (0.25 q)$,
\item $\mu = \frac{0.01}{\Delta(\Delta - 1)}$.
\end{itemize}
It is easy to see that the oracle can draw a sample in polynomial time. Furthermore, $\epsilon^* = 0.001 / n^{\ell} \geq n^{-O(\sqrt{k})}$ and $\gamma^* \geq \Omega(1/\sqrt{k})$, as desired. Hence, we are only left to prove the completeness and the soundness of the reduction, which we will do next.
\subsection{Completeness}
\label{subsec:completeness}
Suppose that the Label Cover instance $\mathcal{L}$ is satisfiable, i.e., that there exists a labeling $\phi^*$ that covers all right vertices. Let $\mathbf{w}^*$ be such that $w^{*}_{\star} = 1/2$ and
\begin{align*}
w^*_{(u, \sigma)} =
\begin{cases}
\frac{1}{2k} & \text{ if } \sigma = \phi^*(u), \\
0 & \text{ otherwise}
\end{cases}
\end{align*}
for all $u \in U, \sigma \in \Sigma_U$. It is simple to check that the samples generated in Steps~\ref{sample:constant-lower},~\ref{sample:constant-upper},~\ref{sample:mass-lower},~\ref{sample:exceed-l2-norm} and~\ref{sample:neg-l2-norm} are all correctly labeled with margin $\gamma^*$.
Hence, we are left with computing the probability that the samples generated in Step~\ref{sample:label-cover} are violated. To do this, first notice that, for every $j \in [t], v \in V^j, \sigma_v \in \Sigma_V$, we have
\begin{align*}
(\mathbf{w}^* (\Pi^j)^T)_{(v, \sigma_v)} &= \sum_{(u, \sigma_u) \in U \times \Sigma_U \atop v^j(u) = v, \pi_{(u, v)}(\sigma_u) = \sigma_v} w^*_{(u, \sigma_u)} \\
(\text{From definition of } w^*) &= \frac{1}{2k} \left|\{u \in N(v) \mid \pi_{(u, v)}(\phi^*(u)) = \sigma_v\}\right|.
\end{align*}
Now since every $v \in V^j$ is covered by $\phi^*$, there exists a unique $\sigma_v$ such that $\pi_{(u, v)}(\phi^*(u)) = \sigma_v$ for all $u \in N(v)$. As a result, $\mathbf{w}^* (\Pi^j)^T$ has $|V^j| = k/\Delta$ coordinates exactly equal to $\Delta \cdot \frac{1}{2k} = \frac{\Delta}{2k}$, and the remaining coordinates are equal to zero. Recall that, for the samples in Step~\ref{sample:label-cover}, $\mathbf{s}$ is a random $\{\pm 1\}$ vector. Thus, $\left<\mathbf{w}^*, \mathbf{s} \Pi^j\right> = \left<\mathbf{w}^*(\Pi^j)^T, \mathbf{s}\right>$ has the same distribution as $\frac{\Delta}{2k}$ times a sum of $k/\Delta$ i.i.d. Rademacher random variables. By Lemma~\ref{lem:anti-concen}, we can conclude that $\Pr_{\mathbf{s}}[\left<\mathbf{w}^*, \mathbf{s} \Pi^j\right> \geq 0.5 C\sqrt{\Delta/k}] \geq 0.4$. Since we set $\gamma^* = 0.5 C\sqrt{\Delta/k}$, this implies that $\mathbf{w}^*$ correctly classifies (at least) $0.4$ fraction of the samples from Step~\ref{sample:label-cover}. Hence, we have
\begin{align*}
\err^{\mathcal{D}}_{\gamma}(\mathbf{w}^*) \leq 0.6 \cdot (0.25q) = \epsilon^* \;,
\end{align*}
as desired.
\subsection{Soundness}
\label{subsec:soundness}
We will prove the soundness contrapositively. For this purpose, suppose that there is a halfspace $\mathbf{w} \in \mathbb{B}_1^d$ such that $\err^{\mathcal{D}}_{(1 - \delta)\gamma}(\mathbf{w}) \leq 1.6 \epsilon^* = 0.96 (0.25q)$. We will show that there exists an assignments $\phi'$ with $\wval(\phi') \geq \mu$.
\subsubsection{Some Simple Bounds}
We start by proving a few observations/lemmas that will be useful in the subsequent steps.
First, observe that every distinct sample from Steps~\ref{sample:constant-lower},~\ref{sample:constant-upper},~\ref{sample:mass-lower},~\ref{sample:exceed-l2-norm} and~\ref{sample:neg-l2-norm} has probability mass (in $\mathcal{D}$) at least $\frac{0.125(1 - q)}{n^{\ell}} > q > 1.6 \epsilon^*$. Since we assume that $\err^{\mathcal{D}}_{(1 - \delta)\gamma}(\mathbf{w}) \leq 1.6 \epsilon^*$, it must be the case that all these examples are correctly classified by $\mathbf{w}$ with margin at least $(1 - \delta)\gamma^*$:
\begin{observation} \label{obs:correctly-classify}
$\mathbf{w}$ correctly classifies all samples in Steps~\ref{sample:constant-lower},~\ref{sample:constant-upper},~\ref{sample:mass-lower},~\ref{sample:exceed-l2-norm} and~\ref{sample:neg-l2-norm} with margin $(1 - \delta)\gamma^*$.
\end{observation}
Throughout the remainder of this section, we will use the following notations:
\begin{definition}
For every $u \in U$, let $M_u$ denote $\sum_{\sigma \in \Sigma_u} |w_{(u, \sigma)}|$. Then, let $U_{\text{small}}$ denote $\{u \in U \mid M_u \leq 1/k\}$ and $U_{\text{large}}$ denote $U \setminus U_{\text{small}}$.
\end{definition}
The next observation, which follows almost immediately from Observation~\ref{obs:correctly-classify}, is that the value of the ``constant coordinate'' $w_{\star}$ is roughly $1/2$ (as we had in the completeness case) and that the sum of the absolute values of the negative coordinates is quite small.
\begin{observation} \label{obs:constant-and-non-neg}
The following holds:
\begin{enumerate}
\item (Constant Coordinate Value) $w_{\star} \in [0.5(1 - \delta), 0.5(1 + \delta)]$.
\item (Negative Coordinate Value) $\sum_{j \in (U \times \Sigma_U) \cup \{\star\} \atop w_j < 0} |w_j| \leq \delta$.
\end{enumerate}
\end{observation}
\begin{proof}
\begin{enumerate}
\item Since $\mathbf{w}$ correctly classifies the sample from Step~\ref{sample:constant-lower} with margin $(1 - \delta)\gamma^*$, we have $2\gamma^* w_{\star} > (1 - \delta)\gamma^*$. This implies that $w_{\star} \geq 0.5(1 - \delta)$.
Let $a = \left<\mathbf{w}, \mathbf{e}_{U \times \Sigma_U}\right>$. Similarly, from $\mathbf{w}$ correctly classifies the sample from Step~\ref{sample:constant-upper} with margin $(1 - \delta)\gamma^*$, we have $a \geq 0.5(1 - \delta)$. Furthermore, observe that
\begin{align} \label{eq:norm-ineq}
a + w_{\star} \leq \|\mathbf{w}\|_1 \leq 1.
\end{align}
As a result, we have $w_{\star} \leq 0.5(1 + \delta)$ as desired.
\item Since $w_{\star} > 0$, we may rearrange the desired term as
\begin{align*}
\sum_{j \in (U \times \Sigma_U) \cup \{\star\} \atop w_j < 0} |w_j| &= \frac{1}{2} \left(\|\mathbf{w}\|_1 - a - w_{\star}\right) \\
&\leq \frac{1}{2}\left(1 - 0.5(1 - \delta) - 0.5(1 - \delta)\right) \\
&< \delta,
\end{align*}
where the first inequality follows from $a, w^* \geq 0.5(1 - \delta)$ that we had shown above. \qedhere
\end{enumerate}
\end{proof}
Another bound we will use is that $U_{\text{large}}$ is quite small, and the sum of absolute values of the coordinates correspond to $U_{\text{large}}$ is also quite small.
\begin{observation}[Bounds on $U_{\text{large}}$] \label{obs:large-mass}
The following holds:
\begin{enumerate}
\item (Size Bound) $|U_{\text{large}}| \leq 2\delta k$.
\item (Mass Bound) $\sum_{u \in U_{\text{large}}} M_u \leq 2\delta$.
\end{enumerate}
\end{observation}
\begin{proof}
To prove the desired bounds, first notice that, since $\mathbf{w}$ correctly classifies the sample in Step~\ref{sample:mass-lower} with $T = U_{\text{small}}$ with margin $(1 - \delta) \gamma^*$, we must have
\begin{align*}
\left<\mathbf{w}, \mathbf{e}_{U_{\text{small}} \times \Sigma_U}\right> \geq \left(\frac{|U_{\text{small}}|}{k} - 2\gamma^*\right) w_{\star} + (1 - \delta) \gamma^*.
\end{align*}
Now, observe that the term on the left hand side is at most $\sum_{u \in U_{\text{small}}} M_u$ which, from $\|\mathbf{w}\|_1 \leq 1$, is in turn at most $1 - w_{\star} - \sum_{u \in U_{\text{large}}} M_u$. Combining these, we get
\begin{align*}
1 - w_{\star} - \sum_{u \in U_{\text{large}}} M_u \geq \left(\frac{|U_{\text{small}}|}{k} - 2\gamma^*\right) w_{\star} + (1 - \delta) \gamma^* = \left(1 - \frac{|U_{\text{large}}|}{k} - 2\gamma^*\right) w_{\star} + (1 - \delta) \gamma^*
\end{align*}
Recall from Observation~\ref{obs:constant-and-non-neg} that $w_{\star} \geq 0.5(1 - \delta)$. Plugging this into the above, we have
\begin{align}
\sum_{u \in U_{\text{large}}} M_u
&\leq 1 - \left(2 - \frac{|U_{\text{large}}|}{k} - 2\gamma^*\right) \cdot 0.5(1 - \delta) - (1 - \delta)\gamma^* \nonumber \\
&= 1 - \left(2 - \frac{|U_{\text{large}}|}{k}\right) \cdot 0.5(1 - \delta) \nonumber \\
&\leq \delta + \frac{0.5|U_{\text{large}}|}{k} \;. \label{eq:mass-bound-intermediate}
\end{align}
\begin{enumerate}
\item Subtracting $\frac{0.5 |U_{\text{large}}|}{k}$ from both sides, we have
\begin{align*}
\sum_{u \in U_{\text{large}}} \left(M_u - \frac{0.5}{k}\right) \leq \delta.
\end{align*}
By definition, $M_u > 1/k$ for all $u \in U_{\text{large}}$. As a result, we have $|U_{\text{large}}| \leq 2\delta k$, as desired.
\item Plugging the bound on $|U_{\text{large}}|$ back into~\eqref{eq:mass-bound-intermediate}, we get the claimed bound on $\sum_{u \in U_{\text{large}}} M_u$. \qedhere
\end{enumerate}
\end{proof}
\subsubsection{Identifying a ``Nice'' Halfspace}
We will now convert $\mathbf{w}$ into a ``nicer'' halfspace, i.e., one without negative and large coordinates. It will be much more convenient to deal with such a nice halfspace when we ``decode'' back a labeling later in this section.
The ``nice'' halfspace is quite simple: we just zero out all coordinates $w_{(u, \sigma)}$, where $u \in U_{\text{large}}$. More formally, let $\hat{\mathbf{w}} \in \mathbb{R}^{|U| \cdot |\Sigma_U|}$ be such that
\begin{align*}
\hat{w}_{(u, \sigma)} =
\begin{cases}
w_{(u, \sigma)} & u \in U_{\text{small}}, \\
0 & u \in U_{\text{large}},
\end{cases}
\end{align*}
for all $u \in U$ and $\sigma \in \Sigma_U$.
The main lemma needed in our analysis is that, for each $j \in [t]$, $\hat{\mathbf{w}} (\Pi^j)^T$ preserves most of the $L_2$ norm compared to the original $\mathbf{w} (\Pi^j)^T$.
\begin{lemma}[Nice Halfspace Preserves Most of $L_2$ Norm] \label{lem:variance-preserved}
For every $j \in [t]$, we have
\begin{align} \label{eq:variance-preserved}
\|\hat{\mathbf{w}}(\Pi^j)^T\|_2^2 \geq \frac{\|\mathbf{w} (\Pi^j)^T\|_2^2}{2} - \frac{\sqrt[4]{\delta}}{k}.
\end{align}
\end{lemma}
\begin{proof}
For convenience, let $\mathbf{v} = \mathbf{w} - \hat{\mathbf{w}}$ and $\mathbf{b} = \mathbf{v}(\hat{\Pi}^j)^T$. The majority of this proof is spent on bounding $\|\mathbf{b}\|_2^2$. To do this, let us define several new notations:
\begin{itemize}
\item Let $C_{> 0} = |\{(u, \sigma) \in U \times \Sigma_V \mid b_{(u, \sigma)} > 0\}|$ and $C_{< 0} = |\{(u, \sigma) \in U \times \Sigma_V \mid b_{(u, \sigma)} < 0\}|$.
\item Let $\mathbf{b}^{\geq 0} \in \mathbb{R}^{U \times \Sigma_V}$ be defined by
\begin{align*}
b^{\geq 0}_{(u, \sigma)} = \max\{0, b_{(u, \sigma)}\}
\end{align*} for all $(u, \sigma) \in U \times \Sigma_V$. Furthermore, let $\mathbf{b}^{< 0} = \mathbf{b} - \mathbf{b}^{\geq 0}$.
\end{itemize}
Observe that $\|\hat{\Pi}^j\|_1 \leq 1$, because each column has exactly a single entry equal to one and the remaining entries equal to zero. As a result, we have
\begin{align} \label{eq:l1-norm-b}
\|\mathbf{b}\|_1 = \|\mathbf{v}(\hat{\Pi}^j)^T\|_1 \leq \|\mathbf{v}\|_1 = \sum_{u \in U_{large}} M_u \leq 2\delta \;,
\end{align}
where the last inequality follows from Observation~\ref{obs:large-mass}.
Since $\mathbf{b} = \mathbf{b}^{\geq 0} + \mathbf{b}^{< 0}$, we may bound $\|\mathbf{b}^{\geq 0}\|_2, \|\mathbf{b}^{< 0}\|_2$ separately, starting with the former.
\paragraph{Bounding $\|\mathbf{b}^{\geq 0}\|_2$.} Let us sort the coordinates of $\mathbf{b}^{\geq 0}$ from largest to smallest entries as $b^{\geq 0}_{(u^1, \sigma^1)}, \dots,$ $b^{\geq 0}_{(u^{|U| \times |\Sigma_V|}, \sigma^{|U| \times |\Sigma_V|})}$ (tie broken arbitrarily). For every $j \leq \min\{C_{> 0}, \ell\}$, consider the sample from Step~\ref{sample:exceed-l2-norm} when $S = \{(u^1, \sigma^1), \dots, (u^j, \sigma^j)\}$. Since $\mathbf{w}$ correctly classifies this sample with margin $(1 - \delta)\gamma^*$, we have
\begin{align*}
(1 - \delta)\gamma^* &\leq \left<\mathbf{w}, \left(\frac{j}{k} + 2\gamma^*\right) \mathbf{e}_{\star} - \mathbf{e}_S \hat{\Pi}^j\right> \\
&= \left(\frac{j}{k} + 2\gamma^*\right) w_{\star} - \mathbf{w}(\hat{\Pi}^j)^T(\mathbf{e}_S)^T \\
&= \left(\frac{j}{k} + 2\gamma^*\right) w_{\star} - \left(\sum_{i \in [j]} (\mathbf{w}(\hat{\Pi}^j)^T)_{(u^i, \sigma^i)}\right) \\
\text{(Observation~\ref{obs:constant-and-non-neg})} &\leq \left(\frac{j}{k} + 2\gamma^*\right)\cdot 0.5(1 + \delta) - \left(\sum_{i \in [j]} (\mathbf{w}(\hat{\Pi}^j)^T)_{(u^i, \sigma^i)}\right) \\
&= \left(\frac{j}{k} + 2\gamma^*\right)\cdot 0.5(1 + \delta) - \left(\sum_{i \in [j]} \left(b^{\geq 0}_{(u^i, \sigma^i)} + (\hat{\mathbf{w}}(\hat{\Pi}^j)^T)_{(u^i, \sigma^i)}\right)\right) \\
&= \left(\frac{j}{k} + 2\gamma^*\right)\cdot 0.5(1 + \delta) - \left(\sum_{i \in [j]} b^{\geq 0}_{(u^i, \sigma^i)}\right),
\end{align*}
where the last equality follows from the fact that, for every $i \leq C_{> 0}$, we must have $u^i \in U_{\text{large}}$ as otherwise $b^{\geq 0}_{(u^i, \sigma^i)}$ would have been equal to zero.
Rearranging the above inequality, we have
\begin{align*}
\left(\sum_{i \in [j]} b^{\geq 0}_{(u^i, \sigma^i)}\right) \leq \frac{0.5(1 + \delta)j}{k} + 2\delta\gamma^* \leq \frac{j}{k} + 2\delta\gamma^*.
\end{align*}
Recall from our assumption that $b^{\geq 0}_{(u^1, \sigma^1)} \geq \cdots \geq b^{\geq 0}_{(u^j, \sigma^j)}$. Plugging this into the above, we get
\begin{align} \label{eq:term-by-term-bound-positive}
b^{\geq 0}_{(u^j, \sigma^j)} \leq \frac{1}{k} + \frac{2\delta \gamma^*}{j}.
\end{align}
Notice that while we have only derived the above inequality for $j \leq \min\{C_{> 0}, \ell\}$, it also extends to all $j \leq \ell$ because $b^{\geq 0}_{(u^j, \sigma^j)} = 0$ for all $j > C_{> 0}$.
We can use this to bound $\|\mathbf{b}^{\geq 0}\|_2^2$ as follows.
\begin{align*}
\|\mathbf{b}^{\geq 0}\|_2^2 &= \sum_{j=1}^{|U| \cdot |\Sigma_V|} \left(b^{\geq 0}_{(u^j, \sigma^j)}\right)^2 \\
&= \sum_{j < \ell} \left(b^{\geq 0}_{(u^j, \sigma^j)}\right)^2 + \sum_{j \geq \ell} \left(b^{\geq 0}_{(u^j, \sigma^j)}\right)^2 \\
&\leq \sum_{j < \ell} \left(b^{\geq 0}_{(u^j, \sigma^j)}\right)^2 + b^{\geq 0}_{(u^\ell, \sigma^\ell)} \cdot \|\mathbf{b}^{\geq 0}\|_1 \\
&\overset{\eqref{eq:term-by-term-bound-positive}}{\leq} \sum_{j < \ell} \left(\frac{1}{k} + \frac{2\delta\gamma^*}{j}\right)^2 + \left(\frac{1}{k} + \frac{2\delta \gamma^*}{\ell}\right) \cdot \|\mathbf{b}\|_1 \\
&\overset{\eqref{eq:l1-norm-b}}{\leq} \sum_{j < \ell} 2\left(\frac{1}{k^2} + \frac{1}{j^2} \cdot 4\delta^2(\gamma^*)^2\right) + \left(\frac{1}{k} + \frac{2\delta\gamma^*}{\ell}\right) \cdot 2\delta \\
&\leq \frac{2(\ell - 1)}{k^2} + \frac{\pi^2}{6} \cdot 8\delta^2(\gamma^*)^2 + \frac{2\delta}{k} + \frac{4\delta^2\gamma^*}{\ell} \\
(\text{From our choice of } \ell \text{ and } \delta\gamma^* \leq 0.1\sqrt{\delta/k}) &\leq \frac{2\delta}{k^{1.5}} + \frac{\delta}{k} + \frac{2\delta}{k} + \frac{\sqrt{\delta}}{k} \\
&\leq \frac{2\sqrt{\delta}}{k}.
\end{align*}
\paragraph{Bounding $\|\mathbf{b}^{< 0}\|_2$.} This is very similar (and in fact slightly simpler) to how we bound $\|\mathbf{b}^{\geq 0}\|_2$ above; we repeat the argument here for completeness. Let us first sort the coordinates of $\mathbf{b}^{< 0}$ from smallest to largest entries as $b^{< 0}_{(u^{-1}, \sigma^{-1})}, \dots, b^{< 0}_{(u^{-|U| \times |\Sigma_V|}, \sigma^{-|U| \times |\Sigma_V|})}$ (tie broken arbitrarily). For every $j \leq \min\{C_{< 0}, \ell\}$, consider the sample from Step~\ref{sample:neg-l2-norm} when $S = \{(u^{-1}, \sigma^{-1}), \dots, (u^{-j}, \sigma^{-j})\}$. Since $\mathbf{w}$ correctly classifies this sample with margin $(1 - \delta)\gamma^*$, we have
\begin{align*}
(1 - \delta)\gamma^* &\leq \left<\mathbf{w}, 2\gamma^* \mathbf{e}_{\star} + \mathbf{e}_S \hat{\Pi}^j\right> \\
&= 2\gamma^* \cdot w_{\star} + \mathbf{b}(\mathbf{e}_S)^T \\
\text{(Observation~\ref{obs:constant-and-non-neg})} &\leq 2\gamma^* \cdot 0.5(1 + \delta) - \left(\sum_{i \in [j]} |b^{< 0}_{(u^{-i}, \sigma^{-i})}|\right).
\end{align*}
Rearranging the above inequality, we have
\begin{align*}
\left(\sum_{i \in [j]} |b^{< 0}_{(u^{-i}, \sigma^{-i})}|\right) \leq 2\delta\gamma^*.
\end{align*}
Recall from our assumption that $|b^{< 0}_{(u^{-1}, \sigma^{-1})}| \geq \cdots \geq |b^{< 0}_{(u^{-j}, \sigma^{-j})}|$. Plugging this into the above, we get
\begin{align} \label{eq:term-by-term-bound-negative}
|b^{< 0}_{(u^{-j}, \sigma^{-j})}| \leq \frac{2\delta \gamma^*}{j}.
\end{align}
Similar to the previous case, although we have derived the above inequality for $j \leq \min\{C_{< 0}, \ell\}$, it also holds for all $j \leq \ell$ simply because $b^{< 0}_{(u^{-j}, \sigma^{-j})} = 0$ for all $j > C_{< 0}$.
We can use this to bound $\|\mathbf{b}^{< 0}\|_2^2$ as follows.
\begin{align*}
\|\mathbf{b}^{< 0}\|_2^2 &= \sum_{j=1}^{|U| \cdot |\Sigma_V|} \left(b^{< 0}_{(u^{-j}, \sigma^{-j})}\right)^2 \\
&= \sum_{j < \ell} \left(b^{< 0}_{(u^{-j}, \sigma^{-j})}\right)^2 + \sum_{j \geq \ell} \left(b^{< 0}_{(u^{-j}, \sigma^{-j})}\right)^2 \\
&\leq \sum_{j < \ell} \left(b^{< 0}_{(u^{-j}, \sigma^{-j})}\right)^2 + |b^{< 0}_{(u^{-\ell}, \sigma^{-\ell})}| \cdot \|\mathbf{b}^{< 0}\|_1 \\
&\overset{\eqref{eq:term-by-term-bound-negative}}{\leq} \sum_{j < \ell} \left(\frac{2\delta\gamma^*}{j}\right)^2 + \left(\frac{2\delta \gamma^*}{\ell}\right) \cdot \|\mathbf{b}\|_1 \\
&\overset{\eqref{eq:l1-norm-b}}{\leq} \frac{\pi^2}{6} \cdot 4\delta^2(\gamma^*)^2 + \left(\frac{2\delta\gamma^*}{\ell}\right) \cdot 2\delta \\
(\text{From our choice of } \ell \text{ and } \delta\gamma^* \leq 0.1\sqrt{\delta/k}) &\leq \frac{\delta}{k} + \frac{\sqrt{\delta}}{k} \\
&\leq \frac{2\sqrt{\delta}}{k}.
\end{align*}
Using our bounds on $\|\mathbf{b}^{\geq 0}\|_2^2, \|\mathbf{b}^{< 0}\|_2^2$, we can easily bound $\|\mathbf{b}\|_2^2$ by
\begin{align} \label{eq:final-b-norm}
\|\mathbf{b}\|_2^2 = \|\mathbf{b}^{\geq 0}\|_2^2 + \|\mathbf{b}^{< 0}\|_2^2 \leq \frac{4\sqrt{\delta}}{k}.
\end{align}
Next observe that $\|\tilde{\Pi}^j\|_1 = 1$, because each column has exactly a single entry equal to one and the remaining entries equal to zero. Furthermore, $\|\tilde{\Pi}^j\|_\infty = \Delta$ because each row has exactly $\Delta$ entries equal to one\footnote{For every row $(v, \sigma_v)$, these 1-entries are the entries $(u, \sigma_v)$ for all $u \in N(v)$.}. As a result, by Holder's inequality, we have $\|\tilde{\Pi}^j\|_2 \leq \sqrt{\|\tilde{\Pi}^j\|_1 \|\tilde{\Pi}^j\|_\infty} = \sqrt{\Delta}$. From this and from~\eqref{eq:final-b-norm}, we arrive at
\begin{align} \label{eq:b-proj-norm}
\frac{4\sqrt{\delta} \cdot \Delta}{k} \geq \|\mathbf{b}(\tilde{\Pi}^j)^T\|_2^2 = \|\mathbf{v}(\Pi^j)^T\|_2^2,
\end{align}
where the latter follows from our definition of $\mathbf{b}$.
Thus, we have
\begin{align*}
\|\mathbf{w} (\Pi^j)^T\|_2^2 = \|\hat{\mathbf{w}} (\Pi^j)^T + \mathbf{v} (\Pi^j)^T\|_2^2 \leq 2\|\hat{\mathbf{w}}(\Pi^j)^T\|_2^2 + 2\|\mathbf{v}(\Pi^j)^T\|_2^2 \overset{\eqref{eq:b-proj-norm}}{\leq} 2\|\hat{\mathbf{w}} (\Pi^j)^T\|_2^2 + \frac{8\sqrt{\delta} \Delta}{k}.
\end{align*}
Finally, recall from our choice of parameter that $\sqrt{\delta} \Delta \leq 0.1 \sqrt[4]{\delta}$. This, together with the above inequality, implies the desired bound.
\end{proof}
\subsubsection{Decoding Label Cover Assignment}
We now arrive at the last part of the proof, where we show that there exists an assignment that weakly covers at least $\mu = \frac{0.01}{\Delta(\Delta - 1)}$ fraction of vertices in $V$, which completes our soundness proof.
\begin{lemma}
There exists an assignment $\phi'$ of $\mathcal{L}$ such that $\wval(\phi') \geq \mu$.
\end{lemma}
\begin{proof}
We define a (random) assignment $\phi$ for $\mathcal{L}$ as follows:
\begin{itemize}
\item For each $u \in U_{\text{small}}$, let $\phi(u)$ be a random element from $\Sigma_U$ where $\sigma_u \in \Sigma_U$ is selected with probability $\frac{|\hat{w}_{(u, \sigma_u)}|}{\sum_{\sigma \in \Sigma_U} |\hat{w}_{(u, \sigma)}|}$.
\item For each $u \in U_{\text{large}}$, let $\phi(u)$ be an arbitrary element in $\Sigma_U$.
\end{itemize}
We will now argue that $\mathbb{E}_{\phi}[\wval(\phi)] \geq \mu$. Since we assume that $\opt_{(1 - \delta)\gamma^*}^{\mathcal{D}}(\mathbf{w}) \leq 0.96(0.25 q)$, we have
\begin{align*}
0.96(0.25 q) &\geq \opt_{(1 - \delta)\gamma^*}^{\mathcal{D}}(\mathbf{w}) \\
&\geq (0.25 q) \Pr_{j \in [t], \mathbf{s} \in \{\pm 1\}^{V \times \Sigma_V}}\left[\left<\mathbf{w}, \mathbf{s}\Pi^j\right> < (1 - \delta)\gamma^*\right] \;,
\end{align*}
where the second inequality is due to the error from the samples from Step~\ref{sample:label-cover}.
Let $J \subseteq [t]$ contain all $j \in [t]$ such that $\Pr_{\mathbf{s} \in \{\pm 1\}^{V \times \Sigma_V}}\left[\left<\mathbf{w}, \mathbf{s}\Pi^j\right> < (1 - \delta)\gamma^*\right] < 0.99$. The above inequality implies that
\begin{align} \label{eq:num-good-j-bound}
\Pr_{j \in [t]}\left[j \in J\right] > 0.01.
\end{align}
Now, let us fix $j \in J$.
By definition of $J$, we have
\begin{align*}
0.01 &\leq \Pr_{\mathbf{s} \in \{\pm 1\}^{V \times \Sigma_V}}\left[\left<\mathbf{w}, \mathbf{s}\Pi^j\right> \geq (1 - \delta)\gamma^*\right] \\
&\leq \Pr_{\mathbf{s} \in \{\pm 1\}^{V \times \Sigma_V}}\left[|\left<\mathbf{w}, \mathbf{s}\Pi^j\right>| \geq (1 - \delta)\gamma^*\right] \\
&= \Pr_{\mathbf{s} \in \{\pm 1\}^{V \times \Sigma_V}}\left[|\left<\mathbf{w}(\Pi^j)^T, \mathbf{s}\right>|^2 \geq ((1 - \delta)\gamma^*)^2\right] \\
(\text{Markov's inequality}) &\leq \frac{\mathbb{E}_{\mathbf{s} \in \{\pm 1\}^{V \times \Sigma_V}}[|\left<\mathbf{w}(\Pi^j)^T, \mathbf{s}\right>|^2]}{((1 - \delta)\gamma^*)^2} \\
&= \frac{\|\mathbf{w}(\Pi^j)^T\|_2^2}{((1 - \delta)\gamma^*)^2}.
\end{align*}
As a result, we must have $\|\mathbf{w}(\Pi^j)^T\|_2^2 \geq 0.01((1 - \delta)\gamma^*)^2$. We now apply Lemma~\ref{lem:variance-preserved}, which yields
\begin{align} \label{eq:good-j-norm-bound}
\|\hat{\mathbf{w}}(\Pi^j)^T\|_2^2 \geq 0.005((1 - \delta)\gamma^*)^2 - \frac{\sqrt[4]{\delta}}{k} \geq \frac{2}{k} \;.
\end{align}
Using the definition of $\Pi^j$, we may now rewrite $\|\hat{\mathbf{w}}(\Pi^j)^T\|_2^2$ as follows.
\begin{align}
&\|\hat{\mathbf{w}}(\Pi^j)^T\|_2^2 \nonumber \\
&= \sum_{(v, \sigma_v) \in V \times \Sigma_V} ((\hat{\mathbf{w}}(\Pi^j)^T)_{(v, \sigma_v)})^2 \nonumber \\
&= \sum_{(v, \sigma_v) \in V_j \times \Sigma_V} ((\hat{\mathbf{w}}(\Pi^j)^T)_{(v, \sigma_v)})^2 \nonumber \\
&= \sum_{(v, \sigma_v) \in V_j \times \Sigma_V}\left(\sum_{u \in N(v), \sigma_u \in \pi_{(u, v)}^{-1}(\sigma_v)} \hat{w}_{(u, \sigma_u)}\right)^2 \nonumber \\
&= \sum_{(v, \sigma_v) \in V_j \times \Sigma_V} \sum_{u \in N(v)}\left(\sum_{\sigma_u \in \pi_{(u, v)}^{-1}(\sigma_v)} \hat{w}_{(u, \sigma_u)}\right)^2 \nonumber \\
& \qquad + \sum_{(v, \sigma_v) \in V_j \times \Sigma_V} \sum_{u, u' \in N(v) \atop u \ne u'}\left(\sum_{\sigma_u \in \pi_{(u, v)}^{-1}(\sigma_v)} \hat{w}_{(u, \sigma_u)}\right)\left(\sum_{\sigma_{u'} \in \pi_{(u', v)}^{-1}(\sigma_v)} \hat{w}_{(u', \sigma_{u'})}\right) \nonumber \\
&= \sum_{(v, \sigma_v) \in V_j \times \Sigma_V} \sum_{u \in N(v) \cap U_{\text{small}}}\left(\sum_{\sigma_u \in \pi_{(u, v)}^{-1}(\sigma_v)} \hat{w}_{(u, \sigma_u)}\right)^2 \nonumber \\
& \qquad + \sum_{(v, \sigma_v) \in V_j \times \Sigma_V} \sum_{u, u' \in N(v) \cap U_{\text{small}} \atop u \ne u'}\left(\sum_{\sigma_u \in \pi_{(u, v)}^{-1}(\sigma_v)} \hat{w}_{(u, \sigma_u)}\right)\left(\sum_{\sigma_{u'} \in \pi_{(u', v)}^{-1}(\sigma_v)} \hat{w}_{(u', \sigma_{u'})}\right), \label{eq:separation-norm-intermediate}
\end{align}
where the last equality follows from the fact that $\hat{w}_{(u, \sigma_u)} = 0$ for all $u \notin U_{\text{small}}$.
We will now bound the two terms in~\eqref{eq:separation-norm-intermediate} separately. For the first term, we have
\begin{align}
\sum_{(v, \sigma_v) \in V_j \times \Sigma_V} \sum_{u \in N(v) \cap U_{\text{small}}}\left(\sum_{\sigma_u \in \pi_{(u, v)}^{-1}(\sigma_v)} \hat{w}_{(u, \sigma_u)}\right)^2
&\leq \sum_{(v, \sigma_v) \in V_j \times \Sigma_V} \sum_{u \in N(v) \cap U_{\text{small}}}\left(\sum_{\sigma_u \in \pi_{(u, v)}^{-1}(\sigma_v)} |\hat{w}_{(u, \sigma_u)}|\right)^2 \nonumber \\
&= \sum_{u \in U_{\text{small}}}\left(\sum_{\sigma_v \in \Sigma_V} \left(\sum_{\sigma_u \in \pi_{(u, v^j(u))}^{-1}(\sigma_v)} |\hat{w}_{(u, \sigma_u)}|\right)^2\right) \nonumber \\
&\leq \sum_{u \in U_{\text{small}}}\left(\sum_{\sigma_u \in \Sigma_U} |\hat{w}_{(u, \sigma_u)}|\right)^2 \nonumber \\
&= \sum_{u \in U_{\text{small}}} M_u^2 \nonumber \\
&\leq \frac{1}{k} \;, \label{eq:separation-norm-intermediate-first-term}
\end{align}
where the last inequality follows from $M_u \leq 1/k$ for all $u \in U_{\text{small}}$ (by definition) and from $\sum_{u \in U_{\text{small}}} M_u \leq \|\mathbf{w}\|_1 \leq 1$.
We now move on to bound the second term of~\eqref{eq:separation-norm-intermediate}. To do so, let us observe that, for every $u \in U_{\text{small}}, v \in N(u)$ and $\sigma_v \in \Sigma_V$, we have
\begin{align*}
\Pr_{\phi}[\pi_{(u, v)}(\phi(u)) = \sigma_v] &= \sum_{\sigma_u \in \pi^{-1}_{(u, v)}(\sigma_v)} \frac{|\hat{w}_{(u, \sigma_u)}|}{M_u} \\
&\geq k \sum_{\sigma_u \in \pi^{-1}_{(u, v)}(\sigma_v)} |\hat{w}_{(u, \sigma_u)}| \;.
\end{align*}
As a result, we have
\begin{align}
&\sum_{(v, \sigma_v) \in V_j \times \Sigma_V} \sum_{u, u' \in N(v) \cap U_{\text{small}} \atop u \ne u'}\left(\sum_{\sigma_u \in \pi_{(u, v)}^{-1}(\sigma_v)} \hat{w}_{(u, \sigma_u)}\right)\left(\sum_{\sigma_{u'} \in \pi_{(u', v)}^{-1}(\sigma_v)} \hat{w}_{(u', \sigma_{u'})}\right) \nonumber \\
&\leq \sum_{(v, \sigma_v) \in V_j \times \Sigma_V} \sum_{u, u' \in N(v) \cap U_{\text{small}} \atop u \ne u'} \frac{\Pr_{\phi}[\pi_{(u, v)}(\phi(u)) = \sigma_v]}{k} \cdot \frac{\Pr_{\phi}[\pi_{(u', v)}(\phi(u')) = \sigma_v]}{k} \nonumber \\
&= \frac{1}{k^2} \sum_{(v, \sigma_v) \in V_j \times \Sigma_V} \sum_{u, u' \in N(v) \cap U_{\text{small}} \atop u \ne u'} \Pr_{\phi}[\pi_{(u, v)}(\phi(u)) = \pi_{(u', v)}(\phi(u')) = \sigma_v] \nonumber \\
&= \frac{1}{k^2} \sum_{v \in V_j} \sum_{u, u' \in N(v) \cap U_{\text{small}} \atop u \ne u'} \sum_{\sigma_v \in \Sigma_V} \Pr_{\phi}[\pi_{(u, v)}(\phi(u)) = \pi_{(u', v)}(\phi(u')) = \sigma_v] \nonumber \\
&= \frac{1}{k^2} \sum_{v \in V_j} \sum_{u, u' \in N(v) \cap U_{\text{small}} \atop u \ne u'} \Pr_{\phi}[\pi_{(u, v)}(\phi(u)) = \pi_{(u', v)}(\phi(u'))] \nonumber \\
&\leq \frac{1}{k^2} \sum_{v \in V_j} \sum_{u, u' \in N(v) \cap U_{\text{small}} \atop u \ne u'} \Pr_{\phi}[\phi \text{ weakly covers } v] \nonumber \\
&\leq \frac{\Delta(\Delta - 1)}{k^2} \sum_{v \in V_j} \Pr_{\phi}[\phi \text{ weakly covers } v] \;, \label{eq:separation-norm-intermediate-second-term}
\end{align}
where the last inequality follows from the fact that each $v \in V$ has degree $\Delta$.
Combining~\eqref{eq:good-j-norm-bound},~\eqref{eq:separation-norm-intermediate},~\eqref{eq:separation-norm-intermediate-first-term} and~\eqref{eq:separation-norm-intermediate-second-term}, we have
\begin{align*}
\sum_{v \in V_j} \Pr_{\phi}[\phi \text{ weakly covers } v] \geq \frac{k}{\Delta(\Delta - 1)} \;.
\end{align*}
By summing over all $j \in J$ and using the bound from~\eqref{eq:num-good-j-bound}, we have
\begin{align*}
0.01t \cdot \frac{k}{\Delta(\Delta - 1)} &\leq \sum_{j \in J} \sum_{v \in V_j} \Pr_{\phi}[\phi \text{ weakly covers } v] \\
&\leq \sum_{v \in V} \Pr_{\phi}[\phi \text{ weakly covers } v] \\
&= |V| \cdot \mathbb{E}_{\phi}[\wval(\phi)] \\
&\leq kt \cdot \mathbb{E}_{\phi}[\wval(\phi)] \;.
\end{align*}
Equivalently, this means that $\mathbb{E}_{\phi}[\wval(\phi)] \geq \mu$, which implies that there exists an assignment $\phi'$ of $\mathcal{L}$ such that $\wval(\phi') \geq \mu$, as desired.
\end{proof}
\section{Introduction} \label{sec:intro}
In recent years, the design of reliable machine learning systems
for secure-critical applications, including in computer vision and natural language processing,
has been a major goal in the field. One of the main concrete goals in this context
has been to develop classifiers that are robust to {\em adversarial examples}, i.e., small
imperceptible perturbations to the input that can result in erroneous
misclassification~\cite{BiggioCMNSLGR13, SzegedyZSBEGF13, GoodfellowSS14}.
This has led to an explosion of research on designing defenses against adversarial
examples and attacks on these defenses. See, e.g.,~\cite{KM-tutorial} for a recent tutorial
on the topic. Despite significant empirical progress over the past few years,
the broad question of designing computationally efficient classifiers that are
provably robust to adversarial perturbations remains an outstanding theoretical challenge.
In this paper, we focus on understanding the {\em computational complexity} of adversarially robust
classification in the (distribution-independent) agnostic PAC model~\cite{Haussler:92, KSS:94}.
Specifically, we study the learnability of {\em halfspaces} (or linear threshold functions) in this model with respect to $L_p$ perturbations.
A halfspace is any
function $h_{\mathbf{w}}: \mathbb{R}^d \to \{ \pm 1\}$ of the form\footnote{The function $\sgn: \mathbb{R} \to \{ \pm 1\}$ is defined as $\sgn(u)=1$ if $u \geq 0$ and $\sgn(u)=-1$ otherwise.}
$h_{\mathbf{w}}(\mathbf{x}) = \sgn \left(\langle \mathbf{w}, \mathbf{x} \rangle \right)$, where $\mathbf{w} \in \mathbb{R}^d$ is the associated weight vector.
The problem of learning an unknown halfspace has been studied for decades --- starting
with the Perceptron algorithm~\cite{Rosenblatt:58} --- and has arguably been one of
the most influential problems in the development of machine learning~\cite{Vapnik:98, FreundSchapire:97}.
Before we proceed, we introduce the relevant terminology.
Let $\mathcal{C}$ be a concept class of Boolean-valued functions
on an instance space $\mathcal{X} \subseteq \mathbb{R}^d$ and
$\mathcal{H}$ be a hypothesis class on $\mathcal{X}$. The set of allowable perturbations is
defined by a function $\mathcal{U}: \mathcal{X} \to 2^{\mathcal{X}}$.
The robust risk of a hypothesis $h \in \mathcal{H}$ with respect to a distribution
$\mathcal{D}$ on $\mathcal{X} \times \{ \pm 1\}$ is defined as
$\mathcal{R}_{\mathcal{U}}(h, \mathcal{D}) =
\Pr_{(\mathbf{x}, y) \sim \mathcal{D}}[\exists z \in \mathcal{U}(\mathbf{x}), h(\mathbf{z}) \neq y].$
The (adversarially robust) agnostic PAC learning problem for $\mathcal{C}$ is the following:
Given i.i.d. samples from an arbitrary distribution $\mathcal{D}$ on $\mathcal{X} \times \{ \pm 1\}$,
the goal of the learner is to output a hypothesis $h \in \mathcal{H}$ such that with high probability it holds
$\mathcal{R}_{\mathcal{U}}(h, \mathcal{D}) \leq \opt+\epsilon$,
where $\opt = \inf_{f \in \mathcal{C}} \mathcal{R}_{\mathcal{U}}(f, \mathcal{D})$ is the robust risk of the best-fitting function in $\mathcal{C}$.
Unfortunately, it follows from known hardness results that
this formulation is computationally intractable for the class
of halfspaces $\mathcal{C} = \{ \sgn(\langle \mathbf{w}, \mathbf{x} \rangle), \mathbf{w} \in \mathbb{R}^d \}$
under $L_p$ perturbations, i.e, for $\mathcal{U}_{p, \gamma}(\mathbf{x}) = \{ \mathbf{z} \in \mathcal{X}: \|\mathbf{z}-\mathbf{x}\|_p \leq \gamma \}$, for some $p \geq 2$. (The reader is referred to Appendix~\ref{app:agnostic-hard} for an explanation.)
To be able to obtain computationally efficient algorithms, we relax the above definition in two ways:
(1) We allow the hypothesis to be robust within a slightly smaller perturbation region,
and (2) We introduce a small constant factor approximation in the error guarantee. In more detail,
for some constants $0< \nu <1$ and $\alpha>1$,
our goal is to efficiently compute a hypothesis $h$ such that with high probability
\begin{equation} \label{eqn:bicriterion-risk}
\mathcal{R}_{\mathcal{U}_{p, (1-\nu) \gamma}}(h, \mathcal{D}) \leq \alpha \cdot \opt_{p, \gamma}+\epsilon \;,
\end{equation}
where $\opt_{p, \gamma} = \inf_{f \in \mathcal{C}} \mathcal{R}_{\mathcal{U}_{p,\gamma}}(f, \mathcal{D})$.
(Note that for $\nu = 0$ and $\alpha = 1$, we obtain the original definition.) An interesting setting
is when $\nu$ is a small constant close to $0$, say $\nu = 0.1$, and $\alpha = 1+\delta$, where
$0< \delta<1$. In this paper, we characterize the computational complexity of this problem
with respect to {\em proper} learning algorithms, i.e., algorithms that output
a halfspace hypothesis.
Throughout this paper, we will assume that the domain of our functions
is bounded in the $d$-dimensional $L_p$ unit ball $\mathbb{B}_p^d$. All our results immediately extend to general domains with a (necessary) dependence on the diameter of the feasible set.
A simple but crucial observation leveraged in our work is the following:
The adversarially robust learning problem of halfspaces under $L_p$ perturbations (defined above) is essentially equivalent to the classical
problem of agnostic proper PAC learning of halfspaces with an $L_p$ margin.
Let $p \geq 2$, $q$ be the dual exponent of $p$, i.e., $1/p+1/q=1$.
The problem of agnostic proper PAC learning of halfspaces with an $L_p$ margin is the following:
The learner is given i.i.d. samples from a distribution
$\mathcal{D}$ over $\mathbb{B}_p^d \times \{\pm 1\}$.
For $\mathbf{w} \in \mathbb{B}_q^d$, its \emph{$\gamma$-margin error} is defined as $\err_\gamma^{\mathcal{D}}(\mathbf{w}) := \Pr_{(\mathbf{x}, y) \sim \mathcal{D}}[\sgn(\left<\mathbf{w}, \mathbf{x}\right> - y \cdot \gamma) \ne y]$. We also define $\opt_\gamma^\mathcal{D} := \min_{\mathbf{w} \in \mathbb{B}_q^d} \err_\gamma^{\mathcal{D}}(\mathbf{w})$.
An algorithm is a proper \emph{$\nu$-robust $\alpha$-agnostic learner} for $L_p$-$\gamma$-margin halfspace if, with probability at least $1 - \tau$, it outputs a halfspace $\mathbf{w} \in \mathbb{B}_q^d$ with
\begin{equation} \label{eqn:agnostic-margin}
\err_{(1 - \nu)\gamma}^{\mathcal{D}}(\mathbf{w}) \leq \alpha \cdot \opt^{\mathcal{D}}_\gamma + \epsilon \;.
\end{equation}
(When unspecified, the failure probability $\tau$ is assumed to be 1/3. It is well-known and easy
to see that we can always achieve arbitrarily small value of $\tau$ at the cost of $O(\log(1/\tau))$ multiplicative factor in the running time and sample complexity.)
We have the following basic observation, which implies that the learning objectives \eqref{eqn:bicriterion-risk} and \eqref{eqn:agnostic-margin} are equivalent. Throughout this paper, we will state our contributions using the margin formulation \eqref{eqn:agnostic-margin}.
\begin{fact} \label{fact:adversarial-robustness-vs-margin}
For any non-zero $\mathbf{w} \in \mathbb{R}^d$, $\gamma \geq 0$ and $\mathcal{D}$ over $\mathbb{R}^d \times \{\pm 1\}$, $\mathcal{R}_{\mathcal{U}_{p,\gamma}}(h_{\mathbf{w}}, \mathcal{D}) = \err^{\mathcal{D}}_{\gamma}(\frac{\mathbf{w}}{\|\mathbf{w}\|_q})$.
\end{fact}
\subsection{Our Contributions} \label{sec:results}
Our main positive result is a robust and agnostic proper learning algorithm for $L_p$-$\gamma$-margin halfspace with near-optimal running time:
\begin{theorem}[Robust Learning Algorithm] \label{thm:learning-algo-main}
Fix $2 \leq p < \infty$ and $0< \gamma <1$. For any $0<\nu, \delta<1$,
there is a proper $\nu$-robust $(1 + \delta)$-agnostic learner for $L_p$-$\gamma$-margin halfspace that draws $O(\frac{p}{\epsilon^2 \nu^2 \gamma^2})$ samples and runs in time $(1/\delta)^{O\left(\frac{p}{\nu^2 \gamma^2}\right)} \cdot \poly(d/\epsilon)$.
Furthermore, for $p = \infty$, there is a proper $\nu$-robust $(1 + \delta)$-agnostic learner for $L_{\infty}$-$\gamma$-margin halfspace that draws $O(\frac{\log d}{\epsilon^2 \nu^2 \gamma^2})$ samples and runs in time $d^{O\left(\frac{\log(1/\delta)}{\nu^2 \gamma^2}\right)} \cdot \poly(1/\epsilon)$.
\end{theorem}
To interpret the running time of our algorithm, we consider the setting $\delta = \nu = 0.1$. We note two different regimes. If $p \geq 2$ is a fixed constant, then our algorithm runs in time $2^{O(1/\gamma^2)} \poly(d/\epsilon)$. On the other hand, for $p = \infty$, we obtain a runtime of $d^{O(1/\gamma^2)} \poly(1/\epsilon)$.
That is, the $L_{\infty}$ margin case (which corresponds to adversarial learning with $L_{\infty}$ perturbations)
appears to be computationally the hardest.
As we show in Theorem~\ref{thm:running-time-lower-bound}, this fact is inherent for proper learners.
Our algorithm establishing Theorem~\ref{thm:learning-algo-main} follows via a simple and unified
approach, employing a reduction from online (mistake bound) learning
~\cite{Lit87}. Specifically, we show that any computationally efficient $L_p$ online learner for halfspaces with margin guarantees and mistake bound $M$ can be used in a black-box manner to obtain an algorithm for our problem with runtime roughly $\poly(d/\epsilon) (1/\delta)^{M}$. Theorem~\ref{thm:learning-algo-main} then follows by applying known results from the online learning literature~\cite{Gentile01}.
For the special case of $p=2$ (and $\nu = 0.1$), recent work~\cite{DiakonikolasKM19} gave a sophisticated algorithm for our problem with running time $\poly(d/\epsilon) 2^{\tilde{O}(1/(\delta \gamma^2))}$. We note that our algorithm has significantly better dependence on the parameter $\delta$ (quantifying the approximation ratio), and better dependence on $1/\gamma$. Importantly, our algorithm is much simpler and immediately generalizes to all $L_p$ norms.
Perhaps surprisingly, the running time of our algorithm is nearly the best possible for proper learning. For constant $p \geq 2$, this follows
from the hardness result of~\cite{DiakonikolasKM19}. Furthermore, we prove a tight running time lower bound
for robust $L_{\infty}$-$\gamma$-margin proper learning of halfspaces.
Roughly speaking, we show that for some sufficiently small constant $\nu > 0$, one cannot hope to significantly speed-up our algorithm for $\nu$-robust $L_{\infty}$-$\gamma$-margin learning of halfspaces. Our computational hardness result is formally stated below.
\begin{theorem}[Tight Running Time Lower Bound] \label{thm:running-time-lower-bound}
There exists a constant $\nu > 0$ such that, assuming the (randomized) Gap Exponential Time Hypothesi
,
there is no proper $\nu$-robust 1.5-agnostic learner for $L_{\infty}$-$\gamma$-margin halfspace that runs in time $f(1/\gamma) \cdot d^{o(1 / \gamma^2)} \poly(1/\epsilon)$ for any function $f$.
\end{theorem}
As indicated above, our running time lower bound is based on the so-called Gap Exponential Time Hypothesis (Gap-ETH),
which roughly states that no subexponential time algorithm can approximate 3SAT to within $(1 - \epsilon)$ factor,
for some constant $\epsilon > 0$. Since we will not be dealing with Gap-ETH directly here,
we defer the formal treatment of the hypothesis and discussions on its application to Section~\ref{subsec:eth}.
We remark that the constant $1.5$ in our theorem is insignificant.
We can increase this ``gap'' to any constant less than 2.
We use the value $1.5$ to avoid introducing an additional variable.
Another remark is that Theorem~\ref{thm:running-time-lower-bound} only applies for a small constant $\nu > 0$.
This leaves the possibility of achieving, e.g., a faster 0.9-robust $L_{\infty}$-$\gamma$-margin learner for halfspaces,
as an interesting open problem.
\subsection{Related Work}
A sequence of recent works~\cite{CullinaBM18, SchmidtSTTM18, BubeckLPR19, MontasserHS19}
has studied the sample complexity of adversarially robust PAC learning for general concept
classes of bounded VC dimension and for halfspaces in particular.
\cite{MontasserHS19} established an upper bound on the sample complexity of PAC learning any concept class with finite VC dimension.
A common implication of the aforementioned works is that, for some concept classes,
the sample complexity of adversarially robust PAC learning
is higher than the sample complexity of (standard) PAC learning.
For the class of halfspaces, which is the focus of the current paper,
the sample complexity of adversarially robust agnostic PAC learning was shown to be
essentially the same as that of (standard) agnostic PAC learning
~\cite{CullinaBM18, MontasserHS19}.
Turning to computational aspects,~\cite{BubeckLPR19, DegwekarNV19} showed that there exist classification tasks
that are efficiently learnable in the standard PAC model, but are computationally hard in the adversarially robust
setting (under cryptographic assumptions). Notably, the classification problems shown hard are
artificial, in the sense that they do not correspond to natural concept classes.
\cite{AwasthiDV19} shows that adversarially robust proper learning of degree-$2$ polynomial threshold
functions is computationally hard, even in the realizable setting. On the positive side, \cite{AwasthiDV19} gives a polynomial-time
algorithm for adversarially robust learning of halfspaces under $L_{\infty}$ perturbations, again in the realizable setting.
More recently,~\cite{MonGDS20} generalized this upper bound to a broad class of perturbations, including $L_p$ perturbations. Moreover,~\cite{MonGDS20} gave an efficient algorithm for learning halfspaces with random
classification noise~\cite{AL88}. We note that all these algorithms are proper.
The problem of agnostically learning halfspaces with a margin
has been studied extensively. A number of prior works~\cite{BenDavidS00, SSS09, SSS10, LS:11malicious, BirnbaumS12, DiakonikolasKM19} studied the case of $L_2$ margin and gave a range of time-accuracy tradeoffs for the problem.
The most closely related prior work is the recent work~\cite{DiakonikolasKM19}, which gave a proper
$\nu$-robust $\alpha$-agnostic learning for $L_2$-$\gamma$-margin halfspace with near-optimal running time when $\alpha, \nu$ are universal constants, and
a nearly matching computational hardness result. The algorithm of the current paper
broadly generalizes, simplifies, and improves the algorithm of~\cite{DiakonikolasKM19}.
\subsection{Organization}
We describe our algorithm and prove Theorem~\ref{thm:learning-algo-main} in Section~\ref{sec:algo}. In Section~\ref{sec:prelim}, we provide further preliminaries needed for our lower bound proof. We then prove our main hardness result (Theorem~\ref{thm:running-time-lower-bound}) in Section~\ref{sec:hardness}. Finally, we conclude with open questions in Section~\ref{sec:open-q}.
\section{Upper Bound: From Online to Adversarially Robust Agnostic Learning}
\label{sec:algo}
\input{algo}
\section{Additional Background for Hardness Result}
\label{sec:prelim}
\input{prelim}
\section{Tight Running Time Lower Bound}
\label{sec:hardness}
\input{hardness-merged}
\section{Conclusions and Open Problems}
\label{sec:open-q}
\input{open}
\bibliographystyle{alpha}
\subsection{Exponential Time Hypotheses}
\label{subsec:eth}
Recall that, in the 3-satisfiability (3SAT) problem, we are given a set of clauses, where each clause is an OR of at most three literals. The goal is to determine whether there exists an assignment that satisfies all clauses. The Exponential Time Hypothesis (ETH)~\cite{IP01,IPZ01} asserts that there is no sub-exponential time algorithm for 3SAT. ETH is of course a strengthening of the famous $P \ne NP$ assumption. In recent years, this assumption has become an essential part of modern complexity theory, as it allows one to prove tight running time lower bounds for many NP-hard and parameterized problems. See, e.g.,~\cite{LokshtanovMS11} for a survey on the topic.
For our lower bound, we use a strengthening of ETH, called Gap-ETH. Roughly speaking, Gap-ETH says that even finding an \emph{approximate} solution to 3SAT is hard. This is stated more precisely below:
\begin{hypothesis}[(Randomized) Gap Exponential Time Hypothesis (Gap-ETH)~\cite{Dinur16,MR17}] \label{hyp:gap-eth}
There exists a constant $\zeta > 0$ such that no randomized $2^{o(n)}$-time algorithm can, given a 3SAT instance on $n$ variables, distinguish between the following two cases correctly with probability $2/3$:
\begin{itemize}
\item (Completeness) There exists an assignment that satisfies all clauses.
\item (Soundness) Every assignment violates at least $\zeta$ fraction of the clauses.
\end{itemize}
\end{hypothesis}
Although proposed relatively recently, Gap-ETH is intimately related to a well-known open question whether linear size probabilistic checkable proofs exist for 3SAT; for more detail, please refer to the discussion in~\cite{Dinur16}. Gap-ETH has been used as a starting point for proving numerous tight running time lower bounds against approximation algorithms (e.g.,~\cite{Dinur16,MR17,BennettGS17,AggarwalS18,JainKR19}) and parameterized approximation algorithms (e.g.,~\cite{ChalermsookCKLM17,DinurM18,BhattacharyyaGS18,Cohen-AddadG0LL19}). Indeed, we will use one such result as a starting point of our hardness reduction.
\subsection{Hardness of Label Cover}
The main component of our hardness result will be a reduction from the \emph{Label Cover}
problem\footnote{Label Cover is sometimes referred to as \emph{Projection Game} or \emph{Two-Prover One-Round Game}.}, which is a classical problem in hardness of approximation literature that is widely used as a starting point for proving strong NP-hardness of approximation results (see, e.g., \cite{AroraBSS97,Hastad96,Hastad01,Feige98}).
\begin{definition}[Label Cover] \label{def:label-cover-full}
A \emph{Label Cover} instance $\mathcal{L} = (U, V, E, \Sigma_U, \Sigma_V, \{\pi_e\}_{e \in \Sigma})$ consists of
\begin{itemize}
\item a bi-regular bipartite graph $(U, V, E)$, referred to as the \emph{constraint graph},
\item \emph{label sets} $\Sigma_U$ and $\Sigma_V$,
\item for every edge $e \in E$, a \emph{constraint} (aka \emph{projection}) $\pi_e: \Sigma_U \to \Sigma_V$.
\end{itemize}
A labeling of $\mathcal{L}$ is a function $\phi: U \to \Sigma_U$.
We say that $\phi$ \emph{covers} $v \in V$ if there exists
$\sigma_v \in \Sigma_V$ such that\footnote{This is equivalent to
$\pi_{(u_1, v)}(\phi(u_1)) = \pi_{(u_2, v)}(\phi(u_2))$ for all neighbors $u_1, u_2$ of $v$.}
$\pi_{(u, v)}(\phi(u)) = \sigma_v$ for all\footnote{For every $a \in U \cup V$,
we use $N(a)$ to denote the set of neighbors of $a$ (with respect to the graph $(U, V, E)$).}
$u \in N(v)$. The \emph{value} $\phi$, denoted by $\val_{\mathcal{L}}(\phi)$,
is defined as the fraction of $v \in V$ covered by $\phi$.
The value of $\mathcal{L}$, denoted by $\val(\mathcal{L})$, is defined as $\max_{\phi: U \to \Sigma_U} \val(\phi)$.
Moreover, we say that $\phi$ \emph{weakly covers} $v \in V$ if there exist distinct neighbors $u_1, u_2$
of $v$ such that $\pi_{(u_1, v)}(\phi(u_1)) = \pi_{(u_2, v)}(\phi(u_2))$. The \emph{weak value} of $\phi$,
denoted by $\wval(\phi)$, is the fraction of $v \in V$ weakly covered by $\phi$.
The weak value of $\mathcal{L}$, denoted by $\wval(\mathcal{L})$, is defined as $\max_{\phi: U \to \Sigma_U} \wval(\phi)$.
For a Label Cover instance $\mathcal{L}$, we use $k$ to denote $|U|$ and $n$ to denote
$|U| \cdot |\Sigma_U| + |V| \cdot |\Sigma_V|$.
\end{definition}
The goal of Label Cover is to find an assignment with maximium value.
In our reduction, we will also need an additional notion of ``decomposability'' of a Label Cover instance. Roughly speaking, an instance is {\em decomposable} if we can partition $V$
into different parts such that each $u \in U$ has exactly
one induced edge to the vertices in each part:
\begin{definition}[Decomposable Label Cover] \label{def:label-cover-decomposable-main-body}
A \emph{Label Cover} instance $\mathcal{L} = (U, V, E, \Sigma_U, \Sigma_V, \{\pi_e\}_{e \in E})$
is said to be \emph{decomposable} if there exists a partition of $V$ into $V_1 \cup \cdots \cup V_t$
such that, for every $u \in U$ and $j \in [t]$, $|N(u) \cap V_j| = 1$.
We use the notation $v^j(u)$ to the denote
the unique element in $N(u) \cap V_j$.
\end{definition}
Several strong inapproximability results for Label Cover
are known~\cite{Raz98,MoshkovitzR10,DinurS14}. To prove a tight
running time lower bound, we require an inapproximability result for Label Cover
with a tight running lower bound as well.
Observe that we can solve Label Cover in time $n^{O(k)}$ by enumerating through all possible
$|\Sigma_U|^{|U|} = n^{O(k)}$ assignments and compute their values. The following result shows that, even if we only aim for a constant approximation ratio, no algorithm that can be significantly faster than this ``brute-force'' algorithm.
\begin{theorem}[\cite{M20}] \label{thm:label-cover-hardness}
Assuming Gap-ETH, for any function $f$ and any constants $\Delta \in \mathbb{N} \setminus \{1\}, \mu \in (0, 1)$, there is no $f(k) \cdot n^{o(k)}$-time algorithm that can, given a decomposable Label Cover instance $\mathcal{L} = (U, V = V_1 \cup \cdots \cup V_t, E, \Sigma_U, \Sigma_V, \{\pi_e\}_{e \in E})$ whose right-degree is equal to $\Delta$, distinguish between
\begin{itemize}
\item (Completeness) $\val(\mathcal{L}) = 1$,
\item (Soundness) $\wval(\mathcal{L}) < \mu$,
\end{itemize}
where $k := |U|$ and $n := |U| \cdot |\Sigma_U| + |V| \cdot |\Sigma_V|$.
\end{theorem}
We remark here that the above theorem is not exactly the same as stated in~\cite{M20}. We now briefly explain how to derive the version above from the one in~\cite{M20}. Specifically, in~\cite{M20}, the decomposability of the instance $\mathcal{L}$ is not stated; rather, the instance there has the following property: $V$ is simply all subsets of size $\Delta$ of $U$, and, for any vertex $\{u_1, \dots, u_{\Delta}\} \in V$, its neighbors are $u_1, \dots, u_{\Delta} \in U$. Now, we can assume w.l.o.g. that $k$ is divisible by $\Delta$ by expanding each vertex $u \in U$ to $\Delta$ new vertices $u^1, \dots, u^\Delta$ and replicate each vertex in $\{u_1, \dots, u_{\Delta}\} \in V$ to $\Delta^{\Delta}$ new vertices $\{u_1^{\xi(1)}, \dots, u_{\Delta}^{\xi(\Delta)}\}$ for all $\xi: [\Delta] \to [\Delta]$. Once we have that $k$ is divisible by $\Delta$, Baranyai's theorem~\cite{Baranyai75} immediately implies the decomposability of the instance.
\subsection{Anti-Concentration}
It is well-known that, if we take $m$ i.i.d. Rademacher random variables, their sum divided by $\sqrt{m}$ converges in distribution to the standard normal distribution (see, e.g.,~\cite{Berry41,Esseen42}). As a consequence, this immediately implies the following ``anti-concentration'' style result:
\begin{lemma} \label{lem:anti-concen}
There exists $C \in (0, 1)$ and $m_0 > 0$ such that, for any $m \geq m_0$, we have
\begin{align*}
\Pr_{X_1, \dots, X_m}[X_1 + \cdots + X_m \geq C \sqrt{m}] \geq 0.4 \;,
\end{align*}
where $X_1, \dots, X_m$ are i.i.d. Rademacher random variables.
\end{lemma}
Note that the constant 0.4 above can be replaced by any constant strictly less than 0.5. We only use 0.4 here to avoid introducing additional variables. | {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,664 |
title: 'Do You Feel Lonely'
date: 20/04/2019
---
**This is Easy Reading Edition of the Sabbath School. For the regular Adult version with Teacher comments and EGW notes please open the top lesson on the main screen**
### Read for this week's lesson
Ecclesiastes 4:9–12; 1 Corinthians 7:25–34; Malachi 2:16; Hosea 2:19, 20.
> <p>Memory verse</p>
> "Then the Lord God said, 'I see that it is not good for the man to be alone. I will make the companion [female helper] he needs, one just right for him' " (Genesis 2:18, ERV).
**MANY PEOPLE FEEL LONELY**. They live alone. They do not have a lot of friends. They are not married. They have no children. Their family does not live close by. Or maybe they have no family at all. So, they spend a lot of time alone. That makes them feel sad.
In 2016, The New York Times newspaper wrote a story about lonely people. The scientists in the news story gave a report about lonely people. The report said that many people felt very lonely in life. Sadly, this problem is worldwide. And it is very real.
Was it ever God's plan for humans to feel alone? No! From the start, God wanted us to feel close to other people. He wanted us to be close to our families, the person we married, our children, and our friends. God planned it that way in Eden. But then sin came. Nothing has been right since then.
Most of us have felt lonely at some time in our lives. During that time, we may wish we had someone to talk to. This week, we will look at the questions: What does it mean to feel lonely? Why are we alone at different times in our lives? And why is having good friends so important? | {
"redpajama_set_name": "RedPajamaGithub"
} | 4,004 |
\section{Introduction}
Studying the convergence of random walks on finite groups, and in particular the problem of generating group elements according to a fixed probability distribution has a long history \cite{scarabotti, diaconis, diacoste2, saloff}. Of particular interest for the purposes of this paper is the important work of Diaconis and Ram \cite{diaram}, who compare systematic scanning techniques with random scanning techniques in the context of generating elements of a finite Coxeter group $W$ using the Metropolis algorithm.
First introduced by Metropolis, Rosenbluth, Rosenbluth, Teller, and Teller \cite{met}, the Metropolis algorithm gives a method for sampling from a probability distribution $\pi$ by modifying an existing Markov chain to produce a new chain with stationary distribution $\pi$. This proves particularly useful for simulating configurations of particles with an associated energy (e.g., the influence that neighboring particles exert on each other). Later applications of the Metropolis algorithm include the simulation of Ising models, initially developed to model a ferromagnet but (surprisingly) also of use in image analysis and Gibbs sampling \cite{cai, fishman}. See \cite{liu} for additional applications.
The Metropolis algorithm has the advantage of being straightforward to construct and implement; however, in analyzing the rate of convergence to $\pi$ (the \textit{mixing time}) rigorous bounds are often dependent on the specific situation (see \cite{pedthesis} for a review of the existing literature for spin systems alone). Further, these methods are most often examples of \textit{random scan Markov chains} in that the process involved is that of selecting a site or set of sites to update at random.
A more intuitively appealing and often more frequently used method in experimental work is that of a \textit{systematic scan Markov chain}: a method to cycle through and update the sites in a deterministic order.
While such scanning strategies may seem intuitive for use in sampling from $\pi$, they have proven difficult to analyze in many situations.
In \cite{diaram} Diaconis and Ram use the Metropolis algorithm construction to produce Markov chains $M_1,M_2,\dots,$ $M_{n-1}$ corresponding to multiplication by the generators $r_1,\cdots, r_{n-1}$ of a Coxeter group $W$.
These Markov chains provide systematic scanning strategies for multiplying by generators of $W$ (for an explicit description of $M_i$ and the corresponding random walk see Section \ref{thewalk}). Diaconis and Ram \cite{diaram} show that convergence of the short systematic scan occurs in the same number of steps as that of a random scan.
The key insight that allows for analysis of the Metropolis scans is the translation of the Markov chains $M_i$ into left multiplication operators in the \textit{Iwahori-Hecke Algebra} corresponding to $W$.
Hecke algebras arise naturally in the extension of Schur-Weyl duality to general centralizer algebras. More relevant for this paper is an alternative definition of the Hecke algebra in terms of \textit{braids}. The thesis \cite{simplylaced} gives
a thorough introduction to braids and their relationship with the Hecke algebra.
Let $b_1,\dots, b_n\in\mathbb{R}$ with $b_1<\cdots <b_n$. An $n$-strand braid is a disjoint union of $n$ smooth curves in $\mathbb{R}^3$ connecting the points $\{(b_1,1,0),(b_2,1,0),\dots, (b_n,1,0)\}$ with $\{(b_1,0,0),(b_2,0,0),\dots, (b_n,0,0)\}$ so that they intersect each parallel plane $y=t$ as $t$ ranges between $0$ and $1$ only once. A braid can be represented by its 2-dimensional projection, its \textit{braid diagram}, and connecting the top strands to the bottom strands of a braid diagram gives rise to a \textit{link}. Two links are \textit{isotopic} if they are related by a sequence of Reidemeister moves (defined in Section \ref{BMWdefs}), and, in fact, every isotopic oriented link can be represented by the closure of a braid \cite{simplylaced}. The braid group has a presentation in terms of generators $T_{r_1},\dots, T_{r_{n-1}}$ corresponding to certain braid diagrams. Remarkably, adding a quadratic relation to this presentation yields the Hecke algebra
Under this definition of the Hecke algebra there is a natural generalization to the \textit{Birman-Murakami-Wenzl (BMW) algebra}.
By now allowing any two points in $\{(b_1,1,0),(b_2,1,0),\dots, (b_n,1,0)\}\cup\{(b_1,0,0),(b_2,0,0),\dots, (b_n,0,0)\}$ to be connected, we have the definition of an \textit{n-tangle}, which gives rise to the idea of a \textit{tangle diagram} by considering its two-dimensional projection. We define tangle diagrams in detail in Section \ref{BMWdefs}. As with the algebra associated to braid diagrams, an algebra is associated to these tangle diagrams. Defined independently as the \textit{Kauffman tangle algebra} by Murakami \cite{murakami} and algebraically by Birman and Wenzl \cite{birmanwenzl}, it was shown in an unpublished paper by Wasserman \cite{wasser} that these two notions are equivalent, giving rise to the single BMW algebra.
In \cite{diaram}, Diaconis and Ram consider the problem of systematically generating elements of a finite Coxeter group $W$. In terms of the group algebra $\mathbb{C}[W]$, this problem is equivalent to generating elements of the basis $W$ of $\mathbb{C}[W]$. We extend these ideas to the BMW algebra. The Metropolis algorithm in this context gives rise to systematic scanning strategies for generating basis elements via multiplication of generators.
As the diagrams forming the BMW monoid basis of the BMW algebra are tangles, scanning strategies for generating BMW monoid elements have applications arising in physics: random generation of links and tangles has been of use in \cite{rangenknots, componentlinks, enumlink}. As in \cite{diaram}, our algorithm gives rise to a natural random walk, in this case on the BMW and Brauer monoids, defined in Section \ref{thewalk}. We translate the random walk into multiplication in the BMW algebra: for $\mathscr{T}_{r_i}, \mathscr{T}_{e_i}$ left multiplication operators in the BMW algebra.
\begin{theorem}\label{leftmult1} The chain $K_i$ arising from the Metropolis algorithm is the same as the matrix of left multiplication by
$$\theta\mathscr{T}_{r_i}+(1-\theta)\mathscr{T}_{e_i}.$$
\end{theorem}
The main tool used in the analysis in \cite{diaram} is Proposition 4.6, which translates the total variation norm into an inner product on the Iwahori-Hecke algebra $H$ arising from a trace on $H$. Plancherel's theorem then allows for bounds using the dimensions and characters of representations of $H$.
We extend the natural trace function on the Hecke algebra to the BMW algebra to provide an analogue of Proposition 4.6 (Theorem \ref{main2}). We develop a trace form $\langle,\rangle_{BMW}$ to study the walk, similarly enabling the use of tools from representation theory to analyze the time to stationarity of such walks. We consider submatrices $\hat{K}$ of $K_i$ with respect to a shifted basis. Let $\hat{\pi}$ denote the stationary distribution of $\hat{K}$.
\begin{theorem}\label{main2}
$$\| [\hat{K}^n/\hat{\pi}]_{x}-1\|^2_2 \leq \|[\hat{K}^n]_{x}-1\|_{BMW}^2.$$
\end{theorem}
Thus, studying the time to stationarity of $\hat{K}$ can be achieved by studying $\|[\hat{K}^n]_{x}-1\|_{BMW}^2$. This opens up representation theoretic tools---in particular the dimensions and traces of representations of the BMW algebra---for studying the random walk.
We begin in Sections \ref{probprelim} and \ref{prelim2} with the preliminaries needed from the probability theory and the representation theory of semisimple algebras. We also give a presentation of the Brauer and BMW algebras. In Section \ref{thewalk} we describe the random walk arising from the Metropolis algorithm, and prove Theorem \ref{leftmult1}. We continue in Section \ref{walkanalysis} with analysis of the walk, recasting it in terms of a translated basis, constructing a trace form to bound the time to stationarity, and proving Theorem \ref{main2}.
\section{Preliminaries: Probability Theory}\label{probprelim}
Background on Markov chains can be found in many standard probability texts (see eg \cite{feller}). The book of Levin, Peres, and Wilmer \cite{levinetall} gives a particularly thorough introduction to Markov chains, including classification of states and the Metropolis algorithm, while \cite{diaram} gives a concise introduction to the probabilistic background needed. We will follow the notation and outline of \cite{diaram}.
\subsection{Markov Chains}\label{L2norm}
A finite Markov chain with state space $X$ is a process that moves among states in $X$ such that the conditional probability of moving from state $x$ to state $y$ is independent of the preceding sequence of states. More formally:
\begin{definition}
A \textbf{Markov chain} on a finite set $X$ is a matrix $K=(K(x,y))_{x,y\in X}$ such that $K(x,y)\in[0,1]$ and for all $x\in X$, $$\sum_{y\in X} K(x,y)=1.$$ We call $X$ the \textbf{state space}.
\end{definition}
Note that $K(x,y)$ gives the probability of moving from $x$ to $y$ in one step, while $K^m(x,y)$ gives the probability of moving from $x$ to $y$ in $m$ steps.
\begin{definition} A Markov chain $K$ is \textbf{irreducible} if for each $x,y\in X$, there exists an integer $m$ such that $K^m(x,y)>0$. Let $T(x)$ denote the minimum $t$ such that $K^t(x,x)>0$. Then $K$ is \textbf{aperiodic} if $$\gcd_x(T(x))=1.$$
\end{definition}
Note that if $K$ is irreducible and aperiodic, there exists an integer $r$ such that $K^r(x,y)>0$ for all $x,y\in X$ \cite[Proposition 1.7]{levinetall}.
\begin{definition} A Markov chain is \textbf{reversible} if there exists a probability distribution $\pi:X\rightarrow [0,1]$ such that for all $x,y\in X$,
$$\pi(x)K(x,y)=\pi(y)K(y,x).$$ We call $\pi$ the \textbf{stationary distribution} of $K$.
\end{definition}
An irreducible, aperiodic, reversible Markov chain $K$ converges to its stationary distribution:
$$\lim_{m\rightarrow\infty} K^m(x,y)=\pi(y).$$
The Metropolis construction introduced in Section \ref{met} produces a reversible Markov chain with a chosen stationary distribution. Our interest is in the time to stationarity of such chains.
\begin{definition}
Let $K^m_x$ denote the probability distribution $K^m(x,\cdot)$. The \textbf{total variation distance} from $K^m_x$ to $\pi$ is
$$\vert K^m_x-\pi\vert_{TV}:=\max_{A\subseteq X}\vert \sum_{y\in A} K^m(x,y)-\pi(y)\vert.$$
\end{definition}
For $L^2(\pi)$ the space of functions $f:X\rightarrow\mathbb{R}$, equipped with the inner product
$$\langle f,g\rangle_2=\sum f(x)g(x)\pi(x),$$ the total variation distance is bounded by the $L^2(\pi)$ norm:
\begin{lemma}\label{tvl2}\cite[Lemma 2.3]{diaram}
For $f\in L^2(\pi)$, $$|f|_{TV}^2\leq \frac{1}{4}\|f/\pi\|_2^2,$$ where $f/\pi(x)=0$ if $\pi(x)=0$.
\end{lemma}
\subsection{The Metropolis Algorithm}\label{met}
Given a symmetric Markov chain $P$ and a probability distribution $\pi$, the Metropolis algorithm modifies $P$ to produce a reversible Markov chain $M$ with stationary distribution $\pi$:
$$M(x,y)=\left\{
\begin{array}{ll}
\displaystyle P(x,y)& \text{if } x\neq y \text{ and } \pi(y)\geq \pi(x),\\
\displaystyle P(x,y)\frac{\pi(y)}{\pi(x)}& \text{if } x\neq y \text{ and } \pi(y)<\pi(x),\\
\displaystyle P(x,x)+\sum_{\pi(z)<\pi(x)} P(x,z)\left(1-\frac{\pi(z)}{\pi(x)}\right)& \text{if } x=y.
\end{array}\right.$$
While $M(x,y)$ is reversible with stationary distribution $\pi$, irreducibility and aperiodicity are not guaranteed. In particular, the Markov chains we consider in Section \ref{thewalk} are aperiodic but not irreducible. To analyze these chains we consider their \textit{closed communication classes}.
\begin{definition}
Let $K$ be a Markov chain with state space $X$. For $x,y\in X$, $y$ \textbf{is accessible} from $x$, denoted $x\rightarrow y$, if $x$ can reach $y$ in finitely many steps. We say $x$ \textbf{communicates with} $y$, denoted $x\leftrightarrow y$, if $x\rightarrow y$ and $y\rightarrow x$. The equivalence classes under the relation $\leftrightarrow $ are the \textbf{communication classes} of $K$. A communication class $C$ is \textbf{closed} if for $x\in C$ and for all $y\notin C$, $y$ is not accessible from $x$.
\end{definition}
Note that studying the time to stationarity of a reversible, aperiodic Markov chain $K$ reduces to studying the time to stationarity of the closed communication classes of $K$.
\subsection{Systematic Scans}
The Metropolis algorithm, in the context of generating elements of a group, provides systematic and random scanning strategies. For example, for each generator $r_i=(i\;i+1)$ of $S_n$, let
$$ P_i(x,y) = \left\{
\begin{array}{ll}
1 & \text{if}\;y=r_ix,\\
0 & \text{else}.
\end{array}
\right.
$$
Then for $l_{S}$ the length function on words in $S_n$, let $\pi$ be the probability distribution $$\pi(x)=\frac{\theta^{-l_{S}(x)}}{\displaystyle\sum_{w\in S_n} \theta^{-l_{S}(w)}}.$$
The Metropolis algorithm construction then produces Markov chains $M_1,M_2,\dots,$ $M_{n-1}$ corresponding to multiplication by the generators $r_1,\cdots, r_{n-1}$. For an explicit description see Section \ref{thewalk}.
A choice of infinite sequence $\{i_l\}_{l=1}^\infty$ gives a scanning strategy: $$\cdots M_{i_l}M_{i_{l-1}}\cdots M_{i_1}.$$ For $M_i$ reversible, each with stationary distribution $\pi$, the following systematic scans produce reversible Markov chains with stationary distribution $\pi$ (see, eg \cite{diaram}):
$$\begin{array}{ll}
\displaystyle \frac{1}{n-1}\sum_{i=1}^{n-1} M_i & \text{(random scan)},\\
M_1M_2\cdots M_{n-1}M_{n-1}\cdots M_2M_1 & \text{(short systematic scan)},\\
(M_1\cdots M_{n-1}M_{n-1}\cdots M_1)\cdots (M_1M_2M_2M_1)(M_1M_1) & \text{(long systematic scan)}.\\
\end{array}
$$
While such scanning strategies may seem intuitive for sampling from $\pi$, they have proven difficult to analyze in many situations. In the context of generation of Coxeter group elements, Diaconis and Ram \cite{diaram} show that convergence of the short systematic scan for the distribution $\pi$ above, with $l_S$ replaced by the length function on the Coxeter group coming from writing words as a product of simple reflections, occurs in the same number of steps as that of a random scan, i.e., choosing a random sequence of indices $\{i_\ell\}_{\ell=1}^\infty$. However, results for different scanning techniques or probability distributions remain open. In the context of graph colorings, Dyer et al. compare systematic scans with random scans for sampling proper $q$-colorings of paths for $q\geq 4$, in which a vertex is assigned a new color $c$ only if none of its neighbors are colored by $c$ \cite{dyeretall}. However, results for more general graphs have resisted analysis.
Fishman \cite{fishman} gives an overview of scanning strategies, while Diaconis and Saloff-Coste's survey \cite{diacoste} provides further applications of the Metropolis algorithm.
\section{Preliminaries: Semisimple Algebras}\label{prelim2}
\subsection{Fourier Inversion and Plancherel}\label{planchinv}
Random walks on groups are frequently studied using Fourier analysis. For example, for a group $G$ and a function $Q:G\rightarrow \mathbb{C}$, let $\hat{Q}$ denote the Fourier transform of $Q$.
\begin{theorem}[Diaconis, \cite{diaconis}]\label{diacthm} For $G$ a group, $Q$ a probability distribution on $G$, and $U$ the uniform distribution on $G$,
$$\vert Q-U\vert_{TV}^2\leq\frac{1}{4}\sum_\rho d_\rho \Tr(\hat{Q}(\rho)\hat{Q}(\rho)^*),$$
where $*$ denotes conjugate transpose and the sum is over all nontrivial irreducible representations $\rho$ of $G$.
\end{theorem}
The Fourier transform of a complex valued function on a finite group arises as a special case of Fourier transforms on semisimple algebras. Here we review the basic concepts and definitions. For more background on the representation theory of semisimple algebras see \cite{ram}.
\begin{definition} A \textbf{matrix representation} of a $\mathbb{C}$-algebra $A$ is an algebra homomorphism $$\rho: A\rightarrow M_d(\mathbb{C}),$$ where $M_{d}(\mathbb{C})$ denotes the complex algebra of $d\times d$ matrices with entries in $\mathbb{C}$. We call $d$ the \textbf{dimension} of $\rho$.
An algebra $A$ is \textbf{simple} if $A\cong M_n(\mathbb{C})$ for some $n\geq 1$ and \textbf{semisimple} if it decomposes as a direct sum of simple algebras: $$A\cong \bigoplus_{\lambda\in \Lambda}M_{\lambda}(\mathbb{C}),$$ for a finite index set $\Lambda$.
\end{definition}
\begin{definition}\label{alg}
Let $A$ be a semisimple algebra, $\{a_i\}_{i\in I}$ a basis for $A$ and $\displaystyle f=\sum_{i\in I}f(a_i)a_i\in A$.
\begin{itemize}
\item[(i)] Let $\rho$ be a matrix representation of $A$. Then the \textbf{Fourier transform of} $f$ \textbf{at} $\rho$, denoted $\hat{f}(\rho)$, is the matrix sum
$$\hat{f}(\rho)=\sum_{i\in I} f(a_i)\rho(a_i).$$
\end{itemize}
\end{definition}
\begin{definition} For $A$ a semisimple algebra, a \textbf{trace} function on $A$ is a $\mathbb{C}$-linear function $\tau:A\rightarrow\mathbb{C}$ such that for all $a,b\in A$,
$$\tau(ab)=\tau(ba).$$
\end{definition}
Note by linearity that the usual trace function on $M_d(\mathbb{C})$ is unique up to multiplication by a constant. Hence, for any trace $\tau$ on $A$ and set $R$ of inequivalent irreducible representations of $A$, there exist constants $t_\rho\in\mathbb{C}$ such that:
$$\tau=\sum_{\rho\in R}t_\rho T_\rho,$$
where for $a\in A$, $T_\rho(a)=\Tr(\rho(a))$.
A trace function $\tau$ gives rise to a symmetric bilinear form $\langle \cdot,\cdot\rangle_\tau:A\times A\rightarrow \mathbb{C}$ by letting $$\langle a,b\rangle_\tau=\tau(ab),$$
for $a,b\in A$.
Both Theorem \ref{diacthm} and the results of \cite{diaram} require the notion of Fourier inversion and Plancherel's Theorem.
\begin{theorem}[Fourier Inversion, Plancherel]\label{planch}
Let $A$ be a semisimple algebra with basis $\{a_i\}$ and $\tau$ a nondegenerate trace on $A$. Let $\{a_i^*\}$ be the dual basis to $\{a_i\}$ with respect to the trace form $\langle \cdot,\cdot\rangle_\tau$. Then for $f, f_1, f_2$ complex-valued functions on $A$,
\begin{equation}
f(a_i)=\sum_{\rho} t_\rho \Tr(\hat{f}(\rho)\rho(a_i^*)),
\end{equation}
\begin{equation}
\langle f_1, f_2\rangle_\tau=\sum_{\rho} t_\rho \Tr(\hat{f_1}(\rho)\hat{f_2}(\rho)).
\end{equation}
\end{theorem}
\subsection{The Brauer Algebra}\label{brauerdefs2}
Elements of the \textbf{Brauer monoid}, $Br_n$, are realized as generalized symmetric group diagrams: consider diagrams on $2$ rows of $n$ points each, with edges connecting pairs of points regardless of row and each point part of exactly one edge. Multiplication is realized as concatenation of diagrams. Note that in some cases, concatenation introduces a closed loop. For a parameter $q$ and two diagrams $x,y\in Br_n$, let $c$ denote the number of closed loops in the multiplication $xy$ and let $z$ be the diagram of this product with the closed loops removed. Then $xy=q^c z$.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\node[pnt] at (-3,0) (v_1) {};
\node[pnt] (v_2) [right of=v_1] {};
\node[pnt] (v_3) [right of=v_2] {};
\node[pnt] (v_4) [right of=v_3] {};
\node[pnt] (v_5) [below of=v_1] {};
\node[pnt] (v_6) [below of=v_2] {};
\node[pnt] (v_7) [below of=v_3] {};
\node[pnt] (v_8) [below of=v_4] {};
\node[pnt] (w_5) [below of=v_5] {};
\node[pnt] (w_6) [below of=v_6] {};
\node[pnt] (w_7) [below of=v_7] {};
\node[pnt] (w_8) [below of=v_8] {};
\node (q) [right of=v_8] {=};
\node at (-3.5,-.5) (x) {$x$};
\node at (-3.5,-1.5) (x) {$y$};
\node (q) at (1.5,-1) {$q$};
\node[pnt] at (2,-.5) (u_1) {};
\node[pnt] (u_2) [right of=u_1] {};
\node[pnt] (u_3) [right of=u_2] {};
\node[pnt] (u_4) [right of=u_3] {};
\node[pnt] (u_5) [below of=u_1] {};
\node[pnt] (u_6) [below of=u_2] {};
\node[pnt] (u_7) [below of=u_3] {};
\node[pnt] (u_8) [below of=u_4] {};
\node at (5.5,-1) (x) {$z$};
\path[-]
(v_1) edge [bend right=25] node {} (v_4)
(v_2) edge node {} (v_8)
(v_3) edge node {} (v_6)
(v_7) edge [bend right=25] node {} (v_5)
(v_7) edge [bend left=25] node {} (v_5)
(v_6) edge node {} (w_8)
(w_6) edge [bend right=25] node {} (w_5)
(v_8) edge node {} (w_7)
(u_1) edge [bend right=25] node {} (u_4)
(u_2) edge node {} (u_7)
(u_3) edge node {} (u_8)
(u_5) edge [bend left=25] node {} (u_6);
\end{tikzpicture}
\caption{ $xy=q^1z$}\label{brauers1}
\end{center}
\end{figure}
Two Brauer diagrams $d_1$ and $d_2$ are equivalent if they differ only in the number of closed loops, i.e., if when $q=1$, $d_1=d_2$. For example, for $x,y,z$ as in Figure \ref{brauers1}, the product $xy$ is equivalent to $z$. The Brauer monoid, $Br_n$ consists of the set of equivalence classes of such diagrams and is generated by $\{r_i, e_i \mid 1\leq i\leq n-1\}$ (see Figure \ref{figg}). The symmetric group $S_n$, generated by the transpositions $\{r_i\mid 1\leq i \leq n-1\}$, sits inside of $Br_n$. As in the symmetric group, a natural length function $l_{Br}:Br_n\longrightarrow\mathbb{N}$ exists for the Brauer monoid: for $w\in Br_n$, define $l_{Br}(w)$ to be the minimum number of generators ($\{r_i,e_i\}$) needed to express $w$.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\node[pnt] at (-3,0) (v_1) {};
\node[pnt] (v_2) [right of=v_1] {};
\node[pnt] (v_3) [right of=v_2] {};
\node[pnt] (v_4) [right of=v_3] {};
\node[pnt] (v_5) [right of=v_4] {};
\node[pnt] (v_6) [right of=v_5] {};
\node at (-2.5,-.5) {$\dots$};
\node at (1.5,-.5) {$\dots$};
\node at (-1,.5) {\small$i$};
\node at (0,.5) {\small$i+1$};
\node at (-.5,-2) {\large${r_i}$};
\node[pnt] (v_7) [below of=v_1] {};
\node[pnt] (v_8) [below of=v_2] {};
\node[pnt] (v_9) [below of=v_3] {};
\node[pnt] (v_10) [below of=v_4] {};
\node[pnt] (v_11) [below of=v_5] {};
\node[pnt] (v_12) [below of=v_6] {};
\node[pnt] at (4,0) (w_1) {};
\node[pnt] (w_2) [right of=w_1] {};
\node[pnt] (w_3) [right of=w_2] {};
\node[pnt] (w_4) [right of=w_3] {};
\node at (4.5,-.5) {$\dots$};
\node at (8.5,-.5) {$\dots$};
\node at (6,.5) {\small$i$};
\node at (7,.5) {\small$i+1$};
\node at (6.5,-2) {\large${e_i}$};
\node[pnt] (w_5) [right of=w_4] {};
\node[pnt] (w_6) [right of=w_5] {};
\node[pnt] (w_7) [below of=w_1] {};
\node[pnt] (w_8) [below of=w_2] {};
\node[pnt] (w_9) [below of=w_3] {};
\node[pnt] (w_10) [below of=w_4] {};
\node[pnt] (w_11) [below of=w_5] {};
\node[pnt] (w_12) [below of=w_6] {};
\path[-]
(v_1) edge node {} (v_7)
(v_2) edge node {} (v_8)
(v_3) edge node {} (v_10)
(v_5) edge node {} (v_11)
(v_6) edge node {} (v_12)
(v_4) edge node {} (v_9)
(w_1) edge node {} (w_7)
(w_2) edge node {} (w_8)
(w_3) edge [bend right=50] node {} (w_4)
(w_10) edge [bend right=50] node {} (w_9)
(w_5) edge node {} (w_11)
(w_6) edge node {} (w_12);
\end{tikzpicture}
\caption{$r_i, e_i\in Br_n$}\label{figg}
\end{center}
\end{figure}
The Brauer algebra, $\mathcal{B}r_n$, is the $\mathbb{C}(q)$-algebra with basis $Br_n$. Equivalently (see, for example \cite{BenkhartShaderRam}), $\mathcal{B}r_n$ has algebraic presentation given by generating set $$\{r_i, e_i \mid 1\leq i\leq n-1\},$$ along with relations:
$$\begin{array}{llll}
(B1)& r_i^2=1, &(B2)& r_ir_j=r_jr_i, \;\;\;\;r_ie_j=e_jr_i,\;\;\;\; e_ie_j=e_je_i,\;\; \\
&&&|i-j|>1\\
(B3)& e_i^2=qe_i, &(B4)& e_ir_i=r_ie_i=e_i, \\
(B5)&r_ir_{i+1}r_i=r_{i+1}r_ir_{i+1},& (B6)&e_ie_{i+1}e_i=e_i,\;\;\;\; e_{i+1}e_ie_{i+1}=e_{i+1},\\
(B7)& r_ie_{i+1}e_i=r_{i+1}e_i, & (B8)& e_{i+1}e_ir_{i+1}=e_{i+1}r_i.\\
\end{array}$$
\subsection{The BMW Algebra}\label{BMWdefs}
Elements of the BMW monoid are realized as generalized Brauer diagrams called \textbf{tangles}. A tangle is again a diagram on $2$ rows of $n$ points each with edges connecting pairs of points regardless of row and each point part of exactly one edge. At each crossing of two edges we distinguish which edge passes above and which passes below (see Figure \ref{BMW}). As in the Brauer monoid, multiplication is concatenation of diagrams and two tangles are equivalent if they differ only in their number of closed loops.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\node[pnt] at (-3,0) (v_1) {};
\node[pnt] (v_2) [right of=v_1] {};
\node[pnt] (v_3) [right of=v_2] {};
\node[pnt] (v_4) [right of=v_3] {};
\node[pnt] (v_5) [below of=v_1] {};
\node[pnt] (v_6) [below of=v_2] {};
\node[pnt] (v_7) [below of=v_3] {};
\node[pnt] (v_8) [below of=v_4] {};
\node (v) at (-.7,-.3) {};
\node (w) at (-.65,-.35) {};
\node (u) at (-1.6,-.8) {};
\node (vv) at (-1.65,-.8) {};
\path[-]
(v_1) edge [bend right=25] node {} (v_2)
(u) edge node {} (v_6)
(v) edge node {} (v_8)
(w) edge node {} (v_3)
(vv) edge node {} (v_4)
(v_7) edge [bend right=25] node {} (v_5);
\end{tikzpicture}
\caption{A Tangle}\label{BMW}
\end{center}
\end{figure}
Further, two tangles are equivalent if they are related by a sequence of Reidemeister moves of type II and III:
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\node at (-3,0) (v_1) {};
\node at (-4,-.5)(r2) {$R_{II}$:};
\node at (-4,-2.5)(r3) {$R_{III}$:};
\node (v_2) [right of=v_1] {};
\node (v_5) at (-3,-1) {};
\node (q) at (-1,-.5) {$\longleftrightarrow$};
\node (q) at (-1,-2.5) {$\longleftrightarrow$};
\node (v_6) [right of=v_5] {};
\node at (0,0) (w_1) {};
\node at (.3,-.65) (w) {};
\node at (.7,-.65) (v) {};
\node at (.25,-.7) (w1) {};
\node at (.8,-.7) (w2) {};
\node (w_2) [right of=w_1] {};
\node (w_5) at (0,-1) {};
\node (w_6) [right of=w_5] {};
\node (u_1) [below of=v_5] {};
\node (u_3) [right of=u_1] {};
\node (u_2) at (-2.5,-2) {};
\node (u_4) [below of=u_1] {};
\node (u_5) [below of=u_2] {};
\node (u_6) [below of=u_3] {};
\node (d_1) [below of=w_5] {};
\node (d_3) [right of=d_1] {};
\node (d_2) at (.5,-2) {};
\node (d_4) [below of=d_1] {};
\node (d_5) [below of=d_2] {};
\node (d_6) [below of=d_3] {};
\node at (-2.65,-2.4) (u1) {};
\node at (-2.68,-2.35) (u2) {};
\node at (-2.55,-2.5) (u3) {};
\node at (-2.65,-2.6) (u4) {};
\node at (-2.75,-2.75) (u5) {};
\node at (.55,-2.4) (d1) {};
\node at (.65,-2.35) (d2) {};
\node at (.75,-2.2) (d3) {};
\node at (.4,-2.57) (d4) {};
\node at (.8,-2.8) (d5) {};
\node at (.65,-2.65) (d6) {};
\path[-]
(v_1) edge [bend right=99] node {} (v_2)
(v_6) edge [bend right=99] node {} (v_5)
(w_1) edge [bend right=10] node {} (w)
(w_2) edge [bend left=10] node {} (v)
(w1) edge [bend right=10] node {} (w2)
(w_6) edge [bend right=100] node {} (w_5)
(u_1) edge node {} (u1)
(u3) edge node {} (u_6)
(u3) edge node {} (u2)
(u4) edge node {} (u_4)
(u_3) edge node {} (u5)
(u_2) edge [bend right=40] node {} (u_5)
(d_1) edge node {} (d5)
(d6) edge node {} (d_6)
(d3) edge node {} (d4)
(d1) edge node {} (d_4)
(d_3) edge node {} (d2)
(d_2) edge [bend left=40] node {} (d_5);
\end{tikzpicture}
\caption{Reidemeister Moves II and III}
\end{center}
\end{figure}
Consider the elements $T_{r_i}$, $T_{r_i}^{-1}$, and $T_{e_i}$ of Figure \ref{gen}.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\node[pnt] at (-3,0) (v_1) {};
\node[pnt] (v_2) [right of=v_1] {};
\node[pnt] (v_3) [right of=v_2] {};
\node[pnt] (v_4) [right of=v_3] {};
\node[pnt] (v_5) [right of=v_4] {};
\node[pnt] (v_6) [right of=v_5] {};
\node at (-2.5,-.5) {$\dots$};
\node at (1.5,-.5) {$\dots$};
\node at (-1,.5) {\small$i$};
\node at (0,.5) {\small$i+1$};
\node at (-.5,-2) {\large$T_{r_i}$};
\node[pnt] (v_7) [below of=v_1] {};
\node[pnt] (v_8) [below of=v_2] {};
\node[pnt] (v_9) [below of=v_3] {};
\node[pnt] (v_10) [below of=v_4] {};
\node[pnt] (v_11) [below of=v_5] {};
\node[pnt] (v_12) [below of=v_6] {};
\node[pnt] at (4,0) (w_1) {};
\node[pnt] (w_2) [right of=w_1] {};
\node[pnt] (w_3) [right of=w_2] {};
\node[pnt] (w_4) [right of=w_3] {};
\node at (4.5,-.5) {$\dots$};
\node at (8.5,-.5) {$\dots$};
\node at (6,.5) {\small$i$};
\node at (7,.5) {\small$i+1$};
\node at (6.5,-2) {\large$T_{e_i}$};
\node[pnt] (w_5) [right of=w_4] {};
\node[pnt] (w_6) [right of=w_5] {};
\node[pnt] (w_7) [below of=w_1] {};
\node[pnt] (w_8) [below of=w_2] {};
\node[pnt] (w_9) [below of=w_3] {};
\node[pnt] (w_10) [below of=w_4] {};
\node[pnt] (w_11) [below of=w_5] {};
\node[pnt] (w_12) [below of=w_6] {};
\node (v) at (-.5,-.5) {};
\path[-]
(v_1) edge node {} (v_7)
(v_2) edge node {} (v_8)
(v_3) edge node {} (v_10)
(v_5) edge node {} (v_11)
(v_6) edge node {} (v_12)
(v) edge node {} (v_4)
(v) edge node {} (v_9)
(w_1) edge node {} (w_7)
(w_2) edge node {} (w_8)
(w_3) edge [bend right=50] node {} (w_4)
(w_10) edge [bend right=50] node {} (w_9)
(w_5) edge node {} (w_11)
(w_6) edge node {} (w_12);
\end{tikzpicture}
\end{center}
\end{figure}
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\node[pnt] at (-3,0) (v_1) {};
\node[pnt] (v_2) [right of=v_1] {};
\node[pnt] (v_3) [right of=v_2] {};
\node[pnt] (v_4) [right of=v_3] {};
\node[pnt] (v_5) [right of=v_4] {};
\node[pnt] (v_6) [right of=v_5] {};
\node at (-2.5,-.5) {$\dots$};
\node at (1.5,-.5) {$\dots$};
\node at (-1,.5) {\small$i$};
\node at (0,.5) {\small$i+1$};
\node at (-.5,-2) {\large$T_{r_i}^{-1}$};
\node[pnt] (v_7) [below of=v_1] {};
\node[pnt] (v_8) [below of=v_2] {};
\node[pnt] (v_9) [below of=v_3] {};
\node[pnt] (v_10) [below of=v_4] {};
\node[pnt] (v_11) [below of=v_5] {};
\node[pnt] (v_12) [below of=v_6] {};
\node (v) at (-.5,-.5) {};
\path[-]
(v_1) edge node {} (v_7)
(v_2) edge node {} (v_8)
(v_4) edge node {} (v_9)
(v_5) edge node {} (v_11)
(v_6) edge node {} (v_12)
(v) edge node {} (v_3)
(v) edge node {} (v_10);
\end{tikzpicture}
\caption{$T_{r_i}, T_{e_i}, T_{r_i}^{-1}$}\label{gen}
\end{center}
\end{figure}
A tangle is \textbf{reachable} if it can be obtained as a finite product of elements from $\{T_{r_i},T_{e_i},T_{r_i}^{-1}\mid 1\leq i \leq n-1\}$. The BMW monoid, $BMW_n$, consists of the set of equivalence classes of reachable tangles on $2n$ points.
For $m,\ell,q$ parameters satisfying $q=(\ell-\ell^{-1})(m-m^{-1})^{-1}+1$, the $BMW$ algebra, $\mathcal{BMW}_n$, is the $\mathbb{C}(q,m,\ell)$-algebra with basis $BMW_n$ and the following untangling relations:
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\node[pnt] (v_3) at (-1,0) {};
\node[pnt] (v_4) [right of=v_3] {};
\node[pnt] (v_9) [below of=v_3] {};
\node[pnt] (v_10) [below of=v_4] {};
\node (v) at (-.5,-.5) {};
\node (e) [right of=v] {$=$};
\node[pnt] (w_3) at (1,0) {};
\node[pnt] (w_4) [right of=w_3] {};
\node[pnt] (w_9) [below of=w_3] {};
\node[pnt] (w_10) [below of=w_4] {};
\node (w) at (1.5,-.5) {};
\begin{scope}[shift={(.2,0)}]
\node (m) at (2.5,-.5) {$+m$};
\node[pnt] (u_3) at (3,0) {};
\node[pnt] (u_4) at (3.5,0) {};
\node[pnt] (u_9) [below of=u_3] {};
\node[pnt] (u_10) [below of=u_4] {};
\end{scope}
\node (u) at (3.5,-.5) {};
\node (m) [right of=u] {$-m$};
\node[pnt] (q_3) at (5,0) {};
\node[pnt] (q_4) [right of=q_3] {};
\node[pnt] (q_9) [below of=q_3] {};
\node[pnt] (q_10) [below of=q_4] {};
\path[-]
(v_3) edge node {} (v_10)
(v) edge node {} (v_4)
(v) edge node {} (v_9)
(w_4) edge node {} (w_9)
(w) edge node {} (w_3)
(w) edge node {} (w_10)
(u_9) edge node {} (u_3)
(u_10) edge node {} (u_4)
(q_4) edge [bend left=50] node {} (q_3)
(q_9) edge [bend left=50] node {} (q_10);
\end{tikzpicture}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}\includegraphics[scale=.65]{BMWTangle2.png}
\end{center}
\begin{center}\includegraphics[scale=.65]{BMWTangle3.png}
\end{center}
\begin{center}\includegraphics[scale=.65]{BMWTangle4.png}
\end{center}
\caption{Untangling Relations}
\end{figure}
Equivalently (see, for example \cite{GoodmanHauschild}), the BMW algebra has algebraic presentation given by generating set $\{T_{e_i},T_{r_i}, T_{r_i}^{-1}\mid1\leq i \leq n-1\}$, along with relations:
$$\begin{array}{rlrl}
(A1)&T_{e_i}^2=qT_{e_i} , & (A2)&T_{e_i}T_{r_i}=T_{r_i}T_{e_i}=\ell^{-1}T_{e_i}\\
(A3)&T_{e_i}T_{e_{i\pm1}}T_{e_i}=T_{e_i},&(A4) & T_{e_i}T_{r_{i\pm1}}T_{e_i}=\ell T_{e_i},\\
(A5)&T_{r_i}T_{r_{i+1}}T_{r_i}=T_{r_{i+1}}T_{r_i}T_{r_{i+1}},&(A6)&T_{r_i}T_{r_{i\pm1}}T_{e_i}=T_{e_{i\pm 1}}T_{e_{i}}=T_{e_{i\pm1}}T_{r_i}T_{r_{i\pm1}},\\
(A7)&T_{r_i}=T_{r_i}^{-1}+mT_{id}-mT_{e_i}&(A8)&T_{r_i}T_{r_j}=T_{r_j}T_{r_i},\;\;\;\;T_{r_i}T_{e_j}=T_{e_j}T_{r_i},\\
&&&\;\;\;\;T_{e_i}T_{e_j}=T_{e_j}T_{e_i},\;\; \vert i-j\vert>1 ,
\end{array} $$
for $q=(\ell-\ell^{-1})(m-m^{-1})^{-1}+1$ and $T_{id}$ the identity element. For all that follows we let $l=1$.
We map an element of the BMW monoid to the Brauer monoid by `forgetting' crossing information. Denote this map by $\phi:BMW_n\longrightarrow Br_n$.
\begin{example}\label{Brauerimage} For $x$ the tangle of Figure \ref{BMW}, $\phi(x)$ has form:
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\node[pnt] at (-3,0) (v_1) {};
\node[pnt] (v_2) [right of=v_1] {};
\node[pnt] (v_3) [right of=v_2] {};
\node[pnt] (v_4) [right of=v_3] {};
\node[pnt] (v_5) [below of=v_1] {};
\node[pnt] (v_6) [below of=v_2] {};
\node[pnt] (v_7) [below of=v_3] {};
\node[pnt] (v_8) [below of=v_4] {};
\node (v) at (-.7,-.3) {};
\node (w) at (-.65,-.35) {};
\node (u) at (-1.6,-.8) {};
\node (vv) at (-1.65,-.8) {};
\path[-]
(v_1) edge [bend right=25] node {} (v_2)
(v_4) edge node {} (v_6)
(v_3) edge node {} (v_8)
(v_7) edge [bend right=25] node {} (v_5);
\end{tikzpicture}
\end{center}
\caption{$\phi(x)$}
\end{figure}
\end{example}
Further, each element of the Brauer monoid lifts to the BMW algebra: for $d\in Br_n$, the \textbf{BMW image} of $d$, $T_d$, realizes $d$ as a tangle by redrawing the edges of $d$ from right to left across the first $\lceil\frac{n}{2}\rceil$ points in the bottom row, lifting the pen when crossing an edge that has already been drawn, then moving to the top row of points and drawing all horizontal edges in this row, again lifting the pen when crossing an edge that has already been drawn, and finally drawing the remaining edges of $d$ from right to left across the bottom row of points.
\begin{example} For $d$ the Brauer diagram of Example \ref{Brauerimage}, the BMW image of d is:
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\node[pnt] at (-3,0) (v_1) {};
\node[pnt] (v_2) [right of=v_1] {};
\node[pnt] (v_3) [right of=v_2] {};
\node[pnt] (v_4) [right of=v_3] {};
\node[pnt] (v_5) [below of=v_1] {};
\node[pnt] (v_6) [below of=v_2] {};
\node[pnt] (v_7) [below of=v_3] {};
\node[pnt] (v_8) [below of=v_4] {};
\node (v) at (-.7,-.3) {};
\node (w) at (-.65,-.35) {};
\node (u) at (-1.6,-.8) {};
\node (vv) at (-1.65,-.8) {};
\path[-]
(v_1) edge [bend right=25] node {} (v_2)
(u) edge node {} (v_6)
(v_8) edge node {} (v_3)
(v_4) edge node {} (w)
(v) edge node {} (vv)
(v_7) edge [bend right=25] node {} (v_5);
\end{tikzpicture}
\end{center}
\caption{$T_d$}
\end{figure}
\end{example}
Note that when $\ell=1$ the BMW image of $d$ has a simple algebraic description.
\begin{definition} For $d\in Br_n$ and $s_i\in\{r_i,e_i\}$, a \textbf{reduced expression} for $d$ is a minimum length expression $d=s_{i_1}s_{i_2}\cdots s_{i_k}$ that has no occurrence of $e_{i+1}r_i$.
\end{definition}
Then the \textbf{BMW image} of $d$, $T_d$, realizes $d$ as a tangle by setting $$T_d:=T_{s_{i_1}}T_{s_{i_2}}\cdots T_{s_{i_k}},$$
for $d=s_{i_1}s_{i_2}\cdots s_{i_k}$ a reduced expression
\begin{definition} For $d\in Br_n$ and $e(d)$ the number of $e_i$ terms in a reduced expression for $d$, The \textbf{BMW length} of $T_d$ $L:\mathcal{T}_n\longrightarrow \mathbb{N}$ is given by $$L(T_d)=l'_{Br}(d)+e(d), $$ where $l'_{Br}(d)$ gives the minimum number of generators needed for a reduced expression of $d$.
\end{definition}
\begin{note}\label{refrel} The relations in the Brauer algebra together with the definition of reduced expression ensure that $e(d)$ is well defined. See Table 4 in \cite{brsimplylaced} for the possible rewrites in the Brauer algebra.
\end{note}
\begin{example}
Let $d=r_3e_2e_1r_3$. Then $T_d=T_{r_3}T_{e_2}T_{e_1}T_{r_3}$ and $L(T_d)=l'_{Br}(d)+2=6$. An alternate reduced expression for $d$ is $d=r_3e_2r_3e_1$, which has the same BMW image by BMW relation (A8): $$T_{r_3}T_{e_2}T_{r_3}T_{e_1}=T_{r_3}T_{e_2}T_{e_1}T_{r_3}.$$
An additional expression for $d$ is $d=r_2e_3r_2e_1$. However, to have a reduced expression we must replace $e_3r_2$:
$$d=r_2e_3r_2e_1=r_2e_3e_2r_3e_1,$$ but then using Brauer relation (B7),
$d=r_3e_2r_3e_1,$ as before.
\end{example}
Theorem 3.12 of \cite{hal} shows that the BMW images of the Brauer monoid elements form a basis for $\mathcal{BMW}_n$. Denote this basis by $\mathcal{T}_n:=\{T_d\mid d\in Br_n\}$.
We consider generation of elements in $\mathcal{T}_n$ via random walks on $\mathcal{T}_n$ and translate these walks into left multiplication in the BMW algebra.
\section{The Random Walk}\label{thewalk}
In the finite group case, left multiplication by a generating set gives rise to a random walk on the group. For example, for each generator $r_i$ of $S_n$, consider the probability distribution
$$ P_i(x,y) = \left\{
\begin{array}{ll}
1 & \text{if}\;y=r_ix,\\
0 & \text{else}.
\end{array}
\right.
$$
Then for $l_S$ the length function on the symmetric group and $\pi$ given by $$\pi(x)=\frac{\theta^{-l_{S}(x)}}{\displaystyle\sum_{w\in S_n} \theta^{-l_{S}(w)}},$$
the Metropolis algorithm construction yields a chain which
interpreted as a random walk on $S_n$ is given by (see \cite{diaram}):
\begin{equation} \tag{$*$}\label{snwalk}
\begin{split}
& \text{From}\;x\in S_n \;\text{multiply by}\; r_i.\;\text{ If the length increases, move to} \\
& \text{$r_ix$. If the length decreases, flip a}\; \theta\text{-coin and if heads move}\\
& \text{to $r_ix$. If tails, remain at } x.
\end{split}
\end{equation}
\noindent We generalize this walk to the basis of tangles $\mathcal{T}_n$ of the BMW algebra. For $T_d\in\mathcal{T}_n$ and $L$ the length function on $\mathcal{T}_n$ defined in Section \ref{BMWdefs}, let
$$\pi(T_d)=\frac{\theta^{-L(T_d)}}{\displaystyle\sum_{w\in\mathcal{T}_n}\theta^{-L(w)}},$$
and for $y\in\mathcal{T}_n$ let
$$P'_i(T_d, y)=\left\{\begin{array}{ll} 1& y=T_{r_id}\\ 0&\text{else}.\end{array}\right.$$
Then the Metropolis algorithm applied to $P'$ with probability distribution $\pi$ yields:
$$ K_i(T_d,y) = \left\{
\begin{array}{ll}
1 & \text{if}\;y=T_{r_id}\;\text{and}\; L(y)\geq L(T_d) ,\\
\theta & \text{if}\;y=T_{r_id}\;\text{and}\; L(y)<L(T_d),\\
1-\theta & \text{if}\; y=T_d.
\end{array}
\right.
$$
\begin{remark}\label{Snremark}
Recall that $S_n\subseteq Br_n$ and note that for $d\in S_n$, $L(T_d)=l'_{Br}(d)=l_{Br}(d)=l_{S}(d)$, where $L,l_{Br}$, and $l_{S}$ denote the length functions on $\mathcal{T}_n,$, $Br_n$, and $S_n$. Then the submatrix of $K_i$ corresponding to states $\{T_d\mid d\in S_n\}$ is exactly the chain $M_i$ of \cite{diaram}.
\end{remark}
\noindent Interpreted as a random walk on $\mathcal{T}_n$, the chain $K_i$ describes the process:
\begin{equation} \tag{$\dagger$}
\begin{split}
& \text{From}\;T_d\in\mathcal{T}_n\text{ consider }d\in Br_n\text{ and multiply by } r_i.\text{ If the}\\
&\text{length of the BMW image $T_{r_id}$ increases, move to it. If the}\\
&\text{length decreases, flip a}\; \theta\text{-coin and if heads move to }T_{r_id}. \text{ If}\\ &\text{tails, remain at}\; T_d.
\end{split}
\end{equation}
\noindent In light of Proposition \ref{translate walk} below, this walk can be rephrased as:
\begin{equation} \label{rewalk} \tag{$\dagger\dagger$}
\begin{split}
& \text{From}\;T_d\in\mathcal{T}_n\text{ multiply by }T_{r_i}. \text{ If the result is an element of } \\
&\mathcal{T}_n,\text{ move to $T_{r_i}T_d$. Else,} \text{ flip a}\; \theta\text{-coin and if heads move to }\\
& T_{r_i}^{-1}T_d. \text{ If tails, remain at}\; T_d.\\
\end{split}
\end{equation}
\noindent Rephrasing in this way yields the equivalent corresponding Markov chain:
$$ K_i(x,y) = \left\{
\begin{array}{ll}
1 & \text{if}\;y=T_{r_i}x,\\
\theta & \text{if}\;y=T_{r_i}^{-1}x,\\
1-\theta & \text{if}\; y=x.
\end{array}
\right.$$
An example of $K_i$ can be found in Appendix \ref{App}.
\begin{proposition}\label{translate walk} For $T_d\in\mathcal{T}_n$, $$L(T_{r_id})<L(T_d)\iff T_{r_i}T_d\notin\mathcal{T}_n.$$
Further, if $T_{r_i}T_d\notin\mathcal{T}_n$, then $T_{r_i}^{-1}T_d=T_{r_id}\in\mathcal{T}_n$, while if $T_{r_i}T_d\in\mathcal{T}_n$, then $T_{r_i}T_d=T_{r_id}$.
\end{proposition}
\begin{proof}
First write $T_d=T_{s_{i_1}}T_{s_{i_2}}\cdots T_{s_{i_k}}$, for $s_{i_1}\cdots s_{i_k}$ a reduced expression for $d$ with maximum number of $e$ terms. Then $$T_{r_i}T_{d}=T_{r_i}T_{s_{i_1}}T_{s_{i_2}}\cdots T_{s_{i_k}},$$ which, after possibly rearranging using BMW relations (A5) and (A8), has one of the following forms, for some $1\leq j\leq k-2$:
\begin{enumerate}
\item $T_{r_i}T_{d}=T_{s_{i_1}}T_{s_{i_2}}\cdots T_{s_{i_j}}T_{r_i}T_{s_i}T_{s_{i_{j+2}}}T_{s_{i_{j+3}}}\cdots T_{s_{i_k}}$
\item $T_{r_i}T_{d}=T_{s_{i_1}}T_{s_{i_2}}\cdots T_{s_{i_j}}T_{r_i}T_{s_{i\pm 1}}T_{s_i}T_{s_{i_{j+3}}}\cdots T_{s_{i_k}},$
\item $T_{r_i}T_{d}=T_{r_i}T_{s_{i_1}}T_{s_{i_2}}\cdots T_{s_{i_k}}, \vert i_1-i\vert >1.$
\end{enumerate}
The proof reduces to checking each possible case. For example, if in case (1) with $s_is_{i_{j+2}}=r_ie_{i\pm 1}$,
$$T_{r_i}T_d=T_{s_{i_1}}T_{s_{i_2}}\cdots T_{s_{i_j}}T_{r_i}T_{r_i}T_{e_{i\pm 1}}T_{s_{i_{j+3}}}\cdots T_{s_{i_k}}.$$
Since $T_{r_i}T_{r_i}\notin\mathcal{T}_n$, we see that $T_{r_i}T_d\notin \mathcal{T}_n$. Further, since BMW relations (A5) and (A8) hold in the Brauer monoid,
$$r_id=r_is_{i_1}\cdots s_{i_k}=s_{i_1}s_{i_2}\cdots s_{i_j}r_ir_ie_{i\pm 1}s_{i_{j+3}}\cdots s_{i_k},$$
which by Brauer relation (B1) gives
$$r_id=s_{i_1}\cdots s_{i_j}e_{i\pm 1}s_{i_{j+3}}\cdots s_{i_k},$$ a reduced expression for $r_id$. Thus $l'_{Br}(r_id)=k-1.$ By Note \ref{refrel} all reduced expressions have the same number of $e$ terms, so $e(r_id)=e(d)$. Hence, $$L(T_{r_id})<L(d).$$
For the second statement, note that $$T_{r_i}^{-1}T_d=T_{s_{i_1}}T_{s_{i_2}}\cdots T_{s_{i_j}}T_{r_i}^{-1}T_{r_i}T_{e_{i\pm 1}}T_{s_{i_{j+3}}}\cdots T_{s_{i_k}}=T_{r_id}.$$
The remaining cases are checked similarly.
\end{proof}
In \cite{diaram}, Diaconis and Ram translate the Markov chain arising from (\ref{snwalk}) into left multiplication by Hecke algebra elements on a suitably chosen basis. Similarly, we translate the chains $K_i$ arising from the Metropolis construction into left multiplication by BMW algebra elements on the basis $\mathcal{T}_n$.
Define $\mathscr{T}_{r_i}, \mathscr{T}_{e_i}:\mathcal{T}_n\longrightarrow\mathcal{BMW}_n$ as follows: for $x\in\mathcal{T}_n$,
$$\begin{array}{l}
\mathscr{T}_{r_i}(x)=T_{r_i}x\\
\\
\\\mathscr{T}_{e_i}(x)=\left\{
\begin{array}{ll}
T_{e_i}x & \text{if}\;T_{r_i}x\notin\mathcal{T}_n,\\
T_{r_i}x & \text{else}.
\end{array}
\right.\\
\end{array}$$
\begin{theorem}\label{leftmult}[Theorem \ref{leftmult1}] Let $Br_n$ be the Brauer monoid and $\mathcal{BMW}_n(m,l)$ the BMW algebra with basis $\mathcal{T}_n=\{T_d\mid d\in Br_n\}$. Let $m=(1-\theta)(\theta)^{-1}$ and $\ell=1$. Then the chain $K_i$ is the same as the matrix of left multiplication by
$$\theta\mathscr{T}_{r_i}+(1-\theta)\mathscr{T}_{e_i},$$
with respect to the basis $\mathcal{T}_n$ of $\mathcal{BMW}_n$.
\end{theorem}
\begin{proof}
Let $x\in\mathcal{T}_n$ and consider left multiplication by $T_{r_i}$. If $T_{r_i}x\in\mathcal{T}_n$,
$$(\theta\mathscr{T}_{r_i}+(1-\theta)\mathscr{T}_{e_i})x= \theta T_{r_i}x+(1-\theta)T_{r_i}x=T_{r_i}x.$$
If $T_{r_i}x\notin\mathcal{T}_n$ then by BMW Relation (A7),
$$T_{r_i}x=(T_{r_i}^{-1}+m T_{id}- ml^{-1}T_{e_i})x=T_{r_i}^{-1}x+(1-\theta)(\theta)^{-1} x- (1-\theta)(\theta)^{-1} T_{e_i}x.$$
By Proposition \ref{translate walk}, $T_{r_i}^{-1}x\in\mathcal{T}_n$, and
$$\begin{array}{ll}
(\theta\mathscr{T}_{r_i}+(1-\theta)\mathscr{T}_{e_i})x&= \theta T_{r_i}^{-1}x+(1-\theta) x.
\end{array}$$
\end{proof}
The chains $K_i$ provide scanning strategies for generating elements of the BMW and Brauer monoids:
$$\begin{array}{ll}
\displaystyle \frac{1}{n-1}\sum_{i=1}^{n-1} K_i & \text{(random scan)},\\
K_1K_2\cdots K_{n-1}K_{n-1}\cdots K_2K_1 & \text{(short systematic scan)},\\
(K_1\cdots K_{n-1}K_{n-1}\cdots K_1)\cdots (K_1K_2K_2K_1)(K_1K_1) & \text{(long systematic scan)}.\\
\end{array}
$$
Theorem \ref{leftmult}, coupled with the results of Section \ref{walkanalysis}, allows for the study of the rate of convergence of the systematic scans arising from the chains $K_i$ using Fourier analysis on the BMW algebra.
\section{Analysis of the Walk}\label{walkanalysis}
Let $K$ denote the matrix corresponding to any of the three scans (random, short systematic, long systematic), as the results of this section hold true for all three scans.
Note that $K$ is Markov and recall that a communication class $C$ of a Markov chain is closed if for each state $x\in C$ and for all $y\notin C$, $y$ is not accessible from $x$. We determine the closed communication classes of $K$ and analyze the stationary distribution of each closed communication class.
The communication classes of $K$ depend on the number of \emph{lower horizontal edges} in the tangle diagrams for the states.
\begin{definition} Let $x\in\mathcal{T}_n$. An edge of $x$ is \textbf{lower (respectively, upper) horizontal} if it connects two points that are both on the bottom (respectively, top) row of the diagram of $x$.\end{definition}
\begin{example} In Figure \ref{e's}, $E_3$ is the only lower horizontal edge and $E_1$ is the only upper horizontal edge.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
\node[pnt] at (-3,0) (v_1) {};
\node[pnt] (v_2) [right of=v_1] {};
\node[pnt] (v_3) [right of=v_2] {};
\node[pnt] (v_4) [right of=v_3] {};
\node[pnt] (v_5) [below of=v_1] {};
\node[pnt] (v_6) [below of=v_2] {};
\node[pnt] (v_7) [below of=v_3] {};
\node[pnt] (v_8) [below of=v_4] {};
\node (v) at (-.7,-.3) {};
\node (w) at (-.65,-.35) {};
\node (u) at (-1.6,-.8) {};
\node (vv) at (-1.65,-.8) {};
\path[-]
(v_1) edge [bend right=25] node {$E_1$} (v_2)
(u) edge node {} (v_6)
(v) edge node {$E_2$} (v_8)
(w) edge node {} (v_3)
(vv) edge node {$E_4$} (v_4)
(v_7) edge [bend right=25] node {$E_3$} (v_5);
\end{tikzpicture}
\caption{}\label{e's}
\end{center}
\end{figure}
\end{example}
Note that left multiplication by $T_{r_i}, T_{r_i}^{-1}$ does not affect existing lower horizontal edges in a tangle diagram, nor can it create new ones. As $K$ is determined by left multiplication by $T_{r_i}, T_{r_i}^{-1}$, the communication classes of $K$ consist of states with common lower horizontal edges. For $x_i\in\mathcal{T}_n$, let $\mathbf{X}_i$ denote its communication class:
$$\mathbf{X}_i:=\{y\in\mathcal{T}_n\mid\text{lower horizontal edges of $y$ the same as those of $x_i$}\}.$$
For each communication class $\mathbf{X}_i$, let $[K]_i$ denote the corresponding submatrix of $K$. Note that the communication class for $x_0:=T_{id}$ consists of the states $\{T_d\mid d\in S_n\}$. Then by Remark \ref{Snremark}, $[K_i]_0=M_i$, and so $[K]_0$ can be analyzed using the methods of \cite{diaram}. For the remainder of the paper we consider the remaining communication classes of $K$.
To analyze the time to stationarity of the submatrix $[K]_1$ corresponding to a communication class $\mathbf{X}_1$, we pair $\mathbf{X}_1$ with a communication class, $\mathbf{X}_2$, whose states have the same number of lower horizontal edges as those in $\mathbf{X}_1$. For $w\in\mathbf{X}_1$, let $w^*$ denote the element of $\mathbf{X}_2$ with the same upper configuration as $w$. Define the matrix:
$$\tilde{K}(x,y) = \left\{
\begin{array}{cl}
K(x,y) & \text{if}\;x,y\in\mathbf{X}_1\;,\\
K(x,y) & \text{if } x=w^*,y=z^* \text{ for } w,z\in\mathbf{X}_1,\\
1 & \text{if}\;x=y, x\notin\mathbf{X}_1\cup\mathbf{X}_2,\\
0 & \text{else}.\\
\end{array}
\right.
$$
\begin{example}\label{exmat} For $\mathcal{T}_3\subseteq \mathcal{BMW}_3$, let $x_1=T_{e_1}$ and $x_2=T_{e_1}T_{r_2}$, so $\mathbf{X}_1=\{T_{e_1}, T_{r_2}T_{e_1}, T_{e_2}T_{e_1}\}$ and $\mathbf{X}_2=\{T_{e_1}T_{r_2}, T_{r_2}T_{e_1}T_{r_2}, T_{e_2}T_{e_1}T_{r_2}\}.$ Note that $T_{e_1}^*=T_{e_1}T_{r_2}$, while $T_{r_2}T_{e_1}^*=T_{r_2}T_{e_1}T_{r_2}$ and $T_{e_2}T_{e_1}^*=T_{e_2}T_{e_1}T_{r_2}$.
Then for $K=\frac{1}{2}(K_1+K_2)$,
$$2[K]_1=\bordermatrix{~ & T_{e_1} & T_{r_2}T_{e_1}&T_{e_2}T_{e_1} \cr
~ & 1 & \theta&0 \cr
~ & 1 & 1-\theta &\theta \cr
~& 0 & 1& 2-\theta \cr}, $$
$$2[K]_2=\bordermatrix{~ &T_{e_1}T_{r_2}&T_{r_2}T_{e_1}T_{r_2}&T_{e_2}T_{e_1}T_{r_2}\ \cr
~ & 1 & \theta&0 \cr
~ & 1 & 1-\theta &\theta \cr
~& 0 & 1& 2-\theta \cr}. $$
\vspace{.5cm}
Then $\tilde{K}=[K]_1\bigoplus [K]_2\bigoplus I_9$, for $I_9$ the $9\times 9$ identity matrix.
\end{example}
Let $\pi$ denote the stationary distribution of $\tilde{K}$ and for $T_x\in\mathcal{T}_n$ let $[\pi]_x$ denote the column of $\pi$ corresponding to $T_x$:
$$[\pi]_x:=\sum_{T_y\in\mathcal{T}_n}\pi_x(y)T_y.$$
Note that $\pi_x(y)$ represents the probability of ending at state $T_y$ after starting at $T_x$. To analyze the time to stationarity of $\tilde{K}$ we consider the total variation norm:
\begin{equation}\label{number} |\tilde{K}^m_x-\pi|_{TV}.\end{equation}
We bound the total variation norm using a trace norm on $\mathcal{BMW}_n$.
\begin{definition} Define $\tilde{\tau}:\mathcal{T}_n\rightarrow\mathbb{C}$ as follows: for $x\in\mathcal{T}_n$,
$$\tilde{\tau}(x) = \left\{
\begin{array}{cl}
1 & \text{if}\;x=T_{id} ,\\
0 & \text{else},\\
\end{array}
\right.
$$ The \textbf{restricted trace}, $\tau: \mathcal{BMW}_n\rightarrow \mathbb{C}$, is the linear extension of $\tilde{\tau}$ to $\mathcal{BMW}_n$.
\end{definition}
\begin{proposition}\label{oftenzero} For $T_x,T_y\in\mathcal{T}_n$, $\tau(T_xT_y) = \left\{
\begin{array}{cl}
1 & \text{if}\;x=y^{-1} ,\\
0 & \text{else}.\\
\end{array}
\right.
$
\end{proposition}
\begin{corollary}\label{istrace} $\tau$ is a trace function on $\mathcal{BMW}_n$.
\end{corollary}
\begin{proof}[Proof of Proposition \ref{oftenzero}] Let $T_x, T_y\in \mathcal{T}_n$. Then $T_x=T_{s_{j_1}}\cdots T_{s_{j_k}}$, where for each $1\leq j\leq k$, $T_{s_{i_j}}\in\{T_{r_i},T_{e_i}\mid 1\leq i\leq n-1\}$. First note by the BMW relations (A1)-(A8) that if for some $1\leq i\leq n-1$, $T_{e_i}$ is a factor of $T_x$, then each term of the product $T_xT_y$ has at least one $T_{e_i}$ factor. Hence, no term in the product $T_xT_y$ is the identity, so $\tau(T_xT_y)=0$. Similarly, $\tau(T_yT_x)=0$.
Thus, if $T_{s_{j_l}}=T_{e_i}$ for some $1\leq l\leq k$, $1\leq i\leq n-1$, then $\tau(T_xT_y)=\tau(T_yT_x)=0$ for all $T_y\in\mathcal{T}_n$. Equivalently, $\tau(T_xT_y)=0$ for all $x\in Br_n- S_n$, $y\in Br_n$.
Next note that $T_x\in \mathcal{T}_n$ has an inverse iff $x\in S_n\subset Br_n$. Hence we need show for $x,y\in S_n$ that $$\tau(T_xT_y) = \left\{
\begin{array}{cl}
1 & \text{if}\; x=y^{-1} ,\\
0 & \text{else}.\\
\end{array}
\right.
$$
But note that $\tau|_{S_n}$ is just a scalar multiple of the trace function $\vec{t}$ on the Iwahori Hecke algebra of $S_n$ (See e.g. \cite{diaram}[Section 3]).
\end{proof}
Thus $\tau$ is a trace function on $\mathcal{BMW}_n$ with $\tau(T_xT_y)=0$ for all $x,y\in Br_n- S_n$. In fact, $\tau$ extends the natural trace function of the Hecke algebra, $\mathcal{H}_n$, viewing $\mathcal{H}_n$ as a subalgebra of $\mathcal{BMW}_n$. We analyze $\tilde{K}$ using the bilinear form
arising from $\tau$, which reformulates questions about the time to stationarity in terms of the representation theory of the underlying Hecke subalgebra of $\mathcal{BMW}_n$.
Recall that $\tilde{K}$ consists of two submatrices corresponding to two communication classes $\mathbf{X}_1$ and $\mathbf{X}_2$ of $K$. Note that for each $T_x\in\in \mathbf{X}_1\cup\mathbf{X}_2$, $x\in Br_n-S_n$. Thus, $\tau(T_xT_y)=0$ for all $T_y\in \mathcal{T}_n$. In order for $\tau$ to be nontrivial on the communication classes of $\tilde{K}$, we rewrite $\tilde{K}$ with respect to a shifted basis for $\mathcal{BMW}_n$.
\begin{definition}\label{basisshift} Let $\pi$ denote the stationary distribution of $\tilde{K}$. To each $T_x\in\mathbf{X}_1$, associate a distinct $s_x\in S_n$ such that $s_x\neq s_y^{-1}$ for all $T_y\in\mathbf{X_1}$ and $s_x$ has order greater than 2. For $T_x\in \mathbf{X}_1$ and for $T_y\notin\mathbf{X}_1\cup\mathbf{X}_2$, let
\begin{equation}
\begin{split}
&\hat{T}_x:=T_x+\pi_x(x)^{-\frac{1}{2}}T_{s_x},\\
&\hat{T}_{x^*}:=T_{x^*}+\pi_x(x)^{-\frac{1}{2}}T_{{s_x}^{-1}}=T_{x^*}+\pi_{x^*}(x^*)^{-\frac{1}{2}}T_{{s_x}^{-1}},\\
&\hat{T}_y:=T_y.
\end{split}
\end{equation}
\end{definition}
\begin{note}
By construction, $\pi_x(x)=\pi_{x^*}(x^*)$ for all $x\in \mathbf{X}_1$.
\end{note}
\begin{note} In Appendix \ref{appcom} we show that $S_n$ contains enough distinct elements to make the associations of Definition \ref{basisshift} for all communication classes corresponding to elements with at least two lower horizontal edges. The remaining communication classes are analyzed separately through techniques discussed in Appendix \ref{appcom}.
\end{note}
For the remainder of this section let $\mathbf{X}_1$ be a communication class whose elements contain at least two lower horizontal edges.
\begin{lemma} $\hat{\mathcal{T}}_n:=\{\hat{T}_x\mid x\in Br_n\}$ is a basis for $\mathcal{BMW}_n$.
\end{lemma}
Now let $\langle \;,\;\rangle_{\mathcal{BMW}}$ denote the trace form of Section \ref{planchinv}
\begin{lemma}\label{innerprodeval} For $T_x\in\mathbf{X}_1\cup\mathbf{X}_2$ and $y\in Br_n$,
$$\langle \hat{T}_x,\hat{T}_y\rangle_{\mathcal{BMW}} = \left\{
\begin{array}{cl}
\pi_x(x)^{-1} & \text{if}\; y=x^* ,\\
\pi_x(x)^{-\frac{1}{2}} & \text{if}\; y=(s_x)^{-1},\\
0 & \text{else},\\
\end{array}
\right. $$
while
$$\langle \hat{T}_{s_x},\hat{T}_y\rangle_{\mathcal{BMW}} = \left\{
\begin{array}{cl}
\pi_x(x)^{-\frac{1}{2}} & \text{if}\; y=x^* ,\\
1 & \text{if}\; y=(s_x)^{-1},\\
0 & \text{else}.\
\end{array}
\right. $$\end{lemma}
\begin{proof} Follows from Proposition \ref{oftenzero} and the linearity of trace.
\end{proof}
Let $\hat{K}$ be the matrix of $\tilde{K}$ with respect to $\hat{\mathcal{T}}_n$. Note that time to stationarity is invariant under change of basis.
\begin{lemma} \label{Khatstruct} For $\hat{T}_x\in\hat{\mathcal{T}}_n$,
\begin{enumerate}
\item If $\hat{T}_x\in\hat{\mathbf{X}}_1$,
$$ \hat{K}(\hat{T}_x,\hat{T}_y) = \left\{
\begin{array}{ll}
K(T_x,T_y) & \text{if}\;\hat{T}_y\in\hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2,\\
(1-K(T_x,T_x))\pi_x(x)^{-\frac{1}{2}}&\text{if}\; y=s_x,\\
-K(T_x,T_y)\pi_y(y)^{-\frac{1}{2}} & \text{if}\;y=s_z, z\neq x, z\in\mathbf{X}_1\\
0& \text{else}, \\
\end{array}
\right.
$$ and similarly for $\hat{T}_x\in\hat{\mathbf{X}}_2$.
\item If $\hat{T}_x\notin\hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2$,
$$ \hat{K}(\hat{T}_x,\hat{T}_y) = \left\{
\begin{array}{ll}
1 & \text{if}\;y=x,\\
0& \text{else}. \\
\end{array}
\right.
$$
\end{enumerate}
\end{lemma}
\begin{proof} Follows from definition of $\hat{K}$ and $\hat{\mathcal{T}}_n$.
\end{proof}
Lemma \ref{Khatstruct} shows that $\hat{K}$ is a direct sum $\hat{K}_1\bigoplus\hat{K}_2\bigoplus I_{\hat{m}}$, where for $i=1,2$, the matrix $\hat{K}_i$ corresponds to $\{\hat{T}_x,\hat{T}_{s_x}\mid T_x\in\mathbf{X}_i\}$, and $\hat{m}=|\mathcal{T}_n|-4|\mathbf{X}_1|$. Further,
$$\hat{K}(\hat{T}_x,\hat{T}_y)=\tilde{K}(T_x,T_y)=K(T_x,T_y),$$
for all $T_x,T_y\in\mathbf{X}_1\cup\mathbf{X}_2$.
Recall that $\pi$ denotes the stationary distribution of $\tilde{K}$. For $T_x\notin\mathbf{X}_1\cup\mathbf{X}_2$, $\pi_x(y)=0$ for all $T_y\neq T_x$, and so \begin{equation}\label{eqn1}[\pi]_x=T_x.\end{equation}
Further, for $T_x\in\mathbf{X}_1$, $\pi_x(y)=0$ for all $T_y\notin\mathbf{X}_1$, and so
\begin{equation}\label{eqn2}[\pi]_{x}=\sum_{T_y\in\mathbf{X}_1} \pi_{x}(y)T_{y},\end{equation}
and similarly for $\mathbf{X}_2$.
Let $\hat{\pi}$ denote the stationary distribution of $\hat{K}$ and $[\hat{\pi}]_x$ the stationary distribution of $\hat{K}$ corresponding to column $\hat{T}_x$. Let $\hat{\mathbf{X}}_i=\{\hat{T}_x\mid T_x\in \mathbf{X}_i\}$.
\begin{lemma}\label{khatstat} Let $\pi$ be the stationary distribution of $\tilde{K}$ and $\hat{\pi}$ the stationary distribution of $\hat{K}$.
\begin{enumerate}
\item For $\hat{T}_{x}\in\hat{\mathbf{X}}_1$,
$$ [\hat{\pi}]_{x} = \sum_{\hat{T}_y\in\mathbf{X}_1}(\pi_x(y)\hat{T}_y-\pi_x(y)^{\frac{1}{2}}\hat{T}_{s_y})+\pi_x(x)^{-\frac{1}{2}}\hat{T}_{s_x},$$
and similarly for $\hat{T}_{x}\in\hat{\mathbf{X}}_2$.
\item If $\hat{T}_y\notin \hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2$, $[\hat{\pi}]_y= \hat{T}_y$
\end{enumerate}
\end{lemma}
\begin{proof} Part (2) follows from Lemma \ref{Khatstruct}. To prove (1), note that for $\hat{T}_x\in\hat{\mathbf{X}}_1$, $[\hat{\pi}]_{x}=[\pi]_{x}+\pi_{x}(x)^{-\frac{1}{2}}[\pi]_{s_{x}}$. Then by equations (\ref{eqn1}) and (\ref{eqn2}),
\begin{equation*}
\begin{split}
[\hat{\pi}]_{x}&=[\pi]_{x}+\pi_{x}(x)^{-\frac{1}{2}}[\pi]_{s_{x}}\\
&=\sum_{T_{y}\in\mathbf{X}_1} \pi_{x}(y)T_{y}+\pi_{x}(x)^{-\frac{1}{2}}T_{s_{x}}\\
&=\sum_{T_{y}\in\mathbf{X}_1} \left( \pi_{x}(y)(T_{y}+\pi_{y}(y)^{-\frac{1}{2}}T_{s_{y}})-\pi_{x}(y)\pi_y(y)^{-\frac{1}{2}}T_{s_{y}}\right)+\pi_{x}(x)^{-\frac{1}{2}}T_{s_{x}}.\\
&=\sum_{\hat{T}_{y}\in\hat{\mathbf{X}}_1 } \left( \pi_{x}(y)\hat{T}_{y}-\pi_{x}(y)^{\frac{1}{2}}\hat{T}_{s_{y}}\right)+\pi_{x}(x)^{-\frac{1}{2}}\hat{T}_{s_{x}}.
\end{split}
\end{equation*}
\end{proof}
For $T_x\in\mathbf{X}_1\cup\mathbf{X}_2$, Lemma \ref{khatstat} shows that $\pi_x(y)=\hat{\pi}_x(y)$ for all $T_{y}\in\mathbf{X}_1\cup\mathbf{X}_2$. However, $\pi_x(s_{y})=0$, but $\hat{\pi}_{x}(s_{x})=\pi_{x}(x)^{-\frac{1}{2}}-\pi_{x}(x)^{\frac{1}{2}}$ and for $y\neq x$, $\hat{\pi}_{x}(s_{y})=-\pi_{x}(y)^{\frac{1}{2}}$.
Let $\hat{\mathcal{S}}:=\{\hat{T}_{s_x}\mid s_x\in\mathcal{S}\}$. Consider the $L^2(\hat{\pi})$-norm restricted to the subspace generated by $\hat{\mathbf{X}}_1\cup \hat{\mathbf{X}}_2\cup\hat{\mathcal{S}}$:
\begin{definition} For functions $f,g:\hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2\cup \hat{\mathcal{S}}\rightarrow\mathbb{C}$, let
$$\langle f,g\rangle_2:=\sum_{\hat{T}_x\in\hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2\cup\hat{\mathcal{S}}} f(x)g(x)\hat{\pi}_x(x).$$
\end{definition}
For $m\in\mathbb{N}$, let $[\hat{K}^m]_x$ denote the column of $\hat{K}^m$ corresponding to $\hat{T}_x$:
$$[\hat{K}^n]_x=\sum_{{\hat{T}_{x_i}}\in\hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2\cup\hat{\mathcal{S}}}K_x^n(x_i)\hat{T}_{x_i}.$$
To find the time to stationarity of $\hat{K}$ (and hence $\tilde{K}$ and $K$), we analyze $\| [\hat{K}^m]_x-[\hat{\pi}]_x\|_2$.
\begin{lemma}\label{translateinnerprod} Let $f,g$ be complex-valued functions on $\hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2\cup\hat{\mathcal{S}}$ and let $*:\hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2\cup\hat{\mathcal{S}}\rightarrow\hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2\cup\hat{\mathcal{S}}$ be the involution that sends $\hat{T}_x$ to $\hat{T}_{x^*}$ for $\hat{T}_x\in\hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2$, and $\hat{T}_{s_x}$ to $\hat{T}_{s_x^{-1}}$ for $\hat{T}_{s_x}\in\hat{\mathcal{S}}$. Then for $\langle\;,\;\rangle_{BMW}$ the bilinear form arising from the trace $\tau$,
$$\langle f/\hat{\pi},g/\hat{\pi}\rangle_2=\langle f,g^*\rangle_{BMW}-\sum_{\hat{T}_{s_x}\in\hat{\mathcal{S}}}\frac{f(x)g(s_x^{-1})+f(s_x^{-1})g(x^*)}{\hat{\pi}_x(x)^{\frac{1}{2}}}.$$
\end{lemma}
\begin{proof} By Lemma \ref{innerprodeval},
\begin{equation*}
\begin{split}
\langle f/\hat{\pi},g/\hat{\pi}\rangle_2&= \sum \frac{f(x)g(x)}{\hat{\pi}_x(x)}\\
&=\sum_{\hat{T}_x\in \hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2\cup\hat{\mathcal{S}}} f(x)g(x)\langle \hat{T}_x, (\hat{T}_{x})^*\rangle_{BMW}\\
&=\sum_{\hat{T}_x,\hat{T}_y\in \hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2\cup\hat{\mathcal{S}}} f(x)g(y)\langle \hat{T}_x, (\hat{T}_{y})^*\rangle_{BMW}-\sum_{\hat{T}_{s_x}\in\hat{\mathcal{S}}}\frac{f(x)g(s_x^{-1})}{\hat{\pi}_x(x)^{\frac{1}{2}}}\\
&\;\;\;\;\;\;\;\;-\sum_{\hat{T}_{s_x}\in\hat{\mathcal{S}}} \frac{f(s_x^{-1})g(x^*)}{\hat{\pi}_x(x)^{\frac{1}{2}}}\\
&=\langle f,g^*\rangle_{BMW}-\sum_{\hat{T}_{s_x}\in\hat{\mathcal{S}}}\frac{f(x)g(s_x^{-1})+f(s_x^{-1})g(x^*)}{\hat{\pi}_x(x)^{\frac{1}{2}}}.
\end{split}
\end{equation*}
\end{proof}
\begin{corollary} For $\hat{T}_x\in\hat{\mathbf{X}}_1$,
$$\langle [\hat{K}^m/\hat{\pi}]_{x},[\hat{K}^m/\hat{\pi}]_{x}\rangle_2=\langle [\hat{K}^m]_{x},[\hat{K}^m]_{x}\rangle_{BMW}-\sum_{\hat{T}_y\in\mathbf{X}_2}\frac{\hat{K}^m_{x}(s_y^{-1})\hat{K}^m_{x}(y^*)}{\hat{\pi}_{y}(y)^{\frac{1}{2}}}.$$
\end{corollary}
$K$ is Markov, so there exists $N\in\mathbb{N}$ with $\hat{K}^m_{x}\geq 0$ for all $m>N$. Further, $\hat{\pi}$ is the stationary distribution of a Markov chain, so $\hat{\pi}_{y}(y)\geq 0$. We can thus bound the time to stationarity by the BMW trace.
\begin{theorem}[Theorem \ref{main2}] For $\hat{T}_x\in \hat{\mathbf{X}}_1\cup\hat{\mathbf{X}}_2$,
$$\langle [\hat{K}^n/\hat{\pi}]_{x},[\hat{K}^n/\hat{\pi}]_{x}\rangle_2\leq \langle [\hat{K}^n]_{x},[\hat{K}^n]_{x}\rangle_{BMW}.$$ Hence, $$\| [\hat{K}^n/\hat{\pi}]_{x}-1\|^2_2 \leq \|[\hat{K}^n]_{x}-1\|_{BMW}^2.$$
\end{theorem}
Thus, studying the time to stationarity of $\hat{K}$ can be achieved by studying $$\|[\hat{K}^n]_{x}-1\|_{BMW}^2.$$
\section*{Acknowledgments}
The author would like to especially thank Arun Ram and Dan Rockmore for many helpful and encouraging conversations.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,308 |
Q: Why do I get input from console in R? I have run R code in VS code. I want to get in put from console window by using readline() function. But it doesn't work: it doesn't allow me to input from keyboard. Please help with this problem. Thank you so much.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,360 |
My big brother got married a few weeks over a month ago, and now that we have talked all about the rehearsal dinner, the wedding, and my brother getting ready so it is time to talk about my dresses!
I got two great dresses for the weekend and I loved them both so much! My dress(on the left) for the Rehearsal dinner was very different than anything else that I currently have in my closet. The print was outside of the box for me and I am so happy I convinced myself to get it! It was a great find at Dress Barn, and even though I have yet to wear it again since the big day I am looking forward to wearing it again hopefully soon!
My dress for the wedding(on the right and below) was perfect for the big day and comfortable to be in all day. I loved the fit and look of it but my favorite part of my outfit was my mint heels. Caitlin told us she wanted us to wear grey dresses and pastel heels and when I saw the dress and heels I fell in love with them.
The best part of both of these outfits is that they can easily be worn again, and you better believe I will find a reason to wear those fabulous mint heels again! For now this is the end of the wedding posts, but I am not making any promises that you will never see these pictures again... just my disclaimer for not being called out in the future for using them over and OVER!
Love love love all of these!! Those mint shoes, ahh, just adore them!
So so so cute!!! Just love it all!
Hot, HOT, HOTTTT!!!! You look great in both of them!!!!!!!!
And those shoes. Minty perfection!!! | {
"redpajama_set_name": "RedPajamaC4"
} | 6,073 |
require 'test_helper'
class CustomerFlowsHelperTest < ActionView::TestCase
end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,277 |
Shelley is a Conveyancing Executive (ILEX Level 6 - Conveyancing) in the Residential Property Team. Shelley has been qualified for 7 years.
She advises private clients on their conveyancing matters. Shelley advises private individuals on their residential conveyancing matters whilst building on existing relationships with Estate Agents and Financial Advisors. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,203 |
{"url":"http:\/\/www.skyrimforge.com\/wiki\/projects\/encodings-or-validating-utf8\/r5\/","text":"# r5\n\nContrary to what some people expect, not every file is valid UTF8. This means that if you mess up and mangle your encodings, the files may get rejected or cause problems with various software, such as the CurseForge packager or WoW.\n\nNote: The repository hooks validate the encoding of Lua files before accepting them. WoW addons must encode their files in UTF-8. WAR addons must encode their files in either little-endian UTF-16 with a BOM, or plain ASCII.\n\n## Notation\n\nMost numbers are in decimal; byte values displayed as AB CD are hexadecimal.\n\n## Encodings in a nutshell\n\nAt the most fundamental level, computers encode information in chunks called bytes (or more precisely, octets; we shall assume that bytes are 8 bits). Most encodings just use one byte per character. This means there are 256 code points that can be mapped to characters (or digits, punctuation, Kanji, runes, whatever -- we'll call them characters).\n\nWhile there are dozens of encodings still in use, the following are especially important to this discussion:\n\n\u2022 ASCII is the basis of most western encodings today. Due to its 7-bit heritage, it only maps 128 code points: 0-127.\n\u2022 The ISO 8859 family of encodings are the most frequently used western encodings. Its parts 1\u201315, known also as \"Latin-1\" etc., are very widely used. Code points 0-127 agree with ASCII; the rest are used for various characters, such that it is possible, e.g., to type most western European languages in Latin-1.\nAnd here's the first catch: the Latin-N encodings do not cover all 256 code points. Specifically, 128-159 are unassigned. The encodings formally known as ISO-8859-1 etc. (note the extra dash) assign control characters to this range, but see below.\n\u2022 Windows-1252 in turn is a superset of ISO Latin-1 (ISO 8859-1), which maps printable characters into the range 128-159. This means it is not ISO-8859-1 (note the dash) compatible! As the name suggests, it is in wide use on western European Windows installations.\n\nBut 256 possible characters are quite restrictive. For example, it is not possible to write Chinese or Japanese in their native writing systems (by a wide margin). Along came Unicode and the UCS (Universal Character Set, formally ISO 10146), which intends to provide characters for all writing systems in use on the Earth.\n\nThis also means that a different mapping to bytes has to be used. Common variants include:\n\n\u2022 UTF-16, which uses two bytes (16 bits) per character and thus limited the original standard to 65536 code points (more are possible with escape codes).\n\u2022 UTF-32, which uses four bytes per character and allowed the second version of the Unicode standard to move beyond the 65k limit. (Unicode now has \"room\" for 1.1 million characters.)\n\u2022 The UTF-1\/7\/8 transformations, of which UTF-8 is by far the most commonly used.\n\nUnicode notably includes all of Latin-1 on the 256 lowest-numbered code points, which also means that it contains all ASCII characters on the same code points as in ASCII itself.\n\nUTF-16\/32 unfortunately require special support from software to be handled sanely. For example, the 'A' character would be encoded in (little-endian) UTF-32 as the bytes 41 00 00 00, and a lot of unprepared software will choke on the null bytes. Furthermore, a program cannot reliably tell an ASCII file from an UTF-16 one, because almost all even-sized ASCII files are valid (though probably nonsensical) UTF-16.\n\nUTF-8 circumvents this problem by guaranteeing the following properties:\n\n\u2022 All ASCII code points (i.e. 0\u2013127) map to the corresponding ASCII bytes (i.e. 0\u2013127).\n\u2022 All other code points map to a series of bytes, all of which have values in the range 128\u2013255.\n\u2022 No encoding of a character is contained in a (longer) encoding of another character.\n\u2022 No encoding contains FE or FF.\n\nThis means that it is ASCII-compatible in the same sense that Latin-1 is: software that expects input to be \"ASCII and maybe some higher bytes\" will be able to cope with it.\n\nHowever, not all byte sequences are valid UTF-8. The details are linked to the last three properties: roughly speaking, 007F correspond to ASCII, C2F4 are legal at the start of a multibyte sequence, and 80BF constitute the rest of the multibyte sequence. (There are some exceptions.) F5FD are currently invalid, but reserved for 5- and 6-byte sequence leaders if UCS ever introduces more characters. Other combinations of these bytes are invalid.\n\nThis concludes the general remarks. You should remember the following points:\n\nSummary: ASCII only maps half the possible byte values. Latin-1, Win-1252 and UTF-8 are all different supersets of ASCII. UTF-8\/16 can encode every language in use. Not every sequence of bytes is valid UTF-8. UTF-16 needs special support and is not a superset of ASCII.\n\n## Encoding detection and BOM\n\nUnicode provides a special means to detect the specific encoding (and endianness) that a file was written in: the Byte-Order Mark (BOM). This is just the special code point (U+FEFF) which is invisible as a character (it is a zero-width non-breaking space), but serves to distinguish the encodings via its byte representation:\n\n\u2022 EF BB BF for UTF-8,\n\u2022 FF FE for little-endian UTF-16,\n\u2022 FF FE 00 00 for little-endian UTF-32, etc.\n\nThus, a file can be marked as UTF-8 simply by putting a BOM at the very beginning. When the file is erroneously interpreted as a different encoding, the BOM will usually appear as garbage bytes.\n\nMany programs also use a heuristic to detect UTF-8, exploiting the property that not all byte sequences are valid: simply attempt to decode the file as UTF-8; if that fails, it's probably Latin-1.\n\n## How does this affect my addon?\n\nBecause Unicode can express all languages, WoW and WAR support Unicode representations for strings in Lua code.\n\n\u2022 WoW expects .lua files to be encoded in UTF-8.\n\u2022 WAR expects .lua files to be encoded in ASCII or little-endian UTF-16.\n\nUnfortunately there are still many editors around that cannot correctly handle some encodings. Before the advent of UTF, this wasn't so bad; many files would simply look wrong if read with the wrong encoding. But in any half sane editor, the damage was limited.\n\nBut with UTF-8, it's a different story: suppose you open a file encoded in UTF-8 in an editor which treats the contents as Latin-1. Now you cut&paste some text, perhaps containing umlauts, into it. What happens?\n\nMost likely the editor will happily save the Latin-1 bytes into the file. Remember that not all byte sequences are valid UTF-8? The file is now most likely corrupted\u2014the author has yet to see a UTF-8 (non-ASCII) byte sequence that makes sense in Latin-1 (in any language); conversely, nothing that makes sense is valid UTF-8!\n\nSimilar remarks hold for UTF-16, but there the file looks so damaged when read as a Latin encoding (it will have a null byte at almost every odd byte offset in the file) that few people would attempt to edit it under such conditions. However, if one does insert two bytes at different positions, everything between them will be garbled!\n\nTo fix encoding corruption, you need to identify the offending bytes (see the next section). Then attempt to guess the encoding they are in, usually Win-1252 is a good starting point, look up what characters they represented, then insert those characters with a UTF aware editor. Of course, opening the file in the right encoding may take some convincing because the editor may detect it as damaged.\n\nOther sources of badly encoded files are more obvious; for example, the author helped debug one case where the Lua sources were generated by a PHP script that simply used Latin-1 output.\n\nSummary: Editing a UTF-8 (UTF-16) encoded file in a non-UTF-aware editor will most likely leave it invalid (garbled, resp.).\n\n## Checking for valid UTF-8\/16\n\nThis is not easily possible with tools that are provided with Windows. You can, however, install the Python programming language, open an interactive Python window and use the following commands:\n\n>>> s = open(r\"c:\\path\\to\\file\").read()\n>>> u = s.decode(\"utf8\") # or \"utf-16-le\"\n\n\nThis loads the entire file into RAM and attempts to decode it as UTF-8 (or little-endian UTF-16). If the file is not valid, you will get a message along these lines:\n\nTraceback (most recent call last):\nFile \"<stdin>\", line 1, in <module>\nFile \"\/usr\/lib64\/python2.5\/encodings\/utf_8.py\", line 16, in decode\nreturn codecs.utf_8_decode(input, errors, True)\nUnicodeDecodeError: 'utf8' codec can't decode bytes in position 13-14: invalid data\n\n\nUnless you have an editor that can jump to a certain byte position, you can slice a bit of the string to get some context:\n\n>>> s[10:20]\n'kc\\x1c\\xd6\\x00\\x82^\\xff\\xff\\xff'\n\n\nIn the case of UTF-16 for WAR, we also mandate a BOM. To check for its presence, you can look at the first Unicode character in the decoded string:\n\n>>> u[0]\nu'\\ufeff'\n\n\nAny other result is not the BOM, and you need to insert it into the file.\n\n## Closing remarks\n\nNote that it is generally a bad idea to debug encoding problems over the internet. Pastebins are especially useless: the posting browser, the pastebin software and the viewing browser all have a chance to switch encodings, and they usually do. Mails are slightly better, but some MUAs are broken too. Similarly, IRC provides few guarantees, though many clients (with the notable exception of the widely used mIRC) now default to UTF-8.\n\nIf you must discuss such byte-level issues, the most reliable tool is a hex dump. OS X and Linux users can use the powerful xxd utility. 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Der Letzigraben ist heute eine Strasse im Westen der Stadt Zürich. Sie führt vom Letzigrund bergwärts zu den Abhängen des Uetlibergs und folgt somit dem früheren Verlauf der Letzimauer, welche ein Vorwerk der früheren Stadtbefestigung der Reichsvogtei Zürich war und damals weit ausserhalb der eigentlichen Stadtmauer lag.
Diese Verteidigungslinie führte vom Hardturm an den linksseitigen Ufern der Limmat bis zu der Ruine Friesenberg, welche an den Abhängen des Uetlibergs liegt. Sie wurde im Mittelalter erstellt. In Zürich war die Letzimauer wohl ebenfalls mit einem Wehrgraben gepaart, was der Strasse letztendlich zu ihrem Namen verholfen haben dürfte. An die auf der rechten Seite der Limmat gelegene Verlängerung dieser Verteidigungslinie zum Zürichberg erinnert auch heute noch die Letzistrasse.
Römischer Gutshof
Vom 1. bis zum 4. Jahrhundert nach Christus stand am Letzigraben, auf demselben Moränenhügel wie später der Zürcher Galgen, ein römischer Gutshof. Das Hauptgebäude war 40 Meter lang und wurde bereits 1838 von der Antiquarischen Gesellschaft ein erstes Mal bei Ausgrabungen untersucht. Damals wurden zahlreiche römische Gegenstände gefunden, welche heute im Landesmuseum zu besichtigen sind.
Galgen am Letzigraben
Am Letzigraben stand ab dem 14. Jahrhundert auch ein von der Stadt Zürich errichteter Galgen. Dieser befand sich auf einem Moränenhügel und stand damals ausserhalb der Stadt. Hier wurden Verurteilte aufgehängt oder gerädert, nachdem man ihnen im Stadthaus den Prozess gemacht hatte. Das Volk, welches am Hinrichtungstag in Scharen am Letzigraben auftauchte, wurde an Verpflegungsständen bewirtet.
Heute gehört der damalige Galgenhügel zum Gebiet des Freibades Letzigraben, welches 1947–49 vom Architekten Max Frisch erbaut wurde.
Die letzte Hinrichtung am Albisrieder Galgen fand 1810 statt. Der heimatlose Melchior Dürr, Mitglied einer Gaunerbande, wurde "mit dem Strang vom Leben zum Tode gebracht", wie eine zeitgenössische Quelle zu berichten weiss. Der Galgen wurde 1831 in einer Nacht-und-Nebel-Aktion durch Sträflinge des Zuchthauses Oetenbach im Auftrag der Obrigkeit abgebrochen. Bei den Grabungen nach römischen Überresten im 19. Jahrhundert wurden wohl auch zahlreiche Fundstücke zerstört, die zum Galgen gehörten.
In den Jahren 2005 bis 2007 wurde das unter Denkmalschutz stehende Freibad am Letzigraben saniert. Im Rahmen dieser Sanierungsarbeiten wurden von der Stadtarchäologie Grabungen auf dem Moränenhügel, dem ehemaligen Standort des Galgens, vorgenommen. Erste Resultate wurden am 10. Februar 2006 über die Medien kommuniziert. So wurden drei Skelette von vermutlich ehem. Hingerichteten nur wenige dutzend Zentimeter unter der heutigen Liegewiese des Bades gefunden, teilweise mit überstreckten Halswirbelsäulen, verdrehten Armen und zertrümmertem Schädeldach.
Weblinks
Burgenwelt: Letzi Zürich
Gang dur Alt-Züri: Der Letzigraben
Einzelnachweise
Innerortsstrasse in Zürich | {
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When this album had been released, I did not like it at all, but when I put in in my CD-player some years later, I began to love it. No idea how I could reject this one.
This song sounds funny and happy, if you do not listen carefully to the lyrics. When I did, I thought, wow, how right ,everybody feels like this somedays. A very nice song.
Great tone and very nice melody, I especially like the soft ending and the chorus. But not as good as Always or Bed of Roses.
Wow! So quiet and beautiful in the beginning and then so powerful. This song has something very special in it, it is a real Bon Jovi classic.
A beautiful ballad with big hardache in it. Very emotional and at the same time sounding warm in a way, not easy to explain why.
A real rocker after a more calm song, but this one is quite forgetable and does not catch me at all, it is one of the weakest songs from this album.
Wow! Unbelievable sturff! One of the best pieces of art, Bon Jovi ever produced, great guitar, emotional vocals and it just has "it".
Oner of these songs, which you like to listen to, when you are just about to gop to bed, calm, but also warm again, a lullaby.
Nice song, but not more. Not a ballad, not a rocker. It is still a good song, but inferior to most others.
This song is a masterpiece. Dark, but catchy, a great song to listen to and Jon scraems like he never did before. Also great in concerts.
This song is better than many fans say. I like the melody. I wonder why this is not very popular.
An accoustic song. Why didn't they put more of songs like this on their albums? Great piece of music.
Much better than Damned, but still not strong enough to be a classic rocker. One of the weaker songs, but the Talkbox fits just perfectly.
A good song to end the album, nice, but a little to ordinary to be a great song.
All in all this is a dark and calm album, but still very catchy and I really like this one now. | {
"redpajama_set_name": "RedPajamaC4"
} | 2,523 |
Cicârd – wieś w Rumunii, w okręgu Alba, w gminie Lopadea Nouă. W 2011 roku pozostawała niezamieszkała.
Przypisy
Wsie w okręgu Alba | {
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} | 8,652 |
\section{\label{sec:intro}Introduction}
The air-water interface has been a subject of extensive studies due to its ubiquity in nature and its unusual macroscopic properties as a model system for aqueous hydrophobic interfaces.\cite{BJ,Jungwirth2001,Wilson2002,Stiopkin2011,Jeon2017}
It is widely accepted that water molecules behave differently at the interface than in bulk phase.\cite{DangLX1997,Hsieh2013}
The air-water interface thickness has experimentally been measured via both ellipsometry\cite{Raman1927,McBain1939,Kinosita1965} and X-ray reflectivity.\cite{Braslau1985,Braslau1988}
Sum frequency generation (SFG) spectroscopy, in which the interface vibrational spectrum can be obtained, can provide detailed
insights into the thickness of the air-water interface.\cite{RS91,VO05}
There are some general consensus on the fact that the thickness is about 3--10 \AA.\cite{Peng2021}
Nonetheless, accurately determining the thickness of the air-water interface remains experimentally challenging.
Molecular dynamics (MD) and Monte Carlo (MC) simulations of the air-water interface yield molecular-level information not readily available in experiments.
Such simulations, employing a variety of intermolecular potential
functions, have been reported for the air-water interface,
and have been used to estimate the air-water interface thickness.\cite{Townsend1985,Wilson1987,Matsumoto1988,Townsend1991,Lie1993,Alejandre1995,Taylor1996,Dang1997}
In previous computer simulations, the defined thickness of the interface depends on the chosen parameters, for example, the 10--90 thickness
\cite{Townsend1991,Beaglehole1993,Lie1993,Alejandre1995,Taylor1996,Sokhan1997,Morita2000, Paul2004, Morita2006, ZhaoJin2018}.
However, the density-based definition of interface thickness faces a debate over the reality of the oscillations of the density profile.\cite{Evans1993}
In addition, interface thickness is a sensitive quantity to the intermolecular potential and
treatment of long-range correction,\cite{Sokhan1997,Morita2000}
and the accuracy of MD simulations are limited by the accuracy of the potential functions being used.\cite{Taylor1996}
Density functional theory-based molecular dynamics (DFTMD) simulations\cite{CP,Marx2000,RC02,Kuo2004b,Kuehne2011,sulpizi2013,Khaliullin2013,Pezzotti2017} offer a predictive platform for density profiles and the thickness of the air-liquid interfaces.\cite{Blas2001}
Using DFTMD simulations,
Sulpizi \emph{et al.}\cite{sulpizi2013} (Pezzotti \emph{et al.}\cite{Pezzotti2017}) have calculated the second-order nonlinear susceptibility for the pre-defined
instantanously interfacial layers\cite{Willard2010} and deduced a thickness of 3 \AA\ (3.5 \AA) of air-water interface at 330 K (315 K).
The existing simulation methods for determining the air-water interface thickness mainly fall into two categories:
(a) Using classical MD or MC simulations and artificially selected parameters to determine the interface thickness;\cite{Townsend1991,Beaglehole1993,Alejandre1995,Taylor1996,Sokhan1997,Morita2000, Paul2004, Morita2006}
(b) Using first-principles molecular dynamics simulations and interfacial layer difference in properties to infer the interface thickness.\cite{sulpizi2013,Pezzotti2017}
A more natural method for estimating the air-water interface thickness, which has the following features
is needed:
(\romannum{1}) It does not depend on the interaction potential functions;
(\romannum{2}) It avoids the direct definition of interface thickness in terms of any parameters, e.g., density.
In this paper, DFTMD simulations are used to satisfy (\romannum{1}), and a combination of interfacial molecule sampling (IMS) and a newly defined interface HB (IHB) population methods was used to satisfy (\romannum{2}).
The paper is organized as follows.
In Sec. \ref{sec:hb_dynamics} we review the HB population operator and related correlation functions,
and the method to obtain the HB breaking and reforming rate constants.
Section \ref{sec:hbd_interface} then introduces the ideas of IHB and IMS to identify interface H-bonds.
The main results and discussions are presented in Sec. \ref{sec:discussions}.
Finally, the conclusions are presented in Sec. \ref{sec:conclusions}.
\section{\label{sec:hb_dynamics}Hydrogen bond dynamics}
Using a geometric criterion of HB, Luzar and Chandler\cite{AL96}
have pioneered the analysis of HB dynamics of pure water, and
subsequently such analysis has been extended to more complex systems, e.g., electrolytes,\cite{Chandra2000} protein\cite{Tarek2002} and micellar surfaces.\cite{SP05}
In the criterion, two water molecules are H-bonded if their interoxygen distance $r_{\text{OO}}$ is less than cutoff radius\cite{Sciortino1989} $r^{\mathrm{c}}_{\mathrm{OO}}=3.5$ \AA \ and
the H-O$\cdots$O angle $\phi$
is less than cutoff angle $\phi^{\mathrm{c}}=\pi/6$.\cite{AKS86,JT90,Luzar1993,SB02}
We denote it as acceptor-donor-hydrogen (ADH) criterion.
For comparison, we also use another HB criterion: $r_{\text{OO}}$ is less than $r^{\mathrm{c}}_{\mathrm{OO}}$,
and the O-H$\cdots$O included angle $\theta$ is greater than cutoff angle $\theta^{\mathrm{c}}=2\pi/3$.\cite{Jeon2017}
We denote this HB criterion as the acceptor-hydrogen-donor (AHD) criterion.
We use a configuration $r(t)$ denotes the positions
of all the atoms in the system at time $t$.
Either of the criteria above allows one to define a HB population $h[r(t)] = h(t)$,
which equals 1 when a tagged pair of molecules are H-bonded, and 0 otherwise.
The fluctuation in $h(t)$ from its time-independent equilibrium average is defined by\cite{DC87}
$\delta h = h(t) - \langle h\rangle$.
The probability that a specific pair of molecules is H-bonded in a large system is extremely small,
then $\delta h(t) = h(t)$. Therefore, the correlation of $\delta h(t)$ can be written as
\[ \langle \delta h(0) \delta h(t)\rangle= \langle h(0)h(t)\rangle, \]
where the averaging $\langle\cdots\rangle$ is to be performed over the ensemble of initial conditions.
\subsection{\label{sec:c_hb}Hydrogen bond correlation functions}
The correlation function\cite{Chandra2000,Benjamin2005}
\begin{eqnarray}
c(t)=\langle h(0)h(t) \rangle/\langle h\rangle
\label{eq:c_HB}
\end{eqnarray}
describes the structural relaxation of H-bonds. Here
the average $\langle h\rangle$ of the HB population is the probability that a pair of randomly chosen water molecules in the system is
H-bonded at any time $t$.
The function $c(t)$ measures correlation in $h(t)$ independent of any possible bond breaking events,
and it relaxes to zero, when $t$ is large.\cite{Rapaport1983}
Because the thermal motion can cause distortions of H-bonds from the perfectly tetrahedral configuration,
water molecules show a librational motion on a time scale of $\sim$ 0.1 ps superimposed to rotational and diffusional motions ($> 1$ ps),
which causes a time variation of interaction parameters.
A new HB population $h^\text{(d)}(t)$ was also defined to obviate the distortion of real HB dynamics
due to the above geometric definition.\cite{Sciortino1989,Chandra2000}
It is 1 when the interoxygen distance of a particular tagged pair of water molecules is less than $r^{\mathrm{c}}_{\mathrm{OO}}$ at time $t$,
and 0 otherwise.
The H-bonds between a tagged molecular pair that satisfy the condition $h^\text{(d)}(t)=1$ may have been broken, but they may more easily form H-bonds again.
The correlation function
\begin{eqnarray}
n(t)=\langle h(0)[1-h(t)]h^\text{(d)}(t) \rangle/\langle h\rangle
\label{eq:n_HB}
\end{eqnarray}
represents the probability at time $t$
that a tagged pair of initially H-bonded water molecules are unbonded but remain separated by less than $r_{\mathrm{OO}}^{\mathrm{c}}$.\cite{Chandra2000}
The rate of HB relaxation to equilibrium is characterized by the reactive flux\cite{AL00}
\begin{eqnarray}
k(t) = -\frac{\mathrm{d}c(t)}{\mathrm{d}t},
\label{eq:k}
\end{eqnarray}
which quantifies the rate that an initially present HB breaks at time $t$,
independent of possible breaking and reforming events in the interval from 0 to $t$.
Therefore, $k(t)$ measures the effective decay rate of an
initial set of H-bonds.\cite{DC87}
For bulk water, there exists a $\sim 0.2$-ps transient period,
during which $k(t)$ changes quickly from its initial value.\cite{FWS00}
However, at longer times, $k(t)$ is independent of the HB definitions.
\subsection{\label{sec:rate_constants_2} Hydrogen bond breaking and reforming rate constants}
Assume that each HB acts independently of other H-bonds,\cite{AL96,AL00}
and due to detailed balance condition, one obtain $\tau_{\mathrm{HB}} = {(1- \langle h\rangle)}/{k}$,
where $k$ is the rate constant of breaking a HB, i.e., the forward rate constant. \cite{Chandler1986,Chandler1978}
Correspondingly, the backward rate constant $k'$ is represented by the rate constant from the HB \emph{on} state to the HB \emph{off} state for a tagged pair of molecules.
Based on the functions $n(t)$, $h(t)$, $h^\text{(d)}(t)$, and $k(t)$,
\citet{Khaliullin2013} have obtained the ratio $k/k'$ of HB breaking
and reforming rate constants in bulk water, and then the lifetime and relaxation time of H-bonds from simulations. Here, for the air-water interface,
we obtain the optimal solution range of $k$ and $k'$ from the relationship
between the reactive flux $k(t)$ and the correlation functions $c(t)$ and $n(t)$:
\begin{eqnarray}
k(t) = kc(t)-k'n(t).
\label{eq:fitting_k_rates}
\end{eqnarray}
We obtain the optimal value of the rate constants, $k$ and $k'$,
by a least squares fit of $k(t)$, $c(t)$ and $n(t)$ beyond the transition phase.
The function $c(t)$ is regarded as a column vector composed by $(c_1,\cdots,c_P)^{\intercal}$, and is denoted as ${\bf c}$,
with $c_i$ representing the value of correlation $c(t)$ at $t=i$.
Similarly, $n(t)$ and $k(t)$ can also be denoted as ${\bf n}$ and ${\bf k}$, respectively.
Then, the rate constants $k$ and $k'$ are determined from the matrix ${\bf A} = \begin{bmatrix} {\bf c} & {\bf n} \end{bmatrix}$:
\begin{eqnarray}
\begin{bmatrix} k\\ -k' \end{bmatrix} = ({\bf A}^{\mathrm{T}} {\bf A})^{-1} {\bf A}^{\mathrm{T}} {\bf k}.
\end{eqnarray}
For bulk water and the air-water interface, the optimal $k$ and $k'$ are reported in Table
~\ref{tab:k_k_prime_128w_pure_1} and ~\ref{tab:k_k_prime_128w_pure_2}.
\begin{table}[h]
\caption{\label{tab:k_k_prime_128w_pure_1} The rate constants $k$ and $k'$ for the bulk water and the air-water interface (the time region 0.2 ps $< t <$ 2 ps).}
\begin{ruledtabular}
\begin{tabular}{ccccccc}
Criterion & $k$ (b)\footnote{The unit for $k$ ($k'$) is ps$^{-1}$. b: bulk; i: interface.} & $k'$ (b) & $\tau_{\mathrm{HB}}$ (b)\footnote{The unit for $\tau_{\mathrm{HB}}$ ($=1/k$) is ps.} & $k$ (i) & $k'$ (i) & $\tau_{\mathrm{HB}}$ (i)\\
\hline
ADH & 0.296 & 0.988 & 3.380 & 0.323 & 0.765 & 3.101 \\
AHD & 0.288 & 1.149 & 3.470 & 0.314 & 0.887 & 3.184 \\
\end{tabular}
\end{ruledtabular}
\end{table}
\begin{table}[h]
\centering
\caption{\label{tab:k_k_prime_128w_pure_2}
The $k$ and $k'$ for the bulk water and the air-water interface (the time region 2 ps $< t <$ 12 ps).}
\begin{ruledtabular}
\begin{tabular}{ccccccc}
Criterion & $k$ (b) & $k'$ (b) & $\tau_{\mathrm{HB}}$ (b) & $k$ (i) & $k'$ (i) & $\tau_{\mathrm{HB}}$ (i)\\
\hline
ADH & 0.115 & 0.039 & 8.718 & 0.157 & 0.068 & 6.372\\
AHD & 0.105 & 0.047 & 9.496 & 0.155 & 0.088 & 6.472 \\
\end{tabular}
\end{ruledtabular}
\end{table}
To obtain $k$ and $k'$,
we perform the fitting in short and long time regions, respectively.
We note that in the long time region ($2 < t < 12$ ps),
the value of HB lifetime $\tau_\mathrm{HB}$ is larger than that in short one ($0.2 < t < 2$ ps),
no matter for the bulk water or for the air-water interface. A larger $\tau_\mathrm{HB}$ value means that the distance between a pair of water molecules
stays within $r_\mathrm{OO}^c$ for a longer time.
\section{Hydrogen bond dynamics for instantaneous air-water interface}\label{sec:hbd_interface}
Due to molecular motions, the identity of molecules that lie at the interface changes with time, and
generally useful procedures for identifying interface must accommodate these motions.
The air-water boundary is modeled with the Willard-Chandler instantaneous surface.\cite{Willard2010,sulpizi2013,Pezzotti2017,Serva2018}
Figure \thinspace\ref{fig:128w_itp_add_z_d_trimed_with_inner_layers}
illustrates the obtained interfaces for one configuration of a slab of
pure water.
\begin{figure}[h]
\centering
\includegraphics [width=0.32\textwidth] {./diagrams/128w_itp_add_z_d_trimed_with_inner_layers.eps}
\setlength{\abovecaptionskip}{0pt}
\caption{\label{fig:128w_itp_add_z_d_trimed_with_inner_layers}
A slab of water with the instantaneous surface $\mathbf{s}$ represented as a blue mesh on the upper and lower phase boundary.
The normal is along the $z$-axis and $d$ is the thickness of the interface.
The grey surface is obtained by translating the surface to the inside of the system along the $z$-axis by $d$.}
\end{figure}
For the slab in the cuboid simulation box,
we can get another surface ${\mathbf s}_0(t)$ by translating the surface ${\mathbf s}(t)$
along the system's normal (into bulk) to a distance $d$.
The region between the two surfaces
${\mathbf s}(t)$ and ${\mathbf s}_0(t)$ is defined as \emph{the air-water interface}.
We have obtained HB dynamics at the instantaneous air-water interface, and there are two extreme cases to be considered.
\subsection{Hydrogen bond dynamics based on interfacial molecule sampling}
As the first method to obtain interface HB dynamics, we use molecule sampling at the instantaneous interface.
Let $T_\text{e}$ be the time it takes for all water molecules in the simulation box to traverse the interface and bulk phase.
If the trajectory length $t_\text{traj}$ satisfies the condition $\tau_\text{HB} \ll t_\text{traj} \ll T_\text{e}$,
one finds $M$ time points $t_i$, $(i=1,\cdots, M)$ that are evenly spaced on the trajectory,
then one obtains interface HB dynamics using the following procedure:
\begin{enumerate}
\item For each observation time $t_i$, define an interface with a thickness $d$ (see Fig.\thinspace\ref{fig:128w_itp_add_z_d_trimed_with_inner_layers}). Then, select a set $S_i$ of water molecules which are at the interface, and calculate the correlation functions $c_i(t)$, $n_i(t)$ and $k_i(t)$ through $t_\text{traj}$ for the molecules that belong to the set $S_i$.
\item Determine average functions $c(t)$, $n(t)$ and $k(t)$, of the correlation functions $c_i(t)$, $n_i(t)$ and $k_i(t)$ over $M$ sampling time points respectively.
\item Determine reaction rate constant of breaking and reforming at the interface with thickness $d$, by Eq. \ref{eq:fitting_k_rates}.
\end{enumerate}
In the IMS method, since the configuration of molecules changes over time,
the contribution of H-bonds in bulk phase is included. Therefore, the IMS method \emph{understimates} the HB breaking rate constant of the interface.
\subsection{\label{sec:ihb}Hydrogen bond dynamics based on interface HB population }
After determining the instantaneous interface,
we introduce an interface HB population operator $h^{(\text{s})}[{\mathbf r}(t)]$ as follows:
It has a value 1 when a tagged molecular pair $i, j$ are H-bonded \emph{and} both molecules are at the interface
with a thickness $d$, and 0 otherwise:
\begin{align}
h^{(\text{s})}[{\mathbf r}(t)]=\left\{
\begin{array}{rcl}
1 & & {i,j\text{ are H-bonded, \emph{and}}}\\
& & {i,j\text{ are at the interface;}} \\ \label{eqn:h_s}
0 & & {\text{otherwise.}}
\end{array} \right.
\end{align}
Then the correlation function $c^{(\text{s})}(t)$ that describes the fluctuation of H-bonds \emph{at the interface}:
\begin{eqnarray}
c^{(\text{s})}(t)=\langle h^{(\text{s})}(0)h^{(\text{s})}(t) \rangle/\langle h^{(\text{s})}\rangle
\label{eq:C_s_HB},
\end{eqnarray}
can be obtained.
Similar to Eq. \ref{eq:n_HB} and Eq. \ref{eq:k}, the corresponding correlation function
\begin{eqnarray}
n^{(\text{s})}(t)=\langle h^{(\text{s})}(0)[1-h^{(\text{s})}(t)]h^{(\text{d,s})} \rangle/\langle h^{(\text{s})}\rangle
\label{eq:n_s_HB},
\end{eqnarray}
and interface reactive flux function
\begin{eqnarray}
k^{(\text{s})}(t)= -\frac{dc^{(\text{s})}(t)}{dt}
\label{eq:k_s_HB}
\end{eqnarray}
are obtained.
The $h^{(\text{d,s})}(t)$ is 1 when the a tagged pair of water molecules $i$, $j$ are \emph{at the interface} and the interoxygen distance between the two molecules is less than $r^{\mathrm{c}}_{\mathrm{OO}}$ at time $t$,
and 0 otherwise, i.e.,
\begin{align}
h^{(\text{d,s})}[{\mathbf r}(t)]=\left\{
\begin{array}{rcl}
1 & & {i,j\text{ are at the interface} } \\
& & {{\text{and }} |\text{O}_i \text{O}_j| < r_{\text{OO}}^{\text{c}}; }\\
0 & & {\text{otherwise.}}
\end{array} \right.
\end{align}
Therefore, $n^{(\text{s})}(t)$ represents the prabability at time $t$ that a tagged pair of initially H-bonded water molecules {at the interface} are unbonded but remain at the interface and separated by less than $r_{\text{OO}}^{\text c}$; $k^{(\text{s})}(t)$ measures the effective decay rate of H-bonds {at the interface}.
The functions defined in Eq.s \ref{eq:C_s_HB}--\ref{eq:k_s_HB} are used to determine the reaction rate constant of breaking and reforming and the lifetimes of H-bonds at the interface by Eq. \ref{eq:fitting_k_rates}, in which $c(t)$, $n(t)$ and $k(t)$ are replaced by $c^{(\text{s})}(t)$, $n^{(\text{s})}(t)$ and $k^{(\text{s})}(t)$ .
In the IHB method, it is accurate to choose the water molecules and H-bonds at the interface.
However, for some special H-bonds, if it connects such two water molecules, one is at the interface
and the other is in bulk water, the HB breaking reaction rate of such H-bonds will be artificially increased.
Therefore, in contrast to the IMS method, the IHB method \emph{overestimates} the HB breaking rate constant.
\section{\label{sec:discussions}Discussions: effects of air-water interface on hydrogen bond dynamics}
\subsection{\label{sec:hb_relaxation} Hydrogen bond relaxation}
\begin{figure}[h]
\centering
\includegraphics [width=0.50\textwidth] {./diagrams/128w_itp_pure_water_pair_c_ihb_aip.eps}
\setlength{\abovecaptionskip}{0pt}
\caption{\label{fig:128w_itp_pure_water_pair_c_ihb}
The $c^{(\text{s})}(t)$ for interface H-bonds with differnt thickness $d$, based on
IHB population $h^\text{(s)}(t)$, as computed from the (a) ADH and (b) AHD criteria of H-bonds.}
\end{figure}
Two geometric criteria of H-bonds are used to calculate the $h^{(\text{s})}(t)$, and the corresponding $c^{(\text{s})}(t)$ from Eq.\ref{eq:C_s_HB} are shown in Fig. \thinspace\ref{fig:128w_itp_pure_water_pair_c_ihb}.
We find that as $d$ increases, $c^{(\text{s})}(t)$ at the interface relaxes more slowly. When $d$ is greater than 4 \AA, $c^{(\text{s})}(t)$
recovers the bulk value.
This feature is independent of the HB definition as shown by the comparison of results in panel a and b of Fig.\thinspace\ref{fig:128w_itp_pure_water_pair_c_ihb}.
\begin{figure}[h]
\centering
\includegraphics [width=0.5\textwidth] {./diagrams/128w_itp_pure_water_pair_c_ihb_scheme1_aip.eps}
\setlength{\abovecaptionskip}{0pt}
\caption{\label{fig:128w_itp_pure_water_pair_c_ihb_scheme1}
The $c(t)$ for H-bonds at the interfaces with different $d$, based on HB population operator $h(t)$,
as computed from the (a) ADH and (b) AHD criteria of H-bonds. These results are based on
the IMS method, in which the sampling is performed every 4 ps.}
\end{figure}
Figure \ref{fig:128w_itp_pure_water_pair_c_ihb_scheme1} shows the $d$-dependence of $c(t)$.
Comparing Fig.s \thinspace\ref{fig:128w_itp_pure_water_pair_c_ihb} and \ref{fig:128w_itp_pure_water_pair_c_ihb_scheme1},
we find that $c(t)$ has the same feature: as $d$ increases, $c(t)$ also relaxes more slowly; when $d$ is greater than 4 \AA, $c^{(\text{s})}(t)$ also approaches a stable function.
Comparing $c(t)$ and $c^{(\text{s})}(t)$, the second feature is that for the same $d$, the value of $c(t)$ is always slightly larger than $c^{(\text{s})}(t)$.
Moreover, regardless of the AHD or ADH criterion of H-bonds, these two features are valid.
The first feature is the general one of interface H-bonds, while the second feature is derived from the difference in the definitions of HB population operators $h(t)$ and $h^{\text{(s)}}(t)$. Using this difference between $h(t)$ and $h^{\text{(s)}}(t)$, we can obtain the HB dynamics of the real interface, especially the thickness of the interface.
\subsection{\label{sec:rate_constants}Hydrogen bond breaking and reforming rate constants}
We further examine how reaction rate constants $k$ and $k'$ of H-bonds at the air-water interface vary with the interface thickness.
Figure \ref{fig:128w_itp_pure_water_pair_k_k_prime_ihb_both_schemes} compares
the rate constants and the lifetime $\tau_\mathrm{HB}$ obtained by
the IHB and IMS methods.
We find that, for all the three quantities $k$, $k'$ and $\tau_\mathrm{HB}$, the behavior as function of the thickness of the interface is hardly affected by the calculation methods.
To illustrate this point more clearly, we compare the $k$, $k'$ and $\tau_\mathrm{HB}$ obtained under the two methods.
\begin{figure}[h]
\centering
\includegraphics [width=0.53\textwidth]{./diagrams/128w_itp_pure_water_pair_k_k_prime_ihb_both_schemes.eps}
\setlength{\abovecaptionskip}{0pt}
\caption{\label{fig:128w_itp_pure_water_pair_k_k_prime_ihb_both_schemes}
The dependence of (a) the rate constants $k$ and $k'$ and (b) the HB lifetime $\tau_\mathrm{HB}$ on the interface thickness,
obtained by the IHB and the IMS method, respectively.
The corresponding $k$, $k'$ and $\tau_\mathrm{HB}$ in bulk water are also represented as dashed lines.
The ADH criterion is used and the fits are carried on the time
region 0.2 ps $< t <$ 12 ps.}
\end{figure}
As shown in Fig.\thinspace\ref{fig:128w_itp_pure_water_pair_k_k_prime_ihb_both_schemes},
when $d$ is larger than 4 \AA, the constants $k$ and $k'$ obtained by the two
methods agree quantitatively.
The forward rate constant $k$ obtained by using the IHB method is relatively larger than that from the IMS method, and backward rate constant $k'$ is relatively smaller.
Since $\tau_\mathrm{HB} \approx 1/k$, this directly leads to a relatively shorter HB lifetime using the IHB method.
This result is related to the definitions of $h(t)$ and $h^{\text{(s)}}(t)$. The definition of $h^{(\text{s})}(t)$ makes the HB break rate at the interface artificially increased.
The IMS method, which is based on $h(t)$, retains the original rate constant of H-bonds,
but it may include the contribution of bulk water molecules to the rate constant.
In Fig.\thinspace\ref{fig:128w_itp_pure_water_pair_k_k_prime_ihb_both_schemes}, the $k$, $k'$ and $\tau_\mathrm{HB}$ for the \emph{bulk} water
are also drawn with dashed lines as a reference.
Comparing the above-mentioned physical quantities at the air-water interface
and bulk water, we find that when the interface thickness is larger than
4 \AA, no matter which statistical method is used, the obtained reaction rate constants of the interface water is \emph{greater} than that in bulk water.
Therefore, the HB lifetime $\tau_\mathrm{HB}=1/k$ in interface water is smaller than that in bulk water.
Furthermore, as $d$ increases, $k$ and $k'$ gradually close to the rates in bulk water at the same condition.
These results are obtained by the least squares method in the same interval (0.2--12 ps). They show that the IHB method
can get results as good as the IMS method when $d$ is larger than 4 \AA.
The above two methods respectively give an extreme case of interface HB dynamics.
In other words, the calculated HB dynamics obtained by IHB method is \emph{accelerated}, compared to the real one.
Real HB dynamics of the interface is between the results of the above two methods.
Naturally, we approximate the true HB dynamics of the interface, by combining both the IHB and IMS methods.
In these two extreme cases the interface HB characteriestics tends to be the same as the thickness of the interface increases.
Properties such as the HB lifetime, HB reaction rate constants, and the
thickness of the air-water interface can be estimated.
For example, as the parameter $d$ increases, when all the $k$, $k'$, $\tau_{\text{HB}}$ of the interface calculated by both methods become consistent, the value of $d$ is the thickness of the air-water interface.
The interface thickness value are comparable to those obtained experimentally (see Table \ref{tab:interface_thickness}). This result shows that our method of determining the interface thickness is reliable and expected to be used in more interface systems.
\section{Conclusions}\label{sec:conclusions}
Based on the DFTMD simulations, the IMS method
{partially} gives information on the HB breaking and reforming reaction
rate constants through the air-water interface and therefore partially shows how much the interface affects dynamics of H-bonds in water.
The IHB method also provides partial information on the HB breaking and reforming reaction rates at the interface.
As the thickness of interface increases, comparing results in the two extreme cases,
we find that the HB breaking and reforming rate constants at the air-water interface tends to be uniform. Therefore, the real HB dynamical characteristics at the air-water interface can be derived.
We conclude that from the perspective of HB dynamics,
the thickness of the air-water interface at room temperature is 4 \AA.
In other words, at this thickness the HB dynamical properties of air-water interface become the same as bulk water.
The idea of combinating IHB and IMS can naturally be extended to the solution interfaces and other interfacial systems.
For systems where statistical properties of the interface and bulk phase differ significantly, these differences will be represented more appropriately.
In addition, the IHB method itself can also be generalize. For example, one can combine the HB population with the hydration shell of an ion, and then study the effects of ions through HB dynamics.
| {
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Сражение при Биберахе (фр. Bataille de Biberach) произошло у Биберах-ан-дер-Рис, в 35 км к юго-западу от Ульма, 9 мая 1800 года во время Войны второй коалиции эпохи французских революционных войн. В нем французские войска под командованием Лорана Гувион Сен-Сира разбили части австрийской армии под командованием Пауля Края.
В конце апреля 1800 года французская армия под командованием Жана-Виктора Моро переправилась через Рейн недалеко от Базеля. Затем 3 мая Моро разбил войска Края при Штоккахе и Энгене и вынудил его отступить. Два дня спустя Край столкнулся со своими преследователями в сражении при Мескирхе, но снова потерпел поражение.
9 мая корпус Гувиона Сен-Сира застиг армию Края у Биберах-ан-дер-Рис, и обе стороны снова сразились. Край разместил основную часть своей армии позади Бибераха; большой овраг, образованный рекой Рис, прикрывал фронт его войск.
Вначале Сен-Сир силами бригад Тарро и Бараге д'Илье атаковал десятитысячный авангард австрийцев, находившийся на левом берегу реки Рис, рассеял его и захватил сами город и расположенные в нём военные склады. После подхода резервных войск генерала Ришпанса, французы по пояс в воде перешли через реку и атаковали главные силы Края. Сначала были заняты первые высоты, а затем атакованы войска противника на плато, ведущем к Митембаху. Австрийский командующий, обескураженный быстрым поражением своего авангарда, не решился продолжить сражение и отступил с главными силами к Ульму. Авангард и другие отдельные отряды отступили к Меммингену, где через день снова были разбиты преследовавшими их французами.
Литература
Голицын Николай Сергеевич. Всеобщая военная история новейших времен. Часть 2, СПб, 1875, 957 с.
Abel Hugo, France militaire, Vol. 3 : Histoire des armées françaises de terre et de mer de 1792 à 1833, Delloye, 1836, p. 139-140.
Arnold, James R. Marengo & Hohenlinden. Barnsley, South Yorkshire, UK: Pen & Sword, 2005. ISBN 1-84415-279-0
Le 9 mai 1800 – La bataille de Biberach
Сражения по алфавиту
Военные сражения революционной Франции
Сражения Австрии
Сражения 1800 года
Май 1800 года
События 9 мая | {
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} | 795 |
Início > Keely Smit... > acordes
Don't Take Your Love From Me Teclado
Keely Smith
Don't Take Your Love From Me Key CC
Don't Take Your Love From Me Key C#C#
Don't Take Your Love From Me Key DD(Disminuir uno tono)
Don't Take Your Love From Me Key D#D#(Disminuir uno semi-tono)
Don't Take Your Love From Me Key EE(tono original)
Don't Take Your Love From Me Key FF(Aumentar uno semi-tono)
Don't Take Your Love From Me Key F#F#(Aumentar uno tono)
Don't Take Your Love From Me Key GG
Don't Take Your Love From Me Key G#G#
Don't Take Your Love From Me Key AA
Don't Take Your Love From Me Key A#A#
E E9 A A7M A9 A E7 E7/9
Tear a star from out the sky, and the sky feels blue
E7 E7/9 E7 E7/13 E7 E7/13 A A6
Tear a petal from a rose, and the rose weeps, too
A A9 Cdim B7 D9 Bm7/4 Bm7 Bm7/E
Take your heart away from mine, and mine will surely break
D9 Bm5-/7 Bm7 Bm7/E A Gdim Edim E7
My life is yours to make, so, please, keep the spark away
E E9 A A7M A9 A E7 E7/9
Would you take the wings from birds so that they can't fly
E7 E7/9 E7 E7/13 E7 E7/13 E7 A A7M A7
Would you take the ocean's roar and leave just a sigh
D6 D9 Bm5-/7 E7 A7M F#m C#m5-/7 F#7
Oh, this your heart won't let you do, this is what I beg of you
D D9 Bm5-/7 E7 A
Lena Horne - Don't Take Your Love From Me Mel Torme - Don't Take Your Love From Me Della Reese - Don't Take Your Love From Me Doris Day - Don't Take Your Love From Me Steve Lawrence - Don't Take Your Love From Me Abbey Lincoln - Don't Take Your Love From Me | {
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{"url":"http:\/\/ptsymmetry.net\/?m=201408","text":"August 2014\nMon Tue Wed Thu Fri Sat Sun\n\u00ab Jul \u00a0 Sep \u00bb\n123\n45678910\n11121314151617\n18192021222324\n25262728293031\n\n## PT-\/non-PT-Symmetric and non-Hermitian Hellmann Potential: Approximate Bound and Scattering States with Any $$\\ell$$-Values\n\nAltug Arda, Ramazan Sever\n\nWe investigate the approximate bound state solutions of the Schrodinger equation for the PT-\/non-PT-symmetric and non Hermitian Hellmann potential. Exact energy eigenvalues and corresponding normalized wave functions are obtained. Numerical values of energy eigenvalues for the bound states are compared with the ones obtained before. Scattering state solutions are also studied. Phase shifts of the potential are written in terms of the angular momentum quantum number $$\\ell$$.\n\nhttp:\/\/arxiv.org\/abs\/1409.0518\nQuantum Physics (quant-ph)\n\n## Three-level $$\\Lambda$$-type atomic systems with a Pseudo-Hermitian PT-symmetric Hamiltonian\n\nAmarendra K. Sarma, Balla Prannay\n\nWe have studied a three-level $$\\Lambda$$-type atomic system with all the energy levels exhibiting decay. The system is described by a pseudo-Hermitian Hamiltonian and subject to certain conditions, the Hamiltonian shows parity-time (PT) symmetry. The probability amplitudes of various atomic levels both below and above the PT-theshold is worked out.\n\nhttp:\/\/arxiv.org\/abs\/1408.6672\nQuantum Physics (quant-ph); Optics (physics.optics)\n\n## Unbreakable PT-symmetry of solitons supported by inhomogeneous defocusing nonlinearity\n\nYaroslav V. Kartashov, Boris A. Malomed, Lluis Torner\n\nWe consider bright solitons supported by a symmetric inhomogeneous defocusing nonlinearity growing rapidly enough toward the periphery of the medium, combined with an antisymmetric gain-loss profile. Despite the absence of any symmetric modulation of the linear refractive index, which is usually required to establish a PT-symmetry in the form of a purely real spectrum of modes, we show that the PT-symmetry is never broken in the present system, and that the system always supports stable bright solitons, fundamental and multi-pole ones. Such phenomenon is connected to non-linearizability of the underlying evolution equation. The increase of the gain-losses strength results, in lieu of the PT-symmetry breaking, in merger of pairs of different soliton branches, such as fundamental and dipole, or tripole and quadrupole ones. The fundamental and dipole solitons remain stable for all values of the gain-loss coefficient.\n\nhttp:\/\/arxiv.org\/abs\/1408.6174\nOptics (physics.optics); Pattern Formation and Solitons (nlin.PS)\n\n## Elementary modes of coupled oscillators with balanced loss and gain\n\nWe provide a reduction of a set of two coupled oscillators with balanced loss and gain in their elementary modes. A possible method of quantization based on these elementary modes, in the framework of PT symmetric quantum mechanics is indicated.\n\nhttp:\/\/arxiv.org\/abs\/1408.5038\nHigh Energy Physics \u2013 Theory (hep-th)\n\n## PT-symmetry in macroscopic magnetic structures\n\nJ. M. Lee, T. Kottos, B. Shapiro\n\nWe introduce the notion of PT-symmetry in magnetic nanostructures and show that they can support a new type of non-Hermitian dynamics. Using the simplest possible set-up consisting of two coupled ferromagnetic films, one with loss and another one with a balanced amount of gain, we demonstrate the existence of a spontaneous PT-symmetry breaking point where both the eigenfrequencies and eigenvectors are degenerate. Below this point the frequency spectrum is real indicating stable dynamics while above this point it is complex signaling unstable dynamics.\n\nhttp:\/\/arxiv.org\/abs\/1408.3285\nMesoscale and Nanoscale Physics (cond-mat.mes-hall); Mathematical Physics (math-ph); Chaotic Dynamics (nlin.CD)\n\n## PT-symmetry in optics beyond the paraxial approximation\n\nChangming Huang, Fangwei Ye, Yaroslav V. Kartashov, Boris A Malomed, Xianfeng Chen\n\nThe concept of the PT-symmetry, originating from the quantum field theory, has been intensively investigated in optics, stimulated by the similarity between the Schr\\\u201dodinger equation and the paraxial wave equation that governs the propagation of light in a guiding structure. We go beyond the bounds of the paraxial approximation and demonstrate, using the solution of the Maxwell\u2019s equations for light beams propagating in deeply subwavelength waveguides and periodic lattices with \u201cbalanced\u201d gain and loss, that the PT symmetry may stay unbroken in this setting. Moreover, the PT-symmetry in subwavelength optical structures may be restored after being initially broken upon the increase of gain and loss. Critical gain\/loss levels, at which the breakup and subsequent restoration of the PT symmetry occur, strongly depend on the scale of the structure.\n\nhttp:\/\/arxiv.org\/abs\/1408.2630\nOptics (physics.optics); Pattern Formation and Solitons (nlin.PS)\n\n## Infinitely many inequivalent field theories from one Lagrangian\n\nCarl M. Bender, Daniel W. Hook, Nick E. Mavromatos, Sarben Sarkar\n\nLogarithmic time-like Liouville quantum field theory has a generalized PT invariance, where T is the time-reversal operator and P stands for an S-duality reflection of the Liouville field $$\\phi$$. In Euclidean space the Lagrangian of such a theory, $$L=\\frac{1}{2}(\\nabla\\phi)^2\u2212ig\\phi \\exp(ia\\phi)$$, is analyzed using the techniques of PT-symmetric quantum theory. It is shown that L defines an infinite number of unitarily inequivalent sectors of the theory labeled by the integer n. In one-dimensional space (quantum mechanics) the energy spectrum is calculated in the semiclassical limit and the $$m$$th energy level in the $$n$$th sector is given by $$E_{m,n}\u223c(m+1\/2)^2a^2\/(16n^2)$$.\n\nhttp:\/\/arxiv.org\/abs\/1408.2432\nHigh Energy Physics \u2013 Theory (hep-th); Mathematical Physics (math-ph); Quantum Physics (quant-ph)\n\n## Light transport in PT-invariant photonic structures with hidden symmetries\n\nM.H. Teimourpour, R. El-Ganainy, A. Eisfeld, A. Szameit, D.N Christodoulides\n\nWe introduce a recursive bosonic quantization technique for generating classical PT photonic structures that possess hidden symmetries and higher order exceptional points. We study light transport in these geometries and we demonstrate that perfect state transfer is possible only for certain initial conditions. Moreover, we show that for the same propagation direction, left and right coherent transports are not symmetric with field amplitudes following two different trajectories. A general scheme for identifying the conservation laws in such PT-symmetric photonic networks is also presented.\n\nhttp:\/\/arxiv.org\/abs\/1408.1561\nOptics (physics.optics); Quantum Physics (quant-ph)\n\n## On Symmetries and Exact Solutions of a Class of Non-local Non-linear Schrodinger Equations with Self-induced PT-symmetric Potential\n\nDebdeep Sinha, Pijush K. Ghosh\n\nA class of non-local non-linear Schrodinger equations(NLSE) is considered in an external potential with space-time modulated coefficient of the nonlinear interaction term as well as confining and\/or loss-gain terms. This is a generalization of a recently introduced integrable non-local NLSE with self induced potential that is PT symmetric in the corresponding stationary problem. Exact soliton solutions are obtained for the inhomogeneous and\/or non autonomous non-local NLSE by using similarity transformation and the method is illustrated with a few examples. It is found that only those transformations are allowed for which the transformed spatial coordinate is odd under the parity transformation of the original one. It is shown that the non-local NLSE without the external potential and a $$d+1$$ dimensional generalization of it, admits all the symmetries of the $$d+1$$ dimensional Schrodinger group. The conserved Noether charges associated with the time-translation, dilatation and special conformal transformation are shown to be real-valued in spite of being non-hermitian. Finally, dynamics of different moments are studied with an exact description of the time-evolution of the \u201cpseudo-width\u201d of the wave-packet for the special case when the system admits a $$O(2,1)$$ conformal symmetry.\n\nhttp:\/\/arxiv.org\/abs\/1408.0954\nExactly Solvable and Integrable Systems (nlin.SI); High Energy Physics \u2013 Theory (hep-th)\n\n## Symmetry breaking of solitons in one-dimensional parity-time-symmetric optical potentials\n\nJianke Yang\n\nSymmetry breaking of solitons in a class of one-dimensional parity-time (PT) symmetric complex potentials with cubic nonlinearity is reported. In generic PT symmetric potentials, such symmetry breaking is forbidden. However, in a special class of PT-symmetric potentials $$V(x)=g^2(x)+\u03b1g(x)+ig\u2032(x)$$, where $$g(x)$$ is a real and even function and \u03b1 a real constant, symmetry breaking of solitons can occur. That is, a branch of non-PT-symmetric solitons can bifurcate out from the base branch of PT-symmetric solitons when the base branch\u2019s power reaches a certain threshold. At the bifurcation point, the base branch changes stability, and the bifurcated branch can be stable.\n\nhttp:\/\/arxiv.org\/abs\/1408.0687\nOptics (physics.optics); Pattern Formation and Solitons (nlin.PS)","date":"2021-10-26 19:09:04","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5722071528434753, \"perplexity\": 1536.0362676468967}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 5, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-43\/segments\/1634323587915.41\/warc\/CC-MAIN-20211026165817-20211026195817-00137.warc.gz\"}"} | null | null |
\section{Introduction}
\label{section:intro}
Several fundamental machine learning tasks in the real world involve intricate interactions between boundedly rational\footnote{\label{brat}Boundedly rational agents are subject to limited cognition and time in making decisions~\cite{Reinhard2002}.}, self-interested agents that can be modeled as a form of repeated games with unknown, complex, and costly-to-evaluate payoff functions for the agents. For example, in adversarial \emph{machine learning} (ML),
the interactions between the \emph{defender} $\mathcal{D}$ and the \emph{attacker} $\mathcal{A}$ of an ML model can be modeled as a repeated game in which the payoffs to $\mathcal{D}$ and $\mathcal{A}$ are the performance of the ML model (e.g., validation accuracy) and its negation, respectively. Specifically, given a fully trained image classification model (say, provided as an online service), $\mathcal{A}$ attempts to fool the ML model into misclassification through repeated queries of the model using perturbed input images. On the other hand, for each queried image that is perturbed by $\mathcal{A}$, $\mathcal{D}$ tries to ensure the correctness of its classification by transforming the perturbed image before feeding it into the ML model. As another example, \emph{multi-agent reinforcement learning} (MARL) in an episodic environment can also be modeled as a repeated game in which the payoff to each agent is its return from the execution of all the agents' selected policies.
Solving such a form of repeated games in a cost-efficient manner is challenging since the payoff functions of the agents are unknown, complex (e.g., possibly noisy, non-convex, and/or with no closed-form expression/derivative), and costly to evaluate. Fortunately, the payoffs corresponding to different actions of each agent tend to be correlated. For example, in adversarial ML, the correlated perturbations performed by the attacker $\mathcal{A}$ (and correlated transformations executed by the defender $\mathcal{D}$) are likely to induce similar effects on the performance of the ML model. Such a correlation can be leveraged to \emph{predict} the payoff associated with any action of an agent using a \emph{surrogate} model such as the rich class of Bayesian nonparametric \emph{Gaussian process} (GP) models~\cite{rasmussen2004gaussian} which is expressive enough to represent a predictive belief of the unknown, complex payoff function over the action space of the agent. Then, in each iteration, the agent can select an action for evaluating its unknown payoff function that trades off between sampling at or near to a likely maximum payoff based on the current GP belief (exploitation) vs. improving the GP belief (exploration) until its cost/sampling budget is expended. To do this, the agent can use a sequential black-box optimizer such as the celebrated \emph{Bayesian optimization} (BO) algorithm~\cite{shahriari2016taking} based on the \emph{GP-upper confidence bound} (GP-UCB) acquisition function~\cite{srinivas2009gaussian}, which guarantees asymptotic no-regret performance and is sample-efficient in practice. How then can we design a BO algorithm to account for its interactions with boundedly rational\cref{brat}, self-interested agents and still guarantee the trademark asymptotic no-regret performance?
Inspired by the cognitive hierarchy model of games~\cite{camerer2004cognitive}, we adopt a recursive reasoning formalism (i.e., typical among humans) to model the reasoning process in the interactions between boundedly rational\cref{brat}, self-interested agents. It comprises $k$ levels of reasoning which represents the cognitive limit of the agent. At level $k=0$ of reasoning, the agent randomizes its choice of actions. At a higher level $k \geq 1$ of reasoning, the agent selects its best response to the actions of the other agents who are reasoning at lower levels $0, 1, \ldots, k - 1$.
This paper presents the first recursive reasoning formalism of BO to model the reasoning process in the interactions between boundedly rational\cref{brat}, self-interested agents with unknown, complex, and costly-to-evaluate payoff functions in repeated games, which we call \emph{\underline{R}ecursive \underline{R}easoning-\underline{B}ased \underline{B}O} (R2-B2) (Section~\ref{r2b2}).
R2-B2 provides these agents with principled strategies for performing effectively in this type of game.
In this paper, we consider repeated games with simultaneous moves and perfect monitoring\footnote{\label{smpm}In each iteration of a repeated game with (a) simultaneous moves and (b) perfect monitoring, every agent, respectively, (a) chooses its action simultaneously without knowing the other agents' selected actions, and (b) has access to the entire history of game plays, which includes all actions selected and payoffs observed by every agent in the previous iterations.}.
Our R2-B2 algorithm is general in that it does not constrain the relationship among the payoff functions of different agents and can thus be applied to various types of games such as constant-sum games (e.g., adversarial ML in which the attacker $\mathcal{A}$ and defender $\mathcal{D}$ have opposing objectives), general-sum games (e.g., MARL where all agents have possibly different yet not necessarily conflicting goals), and common-payoff games (i.e., all agents have identical payoff functions). We prove that by reasoning at level $k\geq 2$ and one level higher than the other agents, our R2-B2 agent can achieve faster asymptotic convergence to no regret than that without utilizing recursive reasoning (Section~\ref{subsec:level_k_policy}).
We also propose a computationally cheaper variant of R2-B2 called R2-B2-Lite at the expense of a weaker convergence guarantee (Section~\ref{lite}). The performance and generality of R2-B2 are demonstrated through extensive experiments using synthetic games, adversarial ML, and MARL (Section~\ref{expt}). Interestingly, we empirically show that by reasoning at a higher level, our R2-B2 defender is able to effectively defend against the attacks from the state-of-the-art black-box adversarial attackers (Section~\ref{exp:comparison_parsimonious}), which can be of independent interest to the adversarial ML community.
\section{Background and Problem Formulation}
\label{background}
For simplicity, we will mostly focus on repeated games between two agents, but have extended our R2-B2 algorithm to games involving \emph{more than two} agents, as detailed in Appendix~\ref{subsec:more_than_two_players}. To ease exposition, throughout this paper, we will use adversarial ML as the running example and thus refer to the two agents as the \emph{attacker} $\mathcal{A}$ and the \emph{defender} $\mathcal{D}$.
For example, the input action space $\mathcal{X}_1\subset \mathbb{R}^{d_1}$ of $\mathcal{A}$ can be a set of allowed perturbations of a test image
while the input action space $\mathcal{X}_2\subset \mathbb{R}^{d_2}$ of $\mathcal{D}$ can represent a set of feasible transformations of the perturbed test image.
We consider both input domains $\mathcal{X}_1$ and $\mathcal{X}_2$ to be discrete for simplicity;
generalization of our theoretical results in Section~\ref{r2b2} to continuous, compact domains can be easily achieved through a suitable discretization of the domains~\cite{srinivas2009gaussian}.
When the ML model is an image classification model, the payoff function $f_1:\mathcal{X}_1 \times \mathcal{X}_2 \rightarrow \mathbb{R}$ of $\mathcal{A}$,
which takes in its perturbation $\mathbf{x}_1 \in \mathcal{X}_1$ and $\mathcal{D}$'s transformation $\mathbf{x}_2 \in \mathcal{X}_2$ as inputs,
can be the maximum predictive probability among all incorrect classes for a test image since $\mathcal{A}$ intends to cause misclassification.
Since $\mathcal{A}$ and $\mathcal{D}$ have opposing objectives (i.e., $\mathcal{D}$ intends to prevent misclassification), the payoff function $f_2: \mathcal{X}_1 \times \mathcal{X}_2 \rightarrow \mathbb{R}$ of $\mathcal{D}$ can be the negation of that of $\mathcal{A}$, thus resulting in a constant-sum game between $\mathcal{A}$ and $\mathcal{D}$.
In each iteration $t=1,\ldots,T$ of the repeated game with simultaneous moves and perfect monitoring\cref{smpm}\footnote{Note that in some tasks such as adversarial ML, the requirement of perfect monitoring can be relaxed considerably. Refer to Section~\ref{exp:comparison_parsimonious} for more details.}, $\mathcal{A}$ and $\mathcal{D}$ select their respective input actions $\mathbf{x}_{1,t}$ and $\mathbf{x}_{2,t}$ simultaneously using our R2-B2 algorithm (Section~\ref{r2b2}) for evaluating their payoff functions $f_1$ and $f_2$.
Then, $\mathcal{A}$ and $\mathcal{D}$ receive the respective noisy observed payoffs $y_{1,t}\triangleq f_1(\mathbf{x}_{1,t}, \mathbf{x}_{2,t})+\epsilon_1$ and $y_{2,t}\triangleq f_2(\mathbf{x}_{1,t}, \mathbf{x}_{2,t})+\epsilon_2$ with i.i.d. Gaussian noises
$\epsilon_i \sim \mathcal{N}(0,\sigma_i^2)$
and noise variances $\sigma_i^2$ for $i=1,2$.
A common practice in game theory is to measure the performance of $\mathcal{A}$ via its \emph{(external) regret}~\cite{nisan2007algorithmic}:
\begin{equation}
\begin{array}{c}
R_{1,T} \triangleq \sum^T_{t=1} [f_1(\mathbf{x}^*_1,\mathbf{x}_{2,t})-f_1(\mathbf{x}_{1,t},\mathbf{x}_{2,t})]
\end{array}
\label{extregret}
\end{equation}
where $\mathbf{x}^*_1\triangleq\mathop{\arg\max}_{\mathbf{x}_1\in \mathcal{X}_1} \sum^T_{t=1} f_1(\mathbf{x}_1, \mathbf{x}_{2,t})$.
The external regret $R_{2,T}$ of $\mathcal{D}$ is defined in a similar manner.
An algorithm is said to achieve asymptotic \emph{no regret} if $R_{1,T}$ grows sub-linearly in $T$, i.e., $\lim_{T\rightarrow \infty} R_{1,T}/T = 0$.
Intuitively, by following a no-regret algorithm, $\mathcal{A}$ is guaranteed to eventually find its optimal input action $\mathbf{x}^*_1$ in hindsight,
regardless of $\mathcal{D}$'s sequence of input actions.
To guarantee no regret (Section~\ref{r2b2}), $\mathcal{A}$ represents a predictive belief of its unknown, complex payoff function $f_1$
using the rich class of \emph{Gaussian process} (GP) models
by modeling $f_1$ as a sample of a GP~\cite{rasmussen2004gaussian}.
$\mathcal{D}$ does likewise with its unknown $f_2$.
Interested readers are referred to Appendix~\ref{app:gp} for a detailed background on GP.
In particular, $\mathcal{A}$ uses the GP predictive/posterior belief of $f_1$ to compute a probabilistic upper bound of $f_1$ called the \emph{GP-upper confidence bound} (GP-UCB)~\cite{srinivas2009gaussian} at any joint input actions $(\mathbf{x}_1, \mathbf{x}_2)$, which will be exploited by our R2-B2 algorithm (Section~\ref{r2b2}):
\begin{equation}
\alpha_{1,t}(\mathbf{x}_1, \mathbf{x}_2)\triangleq\mu_{t-1}(\mathbf{x}_1, \mathbf{x}_2) + \beta^{1/2}_t \sigma_{t-1}(\mathbf{x}_1, \mathbf{x}_2)
\label{eqgpucb}
\end{equation}
for iteration $t$ where $\mu_{t-1}(\mathbf{x}_1, \mathbf{x}_2)$ and $\sigma^2_{t-1}(\mathbf{x}_1, \mathbf{x}_2)$
denote, respectively, the GP posterior mean and variance at $(\mathbf{x}_1, \mathbf{x}_2)$ (Appendix~\ref{app:gp}) conditioned on the history of game plays up till iteration $t-1$ that includes $\mathcal{A}$'s observed payoffs and the actions selected by both agents in iterations $1,\ldots,t-1$.
The GP-UCB acquisition function $\alpha_{2,t}$ for $\mathcal{D}$ is defined likewise.
Supposing $\mathcal{A}$ knows the input action $\mathbf{x}_{2,t}$ selected by $\mathcal{D}$ and chooses an input action $\mathbf{x}_1$ to maximize the GP-UCB acquisition function $\alpha_{1,t}$~\eqref{eqgpucb}, its choice involves trading off between sampling close to an expected maximum payoff (i.e., with large GP posterior mean) given the current GP belief of $f_1$ (exploitation) vs.~that of high predictive uncertainty (i.e., with large GP posterior variance) to improve the GP belief of $f_1$ (exploration) where the parameter $\beta_t$ is set to trade off between exploitation vs.~exploration for bounding its external regret~\eqref{extregret}, as specified later in Theorem~\ref{regret_gp}.
\section{\underline{R}ecursive \underline{R}easoning-\underline{B}ased \underline{B}ayesian Optimization (R2-B2)}
\label{r2b2}
Algorithm~\ref{BORR_attacker} describes the R2-B2 algorithm from the perspective of \emph{attacker} $\mathcal{A}$ which we will adopt in this section. Our R2-B2 algorithm for \emph{defender} $\mathcal{D}$ can be derived analogously.
We will now discuss the recursive reasoning formalism of BO for $\mathcal{A}$'s action selection in step $2$ of Algorithm~\ref{BORR_attacker}.
\subsection{Recursive Reasoning Formalism of BO}
\label{subsec:recursive_reasoning_model}
Our recursive reasoning formalism of BO follows a similar principle as the cognitive hierarchy model~\cite{camerer2004cognitive}:
At level $k=0$ of reasoning, $\mathcal{A}$ adopts some randomized/mixed strategy of selecting its action. At level $k\geq 1$ of reasoning, $\mathcal{A}$ best-responds to the strategy of $\mathcal{D}$ who is reasoning at a lower level.
Let $\mathbf{x}^{k}_{1,t}$ denote the input action $\mathbf{x}_{1,t}$ selected by $\mathcal{A}$'s strategy from reasoning at level $k$ in iteration $t$.
Depending on the (a) degree of knowledge about $\mathcal{D}$ and (b) available computational resource,
$\mathcal{A}$ can choose one of the following three types of strategies of selecting its action with varying levels of reasoning,
as shown in Fig.~\ref{fig:illustration}:
{\bf Level-$k=0$ Strategy.}
Without knowledge of $\mathcal{D}$'s level of reasoning nor its level-$0$ strategy, $\mathcal{A}$ by default can reason at level $0$ and play a mixed strategy $\mathcal{P}^{0}_{1,t}$ of selecting its action by sampling $\mathbf{x}^{0}_{1,t}$ from the probability distribution $\mathcal{P}^{0}_{1,t}$ over its input action space $\mathcal{X}_1$, as discussed in Section~\ref{subsec:level-0}.
{\bf Level-$k=1$ Strategy.}
If $\mathcal{A}$ thinks that $\mathcal{D}$ reasons at level $0$ and has knowledge of $\mathcal{D}$'s level-$0$ mixed strategy $\mathcal{P}^{0}_{2,t}$,
then $\mathcal{A}$ can reason at level $1$ and play a pure strategy that best-responds to the level-$0$ strategy
of $\mathcal{D}$, as explained in Section~\ref{subsec:level_1_reasoning}.
Such a level-$1$ reasoning of $\mathcal{A}$ is general since it caters to \emph{any} level-$0$ strategy of $\mathcal{D}$ and hence does not require $\mathcal{D}$ to perform recursive reasoning.
{\bf Level-$k\geq 2$ Strategy.}
If $\mathcal{A}$ thinks that $\mathcal{D}$ reasons at level $k - 1$,
then $\mathcal{A}$ can reason at level $k$ and play a pure strategy that best-responds to $\mathcal{D}$'s level-$(k-1)$ action, as detailed in Section~\ref{subsec:level_k_policy}.
Different from the level-$1$ reasoning of $\mathcal{A}$, its level-$k$ reasoning assumes that $\mathcal{D}$'s level-$(k-1)$ action is derived using the same recursive reasoning process.
\begin{algorithm}[t]
\begin{algorithmic}[1]
\FOR{$t=1,2,\ldots, T$}
\STATE Select input action $\mathbf{x}_{1,t}$ using its level-$k$ strategy (while defender $\mathcal{D}$ selects input action $\mathbf{x}_{2,t}$)
\STATE Observe noisy payoff $y_{1,t}=f_1(\mathbf{x}_{1,t},\mathbf{x}_{2,t}) + \epsilon_1$
\STATE Update GP posterior belief using $\langle(\mathbf{x}_{1,t}, \mathbf{x}_{2,t}) , y_{1,t}\rangle$
\ENDFOR
\end{algorithmic}
\caption{R2-B2 for attacker $\mathcal{A}$'s level-$k$ reasoning}
\label{BORR_attacker}
\end{algorithm}
\begin{figure}
\centering
\begin{subfigure}[t]{0.313\linewidth}
\includegraphics[width=\linewidth]{figures/level_0.pdf}
\caption{Level $0$}
\end{subfigure}
\begin{subfigure}[t]{0.313\linewidth}
\includegraphics[width=\linewidth]{figures/level_1.pdf}
\caption{Level $1$}
\end{subfigure}
\begin{subfigure}[t]{0.313\linewidth}
\includegraphics[width=\linewidth]{figures/level_2.pdf}
\caption{Level $2$}
\end{subfigure}
\caption{Illustration of attacker $\mathcal{A}$'s strategies of selecting its input action
from reasoning at levels $k=0$, $1$, and $2$.}
\label{fig:illustration}
\end{figure}
\subsubsection{Level-$k=0$ Strategy}
\label{subsec:level-0}
Level $0$ is a conservative, default choice for $\mathcal{A}$ since it does not require \emph{any} knowledge about $\mathcal{D}$'s strategy of selecting its input action and is computationally lightweight.
At level $0$, $\mathcal{A}$ plays a mixed strategy $\mathcal{P}^{0}_{1,t}$ by sampling $\mathbf{x}^{0}_{1,t}$ from the probability distribution $\mathcal{P}^{0}_{1,t}$ over its input action space
$\mathcal{X}_1$: $\mathbf{x}^{0}_{1,t} \sim \mathcal{P}^{0}_{1,t}$.
A mixed/randomized strategy (instead of a pure/deterministic strategy) is considered because
without knowledge of $\mathcal{D}$'s strategy,
$\mathcal{A}$ has to treat $\mathcal{D}$ as a black-box adversary.
This setting corresponds to that of an \emph{adversarial bandit} problem in which any deterministic strategy suffers from linear worst-case regret~\cite{lattimore2018bandit} and
\emph{randomization} alleviates this issue.
Such a randomized design of our level-$0$ strategies is consistent with that of the cognitive hierarchy model in which a level-$0$ thinker does not make any assumption about the other agent and
selects its action via a probability distribution without using strategic thinking~\cite{camerer2004cognitive}.
We will now present a few reasonable choices of level-$0$ mixed strategies.
However, in both theory (Theorems~\ref{theorem_level_1},~\ref{theorem_level_k} and~\ref{theorem_borr_lite}) and practice, \emph{any} strategy of action selection (including existing methods (Section~\ref{exp:comparison_parsimonious})) can be considered as a level-$0$ strategy.
In the simplest setting where $\mathcal{A}$ has no knowledge of $\mathcal{D}$'s strategy,
a natural choice for its level-$0$ mixed strategy is \emph{random search}.
That is, $\mathcal{A}$ samples its action from a uniform distribution over $\mathcal{X}_1$.
An alternative choice is to use the \emph{EXP3 algorithm} for the adversarial linear bandit problem, which requires the GP to be transformed via a random features approximation~\cite{rahimi2008random} into linear regression
with random features as inputs.
Since the regret of EXP3 algorithm is bounded from above by $\mathcal{O}(\sqrt{d'_1 T \log |\mathcal{X}_1|})$~\cite{lattimore2018bandit} where $d'_1$ denotes the number of random features,
it incurs sub-linear regret and can thus achieve asymptotic no regret.
In a more relaxed setting where $\mathcal{A}$ has access to the history of actions selected by $\mathcal{D}$, $\mathcal{A}$ can use the \emph{GP-MW algorithm}~\cite{sessa2019no} to derive its level-$0$ mixed strategy;
for completeness, GP-MW is briefly described in Appendix~\ref{app:gp_mw}.
The result below bounds the regret of $\mathcal{A}$ when using GP-MW for level-$0$ reasoning and
its proof is slightly modified from that of~\citet{sessa2019no} to account for its payoff function $f_1$
being sampled from a GP (Section~\ref{background}):
\begin{theorem}
\label{regret_gp}
Let $\delta\in (0,1)$, $\beta_t\triangleq 2\log (|\mathcal{X}_1|t^2\pi^2/(3\delta))$, and $\gamma_T$ denotes the maximum information gain about payoff function $f_1$ from any history of actions selected by both agents and corresponding noisy payoffs observed by $\mathcal{A}$ up till iteration $T$.
Suppose that $\mathcal{A}$ uses GP-MW to derive its level-$0$ strategy.
Then, with probability of at least $1 - \delta$,
$$
R_{1,T} = \mathcal{O}(\sqrt{T\log |\mathcal{X}_1|} + \sqrt{T\log (2/\delta)} + \sqrt{T \beta_T \gamma_T} )\ .
$$
\end{theorem}
From Theorem~\ref{regret_gp}, $R_{1,T}$
is sub-linear in $T$.\footnote{\label{gammat}The asymptotic growth of $\gamma_T$ has been analyzed for some commonly used kernels: $\gamma_T=\mathcal{O}((\log T)^{d_1+1})$ for squared exponential kernel and $\gamma_T=\mathcal{O}(T^{d_1(d_1+1)/(2\nu + d_1(d_1+1))}\log T)$ for Mat\'ern kernel with parameter $\nu>1$. For both kernels, the last term in the regret bound in Theorem~\ref{regret_gp} grows sub-linearly in $T$.} So, $\mathcal{A}$ using GP-MW for level-$0$ reasoning achieves asymptotic no regret.
\subsubsection{Level-$k=1$ Strategy}
\label{subsec:level_1_reasoning}
If $\mathcal{A}$ thinks that $\mathcal{D}$ reasons at level $0$ and has knowledge of $\mathcal{D}$'s level-$0$ strategy $\mathcal{P}^{0}_{2,t}$, then $\mathcal{A}$ can reason at level $1$.
Specifically, $\mathcal{A}$ selects its level-$1$ action $\mathbf{x}^{1}_{1,t}$ that maximizes the expected value of GP-UCB~\eqref{eqgpucb} w.r.t.~$\mathcal{D}$'s level-$0$ strategy:
\begin{equation}
\begin{array}{c}
\mathbf{x}^{1}_{1,t}\triangleq\mathop{\arg\max}_{\mathbf{x}_1\in \mathcal{X}_1} \mathbb{E}_{\mathbf{x}^{0}_{2,t} \sim \mathcal{P}^{0}_{2,t}}[\alpha_{1,t}(\mathbf{x}_1, \mathbf{x}^{0}_{2,t})]\ .
\end{array}
\label{eq:level_1_defender}
\end{equation}
If input action space $\mathcal{X}_2$ of $\mathcal{D}$ is discrete and
not too large, then~\eqref{eq:level_1_defender} can be solved exactly.
Otherwise,~\eqref{eq:level_1_defender} can be solved approximately via sampling from $\mathcal{P}^{0}_{2,t}$.
Such a level-$1$ reasoning of $\mathcal{A}$ to solve~\eqref{eq:level_1_defender} only requires access to the history of actions selected by $\mathcal{D}$ but not its observed payoffs, which is the same as that needed by GP-MW.
Our first main result (see its proof in Appendix~\ref{app:proof}) bounds the expected regret of $\mathcal{A}$ when using R2-B2 for level-$1$ reasoning:
\begin{theorem}
\label{theorem_level_1}
Let $\delta\in (0, 1)$ and $C_1\triangleq 8/\log(1+\sigma^{-2}_1)$.
Suppose that $\mathcal{A}$ uses R2-B2 (Algorithm~\ref{BORR_attacker}) for level-$1$ reasoning and $\mathcal{D}$ uses mixed strategy $\mathcal{P}^{0}_{2,t}$ for level-$0$ reasoning.
Then, with probability of at least $1 - \delta$,
$\mathbb{E}[R_{1,T}] \leq \sqrt{C_1 T \beta_T \gamma_T}$ where the expectation is with respect to the history of actions selected and payoffs observed by $\mathcal{D}$.
\end{theorem}
It follows from Theorem~\ref{theorem_level_1} that $\mathbb{E}[R_{1,T}]$ is sublinear in $T$.\cref{gammat}
So, $\mathcal{A}$ using R2-B2 for level-$1$ reasoning achieves asymptotic no expected regret, which holds for \emph{any} level-$0$ strategy of $\mathcal{D}$ regardless of whether $\mathcal{D}$ performs recursive reasoning.\vspace{-1mm}
\subsubsection{Level-$k\geq2$ Strategy}\vspace{-0.3mm}
\label{subsec:level_k_policy}
If $\mathcal{A}$ thinks that $\mathcal{D}$ reasons at level $1$, then $\mathcal{A}$ can reason at level $2$ and select its level-$2$ action $\mathbf{x}^{2}_{1,t}$~\eqref{eq:level_2_defender} to best-respond to level-$1$ action $\mathbf{x}^{1}_{2,t}$~\eqref{eq:level_1_attacker} selected by $\mathcal{D}$, the latter of which can be computed/simulated by $\mathcal{A}$ in a similar manner as~\eqref{eq:level_1_defender}:
\begin{equation}
\begin{array}{c}
\mathbf{x}^{2}_{1,t}\triangleq\mathop{\arg\max}_{\mathbf{x}_1\in \mathcal{X}_1} \alpha_{1,t}(\mathbf{x}_1, \mathbf{x}^{1}_{2,t})\ ,
\end{array}
\label{eq:level_2_defender}
\end{equation}
\begin{equation}
\begin{array}{c}
\mathbf{x}^{1}_{2,t}\triangleq\mathop{\arg\max}_{\mathbf{x}_2\in \mathcal{X}_2} \mathbb{E}_{\mathbf{x}^{0}_{1,t} \sim \mathcal{P}^{0}_{1,t}}[\alpha_{2,t}(\mathbf{x}^{0}_{1,t}, \mathbf{x}_2)]\ .
\end{array}
\label{eq:level_1_attacker}
\end{equation}
In the general case, if $\mathcal{A}$ thinks that $\mathcal{D}$ reasons at level $k-1\geq 2$,
then $\mathcal{A}$ can reason at level $k\geq 3$ and
select its level-$k$ action $\mathbf{x}^{k}_{1,t}$~\eqref{eq:determ_best_response} that best-responds to level-$(k-1)$ action $\mathbf{x}^{k-1}_{2,t}$~\eqref{eq:determ_best_response_2} selected by $\mathcal{D}$:\vspace{-0.5mm}
\begin{equation}
\begin{array}{c}
\mathbf{x}^{k}_{1,t}\triangleq\mathop{\arg\max}_{\mathbf{x}_1\in \mathcal{X}_1} \alpha_{1,t}(\mathbf{x}_1, \mathbf{x}^{k-1}_{2,t})\ ,
\end{array}
\label{eq:determ_best_response}
\end{equation}
\begin{equation}
\begin{array}{c}
\mathbf{x}^{k-1}_{2,t}\triangleq\mathop{\arg\max}_{\mathbf{x}_2\in \mathcal{X}_2} \alpha_{2,t}(\mathbf{x}^{k-2}_{1,t}, \mathbf{x}_2)\ .
\end{array}
\label{eq:determ_best_response_2}
\end{equation}
Since
$\mathcal{A}$ thinks that
$\mathcal{D}$'s level-$(k-1)$ action $\mathbf{x}^{k-1}_{2,t}$~\eqref{eq:determ_best_response_2}
is derived using the same recursive reasoning process, $\mathbf{x}^{k-1}_{2,t}$ best-responds to level-$(k-2)$ action $\mathbf{x}^{k-2}_{1,t}$ selected by $\mathcal{A}$,
the latter of which in turn best-responds to level-$(k-3)$ action $\mathbf{x}^{k-3}_{2,t}$ selected by $\mathcal{D}$ and can be computed in the same way as~\eqref{eq:determ_best_response}.
This recursive reasoning process continues until it reaches the base case of the level-$1$ action selected by either
(a) $\mathcal{A}$~\eqref{eq:level_1_defender}
if $k$ is odd (in this case, recall from Section~\ref{subsec:level_1_reasoning} that $\mathcal{A}$ requires knowledge of $\mathcal{D}$'s level-$0$ strategy $\mathcal{P}^{0}_{2,t}$ to compute~\eqref{eq:level_1_defender}),
or (b) $\mathcal{D}$~\eqref{eq:level_1_attacker} if $k$ is even.
Note that $\mathcal{A}$ has to perform the computations made by $\mathcal{D}$ to derive $\mathbf{x}^{k-1}_{2,t}$~\eqref{eq:determ_best_response_2} as well as the computations to best-respond to $\mathbf{x}^{k-1}_{2,t}$ via~\eqref{eq:determ_best_response}.
Our next main result (see its proof in Appendix~\ref{app:proof}) bounds the regret of $\mathcal{A}$ when using R2-B2 for level-$k\geq 2$ reasoning:
\begin{theorem}
\label{theorem_level_k}
Let $\delta\in (0, 1)$.
Suppose that
$\mathcal{A}$ and $\mathcal{D}$ use R2-B2 (Algorithm~\ref{BORR_attacker})
for level-$k\geq 2$ and level-$(k-1)$ reasoning, respectively.
Then, with probability of at least $1 - \delta$,
$R_{1,T} \leq \sqrt{C_1 T \beta_T \gamma_T}$.\vspace{-1mm}
\end{theorem}
Theorem~\ref{theorem_level_k} reveals that $R_{1,T}$ grows sublinearly in $T$.\cref{gammat}
So, $\mathcal{A}$ using R2-B2 for level-$k\geq 2$ reasoning achieves asymptotic no regret
regardless of $\mathcal{D}$'s level-$0$ strategy $\mathcal{P}^{0}_{2,t}$.
By comparing Theorems~\ref{regret_gp} and~\ref{theorem_level_k}, we can observe that if $\mathcal{A}$ uses GP-MW as its
level-$0$ strategy, then
it can achieve faster asymptotic convergence to no regret by using R2-B2 to reason at level $k\geq 2$ and one level higher than $\mathcal{D}$.
However, when $\mathcal{A}$ reasons at a higher level $k$,
its computational cost grows due to an additional optimization of the GP-UCB acquisition function per increase in level of reasoning.
So, $\mathcal{A}$ is expected to favor reasoning at a lower level, which agrees with the observation in the work of~\citet{camerer2004cognitive} on the cognitive hierarchy model that humans usually reason at a level no higher than $2$.
\subsection{R2-B2-Lite}
\label{lite}
We also propose a computationally cheaper variant of R2-B2 for level-$1$ reasoning called R2-B2-Lite at the expense of a weaker convergence guarantee.
When using R2-B2-Lite for level-$1$ reasoning, instead of following~\eqref{eq:level_1_defender},
$\mathcal{A}$ selects its level-$1$ action $\mathbf{x}^{1}_{1,t}$ by sampling $\widetilde{\mathbf{x}}^{0}_{2,t}$ from level-$0$ strategy $\mathcal{P}^{0}_{2,t}$ of $\mathcal{D}$ and best-responding to this sampled action:
\begin{equation}
\begin{array}{c}
\mathbf{x}^{1}_{1,t}\triangleq\mathop{\arg\max}_{\mathbf{x}_1 \in \mathcal{X}_1}\alpha_{1,t}(\mathbf{x}_1,\widetilde{\mathbf{x}}^{0}_{2,t})\ .
\end{array}
\label{eq:borr_light_level_1}
\end{equation}
Our final main result (its proof is in Appendix~\ref{app:proof_borr_lite}) bounds the expected regret of $\mathcal{A}$ using R2-B2-Lite for level-$1$ reasoning:
\begin{theorem}
\label{theorem_borr_lite}
Let $\delta\in (0, 1)$.
Suppose that $\mathcal{A}$ uses R2-B2-Lite for level-$1$ reasoning and $\mathcal{D}$ uses mixed strategy $\mathcal{P}^{0}_{2,t}$ for level-$0$ reasoning.
If the trace of the covariance matrix of $\mathbf{x}^{0}_{2,t} \sim\mathcal{P}^{0}_{2,t}$
is not more than $\omega_t$ for $t= 1,\ldots,T$, then with probability of at least $1-\delta$,
$
\mathbb{E}[R_{1,T}] = \mathcal{O}(\sum^T_{t=1}\sqrt{\omega_t} + \sqrt{T \beta_T \gamma_T})
$
where the expectation is with respect to the history of actions selected and payoffs observed by $\mathcal{D}$ as well as $\widetilde{\mathbf{x}}^{0}_{2,t}$ for $t=1,\ldots,T$.
\end{theorem}
From Theorem~\ref{theorem_borr_lite}, the expected regret bound tightens
if $\mathcal{D}$'s level-$0$ mixed strategy $\mathcal{P}^{0}_{2,t}$ has a smaller variance for each dimension of input action $\mathbf{x}^{0}_{2,t}$.
As a result, the level-$0$ action $\widetilde{\mathbf{x}}^{0}_{2,t}$ of $\mathcal{D}$ that is sampled by $\mathcal{A}$
tends to be closer to the true level-$0$ action $\mathbf{x}^{0}_{2,t}$ selected by $\mathcal{D}$.
Then, $\mathcal{A}$ can select level-$1$ action $\mathbf{x}^{1}_{1,t}$ that best-responds to a more precise estimate $\widetilde{\mathbf{x}}^{0}_{2,t}$ of the level-$0$ action $\mathbf{x}^{0}_{2,t}$ selected by $\mathcal{D}$, hence improving its expected payoff.
Theorem~\ref{theorem_borr_lite} also reveals that $\mathcal{A}$ using R2-B2-Lite for level-$1$ reasoning achieves asymptotic no expected regret
if the sequence $(\omega_t)_{t\in\mathbb{Z}^+}$ uniformly decreases to $0$
(i.e., $\omega_{t+1} < \omega_{t}$ for $t\in\mathbb{Z}^+$ and $\lim_{T \rightarrow \infty} \omega_{T} = 0$).
Interestingly, such a sufficient condition for achieving asymptotic no expected regret has a natural and elegant interpretation in terms of the exploration-exploitation trade-off:
This condition is satisfied if $\mathcal{D}$ uses a level-$0$ mixed strategy $\mathcal{P}^{0}_{2,t}$ with a decreasing variance for each dimension of input action $\mathbf{x}^{0}_{2,t}$, which
corresponds to transitioning from exploration (i.e.,
a large variance results in a diffused $\mathcal{P}^{0}_{2,t}$
and hence many actions being sampled)
to exploitation (i.e., a small variance results in a peaked $\mathcal{P}^{0}_{2,t}$ and hence fewer actions being sampled).
\begin{figure}[t]
\centering
\begin{tabular}{cc}
\hspace{-3mm}
\includegraphics[width=0.49\linewidth]{figures_new/player_1_common_payoff.pdf} & \hspace{-5mm} \includegraphics[width=0.49\linewidth]{figures_new/attack_score_mnist_random.pdf} \\
\hspace{-3mm} (a) common-payoff games & \hspace{-5mm} (d) random search \\
\hspace{-3mm}
\includegraphics[width=0.49\linewidth]{figures_new/player_1_general_sum.pdf} & \hspace{-5mm} \includegraphics[width=0.49\linewidth]{figures_new/attack_score_mnist_gp_mw.pdf}\\
\hspace{-3mm} (b) general-sum games & \hspace{-5mm} (e) GP-MW\\
\hspace{-3mm}
\includegraphics[width=0.49\linewidth]{figures_new/player_1_zero_sum.pdf} & \hspace{-3mm} \includegraphics[width=0.49\linewidth]{figures_new/attack_score_cifar.pdf}\\
\hspace{-3mm} (c) constant-sum games & \hspace{-3mm} (f)
\end{tabular}\vspace{-1mm}
\caption{(a-c) Mean regret of agent $1$ in synthetic games where the legend in (a) represents the levels of reasoning of agents $1$ vs.~$2$. Attack score of $\mathcal{A}$ in adversarial ML for (d-e) MNIST and (f) CIFAR-10 datasets where the legend in (d) represents the levels of reasoning of $\mathcal{A}$ vs. $\mathcal{D}$.}
\label{fig:player_1}\vspace{-1.9mm}
\end{figure}
\section{Experiments and Discussion}
\label{expt}
This section empirically evaluates the performance of our R2-B2 algorithm and demonstrates its generality using synthetic games, adversarial ML, and MARL.
Some of our experimental comparisons can be interpreted as comparisons with existing baselines used as level-$0$ strategies (Section~\ref{subsec:level-0}).
Specifically, we can compare the performance of our level-$1$ agent with that of a baseline method when they are against the same level-$0$ agent.
Moreover, in constant-sum games, we can perform a more direct comparison by playing our level-$1$ agent against an opponent using a baseline method as a level-$0$ strategy (Section~\ref{exp:comparison_parsimonious}).
Additional experimental details and results are reported in Appendix~\ref{app:experiment} due to lack of space.
All error bars represent standard error.
\subsection{Synthetic Games}
\label{subsec:synth_func}
Firstly, we empirically evaluate the performance of R2-B2 using synthetic games with two agents whose payoff functions are sampled from GP over a discrete input domain.
Both agents use GP-MW and R2-B2/R2-B2-Lite for level-$0$ and level-$k\geq1$ reasoning, respectively.
We consider $3$ types of games: common-payoff, general-sum, and constant-sum games.
Figs.~\ref{fig:player_1}a to~\ref{fig:player_1}c show results of the mean regret\footnote{The mean regret $T^{-1}\sum^T_{t=1}(\max_{\mathbf{x}_{1}\in\mathcal{X}_1, \mathbf{x}_{2}\in\mathcal{X}_2} f_1(\mathbf{x}_{1}, \mathbf{x}_{2}) -f_1(\mathbf{x}_{1,t}, \mathbf{x}_{2,t}))$ of agent $1$ pessimistically estimates (i.e., upper bounds) $R_{1,T}/T$~\eqref{extregret} and is thus not expected to converge to $0$. Nevertheless, it serves as an appropriate performance metric here.
}
of agent $1$ averaged over $10$ random samples of GP and $5$ initializations of $1$ randomly selected action with observed payoff per sample:
In all types of games, when agent $1$ reasons at one level higher than agent $2$, it incurs a smaller mean regret than when
reasoning at level $0$ (blue curve), which demonstrates the performance advantage of recursive reasoning and corroborates our theoretical results (Theorems~\ref{theorem_level_1} and~\ref{theorem_level_k}).
The same can be observed for agent $1$ using R2-B2-Lite for level-$1$ reasoning (orange curve) but it does not perform as well as that using R2-B2 (red curve),
which again agrees with our theoretical result (Theorem~\ref{theorem_borr_lite}).
Moreover, comparing the red (orange) and blue curves shows that when against the same level-$0$ agent, our R2-B2 (R2-B2-Lite) level-$1$ agent outperforms the baseline method of GP-MW (as a level-$0$ strategy).
Figs.~\ref{fig:player_1}a and~\ref{fig:player_1}c also reveal the effect of incorrect thinking of the level of reasoning of the other agent on its performance:
Since agent $2$ uses recursive reasoning at level $1$ or more, agent $2$ thinks that it is reasoning at one level higher than agent $1$. However, it is in fact reasoning at one level lower in these two figures.
In common-payoff games, since agents $1$ and $2$ have identical payoff functions, the mean regret of agent $2$ is the same as that of agent $1$ in Fig.~\ref{fig:player_1}a.
So, from agent $2$'s perspective, it benefits from such an incorrect thinking in common-payoff games.
In constant-sum games, since the payoff function of agent $2$ is negated from that of agent $1$,
the mean regret of agent $2$ increases with
a decreasing mean regret of agent $1$ in Fig.~\ref{fig:player_1}c.
So, from agent $2$'s viewpoint, it hurts from such an incorrect thinking in constant-sum games.
Further experimental results on such incorrect thinking are reported
in Appendix~\ref{app:exp_synth_two_player}b.
An intriguing observation from Figs.~\ref{fig:player_1}a to~\ref{fig:player_1}c is that
when agent $1$ reasons at level $k\geq 2$,
it incurs a smaller mean regret than when reasoning at level $1$.
A possible explanation is
that when agent $1$ reasons at level $k\geq 2$,
its selected level-$k$ action~\eqref{eq:determ_best_response} best-responds to the actual level-$(k-1)$ action~\eqref{eq:determ_best_response_2} selected by agent $2$.
In contrast, when agent $1$ reasons at level $1$, its selected level-$1$ action~\eqref{eq:level_1_defender} maximizes the \emph{expected} value of GP-UCB w.r.t.~agent $2$'s level-$0$ \emph{mixed} strategy rather than the actual level-$0$ action selected by agent $2$.
However, as we shall see in the experiments on adversarial ML in Section~\ref{subsec:adv_ml},
when the expectation in level-$1$ reasoning~\eqref{eq:level_1_defender} needs to be approximated via sampling but insufficient samples are used,
the performance of level-$k\geq 2$ reasoning can be potentially diminished due to propagation of the approximation error from level $1$.
Moreover, Fig.~\ref{fig:player_1}c shows another interesting observation that is unique for constant-sum games: Agent $1$ achieves a significantly better performance when reasoning at level $3$ (i.e., agent $2$ reasons at level $2$) than at level $2$ (i.e., agent $2$ reasons at level $1$).
This can be explained by the fact that when agent $2$ reasons at level $2$, it best-responds to the level-$1$ action of agent $1$, which is most likely different from the actual action selected by agent $1$ since agent $1$ is in fact reasoning at level $3$. In contrast, when agent $2$ reasons at level $1$, instead of best-responding to a single (most likely wrong) action of agent $1$, it best-responds to the expected behavior of agent $1$ by attributing a distribution over all actions of agent $1$.
As a result,
agent $2$ suffers from a smaller performance deficit when reasoning at level $1$ (i.e., agent $1$ reasons at level $2$) compared with reasoning at level $2$ (i.e., agent $1$ reasons at level $3$) or higher. Therefore, agent $1$ obtains a more dramatic performance advantage when reasoning at level $3$ (gray curve) due to the constant-sum nature of the game.
A deeper implication of this insight is that although level-$1$ reasoning may not yield a better performance than level-$k\geq2$ reasoning as analyzed in the previous paragraph, it is more robust against incorrect estimates of the opponent's level of reasoning in constant-sum games.
Experimental results on the use of random search and EXP3 (Section~\ref{subsec:level-0}) for level-$0$ reasoning (instead of GP-MW) are reported in Appendix~\ref{app:exp_synth_two_player}c; the resulting
observations and insights are consistent with those presented here.
This demonstrates the robustness of R2-B2 and corroborates the generality of our theoretical results (Theorems~\ref{theorem_level_1} and~\ref{theorem_level_k}) which hold for any level-$0$ strategy of the other agent.
We have also performed experiments using synthetic games involving \emph{more than two} agents (Appendix~\ref{app:exp_multi_player_games}),
which yield some interesting observations that are consistent with our theoretical analysis.
\subsection{Adversarial Machine Learning (ML)}
\subsubsection{R2-B2 for Adversarial ML}
\label{subsec:adv_ml}
We apply our R2-B2 algorithm to black-box adversarial ML for image classification problems with \emph{deep neural networks} (DNNs) using the MNIST and CIFAR-$10$ image datasets.
We consider \emph{evasion attacks}: The attacker $\mathcal{A}$ perturbs a test image to fool a fully trained DNN (referred to as the \emph{target ML model} hereafter) into misclassifying the image,
while the defender $\mathcal{D}$ transforms the perturbed image with the goal of ensuring the correct prediction by the classifier.
To improve query efficiency, dimensionality reduction techniques such as autoencoders have been commonly used for black-box adversarial attacks~\cite{tu2019autozoom}.
In our experiments, \emph{variational autoencoders} (VAE)~\cite{kingma2013auto} are used by both $\mathcal{A}$ and $\mathcal{D}$ to project the images to a lower-dimensional space (i.e., $2$D for MNIST and $8$D for CIFAR-$10$).\footnote{We have detailed in Appendix~\ref{app:adv_ml}a how VAE can be realistically incorporated into our algorithm.}
Following a common practice in adversarial ML, we focus on perturbations with bounded infinity norm as actions of $\mathcal{A}$ and $\mathcal{D}$:
The maximum allowed perturbation to each pixel added by either $\mathcal{A}$ or $\mathcal{D}$ is no more than a pre-defined value $\epsilon$ where $\epsilon=0.2$ for MNIST and $\epsilon=0.05$ for CIFAR-$10$.
We consider \emph{untargeted attacks} whereby the goal of $\mathcal{A}$ ($\mathcal{D}$) is to cause (prevent) misclassification by the target ML model.
So, the payoff function of $\mathcal{A}$ is the maximum predictive probability among all incorrect classes (referred to as \emph{attack score} hereafter) and its negation is the payoff function of $\mathcal{D}$.
As a result, the application of R2-B2 to black-box adversarial ML represents a \emph{constant-sum game}.
An attack is considered \emph{successful} if the attack score is larger than the predictive probability of the correct class, hence resulting in misclassification of the test image.
Both $\mathcal{A}$ and $\mathcal{D}$ use GP-MW/random search\footnote{For CIFAR-$10$ dataset, $\mathcal{A}$ uses only random search for level-$0$ reasoning due to high dimensions, as explained in Appendix~\ref{app:adv_ml}a.} and R2-B2/R2-B2-Lite for level-$0$ and level-$k\geq1$ reasoning, respectively.
Figs.~\ref{fig:player_1}d to~\ref{fig:player_1}f show results of
the attack score of $\mathcal{A}$ in adversarial ML
for both image datasets while Table~\ref{table:mnist_cifar} shows results of the number of successful attacks by $\mathcal{A}$ over $150$ iterations of the game; the results are averaged over $10$ initializations of $5$ randomly selected actions with observed payoffs.\footnote{The results here use a test image from each dataset that can clearly illustrate the effects of both attack and defense.
Refer to Appendix~\ref{app:adv_ml}b for more details and results using more test images; the observations are consistent with those presented here.}
It can be observed from Figs.~\ref{fig:player_1}d to~\ref{fig:player_1}f that
when $\mathcal{A}$ reasons at one level higher than $\mathcal{D}$ (orange, red, and gray curves), its attack score is higher than when
reasoning at level $0$ (blue, green, and purple curves).
Similarly, when $\mathcal{D}$ reasons at one level higher (green, purple, and yellow curves), the attack score of $\mathcal{A}$ is reduced.
These observations demonstrate the performance advantage of using recursive reasoning in adversarial ML.
Such an advantage of recursive reasoning can also be seen from Table~\ref{table:mnist_cifar}:
For MNIST, when random search is used for level-$0$ reasoning and $\mathcal{A}$ reasons at one level higher than $\mathcal{D}$, it achieves a larger number of successful attacks ($12.8$, $10.2$, and $3.0$) than when reasoning at level $0$ ($2.6$, $0.8$, and $1.8$).
Similarly, when $\mathcal{D}$ reasons at one level higher, it reduces the number of successful attacks by $\mathcal{A}$ ($0.8$, $1.8$, and $0.9$)
than when reasoning at level $0$ ($2.6$, $12.8$, and $10.2$).
The observations are similar for MNIST with GP-MW for level-$0$ reasoning as well as for CIFAR-$10$ (Table~\ref{table:mnist_cifar}).
The performance advantage of $\mathcal{A}$ reasoning at level $2$ is observed to be smaller than that at level $1$; this may be explained by the propagation of error of approximating the expectation in level-$1$ reasoning~\eqref{eq:level_1_defender}, as explained previously in Section~\ref{subsec:synth_func}.
We investigate and report the effect of the number of samples for such an approximation in Appendix~\ref{app:adv_ml}c, which reveals that the performance improves with more samples, albeit with higher computational cost.
Moreover, some insights can also be drawn regarding the consequence of an incorrect thinking about the opponent's level of reasoning in constant-sum games.
For example, for the gray curves in Figs.~\ref{fig:player_1}d to~\ref{fig:player_1}f, $\mathcal{D}$ reasons at level $1$ because it thinks that $\mathcal{A}$ reasons at level $0$.
However, $\mathcal{A}$ is in fact reasoning at level $2$. As a result, in this constant-sum game, $\mathcal{D}$'s incorrect thinking about the opponent's level of reasoning negatively impacts
$\mathcal{D}$'s performance since the attack scores are increased.
This is consistent with the corresponding analysis in synthetic games regarding the effect of incorrect thinking about the level of reasoning of the other agent (Section~\ref{subsec:synth_func}).
\begin{table}
\vspace{-2mm}
\scriptsize
\caption{Average number of successful attacks by $\mathcal{A}$ over $150$ iterations in adversarial ML for MNIST and CIFAR-$10$ datasets where
the levels of reasoning are in the form of $\mathcal{A}$ vs. $\mathcal{D}$.}
\centering
\begin{tabular}{c|ccc}
\hline
Levels of reasoning & MNIST (random)\hspace{-1mm} & MNIST (GP-MW)\hspace{-1mm} & CIFAR-$10$\\
\hline
$0$ vs. $0$ & $2.6$ & $4.3$ & $70.1$ \\
$1$ vs. $0$ & $12.8$ & $6.0$ & $113.1$ \\
$1$ vs. $0$ (R2-B2-Lite) & $10.2$ & $6.8$ & $99.7$ \\
$0$ vs. $1$ & $0.8$ & $0.4$ & $25.2$ \\
$0$ vs. $1$ (R2-B2-Lite) & $1.8$ & $1.0$ & $29.7$ \\
$2$ vs. $1$ & $3.0$ & $5.2$ & $70.9$ \\
$1$ vs. $2$ & $0.9$ & $0.4$ & $54.0$\\
\hline
\end{tabular}
\label{table:mnist_cifar}\vspace{-4mm}
\end{table}
\subsubsection{Comparison with State-of-the-art Adversarial Attack Methods}
\label{exp:comparison_parsimonious}
It was mentioned in Section~\ref{subsec:recursive_reasoning_model} that
our theoretical results hold for
\emph{any}
level-$0$ strategy of the other agent.
So, any existing adversarial attack (defense) method can be used the level-$0$ strategy of $\mathcal{A}$ ($\mathcal{D}$).
In this experiment, we perform a direct comparison of R2-B2 with the state-of-the-art black-box adversarial attack method called \emph{Parsimonious}~\cite{moon2019parsimonious}: We use Parsimonious as the level-$0$ strategy of $\mathcal{A}$ and let $\mathcal{D}$ use R2-B2 for level-$1$ reasoning.
We consider a realistic setting where in each iteration,
$\mathcal{D}$ only needs to receive the image perturbed by $\mathcal{A}$ and choose its action that best-responds to this perturbed image.
In this manner, $\mathcal{D}$ naturally has access to the history of actions selected by $\mathcal{A}$ (as required by \emph{perfect monitoring} in our repeated game) since it receives all images perturbed by $\mathcal{A}$.
Additional details of the experimental setting are reported in Appendix~\ref{app:adv_ml_parsimonious}a.
We randomly select $70$ images from the CIFAR-$10$ dataset that are successfully attacked by Parsimonious using $\epsilon=0.05$ over $500$ iterations without the defender $\mathcal{D}$.\footnote{Compared to the work of~\citet{moon2019parsimonious}, we use fewer iterations and a larger $\epsilon$,
which we think is more realistic as attacks with an excessively large no.~of queries may be easily detected.}
Our level-$1$ R2-B2 defender manages to \emph{completely prevent any successful attacks} for $53$ of these images and requires Parsimonious to use \emph{more than $3.5$ times} more queries on average to succeed for $10$ other images.\footnote{The remaining $7$ images are so easy to attack such that the attacks are already successful during the initial exploration phase of our level-$1$ R2-B2 defender.}
Fig.~\ref{fig:parsimonious_some} shows results of the loss incurred by Parsimonious (i.e., its original attack objective) with and without our level-$1$ R2-B2 defender for $4$ of the successfully defended images; results for other images are shown in Appendix~\ref{app:adv_ml_parsimonious}a.
This experiment not only demonstrates the generality of our R2-B2 algorithm, but can also be of significant independent interest to the adversarial ML community
as a defense method
against black-box adversarial attacks.
In addition, as another comparison, we use the same experimental setting with the CIFAR-10 dataset in Section~\ref{subsec:adv_ml} and play Parsimonious against a level-$0$ defender using random search.
The results show that when against the same level-$0$ defender, Parsimonious achieves a significantly smaller average number of successful attacks (27.6) compared with our level-$1$ attacker (113.1, as shown in Table~\ref{table:mnist_cifar}). In other words, our level-$1$ defender can defend effectively against Parsimonious, while our level-$1$ attacker can attack better than Parsimonious.
Note that the unsatisfactory performances of Parsimonious in our experiments might be largely explained the fact that it does not consider the presence of a defender.
Moreover, our level-$1$ R2-B2 defender can also defend against black-box adversarial attacks from standard BO algorithms (Appendix~\ref{app:adv_ml_parsimonious}b)\footnote{The BO attacker here only takes its perturbations as inputs and thus does not consider the defender.}, which have become popular recently~\cite{ru2020bayesopt}.
\begin{figure}
\centering
\includegraphics[width=1\linewidth]{figures_new/losses_some.pdf}\vspace{-2.6mm}
\caption{Loss incurred by Parsimonious with and without our level-$1$ R2-B2 defender on $4$ randomly selected images that are successfully attacked by Parsimonious.}
\label{fig:parsimonious_some}\vspace{-4mm}
\end{figure}
\subsection{Multi-Agent Reinforcement Learning (MARL)}
We apply R2-B2 to policy search for MARL with \emph{more than two} agents.
Each action of an agent represents a particular set of policy parameters controlling the behavior of the agent in an environment.
The payoff to each agent corresponding to a selected set of its policy parameters (i.e., action) is its mean return (i.e., cumulative reward) from the execution of all the agents' selected policies across $5$ independent episodes.
Since the agents interact in the environment, the payoff function of each agent depends on the policies (actions) selected by all agents.
We use the predator-prey game from the widely used multi-agent particle environment in~\cite{lowe2017multi}.
This $3$-agent game (see Fig.~\ref{fig:simple_tag_illu} in Appendix~\ref{app:marl}) contains two predators who are trying to catch a prey.
The prey is rewarded for being far from the predators and penalized for stepping outside the boundary.
The two predators have identical payoff functions and are rewarded for being close to the prey (if the prey stays within the boundary).
So, the predator-prey game represents a \emph{general-sum game}.
All agents use random search\footnote{All agents use only random search for level-$0$ reasoning due to high dimensions, as explained in Appendix~\ref{app:marl}.} and R2-B2 for level-$0$ and level-$k\geq 1$ reasoning, respectively.
Fig.~\ref{fig:simple_tag} shows results of the (scaled) mean return of the agents averaged over $10$ initializations of $5$ randomly selected actions with observed payoffs.
It can be observed from Fig.~\ref{fig:simple_tag}b that
when the prey reasons at level $1$ and both predators reason at level $0$ (orange curve), its mean return is much higher than when reasoning at level $0$ (blue curve);
this results from the prey's ability to learn to stay within the boundary.
Specifically, there exist some ``dominated actions'' in this game, namely, those causing the prey to step beyond the boundary.
Regardless of the predators' policies, such dominated actions never give large returns to the prey and are thus likely to yield small values of GP-UCB for any actions (policies) selected by the predators.
So, by reasoning at level $1$ (i.e., by maximizing the expected value of GP-UCB), the prey is able to eliminate those dominated actions and thus learn to stay within the boundary.
From Fig.~\ref{fig:simple_tag}a, the mean return of the predators is also improved (orange curve)
because the prey's ability to stay within the boundary allows the predators to improve their rewards by being close to the prey despite using random search for level-$0$ reasoning.
In contrast, when the prey reasons at level $0$, the predators rarely get rewarded (blue curve) since the prey repeatedly steps beyond the boundary.
On the other hand, when predator $1$ reasons at level $2$ (purple curve), the mean return of the predators is further increased since predator $1$ is now able to
learn to actively move close to the prey instead of moving around using random search for level-$0$ reasoning (orange curve).
When both predators reason at level $2$ (green curve), their mean return is improved even further.
In both of these scenarios, the mean return of the prey stays close to that associated with the orange curve: Although the predators are able to actively approach the prey,
this also further helps to prevent the prey from moving beyond the boundary, which compensates for the loss in its mean return due to the more strategic predators.
\begin{figure}
\centering
\begin{tabular}{cc}
\hspace{-3mm}\includegraphics[width=0.5\linewidth]{figures_new/simple_tag_return_player_1_adv.pdf} & \hspace{-3mm}\includegraphics[width=0.5\linewidth]{figures_new/simple_tag_return_player_3_agent.pdf}\\
\hspace{-3mm} (a) {predators} & \hspace{-3mm}(b) {prey}
\end{tabular}
\caption{Mean return of predators and prey in predator-prey game where the legend in (b) represents the levels of reasoning of predator $1$ vs. predator $2$ vs. prey.}
\label{fig:simple_tag}\vspace{-3mm}
\end{figure}
\section{Related Work}
The recent work of~\citet{sessa2019no} combines online learning and GP-UCB to derive a no-regret learning algorithm called \emph{GP-multiplicative weight} (GP-MW) for repeated games.
As explained in Section~\ref{subsec:level-0}, GP-MW can be used as a level-$0$ mixed strategy (i.e., no recursive reasoning) in our R2-B2 algorithm.
Moreover, BO has also been recently applied in game theory to find the Nash equilibria~\cite{picheny2019bayesian}.
Humans possess the ability to reason about the mental states of others~\cite{goldman2012theory}.
In particular, a person tends to reason recursively by analyzing the others' thinking about himself, which gives rise to recursive reasoning~\cite{pynadath2005psychsim}.
The recursive reasoning model of humans has inspired the development of the cognitive hierarchy model in behavioral game theory,
which uses recursive reasoning to explain the behavior of players in games~\cite{camerer2004cognitive}.
Moreover, the improved decision-making capability offered by recursive reasoning has motivated its application in ML and sequential decision-making problems such as
interactive partially observable Markov decisionn processes~\cite{gmytrasiewicz2005framework,hoang2013interactive}, MARL~\cite{wen2019probabilistic}, among others.
\emph{Deep neural networks} (DNNs) have recently been found to be vulnerable to carefully crafted adversarial examples~\cite{szegedy2013intriguing}.
Since then, a variety of adversarial attack methods have been developed to exploit this vulnerability of DNNs~\cite{goodfellow2014explaining}.
However, most of the existing attack methods are \emph{white-box} attacks since they require access to the gradient of the ML model.
In contrast, the more realistic \emph{black-box attacks}~\cite{tu2019autozoom,moon2019parsimonious}, which we have adopted in our experiments,
only require query access to the target ML model and have been attracting significant attention recently.
Of note, BO has recently been used for black-box adversarial attacks (without considering defenses) and demonstrated promising query efficiency~\cite{ru2020bayesopt}.
On the other hand, many attempts have been made to design adversarial defense methods~\cite{madry2017towards,tramer2017ensemble} to make ML models robust against adversarial attacks.
In our experiments, we have adopted the input reconstruction/transformation technique~\cite{meng2017magnet,samangouei2018defense} as the defense mechanism,
in which the defender attempts to transform the perturbed input to ensure the correct prediction by the ML model.
Refer to the detailed survey of adversarial ML in~\cite{yuan2019adversarial}.
\section{Conclusion and Future Work}
This paper describes the first BO algorithm called R2-B2 that is endowed with the capability of recursive reasoning to model the reasoning process in the interactions between boundedly rational\cref{brat}, self-interested agents with unknown, complex, and expensive-to-evaluate payoff functions in repeated games.
We prove that by reasoning at level $k\geq 2$ and one level higher than the other agents, our R2-B2 agent can achieve faster asymptotic convergence to no regret than that without utilizing recursive reasoning.
We empirically demonstrate the competitive performance and generality of R2-B2 through extensive experiments using synthetic games, adversarial ML, and MARL.
For our future work, we plan to investigate the connection of R2-B2 to other game-theoretic solution concepts such as Nash equilibrium.
We will also explore the extension of R2-B2 to a more general setting where
a level-$k$ agent selects its best response to the action of the other agent who reasons according to a distribution (e.g., Poisson) over lower levels instead of only at level $k-1$, which is also captured by the cognitive hierarchy model~\cite{camerer2004cognitive}.
We will consider generalizing R2-B2 to nonmyopic BO~\citep{dmitrii20a,ling16},
batch BO~\citep{daxberger17}, high-dimensional BO~\citep{NghiaAAAI18}, differentially private BO~\citep{dmitrii20b}, and multi-fidelity BO~\citep{yehong17,ZhangUAI19} settings and incorporating early stopping~\citep{dai2019}.
For applications with a huge budget of function evaluations, we like to couple R2-B2 with the use of distributed/decentralized~\citep{LowUAI12,Chen13,LowRSS13,LowTASE15,HoangICML16,NghiaAAAI19,HoangICML19,low15,Ruofei18} or online/stochastic~\citep{NghiaICML16,MinhAAAI17,LowECML14a,LowAAAI14,teng20,Haibin19,HaibinAPP} sparse GP models to represent the belief of the unknown objective function efficiently.
\section*{Acknowledgements}
This research/project is supported in part by the Singapore National Research Foundation through the Singapore-MIT Alliance for Research and Technology (SMART) Centre for Future Urban Mobility (FM) and in part by A$^*$STAR under its RIE$2020$ Advanced Manufacturing and Engineering (AME) Industry Alignment Fund -- Pre Positioning (IAF-PP) (Award A$19$E$4$a$0101$).
Teck-Hua Ho acknowledges funding from the Singapore National Research Foundation's Returning Singaporean Scientists Scheme, grant NRFRSS$2014$-$001$.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,071 |
<!DOCTYPE html>
<HTML><head><TITLE>Manpage of BLKIOMON</TITLE>
<meta charset="utf-8">
<link rel="stylesheet" href="/css/main.css" type="text/css">
</head>
<body>
<header class="site-header">
<div class="wrap"> <div class="site-title"><a href="/manpages/index.html">linux manpages</a></div>
<div class="site-description">{"type":"documentation"}</div>
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</header>
<div class="page-content"><div class="wrap">
<H1>BLKIOMON</H1>
Section: (8)<BR>Updated: July 17, 2008<BR><A HREF="#index">Index</A>
<A HREF="/manpages/index.html">Return to Main Contents</A><HR>
<P>
<P>
<A NAME="lbAB"> </A>
<H2>NAME</H2>
blkiomon - monitor block device I/O based o blktrace data
<P>
<P>
<A NAME="lbAC"> </A>
<H2>SYNOPSIS</H2>
<B>blkiomon -I </B><I>interval</I> [ -h <I>file</I> ] [ -b <I>file</I> ]
[ -d <I>file</I> ] [ -D <I>file</I> ] [ -Q <I>path_name</I>
-q <I>msg_queue_id</I> -m <I>msg_id</I> ] [ -V ]
<BR>
<P>
<P>
<A NAME="lbAD"> </A>
<H2>DESCRIPTION</H2>
blkiomon is a block device I/O monitor. It periodically generates per-device
request size and request latency statistics from blktrace data. It provides
histograms as well as data that can be used to calculate min, max, average
and variance. For this purpose, it consumes D and C traces read from stdin.
Note, that this doesn't work for logical volumes, as high-level drivers
don't see the completion of the events (C).
<P>
There are options for binary output and human-readable output to files and
stdout. Output to a message queue is supported as well.
<P>
There is no need to use blkparse with blkiomon. blkiomon is capable of
consuming binary output written to stdout by blktrace.
<P>
<P>
<A NAME="lbAE"> </A>
<H2>OPTIONS</H2>
<P>
-I <I>interval</I>
<BR>
--interval=<I>interval</I>
<DL COMPACT><DT><DD>
Set sample interval
</DL>
<P>
-h <I>file</I>
<BR>
--human-readable=<I>file</I>
<DL COMPACT><DT><DD>
Human-readable output file. Use '-' for stdout.
</DL>
<P>
-b <I>file</I>
<BR>
--binary=<I>file</I>
<DL COMPACT><DT><DD>
Binary output file. Use '-' for stdout.
</DL>
<P>
-d <I>file</I>
<BR>
--dump-lldd=<I>file</I>
<DL COMPACT><DT><DD>
Output file for data emitted by low level device driver.
</DL>
<P>
-D <I>file</I>
<BR>
--debug=<I>file</I>
<DL COMPACT><DT><DD>
Output file for debugging data. Use '-' for stdout.
</DL>
<P>
-Q <I>path_name</I>
<BR>
--msg-queue=<I>path_name</I>
<DL COMPACT><DT><DD>
Sets <I>path_name</I> as path name for existing message queue to be used
for binary output.
</DL>
<P>
-q <I>msg_queue_id</I>
<BR>
--msg-queue-id=<I>msg_queue_id</I>
<DL COMPACT><DT><DD>
Sets <I>msg_queue_id</I> as ID for an existing message queue to be used
for binary output.
</DL>
<P>
-m <I>msg_id</I>
<BR>
--msg-id=<I>msg_id</I>
<DL COMPACT><DT><DD>
Sets <I>msg_id</I> as message identifier to be used for binary output
messages written to an existing message queue.
</DL>
<P>
-V
<BR>
--version
<DL COMPACT><DT><DD>
Print program version.
</DL>
<P>
<P>
<A NAME="lbAF"> </A>
<H2>EXAMPLES</H2>
To get I/O statistics for /dev/sdw every 10 seconds for a period of one hour,
use the following command:
<P>
<BR> % blktrace /dev/sdw -a issue -a complete -w 3600 -o - | blkiomon -I 10 -h -
<P>
<P>
<A NAME="lbAG"> </A>
<H2>AUTHORS</H2>
blkiomon and this man page were written by Martin Peschke.
<P>
<P>
<A NAME="lbAH"> </A>
<H2>REPORTING BUGS</H2>
Report bugs to <<A HREF="mailto:linux-btrace@vger.kernel.org">linux-btrace@vger.kernel.org</A>>
<P>
<P>
<A NAME="lbAI"> </A>
<H2>COPYRIGHT</H2>
Copyright © 2008 IBM Corp.
<BR>
This is free software. You may redistribute copies of it under the terms of
the GNU General Public License <<A HREF="http://www.gnu.org/licenses/gpl.html">http://www.gnu.org/licenses/gpl.html</A>>.
There is NO WARRANTY, to the extent permitted by law.
<P>
<P>
<A NAME="lbAJ"> </A>
<H2>SEE ALSO</H2>
btrace (8), blktrace (8), blkparse (1), verify_blkparse (1), blkrawverify (1),
btt (1)
<P>
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Q: slider bar jquery ui change background for an image I manage to change the background of the progresse bar for an image with an gradient, but the image moves with the handler.
Here its high, so we can see the gradien
But when it's low the image desend with, there's no gradient anymore..
how can i see the darkest part of the gradient as move up??
thanks
A: If you want the gradient to recalculate based off of position so it is obvious that it is still a gradient, then you can do this:
.ui-slider .ui-slider-range {
background-position: inherit;
}
Example of a re-calculated gradient: http://jsfiddle.net/yv6Rt/365/
If you are wanting the gradient to not move at all while moving the slider, you can make it have a fixed position using this:
.ui-slider .ui-slider-range {
background-position: 0 100%;
}
Example of a fixed gradient: http://jsfiddle.net/yv6Rt/366/
| {
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{"url":"https:\/\/math.stackexchange.com\/questions\/1919985\/mistake-in-basic-algebra-i-think","text":"# Mistake in basic algebra, I think?\n\nProblem prove $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2$\n\nInduction proof: base case $n=1$ assume true for all $n$ prove for $n+1$.\n\nThe $n$th or last term becomes $(4(n+1)-1)=4n+3$.\n\nWe also sub $n+1$ in for all $n$ the $n-1$ term is $(4n-1)$ and the first term is $2(n+1)+1=2n+3$\n\nThe right side is $3(n+1)^2 = 3(n^2 + 2n +1 )$\n\nNext thing is I appear to be missing the first term in the sum $(2n+1)$ on the left.\nAdding $(2n+1)$ to both sides and subtracting $4n+3$ from both sides we get the $n$ case that equals $3n^2$ on the left and\n\n$$3n^2 + 6n +3 + 2n+1 -4n -3= 3n^2 + 4n+1$$\n\nWhich leaves me with $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2 +4n +1$\n\nWhich is approximately $4n+1$ on the right bigger than what I started with and is exactly the same on the left algebraically I must of done something wrong.\n\nWhich leads me to my question, what was it?\n\n$\\sum_\\limits{k=1}^{n} (2(n+k)-1) = 3n^2$\n\nBase case: $n = 1$\n\n$(2(1+1) - 1) = 3$\n\nSuppose:\n\n$\\sum_\\limits{k=1}^{n} (2(n+k)-1) = 3n^2$\n\nWe will show that $\\sum_\\limits{k=1}^{n+1} (2((n+1)+k)-1) = 3(n+1)^2$ based on the inductive hypothesis\n\n$\\sum_\\limits{k=1}^{n+1} (2((n+1)+k)-1)\\\\ \\sum_\\limits{k=1}^{n} (2(n+1)+k)-1) + 4n+3\\\\ \\big(\\sum_\\limits{k=1}^{n} (2(n+k)-1)\\big)+\\big(\\sum_\\limits{k=1}^{n}2\\big)+ 4n+3\\\\ 3n^2 + 2n + 4n+3\\\\ 3(n+1)^2$\n\nQED\n\nThe left hand side is the sum of the odd numbers from $2n+1$ up to $4n-1$. Thus when going $n\\to n+1$, the first summand $2n+1$ is dropped, and two summands $4n+1$ and $4n+3$ are appended a the end.\n\n\u2022 i know what generates the 4n+3 term its form the term n which term generates 4n+1 as the n-th term would be the term 4n-1 \u2013\u00a0Faust Sep 9 '16 at 2:56\n\u2022 @Faust7: You should think of it as a range from $2n+1$ to $4n-1$, not that every term gets increased. When you substitute $n \\to n+1$ the range becomes $2(n+1)+1$ to $4(n+1)-1$ or $2n+3$ to $4n+3$. That has one more number in it. \u2013\u00a0Ross Millikan Sep 9 '16 at 2:59\n\u2022 omg you are genius it makes perfect sense now thank you so much! \u2013\u00a0Faust Sep 9 '16 at 3:00\n\nYou can try this approach: Calculate sum of first $k$ odd numbers, namely from 1 to $2k-1$.\n\nDenoting it by $S_{k}$. By induction prove this is $k^2$.\n\nNow the desired sum is $S_{2n}-S_n= 4n^2-n^2=3n^2$.","date":"2020-03-31 23:59:12","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8969707489013672, \"perplexity\": 328.80518792894895}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-16\/segments\/1585370504930.16\/warc\/CC-MAIN-20200331212647-20200401002647-00211.warc.gz\"}"} | null | null |
Абловы (Обловы) — древний дворянский род.
Род внесён в VI часть дворянской родословной книги Московской и Рязанской губерний.
История рода
Происхождение или выезд предков неизвестен. Фёдор Матвеевич служил в детях боярских по Ряжску (1597).
В конце XVII столетия в числе помещиков Зарайского и Переяславль-Залесского уездов числились двадцать представителей рода. Яков, Леонтий и Андрей Григорьевичи владели поместьями в Рязанском уезде (1625). Андрей и Венедикт Степановичи вёрстаны новичными окладами по Рязани (1628), а Арист Иванович помещик в Ряжском уезде. Фёдор Степанович владел поместьями в Рязанском уезде (1628), его потомство внесено в родословную книгу Московской губернии. Рязанец Василий Григорьевич взят в плен татарами под Чудиновым (1660).
Четверо представителей рода владели населёнными имениями (1699).
Известные представители
Облов Фёдор Степанович — убит под Конотопом (1659).
Облов Иван Богданович — стряпчий (1692).
Облов Фёдор Богданович — стряпчий (1692).
Облов Иван Войнович — стольник.
Примечания
Литература | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,584 |
It's been a cold city since a week before Christmas. Lulled by years of uncommonly mild winters, this city has all but forgotten what winter can bring, what it used to be like. Snow started falling in December and at first everyone was enchanted with the idea of having a white Christmas. No one bothered to clean the streets. After all, it's a freak of nature in these times of global warming and it wouldn't last more than a few days. Temperatures continued to drop. News reported very cold weather on the way before New Year and indeed it began to snow, and it got colder and it snowed more. Eventually the ground became so cold, the snow turned to ice, and winter, it seemed, just continued to reinvent itself.
Hardly anyone bothered to clean the streets. Around twenty years ago the laws changed. No side streets were to be cleared of snow, only the main ones. Since the city no longer considered side streets worth the effort, some people living on those streets decided to take the same stance. Why bother shoveling, hacking, throwing down some sand. Even on main streets in front of shops there was far too little effort shown. Small paths were cleared but getting from the car to that path means surviving very thick layers of ice. Hundreds of citizens have been rushed to hospitals with broken limbs. Hundreds more are literally being held hostage, confined to their apartments, fearful of venturing out onto the ice, particularly the elderly. Driving on side streets is next to impossible, biking is not an option and walking is far too risky. Six weeks and counting.
What has the city government done? It finally got around to having a crisis call this past Monday. First, most government agencies here are hardly in any better shape than the rest of the city. Why? A number of years ago, someone, most likely with a MBA, decided Hamburg should sell off lots of its buildings and then rent them back. Most citizens never understood what the city was to gain from this, other than short term profit. What everyone now understands is the new owners of these buildings are not interested in shoveling snow. Certainly not more than a small path. Getting from the subway across large swatches of very thick ice to these buildings is risking serious falls and injuries. Add to this the 'joys' of privatizing winter services instead of using the Sanitation Department. Let's face it: it's much cheaper not to use any one in a union. Pay the private ones six or seven euros an hour and one thought the outcome would be the same. It's not. They often don't bother to show up. Even right in front of City Hall. Subcontracting city services to private firms has proved to be a disaster.
There is, however, one side street fully ice free, compliments of the Sanitation Department. It's the one where the president of the Hamburg Senate lives. It's very cold in Hamburg these days with temperatures dropping even more, on many levels.
According to weather reports this severe cold spell does not disprove global warming. Most of Europe is locked in between a high in Siberia of -35C and a high over Greenland of -20C which is not letting the milder air in. | {
"redpajama_set_name": "RedPajamaC4"
} | 567 |
\section{Introduction}
We consider the Wishart-Laguerre (L) and Jacobi (J) ensembles of random matrix theory characterized by the following joint probability density (jpd) of real eigenvalues \cite{Wishart,James,muir}:
\begin{align}
\mathcal{P}_\beta^{(L)}(\lambda_1,\ldots,\lambda_N) &:=C_{N,\beta,\nu}^{(L)}\prod_{j<k}|\lambda_j-\lambda_k|^\beta \prod_{j=1}^N \lambda_j^{\frac{\beta}{2}(\nu+1)-1} e^{-\frac{1}{2}\lambda_j}\\
\mathcal{P}_\beta^{(J)}(\lambda_1,\ldots,\lambda_N) &:=C_{N,\beta,a,b}^{(J)}\prod_{j<k}|\lambda_j-\lambda_k|^\beta \prod_{j=1}^N (1-\lambda_j)^a (1+\lambda_j)^b.\label{jacobi}
\end{align}
In the above equations, $C_{N,\beta,\nu}^{(L)}$ and $C_{N,\beta,a,b}^{(J)}$ are normalization constants, while the index $\beta=1,2,4$ characterizes the symmetry class of the ensemble (orthogonal, unitary and symplectic respectively).
The Wishart-Laguerre ensemble contains covariance matrices of the form $\mathcal{W}=\mathcal{X}\mathcal{X}^\dagger$, where $\mathcal{X}$ is a $N\times M$ ($M-N:=\nu\geq 0$)
matrix with i.i.d. Gaussian entries (real, complex or quaternionic variables). The matrix $\mathcal{W}$ is symmetric and positive semidefinite, so its $N$ real eigenvalues are non-negative ($\{\lambda_j\}\geq 0$).
Originally introduced by Wishart \cite{Wishart}, matrices from this ensemble have been extensively used in
multivariate statistical data analysis~\cite{Wilks,Johnstone} with
applications in various fields ranging from meteorological
data~\cite{Preisendorfer} to finance~\cite{BP,Burda}. They are also useful when analyzing the capacity of channels with
multiple antennae and receivers~\cite{SP}, in
nuclear physics~\cite{Fyo1}, chiral quantum
chromodynamics~\cite{QCD} and also in statistical physics such as
in a class of $(1+1)$-dimensional directed polymer
problems~\cite{Johansson}. Recently, they have also
appeared in the context of knowledge networks~\cite{MZ1} and new
mathematical results
for the case $\nu<0$ have also been lately
obtained~\cite{Z2,JN}. Large deviation properties of the eigenvalues have been investigated in \cite{vivolarge,castillo,nadalmaj}, while for an excellent review
we refer to \cite{majreview}.
The Jacobi ensemble contains combinations of two $N\times N$ Wishart-Laguerre matrices $\mathcal{W}_1$ and $\mathcal{W}_2$ of the form:
\begin{equation}
\mathcal{J} = (\mathcal{W}_1-\mathcal{W}_2)(\mathcal{W}_1+\mathcal{W}_2)^{-1}
\end{equation}
and its eigenvalues are real and lie on the support $-1\leq \{\lambda_j\}\leq 1$.
Matrices distributed according to the Jacobi weight arise as $\iota)$ truncations of Haar orthogonal, unitary or symplectic matrices (for the case of
unitary matrices, an important application arises in the theory of electronic transport
in mesoscopic systems at low temperatures as detailed in Appendix \ref{appA}); $\iota\iota)$ as composition of projection matrices \cite{collins}.
In order to proceed, we first define a \emph{shifted} version of the Jacobi ensemble with eigenvalues between $0$ and $1$:
\begin{equation}\label{shiftedjacobi}
\mathcal{P}_\beta^{(sJ)}(\lambda_1,\ldots,\lambda_N) :=C_{N,\beta,\mathfrak{a},\mathfrak{b}}^{(sJ)}\prod_{j<k}|\lambda_j-\lambda_k|^\beta \prod_{j=1}^N \lambda_j^{\mathfrak{a}} (1-\lambda_j)^{\mathfrak{b}}.
\end{equation}
which appears more frequently in physical applications (see Appendix \ref{appA}) and numerical algorithms \cite{dumitriu,edeljac}.
By changing variables $\lambda_j=1-2 y_j$ in \eqref{jacobi}, it is easy to see that:
\begin{equation}
C_{N,\beta,a,b}^{(J)}=\frac{C_{N,\beta,a,b}^{(sJ)}}{2^{N(a+b+1)+\frac{\beta}{2}N(N-1)}}.
\end{equation}
We also define the average spectral densities $(\rho(x)=\Big\langle\frac{1}{N}\sum_{i=1}^N\delta(x-\lambda_i)\Big\rangle)$ for the three ensembles above as the marginals of their respective jpds:
\begin{align}
\rho_{N,\beta,\nu}^{(L)}(\lambda_1) &=\int_{[0,\infty]^{N-1}} d\lambda_2\cdots d\lambda_N \mathcal{P}_\beta^{(L)}(\lambda_1,\ldots,\lambda_N)\\
\rho_{N,\beta,a,b}^{(J)}(\lambda_1) &=\int_{[-1,1]^{N-1}} d\lambda_2\cdots d\lambda_N \mathcal{P}_\beta^{(J)}(\lambda_1,\ldots,\lambda_N)\\
\rho_{N,\beta,\mathfrak{a},\mathfrak{b}}^{(sJ)}(\lambda_1) &=\int_{[0,1]^{N-1}} d\lambda_2\cdots d\lambda_N \mathcal{P}_\beta^{(sJ)}(\lambda_1,\ldots,\lambda_N).
\end{align}
It follows immediately from the definition that the above densities are all normalized to $1$, and that the following relation holds between the Jacobi and the shifted-Jacobi densities:
\begin{equation}\label{reljac}
\rho_{N,\beta,a,b}^{(J)}(1-2x)=\frac{1}{2}\rho_{N,\beta,a,b}^{(sJ)}(x).
\end{equation}
The purpose of this paper is twofold:
\begin{itemize}
\item to collect and present in a user-friendly way (well-known, but somehow scattered throughout the literature) explicit formulae for the above densities for finite $N$ and all symmetry classes;
\item to use these formulae to compute integer moments of the eigenvalues $\tau_n=\langle\sum_{i=1}^N\lambda_i^n\rangle$ (where the average is taken w.r.t. any of the three jpds above).
More precisely, we define:
\begin{align}
\tau_n^{(L)}(N,\beta,\nu) &:=\int_{[0,\infty]^N} d\lambda_1\cdots d\lambda_N \mathcal{P}_\beta^{(L)}(\lambda_1,\ldots,\lambda_N)\ \left(\sum_{i=1}^N \lambda_i^n\right)\\
\tau_n^{(J)}(N,\beta,a,b) &:=\int_{[-1,1]^N} d\lambda_1\cdots d\lambda_N \mathcal{P}_\beta^{(J)}(\lambda_1,\ldots,\lambda_N)\ \left(\sum_{i=1}^N \lambda_i^n\right)\\
\tau_n^{(sJ)}(N,\beta,\mathfrak{a},\mathfrak{b}) &:=\int_{[0,1]^N} d\lambda_1\cdots d\lambda_N \mathcal{P}_\beta^{(sJ)}(\lambda_1,\ldots,\lambda_N)\ \left(\sum_{i=1}^N \lambda_i^n\right).
\end{align}
One application to the case of quantum transport in chaotic cavities is detailed in Appendix \ref{appA}. Other interesting mathematical results for the Laguerre case can be found e.g. in \cite{redel}
and references therein. We are also aware that formulae for the Wishart-Laguerre and Jacobi moments for $\beta=1,2,4$ and finite $N$ have been derived by Mezzadri and Simm \cite{simm}. These are different from
the ones provided here and obtained via a different method, but equivalent.
\end{itemize}
It is easy to see that the average of any linear statistics (i.e. a quantity of the form $\mathcal{A}=\sum_{i=1}^N f(\lambda_i)$),
\begin{equation}
\langle\mathcal{A}\rangle=\int d\lambda_1\cdots d\lambda_N \mathcal{P}_\beta(\lambda_1,\ldots,\lambda_N)\ \left(\sum_{i=1}^N f(\lambda_i)\right)
\end{equation}
can be computed as a one-dimensional integral over the corresponding spectral
density of the ensemble as:
\begin{equation}\label{intrho}
\Big\langle\sum_{i=1}^N f(\lambda_i)\Big\rangle=N\int dx \rho(x) f(x).
\end{equation}
The technical achievement we report in this paper is an explicit computation of this integral valid for finite matrix dimension $N$ and all three $\beta$'s for the case $f(x)=x^n$, $n\in\mathbb{N}$.
The integral \eqref{intrho} also elucidates a possible strategy to evaluate a regular asymptotic expansion of moments for large $N$, which has been recently highlighted \cite{kui} as a problem of current
interest in the context of electronic transport in chaotic cavities (see Appendix \ref{appA}): all one has to do is to seek for a regular $(1/N)$ expansion of the macroscopic spectral density of the form:
\begin{equation}
\rho(x)=\rho^{(\infty)}(x)+\frac{1}{N}\rho^{(1)}(x)+\frac{1}{N^2}\rho^{(2)}(x)+\ldots
\end{equation}
in the spirit of high genus correlator expansions \cite{corr}, and then integrate term by term. While this program is far from completion and is thus left for future work, we show in Appendix \ref{appB} that at least the leading order of the asymptotic expansion of moments is well reproduced for the case of quantum transport in cavities with broken time-reversal symmetry ($\beta=2$).
The plan of this paper is as follows: in Section \ref{lagensemble} we consider the Wishart-Laguerre ensemble. We first summarize the densities for all three $\beta$s,
then use the integral formula \eqref{intrho} to compute integer moments for $\beta=1,2,4$ in the three subsections. Then in the last subsection we collect results from numerical
simulations for the density and moments. The same thing is done for the Jacobi ensemble in Section \ref{jacensemble}. A summary and outlook is provided in Section
\ref{conclusion}. In Appendix \ref{appA} we provide a detailed introduction to the problem of quantum transport in chaotic cavities which constitutes the main motivation for this study, while in
Appendix \ref{appB} we show that the leading order term in the expansion of the moments via spectral density is correctly reproduced.
\section{Wishart-Laguerre ensemble}\label{lagensemble}
\subsection{Spectral densities}
The spectral density for the Laguerre ensemble is known for all three symmetry classes \cite{Mehta,nagao,ghosh,kumar} and, after tedious algebraic manipulations can be cast in the form $\rho^{(L)}_{N,\beta,\nu}(x)=\frac{1}{2N}\mathcal{R}_{N,\beta,\nu}^{(L)}(x/2) $, where:
\begin{align}
\mathcal{R}_{N,1,\nu}^{(L)}(x) &=2 \mathcal{R}_{N,2,\nu}^{(L)}(2x)-\frac{\Gamma((N+1)/2)}{\Gamma((N+\nu)/2)}L_{N-1}^{(\nu)}(2x)\left\{\phi_1(x)-\phi_2(x)\right\}\label{R1lag}\\
\mathcal{R}_{N,2,\nu}^{(L)}(x) &=x^{\nu}e^{-x}\sum_{m=0}^{N-1}\frac{\Gamma(m+1)}{\Gamma(m+\nu+1)}\left(L_m^{(\nu)}(x)\right)^2 \label{R2lag}\\
\nonumber \mathcal{R}_{N,4,\nu}^{(L)}(x) &=\frac{1}{2} \mathcal{R}_{2N,2,2\nu}^{(L)}(x)-x^{2\nu}e^{-x}L_{2N}^{(2\nu)}(x)
\frac{\Gamma(N+1)}{2^{2\nu+1}\Gamma(N+\nu+1/2)}\times\\
&\times \sum_{m=0}^{N-1}
\frac{\Gamma(m+1/2)}{\Gamma(m+\nu+1)}L_{2m}^{(2\nu)}(x)\label{R4lag}
\end{align}
where:
\begin{align}
\phi_1(x) &=(2x)^\nu e^{-2x} \sum_{m=0}^{(\kappa+N-2)/2} \mathfrak{d}_m L_{2m+1-\kappa}^{(\nu)}(2x)\\
\phi_2(x) &=x^{(\nu-1)/2}e^{-x}\left[(1-\kappa)\frac{2\Gamma((\nu+1)/2,x)}{\Gamma((\nu+1)/2)}+2\kappa-1\right]\\
\mathfrak{d}_m &=\frac{\Gamma(m+1-\kappa/2)}{2^{\nu-1}\Gamma(m+(\nu-1)/2+2-\kappa/2)}.
\end{align}
In the above equations, $\kappa=N\ \mathrm{mod}\ 2$, $L_N^{(\nu)}(x)$ is a generalized Laguerre polynomial defined by the sum:
\begin{equation}\label{laguerredef}
L_n^{(\lambda)}(z)=\sum_{k=0}^n c_k(n,\lambda)z^k
\end{equation}
where:
\begin{equation}
c_k (n,\lambda)=\frac{\Gamma(\lambda+n+1)(-n)_k }{n!k!\Gamma(\lambda+k+1)}
\end{equation}
and $\Gamma(a,x)=\int_x^\infty t^{a-1}e^{-t}dt$ is the incomplete Gamma function, while $(x)_n = \Gamma(x+n)/ \Gamma(x)$ is the Pochhammer symbol.
In the following subsections, we present the results for integer moments $\tau_n^{(L)}(N,\beta,\nu)=\Big\langle\sum_{j=1}^N\lambda_j^n\Big\rangle$
obtained by integration (eq. \eqref{intrho}) of the densities given above. The task is most easily
accomplished by first defining the following auxiliary function:
\begin{equation}
\mathcal{Q}(r;m,\ell;\alpha):=\int_0^\infty dx\ x^r e^{-x}L_m^{(\alpha)}(x)L_\ell^{(\alpha)}(x)
\end{equation}
which is easily evaluated using \eqref{laguerredef} as:
\begin{equation}\label{Qdef}
\mathcal{Q}(r;m,\ell;\alpha)=\sum_{k=0}^m\sum_{k^\prime=0}^\ell c_k(m,\alpha)c_{k^\prime}(\ell,\alpha)\Gamma(1+r+k+k^\prime).
\end{equation}
Note that, by the orthogonality relation of Laguerre polynomials, one has:
\begin{equation}
\mathcal{Q}(\alpha;m,\ell;\alpha)=\frac{(\ell+\alpha)!}{\ell !}\delta_{m\ell}.
\end{equation}
\subsection{Moments $\beta=1$}
For $\beta=1$, the final result of the integration reads as follows:
\begin{equation}
\boxed{\tau_n^{(L)}(N,1,\nu) =2^{-n}\tau_n^{(L)}(N,2,\nu)+\mathcal{I}_2(n)+\mathcal{I}_3(n)}
\end{equation}
where:
\begin{align}
\mathcal{I}_2(n) &=-\frac{\Gamma((N+1)/2)}{2\Gamma((N+\nu)/2)}\sum_{m=0}^{(\kappa+N-2)/2}\mathfrak{d}_m\mathcal{Q}(n+\nu;N-1,2m+1-\kappa;\nu)\\
\nonumber\mathcal{I}_3(n) &=\frac{\Gamma((N+1)/2)}{2\Gamma((N+\nu)/2)} \left\{(1-\kappa)\left[\frac{2^{(3-\nu)/2}}{\Gamma((\nu+1)/2)}\mathcal{Y}_1(n)-2^{(1-\nu)/2}\mathcal{Y}_2(n)\right]+\right.\\
&\left.+\kappa\ 2^{(1-\nu)/2}\mathcal{Y}_2(n)\right\}\\
\nonumber\mathcal{Y}_1(n) &=\sum_{m=0}^{N-1} \frac{c_m (N-1,\nu) 2^{n + (\nu + 1)/2 + m}
\Gamma(n + \nu + 1 + m)}{n + (\nu + 1)/2 + m}\times\\
&\times\ _2 F_1(n + (\nu + 1)/2 + m, n + \nu + 1 + m; n + (\nu + 3)/2 + m; -1)\label{hyplag}\\
\mathcal{Y}_2(n) &=\sum_{m=0}^{N-1} c_m (N-1,\nu) 2^{n + (\nu + 1)/2 + m}
\Gamma(n + m + (\nu + 1)/2).
\end{align}
In \eqref{hyplag}, we have used the following hypergeometric function:
\begin{equation}
_2 F_1(a_1,a_2;b_1;z):=\sum_{k=0}^\infty \frac{(a_1)_k (a_2)_k}{(b_1)_k \ k!}z^k.
\end{equation}
One can check by direct inspection that $\tau_0^{(L)}(N,1,\nu)=N$ and $\tau_1^{(L)}(N,1,\nu)=N(N+\nu)$ as it should be.
\subsection{Moments $\beta=2$}
Combining \eqref{intrho} with \eqref{R2lag}, one easily obtains:
\begin{equation}
\boxed{\tau_n^{(L)}(N,2,\nu)=2^n\sum_{m=0}^{N-1}\frac{\Gamma(m+1)}{\Gamma(m+\nu+1)}\mathcal{Q}(n+\nu;m,m;\nu)}
\end{equation}
One can check by direct inspection that $\tau_0^{(L)}(N,2,\nu)=N$ and $\tau_1^{(L)}(N,2,\nu)=2N(N+\nu)$ as it should be.
\subsection{Moments $\beta=4$}
Similarly for $\beta=4$ one gets:
\begin{equation}
\boxed{\tau_n^{(L)}(N,4,\nu) =\frac{1}{2}\tau_n^{(L)}(2N,2,2\nu)-\sum_{m=0}^{N-1}\sum_{k=0}^{2N}\sum_{k^\prime=0}^{2m}
\chi_{m,k,k^\prime}}
\end{equation}
where:
\begin{align}
\chi_{m,k,k^\prime} &=f_m(N,\nu,n,k,k^\prime)c_k (2N,2\nu)c_{k^\prime} (2m,2\nu)\\
f_m(N,\nu,n,k,k^\prime) &=\frac{\Gamma(1+k+k^\prime+n+2\nu)\Gamma(N+1)\Gamma(m+1/2)}{2^{2\nu+1-n}
\Gamma(N+\nu+1/2)\Gamma(m+\nu+1)}.
\end{align}
One can check by direct inspection that $\tau_0^{(L)}(N,4,\nu)=N$ and $\tau_1^{(L)}(N,4,\nu)=4N(N+\nu)$ as it should be.
\subsection{Comparison with numerics}
In Fig. \ref{Laguerrefig} we plot the analytical formulae \eqref{R2lag}, \eqref{R1lag} and \eqref{R4lag} together with numerical diagonalization of matrices $\mathcal{W}$ from the Laguerre ensemble
with $\beta=1,2,4$ respectively\footnote{Note that for $\beta=4$ the matrices corresponding to a certain $N$ actually have size $2N\times 2N$ and thus have $2N$ real and positive eigenvalues. Only $N$
of them are distinct though, and only those must be used when carrying out numerical simulations for moments.}, obtained in \textsf{Matlab} as follows\footnote{Alternatively, one can use the tridiagonal algorithm by Dumitriu and Edelman \cite{dumitriu}.}:
\begin{align*}
\beta=1 & &\mbox{\texttt{X=randn(N,M); W = X*X';}}\\
\beta=2 & &\mbox{\texttt{X=randn(N,M)+i*randn(N,M); W = X*X';}}\\
\beta=4 && \mbox{\texttt{X = randn(N,M)+i*randn(N,M);}}\\
&& \mbox{\texttt{Y = randn(N,M)+i*randn(N,M);}}\\
&& \mbox{\texttt{A = [X Y; -conj(Y) conj(X)]; W = A*A';}}
\end{align*}
\begin{figure}[ht]
\begin{center}
\includegraphics[bb=-223 180 836 612,width=\hsize]{RhoWishart1.eps}
\includegraphics[bb=-223 172 836 620,width=\hsize]{RhoWishart2.eps}
\includegraphics[bb= -223 180 836 612,width=\hsize]{RhoWishart4.eps}
\caption{Average spectral density of the Laguerre ensemble for $N=8,\nu=7$ and $\beta=1,2,4$ (top to bottom). Numerical diagonalization is given in red dots, while theoretical
results are in solid black.} \label{Laguerrefig}
\end{center}
\end{figure}
In Table \ref{table:nonlin1} below, we compare the analytical results for moments of the Laguerre ensemble with numerical simulations
via the \textsf{Matlab} algorithm given above.
\begin{table}[ht]
\caption{Comparison between theory and numerics for moments of the Laguerre ensemble $(N=5,M=7\to\nu=2)$ (the numerical
results are obtained averaging over $\mathcal{O}(10^5)$ samples).}
\centering
\begin{tabular}{c c c c}
\hline\hline
$\beta$ & $n$ & Theory & Numerics \\ [0.5ex]
\hline
1 & 1 & 35 & 34.994\\
1 & 2 & 455 & 455.27311\\
1 & 3 & 7665 & 7660.026\\
\hline \\
2 & 1 & 70 & 69.981\\
2 & 2 & 1680 & 1680.238 \\
2 & 3 & 50400 & 50409.428 \\
\hline \\
4 & 1 & 140 & 139.982 \\
4 & 2 & 6440 & 6437.074 \\
4 & 3 & 362880 & 362857.134 \\ [1ex]
\hline
\end{tabular}
\label{table:nonlin1}
\end{table}
\section{Jacobi ensemble}\label{jacensemble}
\subsection{Spectral densities}
The eigenvalue densities for the Jacobi Ensemble ($x\in[-1,1]$) read \cite{ghosh,kumar}\footnote{For $\beta=1$, $N$ is even. The expression for $N$ odd is slightly more complicated \cite{ghosh}.}:
\begin{align}
\label{density1jacobi}\rho^{(J)}_{N,1,a,b}(x) &= \frac{1}{N}\sum_{m=0}^{N/2-1} (g_{2m})^{-1} \left [ \phi_{2m}(x) \psi_{2m+1}(x) - \phi_{2m+1}(x) \psi_{2m}(x) \right ]\\
\label{density2jacobi}\rho^{(J)}_{N,2,a,b}(x) &= \frac{w_{a,b}(x)}{N} \sum_{n=0}^{N-1} (h_n^{a,b})^{-1} \left [ P_n^{(a,b)}(x) \right ]^2\\
\label{density4jacobi}\rho^{(J)}_{N,4,a,b}(x) &= \rho^{(J)}_{2N,2,2\hat{a}+1,2\hat{b}+1}(x)- P_{2N}^{(2\hat{a}+1,2\hat{b}+1)}(x) \sum_{m=0}^{N-1} K_{2N,m}^{(\hat{a},\hat{b})} P_{2m}^{(2\hat{a}+1,2\hat{b}+1)}(x)
\end{align}
where:
\begin{align} \label{hcoeff}
h_{n}^{a,b} &= \frac{2^{a+b+1}}{2n+a+b+1} \frac{\Gamma(n+a+1) \Gamma(n+b+1)}{n! \Gamma(n+a+b+1)}\\
g_{2m} &= g_{2m+1} = h_{2m}^{2a+1,2b+1}\\
\label{phieven}\phi_{2m}(x) &= w_{a,b}(x) P_{2m}^{(2a+1,2b+1)}(x) \\
\label{psiodd}\psi_{2m+1}(x) &= w_{a+1,b+1}(x) P_{2m}^{(2a+1,2b+1)}(x)\\
\label{phiodd}\phi_{2m+1}(x) &= w_{a,b}(x) \left [ A_{2m+1} P_{2m+1}^{(2a+1,2b+1)}(x) - B_{2m-1} P_{2m-1}^{(2a+1,2b+1)}(x) \right ] \\
\label{psieven}\psi_{2m}(x) &= \frac{1}{2} \int_{-1}^{+1} d y \ \mathrm{sign}(x-y) \ \phi_{2m}(y) \\
\nonumber K_{N,m}^{(a,b)} &= \frac{(4m+2a+2b+3) \ \Gamma((N+2)/2) \ \Gamma((N+2a+2b+4)/2)}{2^{2a+2b+3} \ \Gamma((N+2a+2)/2) \ \Gamma((N+2b+2)/2)}\times \\
&\times\frac{\Gamma(m+1/2) \ \Gamma(m+a+b+3/2)}{\Gamma(m+a+3/2) \ \Gamma(m+b+3/2)}\\
\hat{a} &= \frac{a-2}{2}\\
\hat{b} &= \frac{b-2}{2}
\end{align}
where $\mathrm{sign}(z)=z/|z|$ and:
\begin{equation} \label{ABcoeff}
A_n = - \frac{n(n+2a+2b+2)}{2n+2a+2b+1} \ ; \ \ \
B_n = - \frac{(n+2a+2)(n+2b+2)}{2n+2a+2b+5} \ ; \ \ \ (B_{-1}=0).
\end{equation}
In the formulas above, $w_{a,b}(x) = (1-x)^a(1+x)^b$ is the Jacobi weight function, and $P_{n}^{(a,b)}$ is the $n-$th order Jacobi polynomial with parameters $a$ and $b$, defined as:
\begin{align}
\nonumber P_n^{(a,b)}(x) &= \sum_{j=0}^n \binom{n+a}{j} \binom{n+b}{n-j} \left (\frac{x-1}{2} \right )^{n-j} \left ( \frac{x+1}{2} \right )^j = \\
&=\frac{1}{2^n} \sum_{j=0}^n c_j^{(n)}(a,b) (x-1)^{n-j} (x+1)^j\label{jacobiexp}
\end{align}
where we set:
\begin{equation} \label{ccoeff}
c_j^{(n)}(a,b) = \frac{\Gamma(n+a+1)}{\Gamma(j+1) \Gamma(n+a-j+1)}
\frac{\Gamma(n+b+1)}{\Gamma(n-j+1) \Gamma(b+j+1)}.
\end{equation}
We now turn to the problem of computing integer moments of Jacobi matrices. It is easy to derive first the following relations between integer moments in the shifted-Jacobi and ordinary Jacobi ensembles:
\begin{equation}\label{relationmoments1}
\boxed{\tau_n^{(sJ)}(N,\beta,\mathfrak{a},\mathfrak{b}) =2^{-n}\sum_{k=0}^n \binom{n}{k}(-1)^k \tau_k^{(J)}(N,\beta,\mathfrak{a},\mathfrak{b})}
\end{equation}
\begin{equation}\label{relationmoments2}
\boxed{\tau_n^{(J)}(N,\beta,a,b) =\sum_{k=0}^n\binom{n}{k}(-2)^k \tau_k^{(sJ)}(N,\beta,a,b). }
\end{equation}
In the following we will therefore focus on the ordinary Jacobi case.
\subsection{Moments $\beta=1$}\label{subjacobi1}
The $n$-th moment of the distribution is therefore given by:
\begin{equation} \label{nthmoment}
\boxed{\tau^{(J)}_n (N,1,a,b) = \sum_{m=0}^{N/2-1} (g_{2m})^{-1} \left ( I_{2m,2m+1}^{(n)} - I_{2m+1,2m}^{(n)} \right )}
\end{equation}
where:
\begin{equation} \label{Idef}
I_{j,k}^{(n)} = \int_{-1}^{+1} d x \ \phi_j(x) \ \psi_k(x) \ x^n.
\end{equation}
are computed explicitly in \eqref{int1_2} and \eqref{lab} below. As it is clear from equation \eqref{nthmoment}, we only need to compute two different kinds of integrals. Let us then start from the first kind ($I_{2m,2m+1}^{(n)}$).
Using \eqref{phieven}, \eqref{psiodd} and \eqref{jacobiexp}, we therefore get (in the following the $c_j^{(n)}$ coefficients will always depend on the pair $(2a+1,2b+1)$, so we shall omit their explicit dependence on those parameters for the rest of the subsection):
\begin{align}
\nonumber I_{2m,2m+1}^{(n)} &= \int_{-1}^{+1} d x \ \phi_{2m}(x) \ \psi_{2m+1}(x) \ x^n =\frac{1}{2^{4m}} \sum_{i,j=0}^{2m} (-1)^{4m-i-j} c_i^{(2m)} c_j^{(2m)} \times\\
&\times\underbrace{\int_{-1}^{+1} d x \ (1-x)^{4m+2a-i-j+1} (1+x)^{2b+i+j+1} \ x^n}_{\mathcal{L}^{(n)}(4m+2a-i-j+1,2b+i+j+1)} \label{int1_1}
\end{align}
where we have introduced the following integral:
\begin{align}
\nonumber\mathcal{L}^{(n)}(w,z) &= \int_{-1}^{+1} d x \ (1-x)^w (1+x)^z \ x^n =2^{w+z+1} \times\\
&\times\sum_{k=0}^n \binom{n}{k} (-2)^k \mathrm{B}(w+k+1,z+1)
\end{align}\label{L_int}
introducing Euler's Beta function $\mathrm{B}(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$. Inserting this result into equation \eqref{int1_1} we obtain:
\begin{align} \label{int1_2}
\nonumber \blacktriangleright I_{2m,2m+1}^{(n)} &= 2^{2(a+b+1)+1} \sum_{i,j=0}^{2m} (-1)^{4m-i-j} c_i^{(2m)} c_j^{(2m)} \times\\
&\times\sum_{k=0}^n \binom{n}{k} (-2)^k \mathrm{B} \left (2(2m+a+1)+k-i-j,2(b+1)+i+j \right).
\end{align}
In order to compute the second type of integrals ($I_{2m+1,2m}^{(n)}$) in \eqref{nthmoment} and \eqref{Idef}, we use \eqref{phiodd}, \eqref{psieven} and \eqref{jacobiexp} to get:
\begin{align}
\nonumber I_{2m+1,2m}^{(n)} &= \frac{1}{2^{2m+1}} \sum_{i=0}^{2m} c_i^{(2m)} \int_{-1}^{+1} d y \ w_{a,b}(y) (y-1)^{2m-i} (y+1)^i\times\\
&\label{int2_10}\times \underbrace{\int_{-1}^{+1} d x \ \mathrm{sign}(x-y) \ \phi_{2m+1}(x) \ x^n}_{\mathcal{K}_{2m+1}^{(n)}(y)}.
\end{align}
The integral $\mathcal{K}_{2m+1}^{(n)}(y)$ can be rewritten as follows by exploiting \eqref{phiodd}:
\begin{equation} \label{Kint_1}
\mathcal{K}_{2m+1}^{(n)}(y) = A_{2m+1} \mathcal{G}_{a,b}^{(n)}(2m+1;y) - B_{2m-1} \mathcal{G}_{a,b}^{(n)}(2m-1;y)
\end{equation}
where we have:
\begin{align}
\nonumber &\mathcal{G}_{a,b}^{(n)}(m;y)= \int_{-1}^{+1} d x \ \mathrm{sign}(x-y) \ w_{a,b}(x) \ P_m^{(2a+1,2b+1)}(x) \ x^n= \\ \nonumber
&= \frac{1}{2^m} \sum_{j=0}^m (-1)^{m-j} c_j^{(m)} \int_{-1}^{+1} d x \ \mathrm{sign}(x-y) \ (1-x)^{a+m-j} (1+x)^{b+j} \ x^n= \\
&= 2^{a+b+1} \sum_{j=0}^m (-1)^{m-j} c_j^{(m)} \sum_{k=0}^n \binom{n}{k} (-2)^k\ \widetilde{\mathrm{B}}_{\frac{1-y}{2}}\left ( a+m+k-j+1,b+j+1 \right)
\end{align}\label{Gint_1}
where $\widetilde{\mathrm{B}}_y(w,z) = 2 \ \mathrm{B}_y(w,z) - \mathrm{B}(w,z)$ and $\mathrm{B}_y(w,z)=\int_0^y dt\ t^{w-1}(1-t)^{z-1}$ is the incomplete Beta function.
Therefore we get:
\begin{align}
\nonumber &\mathcal{K}_{2m+1}^{(n)}(y) = 2^{a+b+1} \sum_{k=0}^n \binom{n}{k} (-2)^k\times \\ \nonumber
&\times \left [ A_{2m+1} \sum_{j=0}^{2m+1} (-1)^{2m-j+1} c_j^{(2m+1)} \ \widetilde{\mathrm{B}}_{\frac{1-y}{2}} \Big ( 2(m+1)+a+k-j,b+j+1 \Big ) \right. \\
&- \left. B_{2m-1} \sum_{j=0}^{2m-1} (-1)^{2m-j-1} c_j^{(2m-1)} \ \widetilde{\mathrm{B}}_{\frac{1-y}{2}} \Big ( 2m+a+k-j,b+j+1 \Big ) \right ].
\end{align}\label{Kint_2}
Eventually, inserting this result into \eqref{int2_10} and computing the integral in $d y$, we obtain
\begin{align}
\nonumber & \blacktriangleright I_{2m+1,2m}^{(n)} =\frac{2^{a+b}}{2^{2m}} \sum_{i=0}^{2m} c_i^{(2m)} \sum_{k=0}^n \binom{n}{k}
(-2)^k\times \\ \nonumber
&\times \left [ A_{2m+1} \sum_{j=0}^{2m+1} (-1)^{2m-j+1} c_j^{(2m+1)} \ \Omega_{a,b} \Big ( 2m-i,i;2(m+1)+a+k-j,b+j+1 \Big ) \right. \\
&\label{lab} - \left. B_{2m-1} \sum_{j=0}^{2m-1} (-1)^{2m-j-1} c_j^{(2m-1)} \ \Omega_{a,b} \Big (2m-i,i;2m+a+k-j,b+j+1 \Big ) \right ]
\end{align}
where we have:
\begin{align}
\nonumber &\Omega_{a,b}(h,\ell;w,z) = \int_{-1}^{+1} d x \ w_{a,b}(x) (x-1)^h (x+1)^\ell \ \widetilde{\mathrm{B}}_{\frac{1-x}{2}}(w,z) = \\ \nonumber
&=(-1)^h \ 2^{a+b+h+\ell+1} \left [ w^{-1} \ \mathrm{B} \Big (1+b+\ell,1+a+h+w \Big )\times\right.\\
&\nonumber\times ~_3 F_2 \Big (w,1+a+h+w,1-z;1+w,2+a+b+h+\ell+w;1 \Big) \\
&-\left. \mathrm{B}(w,z) \ \mathrm{B} \Big (1+a+h,1+b+\ell \Big ) \right ].\label{Omegasquare}
\end{align}
In \eqref{Omegasquare}, we have used the following generalized hypergeometric function:
\begin{equation}
_3 F_2(a_1,a_2,a_3;b_1,b_2;z):=\sum_{k=0}^\infty \frac{(a_1)_k (a_2)_k (a_3)_k}{(b_1)_k (b_2)_k\ k!}z^k.
\end{equation}
\subsection{Moments $\beta=2$}
We wish now to compute the generic $n-$th moment of the density in \eqref{density2jacobi}, which reads
\begin{equation} \label{moment}
\tau_n^{(J)}(N,2,a,b) = \left \langle \sum_{i=1}^N x_i^n \right \rangle = N\int_{-1}^{+1} d x \ \rho^{(J)}_{N,2,a,b}(x) \ x^n.
\end{equation}
By making use of the representation of Jacobi polynomials in \eqref{jacobiexp}, the $n-$th moment in \eqref{moment} can be written as follows (throughout all the present subsection the $c_j^{(k)}$ coefficients will depend on the pair $(a,b)$):
\begin{align} \label{moment2}
\nonumber\tau_n^{(J)}(N,2,a,b) &= \sum_{k=0}^{N-1} (h_k^{a,b})^{-1} \int_{-1}^{+1} dx \ w_{a,b}(x) \left [ P_k^{(a,b)}(x) \right ]^2 x^n =\\
&= \sum_{k=0}^{N-1} \sum_{i,j=0}^k \frac{c_i^{(k)} c_j^{(k)}}{2^{2k} h_k^{a,b}} I^{(n)}(k,i,j,a,b)
\end{align}
where:
\begin{equation} \label{Iint2}
I^{(n)}(k,i,j,a,b) = (-1)^{2k-i-j} \int_{-1}^{+1} dx \ (1-x)^{2k+a-i-j} (1+x)^{b+i+j} x^n
\end{equation}
and can be computed by changing variables and setting $y = (1-x)/2$. When doing so, one obtains:
\begin{equation} \label{Iint3}
I^{(n)}(k,i,j,a,b) = (-1)^{2k-i-j} 2^{2k+a+b+1} \int_0^1 dy \ y^{2k+a-i-j} (1-y)^{b+i+j} (1-2y)^n,
\end{equation}
and expanding $(1-2y)^n$ according to the binomial Theorem, one gets:
\begin{align} \label{Iint4}
\nonumber I^{(n)}(k,i,j,a,b) &= (-1)^{2k-i-j} 2^{2k+a+b+1} \sum_{\ell=0}^n (-2)^\ell \binom{n}{\ell}\times\\
&\times \int_0^1 dy \ y^{2k+a+\ell-i-j} (1-y)^{b+i+j}.
\end{align}
The integral appearing in the previous equation is of the following kind
\begin{equation} \label{gammaint}
\int_0^1 dt \ t^{x-1} (1-t)^{y-1} = \mathrm{B}(x,y).
\end{equation}
Thus we get:
\begin{align}
\nonumber I^{(n)}(k,i,j,a,b) &= (-1)^{2k-i-j} 2^{2k+a+b+1} \sum_{\ell=0}^n (-2)^\ell \binom{n}{\ell} \times\\
&\times\mathrm{B}(2k+a+\ell-i-j+1,b+i+j+1).
\end{align}
Plugging this result in equation \eqref{moment2}, one can easily see that the $n-$th moment has the compact expression:
\begin{equation} \label{moment3}
\boxed{\tau_n^{(J)}(N,2,a,b) =\sum_{k=0}^{N-1} \sum_{i,j=0}^k \sum_{\ell=0}^n \mathfrak{y}_{k,i,j,\ell} \mathrm{B}(2k+a+\ell-i-j+1,b+i+j+1)}
\end{equation}
where:
\begin{equation}
\mathfrak{y}_{k,i,j,\ell}=\frac{(-1)^{2k-i-j+\ell} c_i^{(k)} c_j^{(k)} 2^{a+b+1+\ell} }{h_k^{a,b}}\binom{n}{\ell}.
\end{equation}
After inserting \eqref{moment3} into \eqref{relationmoments1}, we have checked that the special case $\mathfrak{a}=\mu,\mathfrak{b}=0$ for $\tau_n^{(sJ)}(N,2,\mathfrak{a},\mathfrak{b})$ indeed agrees with
eq. (16) in \cite{vivovivo} and with eq. (13) in \cite{novaes2} as it should (see Appendix \ref{appA} for details).
\subsection{Moments $\beta=4$}\label{subjacobi4}
The $n$-th moment of the density in \eqref{density4jacobi} reads:
\begin{equation} \label{sympmoment}
\boxed{\tau^{(J)}_n(N,4,a,b) = \frac{1}{2} \tau^{(J)}_n(2N,2,2\hat{a}+1,2\hat{b}+1) - \frac{1}{2} \sum_{m=0}^{N-1} \Theta_{N,m}}
\end{equation}
where:
\begin{align} \label{Yint1}
\Theta_{N,m} &= K_{2N,m}^{(2\hat{a}+1,2\hat{b}+1)} Y_{2N,2m}^{(n)}(2\hat{a}+1,2\hat{b}+1)\\
Y_{k,\ell}^{(n)}(a,b) &= \int_{-1}^{+1} dx \ w_{a,b}(x) \ P_k^{(a,b)}(x) \ P_\ell^{(a,b)}(x) \ x^n.
\end{align}
Exploiting the Jacobi polynomial expansion \eqref{jacobiexp} and the binomial Theorem we get (again, we omit the explicit dependence of the $c_i^{(k)}$ coefficients on the pair $(a,b)$):
\begin{align} \label{Yint2}
\nonumber Y_{k,\ell}^{(n)}(a,b) &= \frac{1}{2^{k+\ell}} \sum_{i=0}^k \sum_{j=0}^\ell (-1)^{k+\ell-i-j} c_i^{(k)} \ c_j^{(\ell)} \times\\
\nonumber &\times \int_{-1}^{+1} dx \ (1-x)^{a+k+\ell-i-j} \ (1+x)^{b+i+j} \ x^n= \\
\nonumber &= 2^{a+b+1} \sum_{i=0}^k \sum_{j=0}^\ell \sum_{s=0}^n \binom{n}{s} (-1)^{k+\ell-i-j} (-2)^s c_i^{(k)} \ c_j^{(\ell)} \times\\
&\times \mathrm{B}(a+k+\ell+s-i-j+1,b+i+j+1).
\end{align}
\subsection{Comparison with numerics}
In Fig. \ref{Jacobifig} we plot the analytical formulae \eqref{density1jacobi}, \eqref{density2jacobi} and \eqref{density4jacobi} together with numerical diagonalization of matrices $\mathcal{J}$ from the Jacobi ensemble
with $\beta=1,2,4$ respectively, obtained in \textsf{Matlab} using the algorithm by Edelman and Sutton \cite{edeljac}.
\begin{figure}[ht]
\begin{center}
\includegraphics[bb=-252 149 865 641,width=0.95\hsize]{jacfig1.eps}
\includegraphics[bb=-252 150 865 642,width=0.95\hsize]{jacfig2.eps}
\includegraphics[bb= -252 158 865 634,width=0.95\hsize]{jacfig4.eps}
\caption{Average spectral density of the Jacobi ensemble for $N=6,a=5,b=9$ and $\beta=1,2,4$ (top to bottom). Numerical diagonalization is given in blue triangles, while theoretical
results are in solid black.} \label{Jacobifig}
\end{center}
\end{figure}
In the following tables, we compare the analytical results for moments of the Jacobi and shifted-Jacobi ensembles with numerical simulations
via the algorithm by Edelman and Sutton \cite{edeljac}.
\begin{table}[ht]
\caption{Comparison between theory and numerics for moments of the Jacobi ensemble $(N=6,a=5,b=9)$ (the numerical
results are obtained averaging over $\mathcal{O}(10^5)$ samples).}
\centering
\begin{tabular}{c c c c}
\hline\hline
$\beta$ & $n$ & Theory & Numerics \\ [0.5ex]
\hline
1 & 1 & 1.1428 & 1.1454\\
1 & 2 & 1.1536 & 1.1573\\
1 & 3 & 0.5108 & 0.5125\\
\hline \\
2 & 1 & 0.9230 & 0.9259\\
2 & 2 & 1.4872 & 1.4892 \\
2 & 3 & 0.5470 & 0.5485 \\
\hline \\
4 & 1 & 0.6666 & 0.6674 \\
4 & 2 & 1.9300 & 1.9294 \\
4 & 3 & 0.5602 & 0.5602 \\ [1ex]
\hline
\end{tabular}
\label{table:nonlin}
\end{table}
\begin{table}[ht]
\caption{Comparison between theory and numerics for moments of the shifted-Jacobi ensemble $(N=6,\mathfrak{a}=5,\mathfrak{b}=9)$ (the numerical
results are obtained averaging over $\mathcal{O}(10^5)$ samples).}
\centering
\begin{tabular}{c c c c}
\hline\hline
$\beta$ & $n$ & Theory & Numerics \\ [0.5ex]
\hline
1 & 1 & 2.4286 & 2.4300\\
1 & 2 & 1.2170 & 1.2169\\
1 & 3 & 0.6902 & 0.6902\\
\hline \\
2 & 1 & 2.5385 & 2.5390\\
2 & 2 & 1.4103 & 1.4102 \\
2 & 3 & 0.8932 & 0.8919 \\
\hline \\
4 & 1 & 2.6667 & 2.6669 \\
4 & 2 & 1.6492 & 1.6493 \\
4 & 3 & 1.1605 & 1.1612 \\ [1ex]
\hline
\end{tabular}
\label{table:nonlin}
\end{table}
\section{Conclusions}\label{conclusion}
In conclusions, we have first collected well-known formulae for the spectral density of Laguerre and Jacobi ensembles of random matrices with orthogonal, unitary and symplectic symmetry.
We feel that this paper might serve as a quick reference point for formulae that might be hard to dig out in the literature. Using these results, we have computed the average of integer
moments, reducing the complexity from a $N$-fold integration to a single integral over the spectral density. In all cases, expressions different from ours and derived through different methods already exist \cite{simm,vivovivo,novaes2} (see also Appendix \ref{appA}). It would be interesting to prove mathematically the equivalence of various formulae which are now available for the same objects.
The present paper offers a different and possibly simpler way of deriving transport moments
for all symmetry classes and finite number of open electronic channels. The obtained results have been checked numerically with high accuracy, and the corresponding \textsf{Matlab} codes are freely available
on demand.
\newline\\
{\bf Acknowledgments:} we gratefully acknowledge useful correspondence with Saugata Ghosh, Marcel Novaes, Francesco Mezzadri, Gernot Akemann and Dima Savin. We are indebted to Akhilesh Pandey and Santosh Kumar for valuable correspondence
and for sending us their preprint \cite{kumar} before publication.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,623 |
Q: Image preview in react-native I have an image uploader in my react-native app, once you upload the image it navigates you to another screen and previews the image there, in the image preview screen there is an input for giving a name for this image and a save button, when clicking on save button it should go back to the previous screen and display the image and it's name there inside the flatlist i have, i managed to do the steps until previewing the image but after that i didn't know what to do next, here is the code:
First screen:
state = {
image: null,
previews: []
};
_pickImage = async () => {
await Permissions.askAsync(Permissions.CAMERA_ROLL);
const {navigate} = await this.props.navigation;
let result = await ImagePicker.launchImageLibraryAsync({
allowsEditing: false,
aspect: [4, 4],
});
navigate( 'ImagePreview', { uri : result.uri } );
if (!result.cancelled) {
this.setState({ image: result.uri });
}
};
_keyExtractor (item, index) {
return index.toString();
}
_renderItem ({ item, index }) {
return (
<View>
<Image source={require('')}/>
<Text>Image title</Text>
</View>
);
}
<FlatList style={{ flex: 0.5 }}
data={this.state.previews}
keyExtractor={this._keyExtractor.bind(this)}
renderItem={this._renderItem.bind(this)}
numColumns={2}
/>
Second screen:
const uri = navigation.getParam('uri');
<Image source={{uri:uri}} style={{width: 200, height: 200}} />
<Button title="Save" />
A: From my understanding you are trying to send data back from the second screen to the first screen. One solution is to create a function inside the first screen that you pass to the second screen.
So you declare a function in the first screen where you update the state with the title:
returnTitle(uri, title) {
const {previews} = this.state;
previews.push({uri, title});
this.setState({previews});
}
And you pass this function when you navigate to the second screen:
navigate('ImagePreview', {
uri: result.uri,
returnTitle: this.returnTitle.bind(this)
});
In your second screen you define an onPress handler for your Button where you call the returnTitle function and navigate back:
onSavePress = () => {
const {title} = this.state; // Not sure where you store the title
const {navigation} = this.props;
const uri = navigation.getParam('uri');
navigation.state.params.returnTitle(uri, title);
navigation.goBack();
}
Remember to get the title from your input. The code does not show if you store it in the component state.
Now in the first component change your _renderItem method to fit the previews which is now an array of objects:
_renderItem ({ item, index }) {
return (
<View>
<Image source={{uri: item.uri}}/>
<Text>{item.title}</Text>
</View>
);
}
A: On your second screen, the Image tag should have a source property instead of image, like this:
<Image source={{uri:uri}} style={{width: 200, height: 200}} />
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,944 |
{"url":"https:\/\/2021.help.altair.com\/2021.1\/hwsolvers\/ms\/topics\/solvers\/ms\/grid_motionsolve.htm","text":"# GRID\n\nModel ElementGRID defines an NLFE node in 3D space.\n\n## Description\n\nThe definition of a GRID contains the following information:\n\u2022 Current location in space (X, Y, and Z coordinates)\n\u2022 Relaxed location in space (X, Y, and Z coordinates)\n\u2022 Current gradient vectors in space (X, Y, and Z gradient vectors)\n\u2022 Relaxed gradient vectors in space (X, Y, and Z gradient vectors)\n\nThese are called the nodal co-ordinates for this GRID.\n\n## Format\n\n<GRID\nid = \"integer\"\nx = \"real\"\ny = \"real\"\nz = \"real\"\nrx = \"string\"\nry = \"string\"\nrz = \"string\"\nx0 = \"real\"\ny0 = \"real\"\nz0 = \"real\"\nrx0 = \"string\"\nry0 = \"string\"\nrz0 = \"string\"\nxd = \"real\"\nyd = \"real\"\nzd = \"real\"\nrxd = \"real\"\nryd = \"real\"\nrzd = \"real\"\n\/>\n\n## Attributes\n\nid\nUnique GRID identification number.\nx, y, z\nThe current X, Y, and Z position of the GRID.\nx0, y0, z0\nThe relaxed X, Y, and Z positions of the GRID.\nrx, ry, rz\nThe X, Y, and Z gradient vectors for the GRID. Each vector is specified as a string of three space separated values that represent the vector components.\nrx0, ry0, rz0\nThe X, Y, and Z relaxed position gradient vectors for the GRID. Each vector is specified as a string of three space separated values that represent the vector components.\nxd, yd, zd\nThe X, Y, and Z velocity of the grid.\nDefault for all three is 0.\nrxd, ryd, rzd\nThe time of the X, Y and Z gradient vector.\nDefault for all three is 0.\n\n## Example\n\nThe following is an example of a GRID defined at the origin:\n\n<GRID id=\"303001\" x=\"0.000000\" y=\"0.000000\" z=\"0.000000\" rx=\"1.000000 0.000000 0.000000\" ry=\"0.000000 1.000000 0.000000\" rz=\"0.000000 0.000000 1.000000\" \/>","date":"2023-03-30 15:43:46","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6995020508766174, \"perplexity\": 6937.323705337452}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-14\/segments\/1679296949331.26\/warc\/CC-MAIN-20230330132508-20230330162508-00061.warc.gz\"}"} | null | null |
{"url":"http:\/\/pyphi.readthedocs.io\/en\/0.8.1\/api\/node.html","text":"# node\u00b6\n\nRepresents a node in a subsystem. Each node has a unique index, its position in the network\u2019s list of nodes.\n\nclass pyphi.node.Node(subsystem, index, label=None)\n\nA node in a subsystem.\n\nsubsystem\n\nSubsystem \u2013 The subsystem the node belongs to.\n\nindex\n\nint \u2013 The node\u2019s index in the network.\n\nnetwork\n\nNetwork \u2013 The network the node belongs to.\n\nlabel\n\nstr \u2013 An optional label for the node.\n\nstate\n\nint \u2013 The state of this node.\n\nget_marbl(normalize=True)\n\nGenerate a Marbl for this node TPM.\n\ninput_indices\n\nThe indices of nodes which connect to this node.\n\noutput_indices\n\nThe indices of nodes that this node connects to.\n\ninputs\n\nThe set of nodes with connections to this node.\n\noutputs\n\nThe set of nodes this node has connections to.\n\nmarbl\n\nThe normalized representation of this node\u2019s Markov blanket, conditioned on the fixed state of boundary-condition nodes in the current timestep.\n\nraw_marbl\n\nThe un-normalized representation of this node\u2019s Markov blanket, conditioned on the fixed state of boundary-condition nodes in the current timestep.\n\n__eq__(other)\n\nReturn whether this node equals the other object.\n\nTwo nodes are equal if they belong to the same subsystem and have the same index (their TPMs must be the same in that case, so this method doesn\u2019t need to check TPM equality).\n\nLabels are for display only, so two equal nodes may have different labels.\n\nto_json()","date":"2017-11-24 11:05:51","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7434749007225037, \"perplexity\": 2481.1351944873027}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-47\/segments\/1510934807650.44\/warc\/CC-MAIN-20171124104142-20171124124142-00740.warc.gz\"}"} | null | null |
Q: only django work in presence of mod_wsgi , PHP not working I am using PHP with apache2 and MySQL on a rackspace server, it is unmanaged cluoud hosting. I am hosting multiple sites using virtual host.
Now i want to use Django on it. A site needs django so I have installed django and mod_wsgi using apt-get install . Then I wrote following lines in my httpd.conf
WSGIScriptAlias / /var/www/djangosite/pyproject/mysite/mysite/wsgi.py
WSGIPythonPath /var/www/djangosite/pyproject/mysite/
<Directory /var/www/djangosite/pyproject/mysite/mysite/>
<Files wsgi.py>
Order deny,allow
Allow from all
</Files>
</Directory>
But then only django worked on all sites, and only that specific project appear on all sites. so what I need to if I only want it to work for specific site? so that other sites can work as they work previously working with PHP. Do I need to set server name e.t.c. some where?
A: See the AddHandler/RewriteRule described in:
*
*http://code.google.com/p/modwsgi/wiki/ConfigurationGuidelines#The_Apache_Alias_Directive
That approach will give precedence to static files and/or PHP files under DocumentRoot and only if a match isn't found that way will it fallback to the Django application.
Alternatively, mount your Django application at a sub URL and not at the root of the web site.
WSGIScriptAlias /suburl /var/www/djangosite/pyproject/mysite/mysite/wsgi.py
BTW, your configuration above looks wrong anyway. That could only work if the URL you are using is prefixed by '/wsgi.py'. Did you cut and paste wrong and actually mean:
WSGIScriptAlias / /var/www/djangosite/pyproject/mysite/mysite/wsgi.py
IOW, the last argument should have referenced the wsgi.py file and not the directory.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,662 |
Do Video Games Make You Violent?
For decades there have been periodic spikes of outrage as (sometimes misinformed) groups of people protest games that depict varying levels of violence.
These groups just suffer from being human. Afterall, It's human nature to look for patterns.
That was a necessary survival tactic as we were evolving. If other members of our tribe were getting sick as they ate something, or drank from a specific body of water, we had to recognize that "food" or water as something to avoid.
If the tall plains grass always moved in a certain way before a wooly sabre cut struck, we needed to recognize the signs to detect the danger before it arrived.
That search for patterns is still ingrained in us. Nothing fires it up quite like when we think something is inspiring violence. This tendency to avoid danger and search for patterns means a lot of misplaced outrage.
There's a famous line in statistics (once you take the course, or if you've taken statistics already, this line will never leave you): "Correlation does not imply causation."
What this means in a nutshell is that just because two things seem related, it doesn't mean they are the cause of each other.
For example, ice cream sales increase at the same time that violent crimes and murders increase.
But you can pretty safely assume a delicious frozen treat isn't making everyone more likely to commit acts of serious violence.
Instead, there's an influencer playing behind the scenes. It's called a lurking variable.
In this case, it's heat. Heat means people are out and about. Heat can make people horny, bored, and violent.
And the other side of that is, yeah, of course people will buy more ice cream when it's really hot outside.
Nothing New
Video Game violence is nothing new, but as with the increase in sexual content on television, video game violence has become more mainstream.
The boundaries are continually pushed. In 1976 a game called "Death Race" had to be pulled from shelves.
Why? Because people were outraged at the over-the-top and way-too-extreme violence being exhibited in the game. Sounds nasty, right?!
It was a game in black and white, with graphics landing somewhere in between pong and space invaders, nowhere near the photo-realism of today.
The objective was to run over little gremlins, at which point they'd make a little death noise and a gravestone would pop up to represent their demise.
This outraged people so much forty years ago that the game manufacturer actually pulled the game from market.
Video game culture was in its infancy back then, it wasn't the billion dollar powerhouse that it is now.
That same kind of torch-and-pitchfork mentality still exists today. It's another aspect of human condition. We are predisposed to see facts that support our ideological positions.
You probably remember the Sandy Hook tragedy in 2012. Massachusetts pulled arcade games with plastic guns from its rest stops, replacing them with more "appropriate' games.
They found violent games, among other things, on the shooter's computer, and violent games have always been an easy target.
The recent findings
A recent report by the APA found that there is insufficient evidence to show causation between violent video games and the commission of violent crimes.
The study essentially tasked a team with looking at all the academic papers on the violent video game issue in the time period between 2005 and 2013.
What they determined is that video games could actually have a correlation with aggressive behavior, but not necessarily violent behavior.
There are important caveats here. Researchers know that it's not one particular thing that makes people more aggressive. Rather, it is an accumulation of "risk factors" that seems to lead to more aggressive tendencies.
In this case, they found sufficient evidence to list violent video games as a potential risk factor for aggression.
This means violent video games alone aren't enough to make people aggressive, but in conjunction with other risk factors they pose a threat.
The fact that video games aren't enough to cause aggression on their own should be pretty obvious.
If you google "top ten video games of 2015" you'll see games like Witcher 3, Metal Gear Solid V, Bloodborne, Batman: Arkham Knight, and Fallout 4.
All of these games require you to neutralize opponents, either human characters or humanoid characters.
And with millions of copies sold, played in households that sometimes have more than one person playing the same copy of the game, we don't see millions of people becoming aggressive all of a sudden.
Facts Worth Noting
The article-review task force cited some insufficiency in the studies they looked reviewed.
For example, there's no separation of outcomes for male and female audiences. Is one gender more likely to adopt the aggression from playing aggressive games?
A whole host of the studies look at children aged ten or younger. While this could be valuable for examining early impacts of violent games, it does little to show what happens as children develop.
A child at 7 and the same child at 14 are two very different people, and a lot of the research doesn't account for that development.
New perceptions of importance and new influences are added into a child's life constantly (like puberty and the influence of sexual attraction) in those early days, and these studies do little to examine that.
So the APA has determined violent video games are a risk factor in aggressive behavior.
Games are only one piece of the convoluted puzzle that could, in theory, make for an aggressive child.
Does aggression equate to violence? In short, no. There is no indication that any violence comes from video games. There are so many lurking variables compounding these studies.
The argument goes back to correlation and causation. It could be that aggressive people have an increased interest in violent games.
What about the effect of non-violent games on behavior? Where do games really fit in within the grand scheme of human behavior? Do game designers have an obligation to make different games if research continues to show behavioral effects?
For now all the APA has done is appeal to the ESRB. The APA wants the rating board to adjust the game-rating scale to reflect these recent findings.
Would that be such a bad thing? Is it better to redraw the lines of acceptable content for certain age groups? Will the future see an attempt at serious game censorship?
For now, thankfully, the research doesn't support any extreme moves or regulation against violent video game content.
Questions Related to Gaming:
Will VR Change Game Design in The Future?
What is AbleGamers and Why is it Important?
Does Game Design Spoil the Fun of Playing?
Console vs PC Game Designing: What's the Difference?
Learn Game Design | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,339 |
// Copyright (c) Microsoft. All Rights Reserved. Licensed under the Apache License, Version 2.0. See License.txt in the project root for license information.
using System.Collections.Immutable;
using System.Diagnostics;
using System.Threading;
using Microsoft.CodeAnalysis.CSharp.Symbols;
using Microsoft.CodeAnalysis.CSharp.Syntax;
using Microsoft.CodeAnalysis.PooledObjects;
using Microsoft.CodeAnalysis.Text;
using Roslyn.Utilities;
using System.Collections.Generic;
using System;
namespace Microsoft.CodeAnalysis.CSharp.Symbols
{
internal abstract class SourceDelegateMethodSymbol : SourceMemberMethodSymbol
{
private ImmutableArray<ParameterSymbol> _parameters;
private readonly TypeSymbolWithAnnotations _returnType;
protected SourceDelegateMethodSymbol(
SourceMemberContainerTypeSymbol delegateType,
TypeSymbolWithAnnotations returnType,
DelegateDeclarationSyntax syntax,
MethodKind methodKind,
DeclarationModifiers declarationModifiers)
: base(delegateType, syntax.GetReference(), location: syntax.Identifier.GetLocation())
{
_returnType = returnType;
this.MakeFlags(methodKind, declarationModifiers, _returnType.SpecialType == SpecialType.System_Void, isExtensionMethod: false);
}
protected void InitializeParameters(ImmutableArray<ParameterSymbol> parameters)
{
Debug.Assert(_parameters.IsDefault);
_parameters = parameters;
}
internal static void AddDelegateMembers(
SourceMemberContainerTypeSymbol delegateType,
ArrayBuilder<Symbol> symbols,
DelegateDeclarationSyntax syntax,
DiagnosticBag diagnostics)
{
var compilation = delegateType.DeclaringCompilation;
Binder binder = delegateType.GetBinder(syntax.ParameterList);
RefKind refKind;
TypeSyntax returnTypeSyntax = syntax.ReturnType.SkipRef(out refKind);
var returnType = binder.BindType(returnTypeSyntax, diagnostics);
// reuse types to avoid reporting duplicate errors if missing:
var voidType = TypeSymbolWithAnnotations.Create(binder.GetSpecialType(SpecialType.System_Void, diagnostics, syntax));
// https://github.com/dotnet/roslyn/issues/30079: Should the 'object', IAsyncResult and AsyncCallback parameters be considered nullable or not nullable?
var objectType = TypeSymbolWithAnnotations.Create(binder.GetSpecialType(SpecialType.System_Object, diagnostics, syntax));
var intPtrType = TypeSymbolWithAnnotations.Create(binder.GetSpecialType(SpecialType.System_IntPtr, diagnostics, syntax));
if (returnType.IsRestrictedType(ignoreSpanLikeTypes: true))
{
// Method or delegate cannot return type '{0}'
diagnostics.Add(ErrorCode.ERR_MethodReturnCantBeRefAny, returnTypeSyntax.Location, returnType.TypeSymbol);
}
// A delegate has the following members: (see CLI spec 13.6)
// (1) a method named Invoke with the specified signature
var invoke = new InvokeMethod(delegateType, refKind, returnType, syntax, binder, diagnostics);
invoke.CheckDelegateVarianceSafety(diagnostics);
symbols.Add(invoke);
// (2) a constructor with argument types (object, System.IntPtr)
symbols.Add(new Constructor(delegateType, voidType, objectType, intPtrType, syntax));
if (binder.Compilation.GetSpecialType(SpecialType.System_IAsyncResult).TypeKind != TypeKind.Error &&
binder.Compilation.GetSpecialType(SpecialType.System_AsyncCallback).TypeKind != TypeKind.Error &&
// WinRT delegates don't have Begin/EndInvoke methods
!delegateType.IsCompilationOutputWinMdObj())
{
var iAsyncResultType = TypeSymbolWithAnnotations.Create(binder.GetSpecialType(SpecialType.System_IAsyncResult, diagnostics, syntax));
var asyncCallbackType = TypeSymbolWithAnnotations.Create(binder.GetSpecialType(SpecialType.System_AsyncCallback, diagnostics, syntax));
// (3) BeginInvoke
symbols.Add(new BeginInvokeMethod(invoke, iAsyncResultType, objectType, asyncCallbackType, syntax));
// and (4) EndInvoke methods
symbols.Add(new EndInvokeMethod(invoke, iAsyncResultType, syntax));
}
if (delegateType.DeclaredAccessibility <= Accessibility.Private)
{
return;
}
HashSet<DiagnosticInfo> useSiteDiagnostics = null;
if (!delegateType.IsNoMoreVisibleThan(invoke.ReturnType, ref useSiteDiagnostics))
{
// Inconsistent accessibility: return type '{1}' is less accessible than delegate '{0}'
diagnostics.Add(ErrorCode.ERR_BadVisDelegateReturn, delegateType.Locations[0], delegateType, invoke.ReturnType.TypeSymbol);
}
foreach (var parameter in invoke.Parameters)
{
if (!parameter.Type.IsAtLeastAsVisibleAs(delegateType, ref useSiteDiagnostics))
{
// Inconsistent accessibility: parameter type '{1}' is less accessible than delegate '{0}'
diagnostics.Add(ErrorCode.ERR_BadVisDelegateParam, delegateType.Locations[0], delegateType, parameter.Type.TypeSymbol);
}
}
diagnostics.Add(delegateType.Locations[0], useSiteDiagnostics);
}
protected override void MethodChecks(DiagnosticBag diagnostics)
{
// TODO: move more functionality into here, making these symbols more lazy
}
public sealed override bool IsVararg
{
get
{
return false;
}
}
public sealed override ImmutableArray<ParameterSymbol> Parameters
{
get
{
return _parameters;
}
}
public override ImmutableArray<TypeParameterSymbol> TypeParameters
{
get
{
return ImmutableArray<TypeParameterSymbol>.Empty;
}
}
public override ImmutableArray<TypeParameterConstraintClause> GetTypeParameterConstraintClauses(bool early)
=> ImmutableArray<TypeParameterConstraintClause>.Empty;
public sealed override TypeSymbolWithAnnotations ReturnType
{
get
{
return _returnType;
}
}
public sealed override bool IsImplicitlyDeclared
{
get
{
return true;
}
}
internal override bool IsExpressionBodied
{
get { return false; }
}
internal override bool GenerateDebugInfo
{
get { return false; }
}
protected sealed override IAttributeTargetSymbol AttributeOwner
{
get
{
return (SourceNamedTypeSymbol)ContainingSymbol;
}
}
internal sealed override System.Reflection.MethodImplAttributes ImplementationAttributes
{
get { return System.Reflection.MethodImplAttributes.Runtime; }
}
internal sealed override OneOrMany<SyntaxList<AttributeListSyntax>> GetAttributeDeclarations()
{
// TODO: This implementation looks strange. It might make sense for the Invoke method, but
// not for constructor and other methods.
return OneOrMany.Create(((SourceNamedTypeSymbol)ContainingSymbol).GetAttributeDeclarations());
}
internal sealed override System.AttributeTargets GetAttributeTarget()
{
return System.AttributeTargets.Delegate;
}
private sealed class Constructor : SourceDelegateMethodSymbol
{
internal Constructor(
SourceMemberContainerTypeSymbol delegateType,
TypeSymbolWithAnnotations voidType,
TypeSymbolWithAnnotations objectType,
TypeSymbolWithAnnotations intPtrType,
DelegateDeclarationSyntax syntax)
: base(delegateType, voidType, syntax, MethodKind.Constructor, DeclarationModifiers.Public)
{
InitializeParameters(ImmutableArray.Create<ParameterSymbol>(
SynthesizedParameterSymbol.Create(this, objectType, 0, RefKind.None, "object"),
SynthesizedParameterSymbol.Create(this, intPtrType, 1, RefKind.None, "method")));
}
public override string Name
{
get { return WellKnownMemberNames.InstanceConstructorName; }
}
public override RefKind RefKind
{
get { return RefKind.None; }
}
internal override OneOrMany<SyntaxList<AttributeListSyntax>> GetReturnTypeAttributeDeclarations()
{
// Constructors don't have return type attributes
return OneOrMany.Create(default(SyntaxList<AttributeListSyntax>));
}
internal override LexicalSortKey GetLexicalSortKey()
{
// associate "Invoke and .ctor" with whole delegate declaration for the sorting purposes
// other methods will be associated with delegate's identifier
// we want this just to keep the order of synthesized methods the same as in Dev12
// Dev12 order is not strictly alphabetical - .ctor and Invoke go before other members.
// there are no real reasons for emitting the members in one order or another,
// so we will keep them the same.
return new LexicalSortKey(this.syntaxReferenceOpt.GetLocation(), this.DeclaringCompilation);
}
}
private sealed class InvokeMethod : SourceDelegateMethodSymbol
{
private readonly RefKind _refKind;
private readonly ImmutableArray<CustomModifier> _refCustomModifiers;
internal InvokeMethod(
SourceMemberContainerTypeSymbol delegateType,
RefKind refKind,
TypeSymbolWithAnnotations returnType,
DelegateDeclarationSyntax syntax,
Binder binder,
DiagnosticBag diagnostics)
: base(delegateType, returnType, syntax, MethodKind.DelegateInvoke, DeclarationModifiers.Virtual | DeclarationModifiers.Public)
{
this._refKind = refKind;
SyntaxToken arglistToken;
var parameters = ParameterHelpers.MakeParameters(
binder, this, syntax.ParameterList, out arglistToken,
allowRefOrOut: true,
allowThis: false,
addRefReadOnlyModifier: true,
diagnostics: diagnostics);
if (arglistToken.Kind() == SyntaxKind.ArgListKeyword)
{
// This is a parse-time error in the native compiler; it is a semantic analysis error in Roslyn.
// error CS1669: __arglist is not valid in this context
diagnostics.Add(ErrorCode.ERR_IllegalVarArgs, new SourceLocation(arglistToken));
}
if (_refKind == RefKind.RefReadOnly)
{
var modifierType = binder.GetWellKnownType(WellKnownType.System_Runtime_InteropServices_InAttribute, diagnostics, syntax.ReturnType);
_refCustomModifiers = ImmutableArray.Create(CSharpCustomModifier.CreateRequired(modifierType));
}
else
{
_refCustomModifiers = ImmutableArray<CustomModifier>.Empty;
}
InitializeParameters(parameters);
}
public override string Name
{
get { return WellKnownMemberNames.DelegateInvokeName; }
}
public override RefKind RefKind
{
get { return _refKind; }
}
internal override LexicalSortKey GetLexicalSortKey()
{
// associate "Invoke and .ctor" with whole delegate declaration for the sorting purposes
// other methods will be associated with delegate's identifier
// we want this just to keep the order of synthesized methods the same as in Dev12
// Dev12 order is not strictly alphabetical - .ctor and Invoke go before other members.
// there are no real reasons for emitting the members in one order or another,
// so we will keep them the same.
return new LexicalSortKey(this.syntaxReferenceOpt.GetLocation(), this.DeclaringCompilation);
}
internal override void AfterAddingTypeMembersChecks(ConversionsBase conversions, DiagnosticBag diagnostics)
{
var syntax = (DelegateDeclarationSyntax)SyntaxRef.GetSyntax();
var location = syntax.ReturnType.GetLocation();
Debug.Assert(location != null);
base.AfterAddingTypeMembersChecks(conversions, diagnostics);
if (_refKind == RefKind.RefReadOnly)
{
DeclaringCompilation.EnsureIsReadOnlyAttributeExists(diagnostics, location, modifyCompilation: true);
}
ParameterHelpers.EnsureIsReadOnlyAttributeExists(Parameters, diagnostics, modifyCompilation: true);
if (ReturnType.NeedsNullableAttribute())
{
this.DeclaringCompilation.EnsureNullableAttributeExists(diagnostics, location, modifyCompilation: true);
}
ParameterHelpers.EnsureNullableAttributeExists(Parameters, diagnostics, modifyCompilation: true);
}
public override ImmutableArray<CustomModifier> RefCustomModifiers => _refCustomModifiers;
}
private sealed class BeginInvokeMethod : SourceDelegateMethodSymbol
{
internal BeginInvokeMethod(
InvokeMethod invoke,
TypeSymbolWithAnnotations iAsyncResultType,
TypeSymbolWithAnnotations objectType,
TypeSymbolWithAnnotations asyncCallbackType,
DelegateDeclarationSyntax syntax)
: base((SourceNamedTypeSymbol)invoke.ContainingType, iAsyncResultType, syntax, MethodKind.Ordinary, DeclarationModifiers.Virtual | DeclarationModifiers.Public)
{
var parameters = ArrayBuilder<ParameterSymbol>.GetInstance();
foreach (SourceParameterSymbol p in invoke.Parameters)
{
var synthesizedParam = new SourceClonedParameterSymbol(originalParam: p, newOwner: this, newOrdinal: p.Ordinal, suppressOptional: true);
parameters.Add(synthesizedParam);
}
int paramCount = invoke.ParameterCount;
parameters.Add(SynthesizedParameterSymbol.Create(this, asyncCallbackType, paramCount, RefKind.None, GetUniqueParameterName(parameters, "callback")));
parameters.Add(SynthesizedParameterSymbol.Create(this, objectType, paramCount + 1, RefKind.None, GetUniqueParameterName(parameters, "object")));
InitializeParameters(parameters.ToImmutableAndFree());
}
public override string Name
{
get { return WellKnownMemberNames.DelegateBeginInvokeName; }
}
public override RefKind RefKind
{
get { return RefKind.None; }
}
internal override OneOrMany<SyntaxList<AttributeListSyntax>> GetReturnTypeAttributeDeclarations()
{
// BeginInvoke method doesn't have return type attributes
// because it doesn't inherit Delegate declaration's return type.
// It has a special return type: SpecialType.System.IAsyncResult.
return OneOrMany.Create(default(SyntaxList<AttributeListSyntax>));
}
}
private sealed class EndInvokeMethod : SourceDelegateMethodSymbol
{
private readonly InvokeMethod _invoke;
internal EndInvokeMethod(
InvokeMethod invoke,
TypeSymbolWithAnnotations iAsyncResultType,
DelegateDeclarationSyntax syntax)
: base((SourceNamedTypeSymbol)invoke.ContainingType, invoke.ReturnType, syntax, MethodKind.Ordinary, DeclarationModifiers.Virtual | DeclarationModifiers.Public)
{
_invoke = invoke;
var parameters = ArrayBuilder<ParameterSymbol>.GetInstance();
int ordinal = 0;
foreach (SourceParameterSymbol p in invoke.Parameters)
{
if (p.RefKind != RefKind.None)
{
var synthesizedParam = new SourceClonedParameterSymbol(originalParam: p, newOwner: this, newOrdinal: ordinal++, suppressOptional: true);
parameters.Add(synthesizedParam);
}
}
parameters.Add(SynthesizedParameterSymbol.Create(this, iAsyncResultType, ordinal++, RefKind.None, GetUniqueParameterName(parameters, "result")));
InitializeParameters(parameters.ToImmutableAndFree());
}
protected override SourceMemberMethodSymbol BoundAttributesSource => _invoke;
public override string Name => WellKnownMemberNames.DelegateEndInvokeName;
public override RefKind RefKind => _invoke.RefKind;
public override ImmutableArray<CustomModifier> RefCustomModifiers => _invoke.RefCustomModifiers;
}
private static string GetUniqueParameterName(ArrayBuilder<ParameterSymbol> currentParameters, string name)
{
while (!IsUnique(currentParameters, name))
{
name = "__" + name;
}
return name;
}
private static bool IsUnique(ArrayBuilder<ParameterSymbol> currentParameters, string name)
{
foreach (var p in currentParameters)
{
if (string.CompareOrdinal(p.Name, name) == 0)
{
return false;
}
}
return true;
}
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,953 |
Q: How to send all options from a checkbox list/radio box list/ select option list to server along with the selected one by php? I have a code what has a simple select/options and some check boxes i want to send to server using PHP script, all the options including the option selected ( so i cant make them all selected ) so how do i do it?
A: Keep your dropdown as it is.
<select id="location" name="location">
.
.
.
.
</seelct>
Then add another one dropdown as hidden (display:none;) without any options. This dropdown will be filled with options when clicking submit button.
On submit button click action add below js code,
In js:
$('#location option').clone().appendTo('#near_location');// copied all options
var index = $('#location ').get(0).selectedIndex; // remove selected
$('#near_location option:eq(' + index + ')').remove();
return true;
Now you will get the selected option in $_POST["location"] and un-selected values in $_POST["near_loaction"].
A: You can save generated options on the server via session or database and load it after submit.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,832 |
Our client are a rapidly growing accountancy firm based in Newcastle who are looking to recruit a Business Development Executive who can assist the firm's growth and business development objectives and maintain relationships with existing and new clients.
The role will be office based and provides an excellent opportunity to progress your career in business development.
Developing and growing business relationships with both clients and advisers though account management.
Building and managing a pipeline of advisers and clients and identifying opportunities to discuss appropriate solutions and cross-sell services through initial fact find and outbound activity.
Teaming and liaising with Associate Directors to assist with lead generation, booking new appointments, maintaining regular follow ups and contact, and assisting Associate Directors through to client conversion and closure of sales.
Sourcing new opportunities through lead follow-up and new outbound calls to advisers and clients (including existing contacts and prospect contacts).
Route qualified opportunities to the appropriate Associate Director for further development and conversion to telephone appointments and meetings.
Liaise with Associate Directors to ensure leads are appropriately followed up and additional follow ups are diarised in order to actively keep in contact.
Re-engaging any previous lost opportunities through fact find.
Maintain and expand your database of prospects, working with the Marketing department to ensure there are sufficient leads and firms to call, and suggesting ideas for future campaigns.
Inputting information into appropriate software for meetings and liaise with PAs.
Maintaining new client relationships and developing existing relationships. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,071 |
Looking outside the window, I see grey skies and pouring rain – a sure sign that the school summer holidays have begun! Just what we need in these days of rising unemployment and financial doom and gloom, more commonly referred to as 'the credit crunch' – when more and more families are choosing to holiday at home rather than risk the falling value of the pound and the growing chance of catching swine flu from a fellow airline passenger.
And so we see the return of that staple of the 'holiday on the cheap' which remains a big part of my summer holiday memories from yesteryear – the day out!
The good news is, that come rain or shine, everyone who visits Stratford will find a host of things to do and see, even if they are only coming for the day. Those on a tight budget can enjoy a great day simply walking along the river Avon (with an ice-cream if the funds will stretch) soaking up the history and beautiful scenery which the locals all seems to take for granted.
Boats trips are available; you can take a river cruise or hire a small craft and navigate yourself along the river. Look out for the Holy Trinity Church, where Shakespeare himself is buried, and you won't fail to spot the site of the new theatre, currently being rebuilt on the site of the old one and due for completion in 2010.
The RSC is still in full swing, their operations are centre around the Courtyard Theatre while the building work is still in progress, and their standard is as high as ever, so theatre-lovers must take in one of the many performances.
Whilst on the subject of the Bard, there are, of course, all the Shakespeare properties to visit, three of them in the town itself within easy walking distance, and Mary Aden's house in Wilmcote and Anne Hathaway's cottage in Shottery are just a short bus-ride or drive away. If the weather improves, you might fancy taking the open-top tour bus which visits all the properties regularly through the day.
For young visitors Stratford offers the Butterfly Farm, located close to the river, and the Brass Rubbings Centre. Stratford is, of course, home to Ragdoll Productions, makers of the Teletubbies and Rosie and Jim, so make sure you visit them while you are here. And for children of all ages a great time awaits at the Falstaff's Experience, Creaky Cauldron and Harvard House.
You might choose to browse the many shops in the town centre, or stop for a drink and bite to eat at one of the numerous pubs, tea-rooms and restaurants which can be found all over the town.
With so much to see and do in the town, many visitors want to stay for a few days to make the most of their trip. There are many places to stay, from campsites, caravan parks, B&Bs, guest houses and lovely hotels. And of course, Stratford is close to many other areas of interest, with the Cotswolds on our doorstep, Warwick Castle just a stone's throw away and the city of Birmingham less than twenty miles away with all that you would expect from the country's second city, including one of my personal favourite's Cadbury World!
So what are you waiting for? Don't worry about the weather, we've got lots of shops selling brollies if you forget to bring one, but likely as not, as soon as you get here the sun will come out. Have a lovely day! | {
"redpajama_set_name": "RedPajamaC4"
} | 2,771 |
The Tee Jay Quarter Horse Ranch is located in Central Kansas at the home of Duane and Jo Walker. The ranch was home to many outstanding horses but the most remembered and loved was Jackie Bee.
In the late 1960's and early 1970's Jackie Bee worked his way into the halter horse game by producing horses that carried quite a bit of muscle on a large frame. These horses also had good, stout feet and legs. In the 1970's Jackie Bees offspring, Tee Jay Janie, Tee Jay Badger Bee, and Tee J Black Jill, just to name a few, dominated the halter horse world. Many of these horses by Jackie Bee earn many awards including AQHA World Champion, AQHA High-Point, and Superior halter horse.
Jackie Bee was also known for being a sire of performance horses. In the 1980's the look of halter horse changed and became more refined. Since Jackie Bees were always bred to work the demand for performance horses in the arena grew. Horses such as: Tee J Ro Bee, Tee Jay Roman, Tee Jay Jack Steel, Tee J Robin Bee and many more, went on to earn new titles in the arena. Included in those titles were AQHA Champion, AQHA High-Point and point earners in many performance events such as, calf roping, reining working cow horse western pleasure.
After Jackie Bee passed away in 1990 at the age of 28 the ranch remains to keep Jackie Bee in the there breeding program. With many daughters and granddaughter of Jackie Bee the Tee Jay Quarter Horse Ranch keeps producing the traits that he was known for. When you look at the horses around the ranch you will find mostly grays that are gentle, well muscled and have good bone structure. In the arena, the halter ring or on the ranch they are ready to perform their job.
For more information on these horses please continue looking at the site or contact Duane or Trevin at the e-mail address or numbers listed below. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,081 |
Q: Android: Overlay TextView on LockScreen
Im trying to overlay a TextView on top of the LockScreen (Similar to how Android Overlays the time).
Note: I dont want to by-pass the lockscreen, but just draw on top of it (without interfering with any of the touch events).
I've tried using the following flags (in onCreate):
getWindow().addFlags(WindowManager.LayoutParams.FLAG_NOT_FOCUSABLE);
getWindow().addFlags(WindowManager.LayoutParams.FLAG_NOT_TOUCHABLE);
getWindow().addFlags(WindowManager.LayoutParams.FLAG_SHOW_WHEN_LOCKED);
getWindow().addFlags(WindowManager.LayoutParams.FLAG_KEEP_SCREEN_ON);
getWindow().addFlags(WindowManager.LayoutParams.TYPE_SYSTEM_OVERLAY);
getWindow().addFlags(WindowManager.LayoutParams.FLAG_WATCH_OUTSIDE_TOUCH);
getWindow().addFlags(PixelFormat.TRANSLUCENT);
And applying the following Theme (to the specific activity):
<style name="Transparent">
<item name="android:windowNoTitle">true</item>
<item name="android:windowContentOverlay">@null</item>
<item name="android:windowIsTranslucent">true</item>
<item name="android:windowBackground">@android:color/transparent</item>
<item name="android:windowActionBar">false</item>
<item name="android:backgroundDimEnabled">false</item>
<item name="android:windowIsFloating">true</item>
</style>
But this draws on TOP of the lockscreen hiding the lockscreen and disabling all touch events.
Edit: activity_overlay.xml
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:paddingBottom="@dimen/activity_vertical_margin"
android:paddingLeft="@dimen/activity_horizontal_margin"
android:paddingRight="@dimen/activity_horizontal_margin"
android:paddingTop="@dimen/activity_vertical_margin"
tools:context="com.example.coco.MainActivity" >
<TextView
android:id="@+id/textView1"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:textColor="@android:color/white"
android:layout_alignParentBottom="true"
android:text="TextView" />
</RelativeLayout>
Manifest declaration of Activity (which inflates overlay_activity.xml)
<activity
android:name=".DisplayActivity"
android:label="@string/app_name"
android:theme="@style/Transparent" />
A: Starting with Android Lollipop (5.0), support for lockscreen widgets has been removed (see the very bottom). Instead, you should be using Notifications, which can now appear on the lockscreen.
A: Since you want to display things out of your app activity, you can use a Service and WindowManager for that, the same way how Facebook Messenger and other floating windows apps works ;)
LockScreenTextService.class
import android.app.Service;
import android.content.BroadcastReceiver;
import android.content.Context;
import android.content.Intent;
import android.content.IntentFilter;
import android.graphics.PixelFormat;
import android.os.IBinder;
import android.support.v4.content.ContextCompat;
import android.view.Gravity;
import android.view.WindowManager;
import android.widget.TextView;
/**
* Created on 2/20/2016.
*/
public class LockScreenTextService extends Service {
private BroadcastReceiver mReceiver;
private boolean isShowing = false;
@Override
public IBinder onBind(Intent intent) {
// TODO Auto-generated method stub
return null;
}
private WindowManager windowManager;
private TextView textview;
WindowManager.LayoutParams params;
@Override
public void onCreate() {
super.onCreate();
windowManager = (WindowManager)getSystemService(WINDOW_SERVICE);
//add textview and its properties
textview = new TextView(this);
textview.setText("Hello There!");
textview.setTextColor(ContextCompat.getColor(this, android.R.color.white));
textview.setTextSize(32f);
//set parameters for the textview
params = new WindowManager.LayoutParams(
WindowManager.LayoutParams.WRAP_CONTENT,
WindowManager.LayoutParams.WRAP_CONTENT,
WindowManager.LayoutParams.TYPE_SYSTEM_OVERLAY,
WindowManager.LayoutParams.FLAG_NOT_FOCUSABLE
| WindowManager.LayoutParams.FLAG_NOT_TOUCH_MODAL
| WindowManager.LayoutParams.FLAG_SHOW_WHEN_LOCKED
| WindowManager.LayoutParams.FLAG_WATCH_OUTSIDE_TOUCH,
PixelFormat.TRANSLUCENT);
params.gravity = Gravity.BOTTOM;
//Register receiver for determining screen off and if user is present
mReceiver = new LockScreenStateReceiver();
IntentFilter filter = new IntentFilter(Intent.ACTION_SCREEN_OFF);
filter.addAction(Intent.ACTION_USER_PRESENT);
registerReceiver(mReceiver, filter);
}
@Override
public int onStartCommand(Intent intent, int flags, int startId) {
return START_STICKY;
}
public class LockScreenStateReceiver extends BroadcastReceiver {
@Override
public void onReceive(Context context, Intent intent) {
if (intent.getAction().equals(Intent.ACTION_SCREEN_OFF)) {
//if screen is turn off show the textview
if (!isShowing) {
windowManager.addView(textview, params);
isShowing = true;
}
}
else if(intent.getAction().equals(Intent.ACTION_USER_PRESENT)) {
//Handle resuming events if user is present/screen is unlocked remove the textview immediately
if (isShowing) {
windowManager.removeViewImmediate(textview);
isShowing = false;
}
}
}
}
@Override
public void onDestroy() {
//unregister receiver when the service is destroy
if (mReceiver != null) {
unregisterReceiver(mReceiver);
}
//remove view if it is showing and the service is destroy
if (isShowing) {
windowManager.removeViewImmediate(textview);
isShowing = false;
}
super.onDestroy();
}
}
and add the necessary permission on AndroidManifest.xml and add the service
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.example.textonlockscreenusingservice">
<uses-permission android:name="android.permission.SYSTEM_ALERT_WINDOW"/> //this
<application
android:allowBackup="true"
android:icon="@mipmap/ic_launcher"
android:label="@string/app_name"
android:supportsRtl="true"
android:theme="@style/AppTheme">
<activity
android:name=".MainActivity"
android:label="@string/app_name"
android:theme="@style/AppTheme.NoActionBar">
<intent-filter>
<action android:name="android.intent.action.MAIN" />
<category android:name="android.intent.category.LAUNCHER" />
</intent-filter>
</activity>
<service android:name=".LockScreenTextService" /> //this
</application>
</manifest>
Don't forget to start the service on onCreate() of your activity
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
Intent intent = new Intent(this, LockScreenTextService.class);
startService(intent);
}
TextView showing on lockscreen
TextView removed when unlocked
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,412 |
Drought and the Origin...
Drought and the Origins of the Mexican Revolution
Mikael D. Wolfe
History of Mexico, Environmental History, Revolutions and Rebellions
Meteorology: Porfirian Government Reports and Newspapers versus Modern Historical Climatological Data Reconstructions
The Drought Shortly Preceding the Revolution as Catalyst for Rebellion
What role did drought play in the outbreak of the Mexican Revolution of 1910? Although historians of the Mexican Revolution acknowledge that the effects of drought helped catalyze it, they have not explored in any depth what connects drought to revolution. Instead, they usually subsume it within a more general discussion of agricultural cycles to explain the conduct and fortunes of popular revolutionary armies. In particular, they reference the onset of drought between 1907 and 1909 as exacerbating an economic downturn induced by severe recession in the United States. By then, Mexico had become economically integrated with its northern neighbor through rapidly growing foreign investment, trade, and cross-border migration facilitated by the railroad transportation revolution. These socioeconomic and ecological factors together led to steep declines in wages and earnings, devastating crop failures, spikes in food prices (principally corn and beans), and even famine in the lower and middle classes.
Although suggestive, such passing references to drought in the historiography of the revolution do not furnish a clear picture of its effects and how they may have contributed to social and political conflict. In the 21st century, new technologies, methods, and sources—from historical meteorological reports and climate-related accounts gleaned from archival sources to modern historical climatological data reconstructions—facilitate doing more rigorous climate history. This article provides a sampling of these methods and sources on the role of drought in late 19th- and early 20th-century Mexico that can supplement, elucidate, and even revise our understanding of the origins of the Mexican Revolution.
Keywords: Mexico, Mexican Revolution, Chihuahua, Tomochi, Laguna, climate, weather, drought
Department of History, Stanford University
Access to the complete content on Oxford Research Encyclopedia of Latin American History requires a subscription or purchase. Public users are able to search the site and view the abstracts and keywords for each book and chapter without a subscription. If you are a student or academic complete our librarian recommendation form to recommend the Oxford Research Encyclopedias to your librarians for an institutional free trial. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,034 |
There's a new reason to get up in the morning (or midmorning, or afternoon) if you live near Mattituck. The North Fork Doughnut Company, affectionately known as NOFODOCO, opened its doors on Saturday and sold almost 2,000 doughnuts by Sunday afternoon. Owners Jimmy Lyons and Kelly Briguccia were pleased, if a little stressed out, by the enthusiastic community response. "We were cooking doughnuts from 4 a.m. to 7 p.m. nonstop, with people throwing bottles of water to us every once in a while so we could stay hydrated," Lyons said.
The Yaphank couple decided on Mattituck for their new business after falling in love on the North Fork and coming to appreciate the area's sense of community and slower pace.
When the former Yogurt Expressions space became available on Main Road, it was the perfect fit. Lyons, who worked in the craft beer world before becoming the production manager at Mast Brothers Chocolate in Brooklyn, had plenty of experience with artisan food. Briguccia was an avid home baker.
Together they experimented with some of her mother's doughnut recipes, discovering that they worked well together in the kitchen. Now they are building a family business.Their 6-year-old daughter, Teigen, has already stepped up, taking control of the cold brew coffee dispenser and requesting a special stool behind the counter so she can see and talk to customers. Once the business takes off, the family plans to make Mattituck their home.
For the summer, there will be glazed doughnut ice cream sandwiches made with Magic Fountain ice cream. Hot coffee is from Kelly and Jimmy's friends at North Fork Roasting Company. Sail Away cold brew and nitro coffee are on draft.
North Fork Doughnut Company is open Monday to Friday 9 a.m. to 5 p.m., and Saturday and Sunday from 9 in the morning to sell out, usually between 5 and 7 p.m. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,484 |
NGC 1636 est une galaxie spirale (barrée ?) située dans la constellation de l'Éridan. Sa vitesse par rapport au fond diffus cosmologique est de , ce qui correspond à une distance de Hubble de . Elle a été découverte par l'astronome germano-britannique William Herschel en 1786.
La classe de luminosité de NGC 1636 est I-II et elle présente une large raie HI. C'est aussi une galaxie active de type Seyfert 2.
Notes et références
Voir aussi
Articles connexes
Liste des objets du NGC
Liens externes
NGC 1636 sur spider.seds.org
NGC 1636 sur la base de données LEDA
NGC 1636 sur WikiSky
NGC 1636 sur le site du professeur C. Seligman
1636
Constellation de l'Éridan
Galaxie spirale
Galaxie spirale barrée
Galaxie active
Galaxie de Seyfert
Objet céleste découvert en 1786
Objet céleste découvert par William Herschel | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,737 |
\subsection{The Proposed \ac{CndG} Solver}
To make our approach efficient, we rely on the following Assumption:
\begin{assumption}\label{ass:LMO}
For any $c\in\mathsf{E}^{\ast}$, the auxiliary problem
\begin{equation}\label{eq:LMO}
\mathcal{L}_{g}(c):=\argmin_{s\in\mathsf{X}}\{\inner{c,s}+g(s)\}
\end{equation}
is easily solvable.
\end{assumption}
We can compute the gradient and Hessian of $F$ as
\[
F'(x)=f'(\mathcal{P}(x))\mathcal{P}'(x),\quad F''(x)[s,t]=f''(\mathcal{P}(x))[\mathcal{P}(s),\mathcal{P}(t)]\quad \forall x\in\Int\mathcal{P}^{-1}(\mathsf{K}).
\]
This means that we obtain a local norm on $\mathsf{H}$ given by
\[
\norm{w}_{x}=F''(x)[w,w]^{1/2}\qquad \forall (x,w)\in\Int\mathcal{P}^{-1}(\mathsf{K})\times\mathsf{H}.
\]
Note that in order to evaluate the local norm we do not need to compute the full Hessian $F''(x)$. It only requires a directional derivative, which is potentially easy to do numerically.
\begin{algorithm}[t]
\caption{$\CG(x^{0},\varepsilon,t)$}
\label{alg:CndG}
\begin{algorithmic}
\STATE {\bfseries Input: } $(x^{0},t)\in\mathsf{C}\times(0,\infty)$ initial state; $\varepsilon>0$ accuracy level
\FOR{$k=0,1,\ldots$}
\IF{$\gap_{t}(x^{k})>\varepsilon$}
\STATE Obtain $s^{k}=s_{t}(x^{k})$ defined in \eqref{eq:s}.
\STATE $\alpha_{k}=\alpha_{t}(x^{k})$ defined in \eqref{eq:alpha}.
\STATE Set $x^{k+1}=x^{k}+\alpha_{k}(s^{k}-x^{k})$.
\ELSE
\STATE Return $x^{k}$.
\ENDIF
\ENDFOR
\end{algorithmic}
\end{algorithm}
Define the vector field
\begin{equation}\label{eq:s}
s_{t}(x)\in\mathcal{L}_{g}(t^{-1}F'(x)).
\end{equation}
Note that our analysis does not rely on a specific tie-breaking rule, so any proposal of the oracle \eqref{eq:LMO} will be acceptable. To measure solution accuracy and overall algorithmic progress, we introduce two \emph{merit functions}:
\begin{align}
\gap_{t}(x)&:=t^{-1}F'(x)[x-s_{t}(x)]+g(x)-g(s_{t}(x)),\text{ and }\\
\Delta_{t}(x)&:=V_{t}(x)-V_{t}(z^{\ast}(t)).
\end{align}
Note that $\gap_{t}(x)\geq 0$ and $\Delta_{t}(x)\geq 0$ for all $x\in\dom(F)$. Convexity together with the definition of the point $z^{\ast}(t)$, gives
\begin{align*}
0&\leq \Delta_{t}(x)=t^{-1}[F(x)-F(z^{\ast}(t))]+g(x)-g(z^{\ast}(t))\\
&\leq t^{-1}F'(x)[x-z^{\ast}(t)]+g(x)-g(z^{\ast}(t))\\
&\leq t^{-1}F'(x)[x-s_{t}(x)]+g(x)-g(s_{t}(x))=\gap_{t}(x).
\end{align*}
Define
\begin{equation}
\mathtt{e}_{t}(x):=\norm{s_{t}(x)-x}_{x}.
\end{equation}
Then, for $\alpha\in(0,\min\{1,1/\mathtt{e}_{t}(x)\})$, we get from eq. \eqref{eq:descent}
$$
F(x+\alpha(s_{t}(x)-x))\leq F(x)+\alpha F'(x)[s_{t}(x)-x]+\omega_{\ast}(\alpha\mathtt{e}_{t}(x)).
$$
Together with the convexity of $g$, this implies
$$
V_{t}(x+\alpha(s_{t}(x)-x))\leq V_{t}(x)-\alpha\gap_{t}(x)+t^{-1}\omega_{\ast}(\alpha\mathtt{e}_{t}(x)).
$$
We optimize the r.h.s. in $\alpha$ to obtain the analytic step-size policy
\begin{equation}\label{eq:alpha}
\alpha_{t}(x):=\min\left\{1,\frac{t\gap_{t}(x)}{\mathtt{e}_{t}(x)(\mathtt{e}_{t}(x)+t\gap_{t}(x))}\right\}\in[0,1/\mathtt{e}_{t}(x)].
\end{equation}
Equipped with this step strategy, procedure $\CG(x^{0},\varepsilon,t)$, described in Algorithm \ref{alg:CndG}, constructs a sequence $\{x^{k}_{t}\}_{k\geq 0}$ which produces an approximately-optimal solution in terms of the merit function $\gap_{t}(\cdot)$ and the potential function gap $\Delta_{t}$. Specifically, the following iteration complexity results can be established.
\begin{proposition}\label{prop:gap}
Given $\eta,t>0$. Let $R(x^{0}_{t},\eta,t)$ be the first iterate $k$ of Algorithm $\CG(x^{0}_{t},\eta,t)$ satisfying $\gap_{t}(x^{k}_{t})\leq\eta$. Then
$$
R(x^{0}_{t},\eta,t)\leq \lceil 5.3(\nu+t\Delta_{t}(x^{0}_{t})+t\Omega_{g})\log(10.6t\Delta_{t}(x^{0}_{t}))\rceil+\lceil \frac{24}{t\eta}(\nu+t\Omega_{g})^{2}\rceil,
$$
where $\Omega_{g}:=\max_{x,y\in\dom(g)\cap\mathsf{X}}\abs{g(x)-g(y)}.$
\end{proposition}
\begin{proposition}\label{prop:Delta}
Algorithm $\CG(x^{0}_{t},\eta,t)$ requires at most
\begin{equation}
N(x^{0}_{t},\eta,t):=\lceil 5.3(\nu+t\Delta_{t}(x^{0}_{t})+t\Omega_{g})\log(10.6t\Delta_{t}(x^{0}_{t}))\rceil+\lceil 12(\nu+t\Omega_{g})^{2}(\frac{1}{t\eta}-\frac{1}{t\Delta_{t}(x^{0}_{t})})^{+}\rceil
\end{equation}
iterations, in order to reach a point satisfying $\Delta_{t}(x^{k}_{t})\leq \eta$.
\end{proposition}
We prove these results in Section \ref{sec:inner}.
\subsection{Updating the Homotopy Parameter}
\label{sec:path-following}
Our analysis so far focuses on minimization of the potential function $V_{t}(x)$ for a fixed $t>0$. However, in order to solve the initial problem \eqref{eq:Opt}, one must trace the sequence of approximate solutions as $t_{i}\uparrow \infty$. The construction of such an increasing sequence of homotopy parameters is the purpose of this section.
\begin{algorithm}[t]
\caption{Method $\texttt{Homotopy}(x^{0},\varepsilon,f)$ }
\label{alg:Homotopy}
\begin{algorithmic}
\STATE {\bfseries Input: } $x^{0}\in\mathsf{C}$, $f\in\mathcal{B}_{\nu}(\mathsf{K})$.
\STATE {\bfseries Parameters: } $\varepsilon,t_{0},\eta_{0}>0$, $\sigma\in(0,1)$.
\STATE {\bfseries Initialize: } $x^{0}_{t_{0}}=x^{0}$, $I=\lceil\frac{\log(2\eta_{0}/\varepsilon)}{\log(1/\sigma)}\rceil$.
\FOR{$i=0,1,\ldots,I$}
\STATE Set $R_{i}=R(x^{0}_{t_{i}},\eta_{i},t_{i})$.
\STATE Set $\hat{x}_{i}=x^{R_{i}}_{t_{i}}$ the last iterate of $\CG(x^{0}_{t_{i}},\eta_{i},t_{i})$.
\STATE Update $t_{i+1}=t_{i}/\sigma,\eta_{i+1}=\sigma\eta_{i}$, $x^{0}_{t_{i+1}}=\hat{x}_{i}$.
\ENDFOR
\STATE $\hat{z}=\hat{x}_{I}$ an $\varepsilon$-solution of \eqref{eq:Opt}.
\end{algorithmic}
\end{algorithm}
Our aim is to reach an $\varepsilon$-solution (Definition \ref{def:eps}).
Let $\{\eta_{i}\}_{i\geq 0},\{t_{i}\}_{i\geq 0}$ be a sequence of approximation errors and homotopy parameters. For each run $i$, we activate procedure $\CG(x^{0}_{t_{i}},\eta_{i},t_{i})$ with the given configuration $(\eta_{i},t_{i})$. For $i=0$, we assume to have an admissible initial point $x^{0}_{t_{0}}=x^{0}$ available.
For $i\geq 1$, we restart Algorithm \ref{alg:CndG} using a kind of warm-start strategy by choosing $x^{0}_{t_{i}}=x^{R_{i-1}}_{t_{i-1}}$, where $R_{i-1}\equiv R(x^{0}_{t_{i-1}},\eta_{i-1},t_{i-1})$ is the first iterate $k$ of Algorithm $\CG(x^{0}_{t_{i-1}},\eta_{i-1},t_{i-1})$ satisfying $\gap_{t_{i-1}}(x^{k}_{t_{i-1}})\leq\eta_{i-1}$. Note that $R_{i-1}$ is upper bounded in Proposition \ref{prop:gap}.
After the $i$-th restart, let us call the obtained iterate $\hat{x}_{i}:= x^{R_{i}}_{t_{i}}$. In this way we obtain a sequence $\{\hat{x}_{i}\}_{i\geq 0}$, consisting of candidates for approximately following the central path, as they are $\eta_{i}$-close in terms of the gap $\gap_{t_{i}}$ (and, hence, in terms of the function gap) to the stage $t_{i}$'s optimal point $z^{\ast}(t_i)$. We update the parameters $(t_{i},\eta_{i})$ as follows:
\begin{itemize}
\item The sequence of homotopy parameters is determined as $t_{i}=t_{0}\sigma^{-i}$ for $\sigma\in(0,1)$ until the last round of $\texttt{Homotopy}(x^{0},\varepsilon,f)$ is reached.
\item The sequence of accuracies requested in the algorithm $\CG(x^{0}_{t_{i}},\eta_{i},t_{i})$ is updated by $\eta_{i}=\eta_{0}\sigma^{i}$.
\item The Algorithm stops after $I\equiv I(\sigma,\eta_{0},\varepsilon)=\lceil\frac{\log(2\eta_{0}/\varepsilon)}{\log(1/\sigma)}\rceil$ updates of the accuracy and homotopy parameters and yields an $\varepsilon$-approximate solution of problem \eqref{eq:Opt}.
\end{itemize}
Our updating strategy for the parameters ensures that $t_{i}\eta_{i}=t_{0}\eta_{0}$. This equilibrating choice between the increasing homotopy parameter and the decreasing accuracy parameter is sensible, because the iteration complexity of the \ac{CndG} solver is inversely proportional to $t_{i}\eta_{i}$ (cf. Proposition \ref{prop:gap}). Making the judicious choice $\eta_{0}t_{0}=2\nu$, yields a compact assessment of the total complexity of method $\texttt{Homotopy}(x^{0},\varepsilon,f)$.
\begin{theorem}\label{th:Complexity}
Choose $t_{0}=\frac{\nu}{\Omega_{g}}$ and $\eta_{0}=2\Omega_{g}$. The total iteration complexity of method $\texttt{Homotopy}(x^{0},\varepsilon,f)$ to find an $\varepsilon$-solution of \eqref{eq:Opt} is
$
\texttt{Compl}(x^{0},\varepsilon,f)=
\tilde{O}\left(\frac{384\Omega^{2}_{g}\nu}{\varepsilon^{2}(1-\sigma^{2})}+\Omega_{g}\log(21.2\nu(1+(2/\varepsilon)\Omega_{g}))\frac{21.2\nu}{\varepsilon}\frac{2-\sigma}{1-\sigma}\right),
$
where $\tilde{O}$ hides polylogarithmic factors in $\varepsilon^{-1}$.
\end{theorem}
The theorem leaves open the choice of the parameter $\sigma$. This is a tunable hyperparameter that can be chosen in a tailor-made fashion.
The proof of Theorem \ref{th:Complexity} can be found in Section \ref{sec:outer}.
\subsection{Line Search}
The step size policy employed in Algorithm \ref{alg:CndG} is inversely proportional to the local norm $\mathtt{e}_{t}(x)$. In particular, its derivation is based on the a-priori restriction that we force our iterates to stay within a trust-region defined by the Dikin ellipsoid. This restriction may force the method to take very small steps, and as a result display bad performance in practice. A simple remedy to this is to use the line search. Given $(x,t)\in\dom(F)\times(0,+\infty)$, let
\begin{equation}\label{eq:ExactLS}
\gamma_{t}(x)=\argmin_{\gamma\in[0,1]\text{ s.t. } x+\gamma(s_{t}(x)-x) \in \dom(F)}V_{t}(x+\gamma(s_{t}(x)-x)).
\end{equation}
\begin{algorithm}[t]
\caption{\ac{CndG} with line search: $\LCG(x^{0},\varepsilon,t)$}
\label{alg:LineSearchCndG}
\begin{algorithmic}
\STATE {\bfseries Input: } $(x^{0},t)\in\mathsf{C}\times(0,\infty)$ initial state; $\varepsilon>0$ accuracy level
\FOR{$k=0,1,\ldots$}
\IF{$\gap_{t}(x^{k})>\varepsilon$}
\STATE Same as Algorithm \ref{alg:CndG}, but with step size strategy \eqref{eq:ExactLS}.
\ELSE
\STATE Return $x^{k}$.
\ENDIF
\ENDFOR
\end{algorithmic}
\end{algorithm}
Thanks to the barrier structure of the potential function $V_{t}$, some useful consequences can be drawn from the definition of $\gamma_{t}(x)$. First, $\gamma_{t}(x)\in\{\gamma\geq 0\vert x+\gamma(s_{t}(x)-x)\in\dom(F)\}$. This implies $x+\gamma_{t}(x)(s_{t}(x)-x)\in\mathsf{X}\cap\dom(F)\cap\dom(g)$. Second, since $\alpha_{t}(x)$ is also contained in the latter set, we have
\[
V_{t}(x+\gamma_{t}(x)(s_{t}(x)-x))\leq V_{t}(x+\alpha_{t}(x)(s_{t}(x)-x))\qquad\forall (x,t)\in\dom(F)\times(0,+\infty).
\]
Via a comparison principle, this allows us to deduce the analysis of the sequence produced by $\LCG(x^{0},\varepsilon,t)$ from the analysis of the sequence induced by $\CG(x^{0},\varepsilon,t)$.
Indeed, if $\{\xi^{k}_{t}\}_{k \geq 0}$ is the sequence constructed by the line search procedure $\LCG(x^{0},\eta,t)$, then we have
$V_{t}(\xi^{k+1}_{t})\leq V_{t}(\xi_{t}^{k}+\alpha_{t}(\xi^{k}_{t})(s_{t}(\xi^{k}_{t})-x^{k}_{t}))$ for all $k$.
Hence, we can perform the complexity analysis on the majorizing function as in the analysis of procedure $\CG(x^{0},\varepsilon,t)$.
Consequently, all the complexity-related estimates for the method $\CG(x^{0},\eta,t)$ apply verbatim to the sequence $\{\xi^{k}_{t}\}_{k\geq 0}$ induced by the method $\LCG(x^{0},\eta,t)$.
\subsection{Inexact Implementation}
Algorithm \ref{alg:CndG} assumes that the \ac{LMO} performs exact computation of the search direction $s_t(x)$. This may not be directly available in practice. Just like in \cite{Jag13}, we show that our analysis carries essentially through even in the relaxed \ac{LMO} model with inexact implementation. Let $\gamma\in[0,1)$ be a given parameter, measuring the approximation quality of the oracle \eqref{eq:LMO} relative to the target accuracy $\eta$. Instead of the search direction $s_{t}(x)$, suppose the \ac{LMO} returns us a point $\tilde{s}_t=\tilde{s}_{t}(x)$ satisfying
\begin{equation}\label{eq:inexact}
t^{-1}F'(x)[\tilde{s}_{t}(x)]+g(\tilde{s}_{t}(x))\leq\min_{s\in\mathsf{X}}\left\{t^{-1}F'(x)[s]+g(s)\right\}+\gamma\eta.
\end{equation}
Following the derivations performed in the exact computational model, we are defining the inexact gap function
\begin{equation}
\widetilde{\gap}_{t}(x):=t^{-1}F'(x)[x-\tilde{s}_{t}(x)]+g(x)-g(\tilde{s}_{t}(x)),
\end{equation}
and the associated analytic step size policy, using $\tilde{\mathtt{e}}_{t}(x):=\norm{\tilde{s}_{t}(x)-x}_{x}$,
\begin{equation}\label{eq:inexactstep}
\tilde{\alpha}_{t}(x)=\min\left\{1,\frac{t\widetilde{\gap}_{t}(x)}{\tilde{\mathtt{e}}_{t}(x)(\tilde{\mathtt{e}}_{t}(x)+t\widetilde{\gap}_{t}(x))}\right\}.
\end{equation}
\begin{algorithm}[t]
\caption{\ac{CndG} with inexact oracle: $\ICG(x^{0},\varepsilon,t)$}
\label{alg:ApproxCG}
\begin{algorithmic}
\STATE {\bfseries Input: } $(x^{0},t)\in\mathsf{C}\times(0,\infty)$ initial state; $\varepsilon>0$ accuracy level, $\gamma\in[0,1)$ inexactness parameter
\FOR{$k=0,1,\ldots$}
\IF{$\widetilde{\gap}_{t}(x)>(1-\gamma)\varepsilon$}
\STATE Find $\tilde{s}^{k}= \tilde{s}_{t}(x^{k})\in\mathsf{X}$ such that \eqref{eq:inexact} holds at $x^{k}$.
\STATE Set $\alpha_{k}=\tilde{\alpha}_{t}(x^{k})$ by evaluating eq. \eqref{eq:inexactstep}.
\STATE Update $x^{k+1}=x^{k}+\alpha_{k}(\tilde{s}^{k}-x^{k})$.
\ELSE
\STATE Return $x^{k}$.
\ENDIF
\ENDFOR
\end{algorithmic}
\end{algorithm}
\begin{remark}
Note that as $\eta\to0$, the \ac{LMO}'s accuracy as defined in \eqref{eq:inexact} improves. \cite{Jag13} achieves such an improving oracle by coupling the accuracy imposed on the subproblem solver with the $1/k$-step size policy employed in the vanilla \ac{CndG}.
\end{remark}
The next propositions are the resulting iteration complexity statements of Algorithm $\ICG(x^{0},\eta,t)$, mimicking the statements in Proposition \ref{prop:Delta} and \ref{prop:gap}. Since the analysis of the inexact implementation regime is analogous to the exact \ac{LMO} model, we provide sketches of proofs of these statements in Section \ref{sec:inexactanalysis}.
\begin{proposition}\label{prop:Delta-Inexact}
Algorithm $\ICG(x^{0}_{t},\eta,t)$ requries at most $\tilde{N}_{\gamma}(x^{0}_{t},\eta,t)$ iterations, in order to reach a point satisfying $\Delta_{t}(x^{k}_{t})\leq \eta$, where
\begin{equation}
\tilde{N}_{\gamma}(x^{0}_{t},\eta,t):=K_{t,\gamma}(x^{0}_{t})+\lceil 12(\nu+t\Omega_{g})^{2}\left(\frac{1}{t\eta}-\frac{1}{t(\Delta_{t}(x^{0}_{t})-\gamma\eta)}\right)^{+}\rceil,
\end{equation}
and
\begin{equation}\label{eq:K}
K_{t,\gamma}(x^{0}_{t}):=\left\{\begin{array}{ll}
0 & \text{ if }\gamma\eta \geq\frac{1}{10.6t}\\
\lceil 5.3(\nu+t(\Omega_{g}-\gamma\eta)+t\Delta_{t}(x^{0}_{t}))\rceil \log\left(\frac{10.6t\Delta_{t}(x^{0}_{t})}{1-10.6t\gamma\eta}\right) & \text{if }\gamma\eta<\frac{1}{10.6t}.
\end{array}\right.
\end{equation}
\end{proposition}
\begin{proposition}\label{prop:gap-Inexact}
Given $\eta,t>0$ and $\gamma\in(0,1/3)$. Let $\tilde{R}(x^{0}_{t},\eta,t)$ be the first iterate $k$ satisfying $\gap_{t}(x^{k}_{t})\leq\eta$ for Algorithm $\ICG(x^{0},\eta,t)$. Then
$$
\tilde{R}_{\gamma}(x^{0}_{t},\eta,t)\leq K_{t,\gamma}(x^{0}_{t})+\lceil \frac{24}{t(1-\gamma)\eta}(\nu+t\Omega_{g})^{2}\rceil,
$$
where $K_{t,\gamma}(x^{0}_{t})$ is defined in \eqref{eq:K}.
\end{proposition}
\begin{remark}
Observe that all the complexity estimates reduce to the ones proved for the exact oracle model by setting $\gamma=0$.
\end{remark}
Based on the above estimates, it becomes clear that the worst-case bounds on the overall iteration complexity of $\texttt{Homotopy}(x^{0},\varepsilon,f)$ with $\ICG(x^{0},\eta,t)$ as the subroutine is the same as in Theorem \ref{th:Complexity}, subject to the adjustment of constant factors.
\section{Introduction}
\label{sec:intro}
\input{introduction}
\section{Notation \& Preliminaries}
\label{sec:prelims}
\input{prelims}
\section{Algorithm \& Convergence}
\label{sec:algo}
\input{Algo}
\section{Extensions}
\label{sec:related}
\input{LineSearch}
\section{Examples and Numerical Experiments}
\label{sec:numerics}
\input{experiments}
\section{Complexity Analysis}
\label{sec:compl}
\input{analysis_v3}
\subsection{Modifications Due to Inexact Oracles}
\label{sec:inexactanalysis}
\input{Analysis_inexact}
\section{Conclusion}
\label{sec:conclusion}
\input{conclusion}
\bibliographystyle{plainnat}
\subsection{Analysis of procedure $\CG(x^{0},\varepsilon,t)$}
\label{sec:inner}
Recall $\Omega_{g}:=\max_{x,y\in\dom(g)\cap\mathsf{X}}\abs{g(x)-g(y)}$ and $\mathsf{C}=\dom(F)\cap\mathsf{X}\cap\dom(g)$. The following estimate can be established as in \cite[][Prop. 2.3]{ZhaFre20}. Therefor we omit a proof.
\begin{lemma}\label{lem:e-bound}
For all $(x,t)\in\mathsf{C}\times(0,\infty)$, we have
\begin{equation}\label{eq:eOmega}
t^{-1}\mathtt{e}_{t}(x)\leq \nu/t+\gap_{t}(x)+\Omega_{g}.
\end{equation}
\end{lemma}
Observe that
\begin{align*}
\mathtt{e}_{t}(x)&=\norm{s_{t}(x)-x}_{x}=F''(x)[s_{t}(x)-x,s_{t}(x)-x]^{1/2}\geq \nu^{-1/2}\abs{F'(x)[s_{t}(x)-x]}\\
&=\frac{t}{\sqrt{\nu}}\abs{\gap_{t}(x)-g(x)+g(s_{t}(x))}\\
&\geq \frac{t}{\sqrt{\nu}}(\gap_{t}(x)-\Omega_{g})
\end{align*}
\subsubsection{Proof of Proposition \ref{prop:Delta}}
Suppose that $\gap_{t}(x)>\frac{\nu}{t}+\Omega_{g}$. Then, it readily follows from the previous display that $\mathtt{e}_{t}(x)\geq\sqrt{\nu}\geq 1$. This in turn implies $\frac{t\gap_{t}(x)}{\mathtt{e}_{t}(x)(\mathtt{e}_{t}(x)+t\gap_{t}(x))}<\frac{t\gap_{t}(x)}{\mathtt{e}_{t}(x)+t\gap_{t}(x)}<1$. This suggests a two-phase analysis defined by the threshold $\mathtt{a}_{t}:=\frac{\nu}{t}+\Omega_{g}$. Let $\{x^{k}_{t}\}_{0\leq k\leq K_{t}(x^{0})}$ be the sequence obtained when running $\CG(x^{0}_{t},\varepsilon,t)$ until reaching a point satisfying the stopping criterion $\gap_{t}(x^{K_{t}(x^{0})}) \leq \varepsilon$. To that end, define $\mathbb{K}_{I}(t):=\{k\in\mathbb{N}\vert \gap_{t}(x^{k}_{t})>\mathtt{a}_{t}\}$ and $\mathbb{K}_{II}(t):=\{k\in\mathbb{N}\vert\gap_{t}(x^{k}_{t})\leq \mathtt{a}_{t}\}$.
We start with the analysis of the trajectory on $\mathbb{K}_{I}(t)$. By the previous estimates, we know that on this phase the algorithm chooses the step sizes $\alpha_{k}=\frac{t\gap_{t}(x^{k}_{t})}{\mathtt{e}_{t}(x^{k}_{t})(\mathtt{e}_{t}(x^{k}_{t})+t\gap_{t}(x^{k}_{t}))}$. To reduce notational clutter, we set $\mathtt{e}^{k}_{t}\equiv\mathtt{e}_{t}(x^{k}_{t})$ and $G^{k}_{t}\equiv\gap_{t}(x^{k}_{t})$. In terms of these quantities, the per-iteration reduction of the potential function can be estimated as follows:
\begin{align}
V_{t}(x^{k+1}_{t})&\leq V_{t}(x^{k}_{t})-\frac{G^{k}_{t}}{\mathtt{e}^{k}_{t}}\frac{t G^{k}_{t}}{\mathtt{e}_{t}^{k}+tG^{k}_{t}}+\frac{1}{t}\omega_{\ast}\left(\frac{t G^{k}_{t}}{\mathtt{e}_{t}^{k}+tG^{k}_{t}}\right)\notag\\
&=V_{t}(x^{k}_{t})-\frac{G^{k}_{t}}{\mathtt{e}^{k}_{t}}+\frac{1}{t}\log\left(1+\frac{tG_{t}^{k}}{\mathtt{e}^{k}_{t}}\right) \notag\\
&=V_{t}(x^{k}_{t})-\frac{1}{t}\omega\left(\frac{tG^{k}_{t}}{\mathtt{e}^{k}_{t}}\right), \label{eq:V_t_progress_1}
\end{align}
where $\omega(t)=t-\log(1+t)$. This readily yields $\Delta_{t}(x^{k}_{t})-\Delta_{t}(x^{k+1}_{t})\geq \frac{1}{t}\omega\left(\frac{tG^{k}_{t}}{\mathtt{e}^{k}_{t}}\right).$ This implies that $\{\Delta_{t}(x^{k}_{t})\}_{k}$ is monotonically decreasing. For $k\in\mathbb{K}_{I}(t)$, we also see $\frac{tG^{k}_{t}}{\nu+tG^{k}_{t}+t\Omega_{g}}>\frac{1}{2}$. Together with \eqref{eq:eOmega}, this implies
\[
\frac{tG_{t}^{k}}{\mathtt{e}^{k}_{t}}\geq\frac{G^{k}_{t}}{\nu/t+G_{t}^{k}+\Omega_{g}}=\frac{tG^{k}_{t}}{\nu+tG_{t}^{k}+t\Omega_{g}}>\frac{1}{2}.
\]
Using \eqref{eq:eOmega}, the monotonicity of $\omega(\cdot)$, the fact that the function $x\mapsto \frac{x}{a+tx}$ is strictly increasing for $t,a>0$, and \cite[][Prop. 2.1]{ZhaFre20}, we arrive at
$$
\Delta_{t}(x^{k}_{t})-\Delta_{t}(x^{k+1}_{t})\geq\frac{1}{5.3}\frac{G^{k}_{t}}{\nu+tG^{k}_{t}+t\Omega_{g}}\geq \frac{\Delta_{t}(x^{k}_{t})}{5.3(\nu+t\Omega_{g}+t\Delta_{t}(x^{k}_{t}))}.
$$
Coupled with $\Delta_{t}(x^{k}_{t})\leq\Delta_{t}(x^{0}_{t}$), we conclude
$$
\Delta_{t}(x^{k}_{t})\left[1-\frac{1}{5.3(\nu+t\Omega_{g}+t\Delta_{t}(x^{0}_{t}))}\right]\geq\Delta_{t}(x^{k+1}_{t}).
$$
Define $q_{t}:=\frac{1}{5.3(\nu+t\Omega_{g}+t\Delta_{t}(x^{0}_{t}))}\in(0,1)$ (since $\nu\geq 1$), to arrive at the recursion
$$
\Delta_{t}(x^{k}_{t})\leq(1-q_{t})^{k}\Delta_{t}(x^{0}_{t})\leq\exp(-q_{t}k)\Delta_{t}(x^{0}_{t}).
$$
Furthermore,
\begin{align*}
\Delta_{t}(x^{k}_{t})&\geq \Delta_{t}(x^{k}_{t})-\Delta_{t}(x^{k+1}_{t})\\
&\geq\frac{G_{t}^{k}}{5.3(\nu+tG^{k}_{t}+t\Omega_{g})}\geq \frac{1}{10.6 t}.
\end{align*}
Hence, $\frac{1}{10.6 t\Delta_{t}(x^{0}_{t})}\leq \exp(-q_{t}k)$, so that process $\CG(x^{0}_{t},\varepsilon,t)$ must exit phase I in at most
$K_{t}(x^{0}_{t}):=\lceil 5.3(\nu+t\Delta_{t}(x^{0}_{t})+t\Omega_{g})\log(10.6t\Delta_{t}(x^{0}_{t}))\rceil$ iterations.
We now upper bound the time the process spends in phase II, so that $G^{k}_{t}\leq \frac{\nu}{t}+\Omega_{g}$. When the algorithm enters this phase, we have to distinguish the iterates using large step sizes $(\alpha_{k}=1$), from those using short steps ($\alpha_{k}<1$). Both regimes lead to a universal lower bound on the per-iteration potential reduction achieved, as we illustrate next.
To start with, let us assume $\alpha_{k}=1$. Then, $tG^{k}_{t}\geq \mathtt{e}^{k}_{t}(tG^{k}_{t}+\mathtt{e}^{k}_{t})$, which implies $\mathtt{e}^{k}_{t}\in(0,1)$, and $G^{k}_{t}\geq\frac{(\mathtt{e}^{k}_{t})^{2}}{t(1-\mathtt{e}^{k}_{t})}$. Moreover, using \eqref{eq:boundomega}, we readily obtain
\begin{align*}
V_{t}(x^{k+1}_{t})&\leq V_{t}(x^{k}_{t})-G^{k}_{t}+\frac{1}{t}\omega_{\ast}(\mathtt{e}^{k}_{t})\\
&\leq V_{t}(x^{k}_{t})-G^{k}_{t}+\frac{1}{t}\frac{(\mathtt{e}^{k}_{t})^{2}}{2(1-\mathtt{e}^{k}_{t})}\\
&\leq V_{t}(x^{k}_{t})-\frac{G_{t}^{k}}{2}.
\end{align*}
Therefore,
\begin{align*}
\Delta_{t}(x^{k}_{t})-\Delta_{t}(x^{k+1}_{t})&\geq \frac{G_{t}^{k}}{2}\geq \frac{1}{2}\frac{(G_{t}^{k})^{2}}{\nu/t+\Omega_{g}}=\frac{1}{2}\frac{t(G^{k}_{t})^{2}}{\nu+t\Omega_{g}}\\
&\geq \frac{1}{12}\frac{t(G^{k}_{t})^{2}}{(\nu+t\Omega_{g})^{2}}.
\end{align*}
On the other hand, in the regime where $\alpha_{k}<1$, we obtain the estimate
$\Delta_{t}(x^{k}_{t})-\Delta_{t}(x^{k+1}_{t})\geq \frac{1}{12}\frac{t(G^{k}_{t})^{2}}{(\nu+t\Omega_{g})^{2}}$, by following the same reasoning as in \cite{ZhaFre20}. We conclude
\begin{equation}\label{eq:inverse}
\frac{1}{\Delta_{t}(x^{k+1}_{t})}-\frac{1}{\Delta_{t}(x^{k}_{t})}\geq\frac{t}{12(\nu+t\Omega_{g})^{2}}\quad\forall k\in\mathbb{K}_{II}(t).
\end{equation}
Given the pair $(t,\eta)\in(0,\infty)^{2}$, denote by $N\equiv N(x^{0}_{t},\eta,t)$ an upper bound on the total number of iterations of $\CG(x^{0}_{t},\eta,t)$. We know that at most $K\equiv K_{t}(x^{0}_{t})$ of these iterations are in phase I. To estimate the remaining iterations in phase II, i.e. $M=N-K$, we first telescope the inequality \eqref{eq:inverse}, to get
$$
\frac{1}{\Delta_{t}(x^{N}_{t})}\geq \frac{1}{\Delta_{t}(x^{K}_{t})}+\frac{Mt}{12(\nu+t\Omega_{g})^{2}}\geq\frac{1}{\Delta_{t}(x^{0}_{t})}+\frac{Mt}{12(\nu+t\Omega_{g})^{2}}.
$$
Choosing $M=\lceil 12(\nu+t\Omega_{g})^{2}(\frac{1}{t\eta}-\frac{1}{t\Delta_{t}(x^{0}_{t})})^{+}\rceil$ ensures that $\Delta_{t}(x^{N}_{t})\leq\eta$. This bounds the iteration complexity of algorithm $\CG(x^{0}_{t},\eta,t)$.
\subsubsection{Proof of Proposition \ref{prop:gap}}
Let $\{k_{j}(t)\}_{j}$ denote the increasing set of indices enumerating $\mathbb{K}_{II}(t)$. From the analysis of the trajectory given in the previous section, we know that
\[
\Delta_{t}(x_{t}^{k_{j+1}(t)})-\Delta_{t}(x^{k_{j}(t)}_{t})\leq-\frac{1}{12}\frac{t(G_{t}^{k_{j}(t)})^{2}}{(\nu+t\Omega_{g})^{2}}, \text{ as well as }G_{t}^{k_{j}(t)}\geq\Delta_{t}(x^{k_{j}(t)}_{t}).
\]
Set $d_{j}\equiv \Delta_{t}(x^{k_{j}(t)}_{t})$ and $\frac{1}{M}\equiv \frac{t}{12(\nu+t\Omega_{g})^{2}},\Gamma_{j}\equiv G_{t}^{k_{j}(t)}$. We thus obtain the recursion
$$
d_{j+1}\leq d_{j}-\frac{\Gamma_{j}^2}{M},\quad\Gamma_{j}\geq d_{j}.
$$
We can apply \cite[][Prop. 2.4]{ZhaFre20} directly to the above recursion to obtain the estimates
$$
d_{j}\leq \frac{M}{j},\quad \text{ and } \min\{\Gamma_{0},\ldots,\Gamma_{j}\}\leq \frac{2M}{j}.
$$
Therefore, in order to reach an iterate with $\gap_{t}(x^{k}_{t})\leq\eta$, we need to run the process until the label $j$ is reached satisfying $\frac{2M}{j}\leq\eta$. Solving this for $j$ yields $j=\lceil \frac{24(\nu+t\Omega_{g})^{2}}{t\eta}\rceil$. Combined with the upper bound obtained for the time the process spends in phase I, we obtain the total complexity estimate postulated in Proposition \ref{prop:gap}.
\subsection{Analysis of the outer loop}
\label{sec:outer}
Let $I\equiv I(\eta_{0},\sigma,\varepsilon)=\lceil\frac{\log(2\eta_{0}/\varepsilon)}{\log(1/\sigma)}\rceil$ denote the a-priori fixed number of updates of the accuracy and homotopy parameter. We set $\hat{x}_{i}\equiv x^{R_{i}}_{t_{i}}$, the last iterate of procedure $\CG(x^{0}_{t_{i}},\eta_{i},t_{i})$ satisfying $\Delta_{t_{i}}(\hat{x}_{i})\leq\gap_{t_{i}}(\hat{x}_{i})\leq\eta_{i}$. From Proposition \ref{prop:gap}, we deduce that
\begin{align*}
g(\hat{x}_{i})-g(z^{\ast}(t_{i}))\leq \gap_{t_{i}}(\hat{x}_{i})+\frac{1}{t_{i}}F'(\hat{x}_{i})[z^{\ast}(t_{i})-\hat{x}_{i}]\leq \eta_{i}+\frac{\nu}{t_{i}}.
\end{align*}
Hence, using Lemma \ref{lem:pathfollowing}, we observe
\begin{equation}
\label{eq:final_error}
g(\hat{x}_{i})-\Opt=g(\hat{x}_{i})-g(z^{\ast}(t_{i}))+g(z^{\ast}(t_{i}))-\Opt\leq \eta_{i}+\frac{2\nu}{t_{i}}.
\end{equation}
Since $t_{i}=\frac{2\nu}{\eta_{i}}$, we obtain $g(\hat{x}_{i})-\Opt\leq 2\eta_{i}$. We next estimate the initial gap $\Delta_{t_{i+1}}(x^{0}_{t_{i+1}})=\Delta_{t_{i+1}}(\hat{x}_{i})$ incurred by our warm-start strategy. Observe that
\begin{align*}
V_{t_{i+1}}(x^{0}_{i+1})-V_{t_{i+1}}(z^{\ast}(t_{i+1}))&=V_{t_{i}}(x^{0}_{i})-V_{t_{i}}(z^{\ast}(t_{i+1}))+(\frac{1}{t_{i+1}}-\frac{1}{t_{i}})(F(x^{0}_{t_{i+1}})-F(z^{\ast}(t_{i+1})))\\
&\leq V_{t_{i}}(x^{R_{i}}_{t_{i}})-V_{t_{i}}(z^{\ast}(t_{i}))+(1-\frac{t_{i+1}}{t_{i}})\left(V_{t_{i+1}}(x^{0}_{t_{i+1}})-V_{t_{i+1}}(z^{\ast}(t_{i+1}))\right)\\
&+(1-\frac{t_{i+1}}{t_{i}})(g(z^{\ast}(t_{i+1}))-g(x^{0}_{t_{i+1}})).
\end{align*}
Whence,
$$
\frac{t_{i+1}}{t_{i}}(V_{t_{i+1}}(x^{0}_{t_{i+1}})-V_{t_{i+1}}(z^{\ast}(t_{i+1})))\leq V_{t_{i}}(x^{R_{i}}_{t_{i}})-V_{t_{i}}(z^{\ast}(t_{i}))+(1-\frac{t_{i+1}}{t_{i}})(g(z^{\ast}(t_{i+1}))-g(x^{0}_{t_{i+1}}))
$$
Since $t_{i+1}>t_{i}$ and the definition of the stopping time $R_{i}\equiv R(x^{0}_{t_{i}},\eta_{i},t_{i})$, this implies
\begin{equation}
\label{eq:tDeltat_estimate}
t_{i+1}\Delta_{t_{i+1}}(x^{0}_{t_{i+1}})\leq t_{i}\Delta^{R_{i}}_{t_{i}}-(t_{i+1}-t_{i})\Omega_{g}\leq t_{i}\eta_{i}+(t_{i+1}-t_{i})\Omega_{g}=2\nu+(t_{i+1}-t_{i})\Omega_{g}.
\end{equation}
We are now in the position to estimate the total iteration complexity $\texttt{Compl}(x^{0},\varepsilon,f):=\sum_{i=0}^{I}R(x^{0}_{t_{i}},\eta_{i},t_{i}),$ and thereby prove Theorem \ref{th:Complexity}. We do so by using the updating regime of the sequences $\{t_{i}\}$ and $\{\eta_{i}\}$ explained in Section \ref{sec:algo}. These updating mechanisms imply
\begin{align*}
\sum_{i=1}^{I}R_{i}&\leq \sum_{i=1}^{I}5.3(3\nu+(t_{i}-t_{i-1})\Omega_{g}+t_{i}\Omega_{g})\log(10.6(2\nu+(t_{i}-t_{i-1})\Omega_{g}))+\sum_{i=1}^{I}\frac{12}{\nu}(\nu+t_{i}\Omega_{g})^{2}\\
&\leq \log(10.6(2\nu+t_{I}\Omega_{g}))\sum_{i=1}^{I}5.3(3\nu+t_{i}\Omega_{g}+(t_{i}-t_{i-1})\Omega_{g})+\sum_{i=1}^{I}\frac{24}{\nu}(\nu^{2}+t^{2}_{i}\Omega^{2}_{g})\\
&\leq 15.9\nu\log(10.6(2\nu+t_{I}\Omega_{g}))I+5.3\Omega_{g}\log(10.6(2\nu+t_{I}\Omega_{g}))\sum_{i=1}^{I}t_{i}\\
&+5.3\Omega_{g}\log(10.6(2\nu+t_{I}\Omega_{g}))\sum_{i=1}^{I}(t_{i}-t_{i-1})+24\nu I+\frac{24\Omega_{g}^{2}}{\nu}\sum_{i=1}^{I}t_{i}^{2}\\
&\leq I\left[24\nu+15.9\nu\log(10.6(2\nu+t_{I}\Omega_{g}))\right]+5.3\Omega_{g}\log(10.6(2\nu+t_{I}\Omega_{g}))t_{I}\\
&+\frac{24\Omega^{2}_{g}}{\nu}\sum_{i=1}^{I}t^{2}_{i}+5.3\Omega_{g}\log(10.6(2\nu+t_{I}\Omega_{g}))\sum_{i=1}^{I}t_{i}.
\end{align*}
Let us estimate the terms appearing in this expression. First,
\begin{align*}
t_{I}\leq t_{0}\sigma^{-I}\leq \frac{4\nu}{\varepsilon}.
\end{align*}
Second,
\begin{align*}
\sum_{i=1}^{I}t_{i}=t_{0}\sigma^{-1}\sum_{i=0}^{I-1}(1/\sigma)^{i}\leq t_{0}\frac{\sigma^{-I}}{1-\sigma}\leq \frac{4\nu}{\varepsilon(1-\sigma)}.
\end{align*}
Third,
\begin{align*}
\sum_{i=1}^{I}t_{i}^{2}=t^{2}_{0}\sum_{i=1}^{I}(1/\sigma)^{2i}=\frac{t^{2}_{0}}{\sigma^{2}}\sum_{i=0}^{I-1}(1/\sigma^{2})^{i}\leq \frac{16\nu^{2}}{\varepsilon^{2}(1-\sigma^{2})}.
\end{align*}
This delivers the bound
\begin{align*}
\sum_{i=1}^{I}R_{i}&\leq\frac{\log(2\eta_{0}/\varepsilon)}{\log(1/\sigma)}\left[24\nu+15.9\nu\log(21.2\nu(1+(2/\varepsilon)\Omega_{g}))\right]+5.3\Omega_{g}\log(21.2\nu(1+(2/\varepsilon)\Omega_{g}))\frac{4\nu}{\varepsilon}\\
&+\frac{24\Omega^{2}_{g}}{\nu}\frac{16\nu^{2}}{\varepsilon^{2}(1-\sigma^{2})}+5.3\Omega_{g}\log(21.2\nu(1+(2/\varepsilon)\Omega_{g}))\frac{4\nu}{\varepsilon(1-\sigma)}\\
&=\frac{\nu\log(2\eta_{0}/\varepsilon)}{\log(1/\sigma)}\left[24+15.9\log(21.2\nu(1+(2/\varepsilon)\Omega_{g}))\right]+\frac{384\Omega^{2}_{g}\nu}{\varepsilon^{2}(1-\sigma^{2})}\\
&+\Omega_{g}\log(21.2\nu(1+(2/\varepsilon)\Omega_{g}))\frac{21.2\nu}{\varepsilon}\frac{2-\sigma}{1-\sigma}.
\end{align*}
It remains to bound the complexity at $i=0$. We have
\begin{align*}
R_{0}&\leq 5.3(\nu+t_{0}\Delta_{t_{0}}(x^{0}_{t_{0}})+t_{0}\Omega_{g})\log(10.6t_{0}\Delta_{t_{0}}(x^{0}))+\frac{12}{\nu}(\nu+t_{0}\Omega_{g})^{2}\\
&\leq 5.3(\nu+F(x^{0})-F(z^{\ast}(t_{0}))+2t_{0}\Omega_{g})\log\left(10.6(F(x^{0})-F(z^{\ast}(t_{0}))+t_{0}\Omega_{g})\right)+\frac{24}{\nu}(\nu^{2}+t^{2}_{0}\Omega^{2}_{g}).
\end{align*}
Since $t_{0}=\frac{\nu}{\Omega_{g}}$ and $\eta_{0}=2\Omega_{g}$, we have
$$
R_{0}\leq 5.3(3\nu+F(x^{0})-F(z^{\ast}(t_{0})))\log\left(10.6(\nu+F(x^{0})-F(z^{\ast}(t_{0})))\right)+48\nu.
$$
Adding the two gives the total complexity bound
\begin{align*}
\texttt{Compl}(x^{0},\varepsilon,f)&\leq 5.3(3\nu+F(x^{0})-F(z^{\ast}(t_{0})))\log\left(10.6(\nu+F(x^{0})-F(z^{\ast}(t_{0})))\right)+48\nu\\
&+\frac{\nu\log(2\eta_{0}/\varepsilon)}{\log(1/\sigma)}\left[24+15.9\log(21.2\nu(1+(2/\varepsilon)\Omega_{g}))\right]\\
&+\frac{384\Omega^{2}_{g}\nu}{\varepsilon^{2}(1-\sigma^{2})}+\Omega_{g}\log(21.2\nu(1+(2/\varepsilon)\Omega_{g}))\frac{21.2\nu}{\varepsilon}\frac{2-\sigma}{1-\sigma}\\
&=\tilde{O}\left(\frac{384\Omega^{2}_{g}\nu}{\varepsilon^{2}(1-\sigma^{2})}+\Omega_{g}\log(21.2\nu(1+(2/\varepsilon)\Omega_{g}))\frac{21.2\nu}{\varepsilon}\frac{2-\sigma}{1-\sigma}\right).
\end{align*}
\subsection{A Generic Model Problem}
The following class of SDPs is studied in \cite{AroHazKal05}. This SDP arises in many algorithms such as approximating MAXCUT, approximating the CUTNORM of a matrix, and approximating solutions to the little Grothendieck problem \citep{CharWir04,BanNouVor16}. The program is given by
\begin{equation}\label{eq:MaxQP}\tag{MAXQP}
\begin{split}
& \max\inner{{\mathbf{C}},\mathbf{X}}\\
\text{s.t. }& X_{ii}\leq 1\quad i=1,\ldots,n\\
& X\succeq 0
\end{split}
\end{equation}
Let $\mathsf{Q}:=\{y\in\mathbb{R}^n\vert y_{i}\leq 1,\; i=1,\ldots,n\}$, and consider the conic hull of $\mathsf{Q}$, defined as
$\mathsf{K}:=\{(y,t)\in\mathbb{R}^n\times\mathbb{R}\vert \frac{1}{t}y\in\mathsf{Q},t>0\}\subset\mathbb{R}^{n+1}.$ This set admits the $\nu=n$ logarithmically homogenous barrier
$$
f(y,t)=-\sum_{i=1}^{n}\log(t-y_{i})=-\sum_{i=1}^{n}\log(1-\frac{1}{t}y_{i})-n\log(t).
$$
We can then reformulate \eqref{eq:MaxQP} as
\begin{equation}
\begin{split}
& \max_{\mathbf{X},t}\{g(\mathbf{X},t):=\inner{{\mathbf{C}},\mathbf{X}}\}\\
\text{s.t. }& (\mathbf{X},t)\in\mathsf{X}:=\{\mathbf{X}\in\mathcal{S}^{n}\times\mathbb{R}\vert \mathbf{X}\succeq 0,\;\tr(\mathbf{X})\leq n,t=1\},\\
&\mathcal{P}(\mathbf{X},t)\in\mathsf{K}
\end{split}
\end{equation}
where $\mathcal{P}(\mathbf{X},t):=[X_{11};\ldots;X_{nn};t]^{\top}$ is a linear homogenous mapping from $\mathbb{S}^{n}\times\mathbb{R}\to\mathbb{R}^{n+1}$. Set $F(\mathbf{X},t)=f(\mathcal{P}(\mathbf{X},t))$ for $(\mathbf{X},t)\in\mathbb{S}^{n}\times(0,\infty)$.
\begin{table}[ht]
\centering
\begin{tabular}{lll|r>{\raggedleft\arraybackslash}p{1.2cm}>{\raggedleft\arraybackslash}p{1cm}}
\hline
&&& \multicolumn{3}{c}{SDPT3}\\
Data set & \# Nodes & \# Vertices & Value & \multicolumn{1}{p{1.2cm}}{$\lambda_{\text{min}}$} & Feas. value \\
\hline\hline
G1 & 800 & 19176 & 48333 & 0 & -\\
G2 & 800 & 19176 & 48357 & 0 & -\\
G3 & 800 & 19176 & 48337 & 0 & -\\
G4 & 800 & 19176 & 48446 & 0 & -\\
G5 & 800 & 19176 & 48400 & 0 & -\\
G6 & 800 & 19176 & 10668 & 0 & -\\
G7 & 800 & 19176 & 9977 & 0 & -\\
G8 & 800 & 19176 & 10141 & 0 & -\\
G9 & 800 & 19176 & 48446 & 0 & -\\
G10 & 800 & 19176 & 48446 & 0 & -\\
G11 & 800 & 19176 & 48446 & 0 & -\\
G12 & 800 & 19176 & 48446 & 0 & -\\
G13 & 800 & 19176 & 48446 & 0 & -\\
G14 & 800 & 19176 & 48446 & 0 & -\\
G15 & 800 & 19176 & 48446 & 0 & -\\
G16 & 800 & 4672 & 12700 & 0 & -\\
G17 & 800 & 4667 & 12685 & 0 & -\\
G18 & 800 & 4694 & 4695 & 0 & -\\
G19 & 800 & 4661 & 4364 & 0 & -\\
G20 & 800 & 4672 & 4476 & 0 & -\\
G21 & 800 & 4667 & 4448 & 0 & -\\
G22 & 2000 & 19990 & 73108 & -1 & 48262\\
G23 & 2000 & 19990 & 73157 & -1 & 39951\\
G24 & 2000 & 19990 & 73147 & -1 & 39991\\
G25 & 2000 & 19990 & 73174 & -1 & 39984\\
G26 & 2000 & 19990 & 73083 & -1 & 39982\\
G27 & 2000 & 19990 & 33218 & -1 & 16567\\
G28 & 2000 & 19990 & 33014 & -1 & 16403\\
G29 & 2000 & 19990 & 33511 & -1 & 16836\\
G30 & 2000 & 19990 & 33524 & -1 & 16862\\
\hline
\end{tabular}
\caption{Max-Cut datasets}
\label{tbl:MaxCutDataSetsInfo}
\end{table}
\begin{figure}
\centering
\includegraphics[scale=0.75]{./Figures/MaxCut_small_fgap_vs_iter.pdf}
\caption{Max-Cut datasets G1-G21 relative gap from optimal solution vs. iteration.}\label{fig:MaxCut_Small_GapVsIter}
\end{figure}
\begin{figure}
\centering
\includegraphics[scale=0.75]{./Figures/MaxCut_small_fgap_vs_time.pdf}
\caption{Max-Cut datasets G1-G21 relative gap from optimal solution vs. time.}\label{fig:MaxCut_Small_fVsIter}
\end{figure}
We apply this formulation to the classical MAXCUT problem in which ${\mathbf{C}}=\mathbf{L}$, i.e. the combinatorial Laplace matrix of an undirected graph $\mathcal{G}=(\mathcal{V},\mathcal{E})$ with vertex set $\mathcal{V}=\{1,\ldots,n\}$ and edge set $\mathcal{E}$. To evaluate the performance, we consider the random graphs G1-G30, published online in \cite{Gset}. We implement Algorithm \ref{alg:Homotopy} using procedure $\CG$ and $\LCG$ as solvers for the inner loops. The Max-Cut problem was run using Matlab R2019b on an Intel(R) Xeon(R) Gold 6254 CPU @ 3.10GHz server limited to 4 threads per run and 512G total RAM. We used CVX version 2.2 with SDPT3 version 4.0 to obtain the interior point solution which we take as a reference value in order to benchmark our results. The benchmark solutions are displayed in Table \ref{tbl:MaxCutDataSetsInfo}. The \emph{relative gap} we compute to assess the relative performance of our method is then computed as $\frac{g(x)-g(x^{\text{ref}})}{\abs{g(x^{\text{ref}})}}$. Table~\ref{tbl:MaxCutDataSetsInfo} also displays the size of each dataset, the value obtained by solving the Max-Cut SDP relaxation using CVX with the SDPT3 solver and the minimum eigenvalue of the solution $\lambda_{\text{min}}(X)$. We observe that for larger graphs SDPT3 returns infeasible solutions, featuring negative eigenvalues. If this occurs, the value obtained by a corrected solution is used as a reference value instead. The corrected solution $\tilde{X}$ is obtained as $\tilde{X}=(X-\lambda_{\text{min}}(X)\mathbf{I})/\alpha$ where $\mathbf{I}$ is the identity matrix, and $\alpha$ is the minimal number for which $\tilde{X}_{ii}\leq 1$ for all $i=1,\ldots,n$. Figures \ref{fig:MaxCut_Small_GapVsIter}-\ref{fig:MaxCut_Large_fVsIter} illustrate the performance of our methods for various parameter values, and Table \ref{tbl:max_cut} collects the numerical values obtained for the schemes $\CG$ and $\LCG$, in dependence of the scaling factor $\sigma$.
\begin{figure}
\centering
\includegraphics[scale=0.75]{./Figures/MaxCut_small_f_vs_iter.pdf}
\caption{Max-Cut datasets G1-G21 objective function value vs. iteration.}
\label{fig:MaxCut_Small_fVsIter}
\end{figure}
\begin{figure}
\centering
\includegraphics[scale=0.75]{./Figures/MaxCut_small_f_vs_time.pdf}
\caption{Max-Cut datasets G1-G21 objective function value vs. time.}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[scale=0.75]{./Figures/MaxCut_large_f_vs_iter.pdf}
\caption{Max-Cut datasets G22-G30 objective function value vs. iteration.}\label{fig:MaxCut_Large_GapVsIter}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[scale=0.75]{./Figures/MaxCut_large_f_vs_time.pdf}
\caption{Max-Cut datasets G22-G30 objective function value vs. time.}\label{fig:MaxCut_Large_fVsIter}
\end{figure}
\newpage
\begin{longtable}[l]{lp{0.8cm}p{0.2cm}p{0.2cm}rrrrrr}
\caption{Results for datasets G1-G30 after a number of iterations, including objective function value, relative gap from SDPT3 corrected solution, and time (in seconds).}
\label{tbl:max_cut}\\
\hline
{\bf Dataset} & {\bf Iter.} & & Alg. & \multicolumn{3}{c}{$\CG$} &\multicolumn{3}{c}{$\LCG$} \\
& & &$\sigma$& 0.99 & 0.9 & 0.5 & 0.99 & 0.9 & 0.5 \\
\hline\\
& & {\bf Value} &&&&&&&\\
& & $g(z)$ &&&&&&&\\
& & Gap&&&&&&&\\
& & Time&&&&&&&\\
\hline
\hline
G1 & 100 & & & 29141 & 29500 & 28611 & 33151 & 35604 & 35605 \\
& & & & 3.97e-01 & 3.90e-01 & 4.08e-01 & 3.14e-01 & 2.63e-01 & 2.63e-01 \\
& & & & 14 & 15 & 16 & 34 & 50 & 45 \\
& 1000 & & & 45292 & 45290 & 45175 & 46034 & 46139 & 46204 \\
& & & & 6.29e-02 & 6.29e-02 & 6.53e-02 & 4.76e-02 & 4.54e-02 & 4.40e-02 \\
& & & & 171 & 169 & 188 & 278 & 305 & 298 \\
& 10000 & & & 47785 & 47788 & 47776 & 47813 & 47828 & 47861 \\
& & & & 1.13e-02 & 1.13e-02 & 1.15e-02 & 1.08e-02 & 1.04e-02 & 9.77e-03 \\
& & & & 2013 & 2033 & 2126 & 2619 & 2549 & 2490 \\
\hline G2 & 100 & & & 29127 & 29371 & 28544 & 33314 & 35423 & 35324 \\
& & & & 3.98e-01 & 3.93e-01 & 4.10e-01 & 3.11e-01 & 2.67e-01 & 2.70e-01 \\
& & & & 15 & 16 & 16 & 34 & 54 & 54 \\
& 1000 & & & 45268 & 45268 & 45134 & 46048 & 46148 & 46203 \\
& & & & 6.39e-02 & 6.39e-02 & 6.67e-02 & 4.78e-02 & 4.57e-02 & 4.46e-02 \\
& & & & 165 & 163 & 181 & 270 & 318 & 343 \\
& 10000 & & & 47809 & 47812 & 47797 & 47835 & 47850 & 47884 \\
& & & & 1.13e-02 & 1.13e-02 & 1.16e-02 & 1.08e-02 & 1.05e-02 & 9.79e-03 \\
& & & & 1947 & 1957 & 1990 & 2689 & 2753 & 2742 \\
\hline G3 & 100 & & & 28622 & 28917 & 28167 & 33150 & 35287 & 34566 \\
& & & & 4.08e-01 & 4.02e-01 & 4.17e-01 & 3.14e-01 & 2.70e-01 & 2.85e-01 \\
& & & & 16 & 16 & 16 & 33 & 50 & 48 \\
& 1000 & & & 45258 & 45241 & 45082 & 46015 & 46111 & 46141 \\
& & & & 6.37e-02 & 6.41e-02 & 6.73e-02 & 4.80e-02 & 4.61e-02 & 4.54e-02 \\
& & & & 175 & 172 & 193 & 286 & 310 & 316 \\
& 10000 & & & 47796 & 47800 & 47784 & 47821 & 47833 & 47874 \\
& & & & 1.12e-02 & 1.11e-02 & 1.14e-02 & 1.07e-02 & 1.04e-02 & 9.59e-03 \\
& & & & 2066 & 2090 & 2172 & 2664 & 2574 & 2542 \\
\hline G4 & 100 & & & 28967 & 29132 & 28376 & 33013 & 35264 & 35484 \\
& & & & 4.02e-01 & 3.99e-01 & 4.14e-01 & 3.19e-01 & 2.72e-01 & 2.68e-01 \\
& & & & 15 & 15 & 14 & 29 & 51 & 52 \\
& 1000 & & & 45402 & 45403 & 45235 & 46183 & 46283 & 46355 \\
& & & & 6.28e-02 & 6.28e-02 & 6.63e-02 & 4.67e-02 & 4.46e-02 & 4.32e-02 \\
& & & & 162 & 159 & 173 & 260 & 311 & 323 \\
& 10000 & & & 47911 & 47914 & 47922 & 47939 & 47952 & 47995 \\
& & & & 1.10e-02 & 1.10e-02 & 1.08e-02 & 1.05e-02 & 1.02e-02 & 9.30e-03 \\
& & & & 1837 & 1860 & 1911 & 2624 & 2712 & 2661 \\
\hline G5 & 100 & & & 28920 & 29414 & 28296 & 32808 & 35454 & 35118 \\
& & & & 4.02e-01 & 3.92e-01 & 4.15e-01 & 3.22e-01 & 2.67e-01 & 2.74e-01 \\
& & & & 17 & 15 & 17 & 37 & 54 & 48 \\
& 1000 & & & 45355 & 45365 & 45184 & 46148 & 46234 & 46247 \\
& & & & 6.29e-02 & 6.27e-02 & 6.64e-02 & 4.65e-02 & 4.48e-02 & 4.45e-02 \\
& & & & 181 & 178 & 195 & 306 & 319 & 313 \\
& 10000 & & & 47866 & 47870 & 47868 & 47891 & 47905 & 47945 \\
& & & & 1.10e-02 & 1.09e-02 & 1.10e-02 & 1.05e-02 & 1.02e-02 & 9.40e-03 \\
& & & & 2125 & 2113 & 2106 & 2822 & 2593 & 2581 \\
\hline G6 & 100 & & & 3119 & 6048 & 5803 & 3280 & 7529 & 7565 \\
& & & & 7.08e-01 & 4.33e-01 & 4.56e-01 & 6.93e-01 & 2.94e-01 & 2.91e-01 \\
& & & & 15 & 17 & 21 & 26 & 16 & 36 \\
& 1000 & & & 9819 & 9788 & 9652 & 10088 & 10062 & 10024 \\
& & & & 7.96e-02 & 8.25e-02 & 9.52e-02 & 5.43e-02 & 5.68e-02 & 6.03e-02 \\
& & & & 202 & 208 & 246 & 280 & 134 & 281 \\
& 10000 & & & 10497 & 10486 & 10450 & 10528 & 10505 & 10471 \\
& & & & 1.60e-02 & 1.71e-02 & 2.04e-02 & 1.31e-02 & 1.53e-02 & 1.85e-02 \\
& & & & 3188 & 3029 & 3086 & 3941 & 1649 & 2828 \\
\hline G7 & 100 & & & 3569 & 5641 & 5359 & 3771 & 7098 & 6913 \\
& & & & 6.42e-01 & 4.35e-01 & 4.63e-01 & 6.22e-01 & 2.89e-01 & 3.07e-01 \\
& & & & 18 & 19 & 18 & 28 & 35 & 39 \\
& 1000 & & & 9171 & 9137 & 9020 & 9438 & 9415 & 9335 \\
& & & & 8.08e-02 & 8.42e-02 & 9.59e-02 & 5.40e-02 & 5.63e-02 & 6.43e-02 \\
& & & & 235 & 234 & 232 & 261 & 273 & 289 \\
& 10000 & & & 9817 & 9803 & 9774 & 9839 & 9817 & 9792 \\
& & & & 1.60e-02 & 1.74e-02 & 2.03e-02 & 1.38e-02 & 1.60e-02 & 1.85e-02 \\
& & & & 3315 & 3224 & 2878 & 3509 & 3352 & 3095 \\
\hline G8 & 100 & & & 2788 & 5812 & 5614 & 2931 & 7333 & 7095 \\
& & & & 7.23e-01 & 4.23e-01 & 4.42e-01 & 7.09e-01 & 2.72e-01 & 2.95e-01 \\
& & & & 14 & 17 & 18 & 20 & 24 & 24 \\
& 1000 & & & 9282 & 9263 & 9141 & 9544 & 9520 & 9486 \\
& & & & 7.79e-02 & 7.98e-02 & 9.19e-02 & 5.19e-02 & 5.43e-02 & 5.76e-02 \\
& & & & 208 & 220 & 229 & 215 & 200 & 187 \\
& 10000 & & & 9910 & 9896 & 9859 & 9931 & 9909 & 9876 \\
& & & & 1.56e-02 & 1.69e-02 & 2.05e-02 & 1.34e-02 & 1.57e-02 & 1.89e-02 \\
& & & & 3272 & 3186 & 2561 & 2810 & 2359 & 2045 \\
\hline G9 & 100 & & & 2901 & 5539 & 5214 & 3073 & 6966 & 6708 \\
& & & & 7.14e-01 & 4.54e-01 & 4.86e-01 & 6.97e-01 & 3.13e-01 & 3.39e-01 \\
& & & & 18 & 17 & 18 & 26 & 35 & 36 \\
& 1000 & & & 9322 & 9300 & 9162 & 9588 & 9570 & 9498 \\
& & & & 8.07e-02 & 8.29e-02 & 9.66e-02 & 5.46e-02 & 5.63e-02 & 6.35e-02 \\
& & & & 247 & 236 & 244 & 274 & 289 & 308 \\
& 10000 & & & 9980 & 9968 & 9934 & & 9985 & 9953 \\
& & & & 1.58e-02 & 1.71e-02 & 2.04e-02 & & 1.54e-02 & 1.86e-02 \\
& & & & 3565 & 3381 & 3101 & & 3520 & 3245 \\
\hline G10 & 100 & & & 3378 & 5716 & 5486 & 3557 & 7146 & 6910 \\
& & & & 6.61e-01 & 4.27e-01 & 4.50e-01 & 6.43e-01 & 2.84e-01 & 3.07e-01 \\
& & & & 17 & 17 & 16 & 24 & 28 & 29 \\
& 1000 & & & 9156 & 9126 & 9007 & 9424 & 9405 & 9325 \\
& & & & 8.23e-02 & 8.54e-02 & 9.73e-02 & 5.54e-02 & 5.74e-02 & 6.54e-02 \\
& & & & 224 & 224 & 203 & 235 & 222 & 229 \\
& 10000 & & & 9816 & 9804 & 9775 & 9844 & 9825 & 9796 \\
& & & & 1.62e-02 & 1.73e-02 & 2.02e-02 & 1.34e-02 & 1.52e-02 & 1.81e-02 \\
& & & & 3244 & 3087 & 2545 & 3019 & 2664 & 2414 \\
\hline G11 & 100 & & & 350 & 635 & 665 & 419 & 853 & 864 \\
& & & & 8.62e-01 & 7.50e-01 & 7.38e-01 & 8.35e-01 & 6.64e-01 & 6.60e-01 \\
& & & & 14 & 15 & 16 & 23 & 22 & 27 \\
& 1000 & & & 1922 & 1932 & 1911 & 2027 & 2035 & 2021 \\
& & & & 2.43e-01 & 2.39e-01 & 2.47e-01 & 2.02e-01 & 1.98e-01 & 2.04e-01 \\
& & & & 211 & 212 & 221 & 264 & 270 & 292 \\
& 10000 & & & 2396 & 2392 & 2379 & 2409 & 2407 & 2402 \\
& & & & 5.65e-02 & 5.79e-02 & 6.31e-02 & 5.11e-02 & 5.22e-02 & 5.42e-02 \\
& & & & 2792 & 2826 & 2650 & 3060 & 3053 & 2891 \\
\hline G12 & 100 & & & 369 & 655 & 652 & 422 & 811 & 869 \\
& & & & 8.53e-01 & 7.39e-01 & 7.41e-01 & 8.32e-01 & 6.77e-01 & 6.54e-01 \\
& & & & 15 & 16 & 17 & 18 & 19 & 21 \\
& 1000 & & & 1961 & 1969 & 1938 & 2047 & 2051 & 2024 \\
& & & & 2.20e-01 & 2.17e-01 & 2.29e-01 & 1.86e-01 & 1.84e-01 & 1.95e-01 \\
& & & & 198 & 200 & 232 & 217 & 229 & 233 \\
& 10000 & & & 2393 & 2390 & 2380 & 2402 & 2400 & 2389 \\
& & & & 4.80e-02 & 4.90e-02 & 5.32e-02 & 4.43e-02 & 4.54e-02 & 4.95e-02 \\
& & & & 2751 & 2928 & 2866 & 2619 & 2559 & 2241 \\
\hline G13 & 100 & & & 351 & 665 & 662 & 418 & 829 & 859 \\
& & & & 8.65e-01 & 7.45e-01 & 7.46e-01 & 8.40e-01 & 6.82e-01 & 6.70e-01 \\
& & & & 8 & 20 & 17 & 13 & 25 & 25 \\
& 1000 & & & 2034 & 2050 & 2003 & 2124 & 2135 & 2112 \\
& & & & 2.20e-01 & 2.14e-01 & 2.31e-01 & 1.85e-01 & 1.81e-01 & 1.90e-01 \\
& & & & 111 & 251 & 231 & 160 & 299 & 283 \\
& 10000 & & & 2490 & 2487 & 2476 & 2499 & 2496 & 2486 \\
& & & & 4.45e-02 & 4.57e-02 & 4.98e-02 & 4.13e-02 & 4.21e-02 & 4.60e-02 \\
& & & & 3046 & 3126 & 2741 & 3158 & 3249 & 2849 \\
\hline G14 & 100 & & & 1313 & 2909 & 2794 & 1335 & 2978 & 3092 \\
& & & & 8.97e-01 & 7.72e-01 & 7.81e-01 & 8.95e-01 & 7.67e-01 & 7.58e-01 \\
& & & & 4 & 9 & 7 & 8 & 10 & 8 \\
& 1000 & & & 9101 & 9127 & 8753 & 9381 & 9379 & 9060 \\
& & & & 2.87e-01 & 2.85e-01 & 3.14e-01 & 2.65e-01 & 2.65e-01 & 2.90e-01 \\
& & & & 302 & 416 & 350 & 256 & 260 & 244 \\
& 10000 & & & 12071 & 12056 & 11929 & 12141 & 12129 & 12010 \\
& & & & 5.44e-02 & 5.56e-02 & 6.56e-02 & 4.90e-02 & 4.99e-02 & 5.93e-02 \\
& & & & 4359 & 4129 & 3151 & 2756 & 2776 & 2488 \\
\hline G15 & 100 & & & 1092 & 2786 & 2750 & 1103 & 3182 & 3163 \\
& & & & 9.14e-01 & 7.80e-01 & 7.83e-01 & 9.13e-01 & 7.49e-01 & 7.51e-01 \\
& & & & 3 & 8 & 6 & 9 & 20 & 16 \\
& 1000 & & & 8824 & 8913 & 8456 & 9133 & 9187 & 8766 \\
& & & & 3.04e-01 & 2.97e-01 & 3.33e-01 & 2.80e-01 & 2.76e-01 & 3.09e-01 \\
& & & & 370 & 388 & 325 & 196 & 432 & 365 \\
& 10000 & & & 11961 & 11947 & 11787 & 12022 & 12014 & 11865 \\
& & & & 5.72e-02 & 5.83e-02 & 7.09e-02 & 5.24e-02 & 5.30e-02 & 6.47e-02 \\
& & & & 4233 & 3815 & 3188 & 4010 & 4101 & 3329 \\
\hline G16 & 100 & & & 1474 & 2979 & 2825 & 1451 & 3070 & 2970 \\
& & & & 8.84e-01 & 7.65e-01 & 7.78e-01 & 8.86e-01 & 7.58e-01 & 7.66e-01 \\
& & & & 8 & 8 & 4 & 9 & 19 & 16 \\
& 1000 & & & 9014 & 9014 & 8653 & 9314 & 9270 & 8952 \\
& & & & 2.90e-01 & 2.90e-01 & 3.19e-01 & 2.67e-01 & 2.70e-01 & 2.95e-01 \\
& & & & 474 & 420 & 273 & 213 & 457 & 423 \\
& 10000 & & & 11981 & 11965 & 11851 & 12052 & 12038 & 11931 \\
& & & & 5.66e-02 & 5.79e-02 & 6.69e-02 & 5.10e-02 & 5.21e-02 & 6.06e-02 \\
& & & & 5148 & 4298 & 2691 & 4131 & 4316 & 3339 \\
\hline G17 & 100 & & & 1175 & 2937 & 2948 & 1179 & 3129 & 3313 \\
& & & & 9.07e-01 & 7.68e-01 & 7.68e-01 & 9.07e-01 & 7.53e-01 & 7.39e-01 \\
& & & & 3 & 9 & 6 & 13 & 15 & 16 \\
& 1000 & & & 8869 & 8919 & 8494 & 9110 & 9170 & 8800 \\
& & & & 3.01e-01 & 2.97e-01 & 3.30e-01 & 2.82e-01 & 2.77e-01 & 3.06e-01 \\
& & & & 168 & 391 & 330 & 331 & 351 & 336 \\
& 10000 & & & 11942 & 11924 & 11790 & 12006 & 11999 & 11870 \\
& & & & 5.86e-02 & 6.00e-02 & 7.06e-02 & 5.35e-02 & 5.41e-02 & 6.43e-02 \\
& & & & 3842 & 3932 & 3807 & 3511 & 3477 & 3448 \\
\hline G18 & 100 & & & 435 & 2134 & 2038 & 451 & 2643 & 2548 \\
& & & & 9.07e-01 & 5.45e-01 & 5.66e-01 & 9.04e-01 & 4.37e-01 & 4.57e-01 \\
& & & & 4 & 26 & 22 & 16 & 30 & 31 \\
& 1000 & & & 4075 & 4062 & 3952 & 4230 & 4222 & 4157 \\
& & & & 1.32e-01 & 1.35e-01 & 1.58e-01 & 9.91e-02 & 1.01e-01 & 1.15e-01 \\
& & & & 186 & 430 & 384 & 341 & 358 & 373 \\
\hline G19 & 100 & & & 210 & 2062 & 1921 & 209 & 2460 & 2376 \\
& & & & 9.52e-01 & 5.27e-01 & 5.60e-01 & 9.52e-01 & 4.36e-01 & 4.56e-01 \\
& & & & 3 & 25 & 24 & 12 & 28 & 28 \\
& 1000 & & & 3816 & 3809 & 3731 & 3944 & 3943 & 3881 \\
& & & & 1.26e-01 & 1.27e-01 & 1.45e-01 & 9.62e-02 & 9.67e-02 & 1.11e-01 \\
& & & & 254 & 447 & 451 & 336 & 344 & 337 \\
\hline G20 & 100 & & & 691 & 2141 & 2028 & 702 & 2534 & 2548 \\
& & & & 8.46e-01 & 5.22e-01 & 5.47e-01 & 8.43e-01 & 4.34e-01 & 4.31e-01 \\
& & & & 6 & 26 & 25 & 18 & 32 & 33 \\
& 1000 & & & 3931 & 3908 & 3809 & 4058 & 4037 & 3961 \\
& & & & 1.22e-01 & 1.27e-01 & 1.49e-01 & 9.32e-02 & 9.79e-02 & 1.15e-01 \\
& & & & 340 & 428 & 444 & 350 & 361 & 376 \\
\hline G21 & 100 & & & 396 & 2161 & 2065 & 403 & 2638 & 2485 \\
& & & & 9.11e-01 & 5.14e-01 & 5.36e-01 & 9.09e-01 & 4.07e-01 & 4.41e-01 \\
& & & & 9 & 27 & 32 & 14 & 28 & 26 \\
& 1000 & & & 3890 & 3879 & 3775 & 4025 & 4019 & 3965 \\
& & & & 1.25e-01 & 1.28e-01 & 1.51e-01 & 9.51e-02 & 9.65e-02 & 1.09e-01 \\
& & & & 491 & 535 & 557 & 334 & 353 & 337 \\
\hline
G22 & 100 & & & 25464 & 24946 & 23293 & 34937 & 35727 & 33583 \\
& & & & 4.72e-01 & 4.83e-01 & 5.17e-01 & 2.76e-01 & 2.60e-01 & 3.04e-01 \\
& & & & 23 & 20 & 21 & 162 & 167 & 76 \\
& 1000 & & & 50906 & 50819 & 50279 & 53128 & 53106 & 52908 \\
& & & & -5.48e-02 & -5.30e-02 & -4.18e-02 & -1.01e-01 & -1.00e-01 & -9.63e-02 \\
& & & & 257 & 257 & 259 & 1392 & 1447 & 692 \\
& 10000 & & & 55618 & 55605 & 55516 & 55827 & 55828 & 55790 \\
& & & & -1.52e-01 & -1.52e-01 & -1.50e-01 & -1.57e-01 & -1.57e-01 & -1.56e-01 \\
& & & & 3021 & 3026 & 2870 & 10531 & 10514 & 8783 \\
\hline
G23 & 100 & && 20667 & 25646 & 24530 & 25985 & 36225 & 34699 \\
& & && 4.83e-01 & 3.58e-01 & 3.86e-01 & 3.50e-01 & 9.33e-02 & 1.31e-01 \\
& & && 31 & 34 & 32 & 99 & 140 & 126 \\
& 1000 & & & 51116 & 51217 & 50980 & 52939 & 53089 & 53198 \\
& & & & -2.79e-01 & -2.82e-01 & -2.76e-01 & -3.25e-01 & -3.29e-01 & -3.32e-01 \\
& & & & 394 & 409 & 394 & 1052 & 1088 & 1081 \\
& 10000 & & & 55639 & 55641 & 55670 & 55764 & 55769 & 55843 \\
& & & & -3.93e-01 & -3.93e-01 & -3.93e-01 & -3.96e-01 & -3.96e-01 & -3.98e-01 \\
& & & & 4605 & 4604 & 4074 & 10485 & 9659 & 9384 \\
\hline
G24 & 100 & && 19364 & 25066 & 23552 & 24082 & 34833 & 33722 \\
& & && 5.16e-01 & 3.73e-01 & 4.11e-01 & 3.98e-01 & 1.29e-01 & 1.57e-01 \\
& & && 28 & 33 & 32 & 99 & 123 & 131 \\
& 1000 & & & 51035 & 51119 & 50828 & 52847 & 52957 & 53104 \\
& & & & -2.76e-01 & -2.78e-01 & -2.71e-01 & -3.21e-01 & -3.24e-01 & -3.28e-01 \\
& & & & 362 & 390 & 389 & 1065 & 1057 & 1123 \\
& 10000 & & & 55615 & 55623 & 55646 & 55750 & 55762 & 55826 \\
& & & & -3.91e-01 & -3.91e-01 & -3.91e-01 & -3.94e-01 & -3.94e-01 & -3.96e-01 \\
& & & & 4371 & 4381 & 4323 & 9844 & 9736 & 9322 \\
\hline
G25 & 100 & && 18750 & 25136 & 23712 & 23942 & 34972 & 34383 \\
& & && 5.31e-01 & 3.71e-01 & 4.07e-01 & 4.01e-01 & 1.25e-01 & 1.40e-01 \\
& & && 28 & 33 & 32 & 102 & 129 & 132 \\
& 1000 & & & 51044 & 51159 & 50885 & 52832 & 52984 & 53120 \\
& & & & -2.77e-01 & -2.79e-01 & -2.73e-01 & -3.21e-01 & -3.25e-01 & -3.29e-01 \\
& & & & 380 & 384 & 392 & 1076 & 1066 & 1140 \\
& 10000 & & & 55632 & 55640 & 55661 & 55768 & 55778 & 55842 \\
& & & & -3.91e-01 & -3.92e-01 & -3.92e-01 & -3.95e-01 & -3.95e-01 & -3.97e-01 \\
& & & & 4386 & 4384 & 4311 & 9864 & 9732 & 9311 \\
\hline
G26 & 100 && &18204 & 24203 & 22165 & 22893 & 32934 & 34850 \\
&& & & 5.45e-01 & 3.95e-01 & 4.46e-01 & 4.27e-01 & 1.76e-01 & 1.28e-01 \\
& && &29 & 34 & 33 & 94 & 142 & 126 \\
& 1000 & & & 50949 & 51053 & 50717 & 52774 & 52909 & 53029 \\
& & & & -2.74e-01 & -2.77e-01 & -2.69e-01 & -3.20e-01 & -3.23e-01 & -3.26e-01 \\
& & & & 399 & 419 & 410 & 1040 & 1130 & 1083 \\
& 10000 & & & 55579 & 55585 & 55608 & 55719 & 55725 & 55789 \\
& & & & -3.90e-01 & -3.90e-01 & -3.91e-01 & -3.94e-01 & -3.94e-01 & -3.95e-01 \\
& & & & 4634 & 4644 & 4127 & 10429 & 9631 & 9361 \\
\hline
G27 & 100 && &61 & 6809 & 6827 & 61 & 10089 & 9704 \\
&& & & 9.96e-01 & 5.89e-01 & 5.88e-01 & 7.32e-01 & 3.93e-01 & 4.14e-01 \\
&& & &19 & 34 & 28 & 97 & 96 & 112 \\
& 1000 & & & 14788 & 14900 & 14719 & 15532 & 15474 & 15450 \\
& & & & 1.07e-01 & 1.01e-01 & 1.12e-01 & 6.25e-02 & 6.59e-02 & 6.74e-02 \\
& & & & 484 & 449 & 368 & 856 & 680 & 1058 \\
& 10000 & & & 16302 & 16294 & 16274 & 16367 & 16331 & 16329 \\
& & & & 1.60e-02 & 1.65e-02 & 1.77e-02 & 1.20e-02 & 1.42e-02 & 1.44e-02 \\
& & & & 6601 & 4931 & 4085 & 7575 & 6637 & 10163 \\
\hline
G28 & 100 && &48 & 6691 & 6615 & 49 & 9725 & 9610 \\
& & && 9.97e-01 & 5.92e-01 & 5.97e-01 & 7.39e-01 & 4.25e-01 & 4.14e-01 \\
& && &20 & 33 & 26 & 89 & 95 & 100 \\
& 1000 & & & 14634 & 14743 & 14534 & 15350 & 15296 & 15285 \\
& & & & 1.08e-01 & 1.01e-01 & 1.14e-01 & 6.42e-02 & 6.75e-02 & 6.81e-02 \\
& & & & 480 & 409 & 349 & 677 & 742 & 1043 \\
& 10000 & & & 16156 & 16145 & 16120 & 16204 & 16165 & 16164 \\
& & & & 1.51e-02 & 1.57e-02 & 1.73e-02 & 1.21e-02 & 1.45e-02 & 1.46e-02 \\
& & & & 6620 & 4461 & 3817 & 7461 & 7249 & 10230 \\
\hline
G29 & 100 & & & 179 & 6951 & 6923 & 5356 & 10267 & 9717 \\
& & & & 9.89e-01 & 5.87e-01 & 5.89e-01 & 6.82e-01 & 3.90e-01 & 4.23e-01 \\
& & & & 23 & 30 & 27 & 64 & 70 & 65 \\
& 1000 & & & 14989 & 15085 & 14864 & 15732 & 15662 & 15482 \\
& & & & 1.10e-01 & 1.04e-01 & 1.17e-01 & 6.55e-02 & 6.97e-02 & 8.04e-02 \\
& & & & 484 & 411 & 337 & 644 & 669 & 642 \\
& 10000 & & & 16539 & 16531 & 16503 & 16606 & 16572 & 16527 \\
& & & & 1.76e-02 & 1.81e-02 & 1.97e-02 & 1.36e-02 & 1.56e-02 & 1.83e-02 \\
& & & & 6608 & 4505 & 3819 & 6538 & 6566 & 6082 \\
\hline
G30 & 100 && & 87 & 6814 & 6671 & 89 & 10271 & 10304 \\
&& & & 9.95e-01 & 5.96e-01 & 6.04e-01 & 6.94e-01 & 4.04e-01 & 3.89e-01 \\
& & && 24 & 38 & 23 & 102 & 97 & 109 \\
& 1000 & & & 15049 & 15144 & 14925 & 15802 & 15734 & 15750 \\
& & & & 1.08e-01 & 1.02e-01 & 1.15e-01 & 6.29e-02 & 6.69e-02 & 6.59e-02 \\
& & & & 466 & 441 & 291 & 702 & 709 & 1022 \\
& 10000 & & & 16591 & 16586 & 16566 & 16658 & 16627 & 16619 \\
& & & & 1.61e-02 & 1.64e-02 & 1.76e-02 & 1.21e-02 & 1.40e-02 & 1.44e-02 \\
& & & & 6290 & 4078 & 3374 & 7146 & 6910 & 10328 \\
\hline
\end{longtable}
\subsection{Estimating Probability Distributions Arising in Genetics}
The following SDP arises in the context of estimating haploid frequencies in a population; see \cite{AroHazKal05} for a discussion. The optimization problem reads as
\begin{equation}\label{eq:probability}
\begin{split}
& \max \inner{{\mathbf{C}},\mathbf{X}}\\
\text{s.t. }& \tr(\mathbf{X})=1\\
& X\succeq 0,\mathbf{X}\geq 0
\end{split}
\end{equation}
This defines SDP over the double nonnegative cone. To bring this problem into the form of our convex programming template \eqref{eq:Opt}, we set $\mathsf{X}=\{\mathbf{X}\in \mathbb{S}^{n}_{+}\vert\tr(\mathbf{X})\leq 1\},g(\mathbf{X})=-\inner{{\mathbf{C}},\mathbf{X}},$ and $\mathsf{K}=\{\mathbf{x}\in\mathbb{S}^{n}\vert X_{ij}\geq 0,i,j=1,\ldots,n\}$ the cone of positive matrices. We take $\mathcal{P}(\mathbf{X})=\mathbf{X}$, together with the $\nu=\frac{n(n+1)}{2}$ logarithmically homogeneous barrier
\[
F(\mathbf{X})=-\sum_{i,j}\log(X_{ij}),
\]
to arrive at a problem instance where our algorithm can be applied directly.
\subsection{The Maximum Stable Set Problem}
Let $G=(\mathcal{V},\mathcal{E})$ be a given undirected graph. A set $\mathcal{W}\subset\mathcal{V}$ is an independent set of $G$ if no two vertices in $\mathcal{W}$ are adjacent. The maximum cardinality of an independent set is denoted by $\alpha(G)$, known as the \emph{stability number} of $G$. The maximum independent set problem is to compute $\alpha(G)$. This problem is known to be NP-hard. A classical semidefinite programming upper bound has been derived in \cite{Lov79}, and is based on the following observation. Let $S$ be an independent set in a graph $G$ and let $x\in\{0,1\}^{n}$ be its incidence vector. Define the matrix $\mathbf{X}=\frac{1}{\abs{S}}xx^{\top}$. This matrix satisfies the following conditions: $\mathbf{X}\in\mathbb{S}^{n}_{+},X_{ij}=0$ for all $(i,j)\in\mathcal{E}$ and $\tr(\mathbf{X})=1$. Furthermore, $\tr(\mathbf{X}\mathbf{1}\1^{\top})=\inner{\mathbf{X}\mathbf{1},\mathbf{1}}=\abs{S}$. It is therefore natural to consider the following semidefinite program
\begin{equation}\label{eq:theta}
\begin{split}
& \max \tr(\mathbf{X}\mathbf{1}\1^{\top})\\
\text{s.t.: } & \tr(\mathbf{X})=1,\mathbf{X}\in\mathbb{S}^{n}_{+},\\
& X_{ij}=0\quad\forall (i,j)\in\mathcal{E}
\end{split}
\end{equation}
Clearly, this is an example of our convex programming template \eqref{eq:Opt}, with $\mathsf{X}=\{\mathbf{X}\in\mathbb{S}^{n}_{+}\vert \tr(\mathbf{X})=1,X_{ij}=0\text{ for }(i,j)\in\mathcal{E}\}$ and $\mathcal{P}$ being trivial. A tighter relaxation can be obtained by adding non-negativity constraints to the entries of the matrix. This leads to the doubly nonnegative relaxation of the Max-Stable-Set problem \citep{KlerkPas02,Yoshise:2010tg}:
\begin{equation}\label{eq:DNN-Stable}
\begin{split}
& \max_{\mathbf{X}} \tr(\mathbf{X}\mathbf{1}\1^{\top}) \\
\text{s.t.: } & \tr(\mathbf{X})=1 \\
& X_{ij}=0\quad \forall (i,j)\in \mathcal{E}\\
& \mathbf{X}\succeq 0,\mathbf{X}\geq 0
\end{split}
\end{equation}
In this formulation, we have the added entry-wise restriction $\mathcal{P}(\mathbf{X})=\mathbf{X}\geq 0$, which will be absorbed within the log barrier $F(\mathbf{X})=-\sum_{(i,j)\notin\mathcal{E}}\log(X_{ij})$.
\subsection{Finding the Fastest Mixing Markov Chain}
We next consider the problem of finding the fastest mixing rate of a Markov chain on a graph \citep{SunBoyDiacXia06}. In this problem, a symmetric Markov chain is defined on an undirected graph $G=(\{1,\ldots,n\},\mathcal{E})$. Given $G$ and weights $d_{ij}$ for $\{i,j\}\in\mathcal{E}$, we are tasked with finding the transition rates $w_{ij}\geq 0$ for each $\{i,j\}\in\mathcal{E}$ with weighted sum smaller than 1, that result in the fastest mixing rate. We assume that $\sum_{ij}d^{2}_{ij}=n^{2}$. The mixing rate is given by the second smallest eigenvalue of the graph's Laplacian matrix, described as
$$L(\mathbf{w})_{ij}=\begin{cases}
\sum_{j:\{i,j\}\in\mathcal{E}} w_{ij} & j=i\\
-w_{ij} & \{i,j\}\in\mathcal{E}\\
0 &\text{otherwise}.
\end{cases}.$$
From $L\mathbf{1}=0$ it follows that $\frac{1}{n}\mathbf{1}$ is a stationary distribution of the process The mixing time is defined as
$T_{\text{mix}}:=\sup_{\pi}\norm{\pi{\mathbf P}(t)-\frac{1}{n}\mathbf{1}}_{TV}$, where the supremum is taken with respect to all distrbutions over the set of nodes, and the norm is the total variation distance. It can be shown that $T_{\text{mix}}\leq \frac{1}{2}\sqrt{n}e^{-\lambda_{2}t}$, where $\lambda_{2}$ is the second-largest eigenvalue of the Laplacian $L$ \cite{wilmer2009markov}. To bound this eigenvalue, we follow the strategy laid out in \cite{SunBoyDiacXia06}. Thus, the problem can be written as
\begin{align*}
\max_{\mathbf{w}}\quad &\lambda_2(L(\mathbf{w}))\\
\text{s.t.}\quad &\sum_{\{i,j\}\in\mathcal{E}} d_{ij}^2w_{ij}\leq 1\\
\qquad & \mathbf{w}\geq 0.
\end{align*}
This problem can also alternatively be formulated as
\begin{align*}
\min_{\mathbf{w}}\quad &\sum_{\{i,j\}\in\mathcal{E}} d_{ij}^2w_{ij}\\
\text{s.t.}\quad &\lambda_2(L(\mathbf{w})) \geq 1\\
\qquad & \mathbf{w}\geq 0.
\end{align*}
Due to the properties of the Laplacian, the first constraint can be reformulated as
\begin{align}
\min_{\mathbf{w}}\quad &\sum_{\{i,j\}\in\mathcal{E}} d_{ij}^2w_{ij}\\
\text{s.t.}\quad &L(\mathbf{w})\succeq \mathbf{I}_{n\times n}-\frac{1}{n} \mathbf{1} \mathbf{1}^\top\\
\qquad & \mathbf{w}\geq 0.
\end{align}
The dual problem is then given by
\begin{equation}\label{eq:SDPMixing}
\begin{split}
\min_{\mathbf{X}\in \mathbb{S}^n_{+}}\quad &\inner{\mathbf{I}_{n\times n}-\frac{1}{n} \mathbf{1} \mathbf{1}^\top ,\mathbf{X}}\\
\text{s.t.}\quad &X_{ii}+X_{jj}-2X_{ij}\leq d_{ij}^2\qquad \{i,j\}\in\mathcal{E}
\end{split}
\end{equation}
To obtain an SDP within our convex programming model, we combine arguments from \cite{SunBoyDiacXia06} and \cite{AroHazKal05}. Let $\mathbf{X}$ be a feasible point for \eqref{eq:SDPMixing}. Then, there exists a $n\times n$ matrix ${\mathbf V}$ such that $\mathbf{X}={\mathbf V}\bV^{\top}$. It is easy to see that multiplying each row of the matrix ${\mathbf V}^{\top}=[\mathbf{v}_{1},\ldots,\mathbf{v}_{n}]$ with an orthonormal matrix does not change the feasibility of the candidate solution. Moreover $X_{ij}=\mathbf{v}_{i}^{\top}\mathbf{v}_{j}$ for all $1\leq i,j\leq n$, so that $X_{ii}+X_{jj}-2X_{ij}=\norm{\mathbf{v}_{i}-\mathbf{v}_{j}}^{2}_{2}$. Without loss of generality, we can normalize the vectors $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}$ so that $\sum_{i=1}^{n}\mathbf{v}_{i}=0$. This implies $\inner{\mathbf{I}_{n\times n}-\frac{1}{n} \mathbf{1} \mathbf{1}^\top ,\mathbf{X}}=\tr(\mathbf{X})=\sum_{i=1}^{n}\norm{\mathbf{v}_{i}}^{2}$. This gives the equivalent optimization problem
\begin{equation}\label{eq:L2Embedding}
\begin{split}
\max_{\mathbf{v}_{1},\ldots,\mathbf{v}_{n}}\quad & \sum_{i=1}^{n}\norm{\mathbf{v}_{i}}^{2}\\
\text{s.t.}\quad & \norm{\mathbf{v}_{i}-\mathbf{v}_{j}}^{2}\leq d^{2}_{ij}\qquad\{i,j\}\in\mathcal{E},\\
&\mathbf{v}_{i}\in\mathbb{R}^n, \sum_{i=1}^{n}v_{i}=0.
\end{split}
\end{equation}
This is the geometric dual derived in \cite{SunBoyDiacXia06}, which is strongly connected to the geometric embedding of a graph in the plane \citep{GorHelWap08}. Set $\mathsf{X}=\{\mathbf{X}\in\mathbb{S}^{n}\vert \mathbf{X}\succeq 0,\tr(\mathbf{X})\leq \frac{Dn}{2}\}$, where $D=\max_{ij}d_{ij}^{2}$. Then, we can add this trace constraint to the problem formulation, and obtain the equivalent formulation: Define $\mathcal{D}:\mathbb{S}^{n}\to\mathbb{R}^{\abs{\mathcal{E}}}$ given by $\mathcal{D}(\mathbf{X})_{\{i,j\}}=X_{ii}+X_{jj}-2X_{ij}$. Let $\mathsf{Q}=\{\mathbf{y}\in\mathbb{R}^{\abs{\mathcal{E}}}\vert y_{i,j}\leq d^{2}_{i,j},\{i,j\}\in\mathcal{E}\}$ and $\mathsf{K}=\{(y,t)\in\mathbb{R}^{\abs{\mathcal{E}}}\times\mathbb{R}\vert\frac{1}{t}y\in\mathsf{Q},t>0\}$. This is a closed convex cone with logarithmically homogeneous barrier
\[
f(y,t)=-\sum_{\{i,j\}\in\mathcal{E}}\log(t d^{2}_{i,j}-y_{i,j})=-\sum_{e\in\mathcal{E}}\log(d^{2}_{i,j}-\frac{1}{t}y_{i,j})-\abs{\mathcal{E}}\log(t).
\]
This gives a logarithmically homogeneous barrier $F(\mathbf{X},t)=f(\mathcal{P}(\mathbf{X},t))$, where $\mathcal{P}(\mathbf{X},t)=[\mathcal{D}(\mathbf{X});t]\in\mathbb{R}^{\abs{\mathcal{E}}}\times\mathbb{R}$.
\begin{equation}\label{eq:SDPMixing}
\begin{split}
\min_{\mathbf{X}\in \mathbb{S}^n_{+},t>0}\quad &\inner{\mathbf{I}_{n\times n}-\frac{1}{n} \mathbf{1} \mathbf{1}^\top ,\mathbf{X}}\\
\text{s.t.}\quad &\mathcal{P}(\mathbf{X},t)\in\mathsf{K},t=1\\
& \mathbf{X}\in\mathsf{X}.
\end{split}
\end{equation}
\begin{table}[t]
\centering
\begin{tabular}{lll|r}
\hline
Dataset &
\# Nodes &
\# Edges & SDPT3 Value \\
\hline\hline
1& 100 & 1000 & 15.62\\
2 & 100 & 2000 & 7.93\\
3 & 200 & 1000 & 72.32\\
4 & 200 & 4000 & 17.84\\
5 & 400 & 1000 & 388.52\\
6 & 400 & 8000 & 38.37\\
7 & 800 & 4000 & 333.25\\
8 & 800 & 16000 & 80.03\\
\hline
\end{tabular}
\caption{Mixing datasets characteristics.}
\label{tbl:MixingDataSetsInfo}
\end{table}
\begin{table}[ht!]
\centering
\begin{tabular}{lll|rrrrr}
\hline
Dataset & Iter. & Alg. & $\eta_0$ ($x\cdot\Omega_g$) & $\sigma$ & Obj. Value & Gap & Time \\
\hline
1 & 1000 & CG & 0.01 & 0.99 & 14.18 & 9.23e-02 & 57.33 \\
& & LCG & 2 & 0.99 & 14.67 & 6.07e-02 & 61.07 \\
& 5000 & CG & 1 & 0.25 & 14.34 & 8.23e-02 & 299.84 \\
& & LCG & 2 & 0.99 & 14.99 & 4.08e-02 & 315.06 \\
& 10000 & CG & 2 & 0.9 & 14.36 & 8.06e-02 & 590.61 \\
& & LCG & 2 & 0.99 & 15.12 & 3.22e-02 & 630.42 \\
& 50000 & CG & 2 & 0.99 & 14.56 & 6.80e-02 & 2996.00 \\
& & LCG & 2 & 0.99 & 15.36 & 1.68e-02 & 3172.51 \\
2 & 1000 & CG & 0.01 & 0.99 & 7.24 & 8.65e-02 & 59.50 \\
& & LCG & 2 & 0.99 & 7.40 & 6.65e-02 & 47.08 \\
& 5000 & CG & 1 & 0.25 & 7.27 & 8.33e-02 & 297.11 \\
& & LCG & 2 & 0.99 & 7.54 & 4.89e-02 & 264.89 \\
& 10000 & CG & 1 & 0.25 & 7.27 & 8.26e-02 & 589.45 \\
& & LCG & 2 & 0.99 & 7.60 & 4.12e-02 & 561.60 \\
& 50000 & CG & 2 & 0.99 & 7.33 & 7.47e-02 & 2943.62 \\
& & LCG & 2 & 0.99 & 7.75 & 2.29e-02 & 3037.44 \\
3 & 1000 & CG & 0.01 & 0.99 & 8.78 & 8.79e-01 & 125.78 \\
& & LCG & 2 & 0.99 & 67.95 & 6.04e-02 & 91.18 \\
& 5000 & CG & 0.01 & 0.99 & 67.31 & 6.93e-02 & 717.50 \\
& & LCG & 2 & 0.99 & 69.68 & 3.65e-02 & 485.01 \\
& 10000 & CG & 0.01 & 0.99 & 67.40 & 6.80e-02 & 1512.97 \\
& & LCG & 2 & 0.99 & 70.34 & 2.74e-02 & 1020.81 \\
4 & 1000 & CG & 0.01 & 0.99 & 9.70 & 4.56e-01 & 170.83 \\
& & LCG & 2 & 0.99 & 16.37 & 8.26e-02 & 84.17 \\
& 5000 & CG & 0.01 & 0.99 & 15.98 & 1.04e-01 & 822.57 \\
& & LCG & 2 & 0.99 & 16.82 & 5.72e-02 & 454.63 \\
& 10000 & CG & 0.01 & 0.99 & 15.99 & 1.04e-01 & 1588.48 \\
& & LCG & 2 & 0.99 & 17.02 & 4.59e-02 & 940.77 \\
5 & 1000 & CG & 0.01 & 0.99 & 130.86 & 6.63e-01 & 229.34 \\
& & LCG & 2 & 0.99 & 362.16 & 6.78e-02 & 160.07 \\
& 5000 & CG & 0.01 & 0.99 & 356.87 & 8.15e-02 & 1020.93 \\
& & LCG & 2 & 0.99 & 373.61 & 3.84e-02 & 819.19 \\
& 10000 & CG & 0.01 & 0.99 & 367.95 & 5.29e-02 & 2216.81 \\
& & LCG & 2 & 0.99 & 377.77 & 2.77e-02 & 1668.29 \\
6 & 1000 & CG & 0.01 & 0.99 & 19.71 & 4.86e-01 & 211.59 \\
& & LCG & 1 & 0.99 & 35.29 & 8.03e-02 & 172.43 \\
& 5000 & CG & 0.01 & 0.99 & 34.91 & 9.01e-02 & 1143.19 \\
& & LCG & 1 & 0.99 & 36.17 & 5.72e-02 & 942.42 \\
& 10000 & CG & 0.5 & 0.25 & 35.03 & 8.69e-02 & 2469.83 \\
& & LCG & 1 & 0.99 & 36.57 & 4.67e-02 & 1904.46 \\
7 & 1000 & CG & 0.01 & 0.99 & 89.68 & 7.31e-01 & 669.59 \\
& & LCG & 2 & 0.99 & 311.85 & 6.42e-02 & 441.59 \\
& 5000 & CG & 0.01 & 0.99 & 263.38 & 2.10e-01 & 3589.64 \\
& & LCG & 2 & 0.99 & 320.39 & 3.86e-02 & 2231.72 \\
& 10000 & CG & 0.01 & 0.99 & 312.09 & 6.35e-02 & 6995.37 \\
& & LCG & 2 & 0.99 & 323.44 & 2.95e-02 & 4661.37 \\
8 & 1000 & CG & 0.01 & 0.99 & 10.91 & 8.64e-01 & 667.97 \\
& & LCG & 1 & 0.99 & 72.62 & 9.25e-02 & 427.37 \\
& 5000 & CG & 0.01 & 0.99 & 49.59 & 3.80e-01 & 3859.94 \\
& & LCG & 2 & 0.99 & 75.02 & 6.26e-02 & 2262.00 \\
& 10000 & CG & 0.01 & 0.99 & 71.97 & 1.01e-01 & 7479.36 \\
& & LCG & 2 & 0.99 & 76.04 & 4.98e-02 & 4792.46 \\
\hline
\end{tabular}
\caption{Mixing datasets results. For each dataset, number of iterations, and algorithm provides the best parameter choice for $\eta_0$ and $\sigma$, the function value obtained, the relative gap from the SDPT3 solution, and the time.}
\label{tbl:MixingDataSetsResults}
\end{table}
The Mixing problem was run using Matlab R2021b on an Intel(R) Xeon(R) Gold 6354 CPU @ 3.00GHz server limited to 4 threads per run and 383G total RAM. We generated random connected undirected graphs of various sizes, and for each edge $\{i,j\}$ in the graph we generated a random $d_{ij}^2$ uniformly in $[0,1]$. Table~\ref{tbl:MixingDataSetsInfo} provides the size of each dataset, and the value obtained by solving the Mixing problem SDP using CVX with SDPT3 solver. All datasets were run for both $\CG$ and $\LCG$ options with the following choice of parameters $\eta_0\in\{0.01\Omega_g,0.1\Omega_g,0.5\Omega_g,1\Omega_g,2\Omega_g\}$
and $\sigma\in\{0.99,0.9,0.5,0.25\}$. Each of the runs was terminated after it reached at least $10000$ iterations at least $3600$ seconds running time. Figures \ref{fig:Mixing_GapVsIter}-\ref{fig:Mixing_GapVsTime} illustrate the results for some of the parameter values. Table \ref{tbl:MixingDataSetsResults} displays the numerical values obtained from our experiments with the best configuration of parameters $\eta_{0}$ and $\sigma$.
\begin{figure}[h]
\centering
\includegraphics[scale=0.7]{./Figures/Mixing_gap_vs_iter.pdf}
\caption{Mixing datasets relative gap from SDPT3 solution vs. iteration for various parameter choices.}\label{fig:Mixing_GapVsIter}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[scale=0.7]{./Figures/Mixing_gap_vs_time.pdf}
\caption{Mixing datasets relative gap from SDPT3 solution vs. time for various parameter choices.}\label{fig:Mixing_GapVsTime}
\end{figure}
In the experiments we observe that the line search version $\LCG$ outperforms $\CG$ by orders of magnitude. One explanation of this is that the step size policy employed in $\CG$ is based on global optimization ideas which do not take into account the local structure of the problem. Line search captures these local features of the problem much better, leading to better numerical performance.
\subsection{Self-Concordant Functions}
\label{sec:SCB}
Let $\mathsf{Q}$ be an open and convex subset of $\mathsf{E}$. A function $f:\mathsf{E}\to(-\infty,\infty]$ is \ac{SC} on $\mathsf{Q}$ if $f\in{\mathbf{C}}^{3}(\mathsf{Q})$ and for all $x\in\mathsf{Q},h\in\mathsf{E}$, we have
$\abs{f'''(x)[h,h,h]}\leq 2(f''(x)[h,h])^{3/2}.$ In case where $\cl(\mathsf{Q})$ is a closed convex cone, more structure is available to us. We call $f$ a $\nu$-canonical barrier for $\mathsf{Q}$, denoted by $f\in\mathcal{B}_{\nu}(\mathsf{Q})$, if it is \ac{SC} and
\[
\forall (x,t)\in\mathsf{Q}\times(0,\infty):\quad f(tx)=f(x)-\nu\log(t).
\]
From \citep[][Prop. 2.3.4]{NesNem94}, the following properties are satisfied by a $\nu$-canonical barrier for each $x\in\mathsf{Q},h\in\mathsf{E},t>0$:
\begin{align}
&f'(tx)=t^{-1}f'(x),\\
&f'(x)[h]=-f''(x)[x,h]\quad\forall h\in\mathsf{E},\\
&f'(x)[x]=-\nu\\
&f''(x)[x,x]=\nu.
\end{align}
We define the local norm $\norm{h}_{f''(x)}:=(f''(x)[h,h])^{1/2}$ for all $x\in\dom(f)$ and $h\in\mathsf{E}$.
\ac{SC} functions are in general not Lipschitz smooth. Still, we have access to a version of a descent lemma of the following form \cite[][Thm. 5.1.9]{Nes18}:
\begin{equation}\label{eq:descent}
f(x+h)\leq f(x)+f'(x)[h]+\omega_{\ast}(\norm{h}_{f''(x)})
\end{equation}
for all $x\in\dom(f)$ and $h\in\mathsf{E}$ such that $\norm{h}_{f''(x)}<1$, where $\omega_{\ast}(t)=-t-\log(1-t)$. It also holds true that \cite[][Thm. 5.1.8]{Nes18}
\begin{equation}\label{eq:SC-convex}
f(x+h)\geq f(x)+f'(x)[h]+\omega(\norm{h}_{f''(x)})
\end{equation}
for all $x\in\dom(f)$ and $h\in\mathsf{E}$ such that $x+h\in \dom(f)$, where $\omega(t)=t-\log(1+t)$ for $t\geq 0$.
A classical and useful bound is \cite[][Lemma 5.1.5]{Nes18}:
\begin{equation}\label{eq:boundomega}
\frac{t^{2}}{2-t}\leq\omega_{\ast}(t)\leq\frac{t^{2}}{2(1-t)}\quad\forall t\in[0,1).
\end{equation}
\subsection{The Optimization Problem}
\label{sec:opt}
The following assumptions are made for the rest of this paper.
\begin{assumption}\label{ass:barrier}
$\mathsf{K}\subset\mathsf{H}$ is a closed convex cone with $\Int(\mathsf{K})\neq\emptyset$, admitting a $\nu$-canonical barrier $f\in\mathcal{B}_{\nu}(\mathsf{K})$.
\end{assumption}
The next assumption transports the barrier setup from the codomain $\mathsf{K}$ to the domain $\mathcal{P}^{-1}(\mathsf{K})$. This is a common operation in the framework of the "barrier calculus" developed in \cite[][Section 5.1]{NesNem94}.
\begin{assumption}
The map $\mathcal{P}:\mathsf{E}\to\mathsf{H}$ is linear, and $F(x):=f(\mathcal{P}(x))$ is a $\nu$-canonical barrier on the cone $\mathcal{P}^{-1}(\mathsf{K})$, i.e. $F\in\mathcal{B}_{\nu}(\mathcal{P}^{-1}(\mathsf{K}))$.
\end{assumption}
Note that $\dom(F)=\Int\mathcal{P}^{-1}(\mathsf{K})$.
At this stage a simple example might be useful to illustrate the working of this transportation technique.
\begin{example}
Considering the normalized covering problem presented in Example \ref{ex:covering}.
\cite{ElbMakNaj22} solve this problem via a logarithmic potential function method \cite{NesNem94}, involving the logarithmically homogeneous barrier $f(\mathbf{X})=\log\det(\mathbf{X})$ for $\mathbf{X}\in\mathbb{S}^{n}_{+}$.
This is a typical choice in Newton-type methods to impose the semidefiniteness constraint. However, in our projection-free environment, we use the power of the linear minimization oracle to obtain search directions which leaves the cone of positive semidefinite matrices invariant. Instead, we employ barrier functions to incorporate the additional linear constraints in \eqref{eq:Opt}. Hence, we set $F(x)=f(\mathcal{P}(x))=\log\det(\sum_{i=1}^{m}x_{i}{\mathbf A}_{i}-\mathbf{I})$ to absorb the constraint $\mathcal{P}(x)\in\mathbb{S}^{n}_{+}$. In particular, this frees us from matrix inversions, and related computationally intensive steps coming with Newton and interior-point methods.
\end{example}
\begin{assumption}\label{ass:slater}
$\mathsf{X}$ is a nonempty compact convex set in $\mathsf{E}$, and $\mathsf{C}:=\mathsf{X}\cap\dom(F)\cap\dom(g)\neq\emptyset$.
\end{assumption}
Let $\Opt:=\min\{g(x)\vert x\in\mathsf{X},\mathcal{P}(x)\in\mathsf{K}\}$. Thanks to assumptions \ref{ass:barrier} and \ref{ass:slater}, $\Opt$ is attained. Our goal is to find an $\varepsilon$-solution of problem \eqref{eq:Opt}, defined as follows.
\begin{definition}\label{def:eps}
Given a tolerance $\varepsilon>0$, we say that $z^{\ast}_{\varepsilon}$ is an $\varepsilon$-solution for \eqref{eq:Opt} if
$$
z^{\ast}_{\varepsilon}\in\mathcal{P}^{-1}(\mathsf{K})\cap\mathsf{X}\text{ and } g(z^{\ast}_{\varepsilon})-\Opt\leq \varepsilon.
$$
\end{definition}
We underline that we seek for a \emph{feasible} $\varepsilon$-solution of problem \eqref{eq:Opt}.
Given $t>0$, define
\begin{equation}\label{eq:Optt}
\Opt(t):=\min_{x\in\mathsf{X}}\{V_{t}(x)=\frac{1}{t}F(x)+g(x)\},\text{ and }z^{\ast}(t)\in\{x\in\mathsf{X}\vert V_{t}(x)=\Opt(t)\}.
\end{equation}
The following Lemma shows that the path $t\mapsto z^{\ast}(t)$ traces a trajectory in the interior of the feasible set which can be used to approximate a solution of the original problem \eqref{eq:Opt}, provided the penalty parameter $t$ is chosen large enough.
\begin{lemma}\label{lem:pathfollowing}
For all $t>0$, it holds that $z^{\ast}(t)\in\mathsf{C}$. In particular,
\begin{equation}
g(z^{\ast}(t))-\Opt\leq \frac{\nu}{t}.
\end{equation}
\end{lemma}
\begin{proof}
Since $F\in\mathcal{B}_{\nu}(\mathcal{P}^{-1}(\mathsf{K}))$, it follows immediately that $z^{\ast}(t)\in\dom(F)\cap\mathsf{X}$. By Fermat's principle we have
$$
0\in \frac{1}{t}F'(z^{\ast}(t))+\partial g(z^{\ast}(t))+\NC_{\mathsf{X}}(z^{\ast}(t)).
$$
Convexity and \cite[][Thm. 5.3.7]{Nes18} implies that for all $z\in\dom(F)\cap\mathsf{X}$, we have
$$
g(z)\geq g(z^{\ast}(t))-\frac{1}{t}F'(z^{\ast}(t))[z-z^{\ast}(t)]\geq g(z^{\ast}(t))-\frac{\nu}{t}.
$$
Let $z^{\ast}$ be a feasible point point satisfying $g(z^{\ast})=\Opt$. Choosing a sequence $\{z^{j}\}_{j\in\mathbb{N}}\subset\dom(F)\cap\mathsf{X}$ with $z^{j}\to z^{\ast}$, the continuity on $\dom(g)$ gives
$$
\Opt\geq g(z^{\ast}(t))-\frac{\nu}{t}.
$$
\end{proof}
Lemma \ref{lem:pathfollowing} seems to indicate that our aim of finding an $\varepsilon$-solution is already achieved: It seems to suffice to pick $t=\frac{\nu}{\varepsilon}$, in order to meet our target. Unfortunately this na\"{i}ve choice is usually not working well in practice. Choosing $t$ large (i.e. $\varepsilon$ small) from the outset means that the subproblems involved in any iterative solver are nearly ill-conditioned. In practice this typically manifests in numerical stability issues and slow convergence. Instead, path-following ideas and continuation methods are usually designed in which the homotopy parameter $t$ is increasing.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,375 |
Q: Model attributes from dropdown list not transferring I have a resume table that has (among other things) a position_type_id column that is a foreign key to the position_type table.
On the resume creation form, the options for the dropdown list for position type are pulled correctly from the position_type table, and they are set with their id's as the values for the selections. All seems to be going well.
When the form is submitted, I can verify that the values are going back to the controller in the POST variable, however they are not passed into the model with the attribute population:
$model->attributes=$_POST['Resume'];
Here is the create method on the controller:
public function actionCreate()
{
$model=new Resume;
if(isset($_POST['Resume']))
{
$model->attributes=$_POST['Resume'];
if($model->save())
{
$this->redirect(array('view','id'=>$model->id));
}
}
$this->render('create',array(
'model'=>$model,
));
}
Here is the form that loads the keys from the related model:
echo $form->activeDropDownList(
$model,
'position_type_id',
CHtml::listData(PositionType::model()->findAll(),'id','name')
);
position_type_id is the key on the resume model that carries the foreign key to the PositionType model. It is spelled the same on the form and on the resume model. The relation on the model is: 'positionType' => array(self::BELONGS_TO, 'PositionType', 'position_type_id'),
I suppose I could get each value out of the POST array and set them manually, but it seems like this should 'just work'. The values on the $model after the attributes get set are there for the manually entered fields, and blank for all fields that come from a generated dropdown list.
Here is what actually gets generated:
<select name="Resume[position_type_id]" id="Resume_position_type_id">
<option value="1">English Teacher</option>
<option value="2">School Administrator</option>
</select>
A: I could be wrong but for me the massive assignment is not working. In yii an attribute needs to be declared as safe if we need to perform massive assignment on it.
So In your model Resume do you have the following rule:
public function rules() {
return array(
//the other rules ...
array('position_type_id, others_attributes', 'safe'),
//the rule above indicate the attributes that can be massively assigned
);
}
A link to the wiki about the 'safe' validation rule
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,215 |
Q: Path specs and multiple servlets I'd like to have two servlets (I'm using Jetty) serve URLs like this:
host/aaa/submit
host/bbb/submit
I've tried setting the servlets' pathspecs to aaa and bbb respectively, but then I get an exception along the lines of
two controller methods annotated with @RequestMapping(value = {"/submit"})
(even though the two methods in question are defined in two separate controller classes that are used by two different servlets).
if instead I set both servlets' pathspecs to / and change the @RequestMappings to aaa/submit and bbb/submit, I'm getting 404s. (This, I suppose, is less surprising - not sure how it should work with effectively two 'default' servlets)
How should I map those URLs? (preemptively - they do need to be separate servlets - aaa part should work with or without DB, bbb part should fail without DB)
Just in case, here's the Jetty context:
<property name="servletHandler">
<bean class="org.mortbay.jetty.servlet.ServletHandler">
<property name="servlets">
<list>
<bean name="aaaServlet" class="org.mortbay.jetty.servlet.ServletHolder">
<property name="name" value="aaa" />
<property name="servlet">
<bean class="org.springframework.web.servlet.DispatcherServlet" />
</property>
<property name="initParameters">
<map>
<entry key="contextConfigLocation" value="classpath:aaa-context.xml" />
</map>
</property>
</bean>
<bean name="bbbServlet" class="org.mortbay.jetty.servlet.ServletHolder">
<property name="name" value="bbb" />
<property name="servlet">
<bean class="org.springframework.web.servlet.DispatcherServlet" />
</property>
<property name="initParameters">
<map>
<entry key="contextConfigLocation" value="classpath:bbb-context.xml" />
</map>
</property>
</bean>
</list>
</property>
<property name="servletMappings">
<list>
<bean class="org.mortbay.jetty.servlet.ServletMapping">
<property name="servletName" value="aaa" />
<property name="pathSpec" value="/" />
</bean>
<bean class="org.mortbay.jetty.servlet.ServletMapping">
<property name="servletName" value="bbb" />
<property name="pathSpec" value="/" />
</bean>
</list>
</property>
</bean>
</property>
And the two controllers that look like this:
@Controller
public class AaaController {
@RequestMapping(value = {"/aaa/submit"}, method = (RequestMethod.POST))
public void handleAaaSubmitPostRequest(final HttpServletRequest request,
final HttpServletResponse response,
@RequestBody String body) throws IOException {
}
And
@Controller
public class BbbController {
@RequestMapping(value = {"/bbb/submit"}, method = (RequestMethod.POST))
public void handleBbbSubmitPostRequest(final HttpServletRequest request,
final HttpServletResponse response,
@RequestBody String body) throws IOException {
}
A: Since your controllers seem to be in different mvc-contexts, you need to change your controller according to this:
@Controller
public class AaaController {
@RequestMapping(value = {/*aaa*/"/submit"}, method = (RequestMethod.POST))
public void handleAaaSubmitPostRequest(final HttpServletRequest request,
final HttpServletResponse response,
@RequestBody String body) throws IOException {
}
and
@Controller
public class BbbController {
@RequestMapping(value = {/*bbb*/"/submit"}, method = (RequestMethod.POST))
public void handleBbbSubmitPostRequest(final HttpServletRequest request,
final HttpServletResponse response,
@RequestBody String body) throws IOException {
}
This happens because Spring MVC is always using path without servlet context.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,422 |
Jochū Tengin (如仲天誾, aussi 恕仲天誾; 1363-1437) est un moine Zen Sōtō. Il reçoit sa transmission dharma de Baisan Monpon et l'école Sōtō le considère comme un patriarche.
À l'époque de Jochū, l'institution et l'organisation de la lignée Keizan du Zen Sōtō est achevée. Ses disciples, Kisan Shōsan et Shingan Dōkū, créent des lignées séparées du dharma qui sont honorées dans des temples différents au sein de l'école.
Notes et références
Naissance en 1363
Décès en 1437
Moine bouddhiste zen
Moine bouddhique japonais
Zen
Article dont des informations diffèrent sur Wikidata | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,254 |
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\section*{Introduction}
In this paper we apply for mass composition and hadronic interaction
studies the recently found shower universality property, stating that
the ratio of the muon signal to the EM one \ensuremath{S_\mu/S_\mathrm{em}}\ is the same for
all hadronic showers having the maximum at the same vertical depth
\ensuremath{X_\mathrm{max}^\mathrm{v}}~\cite{ya_PRD2010,ya_icrc2011}. This property provides a very
simple parametrization for the muon signal~\cite{ya_PRD2010}
\begin{equation}
\label{eq:mufit}
S_\mu^\mathrm{fit}=S_{1000}/(1+1/(\left(\ensuremath{X_\mathrm{max}^\mathrm{v}}/A\right)^{1/b}-a)),
\end{equation}
where $A,\,a$ and $b$ are the fit parameters and $S_{1000}$ is the
total ground plane signal in water Cherenkov detectors similar to the
detectors of the Pierre Auger
Observatory~\cite{PAO_proto_NIMA2004}. We use the large set of CORSIKA
showers for interaction models QGSJET~II/Fluka and EPOS~1.99/Fluka,
described in detail in~\cite{ya_icrc2011}, to demonstrate that for
zenith angles above 45 degrees, where EM halo from muon decays and
interactions composes large part of the EM signal, the discussed
universality allows to find the muon signal in almost interaction
model independent way. This is done using QGSJET~II muon signal
parametrizations on EPOS~1.99 simulations which serve in the given
case as a fake data. Once the muon signal is found the independence of
the \ensuremath{S_\mu/S_\mathrm{em}}\ on interaction model properties and independence of the
EM signal on the primary mass are used to determine the scaling factor
between QGSJET~II and EPOS~1.99. It is shown that such scaling
procedure allows to extract with a good accuracy primary mass
composition from EPOS~1.99 samples with the use of the QGSJET~II
model.
\section{$\mathbf{\ensuremath{S_\mu/S_\mathrm{em}}}$ universality in inclined
$\mathbf{\theta>45^\circ}$ showers}
In~\cite{ya_PRD2010} it was demonstrated that the parametrization in
the form~(\ref{eq:mufit}) provides unbiased estimate of the muon
signal with RMS of 8\% and 5\% for protons and iron correspondingly if
the \arange{0}{65} angular range is considered. Here we would like to
study in more detail only inclined $\theta>45^\circ$ showers, since
\ensuremath{S_\mu/S_\mathrm{em}}\ here should be very similar for both QGSJET~II and
EPOS~1.99~\cite{ya_icrc2011}.
From Fig.~\ref{fig:univ45} one can note that \ensuremath{S_\mu/S_\mathrm{em}}\ in proton
showers at some fixed \ensuremath{X_\mathrm{max}^\mathrm{v}}\ is slightly larger than in iron ones for
\arange{45}{65} angular range. As discussed in~\cite{ya_PRD2010}
proton showers having the same average depth of maximum \ensuremath{X_\mathrm{max}^\mathrm{v}}\ with
iron showers should be more inclined since they have deeper \ensuremath{X_\mathrm{max}}. This
in turn means that particles in proton showers should cross larger
slant distance along shower axis from the shower maximum to the
observation level and the fraction of $\pi^0$ EM component in them
should be smaller than in iron showers, bringing to larger \ensuremath{S_\mu/S_\mathrm{em}}.
For this reason, as our calculations show, muon signal for $p$ and
Fe showers obtained with the fit~(\ref{eq:mufit}) on average
slightly ($<2\%$) differ from the simulated signals, and for the EM
signals this difference is within 7\% with RMS deviations from the
simulated signal reaching 15\% and 12\% for $p$ and Fe
correspondingly. It is possible to improve the muon and EM signals
recovery. Taking into account that \ensuremath{S_\mu/S_\mathrm{em}}\ in proton showers at fixed
\ensuremath{X_\mathrm{max}^\mathrm{v}}\ is larger than in iron showers and that at the same time
proton showers are slightly more inclined than iron ones it is enough
to multiply \ensuremath{S_\mu/S_\mathrm{em}}\ by $\cos^\alpha(\theta)$ finding $\alpha$ which
allows to reduce to minimum the difference in
$\ensuremath{S_\mu/S_\mathrm{em}}\cos^\alpha(\theta)$ between $p$ and Fe showers. We have found
that this is achieved for $\alpha\approx1-1.5$ and eventually we have
chosen $\alpha=1.2$. In this case the muon signal parametrization
looks like this
\begin{equation}
\label{eq:smucos}
S_\mu^\mathrm{fit}=\frac{S_{1000}}{1+\cos^\alpha(\theta)/\left((\ensuremath{X_\mathrm{max}^\mathrm{v}}/A)^{1/b}-a\right)}.
\end{equation}
we have performed the fits with~(\ref{eq:smucos}) in three energy
ranges $\logen18.5-19.0,\ 19.0-19.5,\ 19.5-20.0$ and the fit
parameters are given in the Table~\ref{tab:fit}. The use of this
approach brings to unbiased estimates of both muon and EM signals with
RMS around 10\% for EM signals and 5\% (3\%) for muon
signal in proton (iron) showers.
\begin{figure}[b]
\centering\includegraphics[width=0.42\textwidth]{icrc0694_fig01.eps}
\caption{\ensuremath{S_\mu/S_\mathrm{em}}\ vs \ensuremath{X_\mathrm{max}^\mathrm{v}}\ universality for QGSJET~II. Red
squares~---~protons, blue crosses~--~iron nuclei, black
line~---~fit~(\ref{eq:mufit}). $\logen18.8-19.0,\ \theta=45^\circ-65^\circ$.}
\label{fig:univ45}
\end{figure}
\begin{table}[b]
\caption{Fit parameters in~(\ref{eq:smucos}) for QGSJET~II and
EPOS~1.99 interaction models.}
\label{tab:fit}
\begin{center}
\renewcommand{\tabcolsep}{6pt}
\small
\begin{tabular}{crrrrrr}
\hline
&\multicolumn{3}{c}{QGSJET~II}&\multicolumn{3}{c}{EPOS~1.99}\\
$\lg(E/\mathrm{eV})$ & $A$ & $a$ & $b$ & $A$ & $a$ & $b$ \\
\hline
$18.5-19.0$ & 2070 & 2.54 & -1.13 & 10550 & 4.53 & -1.76\\
$19.0-19.5$ & 1045 & 1.52 & -0.81 & 2028 & 2.42 & -1.11\\
$19.5-20.0$ & 742 & 0.94 & -0.62 & 871 & 1.11 & -0.70\\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{figure}[thb]
\centering\includegraphics[width=0.42\textwidth]{icrc0694_fig02a.eps}
\centering\includegraphics[width=0.42\textwidth]{icrc0694_fig02b.eps}
\caption{Means and RMS of distributions of relative difference between
MC simulated EM (top) and muon (bottom) EPOS~1.99 signals and
signals obtained from the fit~(\ref{eq:smucos}) with parameters for
QGSJET~II.}
\label{fig:diffcosQE}
\end{figure}
In Fig.~\ref{fig:diffcosQE} we present the result of application of
formula~(\ref{eq:smucos}) with the fit parameters for QGSJET~II to the
dataset simulated with EPOS~1.99. The increasing role of the EM halo,
that brings to almost the same scaling of total and muon
signals~\cite{ya_icrc2011}, and the similarity of \ensuremath{S_\mu/S_\mathrm{em}}\ behavior on
\ensuremath{X_\mathrm{max}^\mathrm{v}}\ allow to derive muon signal from EPOS~1.99 simulations with
errors below 6\% for protons and 3\% for iron showers applying
formula for QGSJET~II (and vice versa). It is seen that with the
increase of the energy interaction model invariance becomes more
violated due to the increasing fraction of $\pi^0$ EM component
arriving at the observation level.
\section{Determination of the absolute signal scaling factor}
\begin{figure}[thb]
\centering\includegraphics[width=0.42\textwidth]{icrc0694_fig03a.eps}
\centering\includegraphics[width=0.42\textwidth]{icrc0694_fig03b.eps}
\caption{Top: EM signals for QGSJET~II (closed symbols) and EPOS~1.99
(open symbols) normalized by primary energy and
$f(\theta)=1-4.3\cos(\theta)+4.9\cos^2(\theta)$. Bottom: ratio of the
EPOS~1.99 to QGSJET~II EM signals, mean value is
1.23.}
\label{fig:semnorm}
\end{figure}
\begin{figure}[thb]
\centering\includegraphics[width=0.42\textwidth]{icrc0694_fig04a.eps}
\centering\includegraphics[width=0.42\textwidth]{icrc0694_fig04b.eps}
\caption{Top: ratio of the parametrized EPOS~1.99 EM signals to the
EM signal for QGSJET~II oxygen for energies below $10^{19}$
eV fitted with flat line. Mean values are 1.34, 1.28 and 1.27 for
$p$, O and Fe correspondingly. Bottom: scaled by 1.26 QGSJET~II muon
signals and muon signals obtained from EPOS~1.99 simulated dataset
with~(\ref{eq:smucos}) and coefficients for
QGSJET~II.}
\label{fig:smuscaledfitted}
\end{figure}
\begin{figure}[htb]
\centering\includegraphics[width=0.42\textwidth]{icrc0694_fig05a.eps}
\centering\includegraphics[width=0.42\textwidth]{icrc0694_fig05b.eps}
\caption{Top: reconstructed with QGSJET~II and MTA on the basis of
Fisher's variables distributions proton (red squares) and iron (blue
crosses) abundances in the EPOS~1.99 samples with known primaries
content. Bottom: same but for proton--oxygen (brown diamonds)
mixture. Lines mark the true primary fractions. $\logen18.90-19.10$}
\label{fig:mass}
\end{figure}
The muon signal is one of the most powerful shower characteristics for
the mass composition analysis. The procedure of determination of the
muon signal proposed in the previous section provides the way to get
the muon signal from hybrid data in Auger-like experiments, but due to
well-known problem of muon deficit in Monte-Carlo (MC) shower
simulation
codes~\cite{abuzayyad_mass2000,engel_icrc2007,ave_munum_2007} the
retrieved signal can not be used in mass composition studies. The
latter could become possible only if one would be able to find
absolute signal scaling factor
$S(p,\,\mathrm{real\ data})/S(p,\,\mathrm{MC})$. We would like to
propose a possible way to solve this problem using as a fake data
EPOS~1.99 simulations and QGSJET~II as a test interaction model. Of
course, for our fake data produced with EPOS~1.99 we know precisely
the total, muon and EM signals and this will help us to estimate the
accuracy of the proposed procedure.
Since primary mass composition is unknown the only way to determine
the scaling factor is to use mass composition independent shower
characteristics and the only appropriate candidate to this role is the
EM signal. In Fig.~\ref{fig:semnorm} one can see its mass independence
for QGSJET~II and EPOS~1.99 (the signals are divided by
$f(\theta)=1-4.3\cos(\theta)+4.9\cos^2(\theta)$ to reduce spread
caused by the spread of zenith angles) in the energy range
\erange{18.5}{19.5}. At the bottom it is shown the ratio of the
EPOS~1.99 EM signals to QGSJET~II ones and the average of this ratio
is equal to 1.23. Due to increasing role of the EM halo in the
considered angular range \arange{45}{65} \ensuremath{S_\mu/S_\mathrm{em}}\ becomes less
dependent on the hadronic interaction properties~\cite{ya_icrc2011}
providing the following approximate equality which holds true for any
primary nuclei ($p$, O, Fe etc.):
\begin{equation}
\label{eq:allscales}
\frac{\smillam{EPOS}}{\smillam{QGS}}\approx\frac{\smum{EPOS}}{\smum{QGS}}\approx\frac{\semm{EPOS}}{\semm{QGS}}.
\end{equation}
The ratios of the total and muon signals confirm these relation with
quite good accuracy: their average values for all primaries (1.25 and
1.26) are quite the same as the value for the EM signals. The ratios
of the total and muon signals remain constant across the entire
considered energy range.
Now it remains only to try to scale the QGSJET~II model predictions
using EPOS~1.99 set as a fake data. To find EM and muon signals in the
`data' one should simply apply the parametrization~(\ref{eq:smucos})
with coefficient for QGSJET~II to EPOS~1.99 simulations and to find
the ratio of the EM signal from the `data' to the QGSJET~II one. From
Fig.~\ref{fig:diffcosQE} one can see that the error on the EM signal
extracted from EPOS~1.99 can reach 10\% for proton primaries and 6\%
for iron. One can see also that with the increase of the energy the
error in determination of the scaling factor should increase due to
increasing fraction of the $\pi^0$ EM component arriving at ground
level. Hence, it is reasonable to limit the scaling by the energy
range \erange{18.5}{19.0} and in this case one gets the scaling
factors equal to 1.34, 1.28 and 1.27 for $p$, O and Fe correspondingly
(Fig.~\ref{fig:smuscaledfitted}, normalization is done in respect to
QGSJET~II oxygen). The true ratio of the muon signals is 1.26 and in
case of pure iron or mixed primary composition after scaling both
models should give very close predictions of the muon signals, for
pure primary proton flux the discrepancy between the true EPOS~1.99
signals and scaled QGSJET~II signals can exceed 5\%. Evidently, this
error is due to deep proton showers and if one applies
$\ensuremath{X_\mathrm{max}^\mathrm{v}}<500$~g/cm${}^2$\ cut only during scaling procedure this will affect
almost exclusively proton showers and will change the scaling factor
for them to 1.26. Hence, using the cut $\ensuremath{X_\mathrm{max}^\mathrm{v}}<500$~g/cm${}^2$\ the scaling
factor that one gets will be within $1.26-1.28$ range independently on
the mass composition, in very good agreement with the true muon
signals ratio. In Fig.~\ref{fig:smuscaledfitted} the muon signal
obtained from EPOS~1.99 `data' with~(\ref{eq:smucos}) and parameters
for QGSJET~II is compared with the scaled by 1.26 QGSJET~II muon
signal. One can see that after the entire procedure one gets quite
consistent picture with the tendency to underestimation of the `true'
primary mass, since e.g. the `true' proton signals retrieved with the
fit is lower than scaled QGSJET~II signals to which the comparison
should be done.
We have reconstructed a mass composition of proton-oxygen and
proton-iron mixtures prepared with EPOS~1.99 using scaled by 1.26
QGSJET~II model signals. The muon signal from EPOS~1.99 samples was
retrieved with the use of the fit~(\ref{eq:smucos}) with parameters
for QGSJET~II. To discriminate primaries $(\ensuremath{X_\mathrm{max}},\,
\ensuremath{S_\mu}^\mathrm{fit}/(E\ensuremath{X_\mathrm{max}^\mathrm{v}}))$ variables have been used and the
approach has been the same as in~\cite{ya_Lodz}, i.e. with consequent
application of the Fisher's discriminant analysis and Multiparametric
Topological Analysis (MTA)~\cite{durso_mta_2008}. From
Fig.~\ref{fig:mass} one can see that for proton-iron mixtures the
method gives excellent results, while for proton-oxygen mixture the
reconstructed composition is lighter than the original one and errors
grow with the increase of the oxygen fraction from 10 to 17\%. Let us
note that these errors are almost completely due to errors in muon
signal and scaling factor determination (see
Fig.~\ref{fig:smuscaledfitted}), while accuracy provided by MTA itself
is better than 2\%. More precise results for $p$\,--\,Fe samples are
explained by very good separation of the Fisher's variable
distributions for these primaries~\cite{ya_Lodz} and small errors on
the scaling factor do not influence significantly the events
misclassification rate. For the scaling factor of 1.28 the accuracy of
reconstruction is $2-4$\% and $15-20$\% for $p$\,--\,Fe and $p$\,--\,O
mixtures correspondingly.
\section*{Conclusions}
In this paper we have demonstrated that the universality of
\ensuremath{S_\mu/S_\mathrm{em}}\ ratio in \arange{45}{65} angular range in respect to the
interaction model properties allows to get the muon signal from hybrid
data with accuracy of $3-5$\%. The application of this approach to the
data of the Pierre Auger Observatory can be found
elsewhere~\cite{auger_nmu_icrc2011}. Further, using the independence
of the EM signal on the primary mass we have proposed a procedure
giving a possibility to find absolute scaling factor between real data
and MC simulated signals. Using EPOS~1.99 simulations as a fake data
we have found a scaling factor for QGSJET~II signals with an accuracy
of few percents. Application of the scaled QGSJET~II muon signals
allowed to reconstruct mass composition of the samples prepared with
EPOS~1.99 with errors below 4\% for proton-iron mixtures, while for
proton-oxygen ones the accuracy is around $10-20$\%. Hence, the use of
the both models in the proposed way for reconstruction of the real
primary mass composition will give closely agreeing results. The
preference to results obtained with one of the models can be given on
the basis of the comparison with measurements of \ensuremath{X_\mathrm{max}}\ and
\ensuremath{S_\mu}\ distributions.
\subsubsection*{Acknowledgments}
We are very grateful to Maximo Ave and Fabian Schmidt for kind
permission to use their GEANT~4 lookup tables in our calculations of
signal from different particles in Auger water Cherenkov detectors.
\vspace*{-2mm}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,475 |
Q: Android: TextInputLayout with one Edit Text and spinner I want to create a layout like this:
I want to add a spinner in TextInputLayout and show the hint of TextInputEditText like this.
I tried making custom TextInputLayout but still not able to add a spinner to it.
I have added TextInputLayout like this with outlined view:-
<com.google.android.material.textfield.TextInputLayout
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:layout_marginTop="8dp"
android:layout_marginStart="8dp"
android:layout_marginEnd="8dp"
android:textColorHint="@color/colorGreyView"
android:textSize="20dp"
style="@style/Widget.MaterialComponents.Button.OutlinedButton">
<com.google.android.material.textfield.TextInputEditText
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:hint="Mobile Number"/>
</com.google.android.material.textfield.TextInputLayout>
A:
implementation 'com.google.android.material:material:1.1.0-alpha08'
Use TextInputLayout ExposedDropdownMenualongwith EditextText and Try this:
<com.google.android.material.textfield.TextInputLayout
style="@style/Widget.MaterialComponents.TextInputLayout.OutlinedBox.ExposedDropdownMenu"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:hint="Label"
app:errorEnabled="true">
<AutoCompleteTextView
android:id="@+id/filled_exposed_dropdown"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:editable="false" />
<com.google.android.material.textfield.TextInputEditText
android:layout_width="wrap_content"
android:layout_height="wrap_content" />
</com.google.android.material.textfield.TextInputLayout>
Adjust width sizes as per requirement.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,360 |
layout: page
title: Seattle Police Officer 7722 Ryan D. Beecroft
permalink: /information/agencies/city_of_seattle/seattle_police_department/copbook/7722/
---
**Age as of Feb. 24, 2016:** 27
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,207 |
A chef's flame torch? Yes, great for removing air bubbles from your freshly produced domed labels.
Within a few minutes of doming, applying a small amount of heat will force the air bubbles to rise out of the resin. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,864 |
Sigma Theta Psi () is a multicultural, academic, and social sorority. The sorority was founded at San Jose State University (SJSU) in 1991. Sigma Theta Psi is known for their significant contribution to breast cancer awareness throughout sisters' local campuses and communities and even nationwide.
History
The Founding Mothers of Sigma Theta Psi created the sorority on November 13, 1991 at San Jose State University. Over time, the sorority expanded and now has 13 chapters.
Chapters
Sigma Theta Psi has placed 14 chapters at universities and colleges in California and Nevada, six of which are inactive. Active chapters are noted in bold, inactive chapters by italics.
Sorority information
Sigma Theta Psi's colors are purple, black, and gold. Its mascot is a black panther and its flower is an orchid. Each year, chapters organize multicultural, social, and academic events ranging from step shows to guest speakers to professional development events.
Since Sigma Theta Psi is also an academic sorority, members are required to maintain a minimum grade point average (GPA). Thus, each member's primary focus should be on their academic studies.
The sorority's national philanthropy is breast cancer awareness. Their current beneficiary is the American Cancer Society. Sigma Theta Psi's annual event to raise money for this organization and local organizations is their Double D Brunch. This event is a national luncheon hosted in both California and Nevada.
References
External links
National Website
Student societies in the United States
Student organizations established in 1991
1991 establishments in California | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,902 |
Q: Tensorflow: Custom operation used in two networks simultaneously produces nan I have written following custom operation with gradient to binarize a real vector. (this code is inspired from https://gist.github.com/harpone/3453185b41d8d985356cbe5e57d67342)
def py_func(func, inp, Tout, stateful=True, name=None, grad=None):
# Need to generate a unique name to avoid duplicates:
rnd_name = name+'PyFuncGrad' + str(np.random.randint(0, 1E+8))
tf.RegisterGradient(rnd_name)(grad) # see _MyBinarizerGrad for grad example
g = tf.get_default_graph()
with g.gradient_override_map({"PyFunc": rnd_name}):
return tf.py_func(func, inp, Tout, stateful=stateful, name=name)
def mycustombinarizer(x):
if _test_:
return x>0.5
sess_ = tf.Session()
probs = tf.constant(x)
probs = tf.reshape(probs,[-1])
probs = tf.pack([1-probs, probs], axis=1)
probs = tf.log(probs/(1-probs))
indexes = tf.multinomial(probs, 1)
indexes = tf.cast(tf.reshape(indexes, list(x.shape)),tf.float32)
with sess_.as_default():
binary_x = indexes.eval()
return binary_x
def binarizer(x, name=None):
with ops.name_scope(name, "Binarizer", [x]) as name:
sqr_x = py_func(mycustombinarizer,
[x],
[tf.float32],
name=name,
grad=_MyBinarizerGrad) # <-- here's the call to the gradient
return tf.reshape(sqr_x[0], tf.shape(x))
def _MyBinarizerGrad(op, grad):
return grad
This works perfectly fine if there is just one network using this operation. But if I create two copies of the same network and use this binarizer operation and try to optimise the combined cost (cost_net1+cost_net2) then it produces nan cost after a few iterations.
def network_(x, netname):
with tf.variable_scope(netname):
x = someoperation(x)
...
ret_tensor = binarizer(x,netname)
ypred1 = network_(input,'net1')
ypred2 = network_(input,'net2')
cost = costfn(ypred1,ytrue)+costfn(ypred2,ytrue)
Could anyone please tell me what is wrong with the implementation of my custom function? Is it the problem with sessions to evaluate indexes.eval() in mycustombinarizer or is it problem with name_scope/ variable_scope or is it totally something. I am stuck here.
A: Try this maybe.
@function.Defun()
def BinarizerGrad(unused_x, dy):
# Backprop dy directly.
return dy
@function.Defun(grad_func=BinarizerGrad)
def Binarizer(x):
# your whatever forward function here.
return tf.floor(x + tf.random_uniform(tf.shape(x)))
g = tf.Graph()
with g.as_default():
x = tf.placeholder(tf.float32)
y = Binarizer(x)
dy = tf.placeholder(tf.float32)
dx = tf.gradients(y, x, grad_ys=dy)
with tf.Session(graph=g) as sess:
x_val = np.array([[1, 2, 3], [0.5, 0.3, 0.2]])
dy_val = np.array([[1, 0, 1], [0., 0.1, 0.9]])
for v in sess.run([x, y, dx], feed_dict={x : x_val, dy: dy_val}):
print v
A: I don't think building a graph, starting a session, and running is well-supported inside a py_func. In this case you can remove all those things and just use straight-up tensorflow code and everything should work.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,760 |
Een huis voor meneer Biswas (Engels: A House for Mr Biswas) is een roman, geschreven door de Trinidadiaans-Britse schrijver en Nobelprijswinnaar V.S. Naipaul. Het verhaal is fictie maar wel geïnspireerd op het levensverhaal van de vader van de schrijver. De roman staat op de lijst van de Modern Library 100 Beste Romans en wordt gezien als een van de beste werken van Naipaul.
Het thema van het boek is vervreemding. De geboorte van de hoofdpersoon, meneer Biswas, is omgeven met bijgeloof en negatieve voortekenen. Hij wordt min of meer verstoten door zijn omgeving en doet er als volwassene alles aan om hogerop te komen, wat hem niet lukt. Uiteindelijk stelt hij zich tot doel om zij eigen huis te bezitten zodat hij op zichzelf kan zijn en eindelijk zijn eigen plek krijgt.
Een thema dat meer op de achtergrond staat is de koloniale maatschappij van Trinidad en Tobago, gezien met een postkoloniale blik.
Engelse literatuur (Verenigd Koninkrijk) | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 2,523 |
How good is Picasso after all these years?
Good enough to serve ELV the best veal chop he's ever eaten.
Good enough to present perfectly poached oysters in a champagne beurre blanc with extraordinary caviar.
Good enough to understand foie gras better than almost any other kitchen.
Good enough to perfectly roast small, gamy squab and combine them with the perfect accompaniment of nutty wild rice.
Good enough to perfectly capture the spring/summer solstice on a plate with white asparagus with fresh morels in a light cream sauce.
Good enough to excel at wine pairings that never leave you scratching your head (see below).
Good enough to feel as fresh and vibrant and dedicated to excellence as it did when it opened 12+ years ago.
One of the best restaurants anywhere!
Julians column in the sun today was so charming. What an understated classy guy, and also a mensch. Adorable.
Went here three weeks ago for a birthday dinner. This was without question one of the best meals I've had in Las Vegas. The service was impeccable with the master sommelier even helping to clear dinnerware when he was at the table between courses. I wish I could say the same for Chef Serrano's offerings at the Aria.
I am coming to LV for All-Star Cochon, but I made sure to make a reservation to Picasso. It has been 10 months since my last visit. Mr. C., you got me excited to return to one of the best restaurants in the country. Moreover, your recent entries on Beijing Noodle No. 9 have enticed me to try this restaurant during my trip. | {
"redpajama_set_name": "RedPajamaC4"
} | 101 |
\subsubsection{The Casimir equation}\label{iii2i}
The generators of the $d$-dimensional conformal group $\grp{SO}(d+1,1)$ can be taken to be the Lorentz generators $L_{AB}$ of $d+2$ dimensional Minkowski space (with $L_{AB}$ antisymmetric in $A$ and $B$ as usual). The quadratic combination $L^2\equiv{1\over2}L_{AB}L^{AB}$ is a Casimir of the algebra, i.e. it commutes with all the generators $L_{AB}$. As a result, $L^2$ takes a constant value on any irreducible representation of the conformal group, which means all states $|P^{\bf n}\mathcal{O}\rangle$ in the conformal family of a primary state $|\mathcal{O}\rangle$ are eigenstates of $L^2$ with the same eigenvalue. The eigenvalue depends on the dimension $\Delta$ and spin $\ell$ of $|\mathcal{O}\rangle$, and can be shown to be \cite{Dolan:2000ut}
\e{32a}{C_2(\Delta,\ell) = -\Delta(\Delta-d) - \ell(\ell+d-2)~.}
The $\grp{SO}(d+1,1)$ generators are represented on conformal fields by
\e{32b}{[L_{AB},\mathcal{O}_1(x_1)] = L_{AB}^{1}\mathcal{O}(x_1)~.}
where $L^1_{AB}$ is a differential operator built out of the position $x_1$ of $\mathcal{O}_1$ and derivatives with respect to that position. The form of the $L^1_{AB}$ depends on the conformal quantum numbers of $\mathcal{O}_1$. Equation \eqref{32b} together with conformal invariance of the vacuum imply the following identity, which holds for any state $|\alpha\rangle$:
\e{32c}{(L^1_{AB}+L^2_{AB})^2 \langle 0| \mathcal{O}_1 (x_1) \mathcal{O}_2 (x_2) |\alpha\rangle = \langle 0| \mathcal{O}_1 (x_1) \mathcal{O}_2 (x_2) L^2|\alpha\rangle~.}
Consistent with the notation for $L^2$, we have defined
\e{32d}{(L^1_{AB}+L^2_{AB})^2 \equiv \tfrac{1}2 (L^1_{AB}+L^2_{AB})(L^{1\,AB}+L^{2\,AB})~.}
As discussed in section \ref{ii1}, one obtains a conformal partial wave $W_{\Delta,\ell}$ by inserting into a four-point function the projection operator $P_{\Delta,\ell}$ onto the conformal family of a primary $\mathcal{O}$ with quantum numbers $\Delta,\ell$:
\e{32e}{W_{\Delta,\ell}(x_i) = {1\over C_{12 \mathcal{O}}C_{~~34}^{\mathcal{O}}}\sum_{\bf n}
\langle 0|\mathcal{O}_1 (x_1)\mathcal{O}_2 (x_2)|P^{\bf n} \mathcal{O}\rangle \langle P^{\bf n} \mathcal{O} | \mathcal{O}_3 (x_3) \mathcal{O}_4 (x_4) |0\rangle~.}
Applying the identity \eqref{32c} to the equation above and recalling that each state $|P^{\bf n} \mathcal{O}\rangle$ is an eigenstate of $L^2$ with the same eigenvalue $C_2(\Delta,\ell)$, we arrive at the Casimir equation
\e{32f}{(L^1_{AB}+L^2_{AB})^2W_{\Delta,\ell}(x_i) = C_2(\Delta,\ell) W_{\Delta,\ell}(x_i)~.}
One can take this second-order differential equation, plus the corresponding one with $1,2\leftrightarrow 3,4$, supplemented with appropriate boundary conditions, as one's definition of $W_{\Delta,\ell}$ \cite{SimmonsDuffin:2012uy}. Regarding boundary conditions, it is sufficient to require that $W_{\Delta,\ell}$ have the correct leading behavior in the $x_2\to x_1$ and $x_4\to x_3$ limits. The correct behavior in both limits is dictated by the fact that the contribution to $W_{\Delta,\ell}$ of the primary $\mathcal{O}$ dominates that of its descendants since those enter the OPE with higher powers of $x_{12}$ and $x_{34}$.
We will prove that geodesic Witten diagrams $\sW_{\Delta,0}$ are indeed proportional to conformal partial waves $W_{\Delta,0}$ by showing that $\sW_{\Delta,0}$ satisfies the Casimir equation \eqref{32f} and has the correct behavior in the $x_2\to x_1$ and $x_4\to x_3$ limits. The proof is very transparent in the embedding space formalism, which we proceed now to introduce.
\subsubsection{Embedding space}\label{iii2ii}
The embedding space formalism has been reviewed in e.g. \cite{Rychkov:lectures,Costa:2014kfa,SimmonsDuffin:2012uy}. The idea is to embed the $d$-dimensional CFT and the $d+1$ dimensional AdS on which lives the geodesic Witten diagram both into $d+2$ dimensional Minkowski space. We give this embedding space the metric
\e{32g}{ds^2 = -(dY^{-1})^2 + (dY^0)^2 + \sum_{i=1}^d (dY^i)^2~.}
The CFT will live on the projective null cone of embedding space, which is the Lorentz-invariant $d$-dimensional space defined as the set of nonzero null vectors $X$ with scalar multiples identified: $X\equiv aX$. We will use null vectors $X$ to represent points in the projective null cone with the understanding that $X$ and $aX$ signify the same point. The plane $\mathbb{R}^d$ can be mapped into the projective null cone via
\e{32h}{X^+(x) = a|x|^2, \quad X^-(x) = a, \quad X^i(x) = ax^i}
where we have introduced light cone coordinates $X^{\pm}=X^{-1}\pm X^0$. Of course, any nonzero choice of the parameter $a$ defines the same map.
Conformal transformations on the plane are implemented by Lorentz transformations in embedding space. As a specific example, we may consider a boost in the $0$ direction with rapidity $\lambda$. This leaves the $X^i$ coordinates unchanged, and transforms $X^{\pm}$ according to
\e{32i}{X^{+} \to e^{\lambda} X^{+}, \quad X^- \to e^{-\lambda}X^-~.}
A point $X(x)=(|x|^2,1,x^i)$ gets mapped to $(e^{\lambda}|x^2|,e^{-\lambda},x^i)$ which is projectively equivalent to $X(e^{\lambda}x^i)$. Thus boosts in the $0$ direction of embedding space induce dilatations in the plane.
Any field $\hat{\mathcal{O}}$ on the null cone defines a field $\mathcal{O}$ on the plane via restriction: $\mathcal{O}(x) \equiv \hat{\mathcal{O}}(X(x))$. Since $\hat{\mathcal{O}}$ is a scalar field in embedding space, the ${\rm{SO}}(d+1,1)$ generators act on it as
\e{32j}{[L_{AB},\hat{\mathcal{O}}(X)] = (X_A\partial_B-X_B\partial_A)\hat{\mathcal{O}}(X) ~.}
The induced transformation law for $\mathcal{O}$ is the correct one for a primary of dimension $\Delta$ if and only if $\hat{\mathcal{O}}$ satisfies the homogeneity condition
\e{32k}{\hat{\mathcal{O}}(aX) = a^{-\Delta}\hat{\mathcal{O}}(X)~.}
Thus in the embedding space formalism a primary scalar field $\mathcal{O}(x)$ of dimension $\Delta$ is represented by a field $\hat{\mathcal{O}}(X)$ satisfying \eqref{32k}. Below, we drop the hats on embedding space fields. It should be clear from a field's argument whether it lives on the null cone (as $\mathcal{O}(X)$) or on the plane (as $\mathcal{O}(x)$). Capital letters will always denote points in embedding space.
Meanwhile, AdS$_{d+1}$ admits an embedding into $d+2$ dimensional Minkowski space, as the hyperboloid $Y^2 = -1$. Poincare coordinates $(u,x^i)$ can be defined on AdS via
\e{32l}{Y^+ = \frac{u^2 + |x|^2}u,\quad Y^- = \frac{1}u,\quad Y^i = \frac{x^i}u~.}
The induced metric for these coordinates is the standard one, \eqref{22a}.
The AdS hyperboloid sits inside the null cone and asymptotes toward it. As one takes $u\to 0$, the image of a point $(u,x^i)$ in AdS approaches $(Y^+,Y^-,Y^i) = u^{-1}(|x^2|,1,x^i)$ which is projectively equivalent to $X(x^i)$. In this way, the image on the projective null cone of the point $x^i\in\mathbb{R}^d$ marks the limit $u\to 0$ of the embedding space image of a bulk point $(u,x^i)$.
Isometries of AdS are implemented by embedding space Lorentz transformations, and so are generated by
\e{32n}{L_{AB} = Y_A\partial_B - Y_B\partial_A~.}
The Casimir operator $L^2 = {1\over 2}(Y_A\partial_B-Y_B\partial_A)(Y^A\partial^B-Y^B\partial^A)$ is interior to the AdS slice $Y\cdot Y = -1$. That is, for $Y$ belonging to the AdS slice, $L^2 f(Y)$ depends only on the values of $f$ on the slice. In fact, applied to scalar functions on AdS the operator $L^2$ is simply the negative of the Laplacian of AdS:
\e{32m}{L^2 f(Y) = -\nabla_Y^2 f(Y)~}
as long as $Y$ is on the AdS slice. This fact, which is not surprising given that $L^2$ is a second-order differential operator invariant under all the isometries of AdS, can be checked directly from \eqref{32n}.
\subsubsection{Geodesic Witten diagrams satisfy the Casimir equation}\label{iii2iii}
The geodesic Witten diagram ${\cal W}_{\Delta,0}(x_i)$ lifts to a function ${\cal W}_{\Delta,0}(X_i)$ on the null cone of embedding space via a lift of each of the four bulk-to-boundary propagators with the appropriate homogeneity condition
\e{32o}{
G_{b\partial}(y,aX_i) = a^{-\Delta_i}G_{b\partial}(y,X_i),\quad i=1,2,3,4~.
}
The geodesics in AdS connecting the boundary points $X_1$ to $X_2$ and $X_3$ to $X_4$ lift to curves in embedding space which can be parameterized by
\begin{align}
\begin{split}\label{32p}
Y(\lambda) &= {e^{-\lambda}X_1 + e^{\lambda}X_2\over \sqrt{-2X_1\cdot X_2}}\\
Y(\lambda') &= {e^{-\lambda'}X_3 + e^{\lambda'}X_4\over \sqrt{-2X_3\cdot X_4}}
\end{split}
\end{align}
The geodesic Witten diagram is
\e{32q}{{\cal W}_{\Delta,0}(X_i) = \int_{\gamma_{34}} F(X_1,X_2,Y(\lambda');\Delta)G_{b\partial}(Y(\lambda'),X_3)G_{b\partial}(Y(\lambda'),X_4)}
where we have isolated the part that depends on $X_1,X_2$:\footnote{The fact that the bulk-to-bulk propagator satisfies the Laplace equation was used to similar effect in \cite{D'Hoker:1999ni}. In particular, \cite{D'Hoker:1999ni} defines a quantity $A(y', x_1,x_2)$ that is similar to $F(X_1,X_2,Y';\Delta)$, except that the vertex is integrated over all of AdS instead of along a geodesic.}
\e{32r}{F(X_1,X_2,Y';\Delta) = \int_{\gamma_{12}}
G_{b\partial}(Y(\lambda),X_1) G_{b\partial}(Y(\lambda),X_2) G_{bb}(Y(\lambda),Y';\Delta)~.}
$F(X_1,X_2,Y';\Delta)$ is the lift to embedding space of $\varphi^{12}_{\Delta}(y)$ defined in \eqref{31aa}. The bulk arguments of the bulk-to-boundary propagators have been promoted from points $y$ in the bulk to points $Y$ in embedding space. Although the propagators are defined only on the AdS slice, there is no ambiguity because $Y(\lambda)$ and $Y'(\lambda')$ always lie in the AdS slice.
The function $F(X_1,X_2,Y';\Delta)$ is manifestly invariant under simultaneous ${\rm{SO}}(d+1,1)$ rotations of $X_1,X_2,Y'$, and therefore it is annihilated by $(L^{1}+L^{2}+L^{Y'})_{AB}$. This means
\e{32s}{(L^1_{AB}+L^2_{AB})F(X_1,X_2,Y';\Delta)=-L^{Y'}_{AB}F(X_1,X_2,Y';\Delta)}
and so (since of course $L^1_{AB}$ commutes with $L^{Y'}_{AB}$)
\e{32t}{(L^1_{AB} + L^2_{AB})^2F(X_1,X_2,Y';\Delta) = (L^{Y'})^2 F(X_1,X_2,Y';\Delta)~.}
Recall that $(L^{Y'})^2$ is $-\nabla^2_{Y'}$. The function $F(X_1,X_2,Y';\Delta)$, which depends on $Y'$ via the bulk-to-bulk propagator $G_{bb}(Y(\lambda),Y';\Delta)$, is an eigenfunction of $-\nabla^2_{Y'}$ with eigenvalue $-\Delta(\Delta-d)$. Thus we conclude that $F(X_1,X_2,Y';\Delta)$ is an eigenfunction of $(L^1_{AB} + L^2_{AB})^2$ with eigenvalue $C_2(\Delta,0)$, and therefore that
\e{32u}{(L^1_{AB} + L^2_{AB})^2{\cal W}_{\Delta,0}(X_i) = C_2(\Delta,0){\cal W}_{\Delta,0}(X_i)~. }
Note that agreement does not hinge on what the actual eigenvalue is: it is guaranteed by the fact that the bulk-to-bulk propagator and the conformal partial wave furnish the same highest weight representation of SO($d+1,1$).
Furthermore, the behavior in the limit $x_2\to x_1$ of the bulk-to-boundary and bulk-to-bulk propagators guarantees the geodesic Witten diagram to have the power-law behavior ${\cal W}_{\Delta,0}(x_i)\to (\text{constant})\times|x_{12}|^{\Delta-\Delta_1-\Delta_2}$ in that limit, and similarly in the $x_4\to x_3$ limit. This proves ${\cal W}_{\Delta,0}$ is equal to the conformal partial wave $W_{\Delta,0}$ up to a constant factor.
Looking back at the proof, we can see why the bilocal function integrated between the geodesics had to be precisely the bulk-to-bulk propagator $G_{bb}(y,y';\Delta)$. To get \eqref{32u} we needed that function to be the appropriate eigenfunction of the Laplacian, and to get the correct limiting behavior we needed it to be the eigenfunction with normalizable boundary conditions at infinity. It also crucial that the vertices be integrated over geodesics rather than arbitrary curves or over all of AdS. A non-geodesic curve would introduce extra data to specify the curve, which would not be conformally invariant. Integrating the vertices over all of AdS (which would give the full Witten diagram) allows $y$ and $y'$ to collide, but the bulk-to-bulk propagator acted on by the wave operator picks up a source contribution when $y=y'$, hence the diagram would not be an eigenfunction of the Casimir operator in this case; indeed we know that it is a sum of eigenfunctions with different eigenvalues.
\subsection{Comments}
We close this section with a few comments.
\subsubsection{Geodesic versus ordinary Witten diagrams}
A natural question is {\it why}, intuitively, a relation like \eqref{3c} is true. Let us offer two motivational remarks.
The first is that there are two ways to integrate a bulk point while preserving conformal invariance. One is over all of AdS, which defines a Witten diagram, while the other is over a geodesic. The latter is clearly over a smaller range, which makes it seem at least plausible that it represents a conformal partial wave rather than a full correlator. Indeed, the only obvious conformally invariant objects that appear in four-point functions are the correlator itself, and the conformal partial waves.
The second is a heuristic ``derivation'' starting from the exchange Witten diagram, ${\cal A}_4^{\rm Exch}$. Consider taking the following limit of heavy external operators,
\e{34a}{\Delta_{1,2,3,4} \rightarrow\infty~, \quad \Delta_{12}, \Delta_{34}~\text{fixed}~.}
As reviewed in Section \ref{ii} and computed in the next section, the full diagram equals a single trace exchange of ${\cal O}$, plus infinite towers of double trace exchanges of $[{\cal O}_1{\cal O}_2]_{m,0}$ and $[{\cal O}_3{\cal O}_4]_{n,0}$. On the CFT side, the double-trace exchanges are exponentially smaller in this limit than that of the single-trace exchange, simply because the conformal partial waves decay exponentially as the internal operator dimension tends to infinity. So the Witten diagram reduces to the single-trace block in the limit. On the bulk side, the heavy limit restricts the cubic vertices to lie on geodesics, so ${\cal A}_4^{\rm Exch}$ reduces to ${\cal W}_{\Delta,0}$, the geodesic Witten diagram. This establishes equality between ${\cal W}_{\Delta,0}$ and $W_{\Delta,0}$ in the limit \eqref{34a}. To complete the argument we need to use the fact that the conformal block $G_{\Delta,0}$ only depends on $\Delta_i$ through $\Delta_{12}$ and $\Delta_{34}$, as can be seen from the recursion relations in \cite{Dolan:2011dv}. Furthermore, $G_{\Delta,0}$ and $W_{\Delta,0}$ only differ by a prefactor which has exponents linear in $\Delta_i$ (a form which is invariant as $\Delta_i\rightarrow \infty$); see (\ref{21db}). Using these two facts, it follows that if ${\cal W}_{\Delta,0}$ and $W_{\Delta,0}$ agree in the regime \eqref{34a}, then they agree for all values of $\Delta_i$ and $\Delta$.
Note that the geodesic restriction ensures that a cut down the middle of the diagram crosses only the internal line, representing the CFT primary; contrast this with the exchange Witten diagram, where integration over all of AdS ensures that the cut will sometimes cross two external lines, representing the (infinite towers of) double-trace operators.
\subsubsection{Simplification of propagators and blocks}
In even $d$, CFT$_d$ scalar conformal blocks can be resummed into hypergeometric functions. An apparently unrelated simplification occurs for AdS$_{d+1}$ scalar bulk-to-bulk propagators, which are rational functions of $S\equiv e^{-2 \sigma(y,y')}$ rather than hypergeometric. From \eqref{22b}, the even $d$ propagators are, at low $d$,
\es{gbb3}{d=2:&\quad {G_{bb}(y,y';\Delta)}=S^{\Delta/2}\,{1\over 1-S}\\
d=4:&\quad {G_{bb}(y,y';\Delta)}=S^{\Delta/2}\,{(3-\Delta)S+(\Delta-1)\over (\Delta-1)(1-S)^3}\\
d=6:&\quad {G_{bb}(y,y';\Delta)}=S^{\Delta/2}\,\frac{(5-\Delta) (\Delta-4)S^2+2 (\Delta-5) (\Delta-1)S-(\Delta-2) (\Delta-1)}{(\Delta-2) (\Delta-1) (S-1)^5}~.}
The geodesic representation of the scalar conformal blocks reveals that these simplifications have a common origin. Conversely, the lack of simplification of the propagator in odd $d$ gives a new perspective on why generic odd $d$ conformal blocks cannot be reduced to special functions.
\subsubsection{Relation to Mellin space}
It is worth noting that the spin-$\ell$ conformal block has a Mellin representation with exponential dependence on the Mellin parameter: up to normalization \cite{Fitzpatrick:2011hu},
\e{34b}{G_{\Delta,\ell}(s,t) = e^{\pi i({d\over 2}-\Delta)}\left(e^{\pi i(t+\Delta-d)}-1\right){\Gamma\left({\Delta-\ell-t\over 2}\right)\Gamma\left({2d-\Delta-\ell-t\over 2}\right)\over \Gamma\left({\Delta_1+\Delta_2-t\over 2}\right)\Gamma\left({\Delta_3+\Delta_4-t\over 2}\right)}P_{\Delta,\ell}(s,t)}
where $P_{\Delta,\ell}(s,t)$ is a degree-$\ell$ Mack polynomial. (In the scalar case, $\ell=0$.) It has been argued that for holographic CFTs with a gap, the Mellin amplitudes for their correlators are polynomially bounded at large $s,t$. It is interesting that despite its exponential growth at large $t$, the Mellin representation of a conformal block does have a semiclassical AdS description.
In \cite{1208.0337}, it was argued that starting with \eqref{34b}, one recovers the Mellin amplitude for the {\it full} spin-$\ell$ exchange Witten diagram by writing it as a sum over its poles and dropping all other contributions.\footnote{This is true up to polynomial contributions from contact diagrams.} Evidently, this is the Mellin transform, so to speak, of the liberation of bulk vertices from the geodesics to all of AdS.
\section{The conformal block decomposition of scalar Witten diagrams}\label{iv}
We begin our treatment with the technically simplest case: tree-level four-point functions in AdS involving only scalar fields. All of the key steps will be visible in the decomposition of the four-point contact diagram, out of which the geometric representation of the scalar conformal block will naturally emerge. We then move on to the exchange diagram and, in the next section, to fields with spin.
\subsection{An AdS propagator identity}\label{iv1}
The main technical tool that we will employ is an identity obeyed by AdS bulk-to-boundary propagators. Consider two scalar fields dual to gauge-invariant scalar operators ${\cal O}_1, {\cal O}_2$ of conformal dimensions $\Delta_{1},\Delta_2$, respectively. Now consider a product of their bulk-to-boundary propagators, from points $x_1$ and $x_2$ on the boundary to the same point $y$ in the bulk. Then the following identity holds:
\e{41a}{G_{b\partial}(y, x_1)G_{b\partial}(y,x_2) = \sum_{m=0}^{\infty}a^{12}_m\,\varphi^{12}_{\Delta_m}(y)}
where $\varphi^{12}_{\Delta_m}(y)$
is the field solution defined in \eqref{31aa}. The bulk-to-bulk propagator $G_{bb}(y(\lambda),y;\Delta_m)$, running from the geodesic to the original bulk point $y$, is for a scalar field with mass $m_m^2=\Delta_m(\Delta_m-d)$, where
\e{41b}{\Delta_m = \Delta_1+\Delta_2+2m~.}
The $a^{12}_m$ are coefficient functions of $\Delta_1, \Delta_2$ and $d$:
\e{41c}{a^{12}_m = {1\over \beta_{\Delta_m 12}}{(-1)^m\over m!}{(\Delta_1)_m(\Delta_2)_m\over \left(\Delta_1+\Delta_2+m-{d\over 2}\right)_m}~.}
This identity is depicted in Figure \ref{f3}.
\begin{figure}[t!]
\begin{center}
\includegraphics[width = \textwidth]{Prop_ID_final_thin.pdf}
\caption{The identity \eqref{41a} obeyed by AdS scalar propagators. The internal line represents bulk-to-bulk propagator for a scalar field of mass $m^2=\Delta_m(\Delta_m-d)$. $a^{12}_m$ and $\Delta_m$ are defined in \eqref{41b} and \eqref{41c}, respectively.}
\label{f3}
\end{center}
\end{figure}
In words, the original bilinear is equal to an infinite sum of three-point vertices integrated over the geodesic $\gamma_{12}$, for fields of varying masses $m_m^2 = \Delta_m(\Delta_m-d)$. To prove this, we work in global AdS with $t_1\rightarrow -\infty, t_2\rightarrow+\infty$, whereupon $\gamma_{12}$ becomes a worldline at $\rho=0$. We already solved for $\varphi^{12}_{\Delta_m}(y)$ in \eqref{31h}. Plugging that solution into \eqref{41a}, we must solve
\e{41d}{(\cos\rho)^{\Delta_1+\Delta_2} = \sum_{m=0}^{\infty} a^{12}_m \beta_{\Delta_m 12}(\cos\rho)^{\Delta_m}{}_2F_1\left({\Delta_m+\Delta_{12}\over 2},{\Delta_m-\Delta_{12}\over 2};\Delta_m-{d-2\over 2};\cos^2\rho\right)~.
}
Expanding as a power series in $\cos^2\rho$, the unique solution is given by $\Delta_m$ in \eqref{41b} and $a^{12}_m$ in \eqref{41c}.
The identity \eqref{41a} is suggestive of a bulk operator product expansion, where the propagation of two boundary fields to the same bulk point is replaced by an infinite sum over field solutions. Note that the dimensions $\Delta_m$ are those of the scalar double-trace operators $[{\cal O}_1{\cal O}_2]_{m,0}$ at leading order in $1/N$. As we now show, this fact ensures that the decomposition of a given Witten diagram involving $G_{b\partial}(y,x_1)G_{b\partial}(y,x_2)$ includes the exchange of $[{\cal O}_1{\cal O}_2]_{m,0}$, consistent with the generalized free field content of the dual CFT.
\subsection{Four-point contact diagram}
We want to compute the four-point scalar contact diagram \eqref{22g}, for all operator dimensions $\Delta_i$ generic. We reproduce the integral here:
\e{41e}{D_{\Delta_1\Delta_2\Delta_3\Delta_4}(x_i) = \int_y G_{b\partial}(y,x_1)\,G_{b\partial}(y,x_2)\,G_{b\partial}(y,x_3)\,G_{b\partial}(y,x_4)~.}
A helpful pictorial representation of the following calculation is given in Figure \ref{f4}.
\begin{figure}[t!]
\begin{center}
~~~ \includegraphics[width = .44\textwidth]{cd2c1.pdf}
\includegraphics[width = .625\textwidth]{cd2c2f.pdf}%
\includegraphics[width = .98\textwidth]{cd2c33f.pdf}%
\caption{The decomposition of a four-point scalar contact diagram into conformal partial waves disguised as geodesic Witten diagrams. Passage to the second line uses \eqref{41f}, and passage to the last line uses \eqref{41g}. The last line captures the infinite set of CFT exchanges of the double-trace operators $[{\cal O}_1{\cal O}_2]_{m,0}$ and $[{\cal O}_3{\cal O}_4]_{n,0}$. We have suppressed OPE coefficients; the exact result is in equation \eqref{41i}. }
\label{f4}
\end{center}
\end{figure}
%
Using our geodesic toolkit, the evaluation of this diagram is essentially trivial. First, we use the identity \eqref{41a} on the pairs $(12)$ and $(34)$. This yields
\es{41f}{D_{\Delta_1\Delta_2\Delta_3\Delta_4}(x_i) = \sum_{m,n}a^{12}_ma^{34}_n&\int_{\gamma_{12}}\int_{\gamma_{34}}G_{b\partial}(y(\lambda),x_1)\,G_{b\partial}(y(\lambda),x_2)\\
\times &\int_y G_{bb}(y(\lambda),y;\Delta_m)\,G_{bb}(y,y(\lambda');\Delta_n)\\
\times &\,G_{b\partial}(y(\lambda'),x_3)\,G_{b\partial}(y(\lambda'),x_4)~.}
Next, we use
\e{prop}{G_{bb}(y,y';\Delta) = \Big\langle y\Big|{1\over \nabla^2-m^2}\Big|y'\Big\rangle}
to represent the product of bulk-to-bulk propagators integrated over the common bulk point $y$ as
\es{41g}{\int_y G_{bb}(y(\lambda),y;\Delta_m)\,G_{bb}(y,y(\lambda');\Delta_n)&= \,{G_{bb}(y(\lambda),y(\lambda');\Delta_m)-G_{bb}(y(\lambda),y(\lambda');\Delta_n)\over m_m^2-m_n^2}}
where we used completeness, $\int_y |y\rangle \langle y|=1$. The integrated product is thus replaced by a difference of unintegrated propagators from $\gamma_{12}$ to $\gamma_{34}$. This leaves us with
\es{41h}{&D_{\Delta_1\Delta_2\Delta_3\Delta_4}(x_i) = \sum_{m,n}{a^{12}_ma^{34}_n\over m_m^2-m_n^2}\times\\\Bigg(&\int_{\gamma_{12}}\int_{\gamma_{34}}G_{b\partial}(y(\lambda),x_1)G_{b\partial}(y(\lambda),x_2)\times G_{bb}(y(\lambda),y(\lambda');\Delta_m)\times \,G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4)\\-&\int_{\gamma_{12}}\int_{\gamma_{34}}G_{b\partial}(y(\lambda),x_1)G_{b\partial}(y(\lambda),x_2)\times G_{bb}(y(\lambda),y(\lambda');\Delta_n)\times \,G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4)\Bigg)~.}
But from \eqref{gwitt}, we now recognize the last two lines as conformal partial waves\text{!} Thus, we have
\e{41i}{D_{\Delta_1\Delta_2\Delta_3\Delta_4}(x_i) = \sum_{m,n}{a^{12}_ma^{34}_n\over m_m^2-m_n^2}\left({\cal W}_{\Delta_m,0}(x_i) - {\cal W}_{\Delta_n,0}(x_i)\right)~.}
This is the final result. In the CFT notation of section \ref{ii}, we write this as a pair of single sums over double-trace conformal partial waves,
\e{41j}{D_{\Delta_1\Delta_2\Delta_3\Delta_4}(x_i) = \sum_{m} P^{(12)}_1(m,0)\, W_{\Delta_m,0}(x_i) + \sum_{n} P^{(34)}_1(n,0)\,W_{\Delta_n,0}(x_i)}
with squared OPE coefficients
\es{41k}{P_1^{(12)}(m,0) &= \left(\beta_{\Delta_m 12}\,a^{12}_m\right) \left(\beta_{\Delta_m 34}\sum_n{a^{34}_n\over m_m^2-m_n^2}\right)\\
P_1^{(34)}(n,0) &=\left(\beta_{\Delta_n 34}\, a^{34}_n \right)\left( \beta_{\Delta_n 12}\sum_m{a^{12}_m\over m_n^2-m_m^2}\right)}
where $m^2=\Delta(\Delta-d)$ as always. The structure of the answer is manifestly consistent with CFT expectations: only double-trace operators $[{\cal O}_1{\cal O}_2]_{m,0}$ and $[{\cal O}_3{\cal O}_4]_{n,0}$ are exchanged.
We will analyze this result more closely after computing the exchange diagram.
\subsection{Four-point exchange diagram}
Turning to the scalar exchange diagram, we reap the real benefits of this approach: unlike an approach based on brute force integration, this case is no harder than the contact diagram. A pictorial representation of the final result is given in Figure \ref{f5}.
\begin{figure}[t!]
\begin{center}
\includegraphics[width = \textwidth]{ed11f.pdf}
\includegraphics[width = \textwidth]{ed22.pdf}%
\caption{The decomposition of a four-point scalar exchange diagram (upper left) into conformal partial waves, for an exchanged scalar $\phi$ of mass $m^2=\Delta(\Delta-d)$. We have skipped the intermediate steps, which are nearly identical to those of the contact diagram. The term in the upper right captures the single-trace exchange of the scalar operator dual to $\phi$. The second line captures the infinite set of CFT exchanges of the double-trace operators $[{\cal O}_1{\cal O}_2]_{m,0}$ and $[{\cal O}_3{\cal O}_4]_{n,0}$. We have suppressed OPE coefficients; the exact result is in equations \eqref{43d}--\eqref{43e}.}
\label{f5}
\end{center}
\end{figure}
We take all external dimensions $\Delta_i$, and the internal dimension $\Delta$, to be generic. The diagram is computed as
\e{43a}{{\cal A}_4^{\rm Exch}(x_i) = \int_{y} \int_{y'}G_{b\partial}(y,x_1)G_{b\partial}(y,x_2)\times G_{bb}(y,y';\Delta)\times G_{b\partial}(y',x_3)G_{b\partial}(y',x_4)~.}
Expanding in the $s$-channel (12)-(34), the algorithm is the same as the contact case. First, use \eqref{41a} twice to get
\es{43b}{{\cal A}_4^{\rm Exch}(x_i)= \sum_{m,n}a^{12}_ma^{34}_n&\int_{\gamma_{12}}\int_{\gamma_{34}}G_{b\partial}(y(\lambda),x_1)G_{b\partial}(y(\lambda),x_2)\\
\times &\int_{y}\int_{y'} G_{bb}(y(\lambda),y;\Delta_m)G_{bb}(y,y';\Delta)G_{bb}(y',y(\lambda');\Delta_n)\\
\times &\,G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4)~.}
This is of the same form as the contact diagram, only we have three bulk-to-bulk propagators and two integrations. We again use \eqref{prop} to turn the second line into a sum over terms with a single bulk-to-bulk propagator:
\es{43c}{&\int_{y}\int_{y'} G_{bb}(y(\lambda),y;\Delta_m)G_{bb}(y,y';\Delta)G_{bb}(y',y(\lambda');\Delta_n)\\
=~& {G_{bb}(y(\lambda),y(\lambda');\Delta_m)\over (m_m^2-m_{\Delta}^2)(m_m^2-m_n^2)}+{G_{bb}(y(\lambda),y(\lambda');\Delta)\over (m_{\Delta}^2-m_m^2)(m_{\Delta}^2-m_n^2)}+{G_{bb}(y(\lambda),y(\lambda');\Delta_n)\over (m_n^2-m_m^2)(m_n^2-m_{\Delta}^2)}~.}
Recognizing the remaining integrals as conformal partial waves, we reach our final result:
\e{43d}{{\cal A}_4^{\rm Exch}(x_i) = C_{12\Delta}C^{\Delta}_{~~34}W_{\Delta,0}(x_i) + \sum_{m} P^{(12)}_1(m,0)\, W_{\Delta_m,0}(x_i) + \sum_{n} P^{(34)}_1(n,0)\,W_{\Delta_n,0}(x_i)}
where
\es{43e}{C_{12\Delta}C^{\Delta}_{~~34} &=\left(\beta_{\Delta 12}\sum_m {a^{12}_m\over m_{\Delta}^2-m_m^2}\right)\left( \beta_{\Delta 34}\sum_n {a^{34}_n\over m_{\Delta}^2-m_n^2}\right)\\
P_1^{(12)}(m,0) &=\left( \beta_{\Delta_m 12}\,{a^{12}_m\over m_m^2 - m_{\Delta}^2}\right)\left(\beta_{\Delta_m 34}\sum_n {a^{34}_n\over m_m^2-m_n^2}\right)\\
P_1^{(34)}(n,0) &=\left(\beta_{\Delta_n 34}\,{a^{34}_n\over m_n^2-m_{\Delta}^2}\right)\left( \beta_{\Delta_n 12}\sum_m{a^{12}_m\over m_n^2 - m_m^2}\right) ~.}
Its structure is precisely as required by AdS/CFT: in addition to the double-trace exchanges of $[{\cal O}_1{\cal O}_2]_{m,0}$ and $[{\cal O}_3{\cal O}_4]_{n,0}$, there is a single-trace exchange of the operator dual to the exchanged field in the bulk of dimension $\Delta$.
Comparing \eqref{43e} to \eqref{41k}, we can immediately read off a new identity relating the double-trace OPE coefficients of the contact and exchange diagrams:
\e{43f}{{P_1^{(12)}(m,0)\big|_{\rm Contact}\over P_1^{(12)}(m,0)\big|_{\rm Exch}} = m_m^2-m_{\Delta}^2}
and likewise for $P_1^{(34)}(n,0)$. This is quite simple. One can quickly check this against the $d=4$ example in Appendix B of \cite{ElShowk:2011ag}.
\subsection{Further analysis}
\subsubsection{OPE factorization}
Notice that the squared OPE coefficients in \eqref{43e} and \eqref{41k} factorize naturally into terms associated with the (12) and (34) channels. To emphasize this, it is useful to define\footnote{We observe a likeness between $\alpha^{34}_s$ and calculations in \cite{1007.2412} of $\ell=0$ double-trace anomalous dimensions due to heavy operator exchange; see Section 4.3 therein. It is not immediately clear to us whether there is a deeper statement to be made.}
\e{44a}{\alpha^{34}_s \equiv \sum_n{a^{34}_n\over m_s^2-m_n^2}}
for some mass squared $m_s^2=\Delta_s(\Delta_s-d)$, and similarly for $\alpha^{12}_s$. This allows us to write the Witten diagrams in a tidy form as
\e{44b}{D_{\Delta_1\Delta_2\Delta_3\Delta_4}(x_i) = \sum_m a^{12}_m \alpha^{34}_m\, {\cal W}_{\Delta_m,0}(x_i) +\sum_n \alpha^{12}_n a^{34}_n\, {\cal W}_{\Delta_n,0}(x_i)}
and
\e{44c}{{\cal A}_4^{\rm Exch}(x_i) = \alpha^{12}_{\Delta}\alpha^{34}_{\Delta}\,{\cal W}_{\Delta,0}(x_i) + \sum_m {a^{12}_m\alpha^{34}_m\over m_m^2-m_{\Delta}^2}\, \,{\cal W}_{\Delta_m,0}(x_i)+ \sum_n\,{ \alpha^{12}_n a^{34}_n\over m_n^2-m_{\Delta}^2}\, {\cal W}_{\Delta_n,0}(x_i)~.}
For compactness in the above equations we have used ${\cal W}_{\Delta,0}$ in place of $W_{\Delta,0}$. Recall that ${\cal W}_{\Delta,0}$ is a rescaling of the standard conformal partial wave, ${\cal W}_{\Delta,0}(x_i) = \beta_{\Delta 12}\beta_{\Delta 34}W_{\Delta,0}(x_i)$. The coefficient relating ${\cal W}_{\Delta,0}$ to $W_{\Delta,0}$ clearly factorizes.
Writing the OPE coefficients in terms of the coefficients $a^{12}_m, a^{34}_n$ and masses $m_m, m_n, m_{\Delta}$ makes their origin transparent. But the sum defining $\alpha^{34}_s$ can actually be performed, yielding
\es{44d}{\alpha^{34}_s &={\Gamma(\Delta_3+\Delta_4)\over \Gamma(\Delta_3)\Gamma(\Delta_4)} \left(F(\Delta_s,\Delta_3,\Delta_4)+F(d-\Delta_s,\Delta_3,\Delta_4)\right)}
where
\es{}{F(\Delta_s,\Delta_3,\Delta_4) &\equiv{1\over \left(\Delta_s-{d\over 2}\right)(\Delta_s-\Delta_3-\Delta_4)}\\&\quad\times{}_4F_3\Farg{{\Delta_3+\Delta_4\over 2},{\Delta_3+\Delta_4+1\over 2},{\Delta_3+\Delta_4-\Delta_s\over 2},\Delta_3+\Delta_4-{d\over 2}}{{\Delta_3+\Delta_4\over 2}-{d\over 4},{\Delta_3+\Delta_4\over 2}-{d-2\over 4},{\Delta_3+\Delta_4-\Delta_s+2\over 2}}{-1}~.}
\subsubsection{Recovering logarithmic singularities}
Recall from Section \ref{ii} that when the external operator dimensions obey $\Delta_1+\Delta_2-\Delta_3-\Delta_4 \in 2\mathbb{Z}$, logarithms appear in tree-level Witten diagrams due to anomalous dimensions of double-trace operators. In brute force calculation of the AdS integrals, these logarithms are extracted by isolating the relevant integration range. In Mellin space, they appear as double poles in the Mellin amplitude.
In the present approach, these logarithms fall out trivially as algebraic conditions. Considering the scalar four-point contact diagram written in the form \eqref{41i}, for instance, we see that terms for which $m_m^2=m_n^2$ give rise to derivatives of conformal blocks, and hence to logarithms. This is equivalent to the condition $\Delta_m=\Delta_n$ or $\Delta_m = d-\Delta_n$. Since $d\in\mathbb{Z}$, both of these are equivalent to $\Delta_1+\Delta_2-\Delta_3-\Delta_4 \in 2\mathbb{Z}$, which is precisely the integrality condition stated above. Identical structure is visible in \eqref{43c}: logarithms will appear when any of $m_m^2, m_n^2, m_{\Delta}^2$ coincide.
As an explicit example, let us consider $D_{\Delta\D\Delta\D}(x_i)$. Then \eqref{41j} can be split into $m\neq n$ and $m=n$ terms, the latter of which yield logarithms:
\es{44e}{D_{\Delta\D\Delta\D}(x_i) = \sum_{n=0}^{\infty}2 a^{\Delta\D}_n\left(\sum_{m\neq n}{a^{\Delta\D}_m\over m_n^2-m_m^2}\right) {\cal W}_{2\Delta+2n,0}(x_i)+ \left({(a^{\Delta\D}_n)^2\over \partial_nm_n^2}\right)\partial_n{\cal W}_{2\Delta+2n,0}(x_i) ~.}
This takes the form of the $\ell=0$ terms in \eqref{22k}, with
\e{44f}{ P_1(n,0) = 2\beta_{(2\Delta+2n)\,\Delta\D}^2 a^{\Delta\D}_n\left(\sum_{m\neq n}{a^{\Delta\D}_m\over m_n^2-m_m^2}\right)+{(a^{\Delta\D}_n)^2\over \partial_nm_n^2}\partial_n\left(\beta_{(2\Delta+2n)\,\Delta\D}^2\right)}
and
\e{44fa}{{1\over 2} P_0(n,0) \gamma_1(n,0) = {(a^{\Delta\D}_n)^2\over \partial_nm_n^2}\beta_{(2\Delta+2n)\,\Delta\D}^2~.}
As an aside, we note the conjecture of \cite{Heemskerk:2009pn}, proven in \cite{Fitzpatrick:2011dm}, that
\e{44g}{P_1(n,\ell) = {1\over 2} \partial_n(P_0(n,\ell)\gamma_1(n,\ell))~.}
We have checked in several examples that this is obeyed by \eqref{44f}--\eqref{44fa}. It would be interesting to prove it using generalized hypergeometric identities.
\subsection{Taking stock}
We close this section with some perspective. Whereas traditional methods of computing Witten diagrams are technically involved and require explicit bulk integration \cite{D'Hoker:1998mz} and/or solution of differential equations \cite{D'Hoker:1999ni}, the present method skips these steps with a minimum of technical complexity. It is remarkable that for neither the contact nor exchange diagrams have we performed any integration: the integrals have instead been absorbed into sums over, and definitions of, conformal partial waves.
For the contact diagram/$D$-function, we have presented an efficient algorithm for its decomposition into spin-0 conformal blocks in position space. Specific cases of such decompositions have appeared in previous works \cite{ElShowk:2011ag,Heemskerk:2009pn}, although no systematic treatment had been given. Moreover, perhaps the main virtue of our approach is that exchange diagrams are no more difficult to evaluate than contact diagrams.
$D$-functions also appear elsewhere in CFT, including in weak coupling perturbation theory. For example, the four-point function of the {\bf 20'} operator in planar ${\cal N}=4$ SYM at weak coupling is given, at order $\lambda$, by \cite{hep-th/9811155}
\e{44l}{\langle {\cal O}_{\bf 20'}(x_1) {\cal O}_{\bf 20'}(x_2) {\cal O}_{\bf 20'}(x_3) {\cal O}_{\bf 20'}(x_4)\rangle\big|_{\lambda} \propto \overline D_{1111}(z,\overline{z})}
where $\overline D_{1111}(z, \overline{z})$ was defined in \eqref{22h}. The ubiquity of $D$-functions at weak coupling may be related to constraints of crossing symmetry in the neighborhood of free fixed points \cite{1506.04659}.
\section{Spinning exchanges and conformal blocks}\label{v}
The OPE of two scalar primary operators yields not just other scalar primaries but also primaries transforming in symmetric traceless tensor representations of the Lorentz group. We refer to such a rank-$\ell$ tensor as a spin-$\ell$ operator. Thus, for the full conformal block decomposition of a correlator of scalar primaries we need to include blocks describing spin-$\ell$ exchange. The expression for such blocks as geodesic Witten diagrams turns out to be the natural extension of the scalar exchange case. The exchanged operator is now described by a massive spin-$\ell$ field in the bulk, which couples via its pullback to the geodesics connecting the external operator insertion points. This was drawn in Figure \ref{f1}.
In this section we do the following. We give a fairly complete account of the spin-$1$ case, showing how to decompose a Witten diagram involving the exchange of a massive vector field, and establishing that the geodesic diagrams reproduce known results for spin-1 conformal blocks. We also give an explicit treatment of the spin-2 geodesic diagram, again checking that we reproduce known results for the spin-2 conformal blocks. More generally, we use the conformal Casimir equation to prove that our construction yields the correct blocks for arbitrary $\ell$.
\subsection{Known results}
Conformal blocks with external scalars and internal spin-$\ell$ operators were studied in the early work of Ferrara et. al. \cite{Ferrara:1971vh}. They obtained expressions for these blocks as double integrals. It is easy to verify that their form for the scalar exchange block precisely coincides with our geodesic Witten diagram expression (\ref{gwitt}). We thus recognize the double integrals as integrals over pairs of geodesics. Based on this, we expect agreement for general $\ell$, although we have not so far succeeded in showing this due to the somewhat complicated form for the general spin-$\ell$ bulk-to-bulk propagator \cite{Costa:2014kfa,Bekaert:2014cea}. Some more discussion is in section \ref{Fercomp}. We will instead use other arguments to establish the validity of our results.
Dolan and Osborn \cite{Dolan:2003hv} studied these blocks using the conformal Casimir equation. Closed-form expressions in terms of hypergeometric functions were obtained in dimensions $d=2, 4, 6$. For example, in $d=2$ we have
\es{twod}{
G_{\Delta,\ell}(z,\overline{z})&= |z|^{\Delta-\ell }\times \\
&\quad \Big[z^\ell {}_2F_1\left({\Delta-\Delta_{12}+\ell\over 2},{\Delta+\Delta_{34}+\ell\over 2},\Delta+\ell;z\right)\\& \quad\times {}_2F_1\left({\Delta-\Delta_{12}-\ell\over 2},{\Delta+\Delta_{34}-\ell\over 2},\Delta-\ell;\overline{z}\right) + (z\leftrightarrow \overline{z}) \Big]}
and in $d=4$ we have
\es{fourd}{G_{\Delta,\ell}(z,\overline{z})&= |z|^{\Delta-\ell }{1\over z-\overline{z}} \times \\ &\quad\Big[ z^{\ell+1} {}_2F_1\left({\Delta-\Delta_{12}+\ell\over 2}, {\Delta+\Delta_{34}+\ell\over 2};\Delta+\ell;z\right)\\ & \quad\quad\times {}_2F_1\left({\Delta-\Delta_{12}-\ell\over 2}-1, {\Delta+\Delta_{34}-\ell\over 2}-1;\Delta-\ell-2;\overline{z}\right)-(z\leftrightarrow \overline{z})\Big]}
The $d=6$ result is also available, taking the same general form, but it is more complicated. Note that the $d=2$ result is actually a sum of two irreducible blocks, chosen so as to be even under parity. The irreducible $d=2$ blocks factorize holomorphically, since the global conformal algebra splits up as sl(2,$\mathbb{R}$) $\oplus$ sl(2,$\mathbb{R}$). An intriguing fact is that the $d=4$ block is expressed as a sum of two terms, each of which ``almost" factorizes holomorphically. Results in arbitrary dimension are available in series form.
Since the results of Dolan and Osborn are obtained as solutions of the conformal Casimir equation, and we will show that our geodesic Witten diagrams are solutions of the same equation with the same boundary conditions, this will constitute exact agreement. Note, though, that the geodesic approach produces the solution in an integral representation. It is not obvious by inspection that these results agree with those in \cite{Dolan:2003hv}, but we will verify this in various cases to assuage any doubts that our general arguments are valid. As noted above, in principle a more direct comparison is to the formulas of Ferrara et. al. \cite{Ferrara:1971vh}.
\subsection{Geodesic Witten diagrams with spin-$\ell$ exchange: generalities}
Consider a CFT$_d$ primary operator which carries scaling dimension $\Delta$ and transforms in the rank-$\ell$ symmetric traceless tensor representation of the (Euclidean) Lorentz group. The AdS$_{d+1}$ bulk dual to such an operator is a symmetric traceless tensor field $h_{\mu_1 \ldots \mu_{\ell}}$ obeying the field equations
\es{spin_l_eqs}{& \nabla^2 h_{\mu_1 \ldots \mu_{\ell}} - [\Delta(\Delta-d)-\ell]h_{\mu_1 \ldots \mu_{\ell}} =0~,\\
& \nabla^{\mu_1} h_{\mu_1 \ldots \mu_{\ell}} =0~.}
Our proposal is that the conformal partial wave $W_{\Delta,\ell}(x_i)$ is given by the same expression as in (\ref{gwitt}) except that now the bulk-to-bulk propagator is that of the spin-$\ell$ field pulled back to the geodesics. The latter defines the spin-$\ell$ version of the geodesic Witten diagram, ${\cal W}_{\Delta,\ell}(x_i)$: its precise definition is
\es{gwitt2}{&{\cal W}_{\Delta,\ell}(x_i)\equiv\\& \int_{\gamma_{12}}\int_{\gamma_{34}}G_{b\partial}(y(\lambda), x_1)G_{b\partial}(y(\lambda),x_2)\times G_{bb}(y(\lambda),y(\lambda');\Delta,\ell)\times G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4)}
and $G_{bb}(y(\lambda),y(\lambda');\Delta,\ell)$ is the pulled-back spin-$\ell$ propagator,
\e{pbprop}{
G_{bb}(y(\lambda),y(\lambda');\Delta,\ell) \equiv [G_{bb}(y,y';\Delta)]_{\mu_1 \ldots \mu_{\ell}, \nu_1 \ldots \nu_{\ell}}{d y^{\mu_1} \over d\lambda} \ldots {d y^{\mu_\ell} \over d\lambda}{d y'^{\nu_1} \over d\lambda'} \ldots {d y'^{\nu_\ell} \over d\lambda'}\Big|_{y=y(\lambda),\, y'=y(\lambda')}~.}
To explicitly evaluate this we will use the same technique as in section \ref{iii1}. Namely, the integration over one geodesic can be expressed as a normalizable spin-$\ell$ solution of the equations (\ref{spin_l_eqs}) with a geodesic source. Inserting this result, we obtain an expression for the geodesic Witten diagram as an integral over the remaining geodesic. If we call the above normalizable solution $h_{\nu_1 \ldots \nu_{\ell}}$, then the analog of (\ref{31a}) is
\begin{equation} \label{genspin}
{\cal W}_{\Delta,\ell}(x_i) = \int_{\gamma_{34}} h_{\nu_1 \ldots \nu_{\ell}}(y(\lambda')){d y'^{\nu_1} \over d\lambda'} \ldots {d y'^{\nu_\ell} \over d\lambda'} G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4)~.
\end{equation}
As in section \ref{iii1}, we will specifically compute
\es{cpw2}{{\cal W}_{\Delta,\ell}(z,\overline{z})&\equiv {1\over C_{12{\cal O}}C^{{\cal O}}_{~~34}}\,\langle {\cal O}_1(\infty){\cal O}_2(0)\,P_{\Delta,\ell}\,{\cal O}_3(1-z){\cal O}_4(1)\rangle \\&= |z|^{-\Delta_3-\Delta_4}G_{\Delta,\ell}(z,\overline{z})}
%
now written in terms of $(z,\overline{z})$ instead of $(u,v)$ to facilitate easier comparison with \eqref{twod} and \eqref{fourd}. We recall that this reduces $\gamma_{12}$ to a straight line at the origin of global AdS. The form of $\gamma_{34}$ is given in (\ref{31n}), from which the pullback is computed using
\begin{eqnarray}
\cos^2 \rho\big|_{\gamma_{34}}& =& { 1\over 2\cosh\lambda'} { |z|^2\over e^{-\lambda'}+|1-z|^2 e^{\lambda'} }~,\cr
e^{2t}\big|_{\gamma_{34}}&=&{2\cosh \lambda' \over e^{-\lambda'}+|1-z|^2e^{\lambda'} }~,\cr
e^{2i\phi}\big|_{\gamma_{34}}&=& {(1-z) e^{\lambda'}+e^{-\lambda'} \over (1-\overline{z}) e^{\lambda'}+e^{-\lambda'} }~.
\end{eqnarray}
We also recall
\e{qqq}{G_{b\partial}(y(\lambda'),1-z)G_{b\partial}(y(\lambda'),1) = {e^{\Delta_{34}\lambda'}\over |z|^{\Delta_3+\Delta_4}}~.}
Carrying out this procedure for all dimensions $d$ at once presents no particular complications. However, it does not seem easy to find the solution $ h_{\nu_1 \ldots \nu_{\ell}}$ for all $\ell$ at once. For this reason, below we just consider the two simplest cases of $\ell=1, 2$, which suffice for illustrating the general procedure.
\subsection{Evaluation of geodesic Witten diagram: spin-1}\label{v3}
In the global AdS$_{d+1}$ metric
\begin{equation}
ds^2 = {1\over \cos^2 \rho}(d\rho^2+ dt^2+\sin^2 \rho d\Omega_{d-1}^2)
\end{equation}
we seek a normalizable solution of
\begin{equation}
\nabla^2 A_{\mu} - [\Delta(\Delta-d)-1]A_{\mu} =0~,\quad \nabla^{\mu} A_{\mu} =0
\end{equation}
which is spherically symmetric and has time dependence $e^{-\Delta_{12}t}$. A suitable ansatz is
\begin{equation}
A_\mu dx^\mu = A_t(\rho,t) dt + A_\rho(\rho,t)d\rho~.
\end{equation}
Assuming the time dependence $e^{-\Delta_{12}t}$, the divergence free condition implies
\begin{equation}
\partial_\rho\left(\tan^{d-1}\rho A_\rho\right)-\Delta_{12}\tan^{d-1}\rho A_t=0
\end{equation}
and the components of the wave equation are
\es{}{&{\cos^{d-1}\rho \over \sin^{d-1}\rho}\partial_\rho \left({\sin^{d-1}\rho\over \cos^{d-3}\rho}\partial_\rho A_t \right) +\left(\Delta_{12}^2 \cos^2\rho -(\Delta-1)(\Delta-d+1)\right)A_t \\
&\quad-2\Delta_{12}\cos \rho \sin\rho A_\rho =0 \\
& {\cos^{d-1}\rho \over \sin^{d-1}\rho}\partial_\rho \left({\sin^{d-1}\rho\over \cos^{d-3}\rho}\partial_\rho A_\rho \right)+\left(\Delta_{12}^2 \cos^2\rho -{d-1\over \sin^2\rho}-(\Delta-1)(\Delta-d+1)\right)A_\rho \\
&\quad +2\Delta_{12} \cos\rho\sin\rho A_t =0~.}
The normalizable solution is
\begin{eqnarray}\label{vecsol}
A_\rho &=& \Delta_{12} \sin \rho (\cos\rho)^{\Delta} {_2{F_1}}\left({\Delta+\Delta_{12}+1\over 2},{\Delta-\Delta_{12}+1\over 2},\Delta-{d-2\over 2};\cos^2\rho \right)e^{-\Delta_{12}t}\cr
A_t &=& {1\over \Delta_{12}\tan^{d-1}\rho} \partial_\rho(\tan^{d-1} \rho A_\rho )
\end{eqnarray}
where we have inserted a factor of $\Delta_{12}$ in $A_\rho$ to ensure a smooth $\Delta_{12}\rightarrow 0$ limit. In particular, setting $\Delta_{12}=0$ we have $A_\rho=0$ and
\begin{equation}
A_t= (\cos \rho)^{\Delta-1} {_2{F_1}}\left({\Delta+1\over 2},{\Delta-1\over 2},\Delta-{d-2\over 2};\cos^2\rho \right)~.
\end{equation}
It is now straightforward to plug into (\ref{genspin}) to obtain an integral expression for the conformal block. Because the general formula is rather lengthy we will only write it out explicitly in the case $\Delta_{12}=0$. In this case we find (not paying attention to overall normalization factors)
\es{}{{\cal W}_{\Delta,1}(z,\overline{z}) &=|z|^{\Delta-\Delta_3-\Delta_4-1}(1-|1-z|^2)\\
&\quad \int_0^1\! d\sigma \sigma^{{\Delta+\Delta_{34}-1\over 2}}(1-\sigma)^{{\Delta-\Delta_{34}-1\over 2}} \big(1-(1-|1-z|^2)\sigma\big)^{-{\Delta+1\over 2}}
\\ &\quad \quad\times {_2{F_1}}\Bigg({\Delta+1\over 2},{\Delta-1\over 2},\Delta-{d-2\over 2}; {|z|^2\sigma(1-\sigma)\over 1-(1-|1-z|^2)\sigma}\Bigg)~.}
Setting $d=2, 4$, it is straightforward to verify that the series expansion of this integral reproduces the known $d=2, 4$ results in (\ref{twod}),(\ref{fourd}) for $\Delta_{12}=0$. We have also verified agreement for $\Delta_{12}\neq 0$.
\subsection{Evaluation of geodesic Witten diagram: spin-2}
In this section we set $\Delta_{12}=0$ to simplify formulas a bit. We need to solve
\begin{equation}
\nabla^2 h_{\mu\nu} - [\Delta(\Delta-d)-2]h_{\mu\nu} =0~,\quad \nabla^{\mu} h_{\mu\nu} =0~,\quad h^\mu_\mu=0~.
\end{equation}
$h_{\mu\nu}$ should be static and spherically symmetric, which implies the general ansatz
\begin{equation}
h_{\mu\nu}dx^\mu dx^\nu = f_{\rho\rho}(\rho)g_{\rho\rho}d\rho^2+ f_{tt}(\rho)g_{tt}dt^2
+{1\over d-1} f_{\phi\phi}(\rho)\tan^2\rho \,d\Omega^2_{d-1}~.
\end{equation}
We first impose the divergence free and tracelessness conditions. We have
\begin{equation} h^\mu_\mu = f_{\rho\rho}+f_{tt}+f_{\phi\phi}~.
\end{equation}
We use this to eliminate $f_{\phi\phi}$,
\begin{equation}
f_{\phi\phi}=-f_{\rho\rho}-f_{tt}~.
\end{equation}
Moving to the divergence, only the component $\nabla^\mu h_{\mu\rho}$ is not automatically zero. We find
\begin{equation}
\nabla^\mu h_{\mu\rho} = f_{\rho\rho}' + { d+1\over \cos \rho\sin\rho} f_{\rho\rho} -{\cos\rho\over \sin\rho} f_{\rho\rho} +{\cos\rho\over \sin\rho}f_{tt}=0
\end{equation}
which we solve as
\begin{equation} f_{tt} = -\tan\rho f_{\rho\rho}' +\left(1-{d+1\over \cos^2\rho}\right)f_{\rho\rho}~.
\end{equation}
We then work out the $\rho\rho$ component of the field equation,
\begin{equation}
\nabla^2 h_{\rho\rho}-[\Delta(\Delta-d)-2]h_{\rho\rho} = f_{\rho\rho}''+\left({d+3\over \cos\rho\sin\rho}-2\cot\rho \right)f_{\rho\rho}'-{(\Delta+2)(\Delta-d-2)\over \cos^2\rho}f_{\rho\rho}~.
\end{equation}
Setting this to zero, the normalizable solution is
\begin{equation}
f_{\rho\rho} = (\cos \rho)^{\Delta+2} {_2{F_1}}\left({\Delta\over 2},{\Delta+2\over 2},\Delta-{d-2\over 2};\cos^2\rho\right)~.
\end{equation}
This completely specifies the solution, and we now have all we need to plug into (\ref{genspin}). We refrain from writing out the somewhat lengthy formulas. The series expansion of the result matches up with (\ref{twod}) and (\ref{fourd}) as expected.
\subsection{General $\ell$: proof via conformal Casimir equation}\label{v5}
As in the case of scalar exchange, the most efficient way to verify that a geodesic Witten diagram yields a conformal partial wave is to check that it is an eigenfunction of the conformal Casimir operator with the correct eigenvalue and asymptotics.
We start from the general expression \eqref{gwitt2}. A rank-$n$ tensor on AdS is related to a tensor on the embedding space via
\begin{equation}
T_{\mu_1 \ldots \mu_n} = {\partial Y^{M_1} \over \partial y^{\mu_1}}\ldots {\partial Y^{M_n} \over \partial y^{\mu_n}}T_{M_1 \ldots M_n}~.
\end{equation}
In particular, this holds for the bulk-to-bulk propagator of the spin-$\ell$ field, and so we can write
\begin{equation}
G_{bb}(y,y';\Delta,\ell) = [G_{bb}(Y,Y';\Delta)]_{M_1 \ldots M_{\ell}, N_1 \ldots N_{\ell}}{d Y^{M_1} \over d\lambda} \ldots {d Y^{M_\ell} \over d\lambda}{d Y'^{N_1} \over d\lambda'} \ldots {d Y'^{N_\ell} \over d\lambda'}~.
\end{equation}
Now, $[G_{bb}(Y,Y';\Delta)]_{M_1 \ldots M_{\ell}, N_1 \ldots N_{\ell}}$ only depends on $Y$ and $Y'$. Since $Y^M {dY^M\over d\lambda} = {1\over 2} {d\over d\lambda} (Y\cdot Y) =0$, when pulled back to the geodesics the only contributing structure is
\begin{equation} \label{propstruc}
[G_{bb}(Y,Y';\Delta)]_{M_1 \ldots M_{\ell}, N_1 \ldots N_{\ell}} = f(Y\cdot Y') Y'_{M_1} \ldots Y'_{M_\ell}Y_{N_1} \ldots Y_{N_\ell}~.
\end{equation}
We also recall a few other useful facts. Lifted to the embedding space, the geodesic connecting boundary points $X_1$ and $X_2$ is
\begin{equation}
Y(\lambda) = {e^\lambda X_1 + e^{-\lambda} X_2 \over \sqrt{-2 X_1 \cdot X_2} }~.
\end{equation}
The bulk-to-boundary propagator lifted to the embedding space is
\begin{equation} \label{bbform}
G_{b\partial}(X_i,Y)\propto (X_i\cdot Y)^{-\Delta_i}~.
\end{equation}
We follow the same strategy as in the case of scalar exchange. We start by isolating the part of the diagram that contains all the dependence on $X_{1,2}$,
\es{}{&F_{M_1 \ldots M_\ell}(X_1,X_2,Y';\Delta)=\\ &\int_{\gamma_{12}} G_{b\partial}(X_1,Y(\lambda))G_{b\partial}(X_2,Y(\lambda))[G_{bb}(Y(\lambda),Y';\Delta)]_{M_1 \ldots M_{\ell}, N_1 \ldots N_{\ell}}{d Y^{M_1} \over d\lambda} \ldots {d Y^{M_\ell} \over d\lambda}~.}
Here, $Y(\lambda)$ lives on $\gamma_{12}$, but $Y'$ is left arbitrary. This is the spin-$\ell$ generalization of $\varphi^{12}_{\Delta}(y)$ defined in \eqref{31aa}, lifted to embedding space. We now argue that this is annihilated by the SO(d+1,1) generators $L^1_{AB} +L^2_{AB} + L^{Y'}_{AB}$. This generator is the sum of three generators in the scalar representation, plus a ``spin" term acting on the free indices $N_1 \ldots N_\ell$. This operator annihilates any expression of the form $g(X_1 \cdot X_2, X_1 \cdot Y', X_2 \cdot Y') X_{N_1}\ldots X_{N_\ell}$, where each $X$ stands for either $X_1$ or $X_2$. To show this, we just note the SO(d+1,1) invariance of the dot products, along with the fact that $X_N$ is the normal vector to the $X^2=0$ surface and so is also SO(d+1,1) invariant.
From (\ref{propstruc})-(\ref{bbform}) we see that $F_{N_1 \ldots N_\ell}(X_1,X_2,Y';\Delta)$ is of this form, and so is annihilated by $L^1_{AB} +L^2_{AB} + L^{Y'}_{AB}$. We can therefore write
\begin{eqnarray}\label{cas}
(L^1_{AB} +L^2_{AB})^2 F_{N_1 \ldots N_\ell}(X_1,X_2,Y';\Delta) &=& (L^{Y'}_{AB})^2F_{N_1 \ldots N_\ell}(X_1,X_2,Y';\Delta)\cr &=&C_2(\Delta,\ell) F_{N_1 \ldots N_\ell}(X_1,X_2,Y';\Delta)
\end{eqnarray}
where we used that $(L^{Y'}_{AB})^2$ is acting on the spin-$\ell$ bulk-to-bulk propagator, which is an eigenfunction of the conformal Casimir operator\footnote{Note that the conformal Casimir is equal to the spin-$\ell$ Laplacian up to a constant shift: $(L_{AB}^{Y'})^2 = \nabla^2_{\ell}+\ell(\ell+d-1)$ \cite{Pilch:1984xx}.} with eigenvalue \eqref{32c}. The relation \eqref{cas} holds for all $Y'$, and hence holds upon integrating $Y'$ over $\gamma_{34}$ with any weight. Hence we arrive at the conclusion
\begin{equation}
(L^1_{AB} +L^2_{AB})^2{\cal W}_{\Delta,\ell}(x_i) = C_2(\Delta,\ell){\cal W}_{\Delta,\ell}(x_i)
\end{equation}
which is the same eigenvalue equation obeyed by the spin-$\ell$ conformal partial wave, $W_{\Delta,\ell}(x_i)$. The short distance behavior as dictated by the OPE is easily seen to match in the two cases, establishing that we have the same eigenfunction. We conclude that the spin-$\ell$ geodesic Witten diagram is, up to normalization, equal to the spin-$\ell$ conformal partial wave.
\subsection{Comparison to double integral expression of Ferrara et. al.}
\label{Fercomp}
It is illuminating to compare our expression (\ref{gwitt2}) to equation (50) in \cite{Ferrara:1973vz}, which gives the general result (in $d=4$) for the scalar conformal partial wave with spin-$\ell$ exchange, written as a double integral. We will rewrite the result in \cite{Ferrara:1973vz} in a form permitting easy comparison to our formulas. First, it will be useful to rewrite the scalar bulk-to-bulk propagator (\ref{22b}) by applying a quadratic transformation to the hypergeometric function,
\begin{equation} \label{bbmod}
G_{bb}(y,y';\Delta)=\xi^\Delta {}_2F_1\left({\Delta\over 2},{\Delta+1\over 2},\Delta+1-{d\over 2};\xi^2\right)~.
\end{equation}
Next, recall that in embedding space the geodesics are given by (\ref{32p}), from which we compute the quantity $\xi$ with one point on each geodesic
\begin{equation}
\xi^{-1} = -Y(\lambda) \cdot Y(\lambda') = {1\over 2} { e^{\lambda+\lambda'}x_{13}^2 + e^{\lambda-\lambda'}x_{14}^2 +e^{-\lambda+\lambda'}x_{23}^2 + e^{-\lambda-\lambda'}x_{24}^2 \over x_{12} x_{34} }~.
\end{equation}
We also define a modified version as
\begin{equation}
\xi_-^{-1} = -{dY(\lambda)\over d\lambda}\cdot {dY(\lambda')\over d\lambda'}={1\over 2} { e^{\lambda+\lambda'}x_{13}^2 - e^{\lambda-\lambda'}x_{14}^2 -e^{-\lambda+\lambda'}x_{23}^2 + e^{-\lambda-\lambda'}x_{24}^2 \over x_{12} x_{34} }~.
\end{equation}
Comparing to \cite{Ferrara:1973vz}, we have $\xi^{-1}=\lambda_+$ and $\xi_-^{-1}=\lambda_-$.
With these definitions in hand, it is not hard to show that the result of \cite{Ferrara:1973vz} takes the form
\es{}{W_{\Delta,\ell}(x_i)= \int_{\gamma_{12}}\int_{\gamma_{34}}&G_{b\partial}(y(\lambda), x_1)G_{b\partial}(y(\lambda),x_2)\\
& \times C'_{\ell}(2\xi_{-}^{-1})G_{bb}(y(\lambda),y(\lambda');\Delta)\times G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4)~.}
Here $G_{bb}(y(\lambda),y(\lambda');\Delta)$ is the scalar bulk-to-bulk propagator (\ref{bbmod}), and $C'_{\ell}(x)$ is a Gegenbauer polynomial. This obviously looks very similar to our expression (\ref{gwitt2}), and indeed agrees with it for $\ell=0$. The two results must be equal (up to normalization) since they are both expressions for the same conformal partial wave. If we assume that equality holds for the integrand, then we find the interesting result that the pullback of the spin-$\ell$ propagator, as written in (\ref{pbprop}), is equal to $C'_{\ell}(2\xi_{-}^{-1})G_{bb}(y(\lambda),y(\lambda');\Delta)$. The general spin-$\ell$ propagator is very complicated (see \cite{Costa:2014kfa,Bekaert:2014cea}), but apparently has a simple relation to the scalar propagator when pulled back to geodesics. It would be interesting to verify this.
\subsection{Decomposition of spin-1 Witten diagram into conformal blocks}
In the case of scalar exchange diagrams, we previously showed how to decompose a Witten diagram into a sum of geodesic Witten diagrams, the latter being identified with conformal partial waves of both single- and double-trace exchanges. We now wish to extend this to the case of higher spin exchange; we focus here on the case of spin-1 exchange for simplicity. A picture of the final result is given in Figure \ref{f6}.
As discussed in section \ref{ii}, given two scalar operators in a generalized free field theory, we can form scalar double trace primaries with schematic form $[{\cal O}_1{\cal O}_2]_{m,0}\sim {\cal O}_1 \partial^{2m}{\cal O}_2$ and dimension $\Delta^{(12)}(m,0) = \Delta_1 + \Delta_2 +2m+O(1/N^2)$, and vector primaries $[{\cal O}_1{\cal O}_2]_{m,1} \sim {\cal O}_1 \partial^{2m}\partial_{\mu}{\cal O}_2$ with dimension $\Delta^{(12)}(m,1) = \Delta_1 + \Delta_2 +1+2m+O(1/N^2)$. The analysis of \cite{Heemskerk:2009pn}, and later \cite{Fitzpatrick:2011dm, Costa:2014kfa,Bekaert:2014cea} demonstrated that these conformal blocks, and their cousins $[{\cal O}_3{\cal O}_4]_{n,0}$ and $[{\cal O}_3{\cal O}_4]_{n,1}$, should appear in the decomposition of the vector exchange Witten diagram, together with the exchange of a single-trace vector operator. The computations below will confirm this expectation.
\begin{figure}[t!]
\begin{center}
\includegraphics[width = .95\textwidth]{vec_decomp_4f.pdf}
\includegraphics[width = .95\textwidth]{vec_decomp_5.pdf}
\includegraphics[width = .95\textwidth]{vec_decomp_6.pdf}
\caption{The decomposition of a four-point vector exchange diagram (upper left) into conformal partial waves. The term in the upper right captures the single-trace exchange of the dual vector operator. The second line captures the CFT exchanges of the $\ell=0$ double-trace operators $[{\cal O}_1{\cal O}_2]_{m,0}$ and $[{\cal O}_3{\cal O}_4]_{n,0}$. Likewise, the final line captures the CFT exchanges of the $\ell=1$ double-trace operators $[{\cal O}_1{\cal O}_2]_{m,1}$ and $[{\cal O}_3{\cal O}_4]_{n,1}$.}
\label{f6}
\end{center}
\end{figure}
The basic approach is the same as in the scalar case, although the details are more complicated. Before diving in, let us note the main new features. In the scalar case a basic step was to write, in \eqref{41a}, the product of two bulk-to-boundary propagators $G_{b\partial}(y,x_1)G_{b\partial}(y,x_2)$ as a sum over solutions $\varphi^{12}_{\Delta}(y)$ of the scalar wave equation sourced on the $\gamma_{12}$ geodesic. Here, we will similarly need a decomposition of $G_{b\partial}(y,x_1)\nabla_\mu G_{b\partial}(y,x_2)$, where $\nabla_{\mu}$ is a covariant derivative with respect to bulk coordinates $y$. It turns out that this can be expressed as a sum over massive spin-1 solutions and derivatives of massive scalar solutions. This translates into the statement that the spin-1 exchange Witten diagram decomposes as a sum of spin-1 and spin-0 conformal blocks, as noted above.
Now to the computation. We consider a theory of massive scalars coupled to a massive vector field via couplings $\phi_i \nabla_\mu \phi_j A^\mu$. The Witten diagram with vector exchange is then
\begin{equation} \label{witt1}
{\cal A}_4^{\text{Vec}}(x_i) = \int_{y}\int_{y'} G_{b\partial}(y,x_1)\nabla_\mu G_{b\partial}(y,x_2) \times G_{bb}^{\mu\nu}(y,y';\Delta) \times G_{b\partial}(y',x_3)\nabla_\nu G_{b\partial}(y',x_4)~.
\end{equation}
Our first task is to establish the expansion
\begin{equation} \label{spin1eq}
G_{b\partial}(y,x_1)\nabla_\mu G_{b\partial}(y,x_2) = \sum_m \left( c_m A_{m,\mu}(y) + b_m \nabla_\mu \varphi_m(y) \right)
\end{equation}
where $ A_{m,\mu}(y) $ and $\varphi_m(y)$ denote the solutions to the massive spin-1 and spin-0 equations sourced on $\gamma_{12}$, found earlier in sections \ref{v3} and \ref{iv1}, respectively.\footnote{$\varphi_m$ is just $\varphi^{12}_m$, whose superscript we suppress for clarity, and likewise for $\varphi_n$ and $\varphi^{34}_n$.} $m$ labels the masses of the bulk fields, to be determined shortly. We will not attempt to compute the coefficients $c_m$ and $b_m$, which is straightforward but involved, contenting ourselves to determining the spectrum of conformal dimensions appearing in the expansion, and showing how the expansion coefficients can be obtained if desired.
Following the scalar case, we work in global AdS and send $t_1 \rightarrow -\infty$, $t_2 \rightarrow \infty$. Dropping normalizations, as we shall do throughout this section, we have
\es{}{G_{b\partial}(y,x_1)\nabla_\rho G_{b\partial}(y,x_2) &= \sin\rho\,(\cos\rho)^{\Delta_1+\Delta_2-1}e^{-\Delta_{12}t} \\
G_{b\partial}(y,x_1)\nabla_t G_{b\partial}(y,x_2) &= (\cos\rho)^{\Delta_1+\Delta_2}e^{-\Delta_{12}t}~.}
Letting $\Delta_m^{(\ell)}$ denote the dimension of the corresponding spin, we have, from \eqref{vecsol} and \eqref{31h},
\es{}{\varphi_m & = {}_2F_1\Big({\Delta^{(0)}_m+\Delta_{12}\over 2},{\Delta_m^{(0)} -\Delta_{12} \over 2};\Delta_m^{(0)} - {d-2\over 2};\cos^2 \rho\Big)( \cos \rho)^{\Delta_m^{(0)}}~ e^{-\Delta_{12}t }\\
A_{m,\rho}&= \Delta_{12} \sin \rho\, (\cos\rho)^{\Delta_m^{(1)}} {_2{F_1}}\left({\Delta_m^{(1)}+\Delta_{12}+1\over 2},{\Delta_m^{(1)}-\Delta_{12}+1\over 2},\Delta_m^{(1)}-{d-2\over 2};\cos^2\rho \right)e^{-\Delta_{12}t}\\
A_{m,t} &= {1\over \Delta_{12}\tan^{d-1}\rho} \partial_\rho(\tan^{d-1} \rho A_{m,\rho} )}
The various terms have the following powers $(\cos^2 \rho)^k$ in an expansion in powers of $\cos^2 \rho$,
\begin{eqnarray}
G_{b\partial}(y,x_1)\nabla_\rho G_{b\partial}(y,x_2): && \quad k={\Delta_1+\Delta_2-1\over 2} + q\quad
\cr A_{m,\rho}: && \quad k={\Delta^{(1)}_m\over 2} + q \cr
\nabla_\rho \varphi_m:&&\quad k= {\Delta^{(0)}_m-1\over 2}+q \cr
G_{b\partial}(y,x_1)\nabla_t G_{b\partial}(y,x_2): && \quad k={\Delta_1+\Delta_2\over 2} \cr
A_{m,t}: && \quad k={\Delta^{(1)}_m-1\over 2} + q \cr
\nabla_t \varphi_m:&&\quad k= {\Delta^{(0)}_m\over 2}+q
\end{eqnarray}
where $q=0, 1, 2, \ldots$. Comparing, we see that we have the right number of free coefficients for (\ref{spin1eq}) to hold, provided we have the following spectrum of dimensions appearing
\begin{eqnarray}\label{539}
\Delta_m^{(0)}& = & \Delta_1+\Delta_2 + 2m \cr
\Delta_m^{(1)}& = & \Delta_1+\Delta_2+1 + 2m
\end{eqnarray}
with $m=0, 1, 2, \ldots$. The formulas above can be used to work out the explicit coefficients $c_m$ and $b_m$. We noted at the beginning of this subsection that this spectrum of dimensions coincides with the expected spectrum of double-trace scalar and vector operators appearing in the OPE, at leading order in large $N$.
We may now rewrite (\ref{witt1}) as\footnote{Following the precedent of Section \ref{iv}, all quantities with an $m$ subscript refer to the double-trace operators appearing in the ${\cal O}_1{\cal O}_2$ OPE, and those with an $n$ subscript refer to the double-trace operators appearing in the ${\cal O}_3{\cal O}_4$ OPE. }
\begin{equation} \label{witt2}
{\cal A}_4^{\text{Vec}}(x_i) =\sum_{m,n} \int_{y}\int_{y'} \left( c_m A_{m,\mu}(y) + b_m \nabla_\mu \varphi_m(y) \right) G_{bb}^{\mu\nu}(y,y';\Delta) \left( c_n A_{n,\nu}(y') + b_n \nabla_\nu \varphi_n(y') \right)~.
\end{equation}
We expand this out in an obvious fashion as
\begin{equation} \label{4terms}
{\cal A}_4^{\text{Vec}}(x_i) = {\cal A}_{AA}(x_i)+ {\cal A}_{A\phi}(x_i)+ {\cal A}_{\phi A}(x_i)+ {\cal A}_{\phi\phi}(x_i)~.
\end{equation}
The next step is to relate each term to geodesic Witten diagrams, which we now do in turn.
\subsubsection{ ${\cal A}_{AA}$}
We have
\begin{equation}
{\cal A}_{AA} = \sum_{m,n} c_mc_n \int_{y}\int_{y'} A_{m,\mu}(y) G_{bb}^{\mu\nu}(y,y';\Delta)A_{n,\nu}(y')~.
\end{equation}
The solution $A_{m,\mu}(y)$ can be expressed as
\begin{eqnarray}\label{Aint}
A^\mu_{m}(y)&=& \int_{\gamma_{12}} G_{b\partial}(y(\lambda),x_1)\nabla_\nu G_{b\partial}(y(\lambda),x_2)G_{bb}^{\mu\nu}(y(\lambda),y;\Delta^{(1)}_m)\cr
&=&-\Delta_2 \int_{\gamma_{12}} G_{b\partial}(y(\lambda),x_1)G_{b\partial}(y(\lambda),x_2){dy_\nu(\lambda)\over d\lambda} G_{bb}^{\mu\nu}(y(\lambda),y;\Delta^{(1)}_m)~.
\end{eqnarray}
The second equality follows from the relation $\nabla_\mu G_{b\partial}(x,y(\lambda)) = -\Delta {dy_\mu(\lambda)\over d\lambda} G_{b\partial}(x,y(\lambda))$, which is easily verified for a straight line geodesic at the center of global AdS, and hence is true in general. Using this we obtain (dropping the normalization, as usual)
\begin{eqnarray}\label{WAA}
{\cal A}_{AA}&=& \sum_{m,n} c_mc_n \int_{y}\int_{y'} \int_{\gamma_{12}} \int_{\gamma_{34}} \Big[G_{b\partial}(y(\lambda),x_1) G_{b\partial}(y(\lambda),x_2){dy_\mu(\lambda)\over d\lambda} \Big]\cr
&& \quad\quad\quad\quad \times\Big[ G_{bb}^{\mu\nu}(y(\lambda),y;\Delta^{(1)}_m)G_{bb,\nu\alpha}(y,y';\Delta) G_{bb}^{\alpha \beta}(y',y(\lambda');\Delta^{(1)}_n)\Big]\cr
&&\quad\quad\quad\quad \times\Big[ G_{b\partial}(y(\lambda),x_3) G_{b\partial}(y(\lambda),x_4){dy'_\beta(\lambda')\over d\lambda'}\Big]~.
\end{eqnarray}
The bulk-to-bulk propagator for the vector field obeys
\begin{equation}
\big(\nabla^2 -m^2 \big) G_{bb}^{\mu\nu}(y,y';\Delta)= \delta^{\mu\nu}(y-y')
\end{equation}
where $\delta^{\mu\nu}(y-y') $ denotes a linear combination of $g^{\mu\nu} \delta(y-y')$ and $\nabla^\mu \nabla^\nu \delta(y-y')$. Using this, and the fact that the propagator is divergence free at non-coincident points, we can verify the composition law
\begin{equation} \label{comp}
\int_{y'} G_{bb}^{\mu\nu}(y,y';\Delta) G_{bb,\nu\alpha}(y',y'';{\Delta'}) = {1\over m^2-(m')^{2}} \Big( {G_{bb}}^{\mu}_{~\alpha}(y,y'';{\Delta}) - {G_{bb}}^{\mu}_{~\alpha}(y,y'';{\Delta'}) \Big)~.
\end{equation}
We use this relation twice within (\ref{WAA}) to obtain a sum of three terms, each with a single vector bulk-to-bulk propagator. Note also that these propagators appear pulled back to the geodesics. Each term is thus a geodesic Witten diagram with an exchanged vector, that is, a spin-1 conformal partial wave. The spectrum of spin-1 operators that appears is
\begin{equation}
\Delta~,\quad \Delta_1 + \Delta_2 +1 +2m~,\quad \Delta_3 + \Delta_4 +1 +2n~,\quad m,n=0, 1, 2, \ldots
\end{equation}
So the contribution of ${\cal A}_{AA}$ is a sum of spin-1 conformal blocks with internal dimensions corresponding to the original exchanged field, along with the expected spin-1 double trace operators built out of the external scalars.
\subsubsection{${\cal A}_{A \phi}$ and ${\cal A}_{\phi A}$}
We start with
\begin{equation}
{\cal A}_{\phi A}= \sum_{m,n} c_mb_n \int_{y}\int_{y'}\nabla_\mu \varphi_m(y) G^{\mu\nu}_{bb}(y,y';\Delta) A_{n,\nu}(y')~.
\end{equation}
Next we integrate by parts in $y$, use $\nabla_\mu G_{bb}^{\mu\nu}(y,y';\Delta) \propto \nabla^\nu \delta(y-y') $, and integrate by parts again, to get
\begin{equation}
{\cal A}_{\phi A} = \sum_{m,n} b_mc_n \int_{y'}\nabla_\mu \varphi_m(y') A^\mu_n(y')~.
\end{equation}
Now we write $A^\mu_{n}(y')$ as an integral over $\gamma_{34}$ as in (\ref{Aint}) and then again remove the bulk-to-bulk propagator by integrating by parts. This yields
\begin{equation}
{\cal A}_{\phi A} = \sum_{m,n} b_mc_n \int_{\gamma_{34}} G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4){d y^\mu(\lambda')\over d\lambda'}\nabla_\mu \varphi_m(y(\lambda'))~.
\end{equation}
Writing $\varphi_m$ as an integral sourced on $\gamma_{12}$ we obtain
\es{}{&{\cal A}_{\phi A} = \sum_{m,n} b_mc_n\times\\
& \int_{\gamma_{12}} \int_{\gamma_{34}} G_{b\partial}(y(\lambda),x_1)G_{b\partial}(y(\lambda),x_2){d\over d\lambda'}G_{bb}(y(\lambda),y(\lambda');\Delta^{(0)}_m)
G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4)~.}
Integrating by parts and using ${d\over d\lambda'} \big( G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4)\big) \propto G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4)$
we see that ${\cal A}_{\phi A}$ decomposes into a sum of spin-0 exchange geodesic Witten diagrams. That is, ${\cal A}_{\phi A}$ contributes a sum of spin-0 blocks with conformal dimensions
\begin{equation}
\Delta_1 + \Delta_2 + 2m~,\quad m=0, 1, 2, \ldots
\end{equation}
By the same token ${\cal A}_{A \phi}$ yields a sum of spin-0 blocks with conformal dimensions %
\e{}{\Delta_3 + \Delta_4 + 2n~,\quad n=0, 1, 2, \ldots}
\subsubsection{${\cal A}_{\phi\phi}$}
We have
\begin{equation}
{\cal A}_{\phi\phi}= \sum_{m,n} b_mb_n \int_{y}\int_{y'} \nabla_\mu \varphi_m(y)G^{\mu\nu}_{bb}(y,y';\Delta)\nabla_\nu \varphi_n(y')~.
\end{equation}
Integration by parts reduces this to
\begin{equation}
{\cal A}_{\phi\phi}= \sum_{m,n} b_mb_n \int_{y} \nabla^\mu \varphi_m(y) \nabla_\mu \varphi_n(y)~.
\end{equation}
Now rewrite the scalar solutions as integrals over the respective geodesic sources,
\begin{eqnarray}\label{Waa}
{\cal A}_{\phi\phi}&=& \sum_{m,n} b_mb_n\int_{y'} \int_{\gamma_{12}}\int_{\gamma_{34}}G_{b\partial}(y(\lambda),x_1) G_{b\partial}(y(\lambda),x_2) G_{b\partial}(y(\lambda),x_3) G_{b\partial}(y(\lambda),x_4) \cr
&& \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times \nabla^{\mu'}G_{bb}(y(\lambda),y';\Delta^{(0)}_m) \nabla_{\mu'}G_{bb}(y',y(\lambda');\Delta^{(0)}_n)~.
\end{eqnarray}
The composition law analogous to (\ref{comp}) is easily worked out to be
\es{}{\int_{y'} \nabla^{\mu'}G_{bb}(y(\lambda),y';\Delta^{(0)}_m) \nabla_{\mu'}G_{bb}(y',y(\lambda');\Delta^{(0)}_n) &= c_{mn} G_{bb}(y(\lambda),y(\lambda');\Delta^{(0)}_m) \\&+ d_{mn} G_{bb}(y(\lambda),y(\lambda');\Delta^{(0)}_n)}
with some coefficients $c_{mn}$ and $d_{mn}$ that we do not bother to display here. Inserting this in (\ref{Waa}) we see that ${\cal A}_{\phi\phi}$ decomposes into a sum of scalar blocks with conformal dimensions
\begin{equation}
\Delta_1+\Delta_2 +2m~,\quad \Delta_3+\Delta_4 +2n~, \quad m,n=0,1,2,\ldots
\end{equation}
\subsubsection{Summary}
We have shown that the Witten diagram involving the exchange of a spin-1 field of dimension $\Delta$ decomposes into a sum of spin-1 and spin-0 conformal blocks. The full spectrum of conformal blocks appearing in the decomposition is
\begin{eqnarray}
{\rm scalar:}&& \Delta_1+\Delta_2 +2n~,\quad \Delta_3+\Delta_4 +2n \cr
{\rm vector:}&& \Delta~,\quad \Delta_1+\Delta_2+1 +2n~,\quad \Delta_3+\Delta_4 +1+2n
\end{eqnarray}
where $n=0,1,2,\ldots$. This matches the spectrum expected from $1/N$ counting, including single- and double-trace operator contributions. With some patience, the formulas above can be used to extract the coefficient of each conformal block, but we have not carried this out in full detail here.
While we have not explored this in any detail, it seems likely that the above method can be directly generalized to the case of arbitrary spin-$\ell$ exchange. The split \eqref{4terms} will still be natural, and a higher spin version of \eqref{comp} should hold.
\section{Discussion and future work}\label{vi}
In this paper, we have shed new light on the underlying structure of tree-level scattering amplitudes in AdS. Four-point scalar amplitudes naturally organize themselves into geodesic Witten diagrams; recognizing these as CFT conformal partial waves signals the end of the computation, and reveals a transparency between bulk and boundary with little technical effort required. We are optimistic that this reformulation extends, in some manner, to computations of generic holographic correlation functions in AdS/CFT. To that end, we close with some concrete observations and proposals, as well as a handful of future directions.
\vskip .1 in
\noindent
{\bf $\bullet$~Adding loops}
\vskip .1 in
It is clearly of interest to try to generalize our techniques to loop level. We first note that there is a special class of loop diagrams that we can compute already using these methods: namely, those that can be written as an infinite sum of tree-level exchange diagrams \cite{Penedones:2010ue}. For the same reason, this is the only class of loop diagrams whose Mellin amplitudes are known \cite{Penedones:2010ue}. These diagrams only involve bulk-to-bulk propagators that all start and end at the same points; see Figure \ref{f7} for examples. Careful study of the resulting sums would be useful.
\begin{figure}[t!]
\begin{center}
\hspace*{\fill}%
\includegraphics[width = 0.46\textwidth]{Loops1.pdf}\hfill
\includegraphics[width = 0.46\textwidth]{Loops2.pdf}%
\hspace*{\fill}%
\caption{Some examples of loop diagrams that can be written as infinite sums over tree-level diagrams, and hence decomposed into conformal blocks using our methods.}
\label{f7}
\end{center}
\end{figure}
More generally, though, we do not yet know how to decompose generic diagrams into geodesic objects. This would seem to require a ``geodesic identity'' analogous to \eqref{41a} that applies to a pair of bulk-to-bulk propagators, rather than bulk-to-boundary propagators. It would be very interesting to find these, if they exist. Such identities would also help to decompose an exchange Witten diagram in the crossed channel.
\vskip .1 in
\noindent
{\bf $\bullet$~Adding legs}
\vskip .1 in
\begin{figure}[t!]
\begin{center}
\includegraphics[width = 0.47\textwidth]{5pt_block.pdf}
\caption{This is the basic constituent that emerges in applying our technology to the decomposition of a five-point tree-level Witten diagram. However, it is not equal to the five-point conformal partial wave, as discussed in the text.}
\label{f8}
\end{center}
\end{figure}
Consider for example a five-point correlator of scalar operators $\langle {\cal O}_1(x_1)\ldots {\cal O}_5(x_5) \rangle$. We can define associated conformal partial waves by inserting projection operators as
\begin{equation} \label{fiveblock}
W_{\Delta,\ell;\Delta',\ell'}(x_i)= \langle {\cal O}_1(x_1){\cal O}_2(x_2) P_{\Delta,\ell} {\cal O}_5(x_5)P_{\Delta',\ell'}{\cal O}_3(x_3){\cal O}_4(x_4) \rangle~.
\end{equation}
Using the OPE on ${\cal O}_1 {\cal O}_2$ and ${\cal O}_3 {\cal O}_4$, reduces this to three-point functions. The question is, can we represent $ W_{\Delta,\ell;\Delta',\ell'}(x_i)$ as a geodesic Witten diagram?
Suppose we try to dismantle a tree level five-point Witten diagram. For definiteness, we take $\ell'=\ell=0$. All tree level five-point diagrams will lead to the same structures upon using our geodesic identities: namely, they can be written as sums over geodesic-type diagrams, each as in Figure \ref{f8}, which we label $\widehat {\cal W}_{\Delta_a,0;\Delta_b,0}(x_i)$. This is easiest to see starting from a $\phi^5$ contact diagram, and using \eqref{41a} on the pairs of propagators (12) and (34). In that case, $\Delta_a \in \lbrace\Delta_m\rbrace$ and $\Delta_b\in\lbrace \Delta_n\rbrace$. As an equation, Figure \ref{f8} reads
\es{5b}{\widehat {\cal W}_{\Delta_a,0;\Delta_b,0}(x_i)&=\int_{\gamma_{12}}\int_{\gamma_{34}}
G_{b\partial}(y(\lambda),x_1)G_{b\partial}(y(\lambda),x_2)\\
&\times \int_{y_5} G_{bb}(y(\lambda),y_5;\Delta_a)\,G_{b\partial}(y_5,x_5) \,G_{bb}(y_5,y(\lambda');\Delta_b)\\
&\times \,G_{b\partial}(y(\lambda'),x_3)G_{b\partial}(y(\lambda'),x_4)~.}
Note that the vertex at $y_5$, indicated by the orange dot in the figure, must be integrated over all of AdS. Could these diagrams be computing $W_{\Delta_a,0;\Delta_b,0}(x_i)$ as defined above? The answer is no, as a simple argument shows. Suppose we set $\Delta_5=0$ in \eqref{5b}, which requires $\Delta_a=\Delta_b\equiv \Delta$. From (\ref{fiveblock}) it is clear that we must recover the four-point conformal partial wave with the exchanged primary $(\Delta,0)$. So we should ask whether (\ref{5b}) reduces to the expression for the four-point geodesic Witten diagram, ${\cal W}_{\Delta,0}$. Using $G_{b\partial}(y_5,x_5)|_{\Delta_5=0}\propto 1$, the integral over $y_5$ becomes
\begin{equation} \int_{y_5} G_{bb}(y(\lambda),y_5;\Delta)G_{bb}(y_5,y(\lambda');\Delta)\propto {\partial \over \partial m_\Delta^2} G_{bb}(y(\lambda),y(\lambda');\Delta)~.
\end{equation}
Therefore, the $\Delta_5=0$ limit of \eqref{5b} does not give back the four-point partial wave, but rather its derivative with respect to $m_\Delta^2$, which is a different object.
We conclude that although we can decompose a five-point Witten diagram into a sum of diagrams of the type in Figure \ref{f8}, this is not the conformal block decomposition. This raises two questions: what is the meaning of this decomposition in CFT terms, and (our original question) what diagram computes the five-point partial wave?
\vskip .1 in
\noindent
{\bf $\bullet$~External operators with spin}
\vskip .1 in
Another obvious direction in which to generalize is to consider correlation functions of operators carrying spin. As far as the conformal blocks go, partial information is available. In particular, \cite{Costa:2011dw} obtained expressions for such blocks as differential operators acting on blocks with external scalars, but this approach is limited to the case in which the exchanged operator is a symmetric traceless tensor, since only such operators appear in the OPE of two scalar operators. The same approach was taken in \cite{1505.03750}. Explicit examples of mixed symmetry exchange blocks were given in \cite{1411.7351}.
Our formulation in terms of geodesic Witten diagrams suggests an obvious proposal for the AdS computation of an arbitrary conformal partial wave: take our usual expression (\ref{geowit}), now with the bulk-to-boundary and bulk-to-bulk propagators corresponding to the fields dual to the respective operators. Of course, there are many indices here which have to be contracted, and there will be inequivalent ways of doing so. But this is to be expected, as in the general case there are multiple conformal blocks for a given set of operators, corresponding to the multiplicity of ways in which one spinning primary can appear in the OPE of two other spinning primaries. It will be interesting to see whether this proposal turns out to be valid. As motivation, we note that it would be quite useful for bootstrap purposes to know all the conformal blocks that arise in the four-point function of stress tensors.
A related pursuit would be to decompose all four-point scalar contact diagrams, including any number of derivatives at the vertices. This would involve a generalization of \eqref{spin1eq} to include more derivatives.
\vskip .1 in
\noindent $\bullet$~ {\bf Virasoro blocks and AdS$_3$/CFT$_2$}
\vskip .1 in
Our calculations give a new perspective on how to construct the dual of a generic Virasoro conformal block: starting with the geodesic Witten diagram, we dress it with gravitons. Because Virasoro blocks depend on $c$, a computation in semiclassical AdS gravity would utilize a perturbative $1/c\sim G_N$ expansion. In \cite{Hijano:2015qja}, we put the geodesic approach to use in constructing the holographic dual of the heavy-light Virasoro blocks of \cite{Fitzpatrick:2015zha}, where one geodesic essentially backreacts on AdS to generate a conical defect or black hole geometry. It would be worthwhile to pursue a $1/c$ expansion around the geodesic Witten diagrams more generally.
A closely related question is how to decompose an AdS$_3$ Witten diagram into Virasoro, rather than global, blocks. For a tree-level diagram involving light external operators like those considered here, there is no difference, because the large $c$ Virasoro block with light external operators reduces to the global block \cite{zamo}. It will be interesting to see whether loop diagrams in AdS$_3$ are easier to analyze using Virasoro symmetry.
\vskip .1 in
\noindent $\bullet$~ {\bf Assorted comments}
\vskip .1 in
The geodesic approach to conformal blocks should be useful in deriving various CFT results, not only mixed symmetry exchange conformal blocks. For example, the conformal blocks in the limits of large $\tau$, $\ell$ or $d$ \cite{1212.3616, 1212.4103, 1305.4604, 1502.01437, 1504.00772, 1305.0004, Vos:2014pqa}, and subleading corrections to these, should be derivable using properties of AdS propagators. One can also ask whether there are similar structures present in bulk spacetimes besides AdS. For instance, an analog of the geodesic Witten diagram in a thermal spacetime would suggest a useful ingredient for parameterizing holographic thermal correlators. Perhaps the existence of a dS/CFT correspondence suggests similar structures in de Sitter space as well.
It is natural to wonder whether there are analogous techniques to those presented here that are relevant for holographic correlators of nonlocal operators like Wilson loops or surface operators, perhaps involving bulk minimal surfaces.
Let us close by noting a basic fact of our construction: even though a conformal block is not a semiclassical object per se, we have given it a representation in terms of classical fields propagating in a smooth spacetime geometry. In a bulk theory of quantum gravity putatively dual to a finite $N$ CFT, we do not yet know how to compute amplitudes. Whatever the prescription, there is, evidently, a way to write the answer using geodesic Witten diagrams. It would be interesting to understand how this structure emerges.
\section*{Acknowledgments}
We thank Eric D'Hoker, Liam Fitzpatrick, Tom Hartman, Daniel Jafferis, Juan Maldacena, Joao Penedones and Sasha Zhiboedov for helpful discussions. EP wishes to thank the KITP and Strings 2015 for hospitality during this project. P.K. is supported in part by NSF grant PHY-1313986. This research was supported in part by the National Science Foundation under Grant No. NSF PHY11-25915. E.P. is supported by the Department of Energy under Grant No. DE-FG02-91ER40671.
\bibliographystyle{ssg}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,484 |
Q: Build en NextJs 13 no incluye todos los estilos Soy nuevo usando NextJs, ya acabe de hacer una página básica pero cuando ejecuto "npm run build" y despues "npm run start" varios estilos de css no aparecen y unas tipografias, como si no se hubieran incluido en el build.
Puedo pensar que son dos cosas:
1- Como uso React-Bootstrap los estilos quedan más importantes que mi archivo de estilos.
2- Debo configurar algo en Next.config.js para exportar mis estilos.
/** @type {import('next').NextConfig} */
const nextConfig = {
reactStrictMode: true,
images: {
minimumCacheTTL: 60,
dangerouslyAllowSVG: true,
unoptimized: true,
},
};
module.exports = nextConfig;
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,428 |
Q: MS-DOS timestamp in filename I have the following in a batch file.
set timefmt=%TIME:~0,2%%TIME:~3,2%%TIME:~6,2%
dir *.* > logfile_%timefmt%.log
This works perfectly after 10am, but fails before hand because it adds a space to the timestamp instead of a leading 0.
Is there a way in MS-DOS to create a time stamp with a leading 0?
I'd prefer to use fairly standard commands so that it works from Windows XP onward.
A: It's best to get the time once and then parse the elements, too.
The third line will replace a space with a 0
set timefmt=%time%
set timefmt=%TIMEFMT:~0,2%%TIMEFMT:~3,2%%TIMEFMT:~6,2%
set timefmt=%TIMEFMT: =0%
dir *.* > logfile_%timefmt%.log
A: I had this problem too, but I have computers with different locale (some show yyyy-mm-dd for the date, some mm/dd/yyyy). Some show 12-hour clock with AM and PM, others 24-hour clock...
...so I ended up putting together a tool for saving the current date/timestamp in an environment variable. Actually, the tool just prints out the timestamp, but there is an example of how to get it into an environment variable:
for /f %%x in ('@timestamp.exe') do set TIMESTAMP=%%x
...and then you just use %TIMESTAMP% in any way you want.
A: Creating a file name as a timestamp in a batch job has quite a few answers to a problem close to this one. I insert this remark here because Google brought this answer first so that users who are in a hurry might miss the other thread.
A: I had that problem too.
so I did this. this got me over the hurdle.
:STEP_DATESTAMP
::
REM Setting Datestamp to YYYYMMDD
set v_datestampYYYY=%date:~6,4%
set v_datestampMM=%date:~3,2%
set v_datestampDD=%date:~0,2%
set v_datestamp=%v_datestampYYYY%%v_datestampMM%%v_datestampDD%
::
REM Setting Timestamp to HHMMSS
set HH=%time:~0,2%
:: ensure that hour is always 2 digits
if %HH%==0 set HH=00
if %HH%==1 set HH=01
if %HH%==2 set HH=02
if %HH%==3 set HH=03
if %HH%==4 set HH=04
if %HH%==5 set HH=05
if %HH%==6 set HH=06
if %HH%==7 set HH=07
if %HH%==8 set HH=08
if %HH%==9 set HH=09
set MM=%time:~3,2%
set SS=%time:~6,2%
set v_timestamp=%HH%-%MM%-%SS%
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,168 |
$(function() {
// $( "#bbf-tabs" ).tabs();
activateDeletes();
emptyGlobalDialogOnClose();
$("#msnewadd").hide() // dialog box for adding a missing structure
$("#adddamage").hide() // dialog box for adding a damage observation
$("#procImmedYes").hide() // this is the dialog box to show for Q 7 ( ans:Yes)
var brainProcImmed = $('input:radio[name=wasBrainProcImmed]:checked').val();
// $("#procImmedNo").hide();
if(brainProcImmed =="No"){
$("#procImmedNo").show();
}
else{
$("#procImmedNo").hide() // this is the gross evaluation not acceptbal reasons table
}
var grossacceptable = $('input:radio[name=acceptForFurtherProc]:checked').val();
// $("#procImmedNo").hide();
if(grossacceptable =="No"){
$("#grossProcNo").show();
}
else{
$("#grossProcNo").hide() //
}
//add1 is add button for new structures
$("#add1").click(function(){
$("#ca-dialog").html("<label for=\"structName\">Enter other missing structure: </label><input size=\"25\" name=\"structNameTemp\" value=\"\" id=\"structNameTemp\" type=\"text\">");
inputCharLimit();
$("#ca-dialog").dialog({
title: "Add other missing structure",
autoOpen: false,
modal: true,
height:150,
width:400,
buttons : {
"Save" : function() {
$("#ca-dialog .redtext").remove();
if($("#structNameTemp").val().length == 0){
$("#ca-dialog").append("<span class=\"redtext\">Please specify name of the missing Brain structure to be added</span>");
$( "#structNameTemp").focus();
} else {
$("#structName").val($("#structNameTemp").val());
$.ajax({
url: '/cahubdataservices/brainBankFeedback/save?vessel=json',
type: 'POST',
data: $("form.bbfform").serialize(),
success:function(data){
var newEntry = data.success;
var newEntryStr = "";
$(".new-entry").removeClass("new-entry");
if(newEntry != null && newEntry != undefined) {
var newEntryid = newEntry[1].id;
var newEntrystructName = newEntry[1].structName;
newEntryStr = "<li class=\"\" id=\"row" + newEntryid + "\">";
newEntryStr += "<input size=\"30\" name=\"editstructName_" + newEntryid + "\" value=\"" + newEntrystructName + "\"/>";
newEntryStr += " <a class=\"delete-struct ui-button ui-state-default ui-corner-all removepadding\" data-deleteid=\"" + newEntryid + "\" title=\"delete\"><span class=\"ui-icon ui-icon-trash\">Remove</span></a>";
newEntryStr += "</li>";
$("#editstructlist").append(newEntryStr);
$("#structName").val("");
activateDeletes();
$( "#row" + newEntryid ).addClass("new-entry");
inputCharLimit();
}
}
});
$("#ca-dialog").dialog("close");
}
},
"Cancel" : function() {
$(this).dialog("close");
}
}
});
$("#ca-dialog").dialog("open");
});
//add2 is button for damages
$("#add2").click(function(){
$("#ca-dialog").html("<label for=\"damageRegionTemp\">Region of the brain: </label><br /><input type=\"text\" id=\"damageRegionTemp\" name=\"damageRegionTemp\" /><br /><label for=\"damageObservationTemp\">Observation: </label><br /><textarea id=\"damageObservationTemp\" name=\"damageObservationTemp\" />");
inputCharLimit();
$("#ca-dialog").dialog({
title: "Add a damage type",
autoOpen: false,
modal: true,
height:300,
width:400,
buttons : {
"Save" : function() {
$("#ca-dialog .redtext").remove();
if($("#damageRegionTemp").val().length == 0){
$("#ca-dialog").append("<span class=\"redtext\">Please specify damage region upon gross inspection</span>");
$( "#damageRegionTemp").focus();
} else if($("#damageObservationTemp").val().length == 0){
$("#ca-dialog").append("<span class=\"redtext\">Please specify observation of damage done upon gross inspection</span>");
$( "#damageObservationTemp").focus();
} else {
$("#damageRegion").val($("#damageRegionTemp").val());
$("#damageObservation").val($("#damageObservationTemp").val());
$.ajax({
url: '/cahubdataservices/brainBankFeedback/save?vessel=json',
type: 'POST',
data: $("form.bbfform").serialize(),
success:function(data){
var newEntry = data.success;
var newEntryStr = "";
$(".new-entry").removeClass("new-entry");
if(newEntry != null && newEntry != undefined) {
var newEntryid = newEntry[0].id;
var newEntryregion = newEntry[0].region;
var newEntryobservation = newEntry[0].observation;
if($("#damageRegionTable tr").length == 0 ){
newEntryStr = "<thead><tr class=\"prop\" ><th class=\"label\">Region of Brain</th><th>Observations</th><th class=\"action\"></th></tr></thead><tbody></tbody>";
$("#damageRegionTable").html(newEntryStr);
}
newEntryStr = "<tr class=\"prop\" id=\"row" + newEntryid + "\">";
newEntryStr += "<td><input size=\"30\" name=\"editDamRegion_" + newEntryid + "\" value=\"" + newEntryregion + "\"/></td>";
newEntryStr += "<td><textarea name=\"editDamObservation_" + newEntryid + "\" cols=\"30\" rows=\"2\">" + newEntryobservation + "</textarea></td>";
newEntryStr += "<td><a class=\"delete-damage ui-button ui-state-default ui-corner-all removepadding\" data-deleteid=\"" + newEntryid + "\" title=\"delete\" ><span class=\"ui-icon ui-icon-trash\">Remove</span></a></td>";
$("#damageRegionTable tbody").append(newEntryStr);
$("#damageRegion").val("");
$("#damageObservation").val("");
activateDeletes();
$( "#row" + newEntryid ).addClass("new-entry");
inputCharLimit();
}
}
});
$("#ca-dialog").dialog("close");
}
},
"Cancel" : function() {
$(this).dialog("close");
}
}
});
$("#ca-dialog").dialog("open");
});
// add3 is button for abnormalities
$("#add3").click(function(){
$("#ca-dialog").html("<label for=\"hpRegionTemp\">Region: </label><br /><input type=\"text\" id=\"hpRegionTemp\" name=\"hpRegionTemp\" /><br /><label for=\"hpObservationTemp\">Observation: </label><br /><textarea id=\"hpObservationTemp\" name=\"hpObservationTemp\" />");
inputCharLimit();
$("#ca-dialog").dialog({
title: "Add a new Histopathological evaluation observation",
autoOpen: false,
modal: true,
height:300,
width:400,
buttons : {
"Save" : function() {
$("#ca-dialog .redtext").remove();
if($("#hpRegionTemp").val().length == 0){
$("#ca-dialog").append("<span class=\"redtext\">Please specify damage region upon histopathological evaluation</span>");
$( "#hpRegionTemp").focus();
} else if($("#hpObservationTemp").val().length == 0){
$("#ca-dialog").append("<span class=\"redtext\">Please specify observation of damage done upon histopathological evaluation</span>");
$( "#hpObservationTemp").focus();
} else {
$("#hpRegion").val($("#hpRegionTemp").val());
$("#hpObservation").val($("#hpObservationTemp").val());
$.ajax({
url: '/cahubdataservices/brainBankFeedback/save?vessel=json',
type: 'POST',
data: $("form.bbfform").serialize(),
success:function(data){
var newEntry = data.success;
var newEntryStr = "";
$(".new-entry").removeClass("new-entry");
if(newEntry != null && newEntry != undefined) {
var newEntryid = newEntry[0].id;
var newEntryregion = newEntry[0].region;
var newEntryobservation = newEntry[0].observation;
if($("#hpRegionTable tr").length == 0 ){
newEntryStr = "<thead><tr class=\"prop\" ><th class=\"label\">Region of Brain</th><th>Observations</th><th class=\"action\"></th></tr></thead><tbody></tbody>";
$("#hpRegionTable").html(newEntryStr);
}
newEntryStr = "<tr class=\"prop\" id=\"row" + newEntryid + "\">";
newEntryStr += "<td><input size=\"30\" name=\"editHpRegion_" + newEntryid + "\" value=\"" + newEntryregion + "\"/></td>";
newEntryStr += "<td><textarea name=\"editHpObservation_" + newEntryid + "\" cols=\"30\" rows=\"2\">" + newEntryobservation + "</textarea></td>";
newEntryStr += "<td><a class=\"delete-damage ui-button ui-state-default ui-corner-all removepadding\" data-deleteid=\"" + newEntryid + "\" title=\"delete\" ><span class=\"ui-icon ui-icon-trash\">Remove</span></a></td>";
$("#hpRegionTable tbody").append(newEntryStr);
$("#hpRegion").val("");
$("#hpObservation").val("");
activateDeletes();
$( "#row" + newEntryid ).addClass("new-entry");
inputCharLimit();
}
}
});
$("#ca-dialog").dialog("close");
}
},
"Cancel" : function() {
$(this).dialog("close");
}
}
});
$("#ca-dialog").dialog("open");
});
// id s1 c1 is adding/cancelling new structures
$("#s1").click(function(){
var addnewStruct = $('input:radio[name=missingStructExists]:checked').val();
var sname = document.getElementById("structName").value
if(addnewStruct == "Yes"){
if((sname==null || sname.length==0) ){
alert("Please specify name of the missing Brain structure to be added")
$("#addstructure").hide()
return false;
}
}
});
$("#c1").click(function(){
var addnewStruct = $('input:radio[name=missingStructExists]:checked').val();
document.getElementById("structName").value='';
// document.getElementById("d1").style.display = 'none';
if(addnewStruct == "Yes"){
$("#msnewadd").hide()
$("#addstructure").show()
$("#add1td").show()
}
else{
$("#msnewadd").show()
$("#addstructure").hide() // addstructure id is for selecting preselects
$("#add1td").hide()
}
return false;
});
// id s2 c2 is adding/cancelling new damages
$("#s2").click(function(){
var dregion = document.getElementById("damageRegion").value
var ddesc = document.getElementById("damageObservation").value
if((dregion==null || dregion.length==0) ){
alert("Please specify damage region upon gross inspection")
return false;
}
if((ddesc== null || ddesc.length == 0)){
alert("Please specify observation of damage done upon gross inspection")
return false;
}
$("#adddamage").hide()
});
$("#c2").click(function(){
document.getElementById("damageRegion").value='';
document.getElementById("damageObservation").value=''
// document.getElementById("d1").style.display = 'none';
$("#adddamage").hide()
$("#add2").show()
return false;
});
// id s3 c3 is adding/cancelling reasons why structures not acceptable upon gross inspect
$("#s3").click(function(){
var acceptFurtherProc = $('input:radio[name=acceptForFurtherProc]:checked').val();
var leftNoReason = document.getElementById("noLeftHemisProcReason").value
var rightNoReason = document.getElementById("noRightHemisProcReason").value
var wbNoReason = document.getElementById("wholeBrainNoProcReason").value
if(acceptFurtherProc =="No"){
if(((leftNoReason== null || leftNoReason.length == 0) &&(rightNoReason== null || rightNoReason.length == 0)&&(wbNoReason== null || wbNoReason.length == 0))){
alert("Please specify reason why partial/whole Brain not acceptable for processing")
$("#grossProcNo").show()
return false;
}
}
$("#grossProcNo").hide()
});
$("#c3").click(function(){
var grossProc = $('input:radio[name=acceptForFurtherProc]:checked').val();
if(grossProc == "Yes"){
$("#grossProcNo").hide();
}
else{
$("#grossProcNo").show();
}
return false;
});
// id s4 c4 is adding/cancelling reasons if brain not processed immed. and storage info.
$("#s4").click(function(){
var storeImmed =$('input:checkbox[name=storedImmed]:checked').val();
var storeType = $('input:radio[name=storedImmedType]:checked').val();
var storedComments = document.getElementById("storedInComments").value
if(storeImmed!=null && storeType== null ){
alert("Please specify how Brain and Brain contents were stored")
return false;
}
if(((storeType == 'formalin' || storeType == 'other') &&(storedComments== null || storedComments.length == 0))){
alert("Please explain how Brain and Brain contents were stored")
return false;
}
});
$("#c4").click(function(){
var wasBrainProcImmed = $('input:radio[name=wasBrainProcImmed]:checked').val();
// $("#procImmedNo").hide();
if(wasBrainProcImmed =="No"){
$("#procImmedNo").show();
}
else if(wasBrainProcImmed =="Yes"){
$("#procImmedYes").show();
}
else{
$("#procImmedYes").hide();
$("#procImmedNo").hide();
}
return false;
});
// id s5 c5 is adding/cancelling reasons if brain processed immed. and if so by who
$("#s5").click(function(){
var procBy=document.getElementById("processedBy").value
var procDt =document.getElementById("processDate").value
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\section{Introduction}
Due to rapid development and wide application of information technology, an increasing number of datasets with high dimensions and complexity are continually being generated. Much research work focuses on knowledge and pattern extraction using different machine learning models \cite{han2011data}. In the data mining area, feature selection (FS) acts as a pre-processing strategy to reduce the dimensionality and redundancy of data. It has played an essential role for preventing the problems of overfitting, reduction in the computational cost of applying machine learning algorithms to datasets and for enabling comprehensive decision making \cite{shen2018performance}.
In the literature, there are a number of different FS methods. Based on the dependency with a learning algorithm, FS methods can be generally grouped into three types, i.e. filter, wrapper and embedded methods \cite{chandrashekar2014survey}. Filter methods are independent of any learning algorithms, and have high computational efficiency compared with the wrapper and embedded methods. Hence, various filter FS methods are implemented and compared in this research.
One of the main issues for applying different FS methods is that the performance of the various FS methods varies in a manner which is data dependent. That is, it is not possible to state categorically which is the optimal FS method, in the sense of providing the best performance for all kinds of data \cite{hua2009performance}. This situation poses an interesting and challenging problem as to how to select the best FS method to use for any given unseen dataset \cite{parmezan2017metalearning}.
One approach of solving this problem is through ensemble or combination methods \cite{shen2019novel}. By combining the diversity kinds of FS algorithms, ensemble methods result in a better performance by taking advantages of different methods. On the other hand, the combination process may suffer from a high computational cost. In some cases, ensemble methods do not necessarily achieve a better performance than any of the individual methods.
Another approach is using meta-learning method to choose the best algorithm for a given dataset \cite{brazdil2008metalearning}. Meta learning learns the selection of the most appropriate FS method for a given dataset in a meta level. It is normally trained to learn the relationship between the characteristics of training datasets and their corresponding best FS methods \cite{parmezan2017metalearning}. This is quite valuable and of great importance in the decision making area.
In order to deal with the impreciseness and uncertainty in such a decision making context, fuzzy sets and fuzzy methods have been developed to model many practical problems \cite{zadeh1965fuzzy}. A fuzzy similarity based framework has been utilized to solve the classification problems in a flexible and explainable way \cite{luukka2006similarity}. A comprehensive evaluation of the fuzzy similarity based framework has been reported in our previous research \cite{shen2018performance}.
In the current paper, a novel meta-learning method is proposed to achieve automatic selection of the best FS method for a given dataset using a fuzzy similarity based framework. The rest of the paper is organised as follows. Section~II presents a more detailed literature review of feature selection methods; Section~III presents the methodology used; Section~IV presents experiments in which the new methodology is applied to a range of real world datasets, and the paper closes with Conclusions.
\begin{figure*}[tb]
\centering
\includegraphics[width=1.8\columnwidth]{Figures/Overall_Framework.png}
\caption[Framework of Fuzzy Sets Generation using Bootstrap Sampling]{Overall framework of the proposed architecture. Blue lines and red lines show the data flows for training and testing processes respectively.}
\label{overall_framework}
\end{figure*}
\section{Literature Review}
Meta learning, in our scenario, is defined as a process of learning the meta-knowledge to improve model learning using machine learning and data mining methods \cite{brazdil2008metalearning}. Nowadays, it is becoming a hot topic to improve the stability and generalization of the learned models. There are two main aspects of research in meta-learning methods. One is algorithm selection, which aims to choose the best algorithm based on learning the relationship between the characteristics of the datasets and the performances of different algorithms \cite{kalousis2001feature}. The other one focuses on parameter selection, which aims to determine the optimal parameters of a sophisticated FS method \cite{reif2012meta}.
Meta learning methods for algorithm selection have been applied to solve different problems, such as classification, regression, optimization, time series prediction, etc. \cite{lemke2015metalearning, kuck2016meta, lemke2010meta}. Several researchers have also investigated the application of meta learning method for feature selection \cite{parmezan2017metalearning, filchenkov2015datasets}. In general, there are two mainly approaches to algorithm selection, namely feature engineering and neural architecture search. Recent research on meta learning tend to focus on the neural architecture search, which is powerful, generic and versatile to different models. On the other hand, it can also lead to the high computational cost when exploring the suitable configurations. Comparatively, feature engineering is an efficient process which extracts the meta features from the raw dataset \cite{hartmann2019meta}. In this research, a feature engineering based meta learning process has been explored here. Several issues are discussed and investigated as below.
\paragraph{Construction of A Data Repository for Training}
The central concept of meta learning is to learn the knowledge from a data repository. Hence, the selection and construction of the data repository becomes quite essential. In other studies, a large number of real-world datasets have been used to construct a meta database. Parmezan et al. \cite{parmezan2017metalearning} have used 150 real datasets for the meta-learning process. However, it is quite a time consuming process to collect various kinds of datasets from real-world applications and this is often restricted by ethical issues. More importantly, these real datasets may not cover a wide range of characteristics that are similar to the given unseen dataset under consideration. In this paper, we propose to use synthesized datasets to construct a large data repository that covers a variety of characteristics for meta learning.
\paragraph{Selection of Meta Features}
The meta features refer to the features which have certain relationships with the algorithm performance. The selection of meta features is also dependent on the treated problem. In \cite{reif2014automatic}, the meta features are generally classified into five groups: simple, statistic, information theoretic based, model based and land-marking. A number of widely used meta features are implemented in our study.
\paragraph{Choice of A Recommendation Method}
To construct the meta model, different decision making methods have been used, such as decision trees \cite{parmezan2017metalearning}, support vector machine, kNN, etc. Using fuzzy methods to perform the recommendation has been rarely reported before. So on the basis of our previous work \cite{shen2018performance}, we will implement a fuzzy similarity based framework to achieve the decision making in this paper.
The main contribution of the current paper is the implementation of a meta learning method for feature selection using fuzzy similarity measure. More importantly, instead of using real datasets for meta-learning (e.g. \cite{parmezan2017metalearning}), we propose to use data synthesis to generate a large number of datasets that cover a wide range of characteristics, leading to a more generalized meta-learning solution. Our method has been subsequently evaluated on eight public datasets of real-world applications. It achieves a superior performance in recommending the best FS method for each dataset.
\section{Methodology}
As illustrated in Fig. \ref{overall_framework}, the proposed method mainly consists of five steps: (1) Generation of a data repository for training; (2) Meta features extraction; (3) FS methods' performance measures; (4) Meta data construction; (5) Recommendation modeling using fuzzy similarity measure. Blue lines and red lines in Fig. \ref{overall_framework} show the data flows for training and testing processes respectively. In the training phase, the meta features and meta labels have been generated using the synthetic datasets. After implementing different FS methods on the synthetic datasets, the meta label is obtained to represent the FS method with the best performance. A set of meta features are extracted to represent the characteristics of the synthetic datasets. A fuzzy similarity-based classifier is also introduced during the decision making process. In the testing phase, the same meta features are extracted from the test dataset. By applying the recommendation model in the training phase, the recommended optimal method for the test dataset is obtained. The detailed information are described in the following subsections.
\subsection{Generation of a Data Repository for Training}\label{configure_data}
In this section, we aim to construct a data repository that covers a variety of characteristics using data synthesis. Comparing with other synthetic datasets, the Madelon dataset \cite{guyon2008feature} holds the advantages of high flexibility and variability. It consists of relevant, redundant, repeated and useless features. The dataset presents a wide range of values for the number of features and samples. In addition, the values can also be distorted by adding noise, flipping labels, and shifting and rescaling processes \cite{bolon2013review}. By implementing the methodology as first proposed in the NIPS 2003 feature selection challenge \cite{guyon2003design}, different kinds of madelon datasets are generated for our proposed method by varying 11 different parameters, as listed in Table~\ref{madelon_parameters}. The value range for each of these parameters is also listed in Table~\ref{madelon_parameters}. We set the value range for each parameter as large as possible to cover different scenarios. Note that in this paper, we only consider a binary classification problem, but it can be extended to a multiple class situation. Details of our experimental settings are described in Section~\ref{training_repository}.
\begin{table}[tb]\scriptsize
\caption{Parameters for data synthesis using Madelon dataset}
\centering
\begin{tabular}{c|c|c}
\hline
\textbf{Alias} & \textbf{Meaning} & \textbf{Value Range} \\ \hline
P1 & Number of Classes & 2 \\ \hline
P2 & \begin{tabular}[c]{@{}c@{}}Number of Useful Features\\ \textit{(initially drawn to explain the concept)}\end{tabular}
& [4, 5,..., 20] \\ \hline
P3 & \begin{tabular}[c]{@{}c@{}}Number of Redundant Features\\ \textit{(linearly dependent upon the useful features)}\end{tabular}
& [0, 1,..., 20] \\ \hline
P4 & \begin{tabular}[c]{@{}c@{}} Number of Repeated Features\\ \textit{(repeating P2 and P3 at random)}\end{tabular}
& [0, 1,..., 20] \\ \hline
P5 & \begin{tabular}[c]{@{}c@{}} Number of Useless Features\\ \textit{(Drawn at random regardless of class label)}\end{tabular}
& [0, 1,..., 20] \\ \hline
P6 & Number of Samples per Cluster & [10, 11,..., 70] \\ \hline
P7 & Number of Cluster per Class & [2, 3,..., 7] \\ \hline
P8 & Random Seed & [1, 2,..., 1000] \\ \hline
P9 & Factor multiplying the hypercube dimension & [2, 3,..., 10] \\ \hline
P10 & Fraction of y labels to be randomly exchanged & [0.01, 0.02, ..., 0.1] \\ \hline
P11 & Flag to enable or disable random permutations & [0, 1] \\ \hline
\end{tabular}
\label{madelon_parameters}
\end{table}
\subsection{Meta Feature Extraction}
To learn meta features from the synthetic dataset, we extract a set of meta features from $M$ different datasets $D_i$, $i=1,..., M$, each with the number of $S_i$ data samples ($E_1, E_2,..., E_{S_i}$) and $N_i$ features ($F_1, F_2,..., F_{N_i}$). The label information is represented using class $C$ ($c_1, c_2,..., c_{S_i}$) for different data samples. An overall description of the data structure is shown in Table~\ref{description_data}.
\begin{table}[tb]
\centering
\caption{Description of the structure of a generated synthetic dataset $D_i$ for meta feature extraction}
\begin{tabular}{c|c|c|c|c|c}
\hline
\multirow{2}{*}{\textbf{Samples}} & \multicolumn{4}{c|}{\textbf{Features}} & \multirow{2}{*}{\textbf{Class}} \\ \cline{2-5}
& \textbf{$F_1$} & \textbf{$F_2$} & ... & \textbf{$F_{N_i}$} & \\ \hline
\textbf{$E_1$} & $v_{11}$ & $v_{12}$ & ... & $v_{1N_i}$ & $c_1$ \\ \hline
\textbf{$E_2$} & $v_{21}$ & $v_{22}$ & ... & $v_{2N_i}$ & $c_2$ \\ \hline
\textbf{...} & ... & ... & ... & ... & ... \\ \hline
\textbf{$E_{S_i}$} & $v_{S_i1}$ & $v_{S_i2}$ & ... & $v_{S_iN_i}$ & $c_{S_i}$ \\ \hline
\end{tabular}
\label{description_data}
\end{table}
Subsequently the six meta feature extraction methods, which are derived from each of the $D_i$ datasets, are described as below~\cite{parmezana2016supplementary}.
\begin{enumerate
\item \textbf{\textit{Number of Samples (NS)}}: \\
It represents the number of samples for each dataset.
\item \textbf{\textit{Number of Features (NF)}}: \\
It represents the number of features for each dataset.
\item \textbf{\textit{Average Asymmetry of Features (AAF)}}: \\
It measures the average value of the Pearson's asymmetry coefficient. The formulation is used to quantitatively summarize the skewness of a distribution which is shown in (\ref{aaf}).
\begin{equation}\label{aaf}
AAF(D_i) = \frac{3}{N_i} \sum^{N_i}_{j=1} \frac{ Mean(F_j) - Median(F_j)}{Std(F_j)}
\end{equation}
where, $Mean(F_j)$, $Median(F_j)$ and $Std(F_j)$ indicate the average, median and standard deviation values of feature $F_j$ respectively. $j$ is the index of the features.
\item \textbf{\textit{Average Correlation between Features (ACF)}}: \\
It measures the average value of Pearson's correlation coefficient between different features.
\begin{equation}
ACF(D_i) = \frac{2}{N_i(N_i-1)} \sum^{N_i-1}_{j=1} \sum^{N_i}_{k=j+1} Pearson(F_j, F_k)
\end{equation}
where $Pearson(F_j, F_k)$ indicates the Pearson's correlation between feature $F_j$ and feature $F_k$.
\item \textbf{\textit{Average Coefficient of Variation of Features (ACVF)}} \\
It measures the average coefficient of variation by the ratio of the standard deviation and the mean of the feature values.
\begin{equation}
ACVF(D_i) = \frac{1}{N_i} \sum^{N_i}_{j=1} \frac{Std(F_j)}{Mean(F_j)}
\end{equation}
\item \textbf{\textit{Average Entropy of Features (AEF)}} \\
This measures the average amount of the information that each feature provides for the prediction of the class.
\begin{equation}
AEF(D_i) = \frac{1}{N_i} \sum^{S_i}_{k=1}Entropy(F_j)
\end{equation}
where $Entropy(F_j)$ measures the distribution's entropy of feature $F_j$.
\end{enumerate}
\subsection{Performance Measures of FS Methods}\label{FS_Label}
In this step, we generate a label for each of the synthetic datasets, so that the derived meta features and their associated labels can be used for model learning to recommend the best FS method for a given unseen dataset. In this case, the label is the best FS method for a given training dataset. To determine the best FS method for each synthesized dataset, we firstly define a performance measurement metric based on classification accuracy, which is described as below.
FS methods are ultimately used to improve the classification accuracy using reduced number of features by removing redundant features. The FS methods we implemented in our study are able to rank the features from the most significant to the least significant. Then a classification model learning method is used to calculate the classification accuracies by gradually eliminating the least important features one at a time. The detailed procedures are listed as below.
\begin{enumerate}[itemindent=0em, label=\alph*)]
\item Divide the data ($D_i$) into training sets and testing sets in a 10-fold cross validation manner;
\item Implement the candidate FS method to rank the features using the training set;
\item Based on the gradually reduced number of features, model the classifier using the training set and make the prediction on the test set (logistic regression is used as the classifier in our experiments);
\item For each reduced number of features, calculate mean classification accuracy across different folds.
\end{enumerate}
Fig. \ref{demonstration_acc} shows an example of plots of the classification accuracies obtained by increasing the number of removed features using different FS methods. Specific FS methods used in our study are described in Section \ref{FS_methods}.
\begin{figure}[tb]
\centering
\includegraphics[width=\columnwidth]{Figures/Accuracy.png}
\caption[Framework of Fuzzy Sets Generation using Bootstrap Sampling]{Demonstration of classification accuracies using reduced features}
\label{demonstration_acc}
\end{figure}
As shown in Fig. \ref{demonstration_acc} that the classification accuracies vary significantly when features are gradually removed. A single measurement value is desirable to determine the best FS method. Either the mean or maximum value of the classification accuracy can be used, but these are not reliable due to the noisy nature of the curve. Here, we propose a new measure that is a weighted sum (WS) of the classification accuracies based on different numbers of removed features, as expressed by:
\begin{equation}\label{weighted_acc}
WS = \sum Acc. * \% RemovedFeatures
\end{equation}
where $\% RemovedFeatures$ represents the proportion of the removed features and $Acc.$ means the corresponding classification accuracy using the retained features. The classification accuracies are calculated by removing features from the least significant to the most significant, hence the retained features become more and more important. As expressed in (\ref{weighted_acc}), we assign a higher weight (larger $\%RemovedFeatures$) to the classification accuracy that obtained from using more important features. If the features are ranked correctly by a FS method, a higher weighted sum should be achieved. This method is more robust to noise than the mean and maximum values.
\subsection{Meta Data Construction}\label{metadata_construction}
Based on the WS measure, we then select the FS method with the highest WS value as the meta label for the particular input dataset. Subsequently, the meta data is constructed by combining the six different meta features $MF_p$, ($1 \leq p \leq 6$) and the corresponding meta label for each dataset $D_i$.
Based on the meta dataset, a decision making model is then trained to recommend the optimal FS method for a given dataset.
\subsection{Recommendation using Fuzzy Similarity Measure}
Based on our previous work \cite{shen2018performance}, a fuzzy similarity measure based framework is implemented to train a classification model using the generated meta dataset. The overall structure of the classification framework is illustrated in Fig.~\ref{similarity_framework}.
\begin{figure}[tb]
\centering
\includegraphics[width=\columnwidth]{Figures/Similarity_Classifier.png}
\caption[Framework of Fuzzy Sets Generation using Bootstrap Sampling]{Framework of fuzzy similarity based classifier. Blue and red lines show data flows for training and testing processes respectively \cite{shen2018performance}.}
\label{similarity_framework}
\end{figure}
\begin{figure*}[tb]
\centering
\begin{minipage}[t]{0.3\linewidth}
\centering
\includegraphics[width=1.2\textwidth]{Figures/Meta_Features/Feature_1.png}
\parbox{1cm}{\small \hspace{3.5cm}(a){NS}}
\end{minipage}
\hspace{3ex}
\begin{minipage}[t]{0.3\linewidth}
\centering
\includegraphics[width=1.2\textwidth]{Figures/Meta_Features/Feature_2.png}
\parbox{1cm}{\small \hspace{3.5cm}(b)NF}
\end{minipage}
\hspace{3ex}
\begin{minipage}[t]{0.3\linewidth}
\centering
\includegraphics[width=1.2\textwidth]{Figures/Meta_Features/Feature_3.png}
\parbox{1cm}{\small \hspace{3.5cm}(c)AAF}
\end{minipage}
\hspace{3ex}
\begin{minipage}[t]{0.3\linewidth}
\centering
\includegraphics[width=1.2\textwidth]{Figures/Meta_Features/Feature_4.png}
\parbox{1cm}{\small \hspace{3.5cm}(d){ACF}}
\end{minipage}
\hspace{3ex}
\begin{minipage}[t]{0.3\linewidth}
\centering
\includegraphics[width=1.2\textwidth]{Figures/Meta_Features/Feature_5.png}
\parbox{1cm}{\small \hspace{3.5cm}(e)ACVF}
\end{minipage}
\hspace{3ex}
\begin{minipage}[t]{0.3\linewidth}
\centering
\includegraphics[width=1.2\textwidth]{Figures/Meta_Features/Feature_6.png}
\parbox{1cm}{\small \hspace{3.5cm}(f)AEF}
\end{minipage}
\caption{Comparison of the distribution between the training and testing repository}
\label{compare_distribution}
\end{figure*}
The model training process aims to classify a total number of $M$ examples $D_i$ ($1 \leq i \leq M$) into $L$ different classes $FS_l$, ($1 \leq l \leq L$) by their feature vector $\vec{x_q}$. $q$ is the index of the data samples in each class. $Z_l$ is the number of data samples for the $l$th class. Based on the comprehensive performance evaluation reported in \cite{shen2018performance}, the following fuzzy similarity measure based framework is implemented.
\begin{enumerate}[itemindent=1em, label=Step \arabic*:]
\item For the training set, standardize each feature using the Z-score normalization process \cite{grus2019data}.
\item Based on the standardized values from Step 1, calculate the ideal vector $\vec{v}_l$ for the $l^{th}$ class using geometric mean.
\begin{equation}
\vec{v}_l(p) = \sqrt[Z_l]{\prod_{q=1}^{Z_l}\vec{x_q}(p)}, \ 1 \leq p \leq 6
\end{equation}
where $p$ represents the index of meta features.
\item The same standardization process from Step 1 is applied to the meta features extracted from the test dataset. Subsequently, feature vector $\vec{y_r}$ of the meta features is obtained, while $r$ indicates the index of the data samples in the test set.
\item Based on the maximal fuzzy similarity measures proposed in \cite{luukka2001classifier}, a similarity measurement in the form of generalized \L ukasiewicz algebra is used. Geometric mean is used to combine the similarity measures from different features which is expressed in (\ref{eqGeo}).
\begin{equation}\label{eqGeo}
S \langle \vec{y_r}, \vec{v}_l \rangle = \sqrt[6]{\prod_{p=1}^{6} \sqrt{1 - |\vec{y_r}(p)^2- \vec{v}_l(p)^2|}}
\end{equation}
where $S \langle \vec{y_r}, \vec{v}_l \rangle$ represents the fuzzy similarity value between the feature vector of the testing set and the ideal vectors obtained from the training set.
\item Classify the test dataset into the class with the corresponding ideal vector which produces the highest fuzzy similarity value.
\end{enumerate}
Through the steps above, the recommended FS method for a given test dataset is obtained.
\begin{figure*}[tb]
\centering
\begin{minipage}[t]{0.3\linewidth}
\centering
\includegraphics[width=1.2\textwidth]{Figures/Comparison_Accuracy/Acc_PIMA.png}
\parbox{1cm}{\small \hspace{4.5cm}(a)PIMA}
\end{minipage}
\hspace{3ex}
\begin{minipage}[t]{0.3\linewidth}
\centering
\includegraphics[width=1.2\textwidth]{Figures/Comparison_Accuracy/Statlog.png}
\parbox{1cm}{\small \hspace{3.5cm}(b)StatlogHeart}
\end{minipage}
\hspace{3ex}
\begin{minipage}[t]{0.3\linewidth}
\centering
\includegraphics[width=1.2\textwidth]{Figures/Comparison_Accuracy/Acc_WDBC.png}
\parbox{1cm}{\small \hspace{3.5cm}(c)WDBC}
\end{minipage}
\caption{Performance comparison of different FS methods on three test datasets}
\label{acc_reduce_features}
\end{figure*}
\section{Experiments \& Results}
The performance of the proposed method was evaluated for a binary classification problem using eight public datasets.
\subsection{Datasets}
\subsubsection{\textbf{Training Data Repository}}\label{training_repository}
Based on the description in Section \ref{configure_data}, 1000 datasets were generated by using randomly selected parameter values within the defined ranges in Table~\ref{madelon_parameters}. Meta features were then extracted from these data repository. Following the process described in Section~\ref{FS_Label} and \ref{metadata_construction}, the meta dataset for training was constructed, which contained 1000 data samples each had six meta features. The meta label of each sample was one of the four FS methods introduced in Section~\ref{FS_methods}.
\subsubsection{\textbf{Testing Data Repository}}
Eight binary classification datasets which come from the UCI machine learning repository \cite{Dua:2019} were used to evaluate the performance of the proposed method. The detailed information is shown in Table~\ref{test_data}.
In Table~\ref{test_data}, \#Fea. and \#Samples represent the number of features and samples of the dataset, respectively. The distribution over class means the number of samples for each of the binary classes.
\begin{table}[tb]
\caption{Description of the biomedical datasets for testing}
\centering
\begin{tabular}{c c c c c}
\toprule
\textbf{Dataset} & \textbf{\#Fea.} & \textbf{\#Samples} & \textbf{Distribution Over Class} \\
\midrule
\centering
Appendicitis & 7 & 106 & 85 / 21 \\
PIMA & 8 & 768 & 500 / 268 \\
WBC & 9 & 699 & 458 / 241 \\
Statlog Heart & 13 & 270 & 150 / 120 \\
Parkinsons & 22 & 195 & 48 / 147 \\
WDBC & 30 & 569 & 212 / 357 \\
Spectfheart & 44 & 267 & 55 / 212 \\
Sonar & 60 & 208 & 97 / 111 \\
\bottomrule
\end{tabular}
\label{test_data}
\end{table}
\subsection{Feature Selection Methods}\label{FS_methods}
Four filter FS methods which come from different categories~\cite{li2017feature} were implemented in this experiment, i.e. Gini Index FS (GIFS)~\cite{singh2010feature}, ReliefF~\cite{robnik2003theoretical}, Mutual Information FS (MIFS)~\cite{battiti1994using} and Infinite FS (IFS)~\cite{roffo2015infinite}. Logistic regression was used to evaluate the algorithms' classification performance using the generated feature rankings by different FS methods. Through implementing different FS methods using the metric described in Section~\ref{FS_Label} on the training repository, the number of best performances achieved by each FS method was 546, 196, 147 and 111, respectively (total of 1000 datasets).
\subsection{Comparison of the Features' Distribution}
The distributions of the meta features from both 1000 training and eight testing datasets are shown in Fig.~\ref{compare_distribution}. It can be seen that the distributions of the meta features in the training repository cover the value range of the test datasets well for meta feature NS, NF and ACVF. Meta feature AAF, ACF and AEF of the test datasets are slightly higher or lower than the corresponding value ranges in the training datasets. This could be further improved by fine tuning the parameters in the data synthesis procedure.
\subsection{Evaluation Results}
We firstly applied individual FS methods (i.e. GIFS, ReliefF, MIFS and IFS) to the eight test datasets. The classification accuracies by gradually removing the least significant features for PIMA, Statlog Heart and WDBC datasets are shown in Fig.~\ref{acc_reduce_features}.
It can be seen that different FS methods had significantly different behaviours, which are difficult to be quantified and compared. Hence, we used our proposed measurement metric WS in (\ref{weighted_acc}) as the evaluation metric. The evaluation results by applying each of the four FS methods to the test datasets are listed in Table \ref{weighted_sum_acc}. The FS method that produced the highest WS value was treated as the ground truth (`Best Method' column). The recommended FS method using our proposed framework is listed in the last column of Table \ref{weighted_sum_acc}.
\begin{table}[tb]\scriptsize
\centering
\caption{Performance comparison on 8 test datasets}
\begin{threeparttable}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
\textbf{Datasets} & \textbf{GIFS} & \textbf{ReliefF} & \textbf{MIFS} & \textbf{IFS} & \textbf{\begin{tabular}[c]{@{}c@{}}Best\\ Method\end{tabular}} & \textbf{\begin{tabular}[c]{@{}c@{}}Recommend\\ Method\end{tabular}} \\ \hline
\textbf{Appendicitis} & \textbf{2.41} & 2.40 & \textbf{2.41} & 2.40 & MIFS & MIFS \\ \hline
\textbf{PIMA} & $2.63'$ & 2.62 & \textbf{2.64} & 2.56 & MIFS & MIFS \\ \hline
\textbf{WBC} & $3.78'$ & 3.78 & \textbf{3.79} & 3.78 & MIFS & MIFS \\ \hline
\textbf{Statlog Heart} & $4.84'$ & 4.61 & \textbf{4.85} & 4.08 & MIFS & MIFS \\ \hline
\textbf{Parkinsons} & \textbf{8.52} & 8.29 & $8.49'$ & 8.42 & GIFS & ReliefF \\ \hline
\textbf{WDBC} & 13.48 & \textbf{13.60} & $13.51'$ & 13.41 & ReliefF & ReliefF \\ \hline
\textbf{Spectfheart} & 16.03 & $16.07'$ & 15.85 & \textbf{16.11} & IFS & MIFS \\ \hline
\textbf{Sonar} & \textbf{16.07} & $15.96'$ & 15.31 & 12.83 & GIFS & ReliefF \\ \hline
\end{tabular}
\begin{tablenotes}
\footnotesize
\item Bold numbers indicate the best performance;
\item Numbers with $'$ indicate the second best performance.
\end{tablenotes}
\end{threeparttable}
\label{weighted_sum_acc}
\end{table}
It can be observed that our proposed method successfully recommended the best method for the Appendicitis, PIMA, WBC, Statlog Heart and WDBC dataset. In the case of the Sonar dataset, our recommended method ranked the second, which was only slightly lower than the best method. In the other two datasets (i.e. Parkinsons and Spectfheart), the proposed method cannot accurately recommend the best method. This may be due to the fact that the meta feature distributions of these datasets are outside the value ranges of the training dataset. This could be further improved by refining the data synthesis process in Section~\ref{configure_data}.
By counting the number of the achieved best method in the testing data repository, the performance between our proposed method and the individual FS methods are compared and displayed in Fig.~\ref{best_method_achieved}.
\begin{figure}[tb]
\centering
\includegraphics[trim=40 0 40 0, clip, width=\columnwidth]{Figures/Best_Method_Achieved.png}
\caption[Framework of Fuzzy Sets Generation using Bootstrap Sampling]{Performance comparison on testing repository}
\label{best_method_achieved}
\end{figure}
It can be seen that our proposed method achieved the best performance comparing with the other individual FS methods. In the testing data repository, our proposed method successfully recommended the best method on five datasets out of the eight in total. In contrast, the individual FS methods only achieved the best performance in three, one, four and one cases, respectively. Overall, the successful recommendation rate of our proposed method was 62.5\% on the testing repository.
\subsection{Computational Cost}
In this section, we report and compare the execution time using different FS methods and the proposed method. The programs were implemented using Python and ran on a laptop with 2.2GHz, Intel(R) Core(TM) i5-5200U CPU and 8GB RAM. Each method was implemented and ran 10 times. The average execution time (s) of each method is reported in Table \ref{computational_cost}.
\begin{table}[tb]\scriptsize
\caption{Average run time using different methods (/s)}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
\multirow{2}{*}{\textbf{Datasets}} & \multicolumn{4}{c|}{\textbf{Individual Methods}} & \multirow{2}{*}{\textbf{\begin{tabular}[c]{@{}c@{}}Meta\\Learning\end{tabular}}} & \multirow{2}{*}{\textbf{\begin{tabular}[c]{@{}c@{}}Total Run\\Time\end{tabular}}} \\ \cline{2-5}
& \textbf{GIFS} & \textbf{ReliefF} & \textbf{MI} & \textbf{IFS} & & \\ \hline
\textbf{Appendicitis} & 0.37 & 0.36 & 0.32 & 0.09 & 0.07 & 0.39 \\ \hline
\textbf{PIMA} & 1.55 & 7.50 & 0.99 & 0.28 & 0.32 & 1.31 \\ \hline
\textbf{WBC} & 2.71 & 10.19 & 2.61 & 2.68 & 1.30 & 3.91 \\ \hline
\textbf{Statlog Heart} & 0.56 & 1.30 & 5.66 & 6.19 & 0.08 & 5.74 \\ \hline
\textbf{Parkinsons} & 2.55 & 1.04 & 1.21 & 0.43 & 0.38 & 1.42 \\ \hline
\textbf{WDBC} & 15.06 & 6.23 & 3.97 & 1.86 & 2.03 & 8.26 \\ \hline
\textbf{Spectfheart} & 3.40 & 3.16 & 4.18 & 2.23 & 0.20 & 4.37 \\ \hline
\textbf{Sonar} & 7.66 & 1.74 & 3.36 & 1.29 & 0.72 & 2.46 \\ \hline
\end{tabular}
\label{computational_cost}
\end{table}
The figures for the individual methods presented in Table \ref{computational_cost} show the average run time for the implementation of each FS method. The `Meta Learning' column indicates the execution time of our proposed framework. In addition, total run time represents the summation of the execution time using our meta learning framework and the recommended FS method accordingly.
It can be seen that our meta learning framework takes less than one second to run in most cases. By inspecting the total run time, the proposed method has not led to a significantly longer overall execution time, which indicates a high computational efficiency. Comparing with the individual FS methods, our meta learning framework and the recommended FS method has just consumed a moderate amount of time. This indicates that there is comparatively little additional computational cost incurred in implementing our meta learning framework, indicating the applicability of the approach. The use of our meta learning method provides an efficient way to learn the potentially optimal FS method.
\section{Discussion}
The results show that our method has successfully recommended the most appropriate FS method to use in five out of eight evaluation datasets. The overall successful recommendation rate was 62.5\% on the testing repository. From Table \ref{weighted_sum_acc}, it can be seen that the correct recommendations often appear on the datasets with small performance difference between various methods. This may be the misconception caused by the lack of the testing datasets with diverse performance. Further work need to be done to better evaluate it.
As for the computational cost, our proposed method is fast to run and so is sufficiently fast to be able to be widely used (apart from in situations which are very time critical). Rather than choosing one single FS method randomly, the pre-selection process using our meta learning framework has introduced very small additional computational burden. This actually makes it an attractive potential method to be used when a wide variety of candidate algorithms are considered.
Generally speaking, the absolute performance of our proposed method is not very remarkable. However, the point of this current paper is to demonstrate the potential of the method and the overall framework. The results are still provisional and clearly need to be improved in the future, if the approach is to be more widely used.
\section{Conclusion}
In this paper, we have proposed and implemented a meta learning method to recommend the best FS method from four candidate methods using a fuzzy similarity measure based framework.
Instead of using real datasets for meta learning, we generate 1000 different synthetic datasets to form the training repository. Six meta features are extracted from the training data repository. The FS methods' performance is measured by using a novel weighted sum classification accuracy measurement. Based on the constructed meta datasets, a fuzzy similarity measure based framework is then applied to train a classification model. By evaluating the proposed method on eight different datasets from real-world applications, our proposed method successfully recommended the best method for five of these datasets, which is better than any of the individual FS method. Besides, our proposed method is computationally efficient with almost no additional time cost to the feature selection process.
For future work, we will extend our proposed method by generating better training data repository with wider distributions, introducing more meta features and testing other evaluation metrics for FS. More comparisons will be performed to better evaluate the proposed framework, such as using different machine learning methods, various FS methods, datasets with diverse performance, recent meta learning models and etc.
\clearpage
\bibliographystyle{IEEEtran}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,537 |
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