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math/0006227 | We have to check the non-degeneracy axiom. Let us denote by MATH the matrix whose MATH entry is equal to the value of the MATH-framed NAME link with linking -REF and coloring of the components MATH. Then we have that MATH . We deduce that the MATH entry of the matrix MATH is equal to the invariant of the colored link depicted below. MATH . By using REF and the killing property we obtain the formula MATH where MATH is the identity matrix, which proves the invertibility of the MATH matrix. |
math/0006227 | Let MATH be a maximal left ideal of MATH. Suppose that MATH does not contains MATH, then, using maximality of MATH, we get that the left ideal MATH is equal to MATH. We further have that MATH, MATH, MATH, and so MATH is in the ideal MATH, which contradicts the hypothesis. |
math/0006227 | We have to prove the dominating property. The proof is the same in all cases, so we will use the notation MATH, MATH for MATH, MATH where MATH is one of the categories mentioned in the claim. It is enough to show that the identity morphism of the object MATH decomposes using the simple objects in MATH. This is done by induction on MATH. For the step from MATH to MATH, we have to decompose MATH, with MATH. The key point is that any diagram obtained from MATH by adding one cell is in MATH. Hence we have that the branching formula holds and gives the required decomposition, because the idempotents indexed by partitions in MATH are negligible. |
math/0006227 | An object MATH is transparent if and only if for any (non-negligible) MATH in the branching formula for MATH, the braiding coefficient is equal to one. Indeed, if all braiding coefficients are equal to one, by summing over MATH the left hand sides and right hand sides of REF and applying the branching formula we have MATH . Using this equality repeatedly we conclude that MATH is transparent. Conversely, if MATH is transparent, its braiding coefficients are trivial. The object MATH has only one braiding coefficient corresponding to the removal of the last cell, and this coefficient is one. (Two diagrams obtained by adding one cell to MATH are negligible.) It remains to check that any nontrivial MATH distinct from MATH has at least one braiding coefficient distinct from MATH. If MATH is obtained from such MATH by adding a cell in the first row, then MATH is not zero, and the corresponding braiding coefficient in REF is MATH. For a column with MATH cells, the generic quantum dimension reduces to MATH . This gives for MATH . The twist coefficient for MATH is MATH. |
math/0006227 | In the semi-simple category MATH we can decompose the identity of the object MATH as we did in REF . MATH . Here all simple subobjects MATH are transparent and hence have dimension MATH. By comparing the dimensions we see that there is only one such MATH with multiplicity MATH. It should be trivial, because the duality gives a nonzero morphism from the trivial to MATH. We deduce that this duality morphism is an isomorphism, which establishes MATH. We consider the morphism from MATH to MATH depicted below: the strings corresponding to the points in (the expansion of) MATH are joined to the first columns, the points which are not in the first column of MATH and MATH are joined directly. MATH . One wants to show that this morphism is nonzero. We first consider the case where MATH has only one column. Let MATH be the morphism as above and MATH be its mirror image with respect to the target plane. Then MATH. In the general case, if we insert conveniently the isomorphism considered in the particular case between and MATH and MATH we obtain our nontrivial morphism. |
math/0006227 | Let MATH. We show that MATH is a set of dominating simple objects. As in the proof of REF , we decompose the tensor products MATH, for MATH. The sublte point here is that some idempotent in the branching formula for the partition MATH (that is, MATH) is missing. We will avoid this difficulty by using the isomorphism in REF which still holds for MATH. More precisely, if MATH is in MATH, then the branching formula applies. If MATH, then we use the isomorphism between MATH and MATH and we get a decomposition of MATH with subobjects MATH, MATH and MATH. If MATH, then we get an isomorphism between MATH and MATH. |
math/0006227 | It is sufficient to show that the transparent simple objects have order REF, that is, any non-trivial transparent simple object MATH satisfies the equation: MATH. Clearly, MATH contains the trivial object and decomposes into a sum of transparent simple ones. Comparing the quantum dimensions on the left and right hand side of this decomposition formula we get the result. |
math/0006227 | We substitute REF and calculations of REF into REF . Let us consider the MATH case in details. Here MATH. By REF and the calculations of REF we have MATH . Furthermore, MATH . Here we used the bijection MATH sending MATH to MATH. Substituting the last two formulas into REF we get the result. For the third formula we use that MATH . This is because the action of the group MATH of the transparent objects on MATH preserves the quantum dimension and MATH. CASE: By REF we have for any MATH . The second formula can be shown analogously. |
math/0006227 | Let us call the graded sliding property the equality drawn in REF by replacing MATH on the left-hand side by MATH and MATH on the right-hand side by MATH with MATH. The proof of this identity can be adapted from the one of this proposition. Using twice the graded sliding property, we can see that the morphism drawn below is nonzero only if MATH or MATH. MATH . Then MATH where MATH is the NAME link whose MATH-colored component has framing REF and MATH-colored one is MATH-framed. The first equality is due to the graded sliding property. In the second one we use the twist and braiding coefficients for MATH and the fact that MATH . Substituting the values of MATH and MATH into REF we get the result. |
math/0006227 | A colored MATH-component link invariant of a pre-modular category MATH with MATH as a representative set of simple objects can be considered as a multilinear function from MATH to MATH. Here we supply MATH with a ring structure by considering direct sums and tensor products. It is easy to see from the previous discussion, that there exists an isomorphism between such rings in our case. Indeed, for MATH we have MATH, MATH, MATH, MATH, and MATH. After the identification of MATH with MATH, this coincides with MATH. The ring structure is preserved under this identification. Furthermore, it is known that these rings are generated by the fundamental module corresponding to MATH and the object MATH. Therefore, it is sufficient to verify the equality of invariants colored by these two objects. The fact, that the link invariant associated with this fundamental module is a specialization of the NAME polynomial was shown in CITE. In order to identify the parameters, compare the quantum dimensions of simple objects given by REF . We show that MATH. The equivalence between MATH and MATH can be shown analogously. Then we use the level-rank duality. |
math/0006233 | Using REF , an evident inequality introducing an auxiliary object MATH, and twice REF again: MATH . |
math/0006233 | Rewrite the expectation MATH . Define MATH and MATH to obtain MATH . Given the program that computes MATH, we can approximate MATH by a MATH, and similarly for MATH. That is, the distributions MATH REF are lower semicomputable, and by REF , therefore, they are computable. It is known that for every computable probability mass function MATH we have MATH, REF. Hence, MATH (MATH), and MATH. On the other hand, the probabilistic mutual information REF is expressed in the entropies by MATH. By construction of the MATH's above, we have MATH. Since the complexities are positive, substitution establishes the lemma. |
math/0006233 | The lemma follows from the definition of conditional algorithic mutual information, REF , if we show that MATH, where the MATH term implicit in the MATH sign is independent of MATH. Equip the reference universal prefix machine, with a MATH length program to compute a NAME code from the auxiliary table of probabilities. Then, given an input MATH, it can determine whether MATH is the NAME code word for some MATH. Such a code word has length MATH. If this is the case, then the machine outputs MATH, otherwise it halts without output. Therefore, MATH. This shows the upper bound on the expected prefix complexity. The lower bound follows as usual from the Noiseless Coding Theorem. |
math/0006233 | By the triangle inequality, MATH . Thus, MATH . |
math/0006233 | It suffices to prove the case MATH and apply it twice. The proof is by replacing the program MATH that computes a particular string MATH from a particular MATH in REF . There, MATH possibly depends on MATH and MATH. Replace it by a program MATH that first computes MATH from MATH, followed by computing a recursive function MATH, that is, MATH is independent of MATH. Since we only require a MATH-length program to compute MATH from MATH we can choose MATH. By the triangle inequality, MATH . Thus, MATH . |
math/0006233 | We have MATH . The negative logarithm of the left-hand side in the theorem is therefore MATH . Using REF , and the conditional additivity REF , this is MATH . |
math/0006233 | By REF and by typicality MATH. |
math/0006233 | Writing MATH, since MATH by REF , we have MATH. Hence, it suffices to show MATH. Now, from an implicit description MATH we can find the value MATH. To recover MATH we only require an extra MATH bits apart from MATH. Therefore, MATH. This reduces what we have to show to MATH which is asserted by REF . |
math/0006233 | A set MATH is optimal iff REF holds with equalities. Rewriting MATH the first inequality becomes an equality iff MATH, and the second inequality becomes an equality iff MATH (that is, MATH is a typical set). |
math/0006233 | If MATH is optimal for MATH, then MATH. From MATH we can find both MATH and MATH and hence MATH, that is, MATH. We have MATH by REF , respectively. This proves the first property. Substitution of MATH in the expression of REF proves the second property. |
math/0006233 | The lower bound is easiest. Denote by MATH of length MATH a shortest program for MATH. Every string MATH of length MATH can be described in a self-delimiting manner by prefixing it with MATH, hence MATH. For a large enough constant MATH, we have MATH and hence there are MATH strings that are in MATH. For the upper bound: by REF , all MATH satisfy MATH, and there can only be MATH of them. |
math/0006233 | From MATH we can compute both MATH and MATH and recursively enumerate MATH. Since also MATH REF , the string MATH plus a fixed program is an implicit description of MATH so that MATH. Hence, MATH and since MATH is the shortest description by definition equality REF holds. That is, MATH is optimal for MATH. By REF MATH which together with the reverse inequality above yields MATH which shows the theorem. |
math/0006233 | The set MATH is a sufficient statistic for MATH since MATH. It is minimal since by REF we must have MATH for implicit, and hence for explicit sufficient statistics. It is evident that MATH is explicit: MATH. |
math/0006233 | We can describe MATH by MATH where MATH is the index of MATH in the enumeration of MATH. Moreover, MATH explicitly describes the set MATH. Namely, using MATH we can recursively enumerate MATH. At some point the first string MATH is enumerated (index MATH). By REF and MATH. Therefore, in the enumeration of MATH eventually string MATH with MATH occurs which is the last string in the enumeration of MATH. Thus, the size of MATH is precisely MATH, where MATH, and MATH is explicitly described by MATH. Since MATH and MATH we have MATH . This shows MATH is explicitly near optimal for MATH (up to an additive logarithmic term). |
math/0006233 | If MATH is explicitly optimal for MATH, then we can find MATH from MATH (as in the proof of REF ), and given MATH and MATH we find MATH as in REF . Hence, given MATH, we can enumerate MATH and determine the maximal index MATH of a MATH. Since also MATH, the numbers MATH have a maximal common prefix MATH. Write MATH with MATH by REF . Given MATH we can determine MATH from MATH. Hence, from MATH, and MATH we can reconstruct MATH. That is, MATH, which yields the lemma. |
math/0006233 | For every MATH in the set defined by the left-hand side of the inequality, we have MATH, and the length of continuation of MATH to the total padded index of MATH is MATH. Moreover, all these indices share the same first MATH bits. This proves the lemma. |
math/0006233 | Let us prove first MATH . By the NAME inequality, we have, with MATH, MATH since MATH is in REF correspondence with the prefix programs of length MATH. Hence MATH . For the statement of the lemma, we have MATH where in the last inequality we used REF . |
math/0006233 | When we talk about complexity with MATH in the condition, we use a NAME machine with MATH as an ``oracle". With the help of MATH, we can compute MATH, and so we can define the following new semicomputable (relative to MATH) function with MATH: MATH . We have, using REF and defining MATH so that MATH for MATH: MATH . Summing over MATH gives MATH. The theorem that MATH is maximal within multiplicative constant among semicomputable semimeasures is also true relative to oracles. Since we have established that MATH is a semicomputable semimeasure, therefore MATH, or equivalently, MATH which proves the theorem. |
math/0006233 | Let MATH be a set with MATH and assume MATH. Tacitly understanding MATH in the conditions, and using the additivity REF , MATH . Therefore MATH . |
math/0006233 | Denote the conditional universal probability as MATH. We write MATH to indicate sets MATH that satisfy MATH. For every MATH, let us define a function over all strings MATH of length MATH as follows: MATH . The following lemma shows that this function of MATH is a semimeasure. We have MATH . We have MATH . There are constants MATH such that for some MATH of length MATH, MATH . Let us fix MATH somehow, to be chosen appropriately later. REF implies that there is a MATH with REF . Let MATH be the first string of length MATH with this property. To prove the right inequality of REF , let MATH be the program of length MATH that terminates last in the standard running of all these programs simultaneously in dovetailed fashion, on input MATH. We can use MATH and its length MATH to compute all programs of length MATH that output finite sets using MATH. This way we obtain a list of all sets MATH with MATH. Using this list, for each MATH of length MATH we can compute MATH, by using REF explicitly. Since MATH is defined as the first MATH with MATH, we can thus find MATH by using MATH and some program of constant length. If MATH is chosen large enough, then this implies MATH. On the other hand, from REF we have MATH . This implies, by the definition of MATH, that either MATH or MATH. Since MATH we get the left inequality of REF in both cases for an appropriate MATH. Consider now a new semicomputable function MATH on all finite sets MATH with MATH. Then we have, with MATH: MATH by REF , respectively, and so MATH with MATH fixed is a lower semicomputable semimeasure. By the dominating property we have MATH. Since MATH is the length of MATH and MATH we can set MATH, and hence MATH. Then, with the first MATH because of REF , MATH . Then, by the additivity REF : MATH . Hence MATH. |
math/0006233 | : Let MATH be one of the non-stochastic objects of which the existence is established by REF . Choose MATH with MATH so that the set MATH has complexity MATH and MATH has randomness deficiency MATH with respect to MATH. Because MATH is non-stochastic, this yields MATH. For every MATH we have MATH. Together it follows that MATH. That is, these non-stochastic objects MATH have complexity MATH. Nonetheless, there is a constant MATH such that MATH is not random, typical, or in general position with respect to any explicitly represented finite set MATH containing it that has complexity MATH, but they are random, typical, or in general position for some sets MATH with complexity MATH like MATH. That is, every explicit sufficient statistic MATH for MATH has complexity MATH, and MATH is such a statistic. Hence MATH is an explicit minimal sufficient statistic for MATH. |
math/0006233 | By REF and by typicality MATH. |
math/0006233 | Writing MATH, since MATH by REF , we have MATH. Hence, it suffices to show MATH. Now, from an implicit description MATH we can find the value MATH. To recover MATH from MATH, we at most require an extra MATH bits. That is, MATH. This reduces what we have to show to MATH which is asserted by REF . This shows the first statement in the theorem. The second statement follows from the first one: rewrite MATH and substitute MATH. |
math/0006233 | A distribution MATH is optimal iff REF holds with equalities. Rewriting MATH the first inequality becomes an equality iff MATH, and the second inequality becomes an equality iff MATH (that is, MATH is a typical distribution). |
math/0006233 | If MATH is optimal for MATH, then MATH. From MATH we can find both MATH and MATH, and hence MATH, that is, MATH. We have MATH by REF , respectively. This proves the first property. Substitution of MATH in the expression of REF proves the second property. |
math/0006233 | Clearly, REF implies REF . We show that both REF implies REF implies REF : By REF we have MATH where we absorb a MATH additive term in the MATH sign. Together with REF implies MATH and vice versa REF together with REF implies REF . |
math/0006233 | (If) By assumption, MATH. Rearrange and add MATH (by typicality) to the right-hand side to obtain MATH. Substitute according to MATH (by sufficiency) in the right-hand side, and subsequently subtract MATH from both sides, to obtain REF . (Only If) Reverse the proof of the (If) case. |
math/0006233 | If MATH is a probabilistic sufficient statistic, then, by REF , equality of MATH-expectations REF holds. However, it is still consistent with this to have large positive and negative differences MATH for different MATH arguments, such that these differences cancel each other. This problem is resolved by appeal to the algorithmic mutual information non-increase law REF which shows that all differences are essentially positive: MATH. Altogether, let MATH be least positive constants such that MATH is always nonnegative and its MATH-expectation is MATH. Then, by NAME 's inequality, MATH that is, MATH . |
math/0006233 | By assumption, using REF , there is a positive constant MATH, such that, MATH . Therefore, MATH . On the other hand, since MATH we obtain MATH . Altogether, this implies REF , and by REF , the theorem. |
math/0006234 | First consider distinct fixed vertices connected by a path of nonfixed vertices. Since fixed vertices are the points where paths move between squares of parity MATH, such a path must be contained in a single unit square of parity MATH; moreover, for a unit square to contain two fixed vertices, it must be contained in MATH instead of being on the boundary. REF shows that, if there is a monochromatic path between two fixed vertices before the application of MATH, then there is one afterwards. Specifically, in two REF , all four edges are the same color and there are no fixed vertices. In two cases (REF and a rotation), the edges alternate colors and there are four fixed vertices. In these cases, MATH does not change the colors, so we may reuse the old path. In the remaining twelve cases (REF and rotations), there are two fixed vertices connected by one blue and one green path. The application of MATH switches these colors, leaving one path of each color. As the endpoints have degree MATH or MATH, they cannot be in paths connecting interior vertices, such as fixed vertices. Given a path between fixed vertices MATH and MATH, we may divide it at each fixed vertex to obtain many paths of nonfixed vertices connecting fixed vertices. From such paths we may construct paths of nonfixed vertices connecting the intermediate fixed vertices after the application of MATH. Concatenating these paths, we produce a path connecting MATH and MATH. Since MATH is an involution, the lemma in one direction implies the converse. |
nlin/0006024 | It suffices to show that there exists some time MATH such that MATH . Then it follows from the maximum principle that MATH for all MATH, and hence MATH satisfies the linear advection-diffusion equation MATH for MATH, which implies REF . We actually show that REF holds at MATH for sufficiently large MATH. Recall that MATH, and hence MATH can be bounded from above using the maximum principle as follows: MATH . Here the function MATH satisfies the linear problem MATH . Furthermore, we have MATH with the function MATH satisfying the degenerate parabolic equation MATH and MATH . We note that if MATH satisfies the NAME REF then the diffusion process defined by REF has a unique smooth transition probability density. Indeed, the NAME algebra generated by the operators MATH and MATH consists of vector fields of the form MATH which span MATH if MATH satisfies REF . Then the theory of CITE, and the results of CITE imply that there exists a smooth transition probability density MATH such that MATH . In particular, the function MATH is uniformly bounded from above for any MATH CITE. Then we have MATH . It is straightforward to observe that MATH with MATH being the transition probability density for REF with MATH. That is, MATH satisfies MATH . Therefore we obtain MATH and in particular at time MATH we have MATH as long as MATH . REF follows from REF as explained in the beginning of this Section. |
nlin/0006024 | Let us define the set MATH . As before, it suffices to show that solution of REF satisfies MATH and that is what we will do. First, we split the initial data for REF into two parts: one supported on a strip MATH, containing the flat part of MATH and another supported outside it. We will choose MATH such that any solution of REF that is independent of MATH and with initial data supported inside MATH will be small at time MATH. The second part is supported away from the strip MATH, where MATH is flat. Therefore for a sufficiently small time it behaves like a solution of REF with advection satisfying the NAME. We choose MATH as follows. Let MATH be a periodic solution of MATH given by MATH . Then we have MATH . A simple estimate shows that MATH and hence MATH . Therefore we have MATH as long as MATH where MATH . Let us choose MATH so that REF is automatically verified for initial data supported on MATH with MATH as in REF . Let us assume that the width of the interval MATH, on which MATH is constant satisfies MATH . This condition determines the constant MATH in the statement of REF . We may now split the initial data for REF as follows: MATH . Here the smooth function MATH is supported in the interval MATH while the function MATH is supported outside the set MATH. Both of these functions satisfy in addition MATH. Then the function MATH satisfies the inequality MATH with the functions MATH and MATH satisfying REF with the initial data MATH and MATH, respectively. It follows from our choice of MATH that MATH so it remains only to estimate MATH. We will do it separately for MATH inside and outside of the strip MATH. For the points MATH we have: MATH for sufficiently small MATH. Here MATH is the one-dimensional Brownian motion with diffusivity MATH, and MATH denotes probability with respect to it. Thus REF holds inside MATH. In order to estimate the function MATH outside MATH we introduce a profile MATH that coincides with MATH outside of the interval MATH, MATH, and satisfies the NAME on the whole interval MATH. We define the process MATH by MATH . Consider the stopping time MATH which is the first time when MATH enters the interval MATH. Then we have MATH . Here MATH denotes probability with respect to the process MATH starting at MATH, while MATH denotes probability with respect to MATH starting at MATH (recall that MATH is independent of MATH). The process MATH for MATH is identical to the process MATH defined by REF with MATH replaced by MATH. Therefore we have MATH . Recall that MATH and MATH. Therefore the point MATH is a fixed distance away from the interval MATH. Hence we may choose MATH sufficiently small so that MATH . Furthermore, the function MATH satisfies REF with the initial data MATH . However, MATH is quenching and thus we may choose MATH so large that MATH . Therefore we have at MATH: MATH and hence the same upper bound holds at MATH. Therefore REF holds also outside MATH, and REF follows. The fact that MATH is strongly quenching follows from this property of MATH . |
nlin/0006024 | The set of all profiles MATH satisfying NAME is dense in MATH so by REF the set of strongly quenching profiles is dense. To complete the proof, we will show that if MATH satisfies NAME, then there exists MATH such that if MATH then MATH is strongly quenching. From the proof of REF , we know that there exists a constant MATH such that the solution MATH of REF with advection MATH satisfies MATH if the initial data MATH is supported on the interval MATH with MATH (recall that such inequality implies quenching of MATH with the same initial data). Let MATH be a solution of REF with advection MATH and initial data supported in MATH . Then by the NAME formula (see, for example, CITE) MATH where MATH is the solution of REF with advection MATH and initial data equal to the characteristic function of the interval MATH for MATH. Now choose MATH. Then MATH provided that MATH and hence MATH is strongly quenching. |
nlin/0006024 | The proof of REF proceeds in several steps. We consider the initial data satisfying REF . First we find MATH such that there exists a MATH function MATH such that MATH, and MATH and MATH vanishes on the boundary of the disc of radius MATH centered at the point MATH: MATH . Then in the system of coordinates that moves with the speed MATH the function MATH is a sub-solution of REF in MATH. Therefore, initial data that start above MATH will not decay to zero. Next we consider the special solution MATH of REF with the initial data given by MATH . We show that MATH satisfies REF and that implies that REF holds for arbitrary initial data MATH, in particular, such as described in REF . Construction of a stationary sub-solution. Choose MATH so that MATH and define MATH by MATH . The function MATH is NAME continuous, and hence we may choose MATH and MATH so that MATH. Therefore if MATH satisfies MATH then MATH satisfies REF . We are going to construct an explicit radial solution MATH of REF with the ``initial" conditions MATH . Indeed, MATH is given explicitly by MATH . Here MATH is the NAME function of order zero, and MATH is its first zero. Furthermore, we have MATH with MATH and MATH determined by matching REF at MATH. Then we get MATH . Then MATH satisfies MATH . Thus we will take the critical size of plateau in the velocity profile to be MATH so that the disc of radius MATH will fit in. REF . A sub-solution. Let us now assume that MATH. We make a coordinate change MATH . In new coordinates we have a function MATH that solves MATH with the initial data given by REF : MATH . Observe that MATH satisfies REF inside MATH since MATH in MATH. Moreover, MATH on MATH, where MATH vanishes. Therefore the maximum principle implies that inside MATH we have MATH . Note that MATH solves REF with the initial data MATH . The inequality in REF follows from REF inside MATH and the fact that MATH outside MATH. Therefore we have MATH and thus the limit MATH exists since MATH. Moreover, the standard parabolic regularity implies that MATH converges to MATH uniformly on compact sets together with its derivatives up to the second order. Therefore MATH satisfies the stationary problem MATH . We also have MATH . It is easy to show using REF and CITE that for any MATH where the right side is any translation of MATH along the MATH axis. Indeed, assume that there exists the smallest (say, positive) MATH such that MATH at some point MATH. Then the strong maximum principle implies that MATH for all MATH inside the translate MATH of the disc MATH. But that contradicts the fact that MATH on the boundary MATH. Then REF implies that MATH . The next two lemmas show that REF imply that MATH. Let MATH be a solution of REF such that REF holds. Then we have MATH . In order to show that integral of MATH is finite we integrate REF over the set MATH with MATH large and MATH. We get MATH and average this equation in MATH: MATH . Therefore we obtain MATH for all MATH, and hence the first inequality in REF holds. In order to obtain the second inequality we multiply REF by MATH and perform the same integration and averaging as above. This leads to MATH and then the second inequality in REF follows from REF and the first inequality in REF . The limit function MATH . Notice that MATH can not achieve local minima in MATH as follows from the maximum principle. Therefore if we define MATH and MATH then MATH or MATH. Furthermore, if MATH at some point, then MATH everywhere by the strong maximum principle. In particular, if MATH, then MATH. Therefore we consider only the case that MATH and argue by contradiction. We have either MATH for any MATH or for any MATH . Otherwise the minimum of MATH over the set MATH would be achieved inside. Let us assume without loss of generality that MATH for any MATH. Consider MATH . For any MATH we have three options. CASE: MATH for any MATH and MATH . In this case, by definition of MATH REF of MATH and since MATH we have MATH . CASE: MATH . REF implies that there exist MATH such that MATH. Then MATH . Therefore, MATH in this case. CASE: The remaining option is that MATH and there exists MATH such that MATH . In this case, an argument identical to the reasoning of option two leads to the same bound REF . Overall, we see that for any MATH where MATH depends only on MATH and MATH . But this contradicts directly REF . Using REF , we can now complete the proof of REF . Notice that even though we showed MATH we still have to show the limiting function is the same in the original coordinates. REF implies that there exists a time MATH so that for all MATH and all MATH we have MATH . Here MATH is defined by REF . Then we may apply REF with MATH in the original coordinates MATH, and initial data MATH, and get REF . The fact that MATH follows from REF and, for instance, results of CITE. |
nlin/0006024 | Let MATH be an initial data as in REF such that MATH with the function MATH constructed in the proof of REF and given explicitly by REF . Then we have MATH and therefore REF implies that MATH because MATH may not go to zero as MATH. |
nlin/0006024 | Let MATH be solutions of MATH and MATH with the same initial data MATH . The function MATH is assumed to be front-like and monotonic. That is, MATH for MATH, MATH for MATH and MATH . Then by maximum principle applied to the equation for MATH we have MATH and hence MATH . Therefore by maximum principle we have MATH . However, by results of CITE similar to REF but for the front-like initial data we have MATH and similarly for MATH. Then MATH would be incompatible with REF , and hence MATH. Naturally, an analogous result is true for left traveling fronts. |
nlin/0006024 | In order to finish the proof of REF we apply REF to MATH given by REF , and observe that MATH satisfies the estimate in REF according to REF . |
nlin/0006024 | The first statement is a direct corollary of REF . The second statement is the content of REF . |
nlin/0006024 | We will show that solution of REF with MATH replaced by a KPP nonlinearity MATH and the same initial data drops below MATH before time MATH. That will imply REF . Choose MATH such that MATH (such MATH exists since MATH is NAME continuous) and let MATH be the solution of MATH with initial data MATH and periodic boundary REF . Let us also define MATH . Note that MATH and hence REF follows from the following Lemma. There exists a constant MATH that depends only on the non-linearity MATH such that given any initial data MATH with MATH and any MATH, there exists a time MATH such that MATH for all MATH. The following inequality holds for MATH: MATH . We define the set MATH . Observe that there exists a constant MATH that depends only on MATH and MATH such that if the NAME measure MATH at some time MATH, then MATH. Indeed, we have then MATH . Therefore we have with MATH as in REF MATH for a sufficiently small MATH. Hence it suffices to consider the case when MATH for all MATH. We claim that then at any time MATH we have MATH with the constant MATH depending only on MATH and MATH. This may be proved similarly to REF. Therefore we obtain MATH . Thus we have MATH . Therefore MATH . However, we have an a priori bound MATH which contradicts the previous inequality if MATH . Therefore MATH has to drop below MATH if the initial data satisfies REF is proved. REF then follows from REF . |
quant-ph/0006076 | First, apply NAME 's orthonormalization to the horizontal lift MATH of MATH, to obtain the orthonormal basis MATH such that MATH is a subset of the real span of MATH. We immerse NAME space MATH into MATH as MATH where MATH is an orthonormal basis in MATH. Then, letting MATH be a projection valued measure which is obtained as a spectral decomposition of the position operator, the pair MATH is one of the best estimators. This assertion is easily proved by calculating the NAME information matrix of the family, MATH of probability distributions. |
quant-ph/0006076 | Suppose that any member of the manifold MATH in MATH is invariant by the antiunitary operator MATH, and let MATH. Then, we have MATH . Conversely, if MATH is real for any MATH, by NAME 's orthonormalization, we can obtain the orthonormal basis MATH such that MATH is subset of the real span of MATH, which means any member of MATH is invariant by the antiunitary operator MATH, which is defined by, MATH . |
quant-ph/0006125 | Consider the real-valued function MATH defined for unit vectors MATH lying in the unit sphere MATH in MATH . As MATH varies over the compact space MATH this function attains a maximum at some point MATH . Let MATH be any orthonormal basis of MATH containing MATH and expand MATH as in the statement of the theorem. Since MATH is stationary at MATH for variations of MATH on the unit sphere, MATH . Next, find the maximum of MATH as MATH vary over unit vectors orthogonal to MATH respectively. Suppose the maximum occurs at MATH . Then, as before, in any expansion of MATH in terms of orthonormal bases of the MATH containing MATH and MATH the coefficients will satisfy MATH . Continuing in this way, we define the basis vectors MATH for MATH . Then the last basis vector MATH of MATH is determined up to phase. The reality conditions can be imposed as follows. Choose the phase of the basis element MATH so that MATH where the index MATH occurs in the MATH-th place, and then fix MATH by choosing the phase of MATH . |
quant-ph/0006125 | Consider the real-valued function MATH defined for unit vectors MATH lying in the unit sphere MATH in MATH. As MATH varies over the compact space MATH, this function attains a maximum at some point MATH. Let MATH be any orthonormal basis of MATH containing MATH, and expand MATH as in the statement of the theorem. Since MATH is stationary at MATH for variations of MATH on the unit sphere, MATH . Next, find the maximum of MATH as MATH vary over unit vectors orthogonal to MATH respectively. Suppose the maximum occurs at MATH. Then, as before, in any expansion of MATH in terms of orthonormal bases of the MATH containing MATH and MATH, the coefficients will satisfy MATH . Continuing in this way, we define the basis vectors MATH for MATH. Then the last basis vector MATH of MATH is determined up to phase. Next, if the dimensions MATH are not all equal, we maximise MATH as MATH vary orthogonally to the basis vectors already determined, to find basis vectors MATH; then we find MATH for MATH, and hence MATH; then MATH for MATH; and so on until we have MATH for MATH. The maximisation at each step implies that the coefficients satisfy REF . Now, to fix the last MATH basis elements of MATH, we choose a set MATH of MATH indices which is not of the form MATH with MATH or MATH with MATH (so that MATH has not yet been set to zero for any MATH), and maximise MATH with respect to vectors MATH orthogonal to MATH, thus finding MATH; then, choosing a different index set MATH, maximise MATH with respect to vectors MATH orthogonal to MATH; and so on until we have either exhausted the possible index sets MATH or run out of space in which to vary the vector MATH. The coefficients will then satisfy REF . The reality REF can be imposed as follows. (By using a basis vector MATH to fix a coefficient MATH we mean changing the phase of MATH to make MATH real and non-negative.) First use MATH to fix MATH; then use MATH to fix MATH; then use MATH to fix the coefficients in the set MATH; then use MATH to fix the coefficients in MATH and MATH; and finally use MATH excluding MATH to fix the remaining coefficients in MATH. |
quant-ph/0006125 | REF for a stationary value of MATH become MATH where MATH, MATH and MATH. Clearly there is a solution MATH. We have to show that any other solution has MATH. Using REF to eliminate MATH and MATH, REF become MATH . For MATH, it follows that MATH since MATH. Now MATH which is positive if MATH (unless MATH), and the gradient of the quadratic MATH at MATH is also positive. It follows that the zeros of MATH, and therefore any stationary values of MATH other than MATH, are less than MATH. |
cond-mat/0007425 | We calculate MATH where we have set MATH. Note that MATH and that MATH satisfies all the assumptions in REF. We then conclude from REF that the NAME transform of the function MATH is non-negative, where MATH is a function such that MATH as MATH. [The detailed bounds from CITE show that we may in fact choose MATH, since MATH has to control the REFth derivative of MATH.] Note, moreover, that MATH. Hence MATH . The lemma follows by writing MATH and by rescaling from boxes of size REF to boxes of size MATH. |
cond-mat/0007425 | If MATH is a symmetric function. Then MATH where MATH . The lemma follows since it is clear that MATH . |
cond-mat/0007425 | Simply note that MATH and now use the same arguments as before. |
cond-mat/0007425 | Choose MATH and MATH. Then we may assume that MATH since MATH is large. From REF we see immediately that MATH . The corollary follows since MATH and with the given choice of MATH and MATH it is easy to see that MATH. |
cond-mat/0007425 | We simply choose MATH in REF . This is allowed since MATH is ensured from the assumption that MATH is large. We then obtain MATH . The bound on MATH follows since the bound on the gap REF implies that MATH. |
cond-mat/0007425 | The terms containing MATH are MATH using the commutation relation MATH. |
cond-mat/0007425 | The terms containing MATH or MATH are MATH . Note that MATH commutes with MATH. We have that MATH . Hence MATH . Since MATH we obtain the operator inequality MATH and the lemma follows. |
cond-mat/0007425 | The operator MATH with integral kernel MATH is a non-negative NAME operator on MATH with norm less than MATH. Denote the eigenvalues of MATH by MATH, MATH and corresponding orthonormal eigenfunctions by MATH. We may assume that these functions are real. The eigenvalues satisfy MATH. We then have MATH . The identity REF thus follows with MATH. If MATH denotes the projection onto the constant functions we may also consider the operator MATH. Denote its eigenvalues and eigenfunctions by MATH and MATH. Then again MATH. Hence we may write MATH . Thus, since all MATH are orthogonal to constants we have MATH . REF follow immediately from this. The fact that MATH follows from the representation REF . Moreover, since the kernel MATH is a continuous function we have that MATH for almost all MATH and hence MATH . We therefore have MATH and the lemma follows since MATH. |
cond-mat/0007425 | The terms containing MATH, MATH,MATH, or MATH are MATH . In the last term we have used the representation REF and the commutation relation MATH. For all MATH we get that the above expression is bounded below by MATH . The bound REF follows from REF . The second bound REF follows in the same way if we notice that the terms containing MATH, MATH,MATH, or MATH may be written as MATH . |
cond-mat/0007425 | The terms containing MATH, MATH,MATH, or MATH are MATH . Using that MATH we may write this as MATH . The lemma now follows from REF . |
cond-mat/0007425 | It is clear that for all MATH and hence MATH . Now simply use this with MATH. |
cond-mat/0007425 | Let MATH, with MATH, be a parameter to be chosen below. Recall that MATH is the parameter used in the definition of MATH in REF. Then since MATH we have MATH where the last inequality follows from REF. We can now repeat this calculation to get MATH . If we therefore use REF and recall that MATH we arrive at MATH . Note that for MATH we have MATH . Thus if we also assume that MATH we have MATH . Thus if MATH is a normalized function on MATH which is orthogonal to constants we have according to the bound on the gap REF that for all MATH . We choose MATH for an appropriately large constant MATH and assume that MATH and MATH are such that MATH is less than MATH. Then MATH . If we now use REF we may write this as MATH where in the last inequality we have used that MATH and MATH. We now choose MATH and we may then write this inequality in second quantized form as MATH using that MATH. Since we consider only states with particle number MATH the inequality still holds if we insert MATH as in the statement of the lemma. |
cond-mat/0007425 | We may complete the square MATH if MATH . We choose the solution MATH . Hence MATH . |
cond-mat/0007425 | We consider a state with MATH. Then MATH. We shall use REF . Note first that MATH by REF and the fact that MATH. We may of course rewrite MATH. By REF we have MATH where MATH . Since MATH we have that MATH . Note that MATH . The lemma now follows from REF by a simple change of variables in the MATH integral. |
cond-mat/0007425 | REF holds for all MATH hence also if we had replaced MATH by MATH in this case we get REF . The integral MATH satisfies the bound MATH . By REF we may assume that MATH. Hence MATH is bounded by a constant as long as MATH and MATH are sufficiently large and MATH is sufficiently small (which also ensures that MATH is close to MATH). Note that we do not have to make any assumptions on MATH. Moreover, if this is true we also have that MATH is large and hence MATH is small. This would give the bound in the corollary except for the first positive term. The above argument, however, also holds (with different constants) if we replace the kinetic energy in the Foldy Hamiltonian by MATH (assuming that MATH). This proves the corollary. |
cond-mat/0007425 | The lower bound follows from REF . To prove the upper bound on MATH we choose MATH (the maximally allowed value) and MATH, where we shall choose MATH sufficiently small, in particular MATH. We then have that MATH is large. Moreover MATH for some constant MATH and we get from REF that MATH where we have again used that MATH, MATH and MATH. Note that MATH and MATH . From REF we know that MATH. By choosing MATH small enough we see immediately that MATH. |
cond-mat/0007425 | Inserting the bound MATH into REF gives MATH . We now choose MATH and MATH, where we shall choose MATH below, independently of MATH, MATH, and MATH. Note that since MATH is small we may assume that MATH as required and since MATH is small we may assume that MATH. Moreover MATH and MATH. Hence, since MATH (see REF), we have MATH . By choosing MATH appropriately (independently of MATH, MATH, and MATH) we immediately get the bound on MATH and the bound MATH, which implies the stated bound on MATH. The bound on MATH follows since we also have MATH and MATH where we have used the bound on MATH which we have just proved. The case when MATH follows in the same way because we may everywhere replace MATH by MATH and use REF instead of REF . Note that in this case we already know the bound on MATH since we still assume the existence of the state such that MATH. |
cond-mat/0007425 | As explained above we choose MATH to be of order MATH. We then choose MATH as explained above. Then REF holds. We also know that the possible MATH values of MATH range in an interval of length MATH. We do not know however, where this interval is located. REF will allow us to say more about the location of the interval. In fact, it follows from REF that MATH. It is then a consequence of REF that MATH. This of course establishes that the allowed MATH values are less than MATH for some constant MATH. |
cond-mat/0007425 | If we replace all the operators MATH and MATH with MATH in the Foldy Hamiltonian by MATH and MATH we get a unitarily equivalent operator. This operator however differs from the Hamiltonian MATH only by a change of sign on the part that we denoted MATH. Since both operators satisfy the bound in REF we conclude that MATH . Note that both sums above define positive operators. This is trivial for the first sum. For the second it follows from REF since MATH commutes with all MATH and MATH with MATH. The lemma now follows from REF and from REF . |
cond-mat/0007425 | We assume MATH. We proceed as in the beginning of REF, but we now use REF instead of REF . We then get MATH . If we now use REF and the facts that MATH, MATH, and MATH we see with appropriate choices of MATH and MATH that MATH . If we finally insert the choices of MATH and MATH and use REF we arrive at the bound in the lemma. |
cond-mat/0007425 | According to REF we may assume that MATH where MATH is at least as big as the constant in REF . We still assume that MATH. We begin by bounding MATH and MATH using REF . We have from REF that MATH . In order to bound MATH we shall use REF . Together with REF this gives (choosing MATH say) MATH . Inserting the choices for MATH and MATH and using REF gives MATH where we have also used that we may assume that MATH is small. REF now reads MATH . If this is not satisfied we see immediately that the bound REF holds. Thus from REF it follows that we can find a normalized MATH-particle wavefunction MATH with MATH such that MATH . In order to analyze MATH we proceed as in the beginning of Sect. CASE: This time we use REF , and REF together with REF instead of REF . We obtain MATH . This time we shall however not choose MATH small, but rather big. Note that since MATH we have MATH, which follows immediately from MATH . We therefore have MATH . If we now insert the choices of MATH and MATH, take the expectation in the state given by MATH, and use REF and the bound on MATH from REF we arrive at MATH . If we now choose MATH we arrive at REF . |
cond-mat/0007425 | We have accumulated various errors and we want to show that they can all be made small. There are basically two parameters that can be adjusted, MATH and MATH. Instead of MATH it is convenient to use MATH. We shall choose MATH as a function of MATH such that MATH as MATH. From REF we know that for some fixed MATH. Hence according to REF with MATH and MATH given in REF we have that MATH as MATH if MATH . The hypotheses of REF are valid if REF , and MATH hold. From REF , for which the hypotheses are now automatically satisfied, we have that MATH and from REF we see that MATH is MATH to leading order. With these conditions we find that the first term on the right side of REF is, in the limit MATH, exactly Foldy's law. The conditions that the other terms in REF are of lower order are MATH together with REF . It remains to show that we can satisfy REF is trivially satisfied since both MATH and MATH tend to infinity. Since MATH for small MATH we see that REF is implied by REF . REF is implied by REF , which is in turn implied by REF . The remaining two REF are easily satisfied by an approriate choice of MATH and MATH as functions for MATH with MATH and MATH as MATH. In fact, we simply need MATH. The bound REF has now been established. Hence Foldy's law REF follows as discussed in the beginning of the section. |
cond-mat/0007425 | It is convenient to extend the matrix MATH to all MATH by defining MATH unless MATH. Similarly, we extend the vector MATH and we define the numbers MATH and the matrix MATH to be zero when MATH. We shall give the construction for MATH odd, the MATH even case being similar. For MATH set MATH if MATH and MATH otherwise. Thus, MATH for precisely MATH values of MATH. Also, MATH. MATH is chosen so that MATH. For each MATH define the vector MATH by MATH. We then define MATH. (The number MATH will be chosen later.) After this, we define MATH. Using the fact that MATH, we have that MATH and MATH . Hence MATH with MATH . Let us choose MATH. Then, MATH. Recalling that not all of the MATH equal zero, we conclude that there is at least one value of MATH such that REF MATH and REF MATH. This concludes the proof of REF except for showing that MATH for all MATH and MATH. This is evident from the easily computable large MATH asymptotics in REF . |
cond-mat/0007425 | We calculate the commutator MATH . Likewise we calculate the double commutator MATH . Note that MATH and thus the first term above is positive. We claim that MATH . To see this we simply calculate MATH . The last two terms are bounded by MATH. For the first term we have by the NAME inequality for operators, MATH, for all MATH, that MATH and this is bounded above by MATH and we get REF . Inserting REF into REF , recalling that the first term is positive, we obtain MATH . Again using the NAME inequality, we have MATH and REF follows. The bound REF is proved in the same way. Indeed, MATH and REF follows from MATH and REF . |
cs/0007002 | Given MATH, MATH, and MATH three subsets of MATH, with MATH, let MATH be an element of MATH and MATH. CASE: Given any real MATH, we have by definition of Inner that MATH. By definition of Outer, we have that MATH, which leads to MATH. Consequently, MATH; CASE: For any MATH, MATH implies MATH, by definition of Inner; since MATH, we have MATH, and finally, by definition of Inner, MATH. Consequently, MATH; CASE: We first prove MATH: Given MATH, we have MATH, by definition of Inner. By contractance of Inner, MATH. We then have MATH, and then MATH, by definition of Inner again. Now, we prove MATH: Given MATH, we have MATH by definition of Inner. Therefore, any element in MATH is in MATH. As a consequence, MATH. Definition of Inner then leads to MATH; CASE: We prove that: MATH . We have: MATH CASE: We prove that: MATH . We have: MATH . |
cs/0007002 | Immediate consequence of Inner and Outer definitions. |
cs/0007002 | Consider Line MATH of REF. By completeness of MATH, MATH. Consequently, MATH. Since the set MATH returned eventually is the union of all the sets computed on Line MATH, we have that MATH. We can use the dual reasoning to prove that MATH. |
cs/0007002 | Immediate. |
cs/0007002 | Termination of REF depends on a reasonable choice for the functions NAME and MATH. More specifically, termination is ensured whenever MATH creates boxes MATH that are all strictly smaller than MATH, and when NAME checks for the size of all dimensions of MATH but the one of MATH being smaller than some threshold that is attainable with the floating-point format at hand. In order to prove the proposition, we have to show that the properties given by REF always hold on lines REF . Provided that boxes put into MATH and MATH on lines REF, respectively, verify REF , it is not necessary to investigate what is returned on line REF since it is the mere union of the sets previously computed. MATH . We first consider the case when MATH. Let MATH. By hypothesis, MATH during the first call of REF. By completeness of MATH, if MATH, there exists a value in MATH such that MATH does not hold when the free variables take their values in MATH. Consequently, MATH. The property does hold for all the subsequent calls of REF since the corresponding domains for MATH are always included in MATH (by contractance of the narrowing operators used). Since MATH, we have proved that the triplet returned on Line REF verifies REF for any call of REF. MATH . We now consider the case when MATH. Let us prove that for any call MATH of REF, the assignments performed on lines REF are such that: MATH . This suffices to prove the proposition since what is returned for the case MATH is either MATH and MATH, or the union, componentwise, of MATH and MATH with sets of boxes MATH and MATH REF computed in subsequent calls, and since we have already proved that what is returned whenever MATH verifies the property. MATH . We first consider Line REF: by completeness of MATH, MATH. Hence, MATH, and then MATH. MATH . We now consider Line REF: we will first prove that for any call MATH of REF: MATH . This is trivially true for the first call REF since, by hypothesis, MATH. MATH . Assume that REF does hold for the MATH-th call of REF. Since MATH and MATH, we have: MATH . By completeness of MATH, we have: MATH . By contractance of MATH, MATH. From REF , we deduce: MATH . The box MATH is split on Line REF in MATH boxes MATH (MATH), with MATH (no splitting on the domain of MATH). Consequently, we obtain from REF : MATH . From Line REF, it follows that REF does hold for MATH whenever it does hold for MATH. Having proved that REF does hold for any call MATH, we reconsider Line REF: by completeness of MATH, it comes: MATH . Hence: MATH since MATH. From REF , we deduce: MATH . We also know from REF that all the boxes in MATH have MATH as their last dimension. From this and REF , we obtain: MATH . That is, MATH for any call of REF. |
cs/0007002 | For each constraint MATH, we consider only the boxes that were put in MATH by the preceding constraint (Lines MATH and MATH). By soundness of REF respectively, REF, when considering constraint MATH, MATH then contains on line MATH only the boxes MATH verifying MATH (respectively, MATH). Once again, by soundness of REF (respectively, REF, when considering a constraint MATH on Line MATH, MATH contains only boxes MATH such that there was a constraint MATH with MATH for which MATH (respectively, MATH). Consequently, MATH) (respectively,MATH). |
cs/0007015 | The invariance of MATH is verified by observing that none of REF - SREF change any variable value at a timer-final state. Convergence is demonstrated by a sequence of claims about an arbitrary computation MATH. Let MATH be a suffix of MATH beginning following the second round of MATH; by definition, MATH is a based computation. We consider two cases for MATH. Case: MATH contains no step executing SREF within MATH rounds. By arguments similar to those given in the proof of REF , some state of MATH satisfies MATH within MATH rounds and continues to hold at least until SREF executes. Let MATH be a suffix of MATH satisfying MATH at its initial state. Observe that MATH is based and reset-free for MATH rounds, so REF is applicable to MATH. Within MATH rounds of MATH, MATH holds, and the state satisfies MATH. Case: MATH contains some step executing SREF within MATH rounds. Execution of REF results in a state satisfying the premise of REF . Therefore MATH either contains a smooth state within MATH rounds, or contains a state satisfying MATH within MATH rounds. The latter possibility is the premise for REF , which shows that a smooth state is subsequently obtained within an additional MATH rounds, so with either possibility, MATH contains a smooth state within MATH rounds. REF implies that MATH contains a timer-final state within MATH rounds following a smooth state. |
cs/0007015 | Provided MATH, arguments in the proof of REF show that for each perturbed region MATH, some process MATH executes SREF within the first round, where MATH satisfies either MATH or MATH for some MATH. REF then implies that within MATH additional rounds, each MATH satisfies MATH. Each unperturbed process clock variable remains larger than MATH until SREF is executed, which resets the clock to zero. Thus within MATH rounds, each MATH is either larger than MATH or is at most MATH. After MATH rounds, unperturbed processes can decrease clock variables to zero, but such a decrease does not falsify the conditions for a time-accurate state. Therefore, to show time accuracy, it suffices to show that increments to MATH imply corresponding increments have executed at distant processes. After a process MATH executes SREF, it does not increment MATH until cEcho holds. If an unperturbed MATH is a neighbor of MATH, then MATH does not increment MATH until MATH has reset MATH and updated the image variables and register fields so that MATH observes cEcho. It is a simple induction to show that MATH cannot increase to a value MATH unless MATH has incremented MATH at least MATH times. Now consider a minimum length path MATH of processes, of length MATH, from MATH to some process MATH, such that each process in MATH is unperturbed. By a double induction, on MATH and MATH, it follows that MATH cannot increase from zero to MATH unless each process MATH has incremented MATH at least MATH times. The same argument shows that processes of MATH complete at least the same number of MATH-rounds in the period where MATH increases from zero to MATH. Returning to the event of MATH executing SREF, we now consider the case of perturbed MATH. As observed in the proof of REF , it is possible that MATH can increment MATH twice before MATH completes a cycle because corrupt values in the initial state enable the cEcho and wEcho conditions. Furthermore, MATH can increment MATH a third time before MATH increments its clock because MATH completes a cycle to enable MATH. However in the case of such a third successive increment by MATH, MATH and MATH hold as a consequence. Thereafter, we reason about the interaction between MATH and MATH as for unperturbed neighbors (note that any subsequent executions of REF by MATH or MATH validate this argument, since we reason about the highest value attained for clock variables after MATH's initial three increments). Therefore, the value of MATH does not increase to MATH unless MATH has incremented MATH at least MATH times. Again, we may consider a minimum length path MATH of processes, of length MATH, from MATH to some process MATH, such that each process in MATH is perturbed (with the possible exception of MATH). By a double induction, on MATH and MATH, it follows that MATH cannot increase from zero to MATH unless each process MATH has incremented MATH at least MATH times. Similar arguments show the completion of the appropriate number of MATH-rounds while MATH increases from zero to MATH. Notice that in the case of a perturbed path of processes, accuracy can diminish by two extra clock units per unit of distance, whereas in the case of an unperturbed path, accuracy corresponds precisely to distance. These observations combined can be used to verify that in any minimum length path MATH from MATH to MATH, after MATH executes SREF, the value of MATH increases to MATH only if for each MATH, the value of MATH has incremented at least MATH times, where MATH is the number of perturbed processes in the subpath of MATH from MATH to MATH. Since MATH, time accuracy is verified for MATH. The arguments above show that time accuracy holds for all unperturbed processes within MATH rounds and that any subsequent state is MATH-accurate for unperturbed processes. For perturbed processes, similar reasoning applies. Instead of relying on SREF to establish the baseline clock value, we use instead a value bound by the construction given in REF 's proof. Within MATH rounds, there is a state MATH where perturbed MATH has a clock value of at most MATH, and MATH is the distance to some unperturbed process that executes SREF in the first round. The value of MATH cannot increase from MATH to MATH unless process MATH has incremented its clock at least MATH times, where MATH is at most MATH. Therefore when MATH at some state following MATH, we infer that MATH has incremented at least MATH times, which verifies time accuracy for unperturbed processes. |
cs/0007015 | REF directly shows that perturbed processes assign clock variables to at most MATH within the first MATH rounds. REF establishes that MATH increases its clock to at least MATH after MATH rounds following the execution of REF establishes that for each perturbed region, some process MATH either within or neighboring the perturbed region executes SREF in the first round. REF establishes that processes within a given distance increase their clock values as MATH increases. Any process MATH within a perturbed region containing or neighboring MATH is at most distance MATH from MATH; simplifying the bound of REF using MATH as a distance upper bound yields a lower bound of MATH after MATH rounds. |
cs/0007015 | Consider a MATH-faulty initial state. If MATH, then from MATH and REF , the system stabilizes to MATH and hence MATH in MATH rounds, which proves the conclusion. The remaining case is MATH for a MATH-faulty initial state. For this case, we first show that any faulty process MATH is time accurate within the first MATH rounds. Requirement REF ensures that some process within distance MATH from any faulty process executes a double-reset in the first round, and REF implies subsequent time accuracy within MATH rounds. The same argument implies that each nonfaulty process within distance MATH from a faulty process has a time-accurate clock after at most MATH rounds. All nonfaulty processes have time-accurate clock variables throughout the computation. Design REF specifies that some nonfaulty processes do not change their output variables by any repair procedure, so the proof obligation is to show that faulty processes and those nonfaulty processes within distance MATH to a faulty process stabilize their output variables in MATH time. After MATH rounds, all such processes have time-accurate clock variables. By definition of time accuracy for a MATH-faulty initial state, a time-accurate MATH variable with value MATH implies that the number of MATH-rounds preceding in the computation is at least MATH. Procedure MATH converges within MATH of the MATH-rounds, so after time accuracy holds, a MATH value of MATH implies that variables of MATH can be copied to MATH's output variables. The conditions of Design REF also justify copying MATH to the output variables when MATH for MATH. Having established the safety of copying output sets to the output variables, the remaining obligation is to show that all such copying either completes within MATH rounds or that any subsequent copying will not affect MATH. REF implies that all processes within distance MATH to a faulty process will, after time accuracy holds, increase their clock variables to MATH within MATH rounds and will not subsequently decrease their clock values below this value. Therefore, within MATH rounds, all processes within distance MATH to a faulty process assign their output variables, while those processes further than distance MATH from a fault do not assign their output variables to falsify MATH by any step of the computation. |
cs/0007015 | The essence of the proof is that neither SREF nor SREF increment a clock to a value two greater than any neighbor. Since REF involves x variables, the effect of REF requires examination. Reading a register to assign an x variable only increases the accuracy of the image variable; in particular, given MATH as a precondition, SREF does not falsify this condition, because MATH and MATH have clock values differing by at most one in the precondition. Therefore it suffices to verify that any change to MATH or MATH also satisfies the lemma. In a reset-free computation, only SREF and SREF change a clock variable. If SREF executes, incrementing MATH, we have MATH as a precondition. Since MATH, we have MATH also as a precondition; thus the increment to MATH results in a state satisfying MATH, verifying MATH. The postcondition also satisfies MATH, since the change to MATH does not alter the relation between MATH and MATH. A similar argument applies to REF, and also to the case of MATH incrementing its clock. |
cs/0007015 | By definition of a rising computation, each process has a lower bound on neighboring clock variables in its x variable, because clock values cannot decrease in a reset-free computation. Suppose MATH is the first of MATH to increment its clock. A precondition for this step is MATH, which implies MATH, which in turn implies MATH. Consider two cases for this last inequality, REF MATH or REF MATH. For REF , process MATH cannot increment MATH, and this situation will persist until MATH increments its clock sufficiently many times so that MATH. It is straightforward to verify that MATH does not increment MATH beyond MATH, so for REF the first increment to MATH establishes MATH. For REF , we deduce from the inequalities above MATH holds as precondition to MATH's first increment step, and the inequalities with regard to the x variables are similar. So for REF holds directly. |
cs/0007015 | By REF , neighboring MATH establish MATH at or before MATH; by REF such processes continue to satisfy this property for the remainder of the reset-free computation segment. |
cs/0007015 | Because we consider a based computation, and not a rising computation in this lemma, the invariance of MATH stated in REF is not applicable. Note that MATH holds at a smooth state. The invariance of smoothness is therefore verified from the conditions of REF - SREF, since no gap exists at a smooth state and SREF preserves smoothness. It is also simple to verify that the least clock value, if smaller than MATH, increments within two rounds from a smooth state, hence at most MATH rounds are needed to obtain a state satisfying MATH. A similar argument shows that all w variables converge to MATH within MATH rounds. |
cs/0007015 | In the first round, MATH reads neighboring clock values and detects local minimality. If MATH increments in this round, the lemma holds; and if MATH does not increment, it writes its clock and detects cEcho in the next round, and local minimality implies MATH will increment MATH either by REF or SREF. |
cs/0007015 | Observe that MATH holds at least until some clock exceeds MATH so that SREF can execute. And MATH, so process MATH does not execute the assignment of REF. This implies that the computation is reset-free until some clock obtains the value exceeding MATH. REF implies that each minimal clock value increments in any pair of rounds, which implies that the maximum of the set of clock values eventually grows as the computation proceeds. Let MATH be the first clock to attain the value MATH at state MATH. Thus MATH holds prior to MATH. More generally, it follows by induction that MATH. Therefore, each clock value has incremented at least once prior to MATH. Let MATH be the first clock to attain the value MATH at state MATH. At state MATH, each process has incremented its clock twice in a reset-free computation, implying that each process has read all of its registers at least once in this reset-free computation. Therefore the computation segment beginning with MATH is by definition a rising computation segment (at least until some clock exceeds MATH). Now let MATH be the first clock to attain the value MATH at state MATH. At state MATH, each process has incremented its clock at least once in a rising computation, and by REF , MATH is a smooth state. |
cs/0007015 | REF shows that the computation contains a smooth state, so the obligation here is to show the MATH time bound. By REF each minimal clock value increments at least once in any two consecutive rounds, so within MATH rounds, some clock attains the value MATH, establishing a smooth state. |
cs/0007015 | The proof begins with a claim on the first MATH rounds of the based computation: within MATH rounds there is a state satisfying MATH . The claim is shown by induction. The first state of the computation satisfies the claim for MATH as the base case. Suppose the claim holds for MATH and consider two processes MATH and MATH such that MATH, MATH, and MATH. Let MATH be a state satisfying REF - REF for MATH. By the cEcho condition of REF, process MATH does not increment MATH beyond MATH until process MATH's image fields in MATH have the appropriate values. In fact, these register fields may initially have the appropriate values, which would allow MATH to increment clock and w variables to MATH by REF - SREF. However process MATH cannot subsequently increment to MATH until the cEcho condition holds, which requires a cycle by MATH (and all other neighbors). Process MATH therefore observes MATH in its cycle and assigns at most MATH to its w and clock variables. Since MATH occurs at least by round MATH, the bound of MATH for MATH variables applies within round MATH, which establishes REF. REF is also shown by induction. For MATH, the base case, REF holds vacuously. Now suppose REF holds for MATH and consider two processes MATH and MATH such that MATH, MATH, and MATH. REF places an upper bound on variables at distance MATH from process MATH within round MATH. Therefore MATH within round MATH. In moving from round MATH to MATH, we consider the possibilities for process MATH and MATH. If MATH and MATH differ by more than one and process MATH executes a cycle, then SREF resets MATH; before any further change to MATH occurs, the cEcho condition requires a full cycle by MATH, which validates REF up to distance MATH within round MATH. If MATH and MATH are equal or differ by one, then MATH could increment. Observe here that no clock or w variable can increment beyond one more than any neighboring value; by another inductive argument, no clock or w variable increments beyond MATH more than any corresponding variable at distance MATH. Therefore MATH does not increment beyond MATH so long as MATH. This observation is generalized by REF for MATH within round MATH. Note that we have assumed that any clock increment is due to REF and not SREF in this argument; this assumption is justified by REF , since MATH, which disables execution of REF. |
cs/0007015 | Let MATH be a state satisfying MATH. By REF such a state MATH occurs with MATH rounds of the based computation. So long as every clock is at most MATH, no step subsequent to MATH decreases a w variable; and if no w variable is reset by REF in a consecutive pair of rounds, then the minimum value of the set of w variables either increases by that pair of rounds or all w variables already have the maximum MATH value (we consider a consecutive pair of rounds to ensure that wEcho will hold for SREF). Therefore, if no clock variable attains the value MATH within MATH rounds, all w variables equal MATH and the lemma holds. On the other hand, if some clock does attain the value MATH, we shall deduce that all w values equal MATH, which also proves the lemma. The argument rests on the following claim: at all states subsequent to MATH satisfying MATH, the implication MATH holds for every MATH and MATH. This claim is verified by induction on MATH. For MATH the result is immediate from the domain of w variables. Now consider MATH and suppose the claim holds for MATH. Let MATH be the first process to assign MATH. If the assignment occurs by REF then MATH and the claim holds; if the assignment occurs by REF, then each neighbor of MATH has a clock value of MATH, hence by hypothesis each neighboring w variable is at least MATH, and MATH by the same hypothesis. The result is that the same cycle assigning MATH also assigns MATH to be at least MATH. Similar arguments treat the general case for MATH (not necessarily the first) assigning MATH to MATH, verifying that MATH as a result. To complete the lemma, consider the first state MATH where some MATH has value MATH. By the induction argument given in the proof of REF , any clock at distance MATH from MATH has had a value of at least MATH prior to state MATH. Therefore every clock has contained a value of at least MATH prior to MATH, implying that each w variable is at least MATH prior to MATH. The state immediately preceding MATH thus satisfies proof obligation. |
cs/0007015 | REF by induction on MATH. For MATH let MATH to satisfy the base case. For MATH, we have MATH by hypothesis. By the Echo conditions of REF, SREF and SREF, the clock and w values of MATH remain at most MATH until all neighbors either REF complete cycles that observe these values and write corresponding images to output registers or REF happen to have these values already in their output registers. Considering REF , for MATH satisfying MATH, the execution of REF assures MATH within one round, and MATH is at most MATH if MATH observes no gap, or assigned some value at most MATH otherwise; either case verifies the inductive hypothesis for MATH. These considerations for REF also verify the second part of the lemma, which concerns a path of unperturbed processes, and the same hypothesis with MATH replaced by MATH. Considering REF , process MATH may increment MATH and MATH because MATH happens already to have values corresponding to MATH and MATH in its output register fields. In this case, MATH may increment its variables to at most MATH immediately. Furthermore process MATH may initially have its program counter at SREF, about to write its image variables in such a way that MATH can observe the cEcho condition (even though MATH would not actually read and write in a full cycle). Therefore, if MATH executes SREF, process MATH can increment variables again to at most MATH. However, here a cEcho condition will not be satisfied at MATH until all neighbors complete full cycles, so MATH's variables cannot exceed MATH until MATH completes a cycle. When MATH does complete a cycle, by the reasoning above for REF we deduce that MATH and MATH for MATH. |
cs/0007015 | Note that the lemma holds trivially if the initial state is MATH-perturbed. For the case MATH we use induction on MATH and nested induction on MATH and suppose a based computation. For the base case MATH consider MATH. Since MATH is perturbed, there is a path MATH from MATH to some perturbed MATH (possibly through MATH) of MATH or fewer processes, which is not smooth. Because MATH, some neighboring pair of processes along path MATH has the property that one clock exceeds MATH while the other is less than MATH. Therefore some process in path MATH executes SREF in the first round. By the arguments of REF it follows that MATH executes SREF within MATH rounds. This completes the base case, but reasoning similar to the nested induction also applies for MATH. Finally, because the initial state may not justify a based computation, two additional rounds are added to conclude a MATH bound. |
cs/0007015 | REF states that a process executes SREF at most once in a computation, so it suffices to show that MATH either does not execute SREF or executes SREF within the first MATH rounds. If MATH is unperturbed, REF implies the result. If MATH is perturbed, then for some perturbed region MATH containing MATH, there is an unperturbed MATH neighboring some process of MATH that executes SREF within the first MATH rounds by REF . Applying REF we deduce that MATH holds after MATH additional rounds, and by arguments of REF process MATH does not execute SREF in the remainder of the computation. Therefore, for perturbed MATH, the distance from MATH to a perturbed process is MATH and after MATH rounds, process MATH does not execute SREF. |
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