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Albuquerque Travel guide
Albuquerque : Michelin's recommendations
Albuquerque is an intriguing mix of old and new, of 14C Indian pueblos, 18C Spanish village and 21C high-technology center. Visitors flock to New Mexico’s largest city to se its classic roadside architecture, neon-lit cafes and motor courts speckled its cultural corridors, particularly on Central Avenue east and west of downtown, as well hot air balloons and beautiful natural scenery.
Tourist attractions Albuquerque : Things to see and do
2000 Mountain Road North West USA - Albuquerque, NM 87104-1459
The evolution of the Rio Grande valley, from pre-Columbian tribes through Coronado to the present, is chronicled in the fine Four Centuries exhibit at this Old Town museum. An extensive collection of Spanish Colonial artifacts includes religious tape..
Rio Grande Blvd. & Mountain Rd. USA - Albuquerque, NM 87104
Some 150 shops and galleries face hidden gardens and cobbled walkways around a quiet, tree-shaded 18C plaza. The modern city has grown from this traditional central core. Today, restored adobes share the plaza with Pueblo Revival architecture, some s..
10 Tramway Loop NE USA - Albuquerque, NM 87122-2017
The world's longest jig-back aerial tramway covers 2.7mi from the northeastern city limits to 10,678ft Sandia Crest. Climbing from urban desert to alpine terrain in just 15min, the tram affords panoramic views of 11,000sq mi, across the entire Albuqu.. |
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abstract: 'We study the growth of polynomials on semialgebraic sets. For this purpose we associate a graded algebra to the set, and address all kinds of questions about finite generation. We show that for a certain class of sets, the algebra is finitely generated. This implies that the total degree of a polynomial determines its growth on the set, at least modulo bounded polynomials. We however also provide several counterexamples, where there is no connection between total degree and growth. In the plane, we give a complete answer to our questions for certain simple sets, and we provide a systematic construction for examples and counterexamples. Some of our counterexamples are of particular interest for the study of moment problems, since none of the existing methods seems to be able to decide the problem there. We finally also provide new three-dimensional sets, for which the algebra of bounded polynomials is not finitely generated.'
address:
- 'Pinaki Mondal, Weizmann Institute of Science, Israel'
- 'Tim Netzer, Universität Leipzig, Germany'
author:
- Pinaki Mondal
- Tim Netzer
title: 'How fast do polynomials grow on semialgebraic sets?'
---
Introduction
============
Let ${\mathbb R}[x]$ be the polynomial algebra in $n$ variables $x=(x_1,\ldots,x_n)$. For $d\in {\mathbb N}$ we denote by ${\mathbb R}[x]_d=\left\{ p\in{\mathbb R}[x]\mid {{\rm deg}}(p)\leq d\right\}$ the finite-dimensional subspace of polynomials of total degree at most $d.$ For a set $S\subseteq {\mathbb R}^n$ let $${\mathcal B}_d(S):=\left\{ p\in {\mathbb R}[x] \mid p^2 \leq q\ \mbox{Êon } S, \mbox{ for some } q\in {\mathbb R}[x]_{2d}\right\}$$ denote the set of those polynomials, that grow on $S$ as if they were of degree at most $d.$ Cleary ${\mathbb R}[x]_d\subseteq {\mathcal B}_d(S)$ for all $d$, and ${\mathcal B}_d(S)$ is closed under addition, which follows for example from the inequality $$(p+p')^2\leq (p+p')^2+(p-p')^2=2p^2+2p'^2.$$ ${\mathcal B}_0(S)$ is the algebra of [*bounded polynomials*]{} on $S$, and each ${\mathcal B}_d(S)$ carries the structure of a ${\mathcal B}_0(S)$-module. More general, ${\mathcal B}_d(S)\cdot {\mathcal B}_{d'}(S)\subseteq {\mathcal B}_{d+d'}(S),$ so we have a graded algebra $${\mathcal B}(S):=\bigoplus_{d\geq 0} {\mathcal B}_d(S).$$ ${\mathcal B}(S)$ can be identified with a subalgebra of ${\mathbb R}[x,t],$ wher e $t$ is a single new variable, by identifying $p\in {\mathcal B}_d(S)$ with $p\cdot t^d.$ Also note that $${\mathcal B}_d(S_1\cup S_2)={\mathcal B}_d(S_1)\cap {\mathcal B}_d(S_2)\ \mbox{Êand }\ {\mathcal B}(S_1\cup S_2)={\mathcal B}(S_1)\cap {\mathcal B}(S_2)$$ holds for $S_1,S_2\subseteq {\mathbb R}^n.$ This follows from the fact that $q$ in the definition of ${\mathcal B}_d(S)$ can always assumed to be globally nonnegative. In fact you can always take some $C+D\Vert x\Vert^{2d}.$ Finally note that ${\mathcal B}_d(S)$ and thus ${\mathcal B}(S)$ do only depend on the behaviour of $S$ at infinity; if $S$ is changed inside of a compact set, no changes in ${\mathcal B}_d(S)$ and ${\mathcal B}(S)$ occur. In this paper, we consider the following questions:
\[one\] Is $\mathcal B(S)$ a finitely generated algebra?
\[three\] Is ${\mathcal B}_0(S)$ a finitely generated algebra?
\[two\] Is every ${\mathcal B}_d(S)$ a finitely generated ${\mathcal B}_0(S)$-module?
\[four\] If ${\mathcal B}_0(S)={\mathbb R},$ is every ${\mathcal B}_d(S)$ a finite-dimensional vector space?
Note that a positive answer to Question \[one\] yields a positive answer to all the other questions. Also note that Question \[four\] Êis just a special case of Question \[two\]. Let us start with some examples.
\[ex1\] (1) Assume $S\subseteq {\mathbb R}^n$ contains a full-dimensional convex cone $K$ (e.g. $S={\mathbb R}^n$ or $S=[0,\infty)^n$). For any $0\neq p\in{\mathbb R}[x]$ there is a point $0\neq a\in K$, on which the highest degree part of $p$ does not vanish. So on the half-ray through $a$, the polynomial $p^2$ cannot be bounded by a polynomial of degree smaller than $2\cdot {{\rm deg}}(p).$ This proves ${\mathcal B}_d(S)={\mathbb R}[x]_d,$ and the answer to all questions is positive. Note that ${\mathcal B}(S)$ is generated by $t,x_1t,\ldots,x_nt.$
\(2) Let $S=\left([0,1]\times {\mathbb R}\right) \cup \left({\mathbb R}\times [0,1]\right)\subseteq{\mathbb R}^2$ be the union of a vertical and a horizontal strip. A polynomial $p$ belongs to ${\mathcal B}_d(S)$ if and only the degree of $p$ as both a polynomial in $x_1$ and in $x_2$ is at most $d$. From this it is easy to see that the answer to all above questions is positive. Note that ${\mathcal B}_d(S)={\mathbb R}[x]_d$ is not true here, since for example $x_1x_2\in {\mathcal B}_1(S),$ since $x_1^2x_2^2\leq x_1^2+x_2^2$ on $S$. However, ${\mathcal B}(S)$ is generated by $t,x_1t,x_1x_2t$ and $x_2t$.
\(3) If $S$ is bounded, then ${\mathcal B}_0(S)={\mathcal B}_d(S)={\mathbb R}[x]$ for all $d$, and the answer to Question \[one\] is obviously yes. In fact $\mathcal B(S)={\mathbb R}[x,t]$ here.
\(4) The set $S$ should be semialgebraic, if a positive answer to the above questions is to be expected. Indeed, consider $S=\left\{ (a,b)\in{\mathbb R}^2\mid 0\leq a, \exp(a)\leq b \right\}.$ Then for $p_d:=\sum_{i=0}^d \frac{1}{i!}x_1^i$ and $(a,b)\in S$ we have $$0\leq p_d(a,b) \leq \exp(a) \leq q(a,b),$$ with $q=x_2$. So $p_d\in {\mathcal B}_1(S)$ for all $d\in{\mathbb N}$. On the other hand, ${\mathcal B}_0(S)={\mathbb R}$ is easily checked, so Question \[four\] has a negative answer.
\(5) Even in the semialgebraic case, the answer to Question \[four\] can be negative. This example is Example 4.2 from [@cimane]. Consider $S=\left\{ (a,b)\in {\mathbb R}^2 \mid b^2(1-a^2)\geq 0\right\}$. So $S$ is the union of a vertical strip and the $x_1$-axis. One checks that ${\mathcal B}_0(S)={\mathbb R}$ holds. On the other hand, $(x_1^dx_2)^2\leq x_2^2$ on $S$, so $x_1^dx_2\in {\mathcal B}_1(S)$ for all $d.$ So ${\mathcal B}_1(S)$ is not finite-dimensional, and Question \[four\] has a negative answer. In particular, ${\mathcal B}(S)$ is not finitely generated. This example is however somewhat pathological, since $S$ has a lower-dimensional part, i.e. is not regular.
\(6) Another pathology arising from non-regular sets is the following. Let $S$ be the $x_1$-axis in ${\mathbb R}^2$ alone. Then ${\mathcal B}_0(S)$ is not a finitely generated algebra, as one easily checks. So also the answer to Question \[one\]Ê is negative. On the other hand, each ${\mathcal B}_d(S)$ is a finitely generated ${\mathcal B}_0(S)$-module, generated by $1,x_1,\ldots,x_1^d.$
We see that we should restrict to regular semialgebraic sets from now on, i.e. sets containing a dense open subset. Let us recall what is actually known concerning the above questions. It seems like Question \[one\] has not been explicitly studied for semialgebraic sets yet. The paper [@plsc] deals with Question \[three\] and shows that the answer is yes in the two-dimensional regular case, and false in general for higher dimensions. The paper [@kr] provides an explicit three-dimensional such counterexample, based on a counterexample to Hilbert’s 14th Problem of Nagata. In the context of [*moment problems*]{}, Question \[four\] has been extensively studied. The works [@kuma], [@ma], [@ne], [@posc], [@vi] all give positive answers, for large classes of sets. To avoid confusion, we note that the question that is anwered in these papers is in fact the following:
\[five\] If ${\mathcal B}_0(S)={\mathbb R}$, does there exist a function $\psi\colon{\mathbb N}\rightarrow{\mathbb N}$, such that for any two polynomials $p,q\in {\mathbb R}[x]$ with $p,q\geq 0 $ on $S$, one has $${{\rm deg}}(p)\leq \psi\left({{\rm deg}}(p+q)\right)?$$
\[four=five\] For any set $S\subseteq {\mathbb R}^n$, Question \[four\] and Question \[five\] are equivalent.
First assume that each ${\mathcal B}_d(S)$ is a finite dimensional vector space, and let $\psi(d)$ be the maximal degree of a polynomial in ${\mathcal B}_d(S)$. If $p,q$ are nonnegative on $S$, then $p\leq p+q$ on $S,$ so $p\in {\mathcal B}_{{{\rm deg}}(p+q)}(S)$. This shows ${{\rm deg}}(p)\leq \psi\left( {{\rm deg}}(p+q)\right).$ Let conversely such a function $\psi$ be given. Obviously $\psi$ can be assumed to be non-decreasing. If $p^2\leq q$ for some $q\in {\mathbb R}[x]_{2d},$ then $q-p^2$ and $p^2$ are nonnegative on $S$. We obtain ${{\rm deg}}(p^2) \leq \psi\left( {{\rm deg}}(p^2 +(q-p^2))\right)\leq \psi(2d).$ This shows that ${\mathcal B}_d(S)$ is finite dimensional.
Let us briefly recall some facts about the moment problem. Given a linear functional $\varphi\colon{\mathbb R}[x]\rightarrow {\mathbb R}$, one wants to determine whether it has a representing measure $\mu$, i.e. whether $$\varphi(p)=\int_{{\mathbb R}^n}p\ d\mu$$ holds for all $p\in{\mathbb R}[x].$ Also the support of $\mu$ is of interest here. A result by Haviland [@havi] states that $\varphi$ has a representing measure supported on a set $S\subseteq {\mathbb R}^n$ if and only if $\varphi(p)\geq 0$ holds for all polynomials $p$ which are nonnegative as functions on $S$. Unfortunately, describing all nonnegative polynomials on $S$ is a hard problem. So Haviland’s theorem becomes particularly helpful if the nonnegativity condition can be weakend. Towards this goal one resctricts to [*basic closed semialgebraic sets*]{} $$S=\{a\in{\mathbb R}^n\mid p_1(a)\geq 0,\ldots,p_r(a)\geq 0\},$$ where $p_1,\ldots,p_r\in{\mathbb R}[x].$ The polynomials which are obviously nonnegative on $S$ are of the form $q^2$ and $q^2p_i$ for some $q\in{\mathbb R}[x].$
We say that $p_1,\ldots,p_r$ [*solve the moment problem for $S$*]{} if the conditions $$\varphi(q^2p_i)\geq 0 \qquad \forall q\in {\mathbb R}[x], \ i=0,\ldots,r$$ (where $p_0=1$) are enough to ensure the existence of a representing measure for $\varphi$ on $S$. Such a weakened positivity condition on $\varphi$ can then for example be checked by a series of semidefinite programs (see for example [@ma]). The celebrated result of Schmüdgen [@sch1] implies that if $S$ is bounded, then the finitely many products $p_1^{e_1}\cdots p_r^{e_r}$ ($e_i\in\{0,1\}$) always solve the moment problem for $S$. Another result of Schmüdgen [@sch2] (see also [@ma; @ne2]) provides a method to reduce the dimension in the moment problem. Given a nontrivial bounded polynomial $p\in{\mathcal B}_0(S)$, it is enough to check the moment problem on all fibres $S\cap \{p=\lambda\}$ of $p$ in $S$. Since the problem is usually easier in lower dimensions, this is very helpful. Now the significance of Questi on \[five\] for the moment problem is the following. If ${\mathcal B}_0(S)= {\mathbb R}$ and Question \[five\] has a positive answer, then the moment problem is not solvable, at least in dimension $\geq 2$, by a result of Scheiderer [@sc]. So it doesn’t matter that the reduction result of Schmüdgen cannot be applied, since the problem is not solvable anyway. Since Question \[five\] has a positive answer in so many cases, it has been asked whether the answer is always yes, for [*regular*]{} semialgebraic sets, i.e. sets containing a dense open subset (see for example [@pl]). This would mean there is no gap between the results of Schmüdgen and Scheiderer. In this paper we show, among other things, that this is false. In particular, deciding the moment problem for our counterexamples seems to call for completely new methods.
Our contribution is the following. In Section \[tent\] we show that Question \[one\] has a positive answer for a large class of sets in arbitrary dimension, built of so-called standard tentacles. In Section \[count\] we provide a first regular two-dimensional example, for which Question \[four\] (and thus any other of the questions as well) has a negative answer. We give a completely elementary and constructive proof. In Section \[plane\] we use more elaborate techniques to examine planar sets. For certain simple sets (namely sets with a single ‘tentacle’), we give complete answers to our questions \[one\]–\[four\], and a partial solution to the moment problem. We provide a systematic way to produce more examples and counterexamples to our questions. The results show that in principle anything can happen, even for regular sets in the plane. The methods are based on the study of key forms for semidegree functions and corresponding (algebraic) compactifications of ${\ensuremath{\mathbb{C}}}^2$, mostly from [@contractibility; @non-negative-valuation; @sub1]. In Section \[bound\] we show that the algebra ${\mathcal B}(S)$ can always be interpreted as the algebra of bounded polynomials on a higher dimensional set. In this way, we get more three-dimensional and explicit counterexamples to Question \[three\].
We also remark that it is possible to extend the analysis in Sections \[single-subsection\] and \[general-subsection\] of the ‘subdegrees’ associated to planar sets to higher dimensional subalgebraic sets. It is however more technical in nature and will be part of a future work.
Standard tentacles {#tent}
==================
In this section we prove that Question \[one\], the strongest of the above questions, has a positive answer for a large class of sets. We recall the definition of a [*standard tentacle*]{} from [@ne].
For $z=(z_1,\ldots,z_n)\in{\mathbb Z}^n,$ a [*standard $z$-tentacle*]{} is a set $$\left\{ \lambda^zb:=(\lambda^{z_1}b_1,\ldots,\lambda^{z_n}b_n)\mid \lambda\geq 1, b\in B\right\}$$ where $B\subseteq ({\mathbb R}\setminus\{0\})^n$ is a compact semialgebraic set with nonempty interior. We call $B$ a [*base*]{} of the tentacle.
Any $z\in {\mathbb Z}^n$ defines a weighted degree ${{\rm deg}}_z$ on ${\mathbb R}[x],$ by assigning degree $z_i$ to the variable $x_i.$ We set $$\varphi_z:= \max\{0, z_1,\ldots,z_n\}$$ and note that $\max \{ {{\rm deg}}_z(p)\mid p\in {\mathbb R}[x]_d\}= d\cdot \varphi_z.$
For finite unions of standard tentacles, the modules ${\mathcal B}_d(S)$ from the last section can be described via these weighted degrees. For this let $z^{(1)},\ldots, z^{(m)}\in {\mathbb Z}^n$ be given. We write ${{\rm deg}}_i$ instead of ${{\rm deg}}_{z^{(i)}}$ and $\varphi_i$ instead of $\varphi_{z^{(i)}}$.
\[tendeg\] Assume $S\subseteq {\mathbb R}^n$ is a finite union of standard tentacles, corresponding to the directions $z^{(1)},\ldots,z^{(m)}\in{\mathbb Z}^n.$ Then for all $d\in{\mathbb N}$ $${\mathcal B}_d(S)= \left\{ p\in {\mathbb R}[x] \mid {{\rm deg}}_i(p) \leq d\cdot \varphi_i \quad \mbox{for } i=1,\ldots,m\right\}.$$ In particular, ${\mathcal B}_d(S)$ is spanned as a vector space by the monomials contained in it.
“$\subseteq$”: Let $p\in {\mathcal B}_d(S)$ with $p^2\leq q$ on $S$, for some $q\in {\mathbb R}[x]_{2d}.$ Since $q(\lambda^{z^{(i)}}b)$ is of degree at most $2d\cdot \varphi_i$ in $\lambda,$ it follows that ${{\rm deg}}_i(p)\leq d\cdot \varphi_i.$ At this point we need that tentacles have nonempty interior; there is some curve $\lambda^{z^{(i)}}b$ in the tentacle, on which $p$ grows with ${{\rm deg}}_i(p).$ “$\supseteq$”: Since ${\mathcal B}_d(S)$ is a vector space, we can assume that $p=x^\beta$ is a monomial. We can also assume $m=1.$ We have for $\lambda\geq 1$ $$p^2(\lambda^{z^{(1)}}b)=b^{2\beta}\lambda^{2z^{(1)}\beta^t}\leq C\lambda^{2\varphi_1 \cdot d},$$ for all $b$ in the base $B$ of the tentacle. Choose $\alpha$ with $\vert\alpha\vert \leq d$ and $z^{(1)}\alpha^t=\varphi_1 \cdot d.$ Then choose $D>0$ such that $C\leq D^2b^{2\alpha}$ for all $b\in B.$ We obtain $p^2\leq (Dx^{\alpha})^2$ on $S$, and thus $p\in {\mathcal B}_d(S)$.
In [@ne] Theorem 5.4 it was shown that Question \[five\] has a positive answer, if $S$ is a finite union of standard tentacles. We improve upon this now, while also simplifying the proof significantly:
If $S$ is a finite union of standard tentacles, then ${\mathcal B}(S)$ is finitely generated.
We use the same notation as before. Consider the set $$M:=\{ (\alpha,d)\in{\mathbb N}^{n+1}\mid z^{(i)}\alpha^t\leq d\cdot \varphi_i \mbox{ for } i=1,\ldots,m\}.$$ By Proposition \[tendeg\], a polynomial from ${\mathbb R}[x,t]$ belongs to ${\mathcal B}(S)$ if and only of all its monomial do, and a monomial $x^\alpha t^d$ belongs to ${\mathcal B}(S)$ if and only if $(\alpha,d)\in M$. The lattice points in a rational convex cone form a finitely generated semigroup, by a well-known result of Hilbert. So if $(\alpha_1,d_1),\ldots,(\alpha_r,d_r)$ generate $M$, then the monomials $x^{\alpha_i}t^{d_i}$ generate ${\mathcal B}(S)$.
A first counterexample {#count}
======================
In this section we construct a first [*regular*]{} semialgebraic set $S$, for which Question \[four\], and thus all the other questions as well, have a negative answer. We will see more examples later, but for this one we give a completely elementary and constructive proof. We consider two sets $$S_1=\left\{ (a,b)\in{\mathbb R}^2\mid 1\leq a, 1\leq a^3b+ a^6 -a \leq 2\right\}$$ $$S_2=\left\{ (a,b)\in{\mathbb R}^2\mid 1\leq a, 1\leq a^3b-a^6-a\leq 2\right\}$$ and set $S:=S_1\cup S_2.$ Note that $S$ is basic closed semialgebraic, i.e. definable by finitely many simultaneous polynomial inequalities. In fact if $p=x^3y+x^6-x$ and $q=x^3y-x^6-x$, then $S=\mathcal S(x-1, -(2-p)(p-1)(2-q)(q-1)).$
In the above example we have ${\mathcal B}_0(S)={\mathbb R}$, but already ${\mathcal B}_1(S)$ is of infinite dimension.
First note that $S_1$ consists precisely of the points $(\lambda, r\lambda^{-3}-\lambda^3+\lambda^{-2})$ for $\lambda\geq 1$ and $r\in [1,2].$ So if for $p\in {\mathbb R}[x_1,x_2]$ the $\lambda$-degree of $p(\lambda, r\lambda^{-3}-\lambda^3+\lambda^{-2})$ is at most $d$, then $p$ belongs to ${\mathcal B}_d(S_1)$. In fact, $p^2$ can then be bounded by some $C+Dx_1^{2d}$ on $S_1$. Note that ${\mathcal B}_0(S_1)$ consists precisely of those $p$ where this $\lambda$-degree is $\leq 0.$
If $q(x,y):=p(x,y-x^3+x^{-2})$, then $p(\lambda, r\lambda^{-3} -\lambda^3+\lambda^{-2})=q(\lambda, r\lambda^{-3}).$ So the $\lambda$-degree of $p(\lambda,r\lambda^{-3}-\lambda^3+\lambda^{-2})$ equals ${{\rm deg}}_{(1,-3)}$ of the Laurent polynomial $p(x,y-x^3+x^{-2}).$ For $S_2$ the same is true with $+x^3$ instead of $-x^3$ everywhere. The claim will thus follow if we construct polynomials $p\in{\mathbb R}[x,y]$ of arbitrarily high degree, such that the $(1,-3)$-degree of both Laurent polynomials $p_+=p(x,y+x^3+x^{-2})$ and $p_-=p(x,y-x^3+x^{-2})$ is at most $1,$ and if we show that degree $\leq 0$ is only possible for constant polynomials $p$.
First consider $q=y^2-x^6.$ We find $$q_\pm=q(x,y\pm x^3+x^{-2})= \pm 2x \pm 2x^3y+x^{-4}+2x^{-2}y+y^2.$$ Using this, we next consider $r=x^ky^l(y^2-x^6)^m$ and the Laurent polynomials $r_\pm=r(x,y\pm x^3+x^{-2})$. We find $$\begin{aligned}
r_\pm = \sum_{a+b+c=l,d+e+f+g+h=m} \frac{l!m!}{a!b!c!d!e!f!g!h!}(\pm1)^{b+d+e}2^{d+e+g}x^{k+3b-2c+d+3e-4f-2g}y^{a+e+g+2h}\end{aligned}$$ From this formula we can read off the following facts:
- The coefficients of $r_+$ and $r_-$ are the same up to signs. In fact, whether $b+d+e$ is even or odd only depends on the monomial $x^{k+3b-2c+d+3e-4f-2g}y^{a+e+g+2h}$ (in fact only on $k+3b-2c+d+3e-4f-2g$).
- The Newton polytope of $r_\pm$ has vertices $$(-4m-2l+k,0),(k,2m+l),(m+3l+k,0),(3m+3l+k,m).$$ There are monomials on the line from $(m+3l+k,0)$ to $(3m+3l+k,m)$, and on parallel lines shifted by $5$ to the left. No other monomials occur. This can be seen by checking that the $(1,-2)$-degree of the monomial $x^{k+3b-2c+d+3e-4f-2g}y^{a+e+g+2h}$ is $$k-2l+m +5(b-f-g-h).$$
- The signs of the coefficients of $r_-$ and $r_+$ obey the following rule. On the line through $(m+3l+k,0)$ and $(3m+3l+k,m)$ the signs differ by $(-1)^{m+l}$. Going through parallel lines in steps of $5$ to the left, the sign change oscillates from $+$ to $-$.
We are now ready to construct the desired polynomials. We start with $$p^{(1)}=(y^2-x^6)^m$$ for an arbitrarily large $m$. Many of the monomials of $p^{(1)}_+$ and $p^{(1)}_-$ have $(1,-3)$-degree $\leq 1$ anyway. However, there are some which don’t. If in the Newton polytope we follow the line from $(3m,m)$ in direction towards $(m,0$), the first monomial $x^{3m}y^m$ is of degree $0$, and the second monomial $x^{3m-2}y^{m-1}$ is of degree $1$. We can tolerate both of them. The next one is however $x^{3m-4}y^{m-2}$, and here we have a $(1,-3)$-degree of $2$. We now modify $p^{(1)}$ by adding $$p^{(2)}=c\cdot x^2(y^2-x^6)^{m-2}$$ to $p^{(1)}$, with a suitable coefficient $c$. Since $p^{(2)}_\pm$ gives rise to a Newton polytope with vertices $$(-4m+10,0), (2,2m-4)),(m,0),(3m-4,m-2),$$ we can choose $c$ to cancel the monomial $x^{3m-4}y^{m-2}$ in [*both*]{} $p^{(1)}_+$ and $p^{(1)}_-$ [*at the same time*]{}. This follows from the above sign considerations: the coefficients in $
p_+^{(1)}$ and $p_-^{(1)}$ differ by $(-1)^{m}$ at this monomial, and the same is true for $p^{(2)}_+$ and $p^{(2)}_-$.
Now the new Laurent polynomials $(p^{(1)}+p^{(2)})_\pm$ both have one less of the bad monomials, namely $x^{3m-4}y^{m-2}$. At the same time, no new monomials arise. The coefficients are still the same in absolute value, and whether the sign changes is determined by the same rule as described above.
In this way one proceeds: Assume that all monomials up to $(3m-2(i-1),m-(i-1))$ have already been cancelled. Write $i=3l+k$ with $0\leq k\leq 4$ and $i-l$ even. Then a term $$p^{(i)}=c\cdot x^ky^l(y^2-x^6)^{m-i}$$ will allow to also cancel the monomial $(3m-2i,m-i)$ in both Laurent polynomials $$(p^{(1)}+\cdots + p^{i-1)})_\pm$$ at the same time. This follows from the fact that $(-1)^{m-i+l}=(-1)^m$ since $i-l$ is even.
Once all bad monomials on this line are cancelled, one resumes with bad monomials on parallel lines to the left in a similar fashion. On the next line to the left, one for example has to write $i-5=3l+k$, this time $i-l$ odd, and so on...
When trying to figure out why the degree $1$ monomials, for example $x^{3m-2}y^{m-1}$, cannot be cancelled in this way, one sees why only constant polynomials $p$ give rise to $p_\pm$ both of degree $\leq 0.$ One checks that the highest degree part in the $(1,3)$-grading of some $p\in{\mathbb R}[x,y]$ with ${{\rm deg}}_{(1,-3)}p_\pm \leq 0$ must be $(y^2-x^6)^m$ for some $m$, up to scaling. In fact the linear equations for monomials on the maximal $(1,3)$-line in the Newton polytope of $p_\pm$ are linearly independent, having Pascal matrices as coefficient matrices. Since there are $2m+1$ coefficients and $2m$ equations, there is a unique solution up to scaling.
Now the monomial $x^{3m-2}y^{m-1}$ in $(y^2-x^6)^m_\pm$ has sign $(\pm 1)^m$, as we have seen. It can only cancel with terms coming from $x(y^2-k^6)^{m-1},$ but here the sign will be $(\pm1)^{m-1}$. So the cancellation cannot work for both substitutions at the same time. This finishes the proof.
Proposition \[almost-single\] below gives an ‘explanation’ of the above counter example in the language of Section \[plane\].
As explained in the introduction, none of the existing methods to decide the moment problem seems to work for sets of this kind. The reduction result from [@sch2] cannot be applied since ${\mathcal B}_0(S)={\mathbb R}$, and since Question \[five\] has a negative answer, the usual way to see that the moment problem is unsolvable is also not successfull. Sets of this kind seem to call for completely new methods.
Sets in the plane {#plane}
=================
In this section we use more elaborate techniques, mostly from [@contractibility; @non-negative-valuation; @sub1], to examine planar sets in more detail. We will obtain many more examples and counterexamples to our question (see Example \[exex\]).
Degree like functions associated to a subset of ${\ensuremath{\mathbb{R}}}^n$
-----------------------------------------------------------------------------
Let $S$ be a subset of ${\ensuremath{\mathbb{R}}}^n$ and $\delta_S: {\ensuremath{\mathbb{R}}}[x_1, \ldots, x_n]\setminus\{0\} \to {\ensuremath{\mathbb{Z}}}$ be the function that maps $f$ to the smallest $d$ such that $f \in {\mathcal B}_d(S)$. Then $\delta_S$ is a [*degree-like function*]{} in the terminology of [@sub1] (or equivalently, $-\delta_S$ is an [*order function*]{} in the terminology of [@szpiro]), i.e. $\delta_S$ satisfies
1. \[ative\] $\delta_S(f+g) \leq \max\{\delta_S(f),\delta_S(g)\}$ with equality if $\delta_S(f) \neq \delta_S(g)$, and
2. \[mtive\] $\delta_S(fg) \leq \delta_S(f) + \delta_S(g)$.
Some trivial observations are ${\mathcal B}(S) = {\bigoplus}_{d \geq 0} \{f: \delta_S(f) \leq d\}$ and $\delta_S \leq {{\rm deg}}$. Now define $\bar \delta_S: {\ensuremath{\mathbb{R}}}[x_1, \ldots, x_n]\setminus\{0\} \to {\ensuremath{\mathbb{R}}}$ as $$\begin{aligned}
\bar \delta_S(f) := \lim_{n \to \infty} \delta_S(f^n)/n.\end{aligned}$$ It is not hard to see that $\bar \delta_S$ is well defined and satisfies $\bar \delta_S(f^k) = k\bar\delta_S(f)$ for all $f$ and $k$. We call $\bar\delta_S$ the [*normalization*]{} of $\delta_S$. In the terminology of [@szpiro], $-\bar \delta_S$ is a [*homogeneous order function*]{}. We will examine the structure of $\bar \delta_S$ in more details for the case $n=2$.
Let $S$ be a semi-algebraic subset of ${\ensuremath{\mathbb{R}}}^2$. For each $r > 0$, let $B_r$ be the ball of radius $r$ centered at the origin. For large enough $r$, the number of connected components of $S \setminus B_r$ becomes stable. Each of these components is called a [*tentacle*]{} of $S$.
The case of a single tentacle {#single-subsection}
-----------------------------
Throughout this subsection we assume that $S$ is a semi-algebraic subset of ${\ensuremath{\mathbb{R}}}^2$ such that
1. \[single-assumption\] $S$ has only one tentacle, and
2. \[regular-assumption\] the tentacle of $S$ is regular, i.e. it contains a dense open subset.
Let $\bar S$ be the closure of $S$ in ${\ensuremath{\mathbb{R}}}{\ensuremath{\mathbb{P}}}^2$ and $L_\infty$ be the line at infinity on ${\ensuremath{\mathbb{R}}}{\ensuremath{\mathbb{P}}}^2$.
\[single-degree-lemma\] If $\bar S$ intersects $L_\infty$ at more than one point, then $\delta_S = \bar \delta_{S} = {{\rm deg}}$.
Since $S$ has only one tentacle it follows that $\bar S \cap L_\infty$ is connected. It follows that $|\bar S \cap L_\infty| = \infty$. Choose coordinates $[X:Y:Z]$ on ${\ensuremath{\mathbb{R}}}{\ensuremath{\mathbb{P}}}^2$ such that $(x,y) := (X/Z,Y/Z)$ are coordinates on ${\ensuremath{\mathbb{R}}}^2$. Choose an infinite sequence of points $P_i := [1:c_i:0] \in \bar S \cap L_\infty$ and curves $C_i \subseteq S$ such that $P_i \in \bar C_i$. Then $(1/x,y/x)$ are coordinates near each $P_i$ and $\bar C_i$ has a Puiseux expansion at $P_i$ of the form $y/x = c_i + \sum_{q \in {\mathbb Q}, q>0} c_{i,q}(1/x)^q$, or equivalently, of the form $y = c_i x + \sum_{q \in {\mathbb Q}, q<1} c_{i,q}x^q$. Now pick two polynomials $g_1, g_2 \in {\ensuremath{\mathbb{R}}}[x,y]$ with $d_1 := {{\rm deg}}(g_1) < d_2 := {{\rm deg}}(g_2)$. Then $$\begin{aligned}
g_i|_{C_j} = g_i(x,y)|_{y = c_j x + \sum_{q \in {\mathbb Q}, q<1} c_{j,q}x^q} = g_{i,d_i}(1,c_j)x^{d_i} + {\text{l.o.t.}},\end{aligned}$$ where $g_{i,d_i}$ is the leading form of $g_i$ and ${\text{l.o.t.}}$ stands for ‘lower order terms’ (in $x$). If $g_{2,d_2}$ is non-negative on ${\ensuremath{\mathbb{R}}}^2$, then it follows that for generic $C_j$, $g_2|_{C_j} > g_1|_{C_j}$ for sufficiently large $|x|$. It therefore follows that $\delta_S = {{\rm deg}}$, as required.
Now assume \[single-assumption\] and \[regular-assumption\] hold and that $\bar S$ intersects $L_\infty$ at only one point $P$. Choose a linear function $u$ such that $P$ is not on the closure of the line $u = 0$. Then (without changing $\delta_S$), we may assume that for sufficiently large values of $|u|$, all points of $S$ are bounded by curves $C_i := \{f_i(x,y) = 0\}$, $1 \leq i \leq 2$. Choose another linear function $v$ such that $(u,v)$ is linearly independent. Then $(1/u,v/u)$ is a set of coordinates (on ${\ensuremath{\mathbb{R}}}{\ensuremath{\mathbb{P}}}^2$) near $P$ and (the closure of) each $C_i$ has a Puiseux expansion at $P$ of the form $v/u = \sum_{j\geq 0} a_{ij} (1/u)^{\tilde \omega_{ij}}$, with $0 \leq \tilde \omega_{i0} < \tilde \omega_{i1} < \cdots$ and $a_{ij} \in {\ensuremath{\mathbb{R}}}$, or $v = \phi_i(u)$, where $\phi_i(u) := \sum_{j\geq 0} a_{ij} u^{\omega_{ij}}$ where $\omega_{ij} := 1 - \tilde \omega_{ij}$.
\[single-tentacle-lemma\] Let $\omega$ be the largest (rational) number such that the coefficients of $u^{\omega'}$ in the expansion of $C_i$’s are equal for all $\omega' > \omega$. Let $\phi(u)$ be the (common) part of $\phi_i$’s consisting of all terms with the exponent of $u$ greater than $\omega$. Let $\xi$ be a new indeterminate and define $\delta^*_S : {\ensuremath{\mathbb{R}}}[x,y]\setminus\{0\} \to {\mathbb Q}$ as $$\begin{aligned}
\delta^*_S(f) := {{\rm deg}}_u(f(u,\phi(u) + \xi u^\omega)). \label{delta^*_S}\end{aligned}$$ Then
1. \[single-1\] $\bar \delta_S = \max\{0,\delta^*_S\}$.
2. \[single-2\] $\delta_S = \lceil \bar\delta_S \rceil$.
3. \[single-3\] For each $f \in {\ensuremath{\mathbb{R}}}[x,y]\setminus \{0\}$, $f/|u|^{\delta^*_S(f)}$ is bounded outside a compact set on $S$.
At first we claim that $\delta_S(u) = \bar \delta_S(u) = 1$. Indeed, since ${{\rm deg}}(u) = 1$, if the claim does not hold, then $\bar\delta_S(u) < 1$ and therefore there is a positive integer $d$ and a polynomial $h \in {\ensuremath{\mathbb{R}}}[u,v] = {\ensuremath{\mathbb{R}}}[x,y]$ with $e := {{\rm deg}}(h) < 2d$ such that $u^{2d} < h$ on $S$. But it is impossible, since $$\begin{aligned}
h|_{C_i} = h(u,v)|_{v = \phi_i(u)} = cu^e + {\text{l.o.t.}}\end{aligned}$$ for some $c\in r$, and therefore $h|_{C_i} < u^{2d}|_{C_i}$ for large enough $|u|$. This proves the claim.\
For each $t \in [0,1]$, let $\phi_t(u) := t\phi_1(u) + (1-t)\phi_2(u)$. Then for sufficiently large $|u|$, for each $t \in [0,1]$, $ v = \phi_t(u)$ defines a branch of real analytic curve $C_t$ in $S$ such that $\lim_{|u| \to \infty} C_t = P$. Now note that $$\begin{aligned}
\phi_t(u) = \phi(u) + (ta_1 + (1-t)a_2)u^\omega + {\text{l.o.t.}}= \phi(u) + u^\omega\psi_t(u),\end{aligned}$$ where $a_i$ is the coefficient of $u^\omega$ in $\phi_i$, and $\psi_t(u)$ is (a Puiseux series in $1/u$) of the form $\sum_{\omega' \leq 0} b_{\omega'}(t)u^{\omega'}$ with $b_0(t) = ta_1 + (1-t)a_2$. Let $f \in {\ensuremath{\mathbb{R}}}[u,v]$ and $d := {{\rm deg}}_u(f(u,\phi(u) + \xi u^\omega))$. Then $$\begin{aligned}
f(u,\phi(u) + \xi u^\omega) = f_0(\xi)u^d + {\text{l.o.t.}}\end{aligned}$$ where $f_0$ is a non-zero polynomial in $\xi$. It follows that $$\begin{aligned}
f|_{C_t} = f(u,\phi(u) + \xi u^\omega)|_{\xi = \psi_t(u)} = f_0(ta_1 + (1-t)a_2)u^d + {\text{l.o.t.}}\end{aligned}$$ We see that $f/|u|^d$ is bounded outside a compact set on $S$ and consequently $\bar \delta_S(f) \leq \max\{0,d\}$ and $\delta_S(f) \leq \max\{0,\lceil d \rceil\}$.\
On the other hand, if $d > 0$ and $h$ is a polynomial in ${\ensuremath{\mathbb{R}}}[u,v]$ with ${{\rm deg}}(h) < dk$ for some integer $k \geq 1$, then clearly $e := {{\rm deg}}_u(h(u, \phi(u) + \xi u^\omega)) \leq {{\rm deg}}(h) < dk$. Since $$\begin{aligned}
h|_{C_t} = h(u,\phi(u) + \xi u^\omega)|_{\xi = \psi_t(u)} = h_0(ta_1 + (1-t)a_2)u^e + {\text{l.o.t.}}\end{aligned}$$ for some polynomial $h_0 \in {\ensuremath{\mathbb{R}}}[\xi]$, it follows that $f^{2k}$ eventually becomes bigger on each $C_t$ than $h_0^2$. It follows that $\bar \delta_S(f) \geq d$, and consequently, $\bar \delta_S(f) = d$, as required to prove the first assertion of the lemma. This same argument with $k = 1$ in fact also proves the second assertion. The last assertion follows from the last sentence of the preceding paragraph.
\[deg-wise-puiseux\] $\phi(u) + \xi u^\omega$ from identity is called the [*generic degree-wise Puiseux series*]{} corresponding to $S$.
\[semi-remark\] Note that $\delta^*_S$ is a [*semidegree*]{} (in the terminology of [@sub1]), i.e. $\delta^*_S$ satisfies Property \[mtive\] of degree-like functions with an equality.
Assume \[single-assumption\] and \[regular-assumption\] hold. Then $\bar \delta_S$ is a [*semidegree*]{} iff $\bar \delta_S = \bar \delta^*_S$ iff $\bar \delta^*_S$ is non-negative on ${\ensuremath{\mathbb{R}}}[x,y]$.
The general regular case {#general-subsection}
------------------------
\[general-2-prop\] Let $S$ be a semialgebraic subset of ${\ensuremath{\mathbb{R}}}^2$ such that every tentacle of $S$ satisfies property \[regular-assumption\]. Then $$\begin{aligned}
\bar \delta_S = \max\{\bar \delta_T: T\ \text{is a tentacle of}\ S\}
= \max\left( \{0\} \cup \{\delta^*_T: T\ \text{is a tentacle of}\ S\} \right).\end{aligned}$$ \[bar-delta-presentation\]
It is clear that $\bar \delta_T \leq \bar \delta_S$ for every tentacle $T$ of $S$, which proves that LHS $\geq$ RHS in . The $\leq$ direction follows from the last assertion of Lemma \[single-tentacle-lemma\].
Let $S$ be as in Proposition \[general-2-prop\]. Then $\bar \delta_S$ is a [*subdegree*]{} (in the terminology of [@sub1]), i.e. $\bar \delta_S$ is the maximum of finitely many semidegrees.
Let $S$ be as in Proposition \[general-2-prop\]. Then
1. \[integral-1\] There exists a positive integer $N$ such that $\bar \delta_S(f) \in \frac{1}{N}{\ensuremath{\mathbb{Z}}}$ for all $f \in {\ensuremath{\mathbb{R}}}[x,y]\setminus\{0\}$.
2. \[integral-2\] Define $$\begin{aligned}
\bar {\mathcal B}_d(S) &:= \{f \in {\ensuremath{\mathbb{R}}}[x,y]: \bar \delta_S(f) \leq d\}, \\
\bar {\mathcal B}(S) &:= {\bigoplus}_{d \geq 0} \bar {\mathcal B}_d(S). \end{aligned}$$ Then ${\mathcal B}_d(S) = \bar {\mathcal B}_d(S)$ for all $d \geq 0$ and ${\mathcal B}(S) =\bar {\mathcal B}(S)$.
Assertion \[integral-1\] follows immediately from the Lemma \[single-tentacle-lemma\], Proposition \[general-2-prop\] and the observation that a semialgebraic set has only finitely many tentacles. Assertion \[integral-2\] follows from Proposition \[general-2-prop\] and Assertion \[single-2\] of Lemma \[single-tentacle-lemma\].
Semidegrees on ${\ensuremath{\mathbb{C}}}[x,y]$ and corresponding compactifications of ${\ensuremath{\mathbb{C}}}^2$
--------------------------------------------------------------------------------------------------------------------
### Background {#background-subsection}
Let $\delta$ be a semidegree (see Remark \[semi-remark\]) on ${\ensuremath{\mathbb{C}}}[x,y]$ defined as: $$\begin{aligned}
\delta(f) := {{\rm deg}}_x(f(x,\phi(x) + \xi x^\omega)). \label{delta-defn}\end{aligned}$$ where $\xi$ is an indeterminate, $\phi \in {\ensuremath{\mathbb{C}}}[x^{1/N}, x^{-1/N}]$ for some positive integer $N$ and $\omega \in {\mathbb Q}$, $\omega < \operatorname{ord}_x(\phi)$. Associated to $\delta$ there is a finite sequence of elements in ${\ensuremath{\mathbb{C}}}[x,x^{-1},y]$ called the [*key forms*]{} of $\delta$ (see [@contractibility Definition 3.17]). The sequence starts with $f_0 := x, f_1 := y$, and continues until there is an element in ${\ensuremath{\mathbb{C}}}[x,x^{-1},y]$ which can be expressed as a polynomial in the computed key forms and whose $\delta$-value is smaller than the ‘expected’ value. An algorithm and detailed example for the computation of key forms of $\delta$ from $\phi$ and $\omega$ appears in [@contractibility Section 3.3].
\
$\phi(x) + \xi x^\omega$ key forms
--------------------------------------------------------------------------------------------- ------------------------------------------------------------------
$\xi x^{p/q}$ $x,y$
$cx^\frac{p}{q} + \xi x^\omega$, $p,q$ rel. prime integers, $q > 0$, $\omega < \frac{p}{q}$ $x,y, y^q - c^qx^p$
$x^{5/2} + x^{-3/2} + \xi x^{-5/2}$ $x,y,y^2 - x^5, y^2 - x^5 - 2x$
$x^{5/2} + x^{-1} + x^{-3/2} + \xi x^{-5/2}$ $x,y,y^2 - x^5, y^2 - x^5 - 2x^{-1}y, y^2 - x^5 - 2x^{-1}y - 2x$
Given a (normal) algebraic variety $Y$ and a codimension one irreducible subvariety $V$ of $Y$, the [*order of pole*]{} along $V$ defines a semidegree on the field of rational functions on $Y$. Given a semidegree $\delta$ on ${\ensuremath{\mathbb{C}}}[x,y]$, the following proposition gives the construction of a compact algebraic variety containing ${\ensuremath{\mathbb{C}}}^2$ which ‘realizes’ $\delta$ (as the order of pole) along some curve.
\[compact-prop\] Let $\delta$ be defined as in . Assume that $\delta \neq {{\rm deg}}$. Pick the smallest positive integer $N$ such that $N\delta$ is integer-valued. Then there exists a unique compactification $\bar X$ of $X := {\ensuremath{\mathbb{C}}}^2$ such that
1. $\bar X$ is projective and normal.
2. $\bar X_\infty := \bar X \setminus X$ has two irreducible components $C_1,C_2$.
3. The semidegree on ${\ensuremath{\mathbb{C}}}[x,y]$ corresponding to $C_1$ and $C_2$ are respectively ${{\rm deg}}$ and $N\delta$.
Moreover, all singularities of $\bar X$ are [*rational*]{}.
\[positive-thm\] Let $\delta$ be a semidegree on ${\ensuremath{\mathbb{C}}}[x,y]$ and let $g_0, \ldots, g_{n+1}$ be the key forms of $\delta$ in $(x,y)$-coordinates. Then
1. \[non-positive-assertion\] $\delta$ is non-negative on ${\ensuremath{\mathbb{C}}}[x,y]\setminus {\ensuremath{\mathbb{C}}}$ iff $\delta(g_{n+1})$ is non-negative.
2. \[positive-assertion\] $\delta$ is positive on ${\ensuremath{\mathbb{C}}}[x,y]\setminus {\ensuremath{\mathbb{C}}}$ iff one of the following holds:
1. $\delta(g_{n+1})$ is positive,
2. \[almost-zero\] $\delta(g_{n+1}) = 0$ and $g_k \not\in {\ensuremath{\mathbb{C}}}[x,y]$ for some $k$, $0 \leq k \leq n+1$, or
3. \[almost-zero’\] $\delta(g_{n+1}) = 0$ and $g_{n+1} \not\in {\ensuremath{\mathbb{C}}}[x,y]$.
Moreover, conditions \[almost-zero\] and \[almost-zero’\] are equivalent.
To a semidegree $\delta$ on ${\ensuremath{\mathbb{C}}}[x,y]$ we associate a graded ring $$\begin{aligned}
{\ensuremath{\mathbb{C}}}[x,y]^\delta := {\bigoplus}_{d \geq 0} \{f: \delta(f) \leq d\}.\end{aligned}$$ In the case $\delta$ is realized (as in Proposition \[compact-prop\]) as the order of pole along a curve on a normal surface, ${\ensuremath{\mathbb{C}}}[x,y]^\delta$ can be interpreted as the graded ring of global sections of a divisor. The following results exploit this connection to study finiteness properties of ${\ensuremath{\mathbb{C}}}[x,y]^\delta$.
\[non-finite-delta\] Let $f_\delta$ be the last key form of $\delta$. Assume $\delta$ is positive on ${\ensuremath{\mathbb{C}}}[x,y]\setminus {\ensuremath{\mathbb{C}}}$ and $f_\delta$ is [*not*]{} a polynomial. Then ${\ensuremath{\mathbb{C}}}[x,y]^\delta$ is [*not*]{} finitely generated over ${\ensuremath{\mathbb{C}}}$. Moreover,
1. If $\delta(f_\delta) > 0$, then ${\ensuremath{\mathbb{C}}}[x,y]^\delta_d := \{f: \delta(f) \leq d\}$ is a finite dimensional vector space over ${\ensuremath{\mathbb{C}}}$ for all $d$.
2. If $\delta(f_\delta) = 0$, then there exists $d > 0$ such that ${\ensuremath{\mathbb{C}}}[x,y]^\delta_d$ is an infinite dimensional vector space over ${\ensuremath{\mathbb{C}}}$.
It follows from the assumptions that $\delta \neq {{\rm deg}}$. Let $\bar X$ be the compactification of $X := {\ensuremath{\mathbb{C}}}^2$ from Proposition \[compact-prop\]. Since $\delta$ is positive on ${\ensuremath{\mathbb{C}}}[x,y]\setminus {\ensuremath{\mathbb{C}}}$, Theorem \[positive-thm\] implies that $\delta(f_\delta) \geq 0$. At first assume $\delta(f_\delta) > 0$. It then follows by the same arguments as in the proof of [@contractibility Theorem 1.14] that ${\ensuremath{\mathbb{C}}}[x,y]^\delta$ is not finitely generated over ${\ensuremath{\mathbb{C}}}$ and that there exists a divisor $D$ on $\bar X$ such that $$\begin{aligned}
{\ensuremath{\mathbb{C}}}[x,y]^\delta = {\bigoplus}_{d \geq 0} H^0(\bar X, \mathcal{O}_{\bar X}(dD)).\end{aligned}$$ Since $H^0(\bar X, \mathcal{O}_{\bar X}(dD))$ is a finite dimensional vector space over ${\ensuremath{\mathbb{C}}}$ for each $d$, this proves the proposition for the case that $\delta(f_\delta) > 0$.\
Now assume $\delta(f_\delta) = 0$. It then follows from [@non-negative-valuation identity (11)] that $(C_1,C_1) = 0$. Since the singularities of $\bar X$ are rational, this implies that $C_1$ and $C_2$ are ${\mathbb Q}$-Cartier divisors. It follows that $D_k := kC_1 + C_2$ is a nef (${\mathbb Q}$-Cartier) divisor on $\bar X$ for all $k \gg 0$. Pick any positive integer $e$ such that $eC_i$ is a Cartier divisor for each $i$. Then $eD_k$ is an ample Cartier divisor on $\bar X$ for all $k \gg 0$. Let $\pi:\tilde X \to \bar X$ be a resolution of singularities of $\bar X$. Let $H$ be a fixed ample divisor and $K_{\tilde X}$ be the canonical divisor on $\tilde X$. W.l.o.g. we may assume that the supports of both $H$ and $K_X$ are contained in $\tilde X \setminus X$. Since $H + \pi^*(eD_k)$ is also ample for each $k$, it follows from a theorem of Reider (see e.g. [@lazarsfeld]) that $\tilde D_{k} := 3H + \pi^*(3eD_k) + K_{\tilde X}$ is base-point free for each $k$. Let $c_1$ (resp. $c_2$) be the coefficient of $C_1$ (resp. $C_2$) in $3H + K_{\tilde X}$. Since $\tilde D_{k}$ is base-point free, there exists $f_k \in {\ensuremath{\mathbb{C}}}[x,y]$ such that ${{\rm deg}}(f_k) = c_1 + 3ek$ and $\delta(f_k) \leq c_2 + 3e$. Let $d := c_2 + 3e$. It follows that the degree $d$ part of ${\ensuremath{\mathbb{C}}}[x,y]^\delta$ is infinite dimensional over ${\ensuremath{\mathbb{C}}}$, as required.
Let $f_\delta$ be the last key form of $\delta$. Assume one of the following conditions hold:
1. [*all*]{} key forms of $\delta$ are polynomials (equivalently, $f_\delta$ is a polynomial), or
2. $\delta(f_\delta) < 0$ (equivalently, there is $f \in {\ensuremath{\mathbb{C}}}[x,y]\setminus\{0\}$ such that $\delta(f) < 0$).
Then ${\ensuremath{\mathbb{C}}}[x,y]^\delta$ is finitely generated over ${\ensuremath{\mathbb{C}}}$.
Let the key forms of $\delta$ be $f_0 = x, f_1 = y, f_2, \ldots, f_l$. At first we assume that all $f_k$’s are polynomials. Pick a positive integer $N$ such that $N\delta$ is integer-valued. It suffices to show that ${\ensuremath{\mathbb{C}}}[x,y]^{N\delta}$ is finitely generated over ${\ensuremath{\mathbb{C}}}$, where $$\begin{aligned}
{\ensuremath{\mathbb{C}}}[x,y]^{N\delta} = {\bigoplus}_{d \geq 0} \{f: N\delta(f) \leq d\},\end{aligned}$$ (since ${\ensuremath{\mathbb{C}}}[x,y]^{N\delta}$ is integral over ${\ensuremath{\mathbb{C}}}[x,y]^{\delta}$). Let $e_j := N\delta(f_j)$, $1 \leq j \leq l$. Let us denote by $(f_j)_{e_j}$ the ‘copy’ of $f_j$ in the $e_j$-th graded component of ${\ensuremath{\mathbb{C}}}[x,y]^{N\delta}$. We now show that ${\ensuremath{\mathbb{C}}}[x,y]^{N\delta}$ is generated as a ${\ensuremath{\mathbb{C}}}$-algebra by $\{(1)_1\} \cup \{(f_j)_{e_j}: 0 \leq j \leq l\}$. Indeed, pick $f \in {\ensuremath{\mathbb{C}}}[x,y]$. Recall (from [@contractibility Proposition 3.28]) that for each $j \geq 1$, $f_j$ is monic in $y$ (as a polynomial in $y$ with coefficients in ${\ensuremath{\mathbb{C}}}[x]$) and ${{\rm deg}}_y(f_j)$ divides ${{\rm deg}}_y(f_{j+1})$ for $1 \leq j \leq l-1$. It follows that given an $f \in {\ensuremath{\mathbb{C}}}[x,y]$, $f$ has an expression of the form $$\begin{aligned}
f = \sum_{\alpha \in {\ensuremath{\mathbb{Z}}}_{\geq 0}^{l+2}} a_\alpha x^{\alpha_0}f_1^{\alpha_1} \cdots f_l^{\alpha_l},\end{aligned}$$ for $a_\alpha \in {\ensuremath{\mathbb{C}}}$ and $\alpha_j < {{\rm deg}}_y(f_{j+1})/{{\rm deg}}_y(f_j)$ for $1 \leq j \leq l-1$. [@maclane-key Theorem 16.1] then implies that $N\delta(f) = \max\{N\delta(x^{\alpha_0}f_1^{\alpha_1} \cdots f_l^{\alpha_l}): a_\alpha \neq 0\}$. It then immediately follows that $$\begin{aligned}
(f)_{N\delta(f)} = \sum_{\alpha \in {\ensuremath{\mathbb{Z}}}_{\geq 0}^{l+2}} a_\alpha \left((x)_{e_0}\right)^{\alpha_0}\left((f_1)_{e_1}\right)^{\alpha_1} \cdots \left((f_l)_{e_l}\right)^{\alpha_l} \left((1)_1\right)^{N\delta(f) - \sum \alpha_je_j},\end{aligned}$$ as required to show that ${\ensuremath{\mathbb{C}}}[x,y]^{N\delta}$ is finitely generated as an algebra over ${\ensuremath{\mathbb{C}}}$.\
Now assume that $\delta(f_l) < 0$. W.l.o.g. we may (and will) also assume that $f_l$ is [*not*]{} a polynomial. Define a map $\nu: {\ensuremath{\mathbb{C}}}[x,y]\setminus\{0\} \to {\ensuremath{\mathbb{Z}}}^2$ as follows: for every $g \in {\ensuremath{\mathbb{C}}}[x,y]\setminus\{0\}$, $g|_{y = \phi(x) + \xi x^\omega}$ is of the form $x^\alpha g_\alpha(\xi) + {\text{l.o.t.}}$ for some $\alpha \in {\mathbb Q}$ and $g_\alpha \in {\ensuremath{\mathbb{C}}}[\xi]$ (where ${\text{l.o.t.}}$ denotes ‘lower-order terms’ in $x$). Then set $\nu(g) := (N\alpha, {{\rm deg}}_\xi(g_\alpha))$.
\[nu-f-claim\] There exists $f \in {\ensuremath{\mathbb{C}}}[x,y]$ such that $\nu(f) = (0,k)$ for some positive integer $k$.
Indeed, [@non-negative-valuation Theorem 1.7] implies that there exists $h \in {\ensuremath{\mathbb{C}}}[x,y]$ such that $\delta(h) < 0$. Then $f := h^a x^b$ for suitable non-negative integers $a,b$ satisfies the claim.
Let $G_+ := \{\nu(g): g \in {\ensuremath{\mathbb{C}}}[x,y],\ \delta(g) \geq 0\}$. Then $G_+$ is a sub-semigroup of ${\ensuremath{\mathbb{Z}}}_{\geq 0}^2$. Moreover, since $\nu(x)$ is of the form $(k_1,0)$ and $\nu(f)$ (where $f$ is as in Claim \[nu-f-claim\]) is of the form $(0,k_2)$ for positive integers $k_1,k_2$, it follows that ${\ensuremath{\mathbb{Z}}}_{\geq 0}^2$ is [*integral*]{} over $G_+$, and therefore $G_+$ is a finitely generated semigroup. Pick $g_1, \ldots, g_k$ such that $\nu(g_j)$’s generate $G_+$. Let $d_j := \delta(g_j)$, $1 \leq j \leq k$. The proposition follows from the following claims.
$(g_j)_{d_j}$, $1 \leq j \leq k$, generated ${\ensuremath{\mathbb{C}}}[x,y]^\delta$ as an algebra over ${\ensuremath{\mathbb{C}}}[x,y]^\delta_0 := \{f: \delta(f) \leq 0\}$.
Let $\prec$ be the lexicographic order on ${\ensuremath{\mathbb{Z}}}_{\geq 0}^2$. The claim follows from the observation that if $\nu(g) = \sum \alpha_j\nu(g_j)$, then there exists (a unique) $c \in {\ensuremath{\mathbb{C}}}\setminus\{0\}$ such that $\nu(g - cg_1^{\alpha_1} \cdots g_k^{\alpha_k}) \prec \nu(g)$.
${\ensuremath{\mathbb{C}}}[x,y]^\delta_0$ is a finitely generated ${\ensuremath{\mathbb{C}}}$-algebra.
Note that $\delta \neq {{\rm deg}}$. Let $\bar X$ be the compactification of ${\ensuremath{\mathbb{C}}}^2$ from Proposition \[compact-prop\]. Since the singularities of $\bar X$ are rational, it follows that $C_1$ is a ${\mathbb Q}$-Cartier divisor. Since $\delta(f_l) < 0$, [@non-negative-valuation Proposition 2.10] implies that $(C_1,C_1) > 0$. Moreover, since $f_l$ is not a polynomial, [@non-negative-valuation Proposition 2.11] implies that every compact curve on $\bar X$ intersects $C_1$. It follows from the Nakai-Moishezon criterion that $C_1$ is an ample divisor and therefore $\bar X \setminus C_1$ is an affine variety. The claim follows from the observation that ${\ensuremath{\mathbb{C}}}[x,y]^\delta_0$ is precisely the ring of regular functions on $\bar X \setminus C_1$.
\[finite-cor\]
1. \[finite-generation\] ${\ensuremath{\mathbb{C}}}[x,y]^\delta$ is not a finitely generated algebra over ${\ensuremath{\mathbb{C}}}$ iff both of the following conditions holds:
1. $\delta$ is non-negative on ${\ensuremath{\mathbb{C}}}[x,y]$, and
2. there is a key form of $\delta$ which is not a polynomial (or equivalently, the last key form of $\delta$ is not a polynomial).
2. \[finite-d\] ${\ensuremath{\mathbb{C}}}[x,y]^\delta_d$ is a finite dimensional vector space over ${\ensuremath{\mathbb{R}}}$ for all $d \geq 0$ iff $\delta(f_\delta) > 0$, where $f_\delta$ is the last key form of $\delta$.
The only assertion which does not follow immediately from the preceding results is that ${\ensuremath{\mathbb{C}}}[x,y]^\delta_d$ is a finite dimensional vector space over ${\ensuremath{\mathbb{R}}}$ for all $d \geq 0$ in the case that $\delta(f_\delta) > 0$ and $f_\delta$ is a polynomial. But this follows by the same argument as in the proof of the first assertion of Proposition \[non-finite-delta\].
### Applications to the single tentacle case
Throughout this subsection $S$ is assumed to be a semi-algebraic subset of ${\ensuremath{\mathbb{R}}}^2$ which satisfies \[single-assumption\] and \[regular-assumption\]. Propositions \[semi-effective\]–\[single-finite\], which are immediate applications of the results from Sections \[single-subsection\] and \[background-subsection\], completely answers Questions \[one\]–\[four\] for $S$. Proposition \[single-moment\] partially solves the moment problem for such $S$.
\[semi-effective\] Let $f_S$ be the last key form of $\delta^*_S$. Then the following are equivalent:
1. $\bar \delta_S$ is a semidegree.
2. $\delta^*_S$ is non-negative on ${\ensuremath{\mathbb{R}}}[x,y]$.
3. $\delta^*_S(f_S) \geq 0$.
\[single-0\] Let $f_S$ be the last key form of $\delta^*_S$. Then ${\mathcal B}_0(S) = {\ensuremath{\mathbb{R}}}$ iff one of the following holds:
1. $\delta^*_S(f_S) > 0$, or
2. $\delta^*_S(f_S) = 0$ and $f_S$ is [*not*]{} a polynomial.
\[single-finite\] ${\mathcal B}(S)$ is not a finitely generated algebra over ${\mathcal B}_0(S)$ iff both of the following conditions holds:
1. $\delta^*_S$ is non-negative on ${\ensuremath{\mathbb{R}}}[x,y]$ (or equivalently, the $\delta^*_S$-value of the last key form of $\delta^*_S$ is non-negative), and
2. there is a key form of $\delta^*_S$ which is not a polynomial (or equivalently, the last key form of $\delta^*_S$ is not a polynomial).
Moreover, if ${\mathcal B}(S)$ is not a finitely generated algebra over ${\mathcal B}_0(S)$, then
1. ${\mathcal B}_0(S) = {\ensuremath{\mathbb{R}}}$.
2. Let $f_S$ be the last key form of $\delta^*_S$. Then
1. If $\delta^*_S(f_S) > 0$, then ${\mathcal B}_d(S)$ is a finite dimensional vector space over ${\ensuremath{\mathbb{R}}}$ for all $d$.
2. If $\delta^*_S(f_S) = 0$, then there exists $d > 0$ such that ${\mathcal B}_d(S)$ is an infinite dimensional vector space over ${\ensuremath{\mathbb{R}}}$.
\[single-moment\]
1. The moment problem for $S$ cannot be solved by finitely many polynomials in the following cases:
1. \[positively-non-solvable\] $\delta^*_S(f_S) > 0$, or
2. \[zero-pol-non-solvable\] $\delta^*_S(f_S) = 0$, $f_S$ is a polynomial, and the curve $f_S = \xi$ for generic $\xi$ for generic $\xi$ has genus $\geq 1$.
2. \[partial-solvable\] Assume one of the following conditions is satisfied:
1. \[negatively-solvable\] $\delta^*_S(f_S) < 0$, or
2. \[zero-pol-solvable\] $\delta^*_S(f_S) = 0$ and $f_S$ is a polynomial, and the curve $f_S = \xi$ for generic $\xi$ is rational.
Then there is a compact subset $V$ of $S$ such that the moment problem for $S\setminus V$ can be solved by finitely many polynomials.
The only case not covered by Proposition \[single-moment\] (for those $S$ which satisfy \[single-assumption\] and \[regular-assumption\]) occurs when $\delta^*_S(f_S) = 0$ and $f_S$ is [*not*]{} a polynomial. As explained at the end of Section \[count\], ‘sets of this kind seem to call for completely new methods.’
Assertion \[positively-non-solvable\] follows from Assertion \[finite-d\] of Corollary \[finite-cor\] and the discussion following Lemma \[four=five\]. Assertion \[zero-pol-non-solvable\] follows from [@posc Corollary 3.10] coupled with the following observations:
1. for all but finitely many $\xi$, the curve $C_\xi := \{f_S = \xi\}$ is smooth (by Bertini’s theorem),
2. $C_\xi$ has only one point at infinity (by [@contractibility Proposition 4.2]), and
3. there exists a non-empty open interval $I \subseteq {\ensuremath{\mathbb{R}}}$ such that for all $\xi \in {\ensuremath{\mathbb{R}}}$, the point at infinity of $C_\xi$ belongs to the closure (in ${\ensuremath{\mathbb{R}}}{\ensuremath{\mathbb{P}}}^2$) of $S\cap C_\xi$.
For Assertion \[partial-solvable\], choose linear coordinates $(u,v)$ on ${\ensuremath{\mathbb{R}}}^2$ such that $u \to \infty$ along the tentacle of $S$ and $(u,v)$ satisfy the hypotheses of Lemma \[single-tentacle-lemma\] (i.e. the point of intersection of the closure $\bar S$ of $S$ in ${\ensuremath{\mathbb{R}}}{\ensuremath{\mathbb{P}}}^2$ and the line at infinity is not on the closure of the line $u = 0$). At first assume we are in the situation of \[negatively-solvable\]. Then Theorem \[positive-thm\] implies that there is $f \in {\ensuremath{\mathbb{R}}}[u,v]$ such that $\delta^*_S(f) < 0$. Choose positive integers $a,b$ such that $h := u^a f^b$ satisfies $\delta^*_S(h) = 0$. For each $\xi \in {\ensuremath{\mathbb{R}}}$, let $C_\xi$ be the curve $h= \xi$ and for all $r \geq 0$, let $S_r := \{(u,v) \in S: u \geq r\}$. Then for sufficiently large $r$, we have that
1. $S_r$ is defined by $\{u \geq r, h_1 \geq 0, h_2 \geq 0\}$, where $h_1, h_2 \in {\ensuremath{\mathbb{R}}}[u,v]$ which ‘define the boundaries of the tentacle of $S_r$’,
2. $h$ is bounded on $S_r$,
3. $C_\xi$ has a real point at infinity (namely the point in the closure of the line $u=0$) which is not in the closure (in ${\ensuremath{\mathbb{R}}}{\ensuremath{\mathbb{P}}}^2$) of $S_r$,
4. $C_\xi \cap S_r$ does not intersect $\{h_i = 0\}$ for $i \in \{1,2\}$.
Then [@sch2 Theorem 1] and [@posc Theorem 3.11] imply that $\{u-r,h_1,h_2\}$ solves the moment problem on $S_r$.\
Now assume the hypotheses of \[zero-pol-solvable\] are satisfied. Since $C_\xi$ has only one point at infinity for all $\xi$ (by [@contractibility Proposition 4.2]), it follows from the Abhyankar-Moh-Suzuki theorem ([@amoh], [@suzuki]) that $f_S$ is a [*polynomial coordinate*]{} on ${\ensuremath{\mathbb{C}}}^2$, i.e. there exists a polynomial $g \in {\ensuremath{\mathbb{C}}}[u,v]$ such that ${\ensuremath{\mathbb{C}}}[f_S,g] = {\ensuremath{\mathbb{C}}}[u,v]$. It is then not hard to show using Jung’s theorem (see e.g. [@friedland-milnor]) on polynomial automorphisms of the plane that $f_S$ is in fact a polynomial coordinate on ${\ensuremath{\mathbb{R}}}^2$, i.e. there exists $g \in {\ensuremath{\mathbb{R}}}[u,v]$ such that ${\ensuremath{\mathbb{R}}}[f_S,g] = {\ensuremath{\mathbb{R}}}[u,v]$. W.l.o.g. we may assume that $g \to \infty$ along the tentacle of $S$. Let $S_r := \{(u,v) \in S: g(u,v) \geq r\}$. There exists $h_1, h_2 \in {\ensuremath{\mathbb{R}}}[u,v]$ such that $S_r = \{g \geq r, h_1 \geq 0, h_2 \geq 0\}$ for all sufficiently large $r$. [@sch2 Theorem 1] and [@kuma Theorem 2.2] then imply that $\{g-r,h_1,h_2\}$ solves the moment problem on $S_r$ for $r$ large enough. This completes the proof of Assertion \[partial-solvable\].
### Explicit construction of single tentacles {#construction}
In this subsection we give an algorithm to construct tentacles $S$ with desired behaviour of ${\mathcal B}(S)$. The algorithm consists of the construction of a sequence of elements $f_0, \ldots, f_l$, $l \geq 1$, in ${\ensuremath{\mathbb{R}}}[x,x^{-1},y]$ which would be the key forms of the corresponding semidegree $\delta^*_S$. It is a straightforward adaptation of Maclane’s construction of key polynomials in [@maclane-key].\
#### Initial step:
Set $f_0 := x$ and $f_1 := y$. Pick $\omega_1 \in {\mathbb Q}$ and set $\omega_0 := 1$.\
#### Inductive step:
Assume $f_j$’s and $\omega_j$’s have been constructed up to some $k \geq 1$. Let $p_k$ be the [*smallest*]{} positive integer such that $p_k\omega_k$ is in the additive group generated by $\omega_0, \ldots, \omega_{k-1}$. Then $p_k\omega_k$ can be [*uniquely*]{} expressed in the form $$\begin{aligned}
p_k\omega_k = \alpha_{k,0}\omega_0 + \alpha_{k,1} \omega_1 + \cdots + \alpha_{k,k-1}\omega_{k-1}\end{aligned}$$ where $\alpha_{k,j}$’s are integers such that $0 \leq \alpha_{k,j} < p_j$ for all $j \geq 1$ (note that there is no restriction on the range of $\alpha_{k,0}$). Pick a non-zero $c_k \in {\ensuremath{\mathbb{R}}}$ and set $$\begin{aligned}
f_{k+1} := f_k^{p_k} - c_k \prod_{j=0}^{k-1}f_j^{\alpha_{k,j}}.\end{aligned}$$ Set $\omega_{k+1}$ to be a rational number less than $p_k\omega_k$.\
#### Construction of $S$ from a finite sequence of $f_k$’s:
Assume $f_k$’s and $\omega_k$’s have been constructed up to some $l \geq 1$. Construct $p_l$ and $\alpha_{l,0}, \ldots, \alpha_{l,l-1}$ as in the inductive step. Define $$\begin{aligned}
f_{l+1,i} := f_l^{p_l} - c_{l,i} \prod_{k=0}^{l-1}f_k^{\alpha_{l,k}},\end{aligned}$$ where $c_{l,1}, c_{l,2}$ are distinct real numbers such that each $f_{l+1,i}$ defines a curve $C_i$ on ${\ensuremath{\mathbb{R}}}^2 \setminus y$-axis. Let $C^{\ensuremath{\mathbb{C}}}_i$ be the curve defined by $f_{l+1,i}$ in ${\ensuremath{\mathbb{C}}}^2 \setminus y$-axis. Then each $C^{\ensuremath{\mathbb{C}}}_i$ has a unique irreducible branch for which $|x| \to \infty$. It follows that either $C_i$ has a unique branch for which $x \to \infty$, or it has two such branches which come from the same irreducible branch of $C^{\ensuremath{\mathbb{C}}}_i$. In any event, there is a unique ‘top’ branch of $C_i$ for which $x \to \infty$; let us denote it by $C^{top}_i$. Pick $r > 0$ and let $S$ be the region to the right of $x = r$ and bounded by $C^{top}_1$ and $C^{top}_2$.\
The following is an immediate corollary of the results of the preceding subsection and the observations that $f_0, \ldots, f_l$ are precisely the key-forms of $\delta^*_S$ and $\omega_l = \delta^*_S(f_l)$. Note that the result does not change if we took $S$ to be the region bounded by the ‘bottom’ branches of $C_i$, or if we took corresponding branches of $C_i$ for which $x \to -\infty$.
\[example-cor\]
1. ${\mathcal B}_0(S) = {\ensuremath{\mathbb{R}}}$ iff one of the following is true:
1. $\omega_l > 0$, or
2. $\omega_l = 0$ and $f_l \not\in {\ensuremath{\mathbb{R}}}[x,y]$ (equivalently, $\omega_l = 0$ and $\alpha_{k,0} < 0$ for some $k$, $1 \leq k \leq l-1$).
2. ${\mathcal B}(S)$ is a finitely generated algebra over ${\ensuremath{\mathbb{R}}}$ iff one of the following conditions hold:
1. $\omega_l < 0$, or
2. $\omega_l \geq 0$ and $f_k \in {\ensuremath{\mathbb{R}}}[x,y]$ for $1 \leq k \leq l$ (equivalently, $\omega_l \geq 0$ and $\alpha_{k,0} \geq 0$ for $1 \leq k \leq l-1$).
3. If ${\mathcal B}(S)$ is not a finitely generated algebra over ${\ensuremath{\mathbb{R}}}$, then ${\mathcal B}_0(S) = {\ensuremath{\mathbb{R}}}$. Moreover, in this case
1. if $\omega_l > 0$, then ${\mathcal B}_d(S)$ is finite dimensional over ${\ensuremath{\mathbb{R}}}$ for all $d \geq 0$, and
2. if $\omega_l = 0$, then there exists $d > 0$ such that ${\mathcal B}_d(S)$ is an infinite dimensional vector space over ${\ensuremath{\mathbb{R}}}$.
\[exex\] Consider a sequence of key-forms starting with $x,y,y^2 - x^5$ (which corresponds to choices $\omega_1 := 5/2$ and $c_1 := 1$).
1. Take $\omega_2 := 1$, $l := 2$, $c_{2,1} :=0$ and $c_{2,2} := 1$. Let $S$ be the region defined by $x \geq 1$, $y \geq 0$, $x \geq y^2 - x^5 \geq 0$. Then ${\mathcal B}_0(S) = {\ensuremath{\mathbb{R}}}$ and ${\mathcal B}(S)$ is a a finitely generated algebra over ${\ensuremath{\mathbb{R}}}$.
2. Take $\omega_2 := 0$, $l := 2$, $c_{2,1} :=0$ and $c_{2,2} := 1$. Let $S$ be the region defined by $x \geq 1$, $y \geq 0$, $1 \geq y^2 - x^5 \geq 0$. Then ${\mathcal B}_0(S) \supsetneq {\ensuremath{\mathbb{R}}}$ and ${\mathcal B}(S)$ is a finitely generated algebra over ${\ensuremath{\mathbb{R}}}$.
3. Take $\omega_2 := 3/2$, $c_2 := 1$, $\omega_3 := 1$, $l := 3$, $c_{3,1} :=0$ and $c_{3,2} := 1$. Let $S$ be the region defined by $x \geq 1$, $y \geq 0$, $x \geq y^2 - x^5 - yx^{-1} \geq 0$. Then ${\mathcal B}_0(S) = {\ensuremath{\mathbb{R}}}$ and ${\mathcal B}(S)$ is not a finitely generated algebra over ${\ensuremath{\mathbb{R}}}$. But ${\mathcal B}_d(S)$ is finite dimensional over ${\ensuremath{\mathbb{R}}}$ for all $d$.
4. Take $\omega_2 := 3/2$, $c_2 := 1$, $\omega_3 := 0$, $l := 3$, $c_{3,1} :=0$ and $c_{3,2} := 1$. Let $S$ be the region defined by $x \geq 1$, $y \geq 0$, $1 \geq y^2 - x^5 - yx^{-1} \geq 0$. Then ${\mathcal B}_0(S) = {\ensuremath{\mathbb{R}}}$ and ${\mathcal B}(S)$ is not a finitely generated algebra over ${\ensuremath{\mathbb{R}}}$. Moreover, there exists $d > 0$ such that ${\mathcal B}_d(S)$ is infinite dimensional over ${\ensuremath{\mathbb{R}}}$.
### Two tentacles which behave like one
In this subsection we assume that
1. $\xi$ is an indeterminate,
2. $\phi(x) = \sum_{j=1}^k a_j x^{m_j/2} \in {\ensuremath{\mathbb{R}}}[x^{1/2}, x^{-1/2}]$ for some positive integer $k$ and integers $m_1 > m_2 > \cdots > m_k$ such that $\gcd(m_1, \ldots, m_k) = 1$, and
3. $\omega \in {\mathbb Q}$, $\omega < m_k/2 = \operatorname{ord}_x(\phi)$.
Define $$\begin{aligned}
\phi_1(x) &:= \sum_{j=1}^k a_j x^{m_j} + \xi x^{2\omega}, \\
\phi_2(x) &:= \sum_{j=1}^k (-1)^{m_j} a_j x^{m_j} + \xi x^{2\omega}.\end{aligned}$$
\[almost-single\] Let $\tilde S, S_1, S_2$ be semialgebraic subsets of ${\ensuremath{\mathbb{R}}}^2$ which satisfy assumptions \[single-assumption\] and \[regular-assumption\]. Assume that the generic degree-wise Puiseux series (see Definition \[deg-wise-puiseux\]) corresponding to $\tilde S, S_1,S_2$ is respectively $\phi(x) + \xi x^\omega$, $\phi_1(x) + \xi x^{2\omega}$, and $\phi_2(x) + \xi x^{2\omega}$. Set $S=S_1\cup S_2$. Then ${\mathcal B}(S)$ is integral over ${\mathcal B}(\tilde S)$. In particular, for each of the Questions \[one\]–\[four\], its answer is positive for $S$ iff it is positive for $\tilde S$.
Consider the map $(x,y) \mapsto (x^2,y)$. Then it is not hard to see that $\delta^*_{S_1}$ and $\delta^*_{S_2}$ extend $\delta^*_{\tilde S}$ via the pull-back by $f$. [@subalsection Lemma A.3] then shows that ${\mathcal B}(S)$ is the integral closure of $f^*{\mathcal B}(\tilde S)$. The proposition follows immediately.
The generic degree-wise Puiseux series corresponding to $S_1,S_2$ of Section \[count\] are respectively $$\begin{aligned}
\tilde \phi_1(x,\xi) &:= - x^3 + x^{-2} + \xi x^{-3},\\
\tilde \phi_2(x,\xi) &:= x^3 + x^{-2} + \xi x^{-3}.\end{aligned}$$ Let $\tilde S$ be a tentacle with generic degree-wise Puiseux series $\tilde \phi(x, \xi) := x^{3/2} + x^{-1} + \xi x^{-3/2}$; we may construct such $\tilde S$ using the procedure in Section \[construction\]; e.g. take $\tilde S$ to be the set defined by $x \geq 1,\ y \geq 0,\ 1 \geq y^2 - x^3 - 2yx^{-1} \geq 0$. Then it follows exactly as in the last case of Example \[exex\] that ${\mathcal B}_0(\tilde S) = {\ensuremath{\mathbb{R}}}$ and there exists $d > 0$ such ${\mathcal B}_d(\tilde S)$ is infinite dimensional over ${\ensuremath{\mathbb{R}}}$. Proposition \[almost-single\] therefore implies that the same is true for $S := S_1 \cup S_2$, as it was indeed shown in Section \[count\].
Bounded polynomials {#bound}
===================
In this last section we show how to interpret the algebra ${\mathcal B}(S)$ as the algebra ${\mathcal B}_0(S')$ of another set $S'$. In this way, we can produce examples and counterexamples to Question \[three\] from counterexamples to Question \[one\].
Let $S\subseteq {\mathbb R}^n$ be a set. Define $$S':=\left\{ (a,s)\in {\mathbb R}^{n+1}\mid a\in S, \Vert a\Vert \geq 1, \Vert a\Vert^2s^2\leq 1\right\}.$$ Then ${\mathcal B}(S) \cong {\mathcal B}_0(S')$ (via the identification $p\in {\mathcal B}_d(S) \leftrightarrow p\cdot t^d $, where $t$ is the last coordinate function on ${\ensuremath{\mathbb{R}}}^{n+1}$).
We can assume that $S$ contains only points with $\Vert a\Vert\geq 1.$ For “$\subseteq$” we start with $p\in {\mathcal B}_d(S)$. This implies $\vert p\vert \leq D\Vert x\Vert^d$ on $S$, for some large enough $D$. So for $(a,s)\in S'$ we have $$\vert p(a)s^d\vert \leq \vert p(a)\vert \frac{1}{\Vert a\Vert^d}\leq D.$$ So $p\cdot t^d\in {\mathcal B}_0(S'),$ which proves the claim. For “$\supseteq$” take $q=\sum_{i=0}^dp_i(x)t^i\in {\mathcal B}_0(S').$ There is some $C$ such that $$\left\vert\sum_{i=0}^d \frac{p_i(a)}{\Vert a\Vert^i}r^i \right\vert=\left\vert q\left(a,\frac{r}{\Vert a\Vert}\right)\right\vert\leq C$$ for all $a\in S, r\in[0,1].$ From Lemma \[help\] below it follows that there is some $D$ such that $$\left\vert\frac{p_i(a)}{\Vert a\Vert^i}\right\vert\leq D$$ for all $a\in S,$ and this implies $p_i\in{\mathcal B}_i(S),$ and thus $q\in {\mathcal B}(S).$
\[help\] If a univariate polynomial $p\in {\mathbb R}[t]$ fulfills $\vert p\vert \leq C$ on $[0,1]$, then the size of the coefficients of $p$ is bounded by a constant depending only on ${{\rm deg}}(p)$ and $C.$
Write $p=\sum_{i=0}^d p_it^i.$ We have $$-C\leq \sum_i p_i \left(\frac{1}{n}\right)^i\leq C$$ for all $n\in {\mathbb N}$. So the coefficient tuple $(p_0,\ldots,p_d)$ of $p$ lies in a polytope whose normal vectors are $\pm\left(1,\frac1n, \frac{1}{n^2},\ldots,\frac{1}{n^d}\right),$ for $n=1,\ldots,d+1$. Since these vectors are linearly independent, this polytope is compact.
So from examples where ${\mathcal B}(S)$ is not finitely generated, we obtain new examples for which ${\mathcal B}_0(S')$ is not finitely generated. For example, we get new regular examples in dimension three from the examples in Section \[count\] and Section \[plane\].
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Chapman Tripp appoints a special counsel and six new senior associates
Chapman Tripp is pleased to announce the appointment of a special counsel and six senior associates across its Auckland, Wellington and Christchurch offices. The appointments are effective from 1 June.
Special counsel
Geoff Carter has been appointed as a special counsel in the Christchurch office.
Geoff has been with the firm since 1999, and has worked in both the Wellington and Christchurch offices. Geoff is a very experienced and versatile commercial litigator, with particular expertise in employment and fisheries law.
Chapman Tripp managing partner Andrew Poole said: “The special counsel designation recognises specialised knowledge, expertise and experience and is used for only a very select group of our most senior lawyers. We are delighted with this recognition of Geoff’s standing in our Christchurch office and in the Christchurch market.”
Adrien Hunter – Corporate
Te Aopare Dewes – Corporate
Te Aopare specialises in corporate and commercial law, particularly structuring and general commercial advice for iwi entities, Māori businesses and private equity clients. She is from Ngāti Porou and Ngāti Rangitihi, is a fluent speaker of Te Reo Māori and is a member of our Māori Legal Group, Te Waka Ture.
Hamish Bolland – Construction and Major Projects
Hamish will join Chapman Tripp in July from Allen & Overy in Dubai. Hamish has worked on major construction projects, both within New Zealand and around the world, in a variety of sectors including infrastructure, energy, oil and gas, commercial and residential. He is very familiar with PPPs and other forms of procurement.
Wellington
Anna Kraack – Litigation
Anna specialises in civil litigation, with a focus on contract disputes, gas and energy, aquaculture, insurance and trusts.
Christchurch
Sarah Lester – Litigation
Sarah specialises in construction, engineering and insurance across all forms of dispute resolution, including mediation, arbitration and litigation.
Emily Whiteside – Litigation
Emily specialises in insurance-related and civil litigation. She advises insurer and insured clients on professional indemnity, statutory liability, public and products liability, and general liability matters. |
Grant to support U.S., China relations
The Bringing China to Arkansas Program recently received a $120,000 grant from the Freeman Foundation to support the groups’ 2012 programs.
The program is beginning its 12th year, providing development of international exchanges between Chinese schools for Arkansas teachers.
BCAP is part of the Arkansas Global Program, with the mission to “facilitate positive, productive, and peaceful international relations between the U.S. and other countries through global education and exchange focused on education, business, government, and the arts,” according to the Arkansas Global Programs section of the UALR website.
BCAP includes a teacher workshop, three-week study tour of China and a traveling culture exhibit among other activities. BCAP supports the mandate of the Arkansas State Department of Education by integrating reading and writing standards into development of a broader curriculum, according to the website.
“In learning about China’s culture and sharing what they’ve learned with others, Arkansas teachers are improving relations between countries, and through intercultural understanding, are learning important ways of improving human relations at home,” said Martha Morton, the director of Arkansas Global Programs at UALR.
Morton said BCAP came as an inspiration to her because she herself was a foreign language teacher for 14 years, but then decided she wanted her life to have more social and and educational impact on others.
“BCAP is important to me because it results in bringing people from different cultural and social backgrounds and ethnicities together for learning, understanding, cooperation, respect, and ongoing communication that fosters these friendly and positive relationships, both during and after their BCAP year. It is my small way of contributing to a more collaborative, peaceful world … [the] teachers and educators are special. Education is essential to any successful society and culture. It opens the mind and heart and spirit and body,” Morton said.
By the end of 2005, BCAP had directly impacted 14,768 students, 1,345 teachers, 119 schools, and 70 Arkansas communities through their efforts.
BCAP’s goals are to promote cultural understanding of China by giving the appropriate hands-on information to Arkansas teachers who share the freshly learned material with their students and community. They also strive to improve relations between the U.S. and China by sharing educational exchanges and partnerships, according to the BCAP website.
“The Chinese are about as different from Americans as you can get. Yet they are this country’s most important partner in the world, make no mistake about it,” Morton said.
Since 2001, Morton said she has brought in around $2 million to provide BCAP to teachers.
“I began by finding grant funding and have been fortunate enough to fund The Bringing China to Arkansas Program every year since 2001, probably because it embodies what I believe is most important for America’s future,” Morton said.
“These are teachers trying to do all this for themselves, their students, and their communities. I applaud them for their commitment, their willingness to take personal leave time to learn about China and the Chinese and travel around China for 3 weeks, and afterward share their learning and motivation with other Arkansans,” Morton said.
Faculty carrying firearms, it is not such a far-fetched idea anymore.
The Arkansas Legislature is in the process of considering a bill that would allow bill proposed would allow faculty of public colleges and universities the ability …
Aqa-via Allen-Smith, also known as Quay, is a business marketing major from Conway, Arkansas who was nominated by her sorority Delta Sigma Theta and the Delta Chi colony for UALR Homecoming Queen.
She enjoys working with …
The success of the House Of Cards has proved that in the ever-growing nest of worldwide phenomena there is room for everyone, even movies and TV shows in the political genre.
But years before there was …
Ice-cold water came pouring down on UALR Chancellor Joel Anderson on Wednesday August 20, after he agreed to take the ALS Ice Bucket Challenge in order to raise awareness and money.
Anderson’s friend Janet Jones, who … |
The serum level and urinary excretion of beta2-microglobulin in health and renal disease.
In healthy subjects with normal renal function beta2-microglobulin (beta2m) is constantly produced in the body. It is eliminated almost exclusively by the kidneys, predominantly by glomerular filtration but possibly also by some direct uptake from the blood. After glomerular filtration more than 99,9% of excreted protein is reabsorbed in the kidney tubules where it is catabolized. The main factor, influencing on the serum level of beta2m is the GFR. Determination of S-beta2m appears to be more effective than analysis of S-creatinine for the detection of a slightly reduced GFR. A relatively high S-beta2m, in comparison with the GFR, may be seen in e.g. malignant proliferative disorders and SLE. This indicates an increased production of the protein. An entirely free passage over the glomerular membranes is not likely for beta2m in healthy subjects but the sieving coefficient might approach 1,0 in renal disease. The increased glomerular elimination of the protein could then possibly be counterbalanced by an increased synthesis, which should explain the pronounced relationship at a log/log scale between S-beta2m and the GFR. An increased excretion of beta2m in the urine is a sensitive indicator of proximal tubular dysfunction in many clinical conditions. In marked renal insufficiency there is, however, an obligatory 100-1 000-fold increase of the normal excretion, not related to the kind of renal disorder. In studies of the protein precautions are necessary to avoid degradation of the protein, in urine with a low pH. |
The present invention relates to a new sunflower hybrid (Helianthus annuus L.) designated NHW11915. All publications cited in this application are herein incorporated by reference.
There are numerous steps in the development of any novel, desirable plant germplasm. Plant breeding begins with the analysis and definition of problems and weaknesses of the current germplasm, the establishment of program goals, and the definition of specific breeding objectives. The next step is selection of germplasm that possess the traits to meet the program goals. The goal is to combine in a single variety or hybrid an improved combination of desirable traits from the parental germplasm. These important traits may include increased head size and weight, higher seed yield, improved color, resistance to diseases and insects, tolerance to drought and heat, and better agronomic quality.
The cultivated sunflower (Helianthus annuus L.) is a major worldwide source of vegetable oil. In the United States, the major sunflower producing states are the Dakotas, Minnesota, Kansas, Colorado, Nebraska, Texas and California, although most states have some commercial acreage.
Sunflowers are considered oilseeds, along with cottonseed, soybeans and canola and the growth of sunflower as an oilseed crop has rivaled that of soybean. The oil accounts for 80% of the value of the sunflower crop, as contrasted with soybean which derives most of its value from the meal. Sunflower oil is generally considered a premium oil because of its light color, high level of unsaturated fatty acids, lack of linolenic acid, bland flavor and high smoke points. The primary fatty acids in the oil are oleic and linoleic with the remainder consisting of palmitic and stearic saturated fatty acids.
Non-dehulled or partly dehulled sunflower meal has been substituted successfully for soybean meal in isonitrogenous (equal protein) diets for ruminant animals, as well as for swine and poultry feeding. Sunflower meal is higher in fiber, has a lower energy value and is lower in lysine but higher in methionine than soybean meal. Protein percentage of sunflower meal ranges from 28% for non-dehulled seeds to 42% for completely dehulled seeds.
In addition to its use in food and food products for humans and animals, sunflower oil also has industrial uses. It has been used in paints, varnishes and plastics because of good semidrying properties without the color modification associated with oils high in linolenic acid. It has also been used in the manufacture of soaps, detergents and cosmetics. The use of sunflower oil (and other vegetable oils) as a pesticide carrier, and in the production agrichemicals, surfactants, adhesives, fabric softeners, lubricants and coatings has been explored. Considerable work has also been done to explore the potential of sunflower as an alternate fuel source in diesel engines because sunflower oil contains 93% of the energy of US Number 2 diesel fuel (octane rating of 37). Sunflower oil has also been proposed as a source of hydrogen for hydrogen fuel cells. (BBC News, Aug. 26, 2004).
Sunflower is an annual, erect, broadleaf plant with a strong taproot and a prolific lateral spread of surface roots. Stems are usually round early in the season, angular and woody later in the season, and normally unbranched. The sunflower head is not a single flower (as the name implies) but is made up of 1,000 to 2,000 individual flowers joined at a common receptacle. The flowers around the circumference are ligulate ray flowers without stamens or pistils; the remaining flowers are perfect flowers with stamens and pistils. Anthesis (pollen shedding) begins at the periphery and proceeds to the center of the head. Since many sunflower varieties have a degree of self-incompatibility, pollen movement between plants by insects is important, and bee colonies have generally increased yields.
At least 30 diseases, caused by various fungi, bacteria and viruses, have been identified on wild or cultivated sunflower, but only a few are of economic significance as far as causing yield losses. The sunflower diseases Phoma Black Stem (Phoma macdonaldii), Phomopsis Stem Canker (Phomopsis helianthi), Verticillium Leaf Mottle (Verticillium dahliae) and Sclerotinia (Scerotinia sclerotiorum) are some of the most significant disease problems in sunflower producing areas of China, United States and Europe.
The development of a cytoplasmic male-sterile and restorer system for sunflower has enabled seed companies to produce high-quality hybrid seed. Most of these have higher yields than open-pollinated varieties and are higher in percent oil. Performance of varieties tested over several environments is the best basis for selecting sunflower hybrids. The choice should consider yield, oil percentage, maturity, seed size (for non-oilseed markets), and lodging and disease resistance.
Therefore, it is desirable to develop new sunflower hybrids having resistance to significant sunflower diseases while also producing high seed yields with excellent seed characteristics.
The foregoing examples of the related art and limitations related therewith are intended to be illustrative and not exclusive. Other limitations of the related art will become apparent to those of skill in the art upon a reading of the specification. |
Most critical to China in entering this war,
this report continues, is the “grave”
national security threat it faces from both the Islamic State (ISIS/ISIL/Daesh)
and Turkey’s National Intelligence Organization (MIT)—and as, perhaps, best described by the
noted award winning American military-intelligence journalist Seymour M. Hersh who in
his latest article warned of this threat by stating:
[Ed. Note: Western governments and their
intelligence services actively campaign against the information found in these
reports so as not to alarm their citizens about the many catastrophic Earth
changes and events to come, a stance that the Sisters of Sorcha Faal
strongly disagrees with in believing that it is every human beings right to
know the truth.Due to our missions
conflicts with that of those governments, the responses of their ‘agents’ against
us has been a longstanding misinformation/misdirection campaign designed to
discredit and which is addressed in the report “Who Is Sorcha
Faal?”.] |
As many as 468 government flats and bungalows in Delhi are in illegal possession or are being misused, the government told the Lok Sabha on Wednesday. Among the residents are government employees and over 20 journalists.
Over 200 of these illegally occupied dwellings are Type-II flats comprising a bedroom and a living room, which are typically allotted to peons and clerks in the government. These flats are mainly in the neighbourhoods of Kasturba Nagar, M B Road, Srinivaspuri, Minto Road, Kalibari, Kidwai Nagar, Lodhi Road, Timarpur, R K Puram and Andrews Ganj.
Most of these allotments have been cancelled, and the occupants have been living illegally for between one and four years, the government said in reply to an unstarred question by Kandhamal MP Rudra Madhab Ray of the Biju Janata Dal. The bigger flats and bungalows are in Pandara Road, R K Puram, Shahjahan Road and New Moti Bagh, etc, and include Type-IV, V and VI accommodation.
Some of these are occupied by journalists. An occupant of a flat in Rabindra Nagar has been living illegally since 2001. The Ministry of Urban Development has been complaining of a major shortage of government housing in the city. The worst crunch is of Types-I, II and III accommodation, an official said.
The government also told the House that the largest number of complaints of subletting of general pool residential accommodation have been received from Delhi.
A total 820 such complaints were received in Delhi in 2013, as compared to 133 in Mumbai and nine in Chennai. There were fewer complaints in Delhi in 2013 than in 2012, when 1,565 complaints were received; in Mumbai, the number of complaints went up in 2013 from 82 in the previous year. In all cases, allotments have been cancelled and occupants debarred from future allotments. “In Delhi such action was taken in 97 cases, in Mumbai it was 60 while in Chennai it was nine in 2013,” says the government’s response.
“Penalties imposed on allottee have been made more stringent. In addition to the penalties imposed under the allotment rules, the concerned ministry of the delinquent allottee also initiated disciplinary action against them,” the government said.
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For all the latest India News, download Indian Express App. |
Q:
Include specific form from template Angular JS
I have an html template with 3 forms in it. I need to reuse the controller and two forms from this template in another page. How can I achieve this?
mainForm.htm
<form ng-show="activeTab == 'form1'" name="form1" id="form1">
<p> this is form1 </p>
</form>
<form ng-show="activeTab == 'form2'" name="form2" id="form2">
<p> this is form2 </p>
</form>
<form ng-show="activeTab == 'form3'" name="form3" id="form3">
<p> this is form3 </p>
</form>
subForm.htm
<div ng-include="mainForm.htm">
<p> need only form1 and form3 from mainForm.htm here</p>
</div>
A:
As I commented in your question, best way is to creating each form in separate view and ng-include as you need.
form1.html
<form method="POST" name="form1" id="form1">
<!-- your form code -->
</form>
(And repeat this example for each form)
mainForm.html
<div ng-show="activeTab == form1">
<div ng-include="form1.html"></div>
</div>
<!-- Each other form you want -->
|
Some people have a Zen garden, I have my Emacs configuration.
I am far from being an Emacs or Elisp expert, but I love investing time into fiddling with the editor, note-taker and task manager of my choice.
I created this documentation in order to become more structured in my approach - and also learn bit about literate programming. I decided to invest even more time into this and move all my setup into .org files. These files serve as source of the actual Elisp configuration files as well as for this documentation and even generate some fancy illustrations.
Of course, this is a work in progress. |
---
abstract: 'We present a for finite-state shared memory concurrent programs: all variables are shared between exactly two processes, and the guards on transitions are conjunctions of conditions over this pairwise shared state. This representation has been used to efficiently (in polynomial time) synthesize and model-check correctness properties of concurrent programs. Our main result is that any finite state concurrent program can be transformed into pairwise normal form. Specifically, if $Q$ is an arbitrary finite-state shared memory concurrent program, then there exists a finite-state shared memory concurrent program $P$ expressed in pairwise normal form such that $P$ is strongly bisimilar to $Q$. Our result is constructive: we give an algorithm for producing $P$, given $Q$.'
bibliography:
- 'BIBFILES/ABBREV.bib'
- 'BIBFILES/DIST.bib'
- 'BIBFILES/IOAUT.bib'
- 'BIBFILES/MODEL.bib'
- 'BIBFILES/SYNTH.bib'
- 'BIBFILES/LOGIC.bib'
- 'BIBFILES/OS.bib'
---
**Finite-state concurrent programs can be expressed pairwise**\
[Paul C. Attie]{}\
0.05in [Department of Computer Science]{}\
[American University of Beirut]{}\
[and]{}\
[Center for Advanced Mathematical Sciences]{}\
[American University of Beirut]{}\
`paul.attie@aub.edu.lb`
Introduction {#sec:intro}
============
The is recognized as a fundamental impediment to the widespread application of mechanical finite-state verification and synthesis methods, in particular, model-checking. The problem is particularly severe when considering finite-state concurrent programs, as the individual processes making up such programs may be quite different (no similarity) and may be only loosely coupled (leading to a large number of global states).
In previous work [@Att99a; @AE98; @AC04a], we have suggested a method of avoiding state-explosion by expressing the synchronization and communication code for each pair of interacting processes separately from that for other (even intersecting) pairs. In particular, all shared variables are shared by exactly one pair of processes. This “pairwise normal form” enables us, for any arbitrarily large concurrent program, to model-check correctness properties for the concurrent compositions of small numbers of processes (so far 2 or 3) and then conclude that these properties also hold in the large program. If $P$ is a concurrent program consisting of $K$ processes each having $O(N)$ local states, then we can verify the deadlock freedom of $P$ in $O(K^3 N^3 b)$ time[^1] or $O(K^4 N^4)$ time, using either of two conservative tests [@AC04a], and we can verify safety and liveness properties of $P$ in $O(K^2 N^2)$ time [@Att99a; @AE98].
A key question regarding the pairwise approach is: does it give up expressive power? That is, in requiring synchronization and communication code to be expressed pairwise, do we constrain the set of concurrent programs that can be represented? In this paper, we answer this question in the negative: we show that for any concurrent program $Q$, we can (constructively) produce a concurrent program $P$ that is in pairwise normal form, and that is strongly bisimilar to $Q$.
The rest of the paper is as follows. Section \[sec:prelims\] presents our model of concurrent computation and defines the global state transition diagram of a concurrnt program. Section \[sec:pairwise\] defines pairwise normal form. Section \[sec:expressiveness\] presents our main result: any finite-state concurrent program can be expressed in pairwise normal form. Section \[sec:related\] discusses related work, and Section \[sec:conc\] concludes.
Technical Preliminaries {#sec:prelims}
=======================
Model of concurrent computation {#sec:model}
-------------------------------
We consider finite-state shared memory concurrent programs of the form $P = P_1 \| \cdots \| P_K$ that consist of a finite number $n$ of fixed sequential processes $P_1, \ldots, P_K$ running in parallel. Each $P_i$ is a [@EC82], that is, a directed multigraph where each node is a (local) state of $P_i$ (also called an and is labeled by a unique name ($s_i$), and where each arc is labeled with a guarded command [@Dij76] $B_i \ar A_i$ consisting of a guard $B_i$ and corresponding action $A_i$. Each node must have at least one outgoing arc, i.e., a skeleton contains no “dead ends.” With each $P_i$, we associate a set $\AP_i$ of , and a mapping $V_i$ from local states of $P_i$ to subsets of $\AP_i$: $V_i(s_i)$ is the set of atomic propositions that are true in $s_i$. As $P_i$ executes transitions and changes its local state, the atomic propositions in $\AP_i$ are updated. Different local states of $P_i$ have different truth assignments: $V_i(s_i) \ne V_i(t_i)$ for $s_i \ne t_i$. Atomic propositions are not shared: $\AP_i \ints \AP_j = \emptyset$ when $i \ne j$. Other processes can read (via guards) but not update the atomic propositions in $\AP_i$. We define the set of all atomic propositions $\AP = \AP_1 \un \cdots \un \AP_K$. There is also a set $\SH = \{x_1,\ldots,x_m\}$ of shared variables, which can be read and written by every process. These are updated by the action $A_i$. A [*global state*]{} is a tuple of the form $(s_1, \ldots, s_K,v_1,\ldots, v_m)$ where $s_i$ is the current local state of $P_i$ and $v_1,\ldots,v_m$ is a list giving the current values of $x_1,\ldots,x_m$, respectively. A guard $B_i$ is a predicate on global states, and so can reference any atomic proposition and any shared variable. An action $A_i$ is any piece of terminating pseudocode that updates the shared variables.[^2] We write just $A_i$ for $\ltrue \ar A_i$ and just $B_i$ for $B_i \ar skip$, where $skip$ is the empty assignment.
We model parallelism as usual by the nondeterministic interleaving of the “atomic" transitions of the individual processes $P_i$. Let $s = (s_1, \ldots , s_i, \ldots, s_K, v_1,\ldots, v_m)$ be the current global state, and let $P_i$ contain an arc from node $s_i$ to $s'_i$ labeled with $B_i \ar A_i$. We write such an arc as the tuple $(s_i, B_i \ar A_i, s'_i)$, and call it a $P_i$- from $s_i$ to $s'_i$. We use just when $P_i$ is specified by the context. If $B_i$ holds in $s$, then a permissible next state is $s' = ( s_1, \ldots, s'_i, \ldots, s_K, v'_1,\ldots, v'_m )$ where $v'_1, \ldots, v'_m$ are the new values for the shared variables resulting from action $A_i$. Thus, at each step of the computation, a process with an enabled arc is nondeterministically selected to be executed next. The $R$ is the set of all such $(s,i,s')$. The arc from node $s_i$ to $s'_i$ is [*enabled*]{} in state $s$. An arc that is not enabled is [*blocked*]{}. Our model of computation is a high-atomicity model, since a process $P_i$ can evaluate the guard $B_i$, execute the action $A_i$, and change its local state, all in one action.
Recall that we define a global state to be a tuple of local states and shared variable values, rather than a “name” together with a labeling function $L$ that gives the associated valuation, A consequence of this definition is that two different global states must differ in either some local state or some shared variable value. Since we require different local states to differ in at least one atomic proposition value, we conclude that two different global states differ in at least one atomic proposition value or one shared variable value.
We define the valuation corresponding to a global state $s = (s_1, \ldots , s_i, \ldots, s_K,$\
$v_1,\ldots, v_m)$ as follows. For an atomic proposition $p_i \in \AP_i$: $s(p_i) = \ltrue$ if $p_i \in V_i(s_i)$, and $s(p_i) = \lfalse$ if $p_i \not\in V_i(s_i)$. For a shared variable $x_\l$, $1 \le \l \le m$: $s(x_\l) = v_\l$. We define $s \pj \AP$ to be the set $\{ p \in \AP ~|~ s(p) = \ltrue\}$ i.e., the set of propositions that are true in state $s$. $s \pj \AP$ is essentially the projection of $s$ onto the atomic propositions. Also, $s \pj i$ is defined to be $s_i$, i.e., the local state of $P_i$ in $s$. We also define $s \pj \SH$ to be the set $\{ \tpl{p, s(x)} ~|~ x \in \SH
\}$, i.e., the set of all pairs consisting of a shared variable $x$ in $\SH$ together with the value that $s$ assigns to $x$.
Let $St$ be a given set of initial states in which computations of $P$ can start. A *computation path* is a sequence of states whose first state is in $St$ and where each successive pair of states is related by $R$. A state is iff it lies on some computation path. Since we must specify the start states $St$ in order for the computation paths to be well-defined, we re-define our notion of a program to be $P = (St, P_1 \pl \cdots \pl P_K)$, i.e., a program consists of the parallel composition of $K$ processes, together with a set $St$ of initial states.
For technical convenience, and without loss of generality, we assume that no synchronization skeleton contains a node with a self-loop. The functionality of a self-loop (e.g., a busy wait) can always be achieved by using a loop containing two local states. Thus, a transition by $P_i$ changes the local state of $P_i$, and therefore the value of at least one atomic proposition in $\AP_i$. Hence, no global state $s$ has a self loop, i.e., a transition by some $P_i$ both starting and finishing in $s$.
For a local state $s_i$, define $\stof{s_i}$ as follows:
\[def:stof\] $$\stof{s_i} = ``(\AND_{p \in V_i(s_i)} p) \;\ \land\;\
(\AND_{p \not\in V_i(s_i)} \neg p)\mbox{\rm ''}
%%% \;\ \land\;\ (\AND_{x \in \SH} x = s(x))\mbox{\rm ''}$$
$\stof{s_i}$ converts a local state $s_i$ into a propositional formula over $\AP_i$.
If $s$ is a global state and $B$ is a guard, we define $s(B)$ by the usual inductive scheme: $s($“$x = c$"$) = \ltrue$ iff $s(x) = c$, $s(B1 \land B2) = \ltrue$ iff $s(B1) = \ltrue$ and $s(B2) = \ltrue$, $s(\neg B1) = \ltrue$ iff $s(B1) = \lfalse$. If $s(B) = \ltrue$, we also write $s \sat B$.
The Global State Transition Diagram of a Concurrent Program {#subsec:gstd}
-----------------------------------------------------------
\[def:gstd\] Given a concurrent program $P = P_1 \| \cdots \| P_K$ and a set $St$ of initial global states for $P$, the is a Kripke structure $M = (St, S, R)$ given as follows: (1) $S$ is the smallest set of global states satisfying (1.1) $St \sub S$ and (1.2) if there exist $s \in S, i \in \onetok$[^3], and $u$ such that $(s,i,u)$ is in the next-state relation defined above in Section \[sec:model\], then $u \in S$, and (2) $R$ is the next-state relation restricted to $S$.
We define strong bisimulation in the standard way.
\[def:bisim\] Let $M = (St, S, R)$ and $M' = (St', S', R')$ be two Kripke structures with the same underlying set $\AP$ of atomic propositions. A relation $B \sub S \times S'$ is a iff:
\[def:bisim:ap\] if $B(s,s')$ then $s \pj \AP = s' \pj \AP$
\[def:bisim:left-trans\] if $B(s,s')$ and $(s,i,u) \in R$ then $\ex u': (s',i,u') \in R' \land B(u,u')$
\[def:bisim:right-trans\] if $B(s,s')$ and $(s',i,u') \in R$ then $\ex u: (s,i,u) \in R \land B(u,u')$
We also define $\sim$ to be the union of all strong bisimulation relations:\
$\sim \;=\; \UN \{B : B \mbox{ is a strong bisimulation}\}$.
We say that $M$ and $M'$ are , and write $M \sim M'$, if and only if there exists a strong bisimulation $B$ such that $\fa s \in St, \ex s' \in St': B(s's')$ and $\fa s' \in St', \ex s \in St: B(s's')$.
Pairwise normal form {#sec:pairwise}
====================
Let $\gd, \gc$ be binary infix operators. A [@AE98] is either a guarded command as given in Section \[sec:model\] above, or has the form $G_1 \gd G_2$ or $G_1 \gc G_2$, where $G_1$, $G_2$ are general guarded commands. Roughly, the operational semantics of $G_1 \gd G_2$ is that either $G_1$ or $G_2$, but not both, can be executed, and the operational semantics of $G_1 \gc G_2$ is that both $G_1$ or $G_2$ must be executed, that is, the guards of both $G_1$ and $G_2$ must hold at the same time, and the bodies of $G_1$ and $G_2$ must be executed simultaneously, as a single parallel assignment statement. For the semantics of $G_1 \gc G_2$ to be well-defined, there must be no conflicting assignments to shared variables in $G_1$ and $G_2$. This will always be the case for the programs we consider. We refer the reader to [@AE98] for a comprehensive presentation of general guarded commands.
\[def:pairwise\] A concurrent program $P = P_1 \| \cdots \| P_K$ is in iff the following four conditions all hold:
\[def:pairwise:arc\] every arc $a_i$ of every process $P_i$ has the form\
$a_i = (s_i,
\gc_{j \in I(i)} \gd_{\l \in \{1,\ldots,n_j\}} \CB{i}{j}{\l} \ar \CA{i}{j}{\l},
t_i)$, where $\CB{i}{j}{\l} \ar \CA{i}{j}{\l}$ is a guarded command, $I$ is an irreflexive symmetric relation over $\onetok$ that defines a “interconnection” (or “neighbors”) relation amongst processes, and $I(i) = \{j ~|~ (i,j) \in I\}$,
\[def:pairwise:shvar\] variables are shared in a pairwise manner, i.e., for each $(i,j) \in I$, there is some set $\SH_{ij}$ of shared variables that are the only variables that can be read and written by both $P_i$ and $P_j$,
\[def:pairwise:guard\] $\CB{i}{j}{\l}$ can reference only variables in $\SH_{ij}$ and atomic propositions in $\AP_j$, and
\[def:pairwise:assig\] $\CA{i}{j}{\l}$ can update only variables in $\SH_{ij}$.
For each neighbor $P_j$ of $P_i$, $\gd_{\l \in [1:n]} \CB{i}{j}{\l} \ar \CA{i}{j}{\l}$ specifies $n$ alternatives $\CB{i}{j}{\l} \ar \CA{i}{j}{\l}$, $1 \le \l \le n$ for the interaction between $P_i$ and $P_j$ as $P_i$ transitions from $s_i$ to $t_i$. $P_i$ must execute such an interaction with each of its neighbors in order to transition from $s_i$ to $t_i$. We emphasize that $I$ is not necessarily the set of all pairs, i.e., there can be processes that do not directly interact by reading each others atomic propositions or reading/writing pairwise shared variables. We do not assume, unless otherwise stated, that processes are isomorphic, or “similar.”
We use a superscript $I$ to indicate the relation $I$, e.g., process $P_i^I$, and $P_I^i$-arc $\mv{i}{I}$. We define $\mv{i}{I}.start = s_i$, $\mv{i}{I}.guard_j = \bigor_{\l \in \{1,\ldots,n_j\}} \CB{i}{j}{\l}$, and $\mv{i}{I}.guard = \ba_{j \in I(i)} a_i.guard_j$. If $P^I = P_{1}^I \pl \ldots \pl P_{K}^I$ is a concurrent program with interconnection relation $I$, then we call $P^I$ an . For the special case when $I = \{ (i,j) ~|~ i,j \in \onetok, i \ne j\}$, i.e., $I$ is the complete interconnection relation, we omit the superscript $I$.
In pairwise normal form, the synchronization code for $\PP{i}{I}$ with one of its neighbors $\PP{j}{I}$ (i.e., $\gd_{\l \in \{1,\ldots,n_j\}} \CB{i}{j}{\l} \ar \CA{i}{j}{\l}$) is expressed separately from the synchronization code for $\PP{i}{I}$ with another neighbor $P_k^I$ (i.e., $\gd_{\l \in \{1,\ldots,n_k\}} \CB{i}{k}{\l} \ar \CA{i}{k}{\l}$) We can exploit this property to define “subsystems” of an $I$-system $P$ as follows. Let $J \sub I$ and $range(J) = \{i ~|~ \ex j: (i,j) \in J\}$. If $\mv{i}{I}$ is a arc of $\PP{i}{I}$ then define $\mv{i}{J} = (s_i,
\gc_{j \in J(i)} \gd_{\l \in \oneton} \CB{i}{j}{\l} \ar \CA{i}{j}{\l},
t_i)$. Then the $P^J$ is $P_{j_1}^J \pl \ldots \pl P_{j_n}^J$ where $\{j_1,\ldots,j_n\} = range(J)$ and $\PP{j}{J}$ consists of the arcs $\{ \mv{i}{J} ~|~ \mbox{ $\mv{I}{I}$ is a arc of $\PP{j}{I}$}\}$. Intuitively, a $J$-system consists of the processes in $range(J)$, where each process contains only the synchronization code needed for its $J$-neighbors, rather than its $I$-neighbors. If $J = \{ \{i,j\}\}$ for some $i,j$ then $P_J$ is a , and if $J = \{ \{i,j\}, \{j,k\} \}$ for some $i,j,k$ then $P_J$ is a . For $J \sub I$, $M_J = (St_J,S_J,R_J)$ is the GSTD of $P^J$ as defined in Section \[sec:model\], and a global state of $P^J$ is a . If $J = \{ \{i,j\}\}$, then we write $M_{ij} = (St_{ij},S_{ij},R_{ij})$ instead of $M_J = (St_J,S_J,R_J)$.
In [@Att99a; @AE98; @Att03a] we give, in pairwise normal form, solutions to many well-known problems, such as dining philosophers, drinking philosophers, mutual exclusion, $k$-out-of-$n$ mutual exclusion, two-phase commit, and replicated data servers. We conjecture that any finite-state concurrent program can be rewritten (up to strong bisimilation) in pairwise normal form. The restriction to pairwise normal form enables us to mechanically verify certain correctness properties very efficiently. Recall that $K$ is the number of processes, $b$ is the maximum branching in the local state transition relation of a single process, and $N$ is the size of the largest process. Then, safety and liveness properties that can be expressed over pairs of processes can be verified in time $O(K^2 N^2)$ by model-checking pair-systems, [@Att99a; @AE98], and deadlock-freedom can be verified in time in $O(K^3 N^3 b)$ or $O(K^4
N^4)$ using either of two conservative tests [@AC04a], which in turn operate by model checking triple-systems. Exhaustive state-space enumeration would of course require $O(N^K)$ time.
The Pairwise Expressiveness Result {#sec:expressiveness}
==================================
Let $Q = (St_Q, Q_1 \| \cdots \| Q_K)$ be an arbitrary finite-state shared memory concurrent program as defined in Section \[sec:model\] above, with each process $Q_i$ having an associated set $\AP_i$ of atomic propositions and with shared variables $x_1,\ldots,x_m$. The transformation of $Q$ to pairwise normal form proceeds in three phases, as given in the sequel.
Phase One {#sec:phase-one}
---------
First, we generate $M_Q$, the GSTD of $Q$, as given by Definition \[def:gstd\]. By construction of Definition \[def:gstd\], all states in $M_Q$ are reachable. We then execute the algorithm given in Figure \[fig:unique-in\] on $M_Q$ which transforms $M_Q$ intro a Kripke structure $M'_Q = (St'_Q, S'_Q, R'_Q)$ which is bisimilar to $M_Q$ and which has the property that all incoming transitions into a state are labeled with the same process index. This is not strictly necessary, but significantly simplifies the transformation to pairwise normal form.
Define $\inprocs(s) = \{i \in \onetok ~|~ \ex s' : (s',i,s) \in R_Q \}$. We also introduce a new shared variable $in$ whose value in a state $s$ will be the process index that labels the transitions incoming into $s$.
aaāaaāaāaaāaaāa= $\trans(M_Q,M'_Q)$\
$St'_Q := St_Q$; $S'_Q := S_Q$; $R'_Q := R_Q$;\
until there is no change in $M'_Q$\
let $s$ be a state in $M'_Q$ such that $|\inprocs(s)| > 1$;\
$i \in \inprocs(s)$\
create a new marked state $s^i$ such that $s^i \up \AP = s \up \AP$, $s^i \up \SH = s \up \SH$\
$s \in St_Q$ $\setinsert{St'_Q}{s^i}$ ;\
$\setinsert{S'_Q}{s^i}$;\
$j,u : (s,j,u) \in R_Q$ $\setinsert{R'_Q}{(s^i,j,u)}$ ;\
$u: (u,i,s) \in R_Q$ $\setinsert{R'_Q}{(u,i,s^i)}$ ;\
$\setdelete{St'_Q}{s}$;\
$\setdelete{S'_Q}{s}$;\
remove all transitions incident on $s$ from $R'_Q$\
\
\[prop:unique-term\] Procedure $\trans$ terminates. Each iteration of the loop (line 2) reduces the number of states $s$ such that $|\inprocs(s)| > 1$ by one. Since $M'_Q$ is initially set to $M_Q$, which is finite, this cannot go on forever.
\[prop:unique-bisim\] $M'_Q \sim M_Q$ is a loop invariant of the loop (line 2) of $\trans$.
*Proof of Proposition \[prop:unique-bisim\]*. Let $n_0$ be the number of iterations that the loop executes. Let $M^n = (St^n, S^n, R^n)$ be the value of $M'_Q$ at the end of the $n$’th iteration, (for all $n \le n_0$) with $M^0$ being the initial value $M_Q$. We will also use the superscript $n$ for states in $M^n$, when needed. We show that $\fa n: 0 < n \le n_0: M^{n-1} \sim M^{n}$.
Consider the $n$’th iteration of the $\REPEAT$ loop. In this iteration, $M^n$ results from $M^{n-1}$ by deleting some state $s$ and adding some states $s^{i_1} \ldots s^{i_\l}$, where $\set{i_1,\ldots i_\l} = \inprocs(s)$. Since each of $s^{i_1} \ldots s^{i_\l}$ have the same successor states as $s$, and agree with $s$ on the values of all atomic propositions, we have $s \sim s^{i_1}, \ldots, s \sim s^{i_\l}$. Let $u$ be an arbitrary predecessor of $s$ in $M^{n-1}$, i.e., $(u^{n-1},j,s) \in R^{n-1}$, where $u^{n-1}$ indicates the occurrence of $u$ in $M^{n-1}$. At the end of the iteration, we have $(u^n,j,s^j) \in R^n$. Since $s \sim s^j$, we have $u^{n-1} \sim u^n$, i.e., the occurrence of $u$ in $M^{n-1}$ is bisimilar to the occurrence of $u$ in $M^n$. Since all other states in $M^{n-1}$ and $M^n$ have an unchanged set of successors, we conclude that $M^{n-1} \sim M^n$.
By a straightforward induction on $n$, and using the transitivity of $\sim$, we can show that $\fa n: 0 < n \le n_0: M^0 \sim M^{n}$. Thus $M^0 = M^{n_0}$. Now $M_Q = M^0$ and $M'_Q = M^{n_0}$, and the proposition is established.
\[prop:unique-correct\] Upon termination of procedure $\trans$,\
(1) $M'_Q \sim M_Q$, and\
(2) every state $s$ in $M'_Q$ satisfies $|\inprocs(s)| \le 1$. (1) follows from Proposition \[prop:unique-bisim\]. (2) follows immediately fom inspecting line 2 of procedure $\trans$.
For all $s \in S'_Q$ such that $|\inprocs(s)| = 1$, define $\inproc(s)$ to be the unique $i$ such that $\ex s' : (s',i,s) \in R'_Q$.
\[prop:unique-global\] Upon termination of procedure $\trans$, for any two states $s,u$ in $M'_Q$, $s \pj \AP \ne u \pj \AP$ or $s \pj \SH \ne u \pj \SH$ or $in(s) \ne in (u)$. Immediate by construction of procedure $\trans$.
Phase Two {#sec:phase-two}
---------
We exploit the unique incoming process index property of $M'_Q$ to extract a program $P = (St_P, P_1 \| \cdots \| P_K)$ from $M'_Q$ such that $P$ is bisimilar to $Q = (St_Q, Q_1 \| \cdots \| Q_K)$ and $P$ is in pairwise normal form. The interconnection relation $I$ for $P$ is the complete relation, and so we omit the superscripts $I$ on $P$ and $P_i$. $P$ operates by emulating the execution of $Q$. In the sequel, let $i,j,k$ implicitly range over $\onetok$, with possible further restriction, e.g., $i \ne j$. With each process $P_i$ we associate the following state variables, with the indicated access permissions and purpose
. These are written by $P_i$ and read by all processes. For each process $P_i$, these enable $P_i$ to emulate the local state of $Q_i$, which is defined by the same set $\AP_i$ of atomic propositions.
. These are written by $P_i$ and read by $P_j$. These enable $P_i$ to emulate the updates that $Q_i$ makes to $x$. When $P_i$ is the last process to have executed, any other process $P_j$ will read $x_{ij}^i$ to find the correct emulated value of $x$, since this value will have been computed by $P_i$ and stored in $x_{ij}^i$ for all $j \in \onetok$. For technical convenience, we admit $x_{ii}^i$. We select some $\l \in \onetok - \{i\}$ arbitrarily and define $x_{ii}^i$ to be shared pairwise between $P_i$ and $P_\l$. This is needed to conform technically to Definition \[def:pairwise\]. $P_\l$ will not actually reference $x_{ii}^i$.
. These are written and read by $P_i$ only. Timestamps have values in $\set{0,1,2}$. We define orderings $\tls$, $\tgt$ on timestamps as follows [@DS97]: $0 \tls 1$, $1 \tls 2$, and $2 \tls 0$, and $t
\tgt t'$ iff $t' \tls t$. Note that $\tls$ is not transitive. The purpose of $t_i^j$ and $t_j^i$ is to enable the pair of processes $P_i$ and $P_j$ to establish an ordering between themselves by computing $t_i^j \tls t_j^i$. If $t_i^j \tgt t_j^i$, then $P_i$ executed a transition more recently than $P_j$, and vice-versa. The timestamp $t_i^i$ is unused, so we do not worry about initializing it, or what is value is in general.
A A $K$-tuple whose value is maintained equal to $\tpl{t_i^1,\ldots,t_i^K}$. It is written by $P_i$ and read by $P_i$ and $P_j$. Its purpose is to allow $P_i$ to communicate to $P_j$ the values of $P_i$’s timestamps w.r.t. all other processes. By reading all $tv_{ij}^i$, $i \in \onetok - \{j\}$, process $P_j$ can correctly infer the index of the last process to execute. This allows $P_j$ to read the correct emulated values of all shared variables. We use $tv_{ij}^i.k$ to denote the $k$’th element of $tv_{ij}^i$, which is the value of $t_i^k$. For technical convenience, we admit $tv_{ii}^i$. We select some $\l \in \onetok - \{i\}$ arbitrarily and define $tv_{ii}^i$ to be shared pairwise between $P_i$ and $P_\l$. This is needed to conform technically to Definition \[def:pairwise\]. $P_\l$ will not actually reference $tv_{ii}^i$.
For all the above, the order of subscripts does not matter, e.g., $tv_{ij}^i$ and $tv_{ji}^i$ are the same variable, etc.
The essence of the emulation is to deal correctly with the shared variables. This depends upon every process being able to compute the index of the last process to execute, as described above. Define the auxiliary (“ghost”) variable $\last$ to be the index of the last process to make a transition. As described above, every process $P_j$ can compute the value of $\last$ ($\last$ is not explicitly implemented, since doing so would violate pairwise normal form). Then, $P_j$ reads the variable $x_{\last,j}^\last$ that it shares with $P_{\last}$ to find an up to date value for the variable $x$ in $Q$. Together with the unique incoming process index property of $M'_Q$, this allows $P_j$ to accurately determine the currently simulated global state of $M'_Q$. $P_j$ can then update its associated shared variables and atomic propositions to accurately emulate a transition in $M'_Q$.
Let $M_P$ be the GSTD of $P$, as given by Definition \[def:gstd\]. We will define $P = (St_P, P_1 \| \cdots \| P_K)$ so that $M'_Q$ and $M_P$ are bisimilar.
We start with $St_P$. For each initial state $u_0$ of $M'_Q$, we create a corresponding initial state $r_0 \in St_P$ so that: $r_0 \pj \AP = u_0 \pj \AP$ $\AND_{x \in \SH, i,j} r_0(x_{ij}^i) = u_0(x)$ Now for the bisimulation between $M'_Q$ and $M_P$ to work properly, we will require that $\inproc(u) = s(\last)$, where $u,s$ are bisimilar states of $M'_Q$, $M_P$, respectively. It is possible, however, that some initial state $u_0$ of $M'_Q$ does not have an incoming transition, and so $\inproc(u_0)$ is undefined. We deal with this as follows.
Call an initial state (of either $M'_Q$ or $M_P$) that does not have an incoming transition a . Since we defined the corresponding $r_0$ above so that $x_{ij}^i$ has the correct value (namely $u_0(x)$) for all $i,j$, we can let any process be the “last”, as determined by the timestamps. Thus, for a source state $u_0$ in $M'_Q$ and its corresponding source state $r_0$ in $M_P$, we set: $$r_0(t_i^j) = \left\{
\begin{array}{l}
1 ~\mathrm{if}~ i = 1 \land j \ne 1 \\
0 ~\mathrm{if}~ i \ne 1 \land j = 1 \\
X ~\mathrm{if}~ i \ne 1 \land j \ne 1\\
\end{array}
\right.$$ where $X$ denotes a “don’t care,” i.e., any value in $\set{0,1,2}$ can be used. This has the effect of making $P_1$ the “last” process to have executed in a source state, i.e., setting $r_0(\last) = 1$. We now extend the definition of $\inproc$ to source states by defining $\inproc(u_0) = 1$ for every source state $u_0 \in St'_Q$. Together with the fact that states in $M'_Q$ are uniquely determined by the atomic proposition and shared variable values, this automatically takes care of the bisimulation matching between source states in $M'_Q$ and source states in $M_P$, without the need for an extra case analysis. Note also that $\inproc(u)$ is now defined for all states $u$ in $M'_Q$.
For an initial state $u_0$ of $M'_Q$ that is not a source state, and its corresponding initial state $r_0$ in $M_P$, we set: $$r_0(t_i^j) = \left\{
\begin{array}{l}
1 ~\mathrm{if}~ i = \inproc(u_0) \land j \ne \inproc(u_0) \\
0 ~\mathrm{if}~ i \ne \inproc(u_0) \land j = \inproc(u_0) \\
X ~\mathrm{if}~ i \ne \inproc(u_0) \land j \ne \inproc(u_0) \\
\end{array}
\right.$$ where again $X$ means “don’t care.” This has the effect of setting $r_0(\last) = \inproc(u_0)$, as required.
For all initial states $r_0 \in St_P$, whether thay are source states or not, we set the timestamp vector values so that: $\AND_{i,j,k} r_0(tv_{ij}^i.k) = r_0(t_i^k)$
For each transition $(u,i,v)$ in $M'_Q$, we generate a single arc $ARC_i^{u,v}$ in $P_i$ as follows. $ARC_i^{u,v}$ starts in local state $u \pj i$ of $P_i$ and ends in local state $v \pj i$ of $P_i$. Let $\inproc(u) = c$. Then the guard $B_i^{u,v}$ of $ARC_i^{u,v}$ is defined as follows: $B_i^{u,v} \;\df\;
(\last = c) \;\land\;
\AND_{j \ne i} \stof{u \pj j} \;\land\;
(\AND_{x \in \SH} x_{ci}^c = u(x))$ The first conjunct checks that the last process that executed is the process with index $\inproc(u)$. The second conjunct checks that all atomic propositions have the values assigned to them by global state $u$. The third conjunct checks that all shared variables have the values assigned to them by global state $u$.
The action $A_i^{u,v}$ of $ARC_i^{u,v}$ is defined to be $$\begin{aligned}
\pl_{j \ne i} \ t_i^j := \step(t_i^j, tv_{ji}^i.j); \\
\pl_{j} \ tv_{ij}^i := \tpl{t_i^1,\ldots,t_i^K};\\
\pl_{j, x \in \SH} \ x_{ij}^i := v(x) \end{aligned}$$ where $\step(t,t')$ is given in Figure \[fig:step\]. This cannot be factored into pairwise actions $\CA{i}{j}{m}$ because all the $t_i^j$ are used to update all the $tv_{ij}^i$. The solution is to make the $t_i^j$ part of the local state of $P_i$. We do this in phase 3 below. For now, we show that program $P$ with the arcs given by $ARC_i^{u,v} = (u \pj i, B_i^{u,v} \ar A_i^{u,v}, v \pj i)$ is bisimilar to program $Q$.
āaaāaaāaaāaaāa= $\step(t, t')$\
Precondition: $0 \le t, t' \le 2$, that is, $t,t'$ are timestamp values\
$t \tgt t'$ $\RETURN(t)$\
\
$t = 0 \land t' = 1$ $\RETURN(2)$ ;\
$t = 1 \land t' = 2$ $\RETURN(0)$ ;\
$t = 2 \land t' = 0$ $\RETURN(1)$ ;\
\
\[prop:invariants\] The following are invariants of $P$:
$\AND_{i, j, k \ne i} tv_{ij}^i.k = t_i^k$
$\AND_i ( (\last = i) \;\equiv\; \AND_{j \ne i} t_i^j \tgt t_j^i )$
$\AND_{i,j,k} x_{ij}^i = x_{ik}^i$ By construction of $P$: $St_P$ is defined so that the initial states all satisfy the above, and the actions $A_i^{u,v}$ of every process $P_i$ of $P$ are defined so that their execution preserves the above.
\[def:correctness-sim\] Define $\bsim \sub S'_Q \times S_P$ as follows. For $u \in S'_Q$, $r \in S_P$, $u \bsim r$ iff:
\[def:correctness-sim:ap\] $u \pj \AP = r \pj \AP$
\[def:correctness-sim:last\] $\inproc(u) = r(\last)$
\[def:correctness-sim:shvar\] $\AND_{x \in \SH, k} r(\last) = k \imp (\AND_i u(x) = r(x_{ki}^{k}))$
\[thm:bisim\] $\bsim$ is a strong bisimulation
*Proof of Theorem \[thm:bisim\]*. Let $u \in S'_Q$, $r \in S_P$, and $u \bsim r$. We must show that all three clauses of Definition \[def:bisim\] hold, that is:
if $u \bsim r$ then $u \pj \AP = r \pj \AP$
if $u \bsim r$ and $(u,i,v) \in R_Q$ then $\ex s: (r,i,s) \in R_P \land v \bsim s$
if $u \bsim r$ and $(r,i,s) \in R_P$ then $\ex v: (u,i,v) \in R_Q \land v \bsim s$
Clause \[def:bisim:ap\] holds by virtue of clause \[def:correctness-sim:ap\] of Definition \[def:correctness-sim\].\
*Proof of clause \[def:bisim:left-trans\]*. Assume $(u,i,v) \in R_Q$, and let $\inproc(u) = c$. We show that there exists $s$ such that $(r,i,s) \in R_P$ and $v \bsim s$. By our construction of $P$ above, the transition $(u,i,v)$ generates the arc $ARC_i^{u,v}$ in $P_i$. By definition, the guard $B_i^{u,v}$ of $ARC_i^{u,v}$ is $(\last = c \;\land\;
\AND_{j \ne i} \stof{u \pj j} \;\land\;
(\AND_{x \in \SH} x_{ci}^c = u(x)))$. Now by Definition \[def:correctness-sim\] and $u \bsim r$, we have $\inproc(u) = r(\last)$. Hence $r \sat \last = c$. Also by Definition \[def:correctness-sim\] and $u \bsim r$, we have $u \pj \AP \; =\; r \pj \AP$. Hence $r \sat \AND_{j \ne i} \stof{u \pj j}$. Again by Definition \[def:correctness-sim\] and $u \bsim r$, we have $\AND_{x \in \SH} r(\last) = c \imp u(x) = r(x_{ci}^{c}$. Hence $\AND_{x \in \SH,} u(x) = r(x_{ci}^{c})$. And so $r \sat (\AND_{x \in \SH} x_{cj}^c = u(x))$.
Since $r$ satisfies all three conjuncts of (a), it follows that the guard of $ARC_i^{u,v}$ is true in state $r$, and therefore $ARC_i^{u,v}$ is enabled in $r$. By Proposition \[prop:invariants\] and inspection of the action $A_i^{u,v}$ of $ARC_i^{u,v}$, executing of $ARC_i^{u,v}$ leads to a state $s$ such that $s(\last) = i$ and $s \pj \AP = v \pj \AP$ and $(\AND_j x_{ij}^i = v(x))$. By Definition \[def:correctness-sim\], we have $v \bsim s$, as required.\
*Proof of clause \[def:bisim:right-trans\]*. Assume $(r,i,s) \in R_P$. We show that there exists $v$ such that $(u,i,v) \in R_Q$ and $v \bsim s$.
By our construction of $P$ above, the transition $(r,i,s)$ results from executing an arc $ARC_i^{w,v}$ in $P_i$, for some $w,v$. Let $\inproc(w) = c$. By definition of $ARC_i^{w,v}$, we have $r \sat \AND_{j \ne i} \stof{w \pj j}$, and also $r \pj i = w \pj i$. Hence, by the definition of $\stof{w}$ (Definition \[def:stof\]), $r \pj \AP = w \pj \AP$. Also by definition of $ARC_i^{w,v}$, we have $r(\last) = \inproc(w) = c \land (\AND_{x \in \SH} r(x_{ci}^c) = w(x))$. Hence: $r(\last) = \inproc(w) = c$ and $r \pj \AP = w \pj \AP$ and $(\AND_{x \in \SH} r(x_{ci}^c) = w(x))$. Since $u \bsim r$, we have $r(\last) = \inproc(u)$ and $u \pj \AP = r \pj \AP$ and $(\AND_{x \in \SH} r(x_{\last,i}^\last) = u(x))$. From (b), $r(\last) = c$. Hence $r(\last) = \inproc(u)$ and $u \pj \AP = r \pj \AP$ and $(\AND_{x \in \SH} r(x_{ci}^c) = u(x))$. From (b,c) we have $\inproc(w) = \inproc(u)$ and $w \pj \AP = u \pj \AP$ and $(\AND_{x \in \SH} w(x) = u(x))$. Since all global states differ in either some atomic proposition or some shared variable, or some incoming transition, by Proposition \[prop:unique-global\], we conclude from (d) that $w = u$.
By Proposition \[prop:invariants\] and inspection of the action $A_i^{u,v}$ of $ARC_i^{u,v}$, executing $ARC_i^{u,v}$ can only lead to a state $s$ such that $s(\last) = i$ and $s \pj \AP = v \pj \AP$ and $(\AND_j x_{ij}^i = v(x))$. By Definition \[def:correctness-sim\], we have $v \bsim s$, as required.
\[cor:bisim\] $M'_Q \sim M_P$. From Definition \[def:correctness-sim\] and our definition of the initial states of $P$, we see that for every initial state $u_0$ of $M'_Q$, there exists an initial state $r_0$ of $M_P$ such that $u_0 \bsim r_0$, and vice-versa. The result then follows from Theorem \[thm:bisim\] and Definition \[def:bisim\].
Phase Three {#sec:phase-three}
-----------
We now express $ARC_i^{u,v}$ in a form that complies with Definition \[def:pairwise\], that is, as $\gc_{j \in I(i)}
\gd_{\l \in \{1,\ldots,n_j\}} \CB{i}{j}{\l} \ar \CA{i}{j}{\l}$, where $\CB{i}{j}{\l}$ can reference only variables in $\SH_{ij}$ and atomic propositions in $\AP_j$, and $\CA{i}{j}{\l}$ can update only variables in $\SH_{ij}$. Recall that $ARC_i^{u,v} = (u \pj i, B_i^{u,v} \ar A_i^{u,v}, v \pj i)$. For the rest of this section, let $\inproc(u) = c$. First consider $B_i^{u,v}$. By definition $B_i^{u,v} =
(\last = c) \;\land\;
\AND_{j \ne i} \stof{u \pj j} \;\land\;
(\AND_{x \in \SH} x_{ci}^c = u(x))$. Now $\stof{u \pj j}$ is a propositional formula over $\AP_j$, and so $\AND_{j \ne i} \stof{u \pj j}$ is a conjunction of propositional formulae over $\AP_j$, and so it poses no problem. Likewise, since $(\AND_{x \in \SH} x_{ci}^c = u(x))$ is a conjunction over pairwise shared variables, it also is unproblematic. $\last = c$ is not in the pairwise form given above since it refers to the ghost variable $\last$. Note that $\inproc(u)$ is a constant, and so is not problematic in this regard.
Now $\last = c$ checks that the last process to execute is $P_c$. In terms of timestamps, it is equivalent to $\AND_{j \ne c} t_c^j \tgt t_j^c$, i.e., $P_c$ has executed more recently than all other processes. However, the timstamps $t_j^c$ are inaccessible to $P_i$, and the $t_c^j$ are accessible to $P_i$ only in the special case that $c=i$, which does not hold generally. The purpose of the timestamp vectors is precisely to deal with this problem. Recall that $tv_{ci}^c.j$ is maintained equal to $t_c^j$, and $tv_{ji}^j.c$ is maintained equal to $t_j^c$. Hence, we replace $\last = c$ by the equivalent $\AND_{j \ne c} tv_{ci}^c.j \tgt tv_{ji}^j.c$. which moreover can be evaluated by $P_i$, since it refers only to timestamp vectors that are accessible to $P_i$.
Now the expression $tv_{ci}^c.j \tgt tv_{ji}^j.c$ refers to $tv_{ci}^c$, which is shared by $P_c$ and $P_i$, and $tv_{ji}^j$, which is shared by $P_j$ and $P_i$. Thus it is not in pairwise form. We fix this as follows. $tv_{ci}^c.j \tgt tv_{ji}^j.c$ is equivalent to $(tv_{ci}^c.j = 0 \land tv_{ji}^j.c = 1) \lor
(tv_{ci}^c.j = 1 \land tv_{ji}^j.c = 2) \lor
(tv_{ci}^c.j = 2 \land tv_{ji}^j.c = 0)$, by definition of $\tgt$. Hence, $(*)$ is equivalent to\
$\AND_{j \ne c} (tv_{ci}^c.j = 0 \land tv_{ji}^j.c = 1) \lor
(tv_{ci}^c.j = 1 \land tv_{ji}^j.c = 2) \lor
(tv_{ci}^c.j = 2 \land tv_{ji}^j.c = 0)$.\
This formula has length in $O(K)$. We convert this to disjunctive normal form, resulting in a formula of length in $O(exp(K))$. Let the result be $D_1 \lor \ldots \lor D_n$ for some $n$. Each $D_m$, $1 \le m \le n$ is a conjunction of literals, where each literal has one of the forms $(tv_{ci}^c.j ~op~ ts)$, $(tv_{ji}^j.c ~op~ ts)$, where $op \in \{=, \ne\}$, and $ts \in \{0,1,2\}$. Specifically, $D_m = LIT_m^c(tv_{ci}^c.j) \land \AND_{j \not\in \set{c,i}} LIT_m^j(tv_{ji}^j.c)$, where $LIT_m^c(tv_{ci}^c.j)$ is a conjunction of literals of the form $tv_{ci}^c.j ~op~ ts$, and $LIT_m^c(tv_{ji}^j.c)$ is a conjunction of literals of the form $tv_{ji}^j.c ~op~ ts$. Moreover, since logical equivalence to (\*) has been maintained, we have $(D_1 \lor \ldots \lor D_n) \equiv (\last = c)$. For $m \in \set{1,\ldots,n}$, define: $B_i^{u,v}(m) \;\df\;
D_m \land \AND_{j \ne i} \stof{u \pj j} \;\land\;
(\AND_{x \in \SH} x_{ci}^c = u(x))$ where we abuse notation by using $B_i^{u,v}$ as the name for the “array” of guards $B_i^{u,v}(m)$, and also as the name for the guard of $ARC_i^{u,v}$, as defined above. The use of the index $(m)$ will always disambiguate these two uses.
We now define the set of arcs $ARCS_i^{u,v}$ to contain $n$ arcs, $a(1),\ldots,a(n)$, where $a(m) \;\df\;
(u \pj i,
B_i^{u,v}(m) \ar A_i^{u,v},
v \pj i)$ for all $m \in 1,\ldots,n$. In particular, all these arcs start in local state $u \pj i$ of $P_i$ and end in local state $v \pj i$ of $P_i$.
\[prop:guard\] $(\OR_{1 \le m \le n} B_i^{u,v}(m)) \equiv B_i^{u,v}$ Immediate from the definitions and distribution of $\land$ through $\lor$.
It remains to show how each $a(m)$ can be rewritten into pairwise normal form. For all $j \not\in \set{i,c}$, define $B_i^{u,v}(m,j) \df LIT_m^j(tv_{ji}^j.c) \land \stof{u \pj j}$ For $j = c$. $B_i^{u,v}(m,c) \df
LIT_m^c(tv_{ci}^c.j) \land
\stof{u \pj c} \;\land\;
(\AND_{x \in \SH} x_{ci}^c = u(x))$ Note that this works for both $c \ne i$ and $c = i$. The case $c=i$ is why we needed to allow $x_{ii}^i$ and $tv_{ii}^i$. Otherwise we would need a special case to deal with $c=i$. In effect, when $c=i$ we include $B_i^{u,v}(m,c)$ as a conjunct of $B_i^{u,v}(m,\l)$, where $P_\l$ is the process arbitrarily chosen to “share” $x_{ii}^i$ and $tv_{ii}^i$ with $P_i$. This allows us to conform to pairwise normal form, and use $(\AND_{j \ne i} B_i^{u,v}(m,j))$ as the guard of the arc:
\[prop:guardc\] $(\AND_{j \ne i} B_i^{u,v}(m,j)) \equiv B_i^{u,v}(m)$
*Proof of Proposition \[prop:guardc\]*. by definition, $B_i^{u,v}(m) =
D_m \land \AND_{j \ne i} \stof{u \pj j} \;\land\;
(\AND_{x \in \SH} x_{ci}^c = u(x))$. We also have, by construction, $D_m = LIT_m^c(tv_{ci}^c.j) \land \AND_{j \not\in \set{c,i}} LIT_m^j(tv_{ji}^j.c)$. Hence $B_i^{u,v}(m) \equiv
LIT_m^c(tv_{ci}^c.j) \land
(\AND_{j \not\in \set{c,i}} LIT_m^j(tv_{ji}^j.c)) \land
(\AND_{j \ne i} \stof{u \pj j}) \land
(\AND_{x \in \SH} x_{ci}^c = u(x))$.
Splitting up conjunctions and rearranging gives us:
$B_i^{u,v}(m) \equiv
(\AND_{j \not\in \set{c,i}} LIT_m^j(tv_{ji}^j.c)) \land
(\AND_{j \not\in \set{c,i}} \stof{u \pj j}) \land
LIT_m^c(tv_{ci}^c.j) \land
\stof{u \pj c} \land
(\AND_{x \in \SH} x_{c,i}^c = u(x))$.
Grouping together the first two conjunctions, and the last three:
$B_i^{u,v}(m) \equiv
(\AND_{j \not\in \set{c,i}} LIT_m^j(tv_{ji}^j.c) \land \stof{u \pj j}) \land
[LIT_m^c(tv_{ci}^c.j) \land
\stof{u \pj c} \land
(\AND_{x \in \SH} x_{c,i}^c = u(x))]$.
Now $LIT_m^j(tv_{ji}^j.c) \land \stof{u \pj j}$ is just $B_i^{u,v}(m,j)$, and $[LIT_m^c(tv_{ci}^c.j) \land
\stof{u \pj c} \land
(\AND_{x \in \SH} x_{c,i}^c = u(x))]$ is just $B_i^{u,v}(m,c)$. Hence
$B_i^{u,v}(m) \equiv
(\AND_{j \not\in \set{c,i}} B_i^{u,v}(m,j)) \land B_i^{u,v}(m,c)$. Thus $B_i^{u,v}(m) \equiv
\AND_{j \ne i} B_i^{u,v}(m,j)$.
The timestamps $t_i^j$ are written and read by $P_i$ and no other process. To achieve pariwise normal form, we now make the $t_i^j$ part of the local state of $P_i$. Thus, we replace each local state $r_i$ of $P_i$ by $3^K$ local states, each of which agrees with $r_i$ on the atomic propositions in $\AP_i$. There is one such state for every different assignment of timestamp values to $t_1^1,\ldots,t_1^K$. Call the new process that results $PP_i$, and let $PP = (St, PP_1 \pl \cdots \pl PP_K)$. Note that $PP$ has the same initial states as $P$. Let $r'_i$ be a local state of $PP_i$, and let $t_i^1,\ldots,t_i^K$ have some values $d_1,\ldots,d_K$ in $r'_i$. Likewise let $s'_i$ agree with $s_i$ on the atomic propositions in $\AP_i$, and let $t_1^1,\ldots,t_1^K$ have some values $d'_1,\ldots,d'_K$ in $s'_i$. Then, the set of arcs $ARCS_i^{u,v}(r'_i,s'_i)$ is defined as follows.
$ARCS_i^{u,v}(r'_i,s'_i)$ contain $n$ arcs, $a'(1),\ldots,a'(n)$, where $a'(m) \;\df\;$ $(r'_i,
\gc_{j \ne i} BB_i^{u,v}(m,j) \ar AA_i^{u,v}(m,j),
s'_i)$ for all $m \in 1,\ldots,n$. In particular, all these arcs start in $r'_i$ and end in $s'_i$. Also:\
For all $j \ne i$, $BB_i^{u,v}(m,j) \df
B_i^{u,v}(m)_i^j \land step(d_j,tv_{ji}^j.i) = d'_j$ For all $j \ne i$, $AA_i^{u,v}(m,j) \;\df\;
(tv_{ij}^i := \tpl{\ldots,step(d_j,tv_{ji}^j.i),\ldots};\
\pl_{x \in \SH} \ x_{ij}^i := v(x))$ The new conjunct $step(d_j,tv_{ji}^j.i) = d'_j$ in effect checks that the values of the timestamps $t_i^j$ for all $j$ in the new local states are exactly those that the operation $step(t_i^j,t_j^i)$ would return, i.e., those values that would indicate that $P_i$ has excecuted later than $P_j$. The timestamp vector $tv_{ij}^i$ can now be updated correctly without violating pairwise normal form, since the update can be performed using the $d_j$ values, which are constants, and the $tv_{ji}^j.i$. which are shared pairwise between $P_i$ and $P_j$, and are therefore permitted by pairwise normal form.
Let $M_{PP} = (St_P, S_P, R_{PP})$ be the state-transition diagram of $PP$. Note that $PP$ and $P$ have the same initial states, and the same global states, by definition.
\[thm:equal-MP-MPP\] $M_{P} \sim M_{PP}$
*Proof of Theorem \[thm:equal-MP-MPP\]* Let $(r,i,s) \in R_P$. $(r,i,s)$ results from executing an arc $ARC_i^{u,v}$. Hence $B_i^{u,v}$ is true in state $r$. By Proposition \[prop:guard\], some $B_i^{u,v}(m)$ is true in state $r$. Hence $\AND_{j \ne i} B_i^{u,v}(m,j)$ is true in state $r$, by Proposition \[prop:guardc\].
Now let $r',s'$ be the states in $M_{PP}$ that correspond to states $r,s$ in $M_P$, that is $r'$ and $r$ agree on all atomic propositions and shared variabled (including timestamps) and likewise $s$ and $s'$.
Let $r'_i = r' \pj i$, $s'_i = s' \pj i$. Let $t_i^1,\ldots,t_i^K$ have values $d_1,\ldots,d_K$ in $r'_i$ (and hence also in $r'$), and values $d'_1,\ldots,d'_K$ in $s'_i$ (and hence also in $s'$). ($r,r'$ are essentially different ways of refereeing to the same state, to indicate whether the containing structure is $M_P$ or $M_{PP}$, and likewise $s,s'$).
Since $(r,i,s)$ results from executing $ARC_i^{u,v}$, $step(d_j,tv_{ji}^j.i) = d'_j$ must hold, since the action $A_i^{u,v}$ of $ARC_i^{u,v}$ contains the assignment $\pl_{j \ne i} \ t_i^j := \step(t_i^j, tv_{ji}^i.j)$. Hence $\AND_{j \ne i} BB_i^{u,v}(m,j)$ is true in state $r'$. Thus, arc $a'(m)$ of the set $ARCS_i^{u,v}(r'_i,s'_i)$ is enabled in state $r'$. Execution of $a'(m)$ in state $r'$ leads to state $s'$, by definition of $AA_i^{u,v}(m,j)$. Hence $(r',i's') \in R_{PP}$.
Now let $(r',i,s') \in R_{PP}$. $(r',i,s')$ results from executing an arc $a'(m)$ of some set $ARCS_i^{u,v}(r'_i,s'_i)$, where $r'_i = r' \pj i$, $s'_i = s' \pj i$. We can run the previous argument “backwards” to show that $ARC_i^{u,v}$ is enabled in state $r$ of $M_P$, and its execution results in state $s$ of $M_P$. Hence $(r,i,s) \in R_P$.
We have in fact showed that $R_P = R_{PP}$, i.e., that the structures $M_P$ and $M_{PP}$ are identical. Hence they are certainly bisimilar.
\[cor:bisim-MQ-MPP\] $M_Q \sim M_{PP}$ Immediate from Proposition \[prop:unique-correct\], Corollary \[cor:bisim\] and Theorem \[thm:equal-MP-MPP\], along with the transitivity of bisimulation.
Since $PP$ is in pairwise normal form by construction, our main result follows immediately: \[thm:main\] Let $Q$ be any finite-state concurrent program. Then there exists a concurrent program $PP$ such that (1) the global state transition diagrams of $Q$ and $PP$ are bisimilar, and (2) $PP$ is in pairwise normal form.
Our result shows that $PP$ and $Q$ have essentially the same behavior, since strong bisimulation is the strongest notion of equivalence between concurrent programs. A consequence of our result is that $PP$ and $Q$ satisfy the same specifications, for many logics of programs. Recall that $M_{PP}$ and $M_Q$ are the global state transition diagrams of $P$ and $Q$, respectively. Let $f$ be a formula of the temporal logic $\CTLS$ [@Em90], and define $M_Q, u \sat f$ to mean $\fa u \in St_Q: M_Q, u \sat f$, and $M_{PP}, s \sat f$ to mean $\fa s \in St_P: M_P, s \sat f$, where $M_Q, u \sat f$ and $M_{PP}, s \sat f$ refer to the usual satisfaction relation of $\CTLS$ [@Em90]. Then we have:
Let $f$ be a formula of $\CTLS$. Then $M_Q \sat f$ iff $M_{PP} \sat f$. Immediate from Corollary \[cor:bisim-MQ-MPP\] and Theorem 14 in [@CGP99 chapter 11].
We could easily establish similar results for other logics, such as the mu-calculus.
Complexity Results {#sec:complexity}
------------------
For a single process $Q_i$, define $|Q_i|$, the size of $Q_i$, to be the size of the representation of $Q_i$ using a standard complexity-theoretic encoding, i.e., enumeration for sets, character strings for guards and actions etc. Likewise define $|PP_i|$. Define $|Q|$, the size of $Q$, to be $|St_Q|$ + $|Q_1| + \cdots + |Q_K|$, and $|PP|$, the size of $PP$, to be $|St_P|$ + $|PP_1| + \cdots + |PP_K|$.
Define the size of a Kripke structure to be the number of states plus the number of transitions.
\[thm:size\] $|PP|$ is in $O(K exp(|Q|+K))$. $|M_Q|$ is in $O(exp(|Q|))$ by Definition \[def:gstd\]. $|M'_Q|$ is in $O(K \cdot |M_Q|)$, since each state and transition in $M_Q$ is “replicated” at most $K$ times. So $|M'_Q|$ is in $O(K exp(|Q|))$.
For each transition in $M'_Q$, $PP$ contains a number of arcs that is in $O(exp(K))$. Hence $|PP|$ is in $O(|M'_Q| \cdot exp(K))$, and so $|PP|$ is in $O(K \cdot exp(|Q|) \cdot exp(K))$. Thus $|PP|$ is in $O(K exp(|Q|+K))$.
Related Work {#sec:related}
============
It has been long known that a multiple-reader multiple writer atomic register can be implemented using a set of single-reader single-writer registers, and three are many such atomic register constructions in the literature [@AW98 chapter 10]. Since, by definition, a single-reader single-writer register is shared by two processes, these constructions may seem to subsume our result. However, the atomic register constructions do not respect pairwise normal form. For example, they may involve the operation of taking the maximum over a set of single-reader single-writer registers that involve many different pairs of processes. This direct use of register values corresponding to many different pairs, in computing a single expression value, is a direct violation of pairwise normal form.
Conclusions and Future Work {#sec:conc}
===========================
We showed that any finite-state shared memory concurrent program can be rewritten in pairwise normal form, up to strong bisimulation, for a high-atomicity model of concurrent computation. A topic of future work is to establish a similar result in a low-atomicity model, for example that presented in [@AE01]. Our results have significant implications for the efficient synthesis and model-checking of finite-state shared memory concurrent programs. In particular, they show that the approaches of [@Att99a; @AE98; @AC04a] do not sacrifice any expressive power by restricting attention to pairwise normal form.
[^1]: $b$ is the maximum branching in the local state transition relation of a single process
[^2]: We will only use straight-line code in this paper, so termination is always guaranteed.
[^3]: We use $\onetok$ for the set consisting of the natural numbers $1,\ldots,K$.
|
931 P.2d 1382 (1997)
122 N.M. 766
1997-NMSC-7
In the Matter of W. Eugene HERKENHOFF, An Attorney Disbarred from Practice Before the Courts of the State of New Mexico.
No. 21718.
Supreme Court of New Mexico.
January 27, 1997.
Virginia L. Ferrara, Chief Disciplinary Counsel, Albuquerque.
W. Eugene Herkenhoff, Silver City, Pro Se.
DISCIPLINARY PROCEEDING
OPINION
PER CURIAM.
This matter came before the Court for contempt proceedings pursuant to the provisions *1383 of NMSA 1978, Section 34-1-2 and Rule 17-206(G) NMRA 1996 of the Rules Governing Discipline. W. Eugene Herkenhoff was ordered to appear before this Court to explain why he should not be sanctioned for the two findings of criminal contempt entered against him on November 14, 1996.
On December 22, 1993, we suspended respondent from the practice of law for violations of several of the Rules of Professional Conduct but deferred the suspension and placed him on supervised probation for a period of twelve months. In re Herkenhoff, 116 N.M. 622, 866 P.2d 350 (1993). When it became apparent the respondent had no intention of cooperating with his supervisor, his probation was revoked and he was suspended May 18, 1994.
In the order of suspension, respondent was directed to comply with Rule 17-212, which requires an attorney who has resigned from practice or who has been suspended or disbarred to notify in writing all clients, opposing counsel, and courts and administrative agencies before which he or she has cases pending of the effective date of his or her resignation, suspension, or disbarment. Additionally, within ten days of the effective date, the attorney must file with this Court an affidavit confirming that he or she has complied fully with Rule 17-212 and attach thereto copies of the letters sent. Respondent failed to do this.
Further proceedings were held in an effort to compel respondent to comply with Rule 17-212 and, when these efforts were met with little cooperation from respondent, he was disbarred on January 9, 1995, effective retroactively to January 4, 1995, and fined $1540. In re Herkenhoff, 119 N.M. 232, 889 P.2d 840 (1995).[1]
On August 28, 1996, chief disciplinary counsel brought to this Court's attention by way of a verified motion for order to show cause certain information which indicated that respondent was continuing to hold himself out as an attorney and engage in the practice of law despite his disbarment.
In particular, respondent was representing Jose P. Lopez in a dispute Mr. Lopez was having with the Secretary of the Army before the Equal Employment Opportunity Commission (EEOC) in Washington, D.C. While we do not quarrel with federal regulations that permit non-lawyers to represent claimants in EEOC matters, a March 25, 1996, letter from respondent to EEOC personnel was written on his attorney letterhead stationary (with the word "RETIRED" typed below the words "W. Eugene Herkenhoff, Attorney at Law."). Respondent clearly was holding himself out as a duly licensed attorney who had simply taken retirement. When disciplinary counsel wrote to the recipient of the letter to clarify respondent's status as a disbarred attorney, respondent reacted angrily and wrote another letter (on his "RETIRED" attorney letterhead) to the EEOC recipient claiming that disciplinary counsel was harassing him and Mr. Lopez and asking that she be sanctioned under 18 U.S.C. Section 1505.
Additionally, respondent involved himself in the legal affairs of a former client, whose case was pending before the United States District Court for the District of New Mexico. The client was represented in the case by John R. Polk, Esq., although she had been represented in the same matter by respondent prior to his disbarment.
In early July 1996, Polk communicated to his client that the plaintiff would drop its claim against her if she, in turn, would drop her counterclaim for attorney fees. In response to this offer, the client (with whom Polk had previously enjoyed an amicable relationship) sent Polk a highly critical letter questioning the amount of his hourly fee and actions taken (or not taken) by him and directing him to file certain specific but ill-advised pleadings in the case. As the client was not legally sophisticated and was not *1384 conversant with the legal terms contained in her letter, it appeared to Polk that the letter had been authored by respondent.
In a subsequent phone conversation with Polk, the client was cordial and made no reference to the legal concepts she had purportedly discussed in her letter. After an explanation by Polk of the various options available to her, the client agreed to the terms of the proposed settlement and promised to send a written authorization that day. Rather than the promised authorization, however, Polk received another letter from his client refusing to settle the case, requesting a refund of fees already paid, and asking that he not contact her by phone. This letter also appeared to have been written by respondent, although signed by the client.
In response, Polk moved the court for permission to withdraw on grounds that respondent's interference in the case had so seriously compromised his ability to communicate with his client that he was no longer able to represent her. Polk then received two more lettersone from his client and one from respondent. In her letter, again apparently produced by respondent, the client went on at some length about how respondent was not actually disbarred but was simply the subject of "an unconstitutional Bill of Attainder" illegally obtained by disciplinary counsel and was actually a retired attorney who was willing to give her advice and counsel when she requested it and charge her nothing for his time. Respondent's letter threatened Polk with legal action unless Polk took specified steps in the client's case, many of which were prohibited by the revised federal procedural rules.
United States Magistrate Lorenzo F. Garcia held a scheduling conference at which time Polk's motion to withdraw was considered. Judge Garcia advised the client that respondent was indeed disbarred and could not advise her or represent her in the case. In an affidavit presented to this Court, Judge Garcia stated that in his opinion, respondent's "secret participation and legal assistance to [the client] interfered with the attorney-client relationship existing between [the client] and her attorney, Mr. Polk, and further interfered with the parties' settlement discussions. This, in turn, needlessly contributed to the increased costs of litigation and court, counsel, and litigants' time."
Based upon these documented allegations in the verified motion for order to show cause, this Court issued an order to show cause directing respondent to respond to the motion on or before September 17, 1996, and to appear before the Court on September 18, 1996, to show cause why he should not be held in contempt of the Court's order of disbarment. Copies of both the motion and the order were served upon respondent by the clerk of this Court by certified mail at his address of record with the Court. The hearing date was subsequently continued by the Court until October 23, 1996, and respondent received notice of the rescheduled hearing.
Respondent filed no response to the motion for order to show cause denying the allegations. He did not appear at the October 23 hearing nor did he file any request for a continuance.[2] The allegations in the motion were deemed admitted. Respondent was found to be in direct criminal contempt of this Court not only for his misrepresentation of his status as a "retired" attorney and his continued practice of law in the face of our January 9, 1995, order of disbarment but also for his failure to appear before the Court on October 23 as ordered.
The record in this case fully supports the finding of direct contempt. The documentation attached to the unrefuted motion shows that respondent was holding himself out as a lawyer to EEOC personnel and was engaging in the practice of law by providing Polk's client with legal advice and assistance. The court file also shows that *1385 respondent received notice of the October 23 hearing and inexplicably failed to appear. Direct contempt may be established where the court's record shows violations of court rules. In re Avallone, 91 N.M. 777, 581 P.2d 870 (1978). It may also be established where the record shows clear violations of a court's orders.
Where criminal contempt is directbut not flagrantthe contemnor must be given prior warning, an opportunity to explain the conduct at issue, and a hearing on the matter before sanctions are imposed. In re Klecan, 93 N.M. 637, 603 P.2d 1094 (1979). There must be proof beyond a reasonable doubt of the conduct constituting criminal contempt. State ex rel. Bliss v. Greenwood, 63 N.M. 156, 315 P.2d 223 (1957). If the punishment to be imposed is imprisonment for less than six months or a fine of less than $1000, however, counsel need not be appointed and no jury trial is required. State v. Case, 100 N.M. 173, 667 P.2d 978 (Ct.App. 1983) (no right to counsel where imprisonment will not exceed six months); Seven Rivers Farm, Inc. v. Reynolds, 84 N.M. 789, 508 P.2d 1276 (1973) (no jury trial right where sole punishment is fine of $1000 or less).
In this instance, respondent was adequately warned by the order of disbarment. A hearing was scheduled whereby he was given the opportunity to explain his conduct. The record provides proof beyond a reasonable doubt that he failed to appear or respond and that he held himself out as a lawyer and engaged in the practice of law after the order of disbarment was entered.
On November 14, 1996, we entered the order finding respondent in direct criminal contempt for his violations of the disbarment order and his failure to appear on October 23 and ordered him to appear before the Court on December 4, 1996, to present any facts that would justify or mitigate either act of contempt.
At the December 4 hearing, respondent appeared to believe that since he had made the decision to retire prior to the entry of our order of disbarment, the disbarment order was of no effect. He claimed that he had notified his clients of his prior suspension, but the Court file is devoid of any indication that respondent complied with the requirements of Rule 17-212. Respondent also advised the Court that because people kept requesting legal advice and assistance from him, he felt obligated to provide services to them at no cost, even though he was retired. He stated that he failed to attend the October 23 hearing because of a prior commitment to be at Lake Powell with friends.
An attorney does not defeat the disciplinary jurisdiction of the Court by simply deciding to retire, particularly where he is under an order of deferred suspension. The Court's disbarment order of January 9, 1995, is a valid order entered after proceedings conducted pursuant to the Rules Governing Discipline. Respondent violated that order and has shown no reason why sanctions should not be imposed against him. An attorney who has been suspended or disbarred may not continue to provide legal services to members of the public, with or without remuneration.[3] Additionally, a cruise on Lake Powell is not the sort of emergency that excuses an attorney's non-appearance at a court proceeding of which he has received notice. Even if an attorney has a conflicting court engagement or other legitimate excuse, he is expected to advise the court of the conflict by filing the appropriate pleading requesting a continuance well in advance of the scheduled hearing.
NOW, THEREFORE, IT IS ORDERED that W. Eugene Herkenhoff is hereby sentenced to five (5) months of incarceration for his contumacious conduct; and
IT IS FURTHER ORDERED that the entire period of incarceration be and hereby is suspended under the following terms and conditions:
*1386 (1) Respondent shall abide by all of the terms and conditions of the amended disbarment order issued on January 9, 1995, including the payment of the $1540 fine, with accrued interest, to this Court and the payment of $78.51 in assessed costs, with accrued interest, to the Disciplinary Board;
(2) Respondent shall adhere to and comply with the requirements of Rule 17-212 NMSA 1996 on or before December 16, 1996; and
(3) Respondent shall cease from using and destroy all stationary that indicates he is a retired attorney.
IT IS FURTHER ORDERED that should any of the terms and conditions not be followed, this Court shall lift the suspension of the period of incarceration and issue a bench warrant for respondent's immediate arrest and confinement.
IT IS FURTHER ORDERED that this opinion be published in the State Bar of New Mexico State Bar Bulletin and the New Mexico Reports.
IT IS SO ORDERED.
MINZNER, J., not participating.
NOTES
[1] We note that there is a typographical error at 119 N.M. at 234, 889 P.2d at 842. The opinion states that respondent was disbarred effective January 4, 1994. He was disbarred effective January 4, 1995.
[2] Respondent wrote to the Chief Justice advising him that he would not be present on October 23 due to a previous commitment. The letter was not filed with the clerk nor was a copy sent to disciplinary counsel. The letter is not a part of the record of this case. Informal ex parte communications with the Court are not considered responsive pleadings.
[3] See discussion of reasons why disbarred attorneys are not permitted to give legal advice to members of the public in In re Schmidt, 121 N.M. 640, 916 P.2d 840 (1996), also decided today.
|
“The man you’re going to see today is just a man in a suit dressed up like Santa.”
The shouting man is David Grisham, an evangelical street preacher from Anchorage, who has shouted his sermons at people across the country for nearly a decade, whether they want to hear them or not.
AD
On Saturday, his unwilling audience was a group of families snaked around Westgate’s seasonal Christmas village at the food court near Hot Topic and the Gap.
AD
Grisham paid no heed to calls to “chill out” from parents — and as his sermon picked up momentum, things were about to get ugly.
“Don’t lie to your children and tell them there’s such a thing as Santa, when you know in reality that there are no flying reindeer, there is no workshop at the North Pole. . . that you buy all the gifts and put them under the tree.”
A girl in a Santa hat stared, wide-eyed. Other children shot confused looks at their moms and dads. And a few parents decided to take action.
A man in a blue T-shirt approaches. He’s so close to Grisham that the pastor’s cellphone camera captures only his torso.
AD
“Stop,” the man says, speaking over Grisham.
“Stop. Stop. Stop. Stop. Stop.”
“Quit talking this mess, you understand me,” the father says. “I got my kids over there, we don’t need you coming over here blabbing whatever the hell you’re blabbing.”
AD
Other parents approach, diverting Grisham’s attention from his message as tensions rise.
This kind of conflict is not foreign to Grisham, the founder of Last Frontier Evangelism. He travels the country with his wife, spreading the gospel to people who are, for the most part, minding their own business.
He told The Washington Post that God spoke to him a decade ago while he and his wife were on vacation in Mexico. He’s been street-preaching most weekends ever since.
AD
On Saturday, he whipped out his cellphone camera to record his Santa truther episode, hoping a viral video would help spread the gospel.
He’s claiming victory. The video has been seen more than 2 million times. As an added bonus, mall officials say he’s not banned from the mall, per se, though they’re on the lookout for future inappropriate behavior.
AD
Grisham said he’s planning to do the same thing at other malls before Christmas.
Usually, Grisham and his wife carry large signs that quote scripture and let people know that God isn’t fond of whatever they happen to be doing.
“Most of the unsaved people are not going to church,” he said. “My wife and I, we go to gay bars, we go to porn shops. If you’re going to be a fisher of men, you’ve got to go where the fish are.”
AD
In August, they struck up conversations with people outside a strip club, the Great Alaskan Bush Company, carrying a large sign that said “THE WAGES SIN PAYS IS DEATH.”
They spoke at “the Pittsburgh homosexual pride parade,” according to a Facebook picture, and at the Republican and Democratic national conventions, venues where people have come to expect a clash of opinions.
The Grishams have also directed their messages at people who have good reason to say they’re not actively sinning, like Catholics who traveled to Philadelphia to see the pope.
AD
But Grisham told The Post that his biggest beef this time of year is with Kriss Kringle.
“If Santa Claus was a cartoon caricature like Mickey Mouse and everybody knew that it was fake, and no one thought it was real, I’d be fine with it,” he said. “But when you start telling kids that Santa Claus is real, it now becomes idolatry. I’m going after it because it’s idolatry.”
AD
He’s not just “going after” Santa Claus metaphorically.
In 2010, he videotaped a mock execution of a Santa Claus piñata, complete with sound effects and a coup de grace to the Santa piñata.
“What you can’t see,” he told his listeners, “is that we shot Santa in the face.”
He realizes the tactics strike emotional chords. So at the mall on Wednesday he wasn’t surprised when he found himself surrounded by angry parents.
One started pushing him, he said, and on the video he tells the man to stop assaulting him. Nearby, he could see mall employees on their radios, asking for security guards to come.
AD
It was time to go.
But first, in case anyone hadn’t been listening for the past three minutes, he offered a parting shot:
AD
“Kids, there is no Santa. Santa’s not real. Your parents are lying to you. Don’t believe it.
“Y’all have a nice day.” |
The Changes Group: A Follow-Up Report on a Group Intervention to Assist the Successful Discharge of Hospital Residents.
This is a follow-up study of a pilot project first reported on in 2006. A group model was developed for a state psychiatric hospital setting to assist residents who had displayed characteristics of "institutionalism." This includes an aversion or ambivalence to discharge efforts and an acceptance of prolonged life in the hospital. The pilot project, while small, was promising, with five of seven people entering the community successfully within a year and a half. The current project expanded the scope to include three groups with a total of 25 participants. Additional refinement included a standardized group curriculum, expanding outcome measures to include participant attitudes toward change, a protocol to inform and meaningfully involve clinical treatment teams in the participants' progress, and enhanced training for group facilitators. Of the original participants, 32% achieved discharge in the first year of participation in the group. In addition, participants who were not discharged within the first year developed more positive attitudes toward making changes in their life. The guiding principles of this model-including personal reflection, a team approach for sharing life experiences, and encouragement from participants and staff-seem conducive for supporting the attitudinal change and motivation necessary for successful discharge after prolonged hospital stays. |
Q:
What are my options for estimating probabilities of success/failure?
So, I want to estimate the probability that a student will succeed or fail in a particular major (suppose success means in this case to graduate with that major) based on data from that student's application. I have a lot of records of past students' application data and success.
I am aware that I could try logistic regression, but I would like to know all my options, including methods which allow more easily for non-linearities.
If I were just trying to classify incoming students, I think I would try an SVM model. But I don't know how easy it is to adapt that method to the task of estimating probabilities.
EDIT:
Much of the data is categorical. Often binary. Some features are real numbers / integers, as well.
A:
You could consider classification trees. You don't say what software you are using, but there are several ways to do this in R, which is free. I personally have used the party() package with good results, but that's not to say other packages aren't just as good.
A classification tree is often very good at finding various interactions. You wind up with different sets of the IVs and the probability of "success" on your DV for different combinations.
|
/*
JPC: An x86 PC Hardware Emulator for a pure Java Virtual Machine
Copyright (C) 2012-2013 Ian Preston
This program is free software; you can redistribute it and/or modify
it under the terms of the GNU General Public License version 2 as published by
the Free Software Foundation.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License along
with this program; if not, write to the Free Software Foundation, Inc.,
51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA.
Details (including contact information) can be found at:
jpc.sourceforge.net
or the developer website
sourceforge.net/projects/jpc/
End of licence header
*/
package org.jpc.emulator.execution.opcodes.pm;
import org.jpc.emulator.execution.*;
import org.jpc.emulator.execution.decoder.*;
import org.jpc.emulator.processor.*;
import org.jpc.emulator.processor.fpu64.*;
import static org.jpc.emulator.processor.Processor.*;
public class fcos extends Executable
{
public fcos(int blockStart, int eip, int prefices, PeekableInputStream input)
{
super(blockStart, eip);
}
public Branch execute(Processor cpu)
{
double freg0 = cpu.fpu.ST(0);
if (Double.isInfinite(freg0))
cpu.fpu.setInvalidOperation();
if ((freg0 > Long.MAX_VALUE) || (freg0 < Long.MIN_VALUE))
cpu.fpu.conditionCode |= 4; // set C2
else
cpu.fpu.setST(0, Math.cos(freg0));
return Branch.None;
}
public boolean isBranch()
{
return false;
}
public String toString()
{
return this.getClass().getName();
}
} |
Oxidized low density lipoprotein increases acetylcholinesterase activity correlating with reactive oxygen species production.
Hyperlipidemia, low density lipoproteins (LDL) and their oxidized forms, and oxidative stress are suspected to be a key combination in the onset of AD and acetylcholinesterase (AChE) plays a part in this pathology. The present study aimed to link these parameters using differentiated SH-SY5Y human neuroblastoma cells in culture. Both mildly and fully oxidized human LDL (mox- and fox-LDL), but not native (non-oxidized) LDL were cytotoxic in dose- and time-dependent patterns and this was accompanied by an increased production of intracellular reactive oxygen species (ROS). Oxidized LDL (10-200 μg/mL) augmented AChE activity after 4 and 24h treatments, respectively while the native LDL was without effect. The increased AChE with oxidized LDLs was accompanied by a proportionate increase in intracellular ROS formation (R=0.904). These findings support the notion that oxidized LDLs are cytotoxic and that their action on AChE may reduce central cholinergic transmission in AD and affirm AChE as a continued rational for anticholinesterase therapy but in conjunction with antioxidant/antihyperlipidemic cotreatments. |
Hitherto, audio signal coding methods for compressing the amount of data of an audio signal have been developed. As one of such coding methods, High-Efficiency Advanced Audio Coding (HE-AAC) is known. This coding method has been standardized as MPEG-2 HE-AAC and MPEG-4 HE-AAC by the Moving Picture Experts Group (MPEG). In HE-AAC, the low frequency band (low frequency components) of an audio signal is coded in accordance with an Advanced Audio Coding (AAC) method, whereas the high frequency band (high-frequency components) of an audio signal is coded in accordance with a Spectral Band Replication (SBR) method. In the SBR method, each frame of an audio signal is divided into a plurality of time-frequency domains, and auxiliary information or the like for reproducing high-frequency components by reproducing corresponding low frequency components on the basis of the signal power within each time-frequency domain are calculated as SBR data. Then, an SBR parameter is coded. This time-frequency domain is called a grid.
In the SBR method, if the time length of a grid is too long with respect to the temporal change of an audio signal, the electric power of the audio signal is averaged in the grid, and thereby the information indicating the temporal change is lost. As a result, the reproduction sound quality of the coded audio signal deteriorates. There is a case where, in particular, as a result of sound in a certain time period being affected by sound later than that sound, sound that differs from the original sound is produced. Such a phenomenon is called a pre-echo. In Japanese National Publication of International Patent Application No. 2003-529787, a technology is disclosed in which a highly transient sound, such as attack sound, is detected with respect to each channel of an audio signal, and a grid is set so that the time resolution increases with respect to the highly transient sound. Such a transient portion of sound is called a transient.
Furthermore, in Japanese Laid-open Patent Publication No. 2006-3580, a technology has been disclosed in which when it is determined that the degree of similarity of a plurality of channels of an audio signal is high, a grouping of frequency data such that an audio signal is frequency-converted in the time direction or in the frequency direction is performed in common with respect to a plurality of channels. |
/*
* Licensed to the Apache Software Foundation (ASF) under one
* or more contributor license agreements. See the NOTICE file
* distributed with this work for additional information
* regarding copyright ownership. The ASF licenses this file
* to you under the Apache License, Version 2.0 (the
* "License"); you may not use this file except in compliance
* with the License. You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing,
* software distributed under the License is distributed on an
* "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY
* KIND, either express or implied. See the License for the
* specific language governing permissions and limitations
* under the License.
*/
/*!
* Copyright (c) 2016 by Contributors
* \file elemwise_binary_op_basic.cc
* \brief CPU Implementation of basic elementwise binary broadcast operators
*/
#include "./elemwise_unary_op.h"
#include "./elemwise_binary_op-inl.h"
#include "../nn/mkldnn/mkldnn_ops-inl.h"
#include "../nn/mkldnn/mkldnn_base-inl.h"
namespace mxnet {
namespace op {
bool SupportMKLDNNSum(const NDArray& input) {
int ndim = input.shape().ndim();
return (input.dtype() == mshadow::kFloat32 || input.dtype() == mshadow::kBfloat16) &&
(ndim >= 1 && ndim <= 4) && input.storage_type() == kDefaultStorage;
}
static void ElemwiseAddEx(const nnvm::NodeAttrs& attrs,
const OpContext& ctx,
const std::vector<NDArray>& inputs,
const std::vector<OpReqType>& req,
const std::vector<NDArray>& outputs) {
CHECK_EQ(inputs.size(), 2U);
CHECK_EQ(outputs.size(), 1U);
#if MXNET_USE_MKLDNN == 1
if (SupportMKLDNNSum(inputs[0]) && SupportMKLDNNSum(inputs[1])) {
MKLDNNRun(MKLDNNSumForward, attrs, ctx, inputs, req, outputs);
return;
} else if (inputs[0].storage_type() == kDefaultStorage
&& inputs[1].storage_type() == kDefaultStorage) {
FallBackCompute(ElemwiseBinaryOp::Compute<cpu, op::mshadow_op::plus>,
attrs, ctx, inputs, req, outputs);
return;
}
#endif
ElemwiseBinaryOp::ComputeEx<cpu, op::mshadow_op::plus>(attrs, ctx, inputs,
req, outputs);
}
static inline bool ElemwiseAddStorageType(const nnvm::NodeAttrs& attrs,
const int dev_mask,
DispatchMode* dispatch_mode,
std::vector<int> *in_attrs,
std::vector<int> *out_attrs) {
CHECK_EQ(in_attrs->size(), 2);
CHECK_EQ(out_attrs->size(), 1);
bool ret = ElemwiseBinaryOp::PreferDenseStorageType<true, true, true>(
attrs, dev_mask, dispatch_mode, in_attrs, out_attrs);
#if MXNET_USE_MKLDNN == 1
if (dev_mask == mshadow::cpu::kDevMask && !MKLDNNEnvSet()) {
*dispatch_mode = DispatchMode::kFComputeFallback;
} else if (dev_mask == mshadow::cpu::kDevMask
&& common::ContainsOnlyStorage(*in_attrs, kDefaultStorage)
&& out_attrs->at(0) == kDefaultStorage) {
*dispatch_mode = DispatchMode::kFComputeEx;
}
#endif
return ret;
}
MXNET_OPERATOR_REGISTER_BINARY(elemwise_add)
.set_attr<FInferStorageType>("FInferStorageType", ElemwiseAddStorageType)
.set_attr<FCompute>("FCompute<cpu>", ElemwiseBinaryOp::Compute<cpu, op::mshadow_op::plus>)
#if MXNET_USE_MKLDNN == 1
.set_attr<bool>("TIsMKLDNN", true)
#endif
.set_attr<FComputeEx>("FComputeEx<cpu>", ElemwiseAddEx)
.set_attr<THasDeterministicOutput>("THasDeterministicOutput", true)
.set_attr<FResourceRequest>("FResourceRequest", /* For Sparse CSR */
[](const NodeAttrs& attrs) {
return std::vector<ResourceRequest>{ResourceRequest::kTempSpace};})
MXNET_ADD_SPARSE_OP_ALIAS(elemwise_add)
.add_alias("_add").add_alias("_plus").add_alias("_Plus")
.set_attr<nnvm::FListOutputNames>("FListOutputNames", [](const NodeAttrs& attrs) {
return std::vector<std::string>{"output"};
})
.describe(R"code(Adds arguments element-wise.
The storage type of ``elemwise_add`` output depends on storage types of inputs
- elemwise_add(row_sparse, row_sparse) = row_sparse
- elemwise_add(csr, csr) = csr
- elemwise_add(default, csr) = default
- elemwise_add(csr, default) = default
- elemwise_add(default, rsp) = default
- elemwise_add(rsp, default) = default
- otherwise, ``elemwise_add`` generates output with default storage
)code")
.set_attr<nnvm::FGradient>("FGradient", CloneGradient{"_backward_add"});
// specialized gradient add function to do add to optimization
// this must differ from elemwise_add to prevent add to optimization in forward pass.
MXNET_OPERATOR_REGISTER_BINARY_WITH_SPARSE_CPU(_grad_add, op::mshadow_op::plus);
static void _backward_ElemwiseAddEx(const nnvm::NodeAttrs& attrs,
const OpContext& ctx,
const std::vector<NDArray>& inputs,
const std::vector<OpReqType>& req,
const std::vector<NDArray>& outputs) {
CHECK_EQ(inputs.size(), 1U);
CHECK_EQ(outputs.size(), 2U);
#if MXNET_USE_MKLDNN == 1
if (inputs[0].IsMKLDNNData()) {
MKLDNNRun(MKLDNNCopy, attrs, ctx, inputs[0], req[0], outputs[0]);
MKLDNNRun(MKLDNNCopy, attrs, ctx, inputs[0], req[1], outputs[1]);
return;
} else if (common::ContainsOnlyStorage(inputs, kDefaultStorage)) {
FallBackCompute(
ElemwiseBinaryOp::BackwardUseNone<cpu, mshadow_op::identity, mshadow_op::identity>,
attrs, ctx, inputs, req, outputs);
return;
}
#endif
ElemwiseBinaryOp::BackwardUseNoneEx<cpu, mshadow_op::identity, mshadow_op::identity>(
attrs, ctx, inputs, req, outputs);
}
static inline bool ElemwiseAddBackwardStorageType(const nnvm::NodeAttrs& attrs,
const int dev_mask,
DispatchMode* dispatch_mode,
std::vector<int> *in_attrs,
std::vector<int> *out_attrs) {
CHECK_EQ(in_attrs->size(), 1);
CHECK_EQ(out_attrs->size(), 2);
bool ret = ElemwiseStorageType<1, 2, true, true, true>(attrs, dev_mask, dispatch_mode,
in_attrs, out_attrs);
#if MXNET_USE_MKLDNN == 1
if (dev_mask == mshadow::cpu::kDevMask && !MKLDNNEnvSet()) {
*dispatch_mode = DispatchMode::kFComputeFallback;
} else if (dev_mask == mshadow::cpu::kDevMask) {
*dispatch_mode = DispatchMode::kFComputeEx;
}
#endif
return ret;
}
NNVM_REGISTER_OP(_backward_add)
.set_num_inputs(1)
.set_num_outputs(2)
.set_attr<nnvm::TIsBackward>("TIsBackward", true)
.set_attr<nnvm::FInplaceOption>("FInplaceOption",
[](const NodeAttrs &attrs) {
return std::vector<std::pair<int, int> >{{0, 0},
{0, 1}};
})
#if MXNET_USE_MKLDNN == 1
.set_attr<FResourceRequest>("FResourceRequest", [](const NodeAttrs& n) {
return std::vector<ResourceRequest>{ResourceRequest::kTempSpace};
})
.set_attr<bool>("TIsMKLDNN", true)
#endif
.set_attr<FCompute>("FCompute<cpu>", ElemwiseBinaryOp::BackwardUseNone<
cpu, mshadow_op::identity, mshadow_op::identity>)
.set_attr<FComputeEx>("FComputeEx<cpu>", _backward_ElemwiseAddEx)
.set_attr<FInferStorageType>("FInferStorageType", ElemwiseAddBackwardStorageType);
MXNET_OPERATOR_REGISTER_BINARY_WITH_SPARSE_CPU_PD(elemwise_sub, op::mshadow_op::minus)
MXNET_ADD_SPARSE_OP_ALIAS(elemwise_sub)
.add_alias("_sub").add_alias("_minus").add_alias("_Minus")
.describe(R"code(Subtracts arguments element-wise.
The storage type of ``elemwise_sub`` output depends on storage types of inputs
- elemwise_sub(row_sparse, row_sparse) = row_sparse
- elemwise_sub(csr, csr) = csr
- elemwise_sub(default, csr) = default
- elemwise_sub(csr, default) = default
- elemwise_sub(default, rsp) = default
- elemwise_sub(rsp, default) = default
- otherwise, ``elemwise_sub`` generates output with default storage
)code")
.set_attr<nnvm::FGradient>("FGradient", ElemwiseGradUseNone{"_backward_sub"});
NNVM_REGISTER_OP(_backward_sub)
.set_num_inputs(1)
.set_num_outputs(2)
.set_attr<nnvm::TIsBackward>("TIsBackward", true)
.set_attr<nnvm::FInplaceOption>("FInplaceOption",
[](const NodeAttrs &attrs) {
return std::vector<std::pair<int, int> >{{0, 0},
{0, 1}};
})
.set_attr<FCompute>("FCompute<cpu>", ElemwiseBinaryOp::BackwardUseNone<cpu,
mshadow_op::identity, mshadow_op::negation>)
.set_attr<FComputeEx>("FComputeEx<cpu>", ElemwiseBinaryOp::BackwardUseNoneEx<cpu,
mshadow_op::identity, mshadow_op::negation>)
.set_attr<FInferStorageType>("FInferStorageType",
ElemwiseStorageType<1, 2, true, true, true>);
MXNET_OPERATOR_REGISTER_BINARY(elemwise_mul)
MXNET_ADD_SPARSE_OP_ALIAS(elemwise_mul)
.describe(R"code(Multiplies arguments element-wise.
The storage type of ``elemwise_mul`` output depends on storage types of inputs
- elemwise_mul(default, default) = default
- elemwise_mul(row_sparse, row_sparse) = row_sparse
- elemwise_mul(default, row_sparse) = row_sparse
- elemwise_mul(row_sparse, default) = row_sparse
- elemwise_mul(csr, csr) = csr
- otherwise, ``elemwise_mul`` generates output with default storage
)code")
.set_attr<FInferStorageType>("FInferStorageType",
ElemwiseBinaryOp::PreferSparseStorageType)
.set_attr<FCompute>("FCompute<cpu>", ElemwiseBinaryOp::Compute<cpu, op::mshadow_op::mul>)
.set_attr<FComputeEx>("FComputeEx<cpu>",
ElemwiseBinaryOp::ComputeDnsLRValueEx<cpu, op::mshadow_op::mul, true, true>)
.set_attr<FResourceRequest>("FResourceRequest", /* For Sparse CSR */
[](const NodeAttrs& attrs) {
return std::vector<ResourceRequest>{ResourceRequest::kTempSpace};
})
.set_attr<THasDeterministicOutput>("THasDeterministicOutput", true)
.add_alias("_mul").add_alias("_Mul")
.set_attr<nnvm::FGradient>("FGradient", ElemwiseGradUseIn{"_backward_mul"});
NNVM_REGISTER_OP(_backward_mul)
.set_num_inputs(3)
.set_num_outputs(2)
.set_attr<nnvm::TIsBackward>("TIsBackward", true)
.set_attr<nnvm::FInplaceOption>("FInplaceOption",
[](const NodeAttrs &attrs) {
return std::vector<std::pair<int, int> >{{0, 1}};
})
.set_attr<FInferStorageType>("FInferStorageType", ElemwiseBinaryOp::BackwardUseInStorageType)
.set_attr<FResourceRequest>("FResourceRequest", /* For Sparse CSR */
[](const NodeAttrs& attrs) {
return std::vector<ResourceRequest>{ResourceRequest::kTempSpace};
})
.set_attr<FCompute>("FCompute<cpu>", ElemwiseBinaryOp::BackwardUseIn<
cpu, mshadow_op::right, mshadow_op::left>)
.set_attr<FComputeEx>("FComputeEx<cpu>", ElemwiseBinaryOp::BackwardUseInEx<
cpu, mshadow_op::right, mshadow_op::left>);
MXNET_OPERATOR_REGISTER_BINARY_WITH_SPARSE_CPU_DR(elemwise_div, op::mshadow_op::div)
MXNET_ADD_SPARSE_OP_ALIAS(elemwise_div)
.describe(R"code(Divides arguments element-wise.
The storage type of ``elemwise_div`` output is always dense
)code")
.add_alias("_div").add_alias("_Div")
.set_attr<nnvm::FGradient>("FGradient", ElemwiseGradUseIn{"_backward_div"});
NNVM_REGISTER_OP(_backward_div)
.set_num_inputs(3)
.set_num_outputs(2)
.set_attr<nnvm::TIsBackward>("TIsBackward", true)
.set_attr<nnvm::FInplaceOption>("FInplaceOption",
[](const NodeAttrs &attrs) {
return std::vector<std::pair<int, int> >{{0, 1}};
})
.set_attr<FCompute>("FCompute<cpu>", ElemwiseBinaryOp::BackwardUseIn<
cpu, mshadow_op::div_grad, mshadow_op::div_rgrad>);
MXNET_OPERATOR_REGISTER_BINARY(_mod)
.add_alias("_Mod")
.set_attr<FCompute>("FCompute<cpu>", ElemwiseBinaryOp::Compute<cpu, mshadow_op::mod>)
.set_attr<nnvm::FGradient>("FGradient", ElemwiseGradUseIn{"_backward_mod"});
NNVM_REGISTER_OP(_backward_mod)
.set_num_inputs(3)
.set_num_outputs(2)
.set_attr<nnvm::TIsBackward>("TIsBackward", true)
.set_attr<nnvm::FInplaceOption>("FInplaceOption",
[](const NodeAttrs &attrs) {
return std::vector<std::pair<int, int> >{{0, 1}};
})
.set_attr<FCompute>("FCompute<cpu>", ElemwiseBinaryOp::BackwardUseIn<
cpu, mshadow_op::mod_grad, mshadow_op::mod_rgrad>);
} // namespace op
} // namespace mxnet
|
Non-Traditional Student Week
National Non-Traditional Student Week takes place the first week of November each
year. To celebrate Stony Brook University's adult learners, the Office of Commuter
Student Services holds our Lunch on Us event, and actively promotes our Driving Force efforts (see
below). In addition to this week in November, Stony Brook also recognizes adult learners
again in the spring semester, typically in early April. Please check back for specific
dates.
During Non-Traditional Student Week, our partners in the Faculty Student Association
and Campus Dining often offer discounts and/or coupons to celebrate the occasion.
Our office also conducts contests during this week for great prizes!
Non-Traditional Orientation Session
The Office of Orientation and Family Programming and the Office of Off-Campus Living
have worked together to develop a comprehensive Orientation session, during Part II
of Orientation. This session approaches topics from the point of view of a returning
student, while also incorporating the required content that all new students must
fulfill.
Incoming undergraduates over the age of 24, as well as veterans, are directly invited
to attend during their Orientation process. However, if a student self-identifies
as a non-traditional student/adult learner, they may be able to register, as well. |
Inborn errors of steroid biosynthesis: detection by a new mass-spectrometric method.
A new mass-spectrometric technique relies on ionization during bombardment of the analyte (dissolved in a liquid matrix, usually glycerol) by an atom beam (e.g., Ar0, Xe0). This technique, termed "fast atom bombardment," is particularly useful in the characterization of polar charged molecules. A neutral beam is not essential, and a primary beam of cesium ions has been successfully used to produce spectra equivalent to those obtained by fast atom bombardment. In this communication I report data on the use of both ion and atom primary beams for producing secondary-ion mass spectra of conjugated steroids. In negative-ion spectra produced for steroid glucuronides and sulfates, the ion [M - H]- is invariably the major high-mass peak, and the lack of substantial fragmentation allows assay of relatively complex mixtures if the analytes differ in mass. I describe here the use of secondary-ion mass spectrometry for distinguishing, by urinary steroid analysis, patients with the four enzyme defects that can affect cortisol synthesis: defects in 17 alpha-hydroxylase, 3 beta-hydroxysteroid dehydrogenase/isomerase, 21-hydroxylase, and 11 beta-hydroxylase. |
A trio of New England-based PDL sides will partake in the 2014 U.S. Open Cup competition, with all three set for first-round action on Wednesday, May 7.
MPS Portland Phoenix will host FC Lehigh Valley United Sonic (NPSL) at 7:00pm at Deering High School in Portland, Me. In 2012, the Phoenix became the first club from the state of Maine to compete in the historic tournament when they hosted the Brooklyn Italians for a first-round clash, which ended 3-2 in favor of the Italians.
A few miles to the south, the Western Mass Pioneers will take on Mass Premier Soccer (USASA) at 7:30pm at Lusitano Stadium in Ludlow, Mass. The Pioneers’ last chance at Open Cup glory came in 2010, which saw them face the FC New York (USL) in a first round tie that went the way of the Empire State side.
“It is great to qualify for the Open Cup,” Pioneers coach Federico Molinari said in a club release. “Every point counts during the season, so we are happy that we got enough to qualify this year.
“Mass Premier Soccer is a good team that has been playing together and knows each other. This is our first competitive game this year. The pressure is on us to win and advance.”
Meanwhile, the Vermont Voltage will face the New York Red Bulls U-23s (NPSL) at the Red Bull Training Facility in Whippany, N.J. at 8:00pm. The Voltage are the oldest New England-based PDL club, having launched in 1997.
The second round of the competition is set for May 14.
2014 Lamar Hunt U.S. Open Cup Schedule First Round Schedule
Date Game Time/Result Venue May 7 FC Lehigh Valley United Sonic (NPSL) at GPS Portland Phoenix (PDL) 7 p.m. ET Deering High School; Portland, Maine May 7 RWB Adria (USASA) at Detroit City FC (NPSL) 7 p.m. ET Stevenson High School; Livonia, Mich. May 7 Mass Premier Soccer (USASA) at Western Mass. Pioneers (PDL) 7:30 p.m. ET Lusitano Stadium; Ludlow, Mass. May 7 Vermont Voltage (PDL) at New York Red Bulls U-23s (NPSL) 8 p.m. ET Red Bull Training Facility; Whippany, N.J. May 7 Red Force (USASA) at Colorado Rovers (USSSA) 8 p.m. MT Shea Stadium; Highlands Ranch, Colo. May 7 CD Aguiluchos (NPSL) at Ventura County Fusion (PDL) 7 p.m. PT Ventura College; Ventura, Calif. May 7 TBD (US Club Soccer) at San Diego Flash (NPSL) 7 p.m. PT Mira Mesa High School; Mira Mesa, Calif. May 7 LA Misioneros FC (PDL) at PSA Elite (USASA) 7:30 p.m. PT StubHub Center Training Field; Carson, Calif. |
= -61. Let c be 0/(-2 - k) - -7. Solve -3*p + 9 = 0, -5*v - c - 6 = -p for v.
-2
Let s be 1 + (-30)/(-9) + (-125)/375. Solve 5*u + v = 1, u + 19 = s*v - 9*v for u.
1
Let g be (-1 - -1) + 24/(-4) + 5. Let f be 86/4 + (-5)/10 - g. Solve -4*i - i + 3*o = -19, o - f = -4*i for i.
5
Let f(d) = 35*d - 37. Let r be f(2). Solve -4*c - 5*k = r, -4*c + 2*k = -2 - 0 for c.
-2
Let c be (-1 - 3/(-3))/(-2). Suppose c = -5*l + 12 + 33. Let d = l + 4. Solve -h - 4 = -5*t + 5, -5*h + d = 4*t for t.
2
Suppose -7*i + 22 - 8 = 0. Solve i*h - 4*m + 0*m = -14, 0 = 3*m - 15 for h.
3
Suppose 903 - 959 = -14*s. Solve -3*j - 2*h + 11 = -s*j, -3*h = 4*j for j.
-3
Let i be 2/(-5) + (-188)/(-20). Suppose -2*o - o + i = 0. Suppose 360 = w + 29*w. Solve 2*f - 4*c = -0*c - 14, -o*f = 5*c - w for f.
-1
Let c(k) = k**3 - 5*k**2 + 8*k - 3. Let l be c(3). Let w(n) = 2*n + 8. Let y be w(l). Let z = 16 - y. Solve -5*t - 3*u = 24, -z*t + u + 9 = -4*t for t.
-3
Let h be (-56)/(-6) + -1 + 2/(-6). Suppose 3*u + i - 27 = 0, 2*u - 3*i + h*i = 31. Solve -3*k + 4*l = 6 + u, 5*k - l = 5 for k.
2
Let l be ((-68)/(-136))/((-2)/(-20)). Solve 2*i - d = 1, -5*i - l*d = -0*d - 25 for i.
2
Let i(r) = -r**2 + 9*r + 12. Let f = -84 + 94. Let z be i(f). Solve 3*p - z*n = -8 - 9, -1 = -n for p.
-5
Suppose 19*n - 263 + 187 = 0. Solve 16 = -n*i + 2*g, 0 = -5*i - g + 6 - 19 for i.
-3
Let s(x) = x**2 + 7*x + 7. Let p be s(-6). Let v = 2 + p. Solve -6*o + v*f - 4 = -2*o, -f - 24 = 5*o for o.
-4
Suppose -2*g + 4*g = 22. Let s = g - 9. Solve 1 = -5*h - s*z - 0, 3*h - 5*z + 13 = 0 for h.
-1
Let x(q) = q**3 + 6*q**2 - 15*q + 14. Let a = -43 - -35. Let w be x(a). Solve 0 = 2*t - t + 2*j, -4*t - 2*j - w = 0 for t.
-2
Suppose 0 = 5*k + 3*o + 27, -4*k + 5*o = 3*o + 4. Let p = k + 3. Solve -i + 15 = -3*t - 6*i, p = -i for t.
-5
Let y = -331 - -334. Solve -17 = 3*z - 5*k - 5, 3 = -y*z + 2*k for z.
1
Let l = -72 - -109. Let x = l + -26. Solve 3*b - 23 = 6*h - x*h, 0 = 4*b - 4 for h.
4
Suppose 13 = 9*c - 14. Let l = 9 + -5. Suppose l = -2*j + 52. Solve -4*g - q + 2*q + 23 = 0, c*g - j = 3*q for g.
5
Let h(u) be the third derivative of u**4/12 + 7*u**2. Let r be h(1). Suppose r + 6 = 4*i. Solve 11 = i*c - 0*c - 3*j, 0 = c - 2*j - 7 for c.
1
Suppose -3*s - 15 = -2*k, 4*k - 4*s - 7 - 17 = 0. Solve -2*f = 3*c + 20, -k*c - 28 = -3*f + 7*f for f.
-4
Let q(u) = -33*u**3 - u**2 + u. Let g be q(1). Let f = 33 + g. Let j = -8 + 13. Solve l - 3*a = -0 - 5, j*a - 15 = f for l.
4
Suppose 5*h - 4*c = -45 + 201, 68 = 3*h + 4*c. Suppose -5*l + 22 = z - 7, 2*l = 5*z - 10. Let w be 0/6*(-2)/z. Solve 5*b - r - h = 2*r, w = -5*r - 5 for b.
5
Suppose -u + 17 = 2*r, -u + 22 = 4*r - 17. Suppose r + 1 = 3*t. Suppose -28*k = -32*k + 8. Solve 2*d + 4 = 0, 4*h - t = 3*d + k for h.
0
Let h = 606 - 585. Solve -3*z - 2*r = -h, z + 5*r = -0*r + 20 for z.
5
Let t = 105 + -100. Suppose 4*b - 2 = -2*d, 2*b - 2 = -3*d + t. Solve 0 = 3*y + 5*l - 9 - 10, -d*y = 4*l - 17 for y.
3
Suppose -2*v + 3*a - 24 = -6*v, -3*v + a = -18. Suppose -v*z = 3*z - 54. Solve 3*r + 1 = -4*y - z, 0 = -2*r - 5*y - 14 for r.
3
Let n(f) be the second derivative of f**4/6 - 11*f**3/3 + 8*f**2 - 16*f + 1. Let m be n(11). Solve 0 = -4*s - 4*u + 12 + m, -3*s - 3 = -3*u for s.
3
Let v be ((-12)/27)/((84/27)/(-14)). Solve 2*t - 4 = -0*g + 2*g, 6 = t - v*g for t.
-2
Let l(h) = -14*h - 1. Let g be l(5). Let a = 71 + g. Solve 2*d + t = -1, d = -t - a*t - 1 for d.
0
Let j = -2 + 4. Let k = 238 + -232. Solve j*y = -5*u, -y + u - k*u = -5 for y.
-5
Let f(h) = -h - 15 + 1 + 5 - 8. Let g be f(-21). Solve -3*o - 9 = 0, 2 - g = 4*z - 2*o for z.
-2
Suppose 2*y = 34 + 4. Let g = y + -18. Solve 3*j + 24 = 4*i, 0 = -2*i + j + g + 9 for i.
3
Let b(a) = -44*a + 356. Let u be b(8). Suppose -3*g - 22 = -97. Solve u*f + 2*o + 20 = 0, o - 6*o - g = 5*f for f.
-5
Let x(y) = y**3 - 6*y**2 + 4*y + 1. Let f be x(5). Let s = f + 6. Suppose -79*z + 84*z - 150 = 0. Solve -2*t - 4 = -j + s, 0 = 5*t + 5*j + z for t.
-4
Let g = 38 - 30. Suppose 5*y - 20 = -j, y - 3*y + g = 3*j. Solve 4*m = -5*x + 45, -9 + y = -2*x + m for x.
5
Let l be -3 - ((-3 - -5) + 1 + -40). Let x = l - 31. Solve -3*u + 5 = 4*f + 18, x*u + f + 1 = 0 for u.
1
Let z be (-5 - -3)*(-1 + -5). Suppose 0 = -55641*v + 55664*v - 161. Solve -4*k - 5*c + z = 0, 5*k + 3*c - 15 = v*c for k.
3
Let l be (-39)/9 + 154/66 - -4. Solve 5*u - l*s = -3, 4*s - 3*s - 9 = -5*u for u.
1
Suppose -48*j = -44*j + 5*p - 35, -4*j + 3*p + 43 = 0. Solve 0 = -2*r + m + 6, 4*m + m = 2*r + j for r.
5
Let x be 4 - (4 - (1 - (3 + -5))). Solve v - 14 = 5*i, -i + x*i + 24 = 5*v for i.
-2
Suppose -d + 3*m = 8*m - 4, -2*m = -3*d + 12. Let q be (0 - d/(-1)) + -1. Let g be 1*(q - 1 - -3). Solve -x - l + 0 = -1, 3*x + l = -g for x.
-3
Let l be ((-3)/2)/((-5)/10). Let z(j) be the first derivative of -j**4/4 - 7*j**3/3 + 2*j - 4. Let h be z(-7). Solve -8 = -h*m, -m - l = b - 2*m for b.
1
Suppose -14*k = 13*k + 15*k. Solve k = -3*n + n + 2, 5*n = z + 3 for z.
2
Let w be 21 + (-1 + 4 - -3). Solve -w = 5*k - 0*k - 2*b, -2*b = k + 3 for k.
-5
Let t = 379 + -376. Solve -4*q + 8 = 2*u - q, t*q - 7 = -u for u.
1
Let t(p) be the third derivative of -p**5/60 - 7*p**4/24 + p**3 - 4*p**2. Let f be t(-9). Let b = -10 - f. Solve 2*h + 19 = 6*k - k, -31 = -5*k - b*h for k.
5
Suppose 233 = 4*x + 177. Solve q - x = -5*z, 0 = -2*z + 2*q - 1 - 3 for z.
2
Let n be 10/15 - (-2)/6. Let y(u) = 86*u**3 + u**2 - 2*u + 1. Let a be y(n). Suppose -5*q = -4 - a. Solve 2*d + 0*z - 5*z = -24, -d + 4*z - q = 0 for d.
-2
Suppose 2*s + 25 = 7*s. Let p = -6 - -9. Suppose -2*u + 0 = -5*r + 2, 0 = 2*r + p*u + 3. Solve -f - v = v, r = -s*v for f.
0
Suppose -2*j = -4*j - 4. Let q be (3/6 + -1)*j. Suppose -11 = -4*l + q. Solve -20 = -4*y, -x - x + l*y - 19 = 0 for x.
-2
Let c = -2810 + 2813. Solve -3*v - i = -3, 3*v - c = 4*i - 8*i for v.
1
Suppose 5*a = 8*a - 8136. Let y be a/90 + (-6)/45. Let r be (y/8)/(9/12). Solve -r*h + 21 = -3*g + 2, 4*g = -h + 13 for h.
5
Suppose 4*n + 3*t = 96, 2*n - 54 = -t - 2*t. Suppose 0*c - n = -7*c. Solve -v - 5*r - 20 = 0, -c*v + 2 = -3*r + 8 for v.
-5
Let k = -48 - -59. Suppose 0 = 15*g - k*g - 8. Solve 5*d = -4*q + 26, -3*q = -g*d - 12 + 4 for q.
4
Suppose -3*f = -8 + 2. Suppose f*h = -2*s, -h - s = 2*h - 10. Suppose 3*l + h*y = 82, 2*l - 6*y - 38 = -y. Solve -4*x = 2*w - 4*w - 14, -5*w - l = x for w.
-5
Let z be (-12)/(-42) + 216/28. Solve 0 = -d + 5, z = -2*w - 2*d + 22 for w.
2
Let d be 2/(-3) - 221/51. Let r be 30/(-75) + (-12)/d. Solve -5*c + 9 = -q, 2*c + 0*c - 2 = r*q for q.
1
Suppose 0 = -5*l + 11*l - 30. Solve 5*i = -4*c + 2*c - 1, c = l*i - 8 for c.
-3
Suppose 0 = -0*h - 16*h + 96. Let k be (-20)/(-6) - (-6)/9. Suppose -h = -6*u + k*u. Solve f - 9 = 5*y + 12, -31 = u*y + 4*f for y.
-5
Suppose 0 = -14*b + 6*b - 144. Let q be -10*(b/(-4)*-1 + 4). Solve -12 = -4*i - 3*t + 7*t, -q*t = -4*i + 14 for i.
1
Let t(k) = -k**3 + k + 4. Let j be (3 + (2 - 5))/(-1). Let v be t(j). Suppose -6 = 2*p - 0, v*b - 5*p - 23 = 0. Solve 2*h + b = -3*y - 2, -5*y + 2*h = 28 for y.
-4
Suppose 11*p = 2*p + 108. Solve i + 4 = -4*m, 2*i = 5*i - 4*m + p for i.
-4
Let f(m) = -m**2 + 8*m - 3. Let h be f(9). Let s = h + 14. Let p be (-7)/s*16/(-28). Solve -z = -5*y + 23, 3*z - 2 + 3 = -p*y for z.
-3
Suppose 96 = 5*b - 4*o, -4*b + 2*o = o - 79. Let k be (-122)/14 - 18/63. Let q = k - -10. Solve -a = 3*w - b, 2*a - 24 - q = -3*w for a.
5
Let j(d) = -20*d - 3. Let h be j(-2). Suppose -5*n = -47 + h. Solve 0*z = 3*a - 5*z - 6, n*a - z = 4 for a.
2
Let f(q) be the third derivative of q**5/60 + q**4/2 + 20*q**3/3 - 9*q**2. Let z be f(-5). Solve 0 = x + 2, 4*x - 21 = -z*n + 2*x for n.
5
Let h be (-1 - 0/(-4))/(3/(-9)). Solve 5*d + h*l = 32, 0 = -3*d - 4*l + 7*l for d.
4
Let z = -276 - -276. Solve z = -4*n - 4*j - 12, 0 = -n - 2*n + 3*j + 21 for n.
2
Suppose -4*f - 5*o + 25 = 0, -3*o + 12 + 3 = f. Solve -2*t - 2*j = -0 + 4, f = 3*t + 5*j + 8 for t.
-1
Let f(z) = 2*z - 16. Let j be (-3546)/(-270) + (13/15 - 1). Let u be f(j). Let g be 1 - 2/4*-2. Solve -3*q = -b - 9, -g*b = -q - 3*q + u for b.
3
Suppose -5*u - q + 76 = -5*q, -q = 4*u - 65. Solve -3*r - 4*f - u = 0, r + 3*f - 2 + 9 = 0 for r.
-4
Suppose -1191*z - 25 = -1196*z. Solve 3*s - 9 = z*y - 4*y, -s = -2 for y.
-3
Let s be -1 + -4*(-3 + 2). Suppose 3*b = -2*k |
Here we have it! Another winner from Synergy soaps today I will be reviewing Synergy's newest and seasonal artisan soap Pumpk'n 3.14 and, just in time too as the season is coming on us fast this year in the north-east. First off this should come with a warning as I opened the puck and literally not figuratively it smelled good enough to eat. As with all of Synergy soaps they are vegan friendly don't be hating on us vegans and contain all organic butters and oils such as Kokum, Mango, Cocoa, Avocado, PUMPKIN, Coconut and Neem oils and butters just to… Review: Synergy soaps Pumpk’n 3.14 Review: Synergy soaps Pumpk’n 3.14 Packaging Scent Lather Slickness 95 Overall- This seasonal limited edition soap from Synergy soaps is a true winner! whether used as a fall time pumpkin pie spiced shave soap or a summer delight paired with a bay rum after shave this soap pleases the senses in every way from scent to the feel on your face. Now read this review and go pick up a puck today!!!! User Rating: Be the first one !
Here we have it! Another winner from Synergy soaps today I will be reviewing Synergy’s newest and seasonal artisan soap Pumpk’n 3.14 and, just in time too as the season is coming on us fast this year in the north-east. First off this should come with a warning as I opened the puck and literally not figuratively it smelled good enough to eat. As with all of Synergy soaps they are vegan friendly don’t be hating on us vegans and contain all organic butters and oils such as Kokum, Mango, Cocoa, Avocado, PUMPKIN, Coconut and Neem oils and butters just to name a few ( there are more in there I have not mentioned as well as glycerin and powders ). This is a true soap once again fully saponified and ready to make those whiskers soft as can be, for that blade to slice through. As the saying goes as easy as slicing pumpkin…. pie….. pun intended. So on to the review we go.
Packaging
What a great spin on Synergy’s already cool classic design !!!! The handlebar mustachioed man in Synergy’s regular line was carefully transformed into his Halloween costume and became a pumpkin just for you! Whereas there was a blue man as you see is a pumpkin surrounded by much of what I see going on around me in the real world the autumn change of the leaves from green to hues of yellow, orange, red, and combinations of all the above. Of course I could not resist. I had to have the 5″ tin which holds 8 oz. but there was also an option to buy a smaller tin containing 5 oz. of this awesome mouth-watering soap. I keep going back to the eat it mode in my mind…… really it smells that good sorry I digress. Unlike the Cavendish, Vanetiver , or Bay Rum or other scent options Synergy offers this tin is a painted, probably baked finish of a gold/bronze color. I know this really makes little difference but, I noticed this did not have that tape seal that I had on previous purchases from Synergy. Deal breaker ? Not even close but I do think that was a nice touch maybe that could be added to this soap as well. Other than that spot on the lid fits perfect the soap filled the tin to the top and it was sent to me in a protective wrapped packaging ensuring my lather gold was safely going to make it to me. Winner !!!
Scent
Anyone say pumpkin pie ?!! There is the classic spice of pumpkin pie, cinnamon, clove, nutmeg, ginger, as well as of course pumpkin. Blended to perfection the scent is spot on for the season and I might find myself hard separated from this soap as fall moves along and turkey day gets nearer and nearer. Another strong point to make and I never thought about it until I was reading another persons SOTD and saw they used a bay rum aftershave with this soap. I got thinking ( rare as that may be ) about the similarities between the two especially Pinaud as it is very clove forward and really does compliment the soap well other bay rum after shaves may not work as well clove forward is the key. To be more specific this soap though marketed for a seasonal soap really would be nice year round when using a bay rum after shave. Again Mr. Smythe spot on !!! the only thing I must warn you again this is a pumpkin pie in a soap so if you do not like pumpkin pie you just may after smelling this soap.
Lather
As with all of Synergy soaps the lather is a force to be reckoned with. This soap incorporates a few more ingredients to it than the standard offerings and I of course as with any soap I use for the first time did a heavy loading into my Semogue 830 which is a great brush but finicky so I basically made this the money part of the review as this is what is the most important part of a soap. DOES IT LATHER ? well I can confidently say yes and as again with all Synergy soaps is an explosive lathering experience. I did a 15 second clockwise 15 second counter-clockwise loading and again this was with a HOT brush as this soap works best between 90-140 degrees. The reason behind this is very simple an oil or butter stays more solid at a cooler temperature and the warmer it is the easier it is for those oils to adhere to your brush and to lather up on your face. I had enough lather for 4 maybe even 5 passes with this method so a 20 second loading with this particular brush is probably perfect for a 3 pass shave with touch ups. Now with my 1305 or my Edwin Jagger or Omega 63185 it would be more like 10 seconds or less as the saying goes YMMV in this department. Another great thing and the last to be mentioned in this portion of the review is Douglas Smythe artisan of this soap made I so this and all of his soaps will work great in hard or soft water so that right there is a huge deal. For the lather portion again, another huge thumbs up !!!
Slickness
Hold on tight you’re in for a ride !! I choose a 1922 Gillette new improved “New Standard” model to shave with today. this happens to be one of my more aggressive open comb razors I find the angle to be pretty unforgiving without proper preparation, a SLICK SOAP, and great technique. Well not a weeper, nick, or cut was to be had by the way did I mention this was with a Voskhod blade on its first use! The glide was perfect and I went through 3 passes and a 4th for fun using the method shaving technique and came out BBS with no tugging or irritation at all. When properly built the lather is second to none and as I expected from using other Synergy soaps was not disappointed at all in the slickness and glide. I even speed buffed on sensitive areas of my neck with ease and that to me says it all. I’m starting to wonder what this soap and brand cannot do as it seems perfect in every area. A lot of love went into this soap and that is very obvious.
This is what was still left after 4 passes and a 5th coating or moisturizing my skin.
Overall
This is a seasonal soap from a superior artisan soap maker. I would argue this could easily be a year round soap especially when paired with the correct after shave or balm but it really shines this time of year. What can I say this soap passes all the major test of a great soap with flying colors…. Packaging , scent , lather and slickness exceed the “best” soaps out there that are four times the cost I paid for this soap $13 on Etsy I will be using this soap again and again and will be in my rotation which is becoming smaller and smaller every time Synergy puts a new scent out. I highly recommend this brand in general and this limited edition particularly. So get it while you can, heck buy two or three as you will not regret it. From packaging to clean up a near perfect soap. To finish this review I actually asked Douglas about the 3.14 …… I understand the Pumpk’n but 3.14? Why? I actually laughed when he gave me the answer as a self-proclaimed math geek it was staring me in the face Pi or 3.14159 or an irrational number also represented 22/7 now in this case I think it has to do with the circumference of the tin and soap itself as the perfect size for a tin and loading a brush. also think of this as a really inventive way of saying pumpkin PI(e). Get it Pumpkin Pi!!!!!! LOL. I think that’s pretty creative. Tell me now that a lot of thought has not gone into this soap. Again Synergy team a huge thumbs up from us here on 365 Shaves.
— Aaron
Now I am ready for my day !!!!! |
Christmas… as an Agnostic by Melanie Noell Bernard – Day 22 Advent 2017
Christmas… as an Agnostic by Melanie Noell Bernard
Day 22 Advent 2017
Like many children, I was raised in my parents’ religion. For me, this meant going to a Presbyterian church once a week on Sunday (or trying. Sometimes life is a little busy.) But I was expected to get up earlier than I wanted, put on nice clothes to appear presentable, and head to church with my parents and older brother. However, my Christian-upbringing was a bit unorthodox.
I have never read the bible. I stopped going to Sunday school as soon as my parents would let me just so I could sit quietly (and impatiently) in service with them. (I was that person drawing on the bulletin during service… Oops.) No one ever really explained the bible verses to me or that there were two testaments. I had no idea who the apostles were. I can’t recount the ten commandments. Despite having been in church my entire life, I had no idea what I was doing there.
Maybe that is the reason (or perhaps it’s my scientific nature striving for proof), but I never really believed. Not to say I didn’t try. Many nights I would pray before bed, asking for a sign, and not getting one. Of course, that’s because you’re not supposed to base it on proof. You are supposed to believe in God without proof and live your life as a good person based on your faith. Unfortunately, the older I became, the more I realized I didn’t actually believe and no matter how much I tried and how much I prayed, believing is harder than you think. As such, over this past year, I finally told my family… I am Agnostic.
To be quite honest, this was probably one of the most difficult things I have ever told my family. Why? I’m not sure. I was finally being true to who I am, but I think there was this worry of disappointing my family if I admit to not believing what they believe. There was this fear of separation or disapproval from them. Funny thing is, none of them even flinched. I had gotten all worked up, spent nearly two years hiding it, only for it to not be a big deal. And life goes on, but to say nothing has changed would be a lie. Things have changed, namely around Christian holidays like Christmas.
Christmas was always the time of year when everyone is preparing and busy and decorating. My family had plenty of traditions for the season. We promptly got our tree the day after (American) Thanksgiving (the first day the tree farm opened). After cutting it down, sipping on hot cocoa and petting reindeer, we would spend the rest of the day decorating the tree and playing Christmas music. My dad would hang lights outside and make a fake tree out of lights on a table that looked really cool at night. My mom would move all the ornaments I hung on the tree to another location and my brother would help her. (I’m not really the artsy one in the family. Hee hee!) We would spend the rest of the month buying and wrapping presents and planning Christmas dinner with a ham and potatoes and all the other yummy goodies. Then came time for Church on Christmas Eve.
This was one of two times a year when everyone (even the non-Church goers) would show up to the one of three services offered on Christmas Eve and listen to the story of Jesus’ birth and sing songs. The only lights during the last song, Silent Night, were candles being held by each of the church-goers filling the pews. I always remember the wax dripping down onto my fingers despite the little white paper that’s supposed to catch it, trying not to wince and keep singing as beautifully as I could (though I’m not really that great. Hahaha!), praying someone (or myself) would NOT light my hair on fire. Then, the song ends, everyone blows out their candles and files silently out of the church just after midnight to head home and go to sleep and wait for Santa to deliver presents. However, this isn’t the way it happens anymore.
Despite being back in my hometown near my parents, I don’t go to church. My brother lives in another state. My parents are divorced. I am Agnostic, and this last piece is the most drastic change of all.
Even as my mother still goes to church and has invited me to go to Christmas Eve service and join her and her church for dinner (aka lunch) on Christmas Day, I’m not sure I should go. I haven’t even put up Christmas decorations this year (nor last year) and anything related to the Christmas season feels… off. I feel like a poser for trying because it’s technically not my religion anymore. That’s not to say I don’t want to spend time with my family, or eat good food, or just bask in the joy and splendor of the holidays, but… I’m just not sure I belong.
What’s more, I’m worried about going to a Christmas service or dinner and having people presume I’m religious or presume I know things about the bible and practice the faith or ask me when I’m coming back to church. I don’t want to partake in something when I don’t fully believe in the very basis of what they are celebrating.
Were it just my family, waking up around a well-lit, gorgeously decorated tree with some snow on the ground and a warm, crackling fire to enjoy, I wouldn’t have a problem. There are no religious obligations behind a nice decoration or enjoying the company of one’s family, but going to church when I have finally been honest with myself about not being part of the church, feels like I am being dishonest to the people who are religious and who do go to church, but mostly it feels like I am being dishonest with myself. I spent too much of my life being afraid of admitting my religious beliefs. I should not have to explain them to anyone and I do not want to pretend again, even for one more day, that I still believe in the Christian God.
Melanie Noell Bernard is a graduate student who explores the blueprint for life: genes. Her scientific background is the inspiration for many of her stories. When she’s not honing her writing craft or researching in the lab, she’s reviewing books and hosting literary discussions on her blog, MNBernard Books.
Welcome to my blog. This place on the internet has been around for a few years now. For a while I almost called it home. Now, I have been on quite a few breaks, but don’t let that keep you from enjoying my content. |
package selector
// Options used to configure a selector
type Options struct{}
// Option updates the options
type Option func(*Options)
// SelectOptions used to configure selection
type SelectOptions struct{}
// SelectOption updates the select options
type SelectOption func(*SelectOptions)
// NewSelectOptions parses select options
func NewSelectOptions(opts ...SelectOption) SelectOptions {
var options SelectOptions
for _, o := range opts {
o(&options)
}
return options
}
|
Assessing the biodegradation of polycyclic aromatic hydrocarbons and laccase production by new fungus Trematophoma sp. UTMC 5003.
Environmental pollution caused by petroleum compounds has become a global concern. The aim of this study was to evaluate the indigenous fungal isolates in Iran for biodegradation of crude oil pollutants. In order to isolate fungal strains, the soil samples were enriched in minimal salts medium (MSM) with 1% crude oil and then the crude oil degradation was measured by total petroleum hydrocarbon (TPH) assay. The degradation of hydrocarbons compounds was also analysed by FT-IR and HPLC, and the activity of peroxidase enzyme and biosurfactant production were also measured. We isolated 40 fungal strains and selected the isolate G-05 with 70% degradation ability of petroleum hydrocarbons as a premium isolate after 15 days. Residual crude oil analysis with FT-IR spectrophotometry and HPLC showed that G-05 is able to degrade 90 and 100% of aliphatic compounds and some polycyclic aromatic hydrocarbons (PAH), respectively. Evaluation of enzymatic activity showed that this isolate can produce 4 U L-1 of Laccase enzyme for oil removal; it is capable of producing biosurfactant and reducing the surface tension of the medium to 25.95 ± 0.1 m Nm-1. This strain was identified as a member of Trematophoma genus and the obtained results showed that this strain is a highly potent strain in bioremediation of soils contaminated by crude oil. |
LiveTrain.nyc
LiveTrain.nyc is a visual representation of subway locations, based on the realtime trip updates provided by the mta.
Why?
Countdown clocks are great, but we wanted a more visual experience of the New York City subway system. So where are the trains? Lets find out.
Who?
Commuters hoping to find out when their train will arrive, and when that train will reach its destination, or anyone curious to see the beautiful complexities of the world's biggest rapid-transit system.
Users of live train can interact with this data in several ways:
Trains can be filtered by route, to show only the relevant trains.
Users can click on a stop to view upcoming departures.
Click on a train to view its up-to-the-minute schedule. Look back to past stops to determine how long a trip took, or look forward to get a clear picture of when a train will arrive.
How?
Using the MTA's GTFS-realtime feed of schedule updates and static datasets, we approximate the location and speed of individual trains.
This information is made available as a RESTful API, and animated using d3.js and Leaflet/Mapbox.
What else?
This project was conceived, developed, and maintained by Michael Prude and Ted Mahoney as their final project for the Web Development Immersive Course at General Assembly, a 12-week program in full-stack development.
We were inspired by Chris Whong's NYC Taxis: A Day in the Life to visualize another iconic NYC transit option. |
Q:
iText flattening issue with watermark text
I am using iTextSharp to add watermarks to existing documents that are in PDF format. I read them in, create a PdfTemplate object, add the watermark text, color, transparency, etc... to this PdfTemplate, then add the template to a PdfContentByte that I get using PdfWriter.DirectContent.
This solution has been working for a while for all documents that I throw at it, but for some reason a few particular PDFs have been printing with an opaque rectangle around the watermark text.
I have tried a number of things already, including setting the PdfGState's OverPrintMode to 1, opening the generated file after it is created and then using PdfStamper's setFormFlattening and setFreeTextFlattening before saving a new copy... Nothing has worked so far.
The issue is that I don't care about flattening form fields or buttons, I just want to flatten all the layers of the document after I have generated the entire thing. Is this even possible with the iText API?
A:
Alright, coming back to close this one out since I solved it a long time ago:
The issue was that the pdfs having the issue were very old (v1.3 of the PDF standard) and did not support the transparency that was a part of the watermark element.
I got around this by printing the old PDF via a virtual printer to a temporary file, then continuing to process that file, which was now a newer version of the PDF standard.
|
Clinical Outcomes of a Team Approach to Thyroidectomy.
The aim of this study is to assess the rates of thyroidectomy complications performed by two attending surgeons operating together. This is a retrospective chart review. This is a retrospective chart review from September 2008 through October 2013 of thyroidectomy cases performed by the head and neck team at Sanford Health. The primary intervention was the presence of two head and neck attendings during each procedure. Outcomes assessed include rates of temporary and permanent recurrent laryngeal nerve paralysis, and of permanent hypocalcemia. There were 282 patients that underwent a thyroid procedure with a total of 449 at-risk nerves. There were five (1.1 percent) cases of transient vocal cord paresis. There was one case (0.22 percent) of permanent vocal cord paresis after planned nerve resection in a patient with anaplastic thyroid carcinoma. There were no other cases of permanent vocal cord paresis. Of 156 total thyroidectomy cases, there was one case of chronic hypocalcemia (0.64 percent). A two-surgeon approach to thyroidectomy produces excellent functional outcomes. Further investigation into cost-effectiveness is warranted. |
Aptos
Aptos is a comparatively large unincorporated town that claims a total of 24,000 residents. It is believed that the Native Americans originally named this area “Awatos,” which means “Where the waters meet.” The name is a good reflection of the area’s unique setting in a place where two creeks join before entering the bay.
Today, the town’s primary centers of residential development include Seascape, Rio del Mar, Seacliff, and the Cabrillo area. A smaller portion of residents also live in the more rural area known as Aptos Hills.
The residents of Aptos enjoy an average household income that is among the highest in the nation. This is, in part, due to the large number of residents who commute to jobs in Santa Clara. With no municipal or city government, the local library, the fire department, chamber of commerce, and sheriff’s service center act as community information centers.
Boasting a pleasing combination of redwood forest and majestic coastline, Aptos offers a diversity of lifestyles. Housing options range from rustic mountain cabins to gracious modern homes bordering golf courses. All homesites enjoy the region’s beautiful wildlife and fauna. This varied terrain and geography offers residents an equally wide range of recreational activities. Newcomers to Aptos will be pleased to find clean, sandy beaches, award-winning golf courses, nearby state parks, and shopping centers–as well as the selection of amenities found in nearby Santa Cruz. |
Suggestions
Lower Dyraaba
New South Wales 2470
Buying, renting or investing in property is a big decision. Knowing that Lower Dyraaba is right for you is just as important as the property itself. We've done some number crunching on Lower Dyraaba's property supply and demand, median property prices, and demographic information to help you make a more informed decision and better understand the Lower Dyraaba lifestyle.
Median Price: The price of a property that falls in the middle of the total number of houses sold over a period of time, no data available as less than 10 sales have been recorded 2015-12-01 ~ 2016-12-05.
2 Bedroom (no data - less than 10 sales).
3 Bedroom (no data - less than 10 sales).
4 Bedroom (no data - less than 10 sales).
Median House Rent
The advertised weekly rent of a property that falls in the middle of the total number of houses listed over a period of time, no data available as less than 10 listings have been recorded 2015-12-01 ~ 2016-12-05.
2 Bedroom (no data - less than 10 listings).
3 Bedroom (no data - less than 10 listings).
4 Bedroom (no data - less than 10 listings).
Median Unit Price
Median Price: The price of a property that falls in the middle of the total number of units sold over a period of time, no data available as less than 10 sales have been recorded 2015-12-01 ~ 2016-12-05.
1 Bedroom (no data - less than 10 sales).
2 Bedroom (no data - less than 10 sales).
3 Bedroom (no data - less than 10 sales).
Median Unit Rent
The advertised weekly rent of a property that falls in the middle of the total number of units listed over a period of time, no data available as less than 10 listings have been recorded 2015-12-01 ~ 2016-12-05.
Contains property sales information provided under licence from the Land and Property Information ("LPI"). RP Data is authorised as a Property Sales Information provider by the LPI.
Supply and demand in Lower Dyraaba
The level of competition in a suburb can affect prices and availability.
0
Visits per property
0
Visits per property
Lower Dyraaba
Average of NSW
How this was calculated
Supply and Demand data.
To give you an indication of the level of competition in Lower Dyraaba, the ratio of demand relative to supply is considered.
Demand is calculated as the average number of visits per listing per month over the last 12 months to realestate.com.au/buy that included at least one property details page view in Lower Dyraaba, NSW 2470. Supply is calculated as the average number of property listings per month on realestate.com.au/buy in Lower Dyraaba, NSW 2470 over the last 12 months.
The level of demand for individual listings varies. The data above shows the average for the suburb. realestate.com.au makes no claim about the statistical significance nor accuracy of the data. Please note: This information is based on realestate.com.au data. The purpose is to give buyers and sellers an indication of interest in properties in the suburb, based on that data. It is not a definitive representation of market supply and demand.
Time to let go?We've got more if you're thinking of selling in Lower Dyraaba.
Lower Dyraaba Report
You may be interested to know within Lower Dyraaba during November 2016 there were a total of properties sold. Prices ranged from for a , through to for the sale of a .Examples at both ends of the market include a that sold for between and a that sold for between .
Get a free property report and connect with a local property professional for a price estimate.
Established Couples & Families make up the largest percentage of people living in Lower Dyraaba followed by Elderly Couples and Elderly Singles.The statistics above have been sourced from Mosaic demographic data that is the copyright property of Experian Australia Pty Limited (or its licensors).
Similar places to Lower Dyraaba in
Love what Lower Dyraaba has to offer? You may also be interested in taking a look at some of the following areas.
How similarity was identified among suburbs
Experian Pty Ltd classifies Australian households into one of 49 Mosaic Types based on geo-demographic segmentation. realestate.com.au calculates the composition of household Mosaic Types within each suburb, and compares this composition across suburbs to determine the similarity between suburbs. The suburbs with the most similarity to Lower Dyraaba, NSW 2470 are shown. No other factors are considered.
Myfun.com, owned and operated by REA Group, is proud to showcase international real estate to property buyers in China.
REA Group is a multinational digital advertising company specialising in property, headquartered in Melbourne, Australia. Listed on the Australian Securities Exchange (ASX:REA), we operate Australia’s leading residential and commercial property websites, realestate.com.au and realcommercial.com.au, Italian site casa.it, atHome.lu, and atOffice.lu in Luxembourg, atHome.de in Germany, immoRegion.fr in France, and Chinese property site myfun.com. We also have significant shareholdings in Move, Inc. in the United States and iProperty Group Ltd (ASX:IPP), which owns a number of property portals in Asia.
Myfun.com, owned and operated by REA Group, is proud to showcase international real estate to property buyers in China.
REA Group is a multinational digital advertising company specialising in property, headquartered in Melbourne, Australia. Listed on the Australian Securities Exchange (ASX:REA), we operate Australia’s leading residential and commercial property websites, realestate.com.au and realcommercial.com.au, Italian site casa.it, atHome.lu, and atOffice.lu in Luxembourg, atHome.de in Germany, immoRegion.fr in France, and Chinese property site myfun.com. We also have significant shareholdings in Move, Inc. in the United States and iProperty Group Ltd (ASX:IPP), which owns a number of property portals in Asia. |
Monday, December 30, 2013
PROMO: Crystal Spears
Teaser from Shadowing Me (Unedited)
A
look into Chapter One
Subject
to change as rewrites may happen.
Chapter One
I slam harder into this nameless face's mouth, willing
and begging her to gag on my cock. I need to inflict some pain. It’s true I’m a
sadist. Not a sick ass sadist. I just need to cause a little pain in order to
cum. I don’t have a fucked up tragic past that makes me the way I am. I just
need it to release. Regular fucking doesn’t do it for me. I found that out when
I was fifteen and I wrapped my arm around my girlfriend’s neck, choking her as
we both came. She was shocked, yet she didn’t run away. She simply got dressed,
hopped on over to her piece of shit computer, and waited for the slowass
internet to dial up. She then Googled my need to choke her to cum. We
researched for hours upon hours until we figured it out. I was indeed a motherfucking
sadist. She was my best friend and gently let me down, telling me she could no
longer see me. That shit broke my heart. We had been fucking one another since
the age of thirteen, and when I choked her, I knew it scared the living shit
outta her. Who could blame her? I sure the fuck couldn’t. Once I figured out
that was exactly what I needed to get off, I started seeking out older women. Women
that knew what I needed. I’ve fucked and hurt countless women, and they enjoyed
that shit. See, this bitch that has my cock in her mouth right now just isn’t
doing it for me. I shove away from her and start cursing at her for being in a
club she doesn’t understand. Everyone knows what I need here.
Dammit.
She cries that she doesn’t understand, and I zip my pants,
pat her gently on the shoulder, and tell her to do some research. I’m clearly
not in the mood to try again with another nameless cunt so I say my goodbyes to
my best friend. He happens to own this club, and my President and his wife are
about to take it out and he has no fuckin’ clue. As I walk out and the fresh
air hits my face, I breathe it in deep, welcoming the feeling into my lungs. I
felt like I was suffocating in there. There isn’t anything worse than being
tortured by your own fuckin’ downfall. I straddle my bike, start her up, and
head to the clubhouse, praying I don’t run into Tatiana. I want that girl bad,
so fucking bad, but not even she can get my dick hard unless I think about
tying her up and putting some welts to her. I just know she’s waiting for me. Something
in my bones can feel it when I pull into the clubhouse parking lot. I light up
a joint and make my way to the building. As soon as I open the door, walk
through the hall, and start taking the steps up to my new room… I get the scent
of sex. I fling open my door and the doobie falls outta my mouth, hitting my
new carpet. I can’t even think about Winter kicking my ass for burning a hole
in my new carpet because the beautiful, sexy, young Tatiana is naked on my bed.
I kick the door shut and stomp on my joint as I watch the show she’s putting on
for me. Her small delicate fingers work her clit around and around as her eyes
stare directly into mine. Her breathing is heavy and by her smell, I know she’s
about to cum. She’s about to cum so hard, but what throws me off is my dick
starts twitching in my jeans and that snaps me outta my Tatiana induced coma.
“T… whatcha doing’ here darlin?” I laugh, barely making
it seem as if she’s affected me.
She grins at me. Her eyes twinkle with mischief and I
know she’s up to no good. I know Tatiana so well; I just know my bitch is
trying to make a point.
“Oh, Shadow baby, I’m about to cum so hard. That’s what
I’m doing’ here.” She answers, her breath ragged, full of sex.
I watch. Fuck do I watch. I’ve wanted this girl since I
watched her plop down on the couch in the old clubhouse and eye fuck me like no
one’s business.
“Say… T… darlin’,” she begs.
I grin and push my shades up to the top of my head and
cross my arms. She wants to play games. I’ll play.
“T… darlin’,” I breathe out huskily.
And that’s when it happens. That’s when my devilish,
curious girl let’s go, cumming all over my bed. Her chest heaves up and down
for a few moments before she stands, snatches up her dress and throws it on. I’m
still fuckin’ dazed as to why she just did this.
“Why?” I smirk.
She gracefully walks over to me, placing her hand on my half-hard
cock and murmurs into my ear. “Because now when you lie down in your bed every
night… You’re gonna think about laying with my cum. You’re gonna think about me
touching myself, and how badly you wanted to be the one to do it.”
Fuck
Me.
Stay
tuned for more info on Shadowing Me, releasing after Resenting Me.
~~~~~~~~~~~~~~~~~~~~
MY ALL TIME FAVORITE SCENE OUT OF MY SERIES SO FAR IS...
Midway through Chapter 31 and Chapter 32 in Withstanding Me. Be advised 18 and older.
“STORM?”
Mason.
“I’m here,” I croak, my voice
hoarse. He enters the doorway and drops down in front of me. I’m gasping for
air, and he surveys the room. “Oh shit baby. You fought; you fucking fought
back.”
I nod my head yes, as I rub my
throat. “Tatiana.” I gasp and take off on my hands and knees until I can get
upright.
“MACE,” I scream out hoarsely.
Oh god, oh god.
“We’re back here,” Mace yells back.
When I stumble through the door of
the room, I stop, and Mason slams into my back.
“It’s not our blood.” Mace answers
quickly and points to the right. “It’s that motherfucker’s.”
Berry, Rap, and Winter. I take off,
bounding down the steps, one floor after another. When I hit the landing, my
ankle almost gives out as I run and bang on the bathroom door. Rap calls back that
everything is okay and I fly towards the utility closet.
“ANGEL,” Braxxon roars.
“She’s in here,” I call back as I
work to remove all the shit I threw on top of the crawl space door.
Both Mason and Braxxon run up to the
door. “Where the fuck is she?” Braxxon curses at me.
“I locked her in the damn crawl
space. She was trying to play fucking Rambo.”
Braxxon pushes me outta the way, and
when Mason picks me up by my arms, I give way and start bawling. Fuck, what a
nightmare.
“It’s okay.” Mason croons.
“No it’s not. I couldn’t figure out
the safety on Sniper’s gun,” I gasp through my tears. “I almost died. That’s
not okay.”
Mason stiffens beneath me.
“WHAT THE FUCK WERE YOU THINKING?”
Winter yells as Braxxon helps her outta the crawlspace. “GOING AROUND SAVING
EVERYONE BUT YOUR OWN FUCKING SELF.”
He stiffens more beneath me.
“I… I promised Mason that I wouldn’t
let anything happen to them.”
Rap and Berry come out of the
bathroom. “Storm threw us into the tub, and took off after Tatiana, Mace, and Winter.”
Mason pulls away and stares at me.
“She tried strangling the Russian
that was trying to rape me too,” says Tatiana while coming down the other side
of the hall.”
“Don’t ask me to make promises, and
then not expect me not to fulfill them, Mason,” I say while rubbing at my neck.
I lift my fingers and see blood.
Hearing the sirens, I cringe. Fuck,
all this, and now the cops?
“What did I fuckin’ tell you about
your life meaning just as much as theirs?”
I look into his brown eyes and say.
“They’re important to you, so they’re important to me.”
He flinches. “You’re important to me
too Storm, not just them.”
I say nothing. I just blink and lay
my head back. I’m too emotional to be sharing feelings right now.
“Wait. Where’s Piper?” I fling
myself up.
Mason puts his hand on my chest.
“She’s fine. Sniper found her hiding in one of the cars.”
I let out a breath before realizing
two other people are missing. “Bom-Bom? Creamy?”
Shadow snorts. “Those bitches took
off.”
Good for them.
“I’m so fuckin’ pissed at you,”
Winter hisses behind me.
“Get over it. Your ass is pregnant.”
I’m tired of this shit.
“LEAVE ME THE FUCK ALONE. ALL OF YOU.
SEND THE FUCKING COPS TO MY ROOM.”
And I walk my pissed off ass up the
stairs to my room.
***
Chapter 32
ZZ’s POV
The following afternoon…
I’ve left Storm alone. She needs
time. I think. But I gave her last night after the cops and coroner left; I
also gave her this morning. If she wasn’t hurting and pissed off at everyone,
we’d be celebrating the fact we had no casualties for the first time. This
woman put her life on the line for my entire family. Not once thinking about
herself. That is straight up fucking love. I knock on her door, and push it
open when I hear her sniffling.
“I’m not mad at you ZZ; you can go
away now.” She mumbles from the bed, and the breath rushes from my lungs. I
feel my heart pound fast when she calls me ZZ.
This isn’t happening. She’s not fucking
pushing my ass away now. I kick off my boots and crawl into the bed beside her.
I wrap my arms and legs around her and nuzzle my chin into the crook of her
neck. “It guts me when you call me that.”
She sniffles some more. “I don’t
know what you want from me. You ask me to protect them, and I did. Then you get
mad at me.”
Jesus Christ she’s killin me. I
loosen my hold, and rolling her onto her back, I lay on top of her. I run my
fingers through her bangs and then look down into her eyes. I try to keep my
focus off of her neck. It’ll just piss me off. “I didn’t think you would take
it that far Storm. No matter what, I always want you to take care of yourself
too. Always.” Her blue eyes blink at me.
“The Russians you went after?”
Of course, she tries changing the
subject. I don’t know what this shit is where she’s not wanting to share her
feelings now, but it’s pissing me off.
What is this shit? I withstand her;
she withstands me? I’m sick of it.
“You only think you love me because
I keep saving our fuckin’ family.”
I start moving for her, and she
starts back away.
“What the fuck do you think you’re
doing?”
I shake my head with a laugh. “That
right there is one of the reasons I love you.”
She stops moving. “What reason?”
“Because you just called my family
yours babe. That’s one of my reasons.”
She gasps and starts backing away
again. What the fuck is this going to take?
“I love you.” I say again.
“STOP FUCKING SAYING IT.”
Oh, fuck it.
I advance on her. If I have to use force,
I will. I slam her back against the wall and cover her mouth with my hand as I
lock her between the wall and me. Her blues eyes widen with shock. “I fuckin’
love you, Storm. I fuckin’ love you hard, baby. Deal with it. I love every
single thing about you. Your tiny little figure, your black and blonde hair,
that cute little nose stud, those pouty little lips, those big blue eyes, the
way you love me, the way you love our family.
You’re not giving up. I won’t let you. You once told me that’d I fall in love,
and it would be with you. You remember that?” She nods under my hand. “You took
me, a man that didn’t do love, and you owned me Storm. You fuckin’ own me.” I
remove my hand and her body flings into mine, her hands moving quickly tugging
off my shirt. “I love you.” I need to hear her fucking say it.
“I love you, too,” she cries,
unbuckling my jeans. Thank fuck.
Our clothes fly everywhere and we
collide, gasping and pulling at one another, bare skin to bare skin. My heart’s
pounding a hundred miles an hour as I toss her onto her dresser. I smack her
thighs and dig my fingers into her skin as I drag her to the edge and ram into
her.
“FUCK.”
Her head leans against the wall
behind her, her tits bouncing as I thrust in and out of her. She looks so
fucking gorgeous with my cock buried deep. “Tell me you love me,” she moans.
SHIT.
I groan as her walls suffocate my
cock. “I love you.”
Her ankles dig into my ass, pulling
me in deeper. I look in between us and watch as her pussy sucks my cock in and
out. It’s the hottest fucking thing I’ve ever seen. My cock covered in her
juices. I pull out and drop to my knees.
“What are you—?”
Her voice muffles out as my tongue
thrusts into her pussy. She tastes so fucking good. “Ride my face, baby,” I
groan biting her clit.
“Oh god, don’t fucking stop. Don’t fucking
stop.”
Her hips start moving in a wave as I
work her with my tongue. I bring two fingers up to my mouth and suck on them. I
return my mouth to her aching pussy and round her puckering lips with my
fingers, shoving them in.
“FUCK.”
I don’t stop; I know she likes ass
play. I continue to pump my fingers in and out of her as she rides my face. She
starts shaking beneath me. She’s about to come and I want to be inside her when
she does. I back up, stand, and replace my mouth with my cock, continuing to
fuck her ass with my fingers as her cunt sucks at my dick. I lean in and take
her mouth with mine. She licks and sucks her juices off my face, and her pussy
grabs ahold of me so fucking tight I can’t hold it in any longer.
“COME,” I groan into her mouth, and
she does, fuck does she. I bite her lower lip as I shoot inside of her. I
remove my fingers and drop my head to her neck, kissing it softly as we both
try to catch our breath.
After a few minutes of silence, she
finally whispers, “You really love me?”
I lean up and look at her as I ram
my limp dick inside of her playfully. “Yeah.” I say, and for the first time today,
she smiles.
Crystal Spears has always had a passion for literature. She fell in love
with reading at an early age. After spending a decade in business
management, she began blogging and book reviewing, only to find that
that didn't satisfy her craving. So, she began to write. Like any Indie
out there,
she had goals and dreams. She wrote Talania - a trip down memory lane,
a beautiful, contemporary romance novel. It almost made it to the top,
just almost. Someone told Crystal to write what she knew. She gave
serious thought to that advice.
For many years, her father had been CEO of a well-known gun and surplus
store in Texas. He then bought a tattoo shop where Crystal found herself
surrounded by bikers. Hardcore bikers. The Bestselling Breakneck Series was born when Crystal decided to write what she knows. Don't get the wrong idea. Her Breakneck series is completely fictional, but based on things seen and heard, she now had a place to start writing from that knowledge. Seize Me
was hardcore, and Crystal became the writer known as the chick with no
boundaries. She took the critics the wrong way and made her next
bestseller Withstanding Me a little tamer. Readers were quick to
point out that they did not like that and demanded that Crystal return
to gritty writing. When asked if she would write gritty from now on, her
only answer was… “I'll write what my readers want me to write, and I'll
also write what I feel.”
Crystal resides in Indiana with her soon-to-be husband and their two
children. They have a pit named Dozer, two fish, and one fancy rat. They
enjoy the outdoors, and camp, fish, and go boating as much as possible.
But, while the soon-to-be husband is out hunting this winter, Crystal
will be writing three novels and planning her wedding.
Crystal is a perfect example of never giving up on a dream or a goal.
Her goal was to publish a book; her dream was to become a bestseller.
She did both of these, and now, she feels that she’ll never regret
leaving business management to become a writer. She stays humble, loves
meeting new people, and is extremely excited to be a part of so many
signings in 2014 with some top notch Authors. |
Apparatus for taking X-ray pictures of a standing human typically include two vertically-extending rails on which a cassette holder is movable to various positions at which the cassette holder can be locked. This permits the X-ray plate to be positioned at various height to accommodate people of different heights and to permit various parts of the body to be X-rayed.
A sheet of X-ray film is placed into a rigid picture-frame-like cassette which holds the X-ray plate flat and facilitates handling of the X-ray plate.
In practice, a device called a grid is normally positioned in the cassette holder immediately in front of the cassette. The grid typically includes a great number of very small baffles resembling the slats of a venetian blind, but far more delicate because they are formed of very thin sheets of lead. These very small and fragile strips serve to prevent stray X-radiation from entering the cassette, particularly at directions inclined to a normal to the plate.
Typically, the cassette holder includes a lower lip on which the lower edges of the cassette and of the grid are supported, and typically the upper edges of the cassette and of the grid are secured to the cassette holder by a clamp, which may be spring-loaded.
Once the cassette and the grid have been thus mounted in the cassette holder, they cannot be freely rotated with respect to the holder. After having thus mounted the cassette in the holder, the operator checks to make sure the picture will include all of the desired portions of the patient's body. Typically, the X-ray plate is rectangular in shape, and it is not unusual for the operator to decide that better coverage can be obtained by rotating the X-ray plate 90 degrees.
When the operator finds it necessary to reorient the plate, he must release the cassette and grid from the cassette holder, manually turn them 90 degrees, and then reinsert the cassette and grid once again into the cassette holder.
It occasionally happens that during this reorientation operation, the technician accidentally drops the expensive and fragile grid, and the grid is thereby destroyed. Such a grid might typically cost on the order of $600 today.
Thus, there exists a need for a simple and relatively inexpensive apparatus that can be used to prevent the grid from being dropped. |
1. Field of the Invention
The present invention relates generally to methods and compositions for monitoring the processing of .beta.-amyloid precursor protein. More particularly, the present invention relates to the use of such methods and compositions for the diagnosis, prognosis, and monitoring response to therapy of Alzheimer's disease, and for screening and evaluation of potential drugs for the treatment of Alzheimer's disease.
Alzheimer's disease is characterized by the presence of numerous amyloid plaques and neurofibrillary tangles (highly insoluble protein aggregates) present in the brains of Alzheimer's disease patients, particularly in those regions involved with memory and cognition. While in the past there was significant scientific debate over whether the plaques and tangles are a cause or are merely the result of Alzheimer's disease, recent discoveries indicate that amyloid plaque is a causative precursor or factor. In particular, it has been discovered that the production of .beta.-amyloid peptide, a major constituent of the amyloid plaque, can result from mutations in the gene encoding amyloid precursor protein, a protein which when normally processed will not produce the .beta.-amyloid peptide. It is presently believed that a normal (non-pathogenic) processing of the .beta.-amyloid precursor protein occurs via cleavage by a putative ".alpha.-secretase" which cleaves between amino acids 16 and 17 of the protein. It is further believed that pathogenic processing occurs via a putative ".beta.-secretase" at the amino-terminus of the .beta.-amyloid peptide within the precursor protein.
The identification of mutations in the amyloid precursor protein gene which cause familial, early onset Alzheimer's disease is the strongest evidence that amyloid metabolism is the central event in the pathogenic process underlying the disease. Four reported disease-causing mutations include with respect to the 770 isoform, valine.sup.717 to isoleucine (Goate et al. (1991) Nature 349:704-706), valine.sup.717 to glycine (Chartier Harlan et al. (1991) Nature 353:844-846, valine.sup.717 to phenylalanine (Murrell et al. (1991) Science 254:97-99) and with respect to the 695 isoform, a double mutation changing lysine.sup.595 -methionine.sup.596 to asparagine.sup.595 -leucine.sup.596 (Mullan et al. (1992) Nature Genet 1:345-347) referred to as the Swedish mutation. Moreover, .beta.-amyloid peptide appears to be toxic to brain neurons, and neuronal cell death is associated with the disease.
Thus, the ability to monitor cellular processing of the amyloid precursor protein would be of significant value in the diagnosis, prognosis, and therapeutic supervision of Alzheimer's disease. In particular, it would be desirable to identify minimally invasive procedures for screening and evaluating detectable diagnostic markers in readily obtainable patient samples, such as serum, cerebrospinal fluid (CSF), and the like.
A number of potential diagnostic markers for Alzheimer's disease have been proposed. Of particular interest to the present invention are certain fragments of the amyloid precursor protein, including carboxy terminal fragments (such as the .beta.-amyloid peptide itself and fragments thereof), and amino-terminal fragments (such as certain 25 kD, 105 kD, and 125 kD fragments). As yet, none of the proposed markers has proved to be definitive for the antemortem diagnosis or monitoring of Alzheimer's disease.
Thus, it would be desirable to identify additional and alternative diagnostic markers for Alzheimer's disease. Such markers should be useful by themselves and/or in combination with other diagnostic markers and procedures. Preferably, the diagnostic markers would be detectable in body fluids, such as CSF, blood, plasma, serum, urine, tissue, and the like, so that minimally invasive diagnostic procedures can be utilized.
Of further interest to the present invention are in vitro and in vivo systems and methods for screening candidate drugs for the ability to inhibit or prevent the production of .beta.-amyloid plaque. It would be desirable to provide methods and systems for screening test compounds for the ability to inhibit or prevent the conversion of amyloid precursor protein to .beta.-amyloid peptide. In particular, it would be desirable to base such methods and systems on metabolic pathways which have been found to be involved in such conversion, where the test compound would be able to interrupt or interfere with the metabolic pathway which leads to conversion. Such methods and systems should be rapid, economical, and suitable for screening large numbers of test compounds.
2. Description of the Background Art
.beta.-amyloid peptide (also referred to as A4, .beta.AP, A.beta., or A.beta.P; see, U.S. Pat. No. 4,666,829 and Glenner and Wong (1984) Biochem. Biophys. Res. Commun. 120:1131-1135) is derived from .beta.-amyloid precursor protein (.beta.APP), which is expressed in differently spliced forms of 695, 751, and 770 amino acids. See, Kang et al. (1987) Nature 325:773-776; Ponte et al. (1988) Nature 331:525-527; and Kitaguchi et al. (1988) Nature 331:530-532. Normal processing of amyloid precursor protein involves proteolytic cleavage at a site between residues Lys.sup.16 and Leu.sup.17 (as numbered for the .nu.AP region where Asp.sup.597 is residue 1 in Kang et al. (1987)), supra, near the transmembrane domain, resulting in the constitutive secretion of an extracellular domain which retains the remaining portion of the .beta.-amyloid peptide sequence (Esch et al. (1990) Science 248:1122-1124). This pathway appears to be widely conserved among species and present in many cell types. See, Weidemann et al. (1989) Cell 57:115-126 and Oltersdorf et al. (1990) J. Biol. Chem. 265:4492-4497. This normal pathway cleaves within the region of the precursor protein which corresponds to the .beta.-amyloid peptide, thus apparently precluding its formation. Another constitutively secreted form of .beta.APP has been noted (Robakis et al. Soc. Neurosci. Oct. 26, 1993, Abstract No. 15.4, Anaheim, Calif.) which contains more of the .beta.AP sequence carboxy terminal to that form described by Esch et al. supra.
Golde et al. (1992) Science 255:728-730, prepared a series of deletion mutants of amyloid precursor protein and observed a single cleavage site within the .beta.-amyloid peptide region. Based on this observation, it was postulated that .beta.-amyloid peptide formation does not involve a secretory pathway. Estus et al. (1992) Science 255:726-728, teaches that the two largest carboxy terminal proteolytic fragments of amyloid precursor protein found in brain cells contain the entire .beta.-amyloid peptide region.
PCT application. WO 92/00521 describes methods for evaluating Alzheimer's disease based on measuring the amounts of certain 25 kD, 105 kD, and 125 kD soluble derivatives of amyloid precursor protein in a patient's cerebrospinal fluid. FIG. 3 of WO 92/00521 suggests that cleavage of amyloid precursor protein may occur adjacent to the amino-terminus of .beta.-amyloid peptide to produce a soluble amino-terminal fragment, but no evidence or discussion of such cleavage is presented in the application. Kennedy et al. (1992) Neurodegeneration 1:59-64, present data for a form of secreted .beta.APP, which was characterized by its reactivity with antibodies to residues 527-540 of .beta.APP and the lack of reactivity with antibodies to the first fifteen residues of .beta.AP. No direct evidence is provided to suggest the cleavage site or identity of the carboxy terminus of the .beta.APP form. PCT application WO 91/16628 describes methods for diagnosing disease based on detection of amyloid precursor proteins and fragments thereof utilizing antibodies to protease nexin-2 or amyloid precursor protein.
Recent reports show that soluble .beta.-amyloid peptide is produced by healthy cells into culture media (Haass et al. (1992) Nature 359:322-325) and in human and animal CSF (Seubert et al. (1992) Nature 359:325-327).
Palmert et al. (1989) Biochm. Biophys. Res. Comm. 65:182-188, describes three possible cleavage mechanisms for .beta.APP and presents evidence that .beta.APP cleavage does not occur at methionine.sup.596 in the production of soluble derivatives of .beta.APP. U.S. Pat. No. 5,200,339, discusses the existence of certain proteolytic factor(s) which are putatively capable of cleaving .beta.APP at a site near the .beta.APP amino-terminus. |
Q:
Enable java remote debug in code
For example, we can enable java remote debug by adding following to command line.
-agentlib:jdwp=transport=dt_socket,server=y,suspend=n,address=5005
But my application is running in yarn, I'm not sure which port is available.
So I want enable java debug in my code.
First I detect a available port and log in my program, then I can use this port to debug my application.
A:
The address property specifies host (optionally) and port (only the port if host is left out). So address=5005 specifies the port 5005 in your case. If you want your program to wait until you connect your debugger, switch suspend=n to suspend=y.
Edit:
Maybe I misunderstood your question. In case you want to enable debugging programmatically, this won't be possible as the debugging facility JPDA is not exposing a Java API nor any other way to start and stop it programmatically.
|
Q:
How to intercept all Hibernate sessions when they're created (Spring / Grails environment)
Is there a way of intercepting all new Hibernate sessions when they're created? I need to access each Session instance to enable a Hibernate filter with a parameter.
The only solution I've gotten working has involved wrapping the SessionFactory, but this involved a lot of semi nasty hacks as well as it required me to implement around 60 methods, where only a few are interesting.
Hibernate's SessionFactory implementation is for some annoying reason declared final so extending it is not an option. I've also tried aspects and Java proxies without any luck.
A:
I was able to create a JDK proxy:
import java.lang.reflect.InvocationHandler;
import java.lang.reflect.Method;
import java.lang.reflect.Proxy;
import java.util.Date;
import org.hibernate.SessionFactory;
import org.hibernate.engine.SessionFactoryImplementor;
public class SessionFactoryProxyCreator {
public static SessionFactory instance;
public static SessionFactory createProxy(final SessionFactory realSessionFactory) {
ClassLoader cl = SessionFactory.class.getClassLoader();
Class<?>[] interfaces = new Class[] { SessionFactory.class, SessionFactoryImplementor.class };
instance = (SessionFactory)Proxy.newProxyInstance(cl, interfaces, new InvocationHandler() {
public Object invoke(Object proxy, Method method, Object[] args) throws Throwable {
if ("openSession".equals(method.getName())) {
System.out.println("NEW SESSION AT " + new Date());
}
return method.invoke(realSessionFactory, args);
}
});
return instance;
}
}
and you would call this from a custom SessionFactoryBean:
import org.codehaus.groovy.grails.orm.hibernate.ConfigurableLocalSessionFactoryBean;
import org.hibernate.HibernateException;
import org.hibernate.SessionFactory;
import org.hibernate.cfg.Configuration;
public class MyConfigurableLocalSessionFactoryBean extends ConfigurableLocalSessionFactoryBean {
public MyConfigurableLocalSessionFactoryBean() {
setCurrentSessionContextClass(MyCurrentSessionContext.class);
}
@Override
protected SessionFactory buildSessionFactory() throws Exception {
setExposeTransactionAwareSessionFactory(false);
return SessionFactoryProxyCreator.createProxy(super.buildSessionFactory());
}
@Override
protected SessionFactory newSessionFactory(Configuration config) throws HibernateException {
setExposeTransactionAwareSessionFactory(false);
return SessionFactoryProxyCreator.createProxy(super.newSessionFactory(config));
}
}
which depends on a modified version of Spring's SpringSessionContext that uses the proxy instead of the real session factory:
import org.hibernate.HibernateException;
import org.hibernate.classic.Session;
import org.hibernate.context.CurrentSessionContext;
import org.hibernate.engine.SessionFactoryImplementor;
import org.springframework.orm.hibernate3.SessionFactoryUtils;
public class MyCurrentSessionContext implements CurrentSessionContext {
public MyCurrentSessionContext(SessionFactoryImplementor sessionFactory) {
// ignore the real sessionFactory, need to use the proxy
}
public Session currentSession() throws HibernateException {
try {
return (org.hibernate.classic.Session)SessionFactoryUtils.doGetSession(
SessionFactoryProxyCreator.instance, false);
}
catch (IllegalStateException e) {
throw new HibernateException(e.getMessage());
}
}
}
This needs to be registered in resources.groovy to replace the standard Grails ConfigurableLocalSessionFactoryBean:
import org.codehaus.groovy.grails.commons.ApplicationHolder as AH
import org.codehaus.groovy.grails.orm.hibernate.events.PatchedDefaultFlushEventListener
beans = {
sessionFactory(MyConfigurableLocalSessionFactoryBean) {
def ds = AH.application.config.dataSource
def hibConfig = AH.application.config.hibernate
dataSource = ref('dataSource')
List hibConfigLocations = []
if (AH.application.classLoader.getResource('hibernate.cfg.xml')) {
hibConfigLocations << 'classpath:hibernate.cfg.xml'
}
def explicitLocations = hibConfig?.config?.location
if (explicitLocations) {
if (explicitLocations instanceof Collection) {
hibConfigLocations.addAll(explicitLocations.collect { it.toString() })
}
else {
hibConfigLocations << hibConfig.config.location.toString()
}
}
configLocations = hibConfigLocations
if (ds?.configClass) {
configClass = ds.configClass
}
hibernateProperties = ref('hibernateProperties')
grailsApplication = ref('grailsApplication', true)
lobHandler = ref('lobHandlerDetector')
entityInterceptor = ref('entityInterceptor')
eventListeners = ['flush': new PatchedDefaultFlushEventListener(),
'pre-load': ref('eventTriggeringInterceptor'),
'post-load': ref('eventTriggeringInterceptor'),
'save': ref('eventTriggeringInterceptor'),
'save-update': ref('eventTriggeringInterceptor'),
'post-insert': ref('eventTriggeringInterceptor'),
'pre-update': ref('eventTriggeringInterceptor'),
'post-update': ref('eventTriggeringInterceptor'),
'pre-delete': ref('eventTriggeringInterceptor'),
'post-delete': ref('eventTriggeringInterceptor')]
}
}
A:
I've solved this problem (at least until Hibernate provides a proper API for things like this). Short version of the solution:
Proxy the session factory
Intercept method invocations to getCurrentSession and use a CurrentSessionContext implementation we've initialized (not Hibernate).
Longer version:
http://www.developer-b.com/blog/entry/1635/2010/oct/07/intercepting-hibernate-sessions
Sources / Github:
http://github.com/multi-tenant/grails-hibernate-hijacker (still very experimental)
Thanks for the input!
|
Pregnancy-associated plasma protein-A-induced inhibition of human leukocyte elastase: an artifact.
Pregnancy-associated plasma protein A (PAPP-A), was reported to be an inhibitor in many in vitro systems. Since it was shown that the inhibition of coagulation and complement activity attributed to PAPP-A was in fact due to a contamination by heparin occurring during the purification process, we undertook the present study to see whether the reported PAPP-A-induced inhibition of human leukocyte elastase (HLE) could also be attributed to heparin contamination. PAPP-A was purified from maternal pregnancy EDTA plasma by a method which was previously shown to eliminate contaminating heparin: this preparation was inactive in the HLE assay. But PAPP-A isolated by heparin-Sepharose chromatography, or a PAPP-A-free washing of the heparin-Sepharose column were both inhibitors of HLE. Furthermore the inactive PAPP-A preparation, when incubated with the PAPP-A-free washing of the heparin-Sepharose column, yielded a high molecular weight preparation which inhibited HLE. It is concluded that PAPP-A is not an inhibitor of HLE and that the inhibition of HLE previously attributed to PAPP-A was due to contaminating heparin. |
Q:
How to use a string id from res/values/strings.xml instead of String or CharSequence?
If I have this piece of code with hardcoded String "New event of importance":
public class MainActivity extends Activity {
private NotificationManager mNManager;
private static final int NOTIFY_ID = 1100;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
String ns = Context.NOTIFICATION_SERVICE;
mNManager = (NotificationManager) getSystemService(ns);
final Notification msg = new Notification(R.drawable.ic_launcher,
"New event of importance",
System.currentTimeMillis());
But would like to move it into res/values/string.xml file instead:
<string name="new_event">New event of importance</string>
Then how to use the R.strings.new_event in the above constructor (and yes I know that the constructor is deprecated)?
And one more question please - why is final keyword used above?
A:
Activity has getString method that takes a string's id as parameter.
Change
"New event of importance"
with
getString(R.string.new_event);
In general to access resources, you need a context object
|
[Deep brain stimulation for essential tremor. Consensus recommendations of the German Deep Brain Stimulation Association].
In Germany, deep brain stimulation (DBS) of the thalamic ventralis intermedius nucleus (VIM) is licensed for treatment of essential tremor in cases unresponsive to pharmacotherapy. Especially a bothersome hand tremor interfering with activities of daily living will improve, whereas head, tongue or vocal tremor shows less response. DBS was proven to be superior to lesional thalamotomy with better functional outcome and less adverse effects. The consensus statement presented here reflects the current recommendations of the German Deep Brain Stimulation Study Group for inclusion and exclusion criteria as well as for peri-, intra- and postoperative neurological management. |
A Brief History of Markup - duck
http://www.alistapart.com/articles/a-brief-history-of-markup/
======
dctoedt
This is really a brief history of HTML / XHMTL. Markup goes back long before
either. In 1980-81 I used Brian Reid's SCRIBE formatting program (and Emacs as
the text editor) to produce the manuscript of my law review note. According to
Wikipedia, Brian won the ACM's Grace Murray Hopper Award for his dissertation,
which was based on SCRIBE -- which itself was based on (and was the first
robust version of) RUNOFF. See
<http://en.wikipedia.org/wiki/Scribe_(markup_language)>.
~~~
studer
And the RUNOFF manual from 1964 mentions that it doesn't implement the
FOOTNOTE and COMMENT markup commands from a predecessor called "DITTO"...
<http://mit.edu/Saltzer/www/publications/CC-244.html>
|
Progress in understanding preferential detection of live cells using viability dyes in combination with DNA amplification.
The ideal scenario in most applications of microbial diagnostics is that only viable cells are detected. Bacteria were traditionally considered viable when they could be cultured, whereas today's viability concept tends to be alternatively based on the presence of some form of metabolic activity, a positive energy status, responsiveness, detection of RNA transcripts that tend to degrade rapidly after cell death, or of an intact membrane. The latter criterion, although conservative, was the focus of one of the most successful recent approaches to detect viable cells in combination with DNA amplification techniques. The technology is based on sample treatment with the photoactivatable, and cell membrane impermeant, nucleic acid intercalating dyes ethidium monoazide (EMA) or propidium monoazide (PMA) followed by light exposure prior to extraction of DNA and amplification. Light activation of DNA-bound dye molecules results in irreversible DNA modification and subsequent inhibition of its amplification. Sample pretreatment with viability dyes has so far been mainly used in combination with PCR (leading to the term viability PCR, v-PCR), and increasingly with isothermal amplification method. The principle is not limited to bacteria, but has also successfully been applied to fungi, protozoa and viruses. Despite the success of the method, some practical limitations have been identified, especially when applied to environmental samples. In part they can be minimized by choice of experimental parameters and conditions adequate for a particular sample. This review summarizes current knowledge and presents aspects which are important when designing experiments employing viability dyes. |
Monday, August 09, 2010
Autism Speaks Supports More Environmental Research? Terrific! Now Please Help Even Out the Funding
Right now, about 10 to 20 times more research dollars are spent on studies of the genetic causes of autism than on environmental ones.
We need to even out the funding.
Dr. Irva Hertz-Picciotto, UC Davis M.I.N.D. Institute
I have been a supporter of Autism Speaks over the course of its brief existence. I appreciate the media savvy and political skills of its leadership. The World Autism Awareness Day that it assisted in bringing into existence is, in my humble opinion, a great accomplishment in itself. The connections and skills of Autism Speaks leadership have been very impressive in bringing in people and events who, by themselves command attention, from NASCAR to Jerry Seinfeld, people and events that are seen and heard focusing on autism. Well done, very well done.
I have though been concerned, rightly or wrongly, about what I thought was a subscription by Autism Speaks to the "it's gotta be genetic" mindset which has dominated autism research and hindered progress in understanding autism disorders and developing treatments and cures. I was pleasantly surprised when I received from Jane Rubenstein of Rubenstein Communications Inc. the Autism Speaks statement "HEARING ON STATE OF RESEARCH ON POTENTIAL ENVIRONMENTAL HEALTH FACTORS WITH AUTISM AND RELATED NEURODEVELOPMENT DISORDERS U.S. Senate Committee on Environment & Public Works, Subcommittee on Children’s Health". In the statement Autism Speaks Chief Science Officer Dr. Geri Dawson states unequivocally Autism Speaks endorsement on the need for more environmentally based autism research:
(NEW YORK, N.Y., August 4, 2010) – Autism Speaks’ Chief Science Officer Geraldine Dawson, Ph.D. emphasized the importance of research on environmental risk factors for autism spectrum disorders as the U.S. Senate Committee on Environment & Public Works, Subcommittee on Children’s Health convened a special hearing yesterday on potential environmental health factors associated with autism spectrum disorders (ASD) and related neurodevelopmental disorders. The hearing examined the latest research on potential environmental factors that may increase the risk for autism spectrum disorders.As this hearing reviewed studies funded by the Environmental Protection Agency and the National Institute of Environmental Health Sciences on environmental factors associated with autism, including toxins and other factors that can influence brain development, Dr. Dawson reiterated that it is important to remember that, “Although genetic factors clearly contribute to the causes of autism, we also need to understand environmental factors and their interactions with genetic susceptibility.”
Dr. Dawson's statement includes examples of what appear to be impressive initiatives undertaken by Autism Speaks in support of environmental autism research. The links to review these initiatives can be found on the Autism Speaks web site, science section. What isn't clear is the level of financial commitment to environmental autism research compared to genetic research. Does, or will, Autism Speaks commit to balanced funding of environmental and genetic autism research as called for by Dr. Irva Hertz-Picciotto of the UC David MIND Institute?
If I have wronged Autism Speaks with my perception of an imbalance on its part in favor of genetic over environmental autism research I would genuinely appreciate being notified of my error. If that is the case then I will apologize but would humbly and respectfully ask Autism Speaks to use its proven and impressive communication skills to convince public health funding authorities to follow the approach recommended by Dr. Hertz-Picciotto.
Much valuable time has been lost with the autism is genetic obsession.
Balanced funding of environmental and genetic autism research is needed now, not tomorrow.
Balancing the funding of environment versus genetics should not be worried about until the imbalance of autism speaks' funding neurodiversity concerns is eliminated. Dr. Mottron will undoubtedly try to renew his half a million dollar grant when it expires next June.
Also, the funding of some neurodiversity kids in their early 20s who are barely affected by their autism if at all for making some films for fun should be stopped.
101 Noteworthy Sites on Asperger's & Autism Spectrum Disorders
Facing Autism on Facebook
Why ABA For Autism?
The effectiveness of ABA-based intervention in ASDs has been well documented through 5 decades of research by using single-subject methodology21,25,27,28 and in controlled studies of comprehensive early intensive behavioral intervention programs in university and community settings.29–40 Children who receive early intensive behavioral treatment have been shown to make substantial, sustained gains in IQ, language, academic performance, and adaptive behavior as well as some measures of social behavior, and their outcomes have been significantly better than those of children in control groups.31–4American Academy of Pediatrics, Management of Children with Autism Spectrum Disorders
"We have to look also at environmental factors, and from my point of view, the interaction between the genetic factors and the environmental factors ... It looks like some shared environmental factors play a role in autism, and the study really points toward factors that are early in life that affect the development of the child"
Joachim Hallmayer, MD, associate professor of psychiatry at Stanford University in California
Even Out Environmental and Genetic Autism Research Funding
Right now, about 10 to 20 times more research dollars are spent on studies of the genetic causes of autism than on environmental ones.
We need to even out the funding.
Irva Hertz-Picciotto, UC Davis M.I.N.D. Institute Researcher
My Autism Pledge For Conor
Today I pledge to continue;I Pledge to continue to fight for the availability of effective autism treatments;I Pledge to continue to fight for a real education for autistic children;I Pledge to continue to fight for decent residential care for autistic adults;I Pledge to continue to fight for a cure for autism;I Pledge to continue finding joy in my son but not in the autism disorder that restricts his life;Today, and every day, I Pledge to continue to hope for a better life for Conor and others with autism, through accommodation, care, respect, treatment, and some day, a cure;Today, and every day, I Pledge to continue to fight for the best possible life for Conor, my son with autistic disorder.
Dr. Jon Poling : Blinders Won’t Reduce Autism
"Fortunately, the ‘better diagnosis’ myth has been soundly debunked. ... only a smaller percentage of this staggering rise can be explained by means other than a true increase.
Because purely genetic diseases do not rise precipitously, the corollary to a true autism increase is clear — genes only load the gun and it is the environment that pulls the trigger. Autism is best redefined as an environmental disease with genetic susceptibilities."
We should be investing our research dollars into discovering environmental factors that we can change, not more poorly targeted genetic studies that offer no hope of early intervention. Pesticides, mercury, aluminum, several drugs, dietary factors, infectious agents and yes — vaccines — are all in the research agenda.
OnTopList
Conor
Facing Autism Visitors
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It's NOT About ME
I am the father of two sons one of whom is severely autistic with intellectual disability. I have advocated for autism services for autistic children, students and adults in New Brunswick, Canada and I blog and comment about autism on the world wide web. And I like to walk .. a lot. |
POPSUGAR Celebrity
Come Party With Me: Classic Thanksgiving — Menu (Potato Sides)
Nov 6 2007 - 4:24pm
Once you have decided on a turkey recipe [1], it's time to think about the sides. Potatoes are the classic companion to the stuffed bird, so serve both mashed white potatoes and baked sweet potatoes. You could flavor the mashed potatoes with a variety of options, but if serving with gravy, keep them simple and make a roasted garlic version. For a delicious sweet potato casserole, mix the cooked flesh with sugar, butter, and bourbon topped with a pecan butter crumble. To look at these recipes,
Place potatoes in a large stockpot, cover with cold salted water, and bring to a boil. Cook until very tender, about 12 minutes. Drain in a colander, and pass through a potato ricer or food mill over a large bowl.
Heat butter and cream in a saucepan until butter has melted and cream is hot.
Pour over riced potatoes, season with salt and pepper, and stir well to combine. Gently stir in the roasted garlic cloves; serve. |
[How to prevent the various neuralgic complaints after operation of maxillary sinus? (author's transl)].
In order to avoid multiple neuralgic complaints after operations of maxillary sinus the following advice for cautions operative procedures is as follows: 1. A vertical incision in the mucosa of the fossa canine behind the eye-tooth instead of the usual horizontal section. 2. A small dorsolateral fenestration in a part of less nervous ramification in the facial bone layer of the antrum of High-more. 3. The limitation of resection of the mucous membrane in pathological areas of the sinus. 450 operations of the maxillary sinus were done in this way, which resulted in a lower rate of postoperative neuralgic complaints. Another advantage was the minimized intraoperative bleeding and the absence of postoperative swelling and haematoma formation of the cheek. |
By the time the authorities caught up with Jeffrey Dahmer, in 1991, even the perpetrator was astonished by his crimes. “It’s hard for me to believe that a human being could have done what I’ve done,” he said, “but I know that I did it.”
A serial killer and sex offender, Dahmer became known in the tabloids as the Milwaukee Cannibal. He raped, murdered and dismembered 17 boys and men, sometimes preserving their body parts as mementoes, sometimes dabbling in cannibalism.
He has subsequently inspired books, plays, documentaries, a novel, by Joyce Carol Oates, and two rather glamorous biopics: The Secret Life, from 1993, and Dahmer, starring Jeremy Renner, from 2002.
Dahmer didn’t have to wind up a monster, if only the adults in his life hadn’t been so inexplicably, unforgivably, incomprehensibly clueless and/or indifferent Marc Meyers’s My Friend Dahmer is a sobering riposte to these earlier, often sensationalised representations. The film is adapted from the autobiographical graphic novel of the same name, from 2012, by the cartoonist John Backderf, aka Derf, who had been friends with Dahmer at high school in the 1970s.
“The idea of a portrait of a serial killer as a young boy was interesting to me,” Meyers says. “I initially thought I was going to create a fictional story using composite details, a kind of road map towards becoming a serial killer. I was also, just separately, looking at graphic novels whenever I could. So when I came across My Friend Dahmer I thought it was a blessing. Here was the concept, but even better: a real story all laid out.”
Meyers eschews Backderf’s troubled omnipresent narration, but he remains rooted in the source material, by working certain panels into shots of the film and by honouring the author’s original intention as laid out in a prose introduction: “It’s my belief that Dahmer didn’t have to wind up a monster, that all those people didn’t have to die horribly, if only the adults in his life hadn’t been so inexplicably, unforgivably, incomprehensibly clueless and/or indifferent,” Backderf writes. “Once Dahmer kills, however – and I can’t stress this enough – my sympathy for him ends. He could have put a gun to his head.” Related Jane Fonda: From Oscar-winning actor to 50 shades of ****
Six of the best films to see at the cinema this weekend
Almost every film in cinemas this week, reviewed and rated
My Friend Dahmer: the serial killer Jeffrey Dahmer, aka the Milwaukee Cannibal, after his arrest in 1991. Photograph: Curt Borgwardt/Sygma via Getty My Friend Dahmer: John Backderf, who wrote the original graphic novel, and Marc Meyers. Photograph: Gilbert Carrasquillo/FilmMagic/Getty Backderf invited Meyers to stay with him after the director finished his first draft of the screenplay. “We drove around where they grew up, in Akron, and visited the high school they went to, and we talked a lot about that time. That stuff helped me figure out how the characters would really move through the exact spaces.”
Ohio locals were surprisingly welcoming of the production. Car collectors loaned period vehicles to the film. Others contributed period furniture and other props. Many were keen to share recollections that were worked into the screenplay.
“It’s 40 years later, so everyone there has pretty much moved on with their lives,” Meyers says. “All of the crew is from Ohio. The teenagers we used as extras had parents who went to school with Jeffrey Dahmer. They knew the book, and, like Backderf, some insisted that they had other friends who were stranger than he was in high school. All of them were horrified by what happened later in his life, but they did understand that there was a moment in time when he was just another teenager. He was maybe more troubled than they realised. But when the news broke, years later, it was a surprise to the friends he grew up with.”
The things they were saying were so mean, throwing around words like ‘faggot’. Teenagers can still be cruel, but there’s an awareness of how damaging words can be Meyers’s film allows for its subject’s unhealthy preoccupation with roadkill and the confused murderous desire he feels for a local jogger, yet it studiously avoids the more horrific aspects of the story. His Dahmer, like Backderf’s, exists in broad daylight, as he puts it.
“I’m not an expert in serial killers,” Meyers says. “Like everyone else, I’m attracted to true crime stories, and I’ve watched Making a Murderer and those sorts of entertainments. But I’m not one of those people who devours books on serial killers. Everything in the movie came out of the book. We really only cared about the guy in high school up to graduation. So I stayed within that framework.”
Consequently, in My Friend Dahmer, Jeff Dahmer – brilliantly played by the former Disney star Ross Lynch – is an awkward teenager struggling to make it through high school with a severely dysfunctional family. He barely copes with his unstable, uncaring mother, who is played by Anne Heche, and at school he drinks and fakes seizures and “spaz-outs”, antics that win over a group of other outsider teens, who form the Dahmer Fan Club, headed by Derf Backderf. But Dahmer remains a lonely and increasingly damaged kid.
“On set I realised we were making a politically incorrect film,” Meyers says. “Because they’re so politically incorrect by our standards. That felt oddly refreshing. The things they were saying to each other were so mean, throwing around words like ‘faggot’. Teenagers can still be cruel, but there’s an awareness of how damaging words can be. We really have progressed as a species, in that we are sensitive to the different kinds of people around us. They aren’t.”
My Friend Dahmer: Ross Lynch as Jeff Dahmer and Alex Wolff as John Backderf There has been much positive chatter about My Friend Dahmer’s countercasting. The actor who plays Derf, Alex Wolff, came to prominence starring alongside his older brother Nat in the Nickelodeon musical comedy television series The Naked Brothers Band. Ross Lynch gained recognition as Austin Moon on the Disney Channel show Austin & Ally and with the family band, R5.
“A Disney kid doing Dahmer? It’s an easy contrast,” Meyers says. “But look at the talent that has emerged through the Disney Channel. Look at Ryan Gosling, for example. The Disney Channel is full of talented workaholics, who are professional from a very early age. I knew it was a big pool of talent. But when we sat down with Ross I just enjoyed talking to him.
“Like most kids that age, unless they’ve heard his name in a rap song, he wasn’t even aware of Jeffrey Dahmer. We met over 100 actors. A handful selected themselves for reasons of likeness. But he was different. He was very present. He’s a dancer originally. He studied Dahmer on YouTube. He knew the posture and the gait. He thought through the spazzing, all the physical stuff. He was perfect.” |
/**
* Copyright © MyCollab
* <p>
* This program is free software: you can redistribute it and/or modify
* it under the terms of the GNU Affero General Public License as published by
* the Free Software Foundation, either version 3 of the License, or
* (at your option) any later version.
* <p>
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU Affero General Public License for more details.
* <p>
* You should have received a copy of the GNU Affero General Public License
* along with this program. If not, see <http://www.gnu.org/licenses/>.
*/
package com.mycollab.module.project.view.bug;
import com.mycollab.common.i18n.FollowerI18nEnum;
import com.mycollab.common.i18n.GenericI18Enum;
import com.mycollab.form.view.LayoutType;
import com.mycollab.form.view.builder.DynaSectionBuilder;
import com.mycollab.form.view.builder.TextDynaFieldBuilder;
import com.mycollab.form.view.builder.type.DynaForm;
import com.mycollab.form.view.builder.type.DynaSection;
import com.mycollab.module.project.i18n.BugI18nEnum;
import com.mycollab.module.project.i18n.MilestoneI18nEnum;
import com.mycollab.module.project.domain.BugWithBLOBs;
import com.mycollab.module.project.domain.SimpleBug;
import com.mycollab.module.project.i18n.TicketI18nEnum;
/**
* @author MyCollab Ltd.
* @since 5.0.1
*/
public class BugDefaultFormLayoutFactory {
private static DynaSection mainAddSection() {
DynaSection mainSection = new DynaSectionBuilder().layoutType(LayoutType.TWO_COLUMN).build();
//Row 1
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.name)
.displayName(BugI18nEnum.FORM_SUMMARY)
.fieldIndex(0).mandatory(true).required(true).colSpan(true).build());
//Row 2
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.priority)
.displayName(GenericI18Enum.FORM_PRIORITY)
.contextHelp(GenericI18Enum.FORM_PRIORITY_HELP)
.required(true)
.fieldIndex(1).build());
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.assignuser)
.displayName(GenericI18Enum.FORM_ASSIGNEE)
.fieldIndex(2).build());
//Row 3
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.severity)
.displayName(BugI18nEnum.FORM_SEVERITY)
.fieldIndex(3).build());
mainSection.fields(new TextDynaFieldBuilder().fieldName(SimpleBug.Field.components)
.displayName(TicketI18nEnum.FORM_COMPONENTS)
.contextHelp(TicketI18nEnum.FORM_COMPONENTS_HELP)
.fieldIndex(4).build());
//Row 4
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.startdate)
.displayName(GenericI18Enum.FORM_START_DATE).fieldIndex(5).build());
mainSection.fields(new TextDynaFieldBuilder().fieldName(SimpleBug.Field.affectedVersions)
.displayName(TicketI18nEnum.FORM_AFFECTED_VERSIONS)
.contextHelp(TicketI18nEnum.FORM_AFFECTED_VERSIONS_HELP)
.fieldIndex(6).build());
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.enddate)
.displayName(GenericI18Enum.FORM_END_DATE)
.fieldIndex(7).build());
mainSection.fields(new TextDynaFieldBuilder().fieldName(SimpleBug.Field.fixedVersions)
.displayName(BugI18nEnum.FORM_FIXED_VERSIONS)
.contextHelp(BugI18nEnum.FORM_FIXED_VERSIONS_HELP)
.fieldIndex(8).build());
//Row 5
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.duedate)
.displayName(GenericI18Enum.FORM_DUE_DATE)
.fieldIndex(9).build());
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.milestoneid)
.displayName(MilestoneI18nEnum.SINGLE)
.fieldIndex(10).build());
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.originalestimate)
.displayName(BugI18nEnum.FORM_ORIGINAL_ESTIMATE)
.contextHelp(BugI18nEnum.FORM_ORIGINAL_ESTIMATE_HELP)
.fieldIndex(11).build());
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.remainestimate)
.displayName(BugI18nEnum.FORM_REMAIN_ESTIMATE)
.contextHelp(BugI18nEnum.FORM_REMAIN_ESTIMATE_HELP)
.fieldIndex(12).build());
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.environment)
.displayName(BugI18nEnum.FORM_ENVIRONMENT)
.fieldIndex(13).colSpan(true).build());
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.description)
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return mainSection;
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private static DynaSection mainReadSection() {
DynaSection mainSection = new DynaSectionBuilder().layoutType(LayoutType.TWO_COLUMN).build();
//Row 1
mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.name)
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mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.resolution)
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mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.severity)
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mainSection.fields(new TextDynaFieldBuilder().fieldName(SimpleBug.Field.components)
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mainSection.fields(new TextDynaFieldBuilder().fieldName(SimpleBug.Field.affectedVersions)
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mainSection.fields(new TextDynaFieldBuilder().fieldName(SimpleBug.Field.fixedVersions)
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mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.milestoneid)
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mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.originalestimate)
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mainSection.fields(new TextDynaFieldBuilder().fieldName(BugWithBLOBs.Field.remainestimate)
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private static DynaSection subTicketSections() {
DynaSection subTicketsSection = new DynaSectionBuilder().layoutType(LayoutType.ONE_COLUMN).header(TicketI18nEnum.FORM_SUB_TICKETS)
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subTicketsSection.fields(new TextDynaFieldBuilder().fieldName("section-subTickets").fieldIndex(0).build());
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attachmentSection.fields(new TextDynaFieldBuilder().fieldName("section-attachments")
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private static DynaSection followersSection() {
DynaSection followersSection = new DynaSectionBuilder().layoutType(LayoutType.ONE_COLUMN).header(FollowerI18nEnum.OPT_SUB_INFO_WATCHERS)
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followersSection.fields(new TextDynaFieldBuilder().fieldName("section-followers")
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return followersSection;
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public static DynaForm getReadForm() {
return new DynaForm(mainReadSection(), subTicketSections(), attachmentSection());
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public static DynaForm getAddForm() {
return new DynaForm(mainAddSection(), followersSection(), attachmentSection());
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|
House Speaker Paul Ryan on Capitol Hill in June. (Photo: J. Scott Applewhite/AP)
House Speaker Paul Ryan attempted to distance himself Friday from any feud with Donald Trump but noted he could un-endorse the Republican nominee if he continues to make controversial statements.
Ryan told a Wisconsin radio station Friday that his support of Trump is not a “blank check” and added he could pull his endorsement if necessary.
“Of course,” Ryan said when asked by Milwaukee radio host Charlie Sykes if there was anything Trump might say that would cause the congressman to retract his endorsement. “But I’m not going to go down the road of litigating his past comments.”
In a separate interview, Ryan said “heck if I know” when asked if the conflict with Trump was over. The speaker added he was more interested in seeking the support of his constituents in Wisconsin’s Aug. 9 primary than in engaging Trump.
“I’m not going to try and psychoanalyze stuff,” Ryan told WISN-12’s Jay Weber in Milwaukee. “I’m going to rise above the stuff, and I’m not going to get involved in some petty back-and-forth. I don’t see a purpose in that.”
Ryan has been critical of Trump’s feud with Khizr and Ghazala Khan, the parents of a Muslim American soldier killed in Iraq. The Khans spoke out against Trump at the Democratic National Convention, and Trump criticized them in response.
In a statement last week, Ryan said attacking Gold Star families was unacceptable and that they deserved greater respect. Trump then told the Washington Post on Tuesday that he would not endorse Ryan in his GOP primary, saying he was “not quite there yet.” Trump also declined to endorse other Republicans, including Arizona Sen. John McCain and New Hampshire Sen. Kelly Ayotte, in their races.
“I like Paul, but these are horrible times for our country. We need very strong leadership. We need very, very strong leadership,” Trump said, echoing the words Ryan used before he endorsed Trump.
However, Fox News is reporting that Trump will endorse Ryan Friday night at a rally in Green Bay, Wis.
Story continues
Ryan’s primary opponent, Paul Nehlen, has also been supportive of Trump, who responded in kind on Monday by thanking Nehlen for his support during the Khan firestorm.
Nehlen garnered headlines Thursday when he called for a deportation of Muslims in the U.S. in an interview with a Chicago radio station.
Ryan said Nehlen represents “alt-conservativism” and was backed by “scam PACs.”
“You can’t make this stuff up,” Ryan said. “It is antithetical to any kind of principle we have as Republicans, as conservatives and as Americans.”
Ryan is leading Nehlen by 66 points, according to a poll released Friday morning. |
OIC Sensor Support
------------------
The sensor framework provides support for an OIC enabled application to
host the sensor devices as OIC resources. The sensor framework provides
the following OIC support:
- Creates OIC resources for each sensor device that is enabled in the
application. It creates an OIC discoverable and observable resource
for each sensor type that the sensor device is configured for.
- Processes CoAP GET requests for the sensor OIC resources. It reads
the sensor data samples, encodes the data, and sends back a response.
The sensor package (``hw/sensor``) defines the following syscfg settings
for OIC support:
- ``SENSOR_OIC``: This setting specifies whether to enable sensor OIC
server support. The setting is enabled by default. The sensor package
includes the ``net/oic`` package for the OIC support when this
setting is enabled. The ``OC_SERVER`` syscfg setting that specifies
whether to enable OIC server support in the ``net/oic`` package must
also be enabled.
- ``SENSOR_OIC_OBS_RATE``: Sets the OIC server observation rate.
An application defines an OIC application initialization handler that
sets up the OIC resources it supports and calls the ``oc_main_init()``
function to initialize the OC server. The application must call the
``sensor_oic_init()`` function from the the OIC application
initialization handler. The ``sensor_oic_init()`` function creates all
the OIC resources for the sensors and registers request callbacks to
process CoAP GET requests for the sensor OIC resources.
See the :doc:`Enabling OIC Sensor Data Monitoring
Tutorials <../../../tutorials/sensors/sensor_oic_overview>` to run and
develop sample OIC sensor server applications.
|
Q:
a checkmark on top of thumbnail for ios
In Apple's Photos app, after one select each image there is a check-mark appear at the corner. Is this a build-in function of iOS SDK? Or this is a diy if one wants to accomplish to same effect.
A:
This looks like a custom UIView implementation. I'd venture to guess that it's an extension of UIImageView (or even a UIView container) that adds a subview containing the check mark when it detects a tap.
|
WorldTimes 1.2 Gets Intel Mac Support
Artisan Codeworks announced the immediate availability of WorldTimes 1.2 on Monday. WorldTimes is a world clock application that displays up to 12 different time zones from around the world. The updated version is a Universal Binary application, allowing it to run natively on PowerPC and Intel-based Macs. It also now sorts locations by name and time zone, properly saves edits if the application was quit while editing, and properly calculates location times if day light savings time changes while the application is running.
WorldTimes 1.2 is free, and available for download at the Artisan Codeworks Web site. |
---
abstract: 'Incentive mechanisms for crowdsourcing have been extensively studied under the framework of all-pay auctions. Along a distinct line, this paper proposes to use Tullock contests as an alternative tool to design incentive mechanisms for crowdsourcing. We are inspired by the conduciveness of Tullock contests to attracting user entry (yet not necessarily a higher revenue) in other domains. In this paper, we explore a new dimension in [*optimal Tullock contest design*]{}, by superseding the contest prize—which is [*fixed*]{} in conventional Tullock contests—with a [*prize function*]{} that is dependent on the (unknown) winner’s contribution, in order to maximize the crowdsourcer’s utility. We show that this approach leads to attractive practical advantages: (a) it is well-suited for rapid prototyping in fully distributed web agents and smartphone apps; (b) it overcomes the [*disincentive*]{} to participate caused by players’ antagonism to an increasing number of rivals. Furthermore, we optimize conventional, fixed-prize Tullock contests to construct the [*most superior*]{} benchmark to compare against our mechanism. Through extensive evaluations, we show that our mechanism significantly outperforms the optimal benchmark, by over three folds on the crowdsourcer’s utility cum profit and up to nine folds on the players’ social welfare.'
author:
-
bibliography:
- 'IEEEabrv.bib'
title: |
Crowdsourcing with Tullock Contests:\
A New Perspective
---
Introduction {#sec:intro}
============
Crowdsourcing represents a new problem-solving model that elicits solutions, ideas, data, etc.—referred to as [*contributions*]{}—from an undefined, generally large group of people. Classic examples include Amazon Mechanic Turk, Yahoo! Answers, GalaxyZoo.org, TopCoder.com, etc. Recently, a new variant of crowdsourcing called participatory sensing emerged as a new data-collection model, which elicits sensor data contributed from user-owned mobile devices such as smartphones. Examples include GreenGPS[@greengps10mobisys], LiveCompare[@livecomp09], ContriSense:Bus[@lau11ucc] and Waze.com, to name a few.
Pivotal to the viability of all such crowdsourcing systems, is whether there is enough [*incentive*]{} to attract sufficient participation. A large body of prior work [@yang12mobicom; @kout13infocom; @luo14infocom; @DV09csallpay; @AS09ICIS; @SODA12; @luo14mass; @luo12secon] has been dedicated to designing incentive mechanisms for such scenarios, where each incentive mechanism essentially determines some reward according to users’ contributions. The commonly adopted approach turns out to be [*auctions*]{}, where each bidder tenders a bid (e.g., planned sensing duration[@yang12mobicom] or desired payment[@kout13infocom]) to the crowdsourcer, who will then choose the highest or lowest bidder(s) as the winner(s) to give out some reward. A widely used form among these auctions is [*all-pay auctions*]{}[@DV09csallpay; @AS09ICIS; @SODA12; @luo14infocom; @luo14mass], which nicely captures the scenario where each bid represents some [*irreversible*]{} effort. In other words, effort has to be [*sunk*]{} at the time of bidding, for example working out a solution to a problem, or sensing and sending data through a smartphone.
A distinctive characteristic of auctions in general, is that they are [*perfectly discriminating*]{}[@Hillman89]: the best (highest or lowest) bidder wins with probability one while the others lose for sure. Thus in all-pay auctions, due to the inevitable sunk cost (every bidder has to pay for his bid regardless of whether he wins the auction or not), all bidders substantially [*shade*]{} (i.e., decrease) their bids for fear of loss[@Krishna09]. Furthermore, if a bidder believes that there exists some other bidder who will bid higher than him, he will choose not to bid at all. Indeed, as Franke et al. [@Franke14] pointed out, all-pay auctions are so discriminative that, under a complete-information setting, only two (strongest) players will enter the auction in equilibria and only one of them will have a positive expected payoff. Clearly, this is not desirable in many crowdsourcing campaigns that favor a large participant pool (yet not necessarily higher revenue) for the sake of more population diversity (e.g., LiveCompare[@livecomp09]) and/or larger geographic coverage (e.g., GreenGPS[@greengps10mobisys] and Waze).
Taking a radically different approach, this paper proposes to use [*Tullock contests*]{} as an alternative framework to design incentive mechanisms for crowdsourcing. Fathered by Tullock’s seminal work [@Tullock80], Tullock contests represent a distinct contest regime that is [*imperfectly discriminating*]{}: every player has a strictly positive probability to win (determined by a contest success function) as long as he bids.[^1] This characteristic makes Tullock contests highly conducive to attracting user entry, especially weak players[@Franke14],[^2] which is of high practical interest because weak players often constitute the majority of potential participants in crowdsourcing. This explains why, in reality, we see [*lotteries*]{}—the simplest form of Tullock contests—much more often, and usually engaging a larger number of participants, than all-pay auctions.
On the other hand, Tullock contests are not necessarily superior to all-pay auctions in terms of [*revenue*]{}, or total user contribution. In fact, there is no conclusive theory as to which contest regime revenue-dominates the other in general[@allpay-lot-wp13; @Fang02lot; @Franke14], and an experimental study also shows that revenue in all-pay auctions may be independent of the number of participants at some stable state[@allpay06exp]. Intuitively, the reason is that (a) the fierce competition induced by all-pay auctions efficaciously incentivizes (a small number of) strong players to exert high effort, while (b) the mild competition in Tullock contests attracts a medium amount of contributions from more (albeit weaker) players. Therefore, all-pay auctions are advantageous in eliciting the highest-quality contribution from the strongest players, such as selecting the best performer for a competition, while Tullock contests are superior in attracting more users and hence are beneficial to population diversity and geographic coverage, such as in lifestyle[@livecomp09] and transport mobile apps[@greengps10mobisys; @lau11ucc].
Thus Tullock contests are complementary to all-pay auctions. We note that these two frameworks have been compared in terms of their respective benefits in other domains such as fundraising[@allpay-lot-wp13], lobbying[@Fang02lot], and general contests[@lottery13apa]. We find that the comparison results therein can apply to crowdsourcing in principle. Therefore, the rest of the paper will focus, within the regime of Tullock contests, on [*optimizing*]{} this framework for crowdsourcing.
The objective, as is most common, is to maximize the crowdsourcer’s revenue. To this end, we explore a new dimension in the space of Tullock contest design, by superseding the contest prize—which is [*fixed*]{} in conventional Tullock contests—with a [*prize function*]{} that is dependent on the (unknown) winner’s contribution. The rationale is to create a [*two-tier incentive*]{} to improve the efficacy of Tullock contests: the first tier, as exists in conventional contests as well, is for a player to win a prize by competing with and outdoing other players; on top of this, the second-tier incentive is for each player to [*outdo himself*]{} in order to [*amplify the prize*]{}. Logically, this approach also leads to a change of the crowdsourcer’s objective: maximizing revenue becomes maximizing [*profit*]{}—revenue minus the (non-fixed) cost (prize)—which is also his utility. To the best of our knowledge, this paper is the first that introduces prize as a function into [*optimal Tullock contest design*]{}, a subject that is being pursued since the 1990’s[@optlot91] following Tullock’s seminal work[@Tullock80] in 1980.
Ultimately, the “new perspective” in the title of this paper has dual interpretations: (a) a new alternative mechanism-design framework for crowdsourcing, and (b) a novel dimension of optimal Tullock contest design.
To find an appropriate benchmark for a new mechanism designed as such to compare against, we need a fixed-prize Tullock contest. However, even this conventional and seemingly simple case turns out to be challenging—a general analytical solution to its equilibria does not exist and only numerical ones are available in the literature[@Fey08; @Ryvkin10; @Wasser13]. Furthermore, we go one significant step beyond prior art, by not only [*solving*]{} equilibria of such conventional contests, but also [*optimizing*]{} the contests by finding the “best” equilibrium in terms of the same (utility-maximizing) objective. This allows us to compare our proposed mechanism with the best possible benchmark.
Extensive performance evaluations reveal that our mechanism outstrips the optimal benchmark by a remarkable margin: a 250% increase in the crowdsourcer’s utility (profit) and a 830% improvement in the players’ aggregate utility (social welfare). The improved performance achieved with these two typically [*competing*]{} metrics reflects a highly desirable “win-win” situation.
Related Work
------------
Even in their simplest form, Tullock contests are analytically more challenging to tackle than most classic auctions. This is particularly true in the incomplete-information setting[^3] which is a more realistic setting for crowdsourcing. Specifically, the equilibria of most classic auctions with complete information, or with incomplete information and symmetric players, can be solved in closed form; but Tullock contests with incomplete information is generally intractable in analytical means, even in the simplest form (lottery)[@Konrad09book]. This can be attributed to the [*double uncertainty*]{}: in auctions with incomplete information, the uncertainty about other players’ types is the only source of uncertainty; but in Tullock contests, the imperfectly discriminating nature—or more specifically the probabilistic winner selection (unlike in auctions the highest bid guarantees winning)—creates another source of uncertainty.
As a consequence, the literature on Tullock contests exclusively deals with the complete-information setting or restrictive versions of the incomplete-information setting (e.g., only two discrete types for two players[@MY04]). It was not until 2008 that Fey[@Fey08] first proved the existence of a (symmetric) equilibrium for a lottery with incomplete information. However, the model is limited to two players and uniform distribution, and the equilibrium strategy is only numerically characterized.
A subsequent breakthrough was made by Ryvkin[@Ryvkin10], who extended Fey’s model[@Fey08] by allowing for more than two players, arbitrary continuous distributions, and a more general contest success function. He also proved the existence of equilibria (leaving uniqueness as future work), following the spirit of [@Fey08]. Still, the equilibrium strategy was only numerically computed, due to the limited analytical tractability of Tullock contests. So far, the only known analytical solution to equilibria with a continuous type distribution, is due to Ewerhart[@Ewer10IPV] who constructed a rather special distribution to obtain a closed-form expression. However, the distribution is rather complex and not generalizable, and the model is still limited to a two-player lottery only.
In this paper, under a general crowdsourcing model with incomplete information, we derive the [*optimal prize function*]{} that maximizes the crowdsourcer’s utility cum profit. Surprisingly, our solution of the unique Bayesian Nash equilibrium (a) can be expressed in a simple and [*closed form*]{} in general cases, and (b) is [*agnostic*]{} to the number of players. These are in stark contrast to prior art, and in practical terms, imply that our mechanism (a) can be easily implemented in web agents and smartphone apps that act in a fully distributed manner, and (b) overcomes the [*disincentive*]{} to participate caused by player’s [*antagonism*]{} to an increasing number of rivals.
Along the line of optimal Tullock contest design, two general directions have been pursued in prior work. One stream of research explores whether the prize should be allocated to a single winner or all the players in a hierarchical manner. For example, [@optlot91] applies a rank-dependent expected utility model to a lottery in which the prize was divided into, according to ranks, a few large prizes and a large number of small prizes. However, [@optSymTullock12] proves that it is optimal to give the entire prize to a single winner in a symmetric equilibrium. The other direction focuses on whether and how to bias players in such a way that induces the maximum revenue. For instance, [@Franke14] allows the crowdsourcer to assign different weights (preferences) to players in a discriminative manner for revenue maximization. [@lottery13apa] proves that a biased lottery (like [@Franke14]) achieves the same revenue as a biased all-pay auction, when both are fully optimized. In the fair case, [@lottery13apa] proves that an optimal lottery is always superior to an optimal all-pay auction.
Our proposed approach represents a new dimension in the design space of Tullock contests. Provisioning contest prize as a function (of the unknown winner’s contribution) sets this work apart from all prior work on Tullock contests in which prizes are fixed and known ex ante.
Contributions
-------------
The main contributions of this paper are summarized below:
1. This work is the first attempt in the crowdsourcing literature that uses Tullock contests as a new framework to design incentive mechanisms.
2. We explore a new dimension of optimal Tullock contest design by provisioning the prize as a function. We demonstrate the simplicity of our approach which makes it particularly well-suited for rapid prototyping in fully distributed web agents and smartphone apps. We also show that our approach overcomes the disincentive caused by players’ antagonism to an increasing number of rivals.
3. As a byproduct of this work, we construct an optimal fixed-prize Tullock contest as the benchmark for comparison and outline a step-by-step algorithm for it. This benchmark precisely falls in line with standard Tullock contests on which extensive studies are based. Therefore, the benchmark, its constructing algorithm and the associated performance analysis, are highly instructive for future research on Tullock contests.
4. Our last contribution, which is [*not*]{} mentioned above, is that we introduce a new parameter—the crowdsourcer’s valuation of user contribution—into the contest model, and show that it has an [*exponential*]{} positive effect on the performance of both our and conventional mechanisms. In practical terms, this means that a crowdsourcer can accrue higher payoff by improving his business processes via a better utilization of the crowdsourced contributions.
The rest of this paper proceeds as follows. presents our model with our proposed mechanism, and analyzes the model to derive the optimal Tullock contest. The optimal benchmark is then constructed in . Following that, an extensive performance evaluation is provided in which demonstrates key results and offers intuition as well. Finally, concludes.
Contest Model {#sec:model}
=============
Sitting at the core of a Tullock contest framework is a [*contest success function*]{} (CSF) which specifies the probability that a player $i=1,2,...,n$ who exerts (or “bids”) effort $b_i$ wins the contest. Our model assumes a very general form of CSF: $$\begin{aligned}
\label{eq:csf}
\Pr(b_i) = \frac{g(b_i)}{\sum_{j=1}^n g(b_j)}\end{aligned}$$ which generalizes the classic Tullock CSF, $b_i^r/\sum_{j=1}^n b_j^r$ where $r>0$, as well as the most-studied form of $r=1$ (also known as [*lottery*]{}). In , $g(\cdot)$ is a nonnegative, strictly increasing function that satisfies $g(0)=0$ and converts player $i$’s effort $b_i$ into his [*contribution*]{} $\xi_i$. For mathematical convenience, we assume that $g(\cdot)$ is twice differentiable and concave, which captures the common phenomenon of diminishing marginal return when exerting effort. When $b_i=0$ for all $i$, i.e., no one exerts effort, we assume $\Pr(b_i)=0$, i.e., no one will win any prize.[^4]
Another crucial component of our contest model is a prize function $V(\xi_w)$ that we specifically introduce in the contest, which is a (monetary) prize of a common value that is dependent on the (unknown) winner’s contribution $\xi_w$. Accordingly, a player $i$ will receive an expected income of $\Pr(b_i) V(\xi_i)$. This function $V(\cdot)$ is common knowledge to all players (e.g., via announcement by the crowdsourcer).
Each player is characterized by his [*type*]{}—his marginal cost of exerting effort—denoted by $c_i\in[\uline c, \ovl c]$, where $0<\uline c<\ovl c$. That is, exerting effort $b_i$ will incur a cost of $c_i b_i$ to player $i$. Thus, if the effort profile of all the players is $\vect b:=(b_1,b_2,...,b_n)$, the payoff of player $i$ can be expressed in the following quasi-linear form: $$\begin{aligned}
\Pr(b_i) V(\xi_i) - c_i b_i.\end{aligned}$$ Since $\xi_i=g(b_i)$, a player’s (ex post) payoff given the contribution strategy profile of all the players $\vect \xi:=(\xi_1,\xi_2,...,\xi_n)$, is $$\begin{aligned}
\label{eq:utdef-expost-inv}
\tilde u_i(c_i, \vect \xi) = \frac{\xi_i}{\sum_{j=1}^n \xi_j} V(\xi_i) - h(\xi_i) c_i\end{aligned}$$ where $h:=g^{-1}$ is the inverse function of $g(\cdot)$. We note that $g^{-1}$ exists because $g(\cdot)$ is strictly monotone.
As crowdsourcing typically involves an undefined group of people, we assume the [*interim*]{} stage which corresponds to an [*incomplete-information*]{} setting: each player $i$ is informed of his own type $c_i$ but not of others’, yet it is common knowledge that all the $c_i$ are independently drawn from a continuum $[\uline c, \ovl c]$ according to a c.d.f. $F(c)$ or p.d.f. $f(c)=F'(c)$.[^5] On the contrary, the [*ex ante*]{} stage corresponds to a no-information setting where players do not know anyone’s type including their own, and the [*ex post*]{} stage corresponds to a complete-information setting where all the players’ types are common knowledge.
The crowdsourcer collects revenue from the aggregate contribution of all the players, and bears the cost of paying for the (variable) prize. The profit cum utility of the crowdsourcer is thus $$\begin{aligned}
\label{eq:orgut-def}
\tilde \pi = \nu \sum_{i=1}^n \xi_i - V(\xi_w)\end{aligned}$$ where $\nu$ is the crowdsourcer’s valuation of per unit user contribution. This parameter $\nu$ does not appear in prior work where a unity value is always implicitly assumed. However, explicitly modeling this parameter not only homogenizes the dimension of the expression , but also allows us to investigate the impact of $\nu$ on key metrics such as the player strategy, prize, profit, and social welfare, which turns out (cf. ) to be an interesting combination of both linearity and nonlinearity.
Proposed Optimal Tullock Contest {#sec:analysis}
================================
The solution concept of a game with incomplete information is a pure strategy [*Bayesian Nash equilibrium*]{}, in which each player plays a strategy that maximizes his expected utility given his belief about other players’ types and that other players also play their respective equilibrium strategies. Formally, it is a strategy profile $\vect \xi^{BNE}=(\xi_1^{BNE},\xi_2^{BNE},...,\xi_n^{BNE})$ that satisfies $$u_i(c_i,\xi_i^{BNE};\xi_{-i}^{BNE}) \ge u_i(c_i,\xi_i;\xi_{-i}^{BNE}), \;\; \forall \xi_i, \forall i,$$ where $u_i$ is the expected utility of player $i$, defined as $$\begin{aligned}
\label{eq:utdef}
u_i(c_i,\xi_i) := \mathbbm E_{\xi_{-i}}[\tilde u_i(c_i, \vect \xi)]\end{aligned}$$ where $\vect \xi=(\xi_i,\xi_{-i})$, and $\tilde u_i$ is defined in .
Definition can be expanded as follows. In a Bayesian Nash equilibrium, each player’s strategy $\xi_i$ is a function of his own type $c_i$ and the common [*prior*]{}, i.e., each player’s belief about all the other players’ types. As our setting is symmetric, in that the prior is a common distribution $F(\cdot)$ for all the players,[^6] we focus on symmetric equilibria in which any player $i$’s equilibrium strategy is specified by a function $\beta:[\uline c,\ovl c]\rightarrow \mathbbm R_+$ as $\xi_i=\beta(c_i), \forall i$. Therefore, definition can be rewritten based on , as $$\begin{aligned}
\label{eq:ut}
u(c,\xi) = p(\xi) V(\xi) - h(\xi) c\end{aligned}$$ for an arbitrary type $c$ and strategy $\xi$, where $$\begin{aligned}
\label{eq:winprob}
p(\xi) := \int_{\Theta^{n-1}} \frac{\xi}{\xi + \sum_{j=1}^{n-1} \beta(\tilde c_j)}
\prod_{j=1}^{n-1} \opd F(\tilde c_j)\end{aligned}$$ in which $\Theta := [\uline c, \ovl c]$.
Thus, for a particular player $i$, his expected utility is $u_i=u(c_i,\xi_i)$ which can be computed from .
\[thm:exists\] Our Tullock contest model admits a unique, monotone decreasing, pure-strategy Bayesian Nash equilibrium.
Due to space constraint, we defer all the proofs of this paper to [@infocom15appendix].
Henceforth, we will exclusively deal with the equilibrium, and thereby drop the superscript BNE for brevity.
\[lem:strategy\] Given an arbitrary prize function $V(\cdot)$, the (symmetric) equilibrium strategy $\beta(\cdot)$ of our Tullock contest, as in $\xi=\beta(c)$, is implicitly given by $$\begin{aligned}
\label{eq:strategy}
p(\xi) V(\xi) - h(\xi) c = \int_{c}^{\ovl c} h(\beta(\tilde c)) \opd \tilde c,\;\;
\forall c\in[\uline c,\ovl c].\end{aligned}$$
We remark on the following:
- Equation has an intuitive interpretation: player $i$’s expected utility, as represented by the l.h.s., is determined by his cost advantage relative to the highest-cost player, modulated by his contribution level.
- There is no closed-form solution to for an arbitrary $V(\cdot)$. In fact, even the fixed-prize case (as in conventional contests) does not have a closed-form solution in general either[@Konrad09book]. However, an interesting and counter-intuitive finding arises from maximizing the crowdsourcer’s utility through an optimal prize function, on which we remark following the next theorem.
\[thm:optimal\] The optimal prize function that maximizes the crowdsourcer’s utility (profit) in our Tullock contest is given by $$\begin{aligned}
\label{eq:opt-prize}
V^*(\xi_w) = \left[ \beta^{-1}(\xi_w) h(\xi_w)
- \int_{\uline \xi}^{\xi_w} h(\tilde \xi) \opd \beta^{-1}(\tilde \xi) \right]
\Big/ p(\xi_w)\end{aligned}$$ where $\uline\xi=\beta(\ovl c)$ and $\beta^{-1}(\cdot)$ is the inverse function of the equilibrium strategy $\beta(\cdot)$ which, as in $\xi=\beta(c)$, is given by $$\begin{aligned}
\label{eq:opt-strategy}
h'(\xi) = \frac{\nu}{c + \frac{F(c)}{f(c)}},\; \forall c\in[\uline c,\ovl c].\end{aligned}$$ The induced maximum profit of the crowdsourcer is $$\begin{gathered}
\label{eq:profit}
\pi^* = n \int_{\uline c}^{\ovl c} \bigg[
\nu \beta(c) - h(\beta(c)) c \\
+ \frac{F(c)}{f(c)} [ h(\beta(\ovl c)) - h(\beta(c)) ] \bigg] \opd F(c).\end{gathered}$$
We will illustrate how to put to use, in with a case study. Here we remark on the following:
- The strategy $\uline\xi$ or $\beta(\ovl c)$, as of the highest-cost or weakest player, is always 0 in all-pay auctions under standard assumptions. However, in Tullock contests, this is not necessarily the case (unless $\ovl c=\infty$), which will be evidenced in . The reason is that any Tullock contestant has a positive winning probability as long as he exerts nonzero effort, whereas all-pay auctions perfectly discriminate the weakest bidders who have no chance to win.
- [**Analytical tractability and Practical implication**]{}: An interesting and somewhat surprising observation is that, while “functionizing” the contest prize would, intuitively, seem to introduce complexity to conventional, fixed-prize contests, the equilibrium strategy turns out to be much simpler compared to the fixed-prize case (cf. in ). In fact, for most and common functions $h(\cdot)$, it can be expressed in closed form. This convenient analytical tractability is in stark contrast to prior art (see [@Fey08; @Ryvkin10; @Wasser13] and a survey [@Konrad09book]) where equilibria do not have analytical solutions in general and can only resort to numerical methods. Theoretically, this lends us a lot of convenience in subsequent technical treatments and other possible future extensions. Practically, this fosters the application of our mechanism via easily-implementable software deployed in web agents and smartphone apps that act on each user’s behalf to determine his contribution strategy in a fully distributed manner.[^7]
- [**Agnosticism of strategy and Practical implication**]{}: Another surprising observation is that the equilibrium strategy is [*agnostic*]{} to $n$. This is counter-intuitive and in direct contrast with prior findings (again see [@Fey08; @Ryvkin10; @Wasser13; @Konrad09book]; also cf. ) where players are [*antagonistic*]{} to an increasing number of rivals: when the number of players increases, each individual player’s chance of winning will be diluted, and hence if the prize is fixed, each player will have to expect a lower utility, resulting in a [*disincentive*]{} to participate. However, now that players can stay agnostic to the participant pool size,[^8] they need not worry about an increasing number of rivals, which certainly strengthens the motivation to participate or stay in the campaign. In more practical terms, participants would even not be averse to spreading the awareness and publicity of a campaign, which helps further expand the participant pool.
Now we state an important condition pertaining to general incentive mechanisms: [*individual rationality*]{} (IR) [@Jackson03md]. It means that any participating player should receive in equilibrium a nonnegative expected utility, or in other words, each player should be better off or at least remain neutral by participating. In the following, we prove that our mechanism possesses a stronger version of IR.
\[thm:IR\] Our Tullock contest with the prize function given by satisfies [*strict individual rationality*]{} (SIR), where all the players receive strictly positive expected utility, except that a player of type $\ovl c$—which happens with probability zero—expects a surplus of zero (and hence is indifferent in participating).
Another often-discussed property in mechanism design is [*incentive compatibility*]{} (IC) [@Jackson03md] or truthfulness, which means that all players report their types truthfully. This property is technically irrelevant to our mechanism because, unlike some other mechanisms such as [@yang12mobicom; @kout13infocom] which determine workers’ wages based on worker-reported types (costs or desired payments), our mechanism determines players’ reward based on [*observable*]{} user contributions rather than [*unobservable*]{} (and private) player types (costs). On the other hand, those other mechanisms can satisfy [*IR*]{} trivially by paying a wage no less than a worker’s reported cost or payment, provided that IC is satisfied; but in our case, satisfying IR requires a carefully designed prize function (which is demonstrated by the proof of ).
Optimal Fixed-prize Tullock Contests {#sec:opt-fixed}
====================================
This section constructs a conventional, i.e., fixed-prize, Tullock contest for the sake of comparison with our mechanism (later in ).[^9] However, we augment the prior art of dealing with conventional contests[@Fey08; @Ryvkin10; @Wasser13], from [*solving*]{} the equilibrium to [*optimizing*]{} (as well as solving) the equilibrium, in order to create the most superior benchmark to challenge our mechanism.
Specifically, we set to find an optimal fixed prize $V_0^*$ that maximizes the crowdsourcer’s utility $\pi_0$ through a particular equilibrium $\vect\xi_0^*$, and such a solution needs to be found for every possible value of $\nu$, the valuation of contribution. Formally, the problem is formulated as $\max_{V_0\ge 0} \pi_0$ where $$\begin{aligned}
\label{eq:profit-fixed}
\pi_0 = \nu \mathbbm E_{\vect c} \left[ \sum_{i=1}^n \xi_{0 i} \right] - V_0
= n\nu \int_{\uline c}^{\ovl c} \beta_0(c) \opd F(c) - V_0. \end{aligned}$$ The equilibrium strategy $\xi_0=\beta_0(c)$ is given by .
\[thm:strategy-fix\] In a Tullock contest with fixed prize $V_0$, the equilibrium strategy $\xi_0=\beta_0(c)$ is implicitly determined by $$\begin{aligned}
\label{eq:strategy-fixed}
\int_{\Theta^{n-1}} \frac{\sum_{j=1}^{n-1} \beta_0(\tilde c_j)}
{[\beta_0(c) + \sum_{j=1}^{n-1} \beta_0(\tilde c_j)]^2 } \prod_{j=1}^{n-1} \opd F(\tilde c_j)
= h'(\xi_0) \frac{c}{V_0}.\end{aligned}$$
Unfortunately, does not have an analytical solution. The special case of $V_0=1$ was numerically tackled by [@Fey08; @Ryvkin10; @Wasser13]. In our case, $V_0$ is not given and we need to find the optimal $V_0$ that maximizes $\pi_0$ . In the meantime, $\pi_0$ contains $\beta_0(c)$ which is analytically intractable. Moreover, the optimal solution $V_0^*$ must not be a value but a function of $\nu$. This problem is technically challenging.
Our solution was inspired by solving the [*Fredholm equations*]{} using a numerical method described in [@numeric90NASA] (Chap. 5). Consider a two-player case for simplicity. The first key idea in our solution is to transform the integral in into a quadrature sum, by supposing that $V_0$ is given: $$\begin{aligned}
\label{eq:strategy-num1}
\sum_{j=1}^m \frac{\beta_0(t_j) f(t_j) \Delta_j} {[\beta_0(c) + \beta_0(t_j)]^2} + R_m(c)
= h_{\beta_0}'(\beta_0(c)) \frac{c}{V_0},\end{aligned}$$ where $t_j, j=1,2,...,m$, are the quadrature points distributed in $\Theta$, $\Delta_j$ is determined by the chosen quadrature scheme (e.g., Gaussian), and $R_m(c)$ is the residual error due to transforming the original integral into the quadrature sum. In our case, a uniform quadrature scheme suffices and thereby $\Delta_j=\Delta_m:=(\ovl c - \uline c)/m$. In addition, since the integrand is atomless ($\beta_0(t_j)=0$ for all $j$ is not an equilibrium because a player will have incentive to deviate by an infinitesimal amount to gain positive utility), the integral can be closely approximated by the quadrature sum for a sufficiently large $m$, in which case $R_m(c)$ can be safely ignored.
The second key idea is to note that, since holds for all $c\in[\uline c,\ovl c]$, it must also hold for all the $c_i$ that equal to the quadrature points $t_j, j=1,2,...,m$. Thus, is further transformed into a system of $m$ nonlinear equations: $$\begin{aligned}
\label{eq:strategy-num2}
\Delta_m \sum_{j=1}^m
\frac{\beta_0(c_j) f(c_j)} {[\beta_0(c_i) + \beta_0(c_j)]^2}
- h_{\beta_0}'(\beta_0(c_i)) \frac{c_i}{V_0} = 0,\; i=1,...,m.\end{aligned}$$ This system can be solved using the Matlab function [*fsolve*]{}.
Similarly but on a much simpler scale, the profit can be approximated by $$\begin{aligned}
\label{eq:profit-num}
\hat\pi_0(\nu, V_0,\beta_0(\vect c)) = n \nu \Delta_m \sum_{j=1}^m \beta_0(c_j) f(c_j) - V_0\end{aligned}$$ where $\beta_0(c_i)|_{i=1}^m$ are the solutions to the system .
The entire solution is outlined in a self-explanatory fashion by the psuedo-code in . In the actual implementation, we also added a testing condition to ensure the range $[\uline{V_0},\ovl{V_0}]$ to be large enough to include the peak point of $\hat\pi_0$ (which can be shown to be concave in $V_0$); we also improved the efficiency by adding a stopping condition to terminate the inner loop faster. These are peripheral and hence omitted in .
$\Delta_m \gets (\ovl c - \uline c)/m$ $\vect c \gets \{\uline c, \uline c+\Delta_m, \uline c+2\Delta_m,..., \uline c+(m-1)\Delta_m$} $\vect V_0 \gets \{\uline{V_0},\uline{V_0}+\delta_2,\uline{V_0}+2\delta_2,...,\ovl{V_0}\}$ Create a $|\vect V_0| \times m$ strategy matrix $\vect\xi^A$
Performance Evaluation {#sec:numeric}
======================
In this section, we compare for the same crowdsourcing campaign a Tullock contest that employs our design against a Tullock contest that adopts a fixed prize, [*ceteris paribus*]{}. The former uses the optimal prize function determined by , which we refer to as [*Tullock-OPF*]{}, and the latter uses the optimal fixed prize determined by , which we refer to as [*OptBenchmark*]{}. We stress that this benchmark is not designed to favor our proposed mechanism, but honestly follows the well-established standard model and, in fact, is fully optimized.
The performance metrics we evaluate are:
1. the [*equilibrium contribution strategy*]{} of players, which corresponds to revenue;
2. the [*prize*]{} that the crowdsourcer provisions, which corresponds to cost;
3. the expected [*utility*]{} of the crowdsourcer, which corresponds to profit;
4. the [*social welfare*]{} of the campaign, which is the aggregate utility of all the players at equilibrium and we denote by $U:=\mathbbm E_{\vect c}[\sum_{i=1}^n u_i]$; it measures the total surplus of the community by participating in the crowdsourcing campaign.
The rationale is that a company is typically profit-driven and thus concerned about the metrics (a)–(c), while a government agency or a non-profit organization may focus more on the metric (d).
The primary scenario in this section consists of two players whose marginal contribution costs are independently drawn from a uniform distribution $F(c)=c-1, c\in[1,2]$. Player effort $b$ is converted to user contribution $\xi$ according to $\xi=g(b)=\sqrt b$, and hence $h(\xi):=g^{-1}(\xi)=\xi^2$. Such a setup is common in the literature such as [@Fey08; @Ryvkin10]. This scenario is then extend to a $n$-player setting.
Tullock-OPF: Analytical Results
-------------------------------
Our mechanism can be solved analytically, and the solving process also illustrates how to put to use. First, the equilibrium contribution strategy can be obtained via as $$\begin{aligned}
\label{eq:opf-strategy}
\xi = \beta(c) = \frac{\nu}{4c-2}.\end{aligned}$$ Hence $\beta^{-1}(\xi)=\nu/(4\xi)+ 1/2$, and the numerator of equals $$\begin{aligned}
\xi^2 (\frac{\nu}{4\xi}+ \inv 2)
+ \int_{\uline \xi}^{\xi} \tilde\xi^2\cdot \frac{\nu}{4 \tilde\xi^2} \opd \tilde\xi
= \frac{\xi^2}{2} + \frac{\nu \xi}{2} - \frac{\nu^2}{24}\end{aligned}$$ where $\uline\xi=\beta(\ovl c)=\nu/6$. Using , the denominator of equals $$\begin{aligned}
p(\xi) &= \int_1^2 \frac{\xi}{\xi + \frac{\nu}{4c-2} }\opd c
= \int_1^2 \left(1 - \frac{\nu}{\xi (4c-2) + \nu} \right) \opd c \\
&= 1 - \frac{\nu}{4\xi} \log(4\xi c +\nu -2\xi)\Big|_1^2
= 1 - \frac{\nu}{4\xi} \log\frac{6\xi +\nu}{2\xi +\nu}.\end{aligned}$$ Therefore, the optimal prize function is obtained as $$\begin{aligned}
\label{eq:opf-prize}
V^*(\xi_w) = \frac{\frac{\xi_w^2}{2} + \frac{\nu \xi_w}{2} - \frac{\nu^2}{24}}
{1 - \frac{\nu}{4\xi_w} \log\frac{6\xi_w +\nu}{2\xi_w +\nu}}.\end{aligned}$$ Finally, the induced maximum profit is obtained via : $$\begin{aligned}
\label{eq:opf-profit}
\pi^* &= 2 \int_1^2 \!\left[ \frac{\nu^2}{4c-2}\! - \!\frac{\nu^2 c}{(4c-2)^2}
\!+\! (c-1) \Big(\frac{\nu^2}{36}\! -\! \frac{\nu^2}{(4c-2)^2}\Big) \right]\! \opd c \nn\\
&= 2 \nu^2 \int_1^2\! \left( \inv{8c-4} + \frac{c-1}{36} \right)\! \opd c
= \left( \frac{\log 3}{4} + \inv{36} \right) \nu^2\end{aligned}$$
OptBenchmark: Numerical Results
-------------------------------
OptBenchmark can only be numerically solved, using , of which the parameters are specified by .
----------- ------------- ----------- --------------- ------------- ----- ------------ ------------
$\uline\nu$ $\ovl\nu$ $\uline{V_0}$ $\ovl{V_0}$ $m$ $\delta_1$ $\delta_2$
\[0.5ex\] 0.5 5 0.01 5 100 0.5 0.01
----------- ------------- ----------- --------------- ------------- ----- ------------ ------------
: Parameters for (OptBenchmark)[]{data-label="tab:param"}
reveals the trajectory of finding the duple $(V_0^*,\pi_0^*)$, i.e., the optimal prize and maximum profit, by evaluating over a range of possible prizes, for each $\nu=1,2,3$. The optimal duple is found at the peak of each curve, and the curves clearly demonstrate the concavity of profit versus prize, which confirms the existence and uniqueness of the optimum. Furthermore, how the optimal duple $(V_0^*,\pi_0^*)$ is affected by $\nu$ is examined by . Interestingly, prize $V_0^*$ coincides with profit $\pi_0^*$ for all the $\nu$’s. This indicates that the revenue of OptBenchmark is [*double of the cost*]{} (prize). This also implies that, if in we draw all the other trajectories (in addition to the in-situ three), the peak points of all the trajectories will all fall onto the same straight line $y=x$.
Another observation in is that both prize and profit are convexly increasing in $\nu$. We will revisit this nonlinear behavior together with other subsequent observations in .
Comparison
----------
Given that both Tullock-OPF and OptBenchmark are solved by now, we proceed to compare them with respect to the four metrics mentioned earlier.
[**Crowdsourcing Revenue**]{}: examines the equilibrium contribution strategy of players as a function of player type, for each $\nu=1,2,3$. We remark on three observations. First, in all the cases (3$\times$2 curves), the strategy is monotone decreasing in type, which conforms to our (which subsumes $V(\cdot)$ being constant). The convex trend is also consistent with the literature: for example, we verified a special case of OptBenchmark with $h(\xi)=\xi$, $\nu=1$, $c\in[0.01,1.01]$, which is the same as [@Fey08], and the result (not reproduced here) exactly matched [@Fey08]. Second, for any $\nu$, Tullock-OPF elicits significantly higher contribution than OptBenchmark for every possible type $c$, by about 150% for high-cost (weak) players and 400% for low-cost (strong) players. Third, in both mechanisms, the highest-cost or weakest player makes strictly positive contribution, i.e., $\uline\xi=\beta(\ovl c)>0$, indicating a sheer contrast between Tullock contests and auctions where $\uline\xi=0$; here we recall the first remark below .
[**Crowdsourcing Cost**]{}: compares the optimal prize function $V^*(\xi_w)$ in Tullock-OPF against the optimal fixed prize $V_0^*$ in OptBenchmark. Note that the support of $V^*(\xi_w)$ is $[\uline\xi,\ovl\xi]=[\nu/6,\nu/2]$ which follows from . While it is intuitive that each of the three $V^*(\xi_w)$ curves increases in winner’s contribution, it is interesting to note that each curve fits a straight line very well, which suggests that the computation of could be remarkably simplified in practice via linear approximation. While this advantage should not be overstated as to how it generalizes, it hints at a possible line of future work. Moreover and noteworthily, the diagram shows that the prize offered by Tullock-OPF is generally well above OptBenchmark. This raises an important question that whether this much higher cost to be borne by the crowdsourcer will eventually pay off, which is answered next.
[**Crowdsourcing Profit**]{}: evaluates the maximum profit that a crowdsourcer garners from the two mechanisms. The salient observation is that Tullock-OPF outperforms OptBenchmark by a large profit margin, which strongly corroborates our proposal of using an optimized prize function to supersede the conventional, fixed prize in Tullock contests. For a closer examination, we collate the results into and calculate the ratio between each pair of profits for all the $\nu$’s. It is interesting to note that the ratio almost remains constant, at about 3.53. This could be explained by the nature of optimization which has pushed the profit to the limit in both mechanisms. In addition, a rigorous analysis as well as investigating to what extent this result can generalize may be worth future exploring.
[**$\nu$**]{} 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
---------------------- -------- -------- -------- -------- -------- -------- -------- -------- -------- --------
[**OptBenchmark**]{} 0.0213 0.0853 0.1927 0.3426 0.5354 0.7710 1.0494 1.3707 1.7347 2.1417
[**Tullock-OPF**]{} 0.0756 0.3024 0.6805 1.2097 1.8902 2.7219 3.7048 4.8389 6.1242 7.5608
[**Ratio**]{} 3.5533 3.5450 3.5315 3.5307 3.5306 3.5303 3.5303 3.5303 3.5303 3.5303
[**Social Welfare**]{}: In Tullock-OPF, this can be analytically obtained. To solve for $U = \mathbbm E_{\vect c}[\sum_{i=1}^n u_i]$, rather than using the definition , we leverage on which lends us much more convenience: $$\begin{aligned}
U &= \mathbbm E_{\vect c}\left[\sum_{i=1}^n u_i\right] = n \int_{\uline c}^{\ovl c} \int_{c}^{\ovl c} h(\beta(\tilde c)) \opd \tilde c \opd F(c) \label{eq:socwf}\\
&= n \int_{\uline c}^{\ovl c} \int_{c}^{\ovl c} \frac{\nu^2}{(4\tilde c -2)^2} \opd \tilde c \opd F(c) \nn\\
&= 2 \int_1^2 \frac{\nu^2}{8} \left(\inv{2c-1} - \inv 3\right) \opd c
= \left(\frac{\log 3}{8} - \inv{12}\right) \nu^2 \nn\end{aligned}$$
One the other hand, the social welfare in OptBenchmark has to resort to numerical methods. To do so, we take the output $\vect\Xi_0^*$ of , and denote the row $\vect\Xi_{0}^*(\nu)$ corresponding to each $\nu$ by a strategy profile $\vect\xi_{0}^*$. Thus, we can compute the social welfare as $$\begin{aligned}
U_0 = n \Delta_m \sum_{i=1}^m \left[ f(c_i)
\left( \Delta_m \sum_{j=i}^m {\xi_{0j}^*}^2 \right) \right],\end{aligned}$$ which can be understood by rewriting as $$U = n \int_{\uline c}^{\ovl c} f(c) \int_{c}^{\ovl c} \xi^2 \opd \tilde c \opd c$$ and noting that applies, without change, to fixed-prize cases where $V(\cdot)$ is a constant.
The results are presented in . The observation is exciting: Tullock-OPF outstrips OptBechmark in an even more striking manner as compared to profit in . The detailed data are collated in , which indicates a remarkable improvement of 7–9.3 folds. On the other hand, the nearly constant ratio in is not duplicated here.
[**$\nu$**]{} 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
---------------------- -------- -------- -------- -------- -------- -------- -------- -------- -------- --------
[**OptBenchmark**]{} 0.0019 0.0058 0.0155 0.0271 0.0407 0.0600 0.0813 0.1065 0.1336 0.1666
[**Tullock-OPF**]{} 0.0135 0.0540 0.1215 0.2160 0.3375 0.4859 0.6614 0.8639 1.0934 1.3498
[**Ratio**]{} 6.9693 9.2924 7.8404 7.9649 8.2968 8.0934 8.1308 8.1097 8.1813 8.1038
It may be puzzling as to why the crowdsourcer can reap higher profit while, at the same time, users altogether also gain higher surplus. This constitutes a “win-win” situation which is highly desirable but typically hard to attain. To explain this, we note the following rationales. First, generally speaking, crowdsourcing is not a [*zero-sum game*]{} like the stock market; rather, it involves a [*wealth creation*]{} process in which users exert effort to create “something” valuable that we have abstracted as “contribution”. Second, there exists a [*value asymmetry*]{} between players and the crowdsourcer, where the crowdsourcer typically values contribution higher than players do. This value asymmetry is a common phenomenon in reality: for example in the worldwide emerging Smart City and Smart Nation initiatives nowadays, citizen-generated data such as ambient noise and GPS traces collected by smartphones do not usually bear much value to the phone owners but can be very valuable to a crowdsourcer such as a noise control bureau or a transport company (like Waze). Thus, it makes perfect business sense for a crowdsourcer to provision certain attractive reward to incentivize more user contribution which in turn bears even more value to the crowdsourcer.
Impact of Crowdsourcer’s Valuation {#sec:orgvalue}
----------------------------------
Recall that we have introduced a new parameter, $\nu$, to our model in (cf. ). Herein, we investigate how it affects the various performance indicators.
This effect is explicitly examined with respect to the optimal fixed prize (), maximum profit (), and social welfare (), where it is clearly shown that the impact of $\nu$ on these three metrics is [*nonlinear*]{} (convex). Furthermore, this effect is also implicitly examined with respect to the equilibrium contribution strategy () and optimal prize function () (in both diagrams one needs to compare across different curves corresponding to different $\nu$’s), and we see that the impact of $\nu$ on these two metrics is approximately [*linear*]{}.
The main message conveyed here is that, if a crowdsourcer increases his valuation of user contribution, e.g. by improving his business processes to better exploit user contribution, his profit and the players’ social welfare will both [*increase faster*]{}. This is an interesting finding uncovered due to our introduction of $\nu$. Moreover, this observation applies to both Tullock-OPF and the conventional case.
Now we dive in deeper to explain the correlation between the above nonlinearity and linearity. On the one hand, the case of OptBenchmark can be nicely explained: (a) the nonlinear profit () results from the linear revenue (contribution; ) and nonlinear cost (prize; ), and (b) the nonlinear social welfare () follows from the nonlinear prize (player’s gain; ) and the linear strategy (player’s cost; ).
On the other hand, the case of Tullock-OPF is not that straightforward, since both revenue (contribution) and cost (prize) seem to be linear across different $\nu$’s. In fact, an overlooked fact was that the prize functions in are [*shifted*]{} horizontally, and thus one should compare prizes across $\nu$’s for the same [*winner*]{} rather than for the same amount of [*contribution*]{}. To do this, a simple way is to compare the maximum (or the minimum) winner contribution $\xi_w$ across different curves, as it can uniquely identify a particular winner. This reveals that the impact of $\nu$ on prize is actually [*nonlinear*]{}. Combined with the linearity on revenue (), this explains why the profit and social welfare in Tullock-OPF are nonlinear in $\nu$.
$n$-Player Case
---------------
In this section, we extend our investigation to $n$ players. We conjecture that the above comparison results will continue to hold in the $n$-player case, and hence we will not repeat the same comparisons. In fact, as Ryvkin [@Ryvkin10] pointed out, it is computationally infeasible to numerically solve fixed-prize Tullock contests for an arbitrary large number of players because of the “curse of dimensionality”. Therefore, this section focuses on Tullock-OPF. In particular, we are interested in how the [*composition*]{} of a participant pool, i.e., the distribution of the player types, affects the two key metrics, profit and social welfare.
We consider two participant pools: [*Population-1*]{} draws player types from the same distribution as above, i.e., $F(c)=c-1, c\in[1,2]$, while [*Population-2*]{} draws from another distribution $G(c)=\frac{c}{2} -\inv 4, c\in[0.5,2.5]$. Thus, Population-2 is more [*diverse*]{}—or is more [*uncertain*]{} in player types—than Population-1, while they statistically share the same mean value (1.5).
Following from , the equilibrium strategy in Population-2 is $\xi_G = \frac{\nu}{4c - 1}$, and thus the maximum profit is $$\begin{aligned}
\label{eq:opf-profit-Gn}
\pi_{n,G}^* \! &=\! \frac{n}{2} \int_{0.5}^{2.5}\! \left[
\frac{\nu^2}{4c-1}\! - \!\frac{\nu^2 c}{(4c-1)^2}\! +\!
(c-\inv 2) \Big(\frac{\nu^2}{81}\! - \!\frac{\nu^2}{(4c-1)^2}\Big) \right]\! \opd c \nn\\
&=\! \frac{n \nu^2}{2} \int_{0.5}^{2.5}\! \left( \inv{8c-2}\! +\! \frac{2c-1}{162} \right) \opd c
= \left( \frac{\log 3}{8} + \frac{2}{81} \right) n \nu^2. \nn \end{aligned}$$ For Population-1, we leverage as a shortcut to obtain $$\begin{aligned}
\pi_{n,F}^* = \left( \frac{\log 3}{8} + \inv{72} \right) n \nu^2.\end{aligned}$$ Social welfare can be solved by referring to : $$\begin{aligned}
U^*_{n,G} &= \frac{n}{2} \int_{\uline c}^{\ovl c} \int_{c}^{\ovl c} \frac{\nu^2}{(4\tilde c -1)^2} \opd \tilde c \opd c
= \left(\frac{\log 3}{16} - \inv{36}\right) n \nu^2, \\
U^*_{n,F} &= \left(\frac{\log 3}{16} - \inv{24}\right) n \nu^2.\end{aligned}$$
Thus it immediately follows from the above that $$\pi_{n,G}^* > \pi_{n,F}^*, \quad U^*_{n,G} > U^*_{n,F}, \quad \forall n\ge 2, \forall \nu>0.$$ This tells that Population-2 is superior to Population-1 in terms of both profit and social welfare. An insight that may be drawn from this set of results is that population [*diversity*]{} or [*uncertainty*]{} is beneficial to Tullock-contest based crowdsourcing for both crowdsourcers and participants.
Conclusion {#sec:conc}
==========
To recap, this work has presented a first attempt to use Tullock contests as a new framework to design incentive mechanisms for crowdsourcing. Furthermore, we have explored a novel dimension in the space of optimal Tullock contest design, by superseding the conventional, fixed prize by an optimal prize function for utility maximization. In stark contrast to prior art, we have obtained an analytical solution to the unique Bayesian equilibrium, and found that the equilibrium is robust to an increasing number of rivals. As our model employs a very general contest success function and assumes incomplete information, the mechanism and results would fit a wide range of practical crowdsourcing applications. For example, WiFiScout[@wifiscout14mass] is a mobile app that aims to profile the performance of citywide WiFi access points by eliciting personal experience on WiFi usage from smartphone users. Similarly, OpenSignal[@opensignal] aims to construct citywide 3G and 4G LTE cell coverage maps through crowdsourcing too.
The superiority of our design has been demonstrated through extensive evaluations by comparing against a fully-optimized benchmark. Constructing this optimal benchmark significantly extends prior art which only [*solves*]{} conventional, fixed-prize Tullock contests. This benchmark would be highly relevant to a wider research community for future performance evaluations.
Moreover, we have introduced the crowdsourcer’s valuation of user contribution which further extends usual contest models (besides our prize function). It is shown to impact two key metrics—the crowdsourcer’s profit and players’ social welfare—in a nonlinear (exponential) manner, which bears practical implication on the worth of improving crowdsourcers’ business processes. Therefore, this new parameter could be included by future studies (e.g., on radio spectrum auctions or heterogeneous networks) in mathematical models to capture [*value asymmetry*]{} and uncover phenomena that are previously unseen.
[^1]: In an asymptotic limiting case, Tullock contests subsume all-pay auctions. However, they are generally classified as two different contest regimes.
[^2]: A “strong” or “weak” player in this paper refers to a player with a strong or weak [*type*]{}; type, as a term in Bayesian games and mechanism design, refers to the private information held by a player, such as his valuation of the auctioned item or his production cost. Therefore, a strong player means a player with high valuation or low cost, and vice versa.
[^3]: Briefly speaking, with complete information all the players are informed of all the others’ types, while with incomplete information each player only knows his own type. explains this in more detail.
[^4]: \[foot:nonzero\]An alternative and more commonly adopted practice in the literature, is assuming $\Pr(b_i)=1/n$. However, the prizes therein are all fixed, and hence if $b_i=0$ for all $i$, a player $j$ will have incentive to deviate by exerting an infinitesimal effort $\eps>0$ to increase his payoff by $(1-\inv{n})v_j - O(\eps)$ where $v_j$ is his valuation of the prize. Therefore, an “all-zero-bid” equilibrium does not exist. In our case, however, the prize is a function $V(\cdot)$ of contribution and $V(\eps)$ can be so small that players lose the incentive to deviate from all-zero bids. Therefore, we impose $\Pr(b_i)=0$ if $b_i=0, \forall i$ to reinstate this incentive. Indeed, we will show later in that our mechanism ensures that a player will receive strictly positive payoff if he exerts non-zero effort.
[^5]: In practice, such a distribution $F(c)$ can be obtained (and published) by the crowdsourcer based on historic data, or—when historic data is not available—assume uniform distribution as widely used in the Bayesian game literature (our model constitutes a Bayesian game with private information being player types and common prior being the type distribution).
[^6]: In an asymmetric setting, player types follow their respective and generally different distributions $F_i(c_i)$, which is common knowledge. Not only is solving such asymmetric equilibria still an open problem with no analytical solution in general[@Konrad09book], but this setting also makes all players [*onymous*]{} and thus may engender privacy concerns in practice.
[^7]: Note that the optimal prize function and maximum profit are computed on the other hand by a [*centralized*]{}, and computationally powerful server, and the computation takes the already-solved $\beta(\cdot)$ as input, unlike in $\beta_0(\cdot)$ is unknown. In fact, in many cases such as demonstrated in , the computation in our mechanism is fairly straightforward.
[^8]: The agnosticism is achieved by the prize function which absorbs $n$ and thereby isolates the player strategy from this number.
[^9]: It is sound and fair to compare our mechanism with a conventional, fixed-prize counterpart, as our prize function is an extension of the conventional (single) fixed prize. This is also in accordance with [@optSymTullock12] which proves that it is optimal to allocate a single (fixed) prize as compared to multiple (fixed) prizes.
|
---
lastmod: 2015-05-25
date: 2013-10-01T07:32:00Z
description: ""
license: MIT
licenseLink: ""
sitelink: http://gophercon.com
tags:
- conference
thumbnail: /img/gophercon-tn.jpg
title: GopherCon
---
|
The first week’s schedule for 7flix has been revealed.
Seven’s new multichannel is designed to utilise the US general entertainment content it has acquired through its output deals, with an emphasis on its primetime movie lineup.
Brook Hall, Seven head programmer digital channels & sport, has advised Mediaweek that “there will be a movie at 8:30pm every night on Flix. That’s our sell in terms of the movies. It isn’t just a movie channel. We will be running seven nights a week at a time that is consistent.”
READ MORE: Mediaweek’s in-depth interview with Seven’s Brook Hall about 7flix
7flix launches on Sunday 28 February 2016 at 6am.
SUN
0600 – 1400 Once Upon A Time
1400 Frenemies (R) ’CC’ (2012)
1600 Geek Charming (R) ’CC’ (2011)
1810 Princess Diaries 2 – Royal Engagement ’CC’ (2004)
2030 Love Actually ’CC’ (2003)
2300 The Unborn ’CC’ (2009)
0100 Born To Kill (R) (1947)
0300 The Saint In London: The Saint In London (R) (1939)
0430 The Falcon’s Adventure: The Falcon’s Adventure (R) (1946)
MON
0600 Fish Hooks
0630 Gravity Falls
0700 Phineas And Ferb
0730 Dog With A Blog
0800 Jake And The Never Land Pirates
0830 Sofia The First
0900 Henry Hugglemonster
0930 Art Attack
1000 – 1200 Once Upon A Time
1200 – 1500 Quantico
1500 Bewitched
1530 I Dream Of Jeannie
1600 – 1700 Who’s The Boss
1700 – 1800 Married With Children
1800 Seinfeld
1830 – 2030 Once Upon A Time
2030 You, Me and Dupree (R) ’CC’
2230 First Dates (R) ’CC’
2330 Cougar Town
1200 – 0100 Married With Children
0100 – 0200 Who’s The Boss?: Pilot (R) ’CC’
0200 I Dream Of Jeannie: The Lady In The Bottle (R) ’CC’
0230 Bewitched
0300 Dance, Girl, Dance (R) (1940)
0500 – 0600 Good Times
TUES
0600 Fish Hooks
0630 Gravity Falls
0700 Phineas And Ferb
0730 Dog With A Blog
0800 Jake And The Never Land Pirates
0830 Sofia The First
0900 Henry Hugglemonster
0930 Art Attack
1000 – 1100 Benson
1100 – 1200 Good Times
1200 Quantico
1500 Bewitched
1530 I Dream Of Jeannie
1600 – 1700 Who’s The Boss?
1700 – 1800 Married With Children
1800 – 1930 Seinfeld: Male Unbonding (R) ’CC’
1930 – 2030 The Muppets
2000 The Muppets: Pig Out ’CC’
2030 Click
10:25 Any Questions For Ben?
1255 – 0200 Married With Children
0200 – 0300 Who’s The Boss?
0300 Bewitched
0330 My Life With Caroline (R) (1941)
0500 – 0600 Benson
WED
0600 Fish Hooks
0630 Gravity Falls
0700 Phineas And Ferb
0730 Dog With A Blog
0800 Jake And The Never Land Pirates
0830 Sofia The First
0900 Henry Hugglemonster
0930 Art Attack
1000 – 1200 Benson
1100 – 1200 Good Times
1200 – 1500 Quantico
1500 Bewitched
1530 I Dream Of Jeannie
1600 – 1700 Who’s The Boss?
1700 – 1800 Married With Children
1800 – 1930 Seinfeld
1930 – 2030 The Mindy Project
2030 Captain Phillips
2300 – 0000 Cougar Town
0000 – 0100 Married With Children
0100 – 0200 Who’s The Boss?
0200 Bewitched
0230 I Dream Of Jeannie
0300 – Behind The Rising Sun (R) (1943)
0500 – 0600 Benson
THURS
0600 Fish Hooks
0630 Gravity Falls
0700 Phineas And Ferb
0730 Dog With A Blog
0800 Jake And The Never Land Pirates
0830 Sofia The First
0900 Henry Hugglemonster
0930 Art Attack
1000 – 1100 Benson
1100 – 1200 Good Times
12:00 – 1300 Quantico
1400 How To Get Away With Murder
1500 Bewitched
1530 I Dream Of Jeannie
1600 – 1700 Who’s The Boss?
1700 – 1800 Married With Children
1800 – 1930 Seinfeld
1930 The Amazing Race
2030 2012
2330 The Amazing Race
0030 – 0130 Married With Children
0130 – 0230 Who’s The Boss?
0230 Bewitched
0300 I Dream Of Jeannie
0330 Action In Arabia (R) (1944)
0500 – 0600 Benson: Conflict Of Interest (R)
FRI
0600 Fish Hooks
0630 Gravity Falls
0700 Phineas And Ferb
0730 Dog With A Blog
0800 Jake And The Never Land Pirates
0830 Sofia The First
0900 Henry Hugglemonster
0930 Art Attack
1000 – 1100 Benson
1100 – 1200 Good Times
12:00 – 1400 How To Get Away With Murder
1400 Grey’s Anatomy
1500 Bewitched
1530 I Dream Of Jeannie
1600 – 1700 Who’s The Boss?
1700 – Tinker Bell and the Great Fairy Rescue (2011)
1830 Friday : Hoel Transylvania ’CC’ (2012)
2030 Friday : Sister Act ’CC’ (1992)
2215 The Lost Valentine ’CC’ (2011)
0015 – 0300 The Amazing Race
0300 Who’s The Boss?
0330 The Falcon And The Co-Eds (R) (1943)
0500 – 0600 Benson
SAT
0600 Fish Hooks
0630 Gravity Falls
0700 – 0830 Bewitched
0830 – 1000 I Dream Of Jeannie
1000 – 1100 Benson
1100-1200 Good Times
12:00 – 1300 Diff’rent Strokes
1300 – 1400 Just Shoot Me!
1500 – 1600 The Nanny
1610 Honey 2
1830 Guarding Tess (R) (1994)
2030 Sister Act 2: Back In The Habit
2225 Cloverfield (R) (2008)
0025 Who’s The Boss?
0055 Bewitched
0130 I Dream Of Jeannie
0200 Having Wonderful Time (R) (1938)
0330 Allegheny Uprising (R) (1939)
0500 – 0600 Benson |
y -193982?
-1/96991
Calculate -337884 divided by -6.
56314
-31 divided by -1937
31/1937
What is 147321 divided by -49107?
-3
Calculate 21500 divided by -250.
-86
Calculate 1234440 divided by -5.
-246888
Divide -516690 by -17223.
30
Calculate -3 divided by -13815.
1/4605
Calculate -143100 divided by -318.
450
Divide 198958 by -31.
-6418
Calculate -1362810 divided by -681405.
2
Calculate 559 divided by -43.
-13
Calculate -2 divided by 2635.
-2/2635
Divide 94782 by 1.
94782
What is 14910 divided by -2982?
-5
What is 22470 divided by -2?
-11235
What is 32214 divided by -21?
-1534
Calculate -10 divided by -1868.
5/934
135 divided by -15
-9
-30142 divided by 15071
-2
What is -35 divided by 151?
-35/151
Divide 0 by -89240.
0
Divide 28 by 435.
28/435
28498 divided by -3
-28498/3
What is 468666 divided by -6?
-78111
47 divided by 179
47/179
What is 4 divided by 79527?
4/79527
13580 divided by 3395
4
Calculate 602228 divided by 22.
27374
Divide -206340 by -2.
103170
What is -418598 divided by -5?
418598/5
470322 divided by -2958
-159
Calculate -132026 divided by 3.
-132026/3
Calculate 957976 divided by 1.
957976
Calculate 115476 divided by 38492.
3
What is 1817008 divided by 5282?
344
What is -9 divided by -17793?
1/1977
What is -31 divided by 9297?
-31/9297
Divide 1 by 3209.
1/3209
What is 56 divided by 805?
8/115
Divide 135572 by 2.
67786
97 divided by 5506
97/5506
Calculate 5553 divided by -20.
-5553/20
Calculate 0 divided by 353996.
0
Calculate 3444 divided by 1.
3444
What is 1500932 divided by -2?
-750466
Divide -342090 by -9774.
35
Calculate 2 divided by -225867.
-2/225867
What is 62751 divided by -1609?
-39
99144 divided by -3
-33048
What is 4526 divided by 2263?
2
Calculate -204428 divided by -4.
51107
Calculate -63880 divided by 5.
-12776
420315 divided by -4003
-105
Calculate 51123 divided by -5.
-51123/5
-1327173 divided by 3
-442391
Divide 0 by -5477.
0
What is 0 divided by 39818?
0
-1574352 divided by 54288
-29
Calculate -65984 divided by 4.
-16496
What is -1580 divided by -221?
1580/221
-88385 divided by -17677
5
Calculate 7741 divided by -3.
-7741/3
Calculate 3589740 divided by 148.
24255
-37 divided by 19651
-37/19651
Divide 31479 by 4.
31479/4
Divide -78501 by -573.
137
Divide -605 by 4914.
-605/4914
Divide 5 by 20292.
5/20292
2 divided by 310897
2/310897
Calculate -307951 divided by 37.
-8323
Divide -11787 by -3929.
3
Calculate 11 divided by 14630.
1/1330
Divide -134750 by 5.
-26950
What is 171881 divided by 3?
171881/3
-110370 divided by 8490
-13
608515 divided by 121703
5
Calculate -8955404 divided by 4.
-2238851
Calculate -150843 divided by 11.
-13713
Calculate 1 divided by 9959.
1/9959
Calculate 4217 divided by 369.
4217/369
Calculate 4 divided by 17752.
1/4438
What is -11608 divided by -2?
5804
12105 divided by 1
12105
-124055 divided by -24811
5
Divide -30792 by -10264.
3
What is -2003240 divided by 50081?
-40
Calculate -7009 divided by 80.
-7009/80
Calculate 39 divided by -6.
-13/2
Calculate 63342 divided by -3726.
-17
60 divided by 68
15/17
2165785 divided by 433157
5
Calculate 6335 divided by 2.
6335/2
-785755 divided by 5
-157151
What is -1928778 divided by -2?
964389
6 divided by 155000
3/77500
What is -360 divided by 221?
-360/221
-14840 divided by -424
35
Divide -273 by 9526.
-273/9526
Calculate 624 divided by 814.
312/407
What is 554868 divided by -15413?
-36
Calculate -3 divided by -9580.
3/9580
Divide -10974 by -3.
3658
What is -2912 divided by 38?
-1456/19
-4 divided by -2545
4/2545
Divide 4406 by -25.
-4406/25
Calculate 609876 divided by -7.
-609876/7
6876 divided by -3
-2292
Calculate -22916 divided by -3.
22916/3
What is 544440 divided by -2?
-272220
What is 468062 divided by 2?
234031
What is 5 divided by 8550?
1/1710
Calculate 36776 divided by -4.
-9194
Divide 148649 by 1.
148649
Calculate -3669265 divided by -5.
733853
1 divided by -49856
-1/49856
What is -1 divided by -55364?
1/55364
Divide -112245 by 35.
-3207
-681 divided by 31
-681/31
Divide 832 by -2773.
-832/2773
-5460 divided by 39
-140
Divide -4430 by -12.
2215/6
Calculate 7163435 divided by -5.
-1432687
Divide 300000 by 300000.
1
Calculate -240642 divided by -261.
922
267 divided by -286
-267/286
What is -190 divided by -1152?
95/576
What is -238464 divided by 2208?
-108
1 divided by -47429
-1/47429
978 divided by -163
-6
Calculate 40320 divided by 6720.
6
Calculate -283457 divided by -6031.
47
Calculate -434846 divided by 41.
-10606
-2060886 divided by 3
-686962
Calculate -7 divided by -2251.
7/2251
Calculate -5612 divided by 244.
-23
-39 divided by 23385
-13/7795
What is 108173 divided by -8321?
-13
Divide -436 by 374.
-218/187
20184 divided by -232
-87
Divide 37716 by -3.
-12572
26113 divided by 1
26113
Calculate -230396 divided by -5.
230396/5
226 divided by -285
-226/285
10949648 divided by -4
-2737412
-7875903 divided by 3
-2625301
-14850 divided by -5
2970
What is 2880 divided by 360?
8
What is 68530 divided by 9790?
7
Calculate -106920 divided by 810.
-132
What is -32863 divided by 6?
-32863/6
-5605 divided by -5605
1
What is 27194 divided by 13597?
2
Calculate -1 divided by -10095.
1/10095
26 divided by -13216
-13/6608
Calculate 118200 divided by -1.
-118200
Divide -4 by -1763.
4/1763
Calculate 4935 divided by -1.
-4935
-402 divided by -2847
134/949
Divide 21424 by -13.
-1648
-12744 divided by -5
12744/5
Calculate -21 divided by 1028.
-21/1028
Divide 29 by 3968.
29/3968
1200380 divided by 600190
2
900716 divided by 2
450358
Divide -289602 by 346.
-837
Calculate 226818 divided by 226818.
1
Calculate -39 divided by 3838.
-39/3838
What is -6 divided by 3510?
-1/585
What is -325424 divided by 81356?
-4
What is -67519 divided by -251?
269
Divide 5 by 656845.
1/131369
121 divided by -440
-11/40
Calculate -265438 divided by 74.
-3587
What is 1243 divided by 672?
1243/672
Divide -52445 by -3.
52445/3
Calculate 4454 divided by -15.
-4454/15
Calculate 481400 divided by 415.
1160
Divide 54747 by -3.
-18249
Calculate 2 divided by -65647.
-2/65647
Divide -147 by 1.
-147
-39226 divided by 2
-19613
Calculate 249457 divided by 619.
403
Divide -2835115 by -5.
567023
What is 1524334 divided by -217762?
-7
What is 777225 divided by 3?
259075
Calculate 1 divided by 8219.
1/8219
What is -158496 divided by 508?
-312
Calculate -4007 divided by 7.
-4007/7
3288615 divided by 95
34617
Calculate -2188 divided by -35.
2188/35
-4 divided by 373839
-4/373839
4609 divided by 190
4609/190
57404 divided by -14351
-4
Divide -8427198 by 78.
-108041
257792 divided by -1007
-256
Calculate 120154 divided by 3.
120154/3
-4320 divided by 160
-27
Divide 110742 by 2.
55371
18000 divided by 72
250
Calculate -6920 divided by -8.
865
-59 divided by 48
-59/48
1220610 divided by -406870
-3
Calculate -89027 divided by 5.
-89027/5
What is -57362 divided by -37?
57362/37
What is -1415 divided by -533?
1415/533
What is -392424 divided by 12?
-32702
Calculate -142807 divided by 142807.
-1
-571 divided by 733
-571/733
What is -210289 divided by -23?
9143
What is -3 divided by -1444533?
1/481511
4 divided by 421976
1/105494
Divide -117249 by -3.
39083
2 divided by -52084
-1/26042
Calculate 571445 divided by -16327.
-35
-4428 divided by 3
-1476
What is -70831 divided by 2?
-70831/2
What is -402544 divided by -1448?
278
-50669 divided by 1
-50669
What is 14 divided by -9838?
-7/4919
What is 5297 divided by -170?
-5297/170
Calculate 1132677 divided by 91.
12447
237 divided by -434
-237/434
Divide -55672 by 13918.
-4
-13960 divided by 349
-40
What is -23459 divided by -3?
23459/3
7619 divided by -15
-7619/15
Calculate 651911 divided by 50147.
13
2 divided by 2197
2/2197
Calculate -40413 divided by 5.
-40413/5
Calculate -7 divided by -30.
7/30
What is -3944735 divided by -1?
3944735
Divide -4428440 by 553555.
-8
What is 7014 divided by -6?
-1169
0 divided by -412677
0
Calculate -54360 divided by 1510.
-36
What is 19 divided by 1229?
19/1229
-2875 divided by -575
5
What is 299 divided by 8235?
299/8235
What is 3786855 divided by -1262285?
-3
1039514 divided by 82
12677
96 divided by 4399
96/4399
What is 9866 divided by 4933?
2
What is -1715890 divided by 418?
-4105
What is -79 divided by -1653?
79/1653
Divide 51387 by 17129.
3
Divide -2306850 by -4394.
525
What is -13776 divided by 41?
-336
-23 divi |
Q:
visual studio 2017 15.9.13 can't use auto c++ templates
I installed visual studio 2017 for c++ desktop and linux development some about an hour ago
I tried this code which uses c++17 auto templates and was surprised that it gives an error saying :
Error C3533 a parameter cannot have a type that contains 'auto'
this is the code causing the problem
template <class T, T null_value, bool no_negative, auto Deleter>
struct HandleHelper
{
using pointer = HandleWrapper<T, null_value, no_negative>;
void operator(pointer p)
{
Deleter(p);
}
};
before in visual studio 2015 I used something like this due to lack of c++17 support :
template <class T, T null_value, bool no_negative, class DelType, DelType Deleter>
struct HandleHelper
{
using pointer = HandleWrapper<T, null_value, no_negative>;
void operator(pointer p)
{
Deleter(p);
}
};
but auto templates looks more elegant
A:
Make sure you have the correct "C++ Language Standard" set in the property pages for your project. The default for VS 2017 is C++14.
Right click on your project and select "Properties".
Then expand the C/C++ node on the tree view on the left hand side.
Select "Language" from the expanded menu options.
Check that the "C++ Language Standard" is set to ISO C++17 Standard (/std:c++17).
If its blank, it will default to C++14.
|
DC Comics
Justice League TP Vol 07 Justice Lost
Relisted
Limt 2 at 40% off. Additional copies available at 40% off.
Trapped aboard the crippled Watchtower, the Justice League and Justice League of America teams clash over leadership and the urgent question of who can be rescued... and who will be left behind. Meanwhile, the Green Lanterns face critical choices over how to deal with the League's two most powerful figures! Collects issues #39-43. |
How to Identify and Troubleshoot an IP Conflict
What’s an IP Conflict?
An IP conflict occurs when two or more hosts in the same subnet are configured with the same IP address. When this happens, communications with the two conflicting hosts are mixed up. One host may receive packets that belong to the other one, and vice versa.
As result, IP conflicts have very unpredictable consequences on the affected hosts. Hosts may experience continuous connections and disconnections. This is something that should be addressed as soon as the problem arises.
What are the causes of an IP conflict?
IP conflicts happen for different reasons. In one scenario, one (inexperienced) user may assign a static IP address that is part of a DHCP pool to his computer. If that same IP address is then dynamically assigned to another computer by the DHCP server, an IP conflict will occur.
Fortunately, this problem can be avoided by denying users permission to set the IP settings on their computer. Unfortunately, this something that is not always possible in all network environments (e.g. Bring Your Own Device).
In another scenario, a computer is assigned an IP from the DHCP server; this computer then goes offline. As the lease timeout expires, the DHCP server may assign that IP address to another computer.
Now, let’s imagine the computer that first received the DHCP lease comes back online; for some reason, the computer isn’t able to reach the DHCP server. The computer will now self-assign the IP address using the DHCP lease saved in the cache (a behavior that is common on Linux hosts). This computer is now causing an IP conflict. This scenario is more difficult to troubleshoot than the previous one, but still possible, in my experience.
The last scenario could be caused by human error. For example, an operator may assign an existing IP address to a network device, like a router or switch. When the misconfigured device is then connected to the network, it creates an IP conflict. This scenario is avoidable by implementing peer review of network configuration changes.
How to detect an IP conflict
Both Windows and Macintosh operating systems notify the user via pop-up notification when an IP conflict is detected with another computer. Here’s an example of a Windows IP conflict notification …
Detecting an IP conflict is even more difficult if it’s affecting remote network devices that are not end-user workstations. It’s more difficult because you don’t have local access to the host and you are not able to have a stable remote session.
In this scenario, one way to detect an IP conflict is to first ping the remote IP address. If the ping test returns high packet loss, then it’s worth the continued troubleshooting.
To troubleshoot this problem you should:
1) Get access to the router that serves as the default gateway of the subnet where the conflict is happening.
2) Inspect the router’s ARP cache and check if the MAC address associated to the conflicting IP changes frequently. To verify this, you have to execute every two or three seconds the command that returns the ARP cache.
If the MAC address does change then, congratulations! … you have an IP conflict.
How to fix an IP conflict?
If you have detected an IP conflict in your network, you’ll need to correct the IP settings of the device that is “squatting”. Ideally, you or someone else on your team has access to the device, to ensure that the correct IP settings can be set. If there’s no way for you to locally access the problematic host, then you have few options.
One option is to remove the host from the network by shutting down the switch port that it’s connected to. Make sure that you do this off hours when there are no users on the network. Also, make sure that your changes won’t cause any further damage to applications or network services. Please don’t shut down a trunk port. In any case, it’s highly recommended that you review your specific case with your team and act accordingly.
I hope this article provided some guidance on how to deal with IP conflicts. If you have any feedback or questions, please do so in the comments section. Cheers! |
# What is RethinkDB?
RethinkDB is an open-source, distributed database built to store JSON documents and effortlessly scale to multiple machines. It's easy to set up and learn and features a simple but powerful query language that supports table joins, groupings, aggregations, and functions.
%%LOGO%%
# How to use this image
## Start an instance with data mounted in the working directory
The default CMD of the image is `rethinkdb --bind all`, so the RethinkDB daemon will bind to all network interfaces available to the container (by default, RethinkDB only accepts connections from `localhost`).
```bash
docker run --name some-rethink -v "$PWD:/data" -d %%IMAGE%%
```
## Connect the instance to an application
```bash
docker run --name some-app --link some-rethink:rdb -d application-that-uses-rdb
```
## Connecting to the web admin interface on the same host
```bash
$BROWSER "http://$(docker inspect --format \
'{{ .NetworkSettings.IPAddress }}' some-rethink):8080"
```
# Connecting to the web admin interface on a remote / virtual host via SSH
Where `remote` is an alias for the remote user@hostname:
```bash
# start port forwarding
ssh -fNTL localhost:8080:$(ssh remote "docker inspect --format \
'{{ .NetworkSettings.IPAddress }}' some-rethink"):8080 remote
# open interface in browser
xdg-open http://localhost:8080
# stop port forwarding
kill $(lsof -t -i @localhost:8080 -sTCP:listen)
```
## Configuration
See the [official docs](http://www.rethinkdb.com/docs/) for infomation on using and configuring a RethinkDB cluster.
|
Sunday, 9 February 2014
It’s not often that I finish an
Alfred Hitchcock picture unable to take something away from it but I feel like
I wasted my time with The Trouble with
Harry. A departure from the type of mystery that made his name, this is a
black comedy with thriller elements. Set during a crisp autumn in Vermont, a retired sea
captain discovers the recently deceased body of a man while out hunting on a
hill. Believing to be responsible for his death, the captain attempts to hide
the body but various passers by happen upon it and react in unusual ways. It
turns out that several people believe themselves responsible and the small
community at the bottom of the hill attempt to discover exactly what happened
to the man and what to do next.
The use of the body, which turns
out to be that of the titular character, is a clever Macguffin which is used to
unite two couples in what turns out to be a romantic black comedy. Ordinarily
when a Hitchcock movie opens on a corpse, you’d be expecting a whodunit but here
that isn’t important to the director. For me, that’s one of the problems. I
wanted more excitement and intrigue from the film. Although billed as a comedy,
I didn’t laugh once and was barely amused. The film just washed over me with a
plot that didn’t grab me in the slightest. More disappointing than the plot is
the cast who are as wooden as the corpse they attempt to cover up. |
Sunday, December 16, 2012
Lagunitas SoCo Stout Background
In a previous post about Lagunitas, I mentioned the SoCo Stout, which Kay and I both responded to with puckered lips: very sour and unlike any previous stouts we had drunk. So, I contacted Lagunitas for an explanation. This is how they responded:
Lagunitas Brewing Co wrote: "Here's the word on SoCo Stout (as in Sonoma County), straight from head brewer, Jeremy Marshall: The barrels are Dirty Pinots & Chards (means not rinsed, freshly emptied barrels, lots of lees---looks like purple pudding) We purge & rack 'em full. Any Brettanomyces or bacterial action comes from the barrel. A lot of oak left from "neutral" barrels will become evident. Aged 2 years then all blended together. Cheers!"
Will certainly be trying it again and I'll post the new tasting notes here. |
Q:
How to change day-format (M > Mo, F > Fr...) in Date/month pickers of Vuetify
I try to change format of days in Date/month picker component. At the moment list of days as "M,T,W,T,F,S,S", but I need "Mo,Tu,We,Th,Fr,Sa,Su". In API I can see that necessary parameter called "day-format" and its value is null (default).
Can anybody please explain: is it possible to change this format of days?
Here is the API of Vuetify:
https://vuetifyjs.com/en/components/date-pickers#date-month-pickers
A:
Use the weekday-format to pass the date to a function;
Convert the date to day of the week (0..6)
Use that as the index for an array of strings ('Mo'..'Su')
code:
<div id="app">
<v-app id="inspire">
<v-row justify="center">
<v-date-picker :weekday-format="getDay" v-model="picker"></v-date-picker>
</v-row>
</v-app>
</div>
const daysOfWeek = ['Mo', 'Tu', 'We', 'Th', 'Fr', 'Sa', 'Su'];
new Vue({
el: '#app',
vuetify: new Vuetify(),
data () {
return {
picker: new Date().toISOString().substr(0, 10),
}
},
methods:{
getDay(date){
let i = new Date(date).getDay(date)
return daysOfWeek[i]
}
}
})
|
/* Copyright 2012-present Facebook, Inc.
* Licensed under the Apache License, Version 2.0 */
#include "watchman.h"
#include "InMemoryView.h"
#include "watchman_error_category.h"
using namespace watchman;
using folly::to;
void watchman_root::syncToNow(std::chrono::milliseconds timeout) {
w_perf_t sample("sync_to_now");
auto root = shared_from_this();
try {
view()->syncToNow(root, timeout);
if (sample.finish()) {
sample.add_root_meta(root);
sample.add_meta(
"sync_to_now",
json_object({{"success", json_boolean(true)},
{"timeoutms", json_integer(timeout.count())}}));
sample.log();
}
} catch (const std::exception& exc) {
sample.force_log();
sample.finish();
sample.add_root_meta(root);
sample.add_meta(
"sync_to_now",
json_object({{"success", json_boolean(false)},
{"reason", w_string_to_json(exc.what())},
{"timeoutms", json_integer(timeout.count())}}));
sample.log();
throw;
}
}
/* Ensure that we're synchronized with the state of the
* filesystem at the current time.
* We do this by touching a cookie file and waiting to
* observe it via inotify. When we see it we know that
* we've seen everything up to the point in time at which
* we're asking questions.
* Throws a std::system_error with an ETIMEDOUT error if
* the timeout expires before we observe the change, or
* a runtime_error if the root has been deleted or rendered
* inaccessible. */
void watchman::InMemoryView::syncToNow(
const std::shared_ptr<w_root_t>& root,
std::chrono::milliseconds timeout) {
try {
cookies_.syncToNow(timeout);
} catch (const std::system_error& exc) {
if (exc.code() == watchman::error_code::no_such_file_or_directory ||
exc.code() == watchman::error_code::permission_denied ||
exc.code() == watchman::error_code::not_a_directory) {
// A key path was removed; this is either the vcs dir (.hg, .git, .svn)
// or possibly the root of the watch itself.
if (cookies_.cookieDir() == root_path) {
// If the root was removed then we need to cancel the watch.
// We may have already observed the removal via the notifythread,
// but in some cases (eg: btrfs subvolume deletion) no notification
// is received.
root->cancel();
throw std::runtime_error("root dir was removed or is inaccessible");
} else {
// The cookie dir was a VCS subdir and it got deleted. Let's
// focus instead on the parent dir and recursively retry.
cookies_.setCookieDir(root_path);
return cookies_.syncToNow(timeout);
}
}
// Let's augment the error reason with the current recrawl state,
// if any.
{
auto info = root->recrawlInfo.rlock();
if (!root->inner.done_initial || info->shouldRecrawl) {
std::string extra = (info->recrawlCount > 0)
? to<std::string>("(re-crawling, count=", info->recrawlCount, ")")
: "(performing initial crawl)";
throw std::system_error(
exc.code(), to<std::string>(exc.what(), ". ", extra));
}
}
// On BTRFS we're not guaranteed to get notified about all classes
// of replacement so we make a best effort attempt to do something
// reasonable. Let's pretend that we got notified about the cookie
// dir changing and schedule the IO thread to look at it.
// If it observes a change it will do the right thing.
{
struct timeval now;
gettimeofday(&now, nullptr);
auto lock = pending_.lock();
lock->add(cookies_.cookieDir(), now, W_PENDING_CRAWL_ONLY);
lock->ping();
}
// We didn't have any useful additional contextual information
// to add so let's just bubble up the exception.
throw;
}
}
/* vim:ts=2:sw=2:et:
*/
|
This instrument provides that the instrument applies to the major disaster being the bushfire occurring in November and December 2019 and January 2020 in Victoria, affecting the local government areas of Alpine, East Gippsland and Towong. |
OAKLAND (KPIX) — Raider Nation is not going down without a fight. The team may go to Vegas, but fans say the Raiders’ name should stay in Oakland.
Monday they announced a coalition of fans is joining forces with city leaders and businesses to keep NFL football in Oakland
“We are breaking out silence today and exploring options to keep football in Oakland,” says Ray Bobbitt of the Legal Action Committee. ”
They’ve even enlisted high-profile New York sports attorney James Quinn to take on the crusade despite plans already in the works to build a $1.9 billion stadium in Las Vegas by 2020 .
“I assure you that we’re going to take a long hard look and do everything that we can to ensure that, in fact, we continue to have NFL football — Raider NFL football,” said Quinn addressing a crowd on the steps of Oakland City Hall.
Even if the team leaves Raiders fans say they are looking info legal options to keep the Raiders name and colors in Oakland.
The move is not without a precedent.
During the 1995 NFL season, legal action by the city of Cleveland and Cleveland Browns ticket holders led to a compromise when the team left to Baltimore to become the Ravens.
Cleveland kept the team name and colors, and a new team was reactivated in 1998.
Fast-forward to 2017. Raiders fans aren’t ready to say ‘game over’ when it comes to keeping the NFL in Oakland. |
The Chinese government has reportedly ordered Video streaming websites in the country to stop showing four popular American TV shows.
A spokeswoman for a leading video site, Youku, said that on Sunday it had received notification not to show sitcom The Big Bang Theory, political and legal drama The Good Wife, crime drama NCIS and legal drama The Practice, News.au reported.
However, she added that the State Administration of Press, Publication, Radio, Film and Television didn't give a reason for its order.
The move is a glaring reminder of the increasing government control on the online industry, which otherwise enjoys far more freedom than the country's state television and cinemas when it comes to showing foreign productions.
Meanwhile, an unnamed senior manager at another site said that it received a surprise order to "clean the website" last week, the report added. |
Categories
Found 25 article(s) for author 'Robert Lawrence'
China Is Vulnerable But Deal Unlikely. Robert Lawrence, May 17, 2019, Audio, “Robert Lawrence, Professor of International Trade and Investment at Harvard Kennedy School and former economic advisor to President Clinton, on why a China trade deal is looking unlikely in the near future. Martin Stephan, the Deputy CEO of Carbios, a French recycling biotech company, on their technology that aims for zero plastic waste. Alex Webb, Bloomberg Opinion technology columnist, discusses his column: “Amazon-Deliveroo Alliance Would Eat Uber For Dinner.” Will Rhind, CEO of GraniteShares, with his mid-year outlook on gold and oil. Hosted by Lisa Abramowicz and Paul Sweeney.” Link
Can the Trading System Survive US–China Trade Friction? Robert Lawrence, September 13, 2018, Paper, “Donald Trump has sought to change US trading relationships by raising protection at home and taxing the offshore activities of US companies abroad. These measures, which both use and violate trade rules, have provoked retaliation from other countries. Such friction has restricted and distorted trade and investment, undermined the rules‐based trading system and perhaps permanently damaged global value chains that depend on stable rules for market access. Trump has justified some of his measures as a response to China’s alleged unfair practices and indeed, China has adopted industrial and technology policies that are formally neutral between domestic and foreign firms but in practice have led foreign firms to complain about discriminatory practices that favor Chinese firms.” Link
Trump’s Goal With China Is Big Tariffs, Not A Deal. Robert Lawrence, August 15, 2018, Audio, “Robert Lawrence, Professor of International Trade and Investment at the Harvard Kennedy School and former economic advisor to Clinton, on the deal that Trump really wants with China. Hosted by Pimm Fox and Lisa Abramowicz.” Link
U.S. Should Use Allies, WTO, To Combat China On Trade. Robert Lawrence, July 9, 2018, Audio, “GUEST: Robert Lawrence, Professor of International Trade and Investment at Harvard Kennedy School and former economic advisor to President Bill Clinton, on the global supply chain and whether trade globalization can be undone at this point.” Link
US-China Trade Frictions and the Global Trading System. Robert Lawrence, 2018, Book Chapter, “Recent trade frictions between the United States and China have violated several rules and practices of the rules-based multilateral trading system established under the General Agreement on Tariffs and Trade (GATT) and its successor, the World Trade Organization (WTO). President Donald Trump’s preoccupation with trade balances in goods, both bilateral and total, has led to protectionist trade policies at home— primarily to minimize imports and offshoring by US firms—and aggressive demands for more market opening abroad. President Trump appears to view trade not as an activity from which all nations can gain but rather as a zero-sum game in which some win and some lose.” Link
Five Reasons Why the Focus on Trade Deficits Is Misleading. Robert Lawrence, March 2018, Opinion, “President Trump has asserted that trade balances are a key measure of a nation’s commercial success and that large US trade deficits prove that past trade approaches have been flawed. But trade deficits are not in fact a good measure of how well a country is doing with respect to its trade policies. Many of the assumptions on which the administration’s beliefs rest are not supported by the evidence. This Policy Brief argues that trade deficits are not necessarily bad, do not necessarily cost jobs or reduce growth, and are not a measure of whether foreign trade policies or agreements with other countries are fair or unfair. Efforts to use trade policy and agreements to reduce either bilateral or overall trade deficits are also unlikely to produce the effects the administration claims they will and instead lead to friction with US trading partners, harming the people the policies claim to help.” Link
Trade deficits caused by foreign borrowings; Harvard economist at Sri Lanka forum. Robert Lawrence, February 9, 2018, Video, “Trade and current account deficits are a result of foreign borrowings, a US economist told a forum in Sri Lanka as US President Donald Trump mistakenly attacked trade deficit based on false Mercantilist doctrine. Exporting more will not reduce a trade deficit, Robert Lawrence, a professor at Harvard University’s Kennedy School of Business told an economic forum organized by Advocata Institute, a Colombo-based think tank.” Link
Recent US Manufacturing Employment: The Exception that Proves the Rule. Robert Lawrence, November 2017, Paper, “This Working Paper challenges two widely held views: first that trade performance has been the primary reason for the declining share of manufacturing employment in the United States, and second that recent productivity growth in manufacturing has actually been quite rapid but is not accurately measured. The paper shows that for many decades, faster productivity growth interacting with unresponsive demand has been the dominant force behind the declining share of employment in manufacturing in the United States and other industrial economies. It also shows that since 2010, however, the relationship has been reversed and slower productivity growth in manufacturing has been associated with more robust performance in manufacturing employment.”Link
Manufacturing and Inclusive Growth: The Experience in the Rest of the World. Robert Lawrence, 2017, Paper, “This report describes some of the results from the second phase of the research project on the role of manufacturing in inclusive growth. The first phase of the project examined the US experience. In the second phase, undertaken by Robert Lawrence and Danial Lashkari – a graduate student in the Harvard Department of Economics, the scope of the analysis has broadened to explore the experience of manufacturing employment growth in the rest of the world.” Link
Making US Trade and Investment Policies Work for Global Development. Robert Lawrence, , Paper, “In an interconnected world, sustainable and inclusive economic growth in less developed countries helps to secure US interests and values. Economic development nurtures peaceful societies, reduces refugee flows, ameliorates humanitarian crises, expands markets for US exports, and safeguards human rights and other core US aspirations. Well-designed trade and investment policies are indispensable tools to achieve these objectives.” Link
GrowthPolicy.org pulls together recent research by Harvard faculty about economic growth. Our goal is to disseminate Harvard research to a broad audience in order to inform debate about the role of policy in achieving stability in our financial systems and shared, sustainable economic prosperity. GrowthPolicy.org is a project of: |
But little is known about the scale or nature of organized crime in Peru. What experts do agree is that repression, interdiction, and coca eradication are not working out as planned, and that the dynamics of the drug trade have changed. Instead of feeding the once-insatiable U.S. market, Peru may now account for as little as five percent of the estimated 300 tons of cocaine Americans snort. Now Brazil, the world's second-biggest market, sucks up much of Peruvian drug production, often not as cocaine but its more addictive and cheaper variants, crack or "bazuco."
Most law enforcement specialists believe that locals run the production and local transportation of cocaine, while Colombian and Mexican intermediaries manage exports, with the recent appearance of the Russian mob to shake things up a bit. The business is supposedly straight-forward. Hundreds of campesinos (farmers) grow the crop, mainly in central and northern Peru. The cocaleros sell the coca leaf or coca base to clanes (small criminal groups often based around families), who ship either coca base or processed cocaine, to a handful of firmas (Peruvian organized crime syndicates), which shift the drugs to departure points (airports, seaports, and border areas) ready to move to Colombia, Ecuador, Bolivia, and Brazil. While some Peruvian capos (drug lords) are known to operate in neighboring countries, groups like the Sinaloa cartel and the Russian mob generally handle export.
In spite of government rhetoric to the contrary, organized crime seems to growing unmolested. Virtually no serious players are known to have been arrested or prosecuted for drug running. The one big case working its way through the courts at the moment, involving the notorious Sanchez Paredes clan, looks set to collapse. To add insult to impunity, former President Alan Garcia pardoned some 400 drug traffickers during his second term of office, citing overcrowded jails. And current President Humala also pardoned as many as 100 criminals convicted of trafficking since his election in 2011.
***
There is remarkably little concern in Peru over all of this. Peruvians are far more worried about the common crime that touches their daily lives, than the organized crime that has shown itself capable of corrupting the police, prosecutors, judges, and it seems, presidents. And this is because, unlike in Colombia and Mexico, the drug trade involves very little violence. While popular perceptions of insecurity are rising slightly, Peru is widely considered one of the safest countries in South America.
There are at least two possible explanations for this paradox.
The first is that analysts (including the present authors) are completely misreading the situation. In other words, it could very well be that there is considerable violence between producers, dealers and exporters as they compete over market share. The fact is that it is almost impossible to know one way or another. There is no reliable baseline data on the situation and even the most basic figures are wildly inconsistent. For example, the Ministry of the Interior claims that there are 10,000 homicides each year, while the National Police argue that there are just 3,000. |
Alexandria, Egypt, August 31 – Mohammad El-Bir is your typical blue-collar Egyptian: struggling economically, hoping for political stability, and certain that Israel is the chief cause of the world’s problems. Pictures of the recent carnage in the Gaza Strip, however, have confronted El-Bir with a dilemma: if, as Hamas leaders claim, Israel’s actions there are Hitler-like, that means Hitler was actually responsible for killing Jews, a notion that flies in the face of the Holocaust-denying ethos that pervades the Muslim world.
“I have no problem believing the claims of Hamas, even though they are wretched pig-dogs betraying Islam and trying to undermine Egypt’s stability,” said the 30-year-old day laborer. “Because those claims are consistent with everything I have been conditioned to believe about Jews and Israel. But the problem is that if Israel is like Hitler and Hitler was right, that means Israel is right, and that just can’t be.”
El-Bir added that it is difficult for him to wrap his head around the possibility that Hitler initiated killings of Jews on a mass scale, since that notion could never be assimilated properly in his mind, but that only such numbers of Jews killed would be congruent with charges of Israeli genocide. If the casualty numbers from the Hamas-run Gaza health authorities are to be believed, approximately 2,000 Gazans died during Operation Protective Edge, meaning that invoking Hitler makes no sense.
“If Hitler didn’t really do anything wrong, then Israel being like Hitler is simply untrue,” continued the father of three. “Only if Hitler actually did terrible things does it make sense to mention him. I’m having a hard time with this one.”
El-Bir is not alone in his confusion, says Joe Goebbels, who studies media manipulation and public opinion in the Arab world. “There is no end to the contradictions an Arab must face in his encounters with news. Jews are supposed to be descended from apes and pigs, but these simian and porcine people have also somehow managed to outfight them despite being outnumbered, over and over again. Israel has a burgeoning high-tech industry, a flourishing democracy, and a successful economy that is simply out of place in the Middle East. So are the Jews subhuman, or are they superhuman, capable of controlling the media, the banks, and whole countries, without leaving a trace of evidence?”
“You know who else confronted those contradictory notions about Jews?” said Goebbels. “Hitler.” |
COURT OF APPEALS OF VIRGINIA
Present: Judges Frank, Humphreys and Agee
Argued at Chesapeake, Virginia
CHARLES RUSSELL GUY
MEMORANDUM OPINION * BY
v. Record No. 2270—01-1 JUDGE ROBERT P. FRANK
AUGUST 6, 2002
COMMONWEALTH OF VIRGINIA
FROM THE CIRCUIT COURT OF NORTHAMPTON COUNTY
Glen A. Tyler, Judge
Lynwood W. Lewis, Jr. (Vincent, Northam &
Lewis, on brief), for appellant.
Donald E. Jeffrey, III, Assistant Attorney
General (Jerry W. Kilgore, Attorney General,
on brief), for appellee.
Charles Russell Guy (appellant) appeals from his jury trial
convictions for aggravated sexual battery, in violation of Code
§ 18.2-67.3, and object sexual penetration, in violation of Code
§ 18.2-67.2. On appeal, he argues the trial court (1) improperly
admitted hearsay testimony and (2) erred in overruling his motion
to strike the evidence. 1 For the reasons stated below, we affirm
the convictions.
* Pursuant to Code § 17.1-413, this opinion is not
designated for publication.
1
Appellant argues the trial court should have granted his
motion to strike the evidence because without inappropriately
admitted hearsay evidence, the evidence was insufficient to
convict him of these offenses. Although appellant's
characterization of his argument is confusing, he clearly raises
BACKGROUND
M.G., an eight-year-old girl, walked over to her neighbors'
house on October 15, 1999. Her neighbors, appellant and his
wife, lived in a house directly behind M.G.'s home. After M.G.
had been gone for thirty minutes, her mother walked down the
lane toward appellant's home, calling out M.G.'s name.
Mother knocked on appellant's door, which was answered by
his stepson. The stepson told M.G.'s mother that the girl had
been at the home, but left, and he did not know anything more
about her. Mother then continued to search for her daughter.
Suddenly, M.G. responded to her mother's calls, sounding
very close and clear. Mother found M.G. in the last shed of
three that were beside appellant's house. M.G. was lying on the
floor of the shed with her pants and underwear down around her
ankles.
Initially, M.G. said she was tired and lying down. Her
mother said, "[T]hat's not what you're doing," and asked, "Who
was in here with you." As M.G. pulled on her clothes, she told
her mother, "You know who he is, Mama. He's not a stranger."
She then walked out of the shed, toward the end of the row, and
indicated, "[H]e's back here." Mother walked to the side of the
shed and saw appellant. When mother confronted appellant, he
denied knowing anything.
both a hearsay argument and a separate sufficiency argument in
his appeal. Therefore, we will address both arguments.
- 2 -
As mother and M.G. walked by the front door of appellant's
home, M.G. became hysterical, saying, "I'm going to get in
trouble," and "He's got a gun." M.G. remained hysterical after
they got home.
Mother testified M.G. told her that appellant pulled her
into the shed, and then licked her pubic area and put his finger
into her vagina. Deputy Sheriff Mike Smith testified, when he
arrived about a half-hour after mother discovered M.G., she
described basically these same incidents. When M.G. testified
at trial, she explained appellant pulled her into the shed,
pulled her pants down, and then put his finger into her vagina.
She said he did nothing else.
Mother also testified on cross-examination that M.G. had
talked to her on two other occasions about the incident in the
shed and was clear each time that appellant had licked her and
put his finger in her. She did admit M.G. also said "Matthew"
had a gun, not appellant. Mother further testified that M.G.
told her appellant had pulled his penis out of his pants while
they were in the shed.
The doctor who examined M.G. at the emergency room
testified that she had bruising on her vagina and some tearing
to her hymen. Both injuries occurred within twenty-four hours
before the examination, according to the doctor. He also
testified the injuries were consistent with a finger inserted
- 3 -
into the vagina, but were not likely self-inflicted or from a
fall.
The SANE 2 nurse, who also examined M.G., testified the
injuries were no more than six hours old. She explained the
injuries could be caused by a man's finger. She also testified,
although a person possibly could injure herself in this way, it
would be painful to M.G. to cause these injuries to herself.
She explained the injuries were inconsistent with a fall.
Appellant's wife and stepson testified that M.G. visited
their home on October 15, 1999. Neither of them heard M.G. cry
out nor did they see anything unusual. Wife testified appellant
was at the shed when she left for work. The stepson was in the
shower before M.G.'s mother knocked on the door, asking about
her daughter.
Appellant's doctor testified appellant was on disability
and prescribed oxygen for eighteen hours a day. The doctor
admitted on cross-examination that appellant will feel better on
some days and could engage in more activity on those days.
Appellant denied to the police and in his testimony at
trial that he ever touched M.G.
2
"SANE" is an acronym for sexual assault nurse examiner, a
discipline that involves training in the medical signs of sexual
assault.
- 4 -
ANALYSIS
Appellant argues the trial court erred by permitting mother
to testify regarding statements made by M.G, which affected both
his conviction for aggravated sexual battery and his sentencing.
Appellant further contends the evidence was insufficient to
convict him of aggravated sexual battery and object sexual
penetration. The Commonwealth argues the evidence was
admissible under the excited utterance exception to the hearsay
rule, 3 appellant waived his objection to this evidence, and the
evidence was sufficient for the convictions.
I. Hearsay
Hearsay is "testimony which consists [of] a
narration by one person of matters told him
by another." Williams v. Morris, 200 Va.
413, 417, 105 S.E.2d 829, 832 (1958). The
strongest justification for the exclusion of
hearsay evidence is that the trier of fact
has no opportunity to view the witness on
cross-examination and to observe the
demeanor of the out-of-court declarant to
determine reliability. C. Friend, [The Law
of Evidence in Virginia] § 224 [(2d ed.
1983)]. . . . [H]earsay evidence is
admissible if it falls into one of the
recognized exceptions to the hearsay rule
which are based on necessity and inherent
trustworthiness. C. Friend, supra, § 230 et
seq.
3
The Commonwealth specifically denies the trial court
admitted the evidence under the recent complaint exception to
the hearsay rule. See Code § 19.2-268.2. Therefore, we do not
discuss this exception.
- 5 -
Evans-Smith v. Commonwealth, 5 Va. App. 188, 197, 361 S.E.2d
436, 441 (1987). See also Jenkins v. Commonwealth, 254 Va. 333,
338, 492 S.E.2d 131, 134 (1997).
Hearsay statements are admissible under the excited
utterance exception when the declaration "is spontaneous and
impulsive, thus guaranteeing its reliability." Goins v.
Commonwealth, 251 Va. 442, 460, 470 S.E.2d 114, 126 (1996). See
also Braxton v. Commonwealth, 26 Va. App. 176, 184, 493 S.E.2d
688, 691 (1997). "The statement must be prompted by a startling
event and be made at such time and under such circumstances as
to preclude the presumption that it was made as the result of
deliberation. In addition, the declarant must have firsthand
knowledge of the startling event." Goins, 251 Va. at 460, 470
S.E.2d at 126 (citations omitted). Admissibility of evidence as
an excited utterance rests within the discretion of the trial
judge. 4 Walker v. Commonwealth, 19 Va. App. 768, 772, 454 S.E.2d
737, 740 (1995).
4
Appellant correctly notes the Commonwealth has the burden
to establish evidence falls within an exception when introducing
the evidence at trial. However, on appeal, we presume the judge
knows and understands the law, applying the appropriate
principles correctly. Yarborough v. Commonwealth, 217 Va. 971,
978, 243 S.E.2d 286, 291 (1977) (appellate court presumes the
trial court correctly applied the law to the facts); Justis v.
Young, 202 Va. 631, 632, 119 S.E.2d 255, 256-57 (1961)
(appellate courts presume a trial court's ruling is correct);
Dunn v. Commonwealth, 20 Va. App. 217, 219-20, 456 S.E.2d 135,
136 (1995) (the judgment of a trial court is presumed correct on
appeal). Therefore, where the trial court rules evidence is
admissible, but does not elaborate, appellate courts must
examine the record for justification of the trial court's
- 6 -
Appellant argues M.G.'s statements to her mother do not
fall within the excited utterance exception because (1) the
evidence does not establish the proximate time between the
startling event and the statement, (2) M.G. was not excited by
the startling event when she made the statements, and (3) M.G.
was responding to questions from her mother rather than making
spontaneous statements. We disagree.
The lapse of time between a startling event and a
statement, while a factor to consider, is not determinative of
whether to admit the statement as an excited utterance. Doe v.
Thomas, 227 Va. 466, 471, 318 S.E.2d 382, 385 (1984); Walker v.
Commonwealth, 19 Va. App. 768, 772, 454 S.E.2d 737, 740 (1995).
Failure of the evidence to indicate a specific length of time
between the event and the statement does not preclude admission
of the utterance. See, e.g., Braxton v. Commonwealth, 26 Va.
App. 176, 185, 493 S.E.2d 688, 692 (1997) (finding trial court
did not err when admitting an excited utterance, even though the
record did "not establish how much time elapsed" between the
event and the statement). Therefore, the fact that the
testimony did not delineate a specific time between the
decision. Additionally, if a defendant believes the trial court
has not justified its ruling, then he must ask the judge to
explain the rationale, especially when, as here, the defense
objection is a one-word statement. Id. (appellant has the
burden to prove the trial court erred).
- 7 -
startling event and M.G.'s statement to her mother did not
prevent the trial court from admitting the evidence. 5
The record did disclose that M.G. made the statement to her
mother a very short time after the incident. M.G. testified the
assault occurred around the time appellant's wife left for work,
at approximately 4:45 p.m. M.G.'s mother began looking for her
around five o'clock. Given the description of events, M.G. was
found and returned to her home within thirty minutes of the
assault. M.G. made her statements at the time she was
discovered in the shed and soon after she returned to her home.
Based on this record, the trial court could conclude the
utterance was "the transaction speaking through the declarant,"
rather than "the declarant speaking about the transaction."
Royal v. Commonwealth, 12 Va. App. 928, 931, 407 S.E.2d 346, 348
(1991).
Appellant also argues the hearsay should not have been
admitted, as M.G. was not excited by the startling event when
she made the statements. However, the record "contains
sufficient evidence to establish" that M.G. was speaking "under
the agitation" of the assault when she made the statements to
her mother. Goins, 251 Va. at 470, 470 S.E.2d at 126.
M.G., who was eight years old at the time, started talking
to her mother after she pulled on her clothes. A reasonable
5
We do not suggest that the criminal event must be the
startling event that precipitates the utterance.
- 8 -
inference is that this conversation occurred almost immediately
after the sexual assault. Instead of telling her mother the
name of the person who had been in the shed with her, M.G.
described him as "not a stranger" and pointed to where he was
hiding. The trial court could find, based on her age, the
immediacy of the statement, and the manner in which she
identified appellant, that M.G. was under the influence of the
event at the time she made her statements outside the shed.
Additionally, when M.G. and mother began to leave
appellant's yard, M.G. became hysterical and overwrought. She
was afraid someone with a gun would try to hurt her. When they
arrived at their home moments later, M.G. would not sit still.
At this point, when M.G. was in the safety of her own home, she
told mother that appellant had licked her pubic area and put his
finger into her vagina. Given all these factors, see Walker, 19
Va. App. at 772-74, 454 S.E.2d at 740, we cannot say the trial
court abused its discretion.
Appellant also argues M.G. was responding to questions from
her mother and, therefore, the trial court erred in admitting
the statement under the excited utterance exception. The
testimony directly contradicts appellant's assertion that M.G.'s
statement about the licking was in response to a question. Her
mother testified, "I did not question her." However, the
initial identification of appellant as the person who was in the
- 9 -
shed with M.G. was in response to her mother's questions, "what
were you doing" and "who was in here with you."
Again, no fixed rules determine whether a statement is
admissible as an excited utterance. Royal, 12 Va. App. at 931,
407 S.E.2d at 348. This exception can apply when statements are
made in response to questions. Martin v. Commonwealth, 4
Va. App. 438, 442, 358 S.E.2d 415, 418 (1987). The key is
whether "the question or questioner suggested or influenced the
response, then the declaration may lack the necessary
reliability to be admitted." Id.
Mother did not frame her questions in such a manner that
they suggested an answer nor did M.G.'s responses directly
answer the questions. When mother asked who had been in the
shed with M.G., the question did not suggest a particular name.
In fact, M.G. refused to say a name, but instead said the person
was someone mother knew. She then pointed in appellant's
direction rather than directly answer her mother's question.
More importantly, mother did not ask questions about what
happened in the shed. She testified, "I did not question her.
I have worked these cases before."
We find the trial court did not abuse its discretion when
it admitted this evidence. We also find, even if the mother's
statements on direct examination were improperly admitted,
appellant waived any objection to this evidence during his
cross-examination of the Commonwealth's witnesses.
- 10 -
"[W]here an accused unsuccessfully objects to evidence
which he considers improper and then on his own behalf
introduces evidence of the same character, he thereby waives his
objection, and we cannot reverse for the alleged error."
Saunders v. Commonwealth, 211 Va. 399, 401, 117 S.E.2d 637, 638
(1970) (citations omitted). While a defendant can cross-examine
a witness without waiving an earlier objection, once "evidence
that is similar to that to which the objection applies" is
introduced by the questioning, the original objection is waived.
Brant v. Commonwealth, 32 Va. App. 268, 278, 527 S.E.2d 476,
480-81 (2000). See also Newton v. Commonwealth, 29 Va. App.
433, 451, 512 S.E.2d 846, 854-55 (1999).
On cross-examination, mother initially answered questions
about M.G.'s statements made on the day of the incident, to
which appellant had previously objected. These questions were
designed to clarify and impeach mother's testimony regarding the
statements and did not waive the previous objections. However,
defense counsel then asked, "Now, you had the occasion over a
number of days to hear further descriptions of what had occurred
from [M.G.]; is that correct?" When mother said they had,
counsel asked, "Were [sic] there more than one version of the
facts that were given to you by your daughter?" Mother
answered, without objection or limitation, "She was very clear
on the three things that she originally told me that he had put
a cigarette in her mouth; that he had put his finger in her
- 11 -
private parts and that he had licked her private parts." Since
this cross-examination was beyond the scope of mother's direct
testimony, her answers were introduced on appellant's "own
behalf."
Additionally, Deputy Sheriff Smith testified on direct
examination, without objection, 6 "[M.G.] was very much upset and
scared that Mr. Guy was going to come and get her. That's what
she told me. [M.G.] then told me that Mr. Guy had basically
licked her vagina and stuck his finger in her hole."
Appellant also argues M.G.'s statement was prejudicial as
to sentencing. However, as we find the evidence was properly
admitted, appellant cannot complain of prejudice from this
testimony.
II. Sufficiency
Appellant argues the trial court should have granted his
motion to set aside the verdict as "a serious credibility issue"
existed about the victim's testimony. We disagree.
Appellant's motion to the trial court argued initially that
double jeopardy prevented conviction on both charges. Counsel
then admitted the evidence was sufficient "for an incident to
have occurred and the jury having found guilt on the particular
6
At oral argument, appellant contended he made a continuing
objection at trial that included this testimony. The record
does not support this contention.
- 12 -
penetration offense it certainly seems that the penetration was
an offense that was committed and there was no further evidence
supported by testimony of any aggravated sexual battery act."
Clearly, appellant conceded sufficiency of the evidence for the
penetration offense, thereby waiving any sufficiency argument
related to this conviction. See Redman v. Commonwealth, 25
Va. App. 215, 220, 487 S.E.2d 269, 272 (1997). See also Rule
5A:18.
Appellant claims several inconsistencies between the
witnesses' testimony made M.G.'s testimony incredible. However,
credibility issues are in the province of the jury. Wilson v.
Commonwealth, 31 Va. App. 495, 508, 525 S.E.2d 1, 7 (2000). The
trier of fact resolves any inconsistencies in the testimony.
See Barker v. Commonwealth, 230 Va. 370, 373-74, 337 S.E.2d 729,
732 (1985). As nothing in this record suggests the witnesses
were inherently incredible, we will not set aside the aggravated
sexual battery conviction.
For the reasons stated above, we affirm appellant's
convictions.
Affirmed.
- 13 -
|
Footwear Size Guide - Portwest
Bought these to go beach fishing actually because one needs non-metal boots which will take a lot of punishment. The price was at least 50% less than specialist dealers were offering so I am well pleased and I don't have to worry about any sharp objects which the boots might encounter.
We are happy to take returns
Total Workwear operate a "no-quibbles" guarantee, which means that if for any reason you are unhappy with your purchase, simply return it to us in its original packaging within 30 days from the date of delivery. We will issue a full refund for the price you paid for the item.
Faulty ProductsIf you think there is a fault with the items you have received please email us first. Please do not return faulty items until we have contacted you.
Exchanging goods. To exchange goods you must return the incorrect item by clicking on the CLICK HERE link below. Then you should reorder again from this web site as normal. Once we have received your returned items we will issue a refund for the value of the goods returned.
Items must be unworn and (in our opinion) in perfect condition and packed carefully in the original packaging.
Proof of posting should be obtained.
You, the customer, are responsible for the cost of the return postage.
Your original postage cost will not be refunded unless we have made a mistake or the goods are faulty.
If the return is due to faulty goods or an error we have made we will gladly refund the delivery charges.
Small 100mm approx Usually for left or right breast on the front of garments.
Large 230mm approx. Usually for large areas on the backs of garments
Prices for printing onto all garments including tops trousers and caps.
You will be advised if a garment is not suitable for printing or embroidery.
Garment Embroidery prices (ex. VAT)
Embroidered Logo or Text set up charge:
Text set up
For all new text
£5
Logo set up
Once only charge for each design or size.
£15
Embroidery application charge per unit:
SIZE
SETUP
1 to 24
25-49
50-99
100-149
150-199
200-249
250+
Smalll
100mm
£5
£2.75
£2.50
£2.25
£2
£1.75
£1.50p
£1.25
Large
230mm
£15
£4.50
£4.25
£4
£3.75
£3.50
Call
Call
Small 100mm approx Usually for left or right breast on the front of garments.
Large 230mm approx. Usually for large areas on the backs of garments
Prices for printing onto all garments including tops trousers and caps.
You will be advised if a garment is not suitable for printing or embroidery.
Embroidery
We offer an embroidery service to have your garments branded with stitched logos or text. This process is an extremely popular choice for high quality, visible branding onto the products of your choice. Stitching embroidery is the perfect option for office uniform, corporate clothing and general office wear. The process of stitching can however interfere with specialised garment properties, such as flame retardant and waterproofing. In this situation, please see below as a printed option would be more suitable to your requirement. If you have any queries please don't hesitate to speak to one of our experts.
Printed Logos and Text
As an alternative to embroidery, we also offer high quality printed logos and text. This process will provide a sharp professional finish and is suitable for most garments. |
Hot Toys’ Batman: Arkham City Figure Teased
Better break out that Bat card from 'Batman & Robin,' because Hot Toys is releasing a figure based on the Caped Crusader's appearance in Batman: Arkham City.
Hot Toys has just released a teaser image on Facebook of its latest figure depicting the Caped Crusader as he was in Batman: Arkham City. For those who do not know, Hot Toys is probably the most intricately detailed action figure creator on the planet. Their 1/6 scale figures often sell for over $200, but what you get is arguably the highest quality a collector could ever get. We have provided pictures of Hot Toys' Iron Man and Joker figures below just so you could see how detailed and accurate these figures can be. Once Hot Toys' figures have their run, they are often found on third-party vendor websites for ridiculously higher rates than their original prices.
Hot Toys will be unveiling this Batman figure during San Diego Comic-Con 2014 later this week. It's a bit odd the figure will be based on Arkham City, and not the upcoming Arkham Knight, but given Hot Toys' history with stupendous Bat-figures in the past, we're eager to see what lies ahead for the video game Bruce Wayne.
Now if only we could get a Harley Quinn figure from the Arkham series done as well. |
Béthune (river)
The Béthune is a river of Normandy, France, in length, flowing through the department of Seine-Maritime and it is a tributary of the Arques River. The French Sandre regulators however, consider the Béthune as the Arques for all its length.
Geography
The river’s source is at the village of Gaillefontaine near to Forges-les-Eaux. Its valley is wholly within the pays de Bray. Its course takes it past the communes of Neufchâtel-en-Bray, Mesnières-en-Bray, Bures-en-Bray, Osmoy-Saint-Valery, Saint-Vaast-d'Équiqueville, Dampierre-Saint-Nicolas, Saint-Aubin-le-Cauf and finally Arques-la-Bataille where it joins the Eaulne and the Varenne Rivers to form the Arques River
Like other rivers in the region, the Béthune is classified as a first class river, offering anglers the chance to catch salmon and trout.
See also
French water management scheme
Notes
This article is based on the equivalent article from the French Wikipedia, consulted on October 14, 2008.
External links
The Béthune on the Sandre website
Accès aux villes et villages de France du Quid
French Geography website
Category:Rivers of France
Category:Rivers of Normandy
Category:Rivers of Seine-Maritime |
Q:
Muliple div creation, jquery/javascript, performance/best practice
Im trying to figure out best practices in regard to performance when creating multiple DIV's at an insane rate. For example, on every .mousemove event...
$('head').append("<style>.draw {width: 20px; height: 20px; position:fixed;</style>");
$(document).mousemove(function(mouseMOVE) {
//current mouse position
var mouseXcurrent = mouseMOVE.pageX;
var mouseYcurrent = mouseMOVE.pageY;
//function to create div
function mouseTRAIL(mouseX, mouseY, COLOR) {
$('body').append("<div class='draw' style='top:" + mouseY + "px; left:" + mouseX + "px; background: " + COLOR + ";'></div>");
}
// function call to create <div> at current mouse positiion
mouseTRAIL(mouseXcurrent, mouseYcurrent, '#00F');
// Remove <div>
setTimeout(function() {
$('.draw:first-child').remove();
}, 250);
});
So, this works all nice and dandy but it's mega inefficient (especially so when I try filling in the space between each mouse move position). Here's an example...
$('head').append("<style>.draw {width: 20px; height: 20px; position:fixed;</style>");
$(document).mousemove(function(mouseMOVE) {
//current mouse position
var mouseXcurrent = mouseMOVE.pageX;
var mouseYcurrent = mouseMOVE.pageY;
// function to create div
function mouseTRAIL(mouseX, mouseY, COLOR) {
$('body').append("<div class='draw' style='top:" + mouseY + "px; left:" + mouseX + "px; background: " + COLOR + ";'></div>");
}
// function call to create <div> at current mouse positiion
mouseTRAIL(mouseXcurrent, mouseYcurrent, '#00F');
// variabls to calculate position between current and last mouse position
var num = ($('.draw').length) - 3;
var mouseXold = parseInt($('.draw:eq(' + num + ')').css('left'), 10);
var mouseYold = parseInt($('.draw:eq(' + num + ')').css('top'), 10);
var mouseXfill = (mouseXcurrent + mouseXold) / 2;
var mouseYfill = (mouseYcurrent + mouseYold) / 2;
// if first and last mouse postion exist, function call to create a div between them
if ($('.draw').length > 2) {
mouseTRAIL(mouseXfill, mouseYfill, '#F80');
}
// Remove <div>
setTimeout(function() {
$('.draw:first-child').remove();
$('.draw:nth-child(2)').remove();
}, 250);
});
I really cant figure out how to improve things. Believe me, Ive tried researching but it hasn't done much good... What I'm looking for is some suggestions, examples, or links to better practices...
Please note that I'm teaching myself to code. I'm a Graphic Design student and this is how I'm spending my summer out of class... Making little projects to teach myself JavasSript, fun stuff :)
Ive set up some jsfiddles to show what Im working on...
Mouse Trail, More Elements - Very Very Slow
Mouse Trail, Less Elements - Very Slow
Mouse Trail, Bare Bones - Slow
A:
There are multiple bad practices going on here:
Using elements instead of Canvas
Using those elements via jQuery
Abusing that jQuery as if you were trying to make it slow on purpose
Stuffing all of the above inside a mousemove handler
The root issue here really is using elements instead of canvas. After fixing that, the interaction with DOM should become minimal and thus
fix the other points as well.
Also, those who claim that this works fine didn't check their CPU usage. On my Core I5-2500K one core is instantly maxed up which is ridiculous and unacceptable for something trivial like rendering a mouse trail on screen.
I can very well imagine this being very very slow on an older computer. So yes, it's smooth but at the cost of using amount of resources enough for 10-20+ tabs to do the same properly.
This takes 7-14% cpu for me when moving mouse around fast, this takes full 25%.
|
Joseph Wash, suspect in Moss Point murder case, dies after hanging himself in April
PASCAGOULA, Mississippi -- Jackson County Coroner Vicki Broadus has confirmed that murder suspect Joseph Dwayne "Bodee" Wash has died from injuries sustained after attempting to hang himself in his jail cell in April.
Wash had remained in a vegetative state since the April 4 suicide attempt, when officers at the Jackson County Adult Detention Center found him hanging by a rolled up blanket tied to a window bar.
Broadus said the official cause of death is "death by hanging."
Wash's identity was not released at the the time of the incident, with Sheriff Charles Britt citing federal privacy laws.
Wash, 34, was facing a 2nd-degree murder charge and a felony gun charge from an Oct. 12, 2013 incident in which Joseph Denell White was fatally shot. After the hanging, the case was sent to the inactive files.
At the time of the shooting, Pascagoula police said White was killed after pulling a gun on Wash's handicapped son. Joshua Liddell, Wash's cousin, was also arrested for the shooting.
Wash was arrested by U.S. Marshals in Jackson on Oct. 16. He had been released from prison earlier in 2013 after spending 12 years in jail for a manslaughter conviction. |
Urinary tract disorders. Clinical comparison of flavoxate and phenazopyridine.
In nine separate clinical trials, 382 patients having symptoms of either prostatitis, acute cystitis, urethritis, and/or trigonitis were randomly assigned to treatment with flavoxate or phenazopyridine. Over-all response was evaluated in 384 patients after five days of therapy. In patients having prostatitis, response was satisfactory in 66 per cent treated with flavoxate and 31 per cent treated with phenazopyridine. In all other patients, satisfactory responses were reported in 80 per cent on flavoxate compared with 56 per cent on phenazopyridine. Similarly, symptom-severity evaluations at two and five days of therapy showed most symptoms improved in more of the patients on flavoxate therapy than on phenazopyridine therapy. Although more adverse effects were reported in patients treated with phenazopyridine than with flavoxate, the difference between medications was not statistically significant. |
Q:
Unexpected behaviour plotting a PDE solution
I'm solving some coupled PDEs (Eb1 and Ef1) and what I plot for Eb1 appears to be correct. However, for some reason, when I go to plot Ef1, I get nothing. MWE is below (beginning is constants and functions needed to solve PDEs, relevant part near end)
(*Constants needed*)
a = 8.2314*10^-7; omega = 3.0318*10^7; Do2 = 2*10^-9; po = 106;
ro = 235*10^-6; micron = 1*10^-6; qm = 10^-4;
mic = 1*10^-6; De = 5.5*10^-11; eo = 100; j = 0.12; kmn = 4.7;
kme = 2.1; con = (a*omega)/(6*Do2); rl = Sqrt[(6*Do2*po)/(omega*a)];
(*Functions needed*)
rn = Piecewise[{{0, ro <= rl}, {ro*(0.5 - Cos[(ArcCos[1 - (2*rl*rl)/(ro^2)] - 2*Pi)/3]), ro > rl}}];
p[r_] = Piecewise[{{po + con*(r^2 - ro^2 + 2*(rn^3)*(1/r - 1/ro)), r > rn} , {0 , r <= rn}}];
q[r_] = qm*((kme/(kme + p[r]))*(p[r]/(kmn + p[r])) + (kmn/(kmn + p[r]))*j);
(*Equations defined*)
eqnDe = D[Ef1[r, t], t] - De*(D[Ef1[r, t], r, r] + (2/r)*(D[Ef1[r, t], r])) + q[r]*Ef1[r, t];
eqnBo = D[Eb1[r, t], t] - (Ef1[r, t])*q[r];
(*Solving equations*)
x = NDSolve[{eqnBo == 0, Eb1[r, 0] == 0, eqnDe == 0, Ef1[r, 0] == 0, Derivative[1, 0][Ef1][rn, t] == 0, Ef1[ro , t] == eo},
Eb1, {r, rn, ro}, {t, 0, 14400}];
Then it's easy to plot Eb1 between rn and ro with this;
Plot[Eb1[r, 14400] /. x, {r, rn, ro}, PlotRange -> Automatic]
and I get this;
Yet when I try to flow Ef1 with a similar command I get absolutely nothing;
Plot[Ef1[r, 14400] /. x, {r, rn, ro}, PlotRange -> Automatic]
Similarly, I can't seem to evaluate Ef1 at a given value of $r$ or $t$ - any ideas what I'm doing wrong and why this is the case?
A:
It seems that you're missing a solution of Ef1. Try
y = NDSolve[{eqnBo == 0, Eb1[r, 0] == 0, eqnDe == 0, Ef1[r, 0] == 0,
Derivative[1, 0][Ef1][rn, t] == 0, Ef1[ro, t] == eo},
Ef1, {r, rn, ro}, {t, 0, 14400}];
Plot[Ef1[r, 14400] /. y, {r, rn, ro}, PlotRange -> Automatic]
|
Role of gut cryptopatches in early extrathymic maturation of intestinal intraepithelial T cells.
Lympho-hemopoietic progenitors residing in murine gut cryptopatches (CP) have been shown to generate intestinal intraepithelial T cells (IEL). To investigate the role of CP in progenitor maturation, we analyzed IEL in male mice with a truncated mutation of common cytokine receptor gamma-chain (CRgamma-/Y) in which CP were undetectable. IEL-expressing TCR-gammadelta (gammadelta-IEL) were absent, and a drastically reduced number of Thy-1highCD4+ and Thy-1highCD8alphabeta+ alphabeta-IEL were present in CRgamma-/Y mice, whereas these alphabeta-IEL disappeared from athymic CRgamma-/Y littermate mice. Athymic CRgamma-/Y mice possessed a small TCR- and alphaEbeta7 integrin-negative IEL population, characterized by the disappearance of the extrathymic CD8alphaalpha+ subset, that expressed pre-Talpha, RAG-2, and TCR-Cbeta but not CD3epsilon transcripts. These TCR- IEL from athymic CRgamma-/Y mice did not undergo Dbeta-Jbeta and Vdelta-Jdelta joinings, despite normal rearrangements at the TCR-beta and -delta loci in thymocytes from euthymic CRgamma-/Y mice. In contrast, athymic severe combined immunodeficient mice in which CP developed normally possessed two major TCR-alphaEbeta7+ CD8alphaalpha+ and CD8- IEL populations that expressed pre-Talpha, RAG-2, TCR-Cbeta, and CD3epsilon transcripts. These findings underscore the role of gut CP in the early extrathymic maturation of CD8alphaalpha+ IEL, including cell-surface expression of alphaEbeta7 integrin, CD3epsilon gene transcription, and TCR gene rearrangements. |
Kinetic adsorption energy distributions of rough surfaces: a computational study.
The adsorption energy distribution usually refers to localized monolayers of adsorbate at thermodynamic equilibrium. Many papers have been published that analyze its influence on adsorption isotherms, heats of adsorption, and adsorption kinetics. However, the adsorption energy distribution, in its classical thermodynamic equilibrium sense, may be not as useful as expected. This is because many important processes involving adsorption have dynamic character and reactant particles have a finite time for penetration of the adsorbent. The above suggests that some adsorption centers located in less accessible fragments of the surface can be invisible in a dynamic process. However, under conditions allowing the thermodynamic equilibrium such adsorption centers could noticeably contribute to the adsorption energy distribution. The aim of this work is to measure the adsorption energy distributions of special rough surfaces using a dynamic method. This method is based on the molecular dynamics simulation of an ideal gas flowing over a sample surface. The ideal gas particles penetrate the surface, and at the moment of collision of a gas particle with the surface the Lennard-Jones potential energy is calculated. This energy can be identified with the adsorption energy at a given point on the surface. The surfaces used in the calculations have been created using two surface growth models (i.e., random deposition and ballistic deposition). The application of these highly disordered surfaces enables us to draw some general conclusions about the properties of real surfaces that are usually far from any deterministic geometry. |
Fibrinolytic serine protease isolation from Bacillus amyloliquefaciens An6 grown on Mirabilis jalapa tuber powders.
In this study, Mirabilis jalapa tuber powder (MJTP) was used as a new complex organic substrate for the growth and production of fibrinolytic enzymes by a newly isolated Bacillus amyloliquefaciens An6. Maximum protease activity (1,057 U/ml) with casein as a substrate was obtained when the strain was grown in medium containing (grams per liter) MJTP 30, yeast extract 6, CaCl(2) 1, K(2)HPO(4) 0.1, and K(2)HPO(4) 0.1. The strain was also found to grow and produce extracellular proteases in a medium containing only MJTP, indicating that it can obtain its carbon, nitrogen, and salts requirements directly from MJTP. The B. amyloliquefaciens An6 fibrinase (BAF1) was partially purified, and fibrinolytic activity was assayed in a test tube with an artificial fibrin clot. The molecular weight of the partially purified BAF1 fibrinolytic protease was estimated to be 30 kDa by sodium dodecyl sulfate polyacrylamide gel electrophoresis and gel filtration. The optimum temperature and pH for the caseinolytic activity were 60 degrees C and 9.0, respectively. The enzyme was highly stable from pH 6.0 to 11.0 and retained 62% of its initial activity after 1 h incubation at 50 degrees C. However, the enzyme was inactivated at higher temperatures. The activity of the enzyme was totally lost in the presence of phenylmethylsulfonyl fluoride, suggesting that BAF1 is a serine protease. |
480 So.2d 669 (1985)
NORTH AMERICAN CATAMARAN RACING ASSOCIATION, INC. (NACRA), Appellant,
v.
John McCOLLISTER, As Personal Representative of the Estate of Christine Wapniarski, Deceased, Appellee.
No. 84-1796.
District Court of Appeal of Florida, Fifth District.
December 5, 1985.
Rehearing Denied January 7, 1986.
*670 Raymond A. Haas and Delia Rose of Haas, Boehm, Brown, Rigdon, Seacrest & Fischer, P.A., Daytona Beach, for appellant.
Frederick S. Jaeger, Jr. of Robinson & Jaeger, Daytona Beach, for appellee.
SHARP, Judge.
North American Catamaran Racing Association, Inc. (NACRA) appeals from a judgment entered against it in a wrongful death case. Wapniarski died in the tragic aftermath of the capsize of a catamaran manufactured by NACRA. Because we find in this case no evidence of negligence on the part of NACRA, and no evidence of any defect in the design of the catamaran which was the cause of Wapniarski's death, we reverse.
This suit arose out of an accident in 1981 which occurred at sea. Perrin, the owner of the catamaran and a co-defendant,[1] invited three other young people to sail with him, including Wapniarski, that fateful afternoon. Perrin was the third owner of the boat, and he was not an expert sailor. The 1977 NACRA catamaran had been sailing poorly that day because one pontoon was leaking. Perrin thought the cause of the leak was a broken inspection cap which had a chipped or broken thread. Perrin tried to repair it with duct tape.
After making this repair on shore, the four sailed out to sea again. A storm came up, and the pontoon was again seen to be filling with water. Perrin tried to drain it by pulling the drain plug while sailing. However, the boat capsized and the leaking pontoon filled with water.
The group clung to the remaining catamaran pontoon, which kept them afloat during the night. When daylight came, they decided to try to swim to shore. Wapniarski was not as good a swimmer as the other three, but she did not want to be left alone. On the journey to shore she was attacked by a shark, and died from loss of blood or drowning. Later, the Coast Guard found the catamaran, with one pontoon still afloat, but they were unable to recover it.
At trial, the appellee sought to prove that NACRA was negligent in its design of the catamaran for two reasons: it did not incorporate a positive flotation system into the design of the hulls; and it made the fiberglass hulls too thin, so they would leak and break up with use. The evidence on these two points, as well as the evidence on causation, even if defective design had been proven, was exceedingly weak and strongly countered by other evidence.
The jury was asked to answer a special verdict form, which included the following two questions:
*671 Was the sailboat defective when sold and, if so, was the defect a legal cause of the death of Christine Wapniarski?
Was there negligence on the part of defendant NACRA which was the legal cause of the death of Christine Wapniarski?
The jury answered the first question "No," but it answered the second question "Yes."
NACRA contends these verdicts are inconsistent and require reversal. The trial court instructed the jury on two theories of NACRA's liability: strict liability and negligence. The first question pertains to the strict liability claim while the second question relates to the negligence claim. The allegation of negligence was the same under both theories: NACRA's negligent design of the boat.
The appellee counters that NACRA waived appellate review of the inconsistency by not objecting before the jury was discharged.[2] True, a party must object to defective verdict forms or inconsistent verdicts before the jury is discharged to preserve the claim, Higbee v. Dorigo, 66 So.2d 684 (Fla. 1953); Papcun v. Piggy Bag Discount Souvenirs Food and Gas Corp., 472 So.2d 880 (Fla. 5th DCA 1985). Here, however, the inconsistency is of a fundamental nature because the only evidence of negligence offered against NACRA at trial related to its alleged negligent design. See Papcun v. Piggy Bag Discount Souvenirs Food and Gas Corp.; Robbins v. Graham, 404 So.2d 769 (Fla. 4th DCA 1981). But, the jury found that there was no design defect. And if that were true, there was no other evidence to sustain the jury's verdict in this case. Cf. Cowart v. Kendall United Methodist Church, 476 So.2d 289 (Fla. 3rd DCA 1985) (while verdict not inconsistent, no evidence to support it). Accordingly, we have no alternative but to reverse the judgment and remand for entry of judgment in NACRA's favor.[3]
REVERSED and REMANDED.
DAUKSCH and UPCHURCH, JJ., concur.
NOTES
[1] The jury found Perrin 20% responsible, the decedent 20% responsible and NACRA 60% responsible for Wapniarski's death.
[2] NACRA moved for a directed verdict, judgment notwithstanding the verdict and new trial on the grounds that the plaintiff failed to prove strict liability or negligence.
[3] The co-defendant, Perrin, did not appeal the final judgment and, therefore, our decision affects only the liability of NACRA.
|
Industrial vacuum equipment has dozens of wet and dry uses such as locating underground utilities (potholing), hydro excavation, air excavation and vacuum excavation. In addition, the equipment can be used for directional drilling slurry removal, industrial clean-up, waste clean-up, lateral and storm drain clean-out, oil spill clean-up and other natural disaster clean-up applications, signs and headstone setting, for example. The vacuum systems may be mounted to a truck or trailer and are typically powered by gas or diesel engines. The material is vacuumed up and stored in a storage tank. From there, the material may be hauled away and disposed or the tank is emptied by opening a rear hatch and dumped in a pile at the site. In addition, outside material may be imported to the location of the hole for backfill. Workers move the material by hand from the pile and fill in the excavation and holes as needed. A shortcoming of the prior art is the inefficiency and difficulty to fill the excavation (or holes) easily, quickly and with precision. Accordingly, what is needed is a method and system to excavate and fill that can both excavate material and selectively return the fill material to a site in a controlled manner. |
Tuesday, July 24, 2012
Ellie's Eyes
This post has been a long time coming. I wanted to make sure things were for sure before writing a post about it. You have probably noticed in pictures that Ellie's eyes are a little crossed. At about four months, we noticed that her eyes just weren't straightening out like they are supposed to. Since things like lazy eye run in my family, I immediately became worried. I asked my mom and other people what they thought, and they suggest we get it looked at. I took her into her pediatrician and told her everything we have noticed and how things like this run in my family. She told me that they don't normally look at their eyes until six months but because of our circumstances we should definitely get a second opinion (thank heavens for good doctors.) We scheduled an appointment with Doctor Petersen who is one of the best in the state and just happens to be my sisters eye doctor.
Our worries were confirmed and he told us at the moment her eyes were just far sided (she can't see things up close) but he wasn't too concerned about lazy eye. He wanted to wait until she was nine months to see how things played out, so we scheduled another appointment and waited. We took her in at nine months and the change was so drastic that he suggested we get glasses. She is still just far sided and luckily we aren't looking at lazy eye.....yet. He gave us a prescription and we went to get her glasses ordered. She got them the weekend I ran Ragnar so I wasn't there to see them the first day but they were sending me pictures to let me know how cute she looked in them. We took her again last week and her eyes are showing so much change with the glasses that we don't need to worry about surgery or patching yet. This makes us very happy. We are so thankful for good doctors who didn't just push us aside and really took the time to look at Ellie's eyes. They sooner we get her in glasses the better her eyes will get. Hopefully they continue to work and we don't have to worry about surgery or patching.
She does look pretty dang cute in glasses:
Since getting glasses, many people have asked us questions so I figured I would answer them on here.
How long will she have to wear them? If everything goes as planned and we don't have to deal with surgery or patching, she could be out of glasses completely by the time she is twelve. Just in time for Junior High.
How did the doctors know that her eyes were bad? I have no idea how it works. They measured her lens and other parts of her eye. One thing I do know is that they are excellent at their jobs. They went to school for it so I figure they know what they are doing.
How do you keep them on her? We don't. She has to be really distracted or they come off the minute we put them on. I was working extremely hard to try and keep them on ALL THE TIME. As you can imagine I was exhausted. I told the doctor this and he told me I was working way to much and that as long as she wore them for an hour each day, they would work. Needless to say I am not as stressed.
How are we handling this? I think I am more emotional about it than A.E. but I have surprised myself. I thought I would be way upset but I guess I figure there could be way worse things wrong with her. On a scale of bad this is minimal so we are just thankful for that. |
Wizards fans probably can recite from memory the list of snubs and undeserved criticisms that John Wall has been subjected to since joining the team in Washington.
They started with Colin Cowherd's absurd rant after Wall's debut ("Who's your daddy?") and continued last year with agent David Falk's rant ("John Wall will never be as good as Kyrie Irving was in his first week in the NBA"). Even Wall's decision to get tattoos was subject to judgment in the local media, with his flip-flopping on the issue being evident that Wall was not a leader deserving of a max contract, or something like that.
There has been no shortage of snubs for Wall either, at least in his mind. There's the recent ranking of No. 31 best player in the league by SI or being excluded from ESPN's top 25 players under 25 in 2013. Being left off the original list of players invited to camp for Team USA and getting cut after a belated invite are further examples.
Rest assured, Wizards fans, Wall is taking note of this disrespect and using it to fuel his fire as he enter the 5th season of his career. Per the Washington Post's Jorge Castillo:
John Wall keeps a list on his phone of the slights and criticism he hears, reads and perceives. He updates it whenever he comes across a new form of disrespect and reads it for motivation before he steps on the floor for every practice and game. [...] "I'm one of the most complete point guards in the league," Wall, 24, said Tuesday after the Washington Wizards concluded their first training camp practice. "I rebound, I assist, play defense, steal, score. I don't get why I'm overlooked. I'm still trying to figure it out."
If you're wondering what Wall is going to do about Cavaliers shooting guard Dion Waiters suggesting that the Wizards being the best backcourt in the league is "nonsense," wonder no more. Wall's phone list has a new entry. |
Q:
How to compare 2 non-stationary time series to determine a correlation?
I have two data series that plot median age at death over time. Both series demonstrate an increased age at death over time, but one much lower than another. I want to determine if the increase in the age at death of the lower sample is significantly different than that of the upper sample.
Here are the data, ordered by year (from 1972 through 2009 inclusive) as rounded to three decimal places:
Cohort A 70.257 70.424 70.650 70.938 71.207 71.263 71.467 71.763 71.982 72.270 72.617 72.798 72.964 73.397 73.518 73.606 73.905 74.343 74.330 74.565 74.558 74.813 74.773 75.178 75.406 75.708 75.900 76.152 76.312 76.558 76.796 77.057 77.125 77.328 77.431 77.656 77.884 77.983
Cohort B 5.139 8.261 6.094 12.353 11.974 11.364 12.639 11.667 14.286 12.794 12.250 14.079 17.917 16.250 17.321 18.182 17.500 20.000 18.824 21.522 21.500 21.167 21.818 22.895 23.214 24.167 26.250 24.375 27.143 24.500 23.676 25.179 24.861 26.875 27.143 27.045 28.500 29.318
Both series are non-stationary - how can I compare the two please? I'm using STATA. Any advice would be gratefully received.
A:
This is a simple situation; let's keep it so. The key is to focus on what matters:
Obtaining a useful description of the data.
Assessing individual deviations from that description.
Assessing the possible role and influence of chance in the interpretation.
Maintaining intellectual integrity and transparency.
There are still many choices and many forms of analysis will be valid and effective. Let's illustrate one approach here that can be recommended for its adherence to these key principles.
To maintain integrity, let's split the data into halves: the observations from 1972 through 1990 and those from 1991 through 2009 (19 years in each). We will fit models to the first half and then see how well the fits work in projecting the second half. This has the added advantage of detecting significant changes that may have occurred during the second half.
To obtain a useful description, we need to (a) find a way to measure the changes and (b) fit the simplest possible model appropriate for those changes, evaluate it, and iteratively fit more complex ones to accommodate deviations from the simple models.
(a) You have many choices: you can look at the raw data; you can look at their annual differences; you can do the same with the logarithms (to assess relative changes); you can assess years of life lost or relative life expectancy (RLE); or many other things. After some thought, I decided to consider RLE, defined as the ratio of life expectancy in Cohort B relative to that of the (reference) Cohort A. Fortunately, as the graphs show, the life expectancy in Cohort A is increasing regularly in a stable fashion over time, so that most of the random-looking variation in the RLE will be due to changes in Cohort B.
(b) The simplest possible model to start with is a linear trend. Let's see how well it works.
The dark blue points in this plot are the data retained for fitting; the light gold points are the subsequent data, not used for the fit. The black line is the fit, with a slope of .009/year. The dashed lines are prediction intervals for individual future values.
Overall, the fit looks good: examination of residuals (see below) shows no important changes in their sizes over time (during the data period 1972-1990). (There is some indication that they tended to be larger early on, when life expectancies were low. We could handle this complication by sacrificing some simplicity, but the benefits for estimating the trend are unlikely to be great.) There is just the tiniest hint of serial correlation (exhibited by some runs of positive and runs of negative residuals), but clearly this is unimportant. There are no outliers, which would be indicated by points beyond the prediction bands.
The one surprise is that in 2001 the values suddenly fell to the lower prediction band and stayed there: something rather sudden and large happened and persisted.
Here are the residuals, which are the deviations from the description mentioned previously.
Because we want to compare the residuals to 0, vertical lines are drawn to the zero level as a visual aid. Again, the blue points show data used for the fit. The light gold ones are the residuals for data falling near the lower prediction limit, post-2000.
From this figure we can estimate that the effect of the 2000-2001 change was about -0.07. This reflects a sudden drop of 0.07 (7%) of a full lifetime within Cohort B. After that drop, the horizontal pattern of residuals shows that the previous trend continued, but at the new lower level. This part of the analysis should be considered exploratory: it was not specifically planned, but came about due to a surprising comparison between the held-out data (1991-2009) and the fit to the rest of the data.
One other thing--even using just the 19 earliest years of data, the standard error of the slope is small: it's only .0009, just one-tenth of the estimated value of .009. The corresponding t-statistic of 10, with 17 degrees of freedom, is extremely significant (the p-value is less than $10^{-7}$); that is, we can be confident the trend is not due to chance. This is one part of our assessment of the role of chance in the analysis. The other parts are the examinations of the residuals.
There appears to be no reason to fit a more complicated model to these data, at least not for the purpose of estimating whether there's a genuine trend in RLE over time: there is one. We could go further and split the data into pre-2001 values and post-2000 values in order to refine our estimates of the trends, but it wouldn't be completely honest to conduct hypothesis tests. The p-values would be artificially low, because the splitting testing were not planned in advance. But as an exploratory exercise, such estimation is fine. Learn all you can from your data! Just be careful not to deceive yourself with overfitting (which is almost sure to happen if you use more than a half dozen parameters or so or use automated fitting techniques), or data snooping: stay alert to the difference between formal confirmation and informal (but valuable) data exploration.
Let's summarize:
By selecting an appropriate measure of life expectancy (the RLE), holding out half the data, fitting a simple model, and testing that model against the remaining data, we have established with high confidence that: there was a consistent trend; it has been close to linear over a long period of time; and there was a sudden persistent drop in RLE in 2001.
Our model is strikingly parsimonious: it requires just two numbers (a slope and intercept) to describe the early data accurately. It needs a third (the date of the break, 2001) to describe an obvious but unexpected departure from this description. There are no outliers relative to this three-parameter description. The model is not going to be substantially improved by characterizing serial correlation (the focus of time-series techniques generally), attempting to describe the small individual deviations (residuals) exhibited, or introducing more complicated fits (such as adding in a quadratic time component or modeling changes in the sizes of the residuals over time).
The trend has been 0.009 RLE per year. This means that with each passing year, the life expectancy within Cohort B has had 0.009 (almost 1%) of a full expected normal lifetime added to it. Over the course of the study (37 years), that would amount to 37*0.009 = 0.34 = one-third of a full lifetime improvement. The setback in 2001 reduced that gain to about 0.28 of a full lifetime from 1972 to 2009 (even though during that period overall life expectancy increased 10%).
Although this model could be improved, it would likely need more parameters and the improvement is unlikely to be great (as the near-random behavior of the residuals attests). On whole, then, we should be content to arrive at such a compact, useful, simple description of the data for so little analytical work.
|
do return end --TODO aby8
-----------------------
-- 战场增强 --
-----------------------
local addonName = ...
local BF = {}
LibStub('AceEvent-3.0'):Embed(BF)
LibStub('AceTimer-3.0'):Embed(BF)
local requestTimer
function requestBoard()
if(select(2, IsInInstance()) == 'pvp') then
return RequestBattlefieldScoreData()
else
BF:CancelRequestTimer()
end
end
function BF:CancelRequestTimer()
if(requestTimer) then
BF:CancelTimer(requestTimer)
requestTimer = nil
end
end
function BF:CreateCountText(f, n)
local txt = f:CreateFontString(nil, 'OVERLAY')
f[n] = txt
txt:SetFont(STANDARD_TEXT_FONT, 10, 'OUTLINE')
txt:SetPoint('CENTER', _G[f:GetName()..'Icon'])
return txt
end
function BF:UPDATE_BATTLEFIELD_SCORE()
if(not AlwaysUpFrame2) then return end
local atxt = AlwaysUpFrame1._txt or self:CreateCountText(AlwaysUpFrame1, '_txt')
local htxt = AlwaysUpFrame2._txt or self:CreateCountText(AlwaysUpFrame2, '_txt')
if(select(2, IsInInstance()) == 'pvp') then
local _, _, _, _, num_horde = GetBattlefieldTeamInfo(0)
local _, _, _, _, num_alliance = GetBattlefieldTeamInfo(1)
htxt:SetText(num_horde)
atxt:SetText(num_alliance)
else
htxt:SetText''
atxt:SetText''
end
end
function BF:Init()
self:RegisterEvent'PLAYER_ENTERING_WORLD'
end
function BF:PLAYER_DEAD()
if(U1GetCfgValue(addonName, 'bfautorelease')) then
return RepopMe()
end
end
do
local no_scoreboard = (U1IsAddonEnabled'capping' or U1IsAddonEnabled'battleinfo')
function BF:PLAYER_ENTERING_WORLD()
if(select(2, IsInInstance()) == 'pvp') then
self:RegisterEvent'PLAYER_DEAD'
if(not no_scoreboard) then
self:RegisterEvent'UPDATE_BATTLEFIELD_SCORE'
RequestBattlefieldScoreData()
if(not requestTimer) then
requestTimer = self:ScheduleRepeatingTimer(requestBoard, 5)
end
end
else
self:UnregisterEvent'PLAYER_DEAD'
self:UnregisterEvent'UPDATE_BATTLEFIELD_SCORE'
if(requestTimer) then
self:CancelRequestTimer()
end
end
end
end
BF:Init()
--------------------------------
-- 队友地图标记颜色 --
--------------------------------
local has_mapster = select(5, GetAddOnInfo'mapster') ~= 'MISSING'
if(not has_mapster) then return end
local RaidColor = {}
LibStub('AceEvent-3.0'):Embed(RaidColor)
LibStub('AceTimer-3.0'):Embed(RaidColor)
U1PLUGIN_ColorRostersOnMap = RaidColor
local path = [[Interface\AddOns\Mapster\Artwork\]]
local _size = 18
local stop_timer, start_timer
local noop = function() end
local function fixUnit(tex, unit)
local _, _, subgroup = GetRaidRosterInfo(string.sub(unit, 5)+0)
local _, class = UnitClass(unit)
tex:SetTexture(path..'Group'..subgroup)
if(not tex._SetVertexColor) then
tex._SetVertexColor = tex.SetVertexColor
tex.SetVertexColor = noop
end
local t = RAID_CLASS_COLORS[class]
if(GetTime() % 1 < .5) then
if UnitAffectingCombat(unit) then
tex:_SetVertexColor(.8, 0, 0)
elseif UnitIsDeadOrGhost(unit) then
tex:_SetVertexColor(.2, .2, .2)
elseif PlayerIsPVPInactive(unit) then
tex:_SetVertexColor(.5, .2, .8)
end
elseif t then
tex:_SetVertexColor(t.r, t.g, t.b)
else
tex:_SetVertexColor(.8, .8, .8)
end
end
local function fixBMunit(i, unit)
local tex = _G['BattlefieldMinimapRaid'..i..'Icon']
fixUnit(tex, unit)
end
local function fixWMunit(i, unit)
local tex = _G['WorldMapRaid'..i..'Icon']
fixUnit(tex, unit)
if(not tex._SetTexCoord) then
tex._SetTexCoord = tex.SetTexCoord
tex.SetTexCoord = noop
end
tex:_SetTexCoord(0, 1, 0, 1)
_G['WorldMapRaid'..i]:SetSize(_size, _size)
end
local function group_icon_update()
local battlefiledmapShown = BattlefieldMinimap and BattlefieldMinimap:IsVisible()
local worldMapShown = WorldMapFrame:IsVisible()
if not (battlefiledmapShown or worldMapShown) then
return stop_timer()
end
--no BattlefieldMinimapRaid1 in 7.1
--[[local numRaid = GetNumGroupMembers()
if(numRaid > 0) then
for i = 1, numRaid do
local bmunit = battlefiledmapShown and _G['BattlefieldMinimapRaid'..i].unit
local wmunit = _G['WorldMapRaid' .. i] and _G['WorldMapRaid' .. i].unit
if(battlefiledmapShown and bmunit) then
fixBMunit(i, bmunit)
end
if(worldMapShown and wmunit) then
fixWMunit(i, wmunit)
end
end
end--]]
end
do
local _timer
function stop_timer()
if(_timer) then
RaidColor:CancelTimer(_timer)
_timer = nil
end
end
function start_timer()
if(not _timer) then
_timer = RaidColor:ScheduleRepeatingTimer(group_icon_update, .5)
end
end
end
function RaidColor:WORLD_MAP_UPDATE()
return start_timer()
end
function RaidColor:Init()
if(U1GetCfgValue(addonName, 'map_raid_color')) then
return self:Enable()
else
return self:Disable()
end
end
function RaidColor:Enable()
if(self._enabled) then return end
self._enabled = true
self:RegisterEvent'WORLD_MAP_UPDATE'
start_timer()
end
function RaidColor:Disable()
if(not self._enabled) then return end
self._enabled = false
self:UnregisterEvent'WORLD_MAP_UPDATE'
stop_timer()
-- return self:Restore()
end
-- local recover = function(icon, texture)
-- if(icon and texture) then
-- return icon:SetTexture(texture)
-- end
-- end
-- function RaidColor:Restore()
-- for i = 1, 40 do
-- local tex = _G['BattlefieldMinimapRaid'..i..'Icon']
-- recover(tex, [[Interface\WorldMap\WorldMapPartyIcon]])
-- local tex = _G['WorldMapRaid'..i..'Icon']
-- recover(tex, [[Interface\Minimap\PartyRaidBlips]])
-- end
-- end
-- done in config
-- RaidColor:Init()
|
Esophageal motility, heartburn, and gastroesophageal reflux: variations in clinical presentation of esophageal dysphagia.
Dysphagia is a potentially important symptom, often leading to the finding of an anatomical or motility disorder of the esophagus. Dysphagia and heartburn represent two of the most common symptoms associated with esophageal motility disorders. To explore the relationship of symptomatic esophageal dysphagia and heartburn and their association with primary esophageal motor disorders, we have performed a retrospective assessment of 1035 patient evaluations performed at our gastrointestinal laboratory. A clear statistical association of symptomatic dysphagia and heartburn was established; however, no pattern diagnostic of a specific motility disorder was discernible. A sizable fraction of our patient population with dysphagia demonstrated normal esophageal motility. A significant portion of dyspeptic patients exhibited both normal motility and acid exposure. The differences observed between the incidence of subjective symptoms and objective dysfunction may be explained in part by an altered or increased esophageal sensitivity of these patients. |
Preview images for The 100 episode "The Warriors Will"
“The Warriors Will” is the title of the tenth episode of The 100 Season 5, and we’re assuming it will be airing on Tuesday, July 17. We don’t have an official description yet, nor do we know if there is supposed to be an apostrophe in the episode title.
Could it be “The Warrior’s Will?” Or “The Warriors’ Will?” We’ll surely find out soon enough.
UPDATE: The official description didn’t clarify, but we do have the description now! Here it is, complete with some new spoilers. Henry Ian Cusick is directing!
HENRY IAN CUSICK DIRECTS THE EPISODE — Monty (Chris Larkin) strives to show Wonkru an alternative to war, and to the valley itself. Meanwhile, Abby’s (Paige Turco) health continues to deteriorate, along with McCreary’s (guest star William Miller) patience. Eliza Taylor, Bob Morley, Paige Turco, Henry Ian Cusick, Marie Avgeropoulos, Lindsey Morgan, Richard Harmon and Tasya Teles also star. Henry Ian Cusick directed the episode written by Julie Benson & Shawna Benson (#510). Original airdate 7/17/2018. |
s - 1/4. Let o = -3 - -2.8. Which is the second biggest value? (a) p (b) o (c) -1/2
b
Let i = 3.98 - 4. Let m = i - 1.98. Which is the third smallest value? (a) m (b) -0.1 (c) 5
c
Let j = -4 - -9. Let x = j + -4.7. Let m = x - 1.3. Which is the second biggest value? (a) -2/25 (b) 0.5 (c) m
a
Let m be 2*(10/(-16) - -1). Let k = 9 - 10. What is the smallest value in m, k, 0?
k
Let k = 0.826 + -0.84. Which is the third smallest value? (a) -5 (b) 0.3 (c) k
b
Let u = -0.6 + 0.71. Let j = u - 5.11. Let z = -0.41 - -0.21. Which is the third smallest value? (a) 3 (b) j (c) z
a
Let h = -2.84 + 3. Let z = 5.84 + h. Let p = z + -6.2. Which is the third biggest value? (a) 0 (b) -2/23 (c) p
c
Let m = -89/21 + 32/7. What is the biggest value in -4, m, -8?
m
Let i be 12/2 + (-8)/1. Let d = 6.3 - -1.7. Let f = -7.7 + d. Which is the smallest value? (a) i (b) 3 (c) f
a
Let j = 0.461 - 0.161. What is the second biggest value in 2, -2, j, -5/3?
j
Let l = -5 + 5.3. Suppose 0 = -4*m + 4*o + 48, -3 = 3*o - 0. Let q be 8/m*2/8. What is the second biggest value in l, 2/13, q?
q
Let r = -5 - -8. Let q = r - 5. Let f be (1 - 14/10)*(-40)/(-24). What is the smallest value in -5, q, f?
-5
Suppose -5*t - 5*t = 20. Which is the second biggest value? (a) -4/5 (b) 3 (c) t
a
Let m = 517/870 - -1/174. Let q = -0.4 - -3.4. What is the third biggest value in m, -1, q?
-1
Suppose -p = -0 - 2. Let m = -0.6 - -0.8. Let f = -0.1 - 0.1. Which is the second smallest value? (a) m (b) p (c) f
a
Let m be 0 + (483/(-55))/7. Let o be 16/(-11) + 1/1. Let g = o - m. What is the second biggest value in -1/9, 4, g?
g
Let y = 3.165 - 0.165. Which is the smallest value? (a) -0.5 (b) y (c) -19
c
Let m be ((-2)/(-8))/((-44)/16). What is the second biggest value in 2/9, m, -7?
m
Let w = 0.07 - -13.93. Let h = 14 - w. Which is the smallest value? (a) -5 (b) h (c) -13
c
Let o = -7 + 11. Let l be 1/(-3) - 11/33. Which is the second biggest value? (a) l (b) -0.3 (c) o
b
Let q(m) = 3*m**2 - 46*m - 29. Let f be q(16). Let l(p) = -p**2 - 3*p. Let u be l(-3). Which is the second biggest value? (a) -3/4 (b) f (c) u
c
Let r = 74 - 104. Let k be (-15)/(-12)*4/r. Let g = 0.25 + 0.75. Which is the biggest value? (a) k (b) g (c) -0.2
b
Let g be (-6)/(-33) - 57/11. Which is the second smallest value? (a) -2 (b) 0.1 (c) g
a
Let b = -1.5 + 8.5. Let m = -5 + b. What is the second smallest value in -4, 1/3, m?
1/3
Let d be (5 + -2)/(-2 + 1). Suppose -5*z = -5*h - 40, 0*h = -5*z + 3*h + 30. Which is the biggest value? (a) d (b) 0.5 (c) z
c
Let m = -30 + 208/7. Which is the fourth biggest value? (a) -2 (b) 0.4 (c) 5 (d) m
a
Let y = 0.8 + -1.8. Let i = -2.5 - -2.3. What is the smallest value in 0.2, i, y?
y
Suppose -1 = -w - 9. Let x = w - -11. Let i be 1/(-10*x/(-18)). Which is the smallest value? (a) 0 (b) i (c) 0.5
a
Let m be 32/18 - (-30)/135. Let g be 1/4*4/6. Which is the biggest value? (a) g (b) m (c) -1
b
Let w be 4/10 - 36/(-10). Let s be (-22)/(-11) + (-17142)/7070. Let h = s + -2/505. What is the biggest value in -1/2, h, w?
w
Let l(z) = z**2 - 6*z - 8. Let d be l(7). Which is the third biggest value? (a) -3 (b) 3 (c) d
a
Let j = 0.266 + 0.034. Let x = 3/34 + 11/68. Let g = -0.3 - -0.2. What is the third biggest value in x, j, g?
g
Let v = 10 + -9.95. Let g = 1 - 1.25. Let n = g - v. Which is the second biggest value? (a) 2 (b) n (c) -1
b
Let f = 0.21 - -0.09. What is the second biggest value in -2, 1/4, f, -0.5?
1/4
Let z = -0.14 + -6.86. Let r = z + 6. What is the second smallest value in -0.5, 0, r?
-0.5
Let d = -1.2 - -13.2. Let l = 5 - d. Let n = 12 + l. Which is the third smallest value? (a) n (b) -3 (c) 1
a
Let u be 3 + 10/(-4) + -1. What is the second biggest value in 2/13, -2/3, u?
u
Let d = -1/11 + -53/22. Let f = -59.5 - -55.43. Let v = -0.07 - f. Which is the third smallest value? (a) d (b) v (c) 0
b
Let q = 31 - 33. Let o = -5 + 5.3. What is the smallest value in o, q, -0.3?
q
Let w = 26 + -26.2. Which is the biggest value? (a) 0.1 (b) w (c) -5
a
Let y = -9.4 - -15.7. Let k = y - 6. Let x = k - 3.3. Which is the third smallest value? (a) 2/3 (b) 5 (c) x
b
Let t = -1.9 - 0.1. Let p = t + 6. Let x = p + -4.4. What is the third biggest value in -2/13, 0.2, x?
x
Let z = 0 - 0.4. Suppose -2*r + 2*g - 92 = 0, -3*r + 4*g - 140 = -0*g. Let o be r/(-36) + 4/(-18). Which is the third biggest value? (a) -4 (b) o (c) z
a
Let n = -0.24 - -0.54. What is the biggest value in n, -3, -0.05?
n
Let j = 30 - 29.6. Which is the fourth biggest value? (a) -0.2 (b) 0.3 (c) 2 (d) j
a
Let j be 2 - ((-64)/(-12))/2. Let l = 8 + -4. Let f = 4 + -3.7. What is the third smallest value in j, l, f?
l
Let y be (-2)/(-8) - (-41)/(-4). Let o = 8 + y. Let h = -1 + -3. What is the third smallest value in h, 0.4, o?
0.4
Let b = 115/2 + -59. What is the second smallest value in -2/11, b, 2/13?
-2/11
Let f be ((-2)/(-6))/((-5)/(-15)). Suppose -21 = -5*q - f. Suppose q*z - 2*z = -10. Which is the second biggest value? (a) z (b) -2 (c) -4
c
Let y(j) = j**2 - 2*j - 5. Suppose 3*z - 2 - 10 = 0. Let p be y(z). Let g be p + (-12)/7 + -1. What is the biggest value in 1/5, g, -2?
g
Let g = -519/5 - -103. Let s = -2/5 - g. Which is the third biggest value? (a) -4 (b) 3/5 (c) s
a
Let d = 1 + 1. Let w = -2.04 - 0.16. Let z = w + d. What is the second biggest value in z, -1, -2?
-1
Let b = -113/5 + 116/5. What is the second smallest value in 1, b, -0.3?
b
Let m = -0.46 - -0.06. Let p = 10.97 + 0.03. Let d = 11 - p. Which is the second smallest value? (a) m (b) d (c) 4
b
Let c = -0.8 - -0.5. What is the smallest value in 0, c, -3?
-3
Let h be ((-4)/(-10))/((-63)/15). Which is the second biggest value? (a) 0.3 (b) -1/5 (c) 1/2 (d) h
a
Let j be 8/(-20) - 1714/(-360). Let l = j - 46/9. Let y(v) = -v**3 + 2*v**2 + v - 2. Let g be y(2). Which is the biggest value? (a) l (b) g (c) 5
c
Let a be (-14)/63 + 24/27. Let c = 34 + -65/2. What is the biggest value in -1, c, a?
c
Let s = -4 - -4. Which is the biggest value? (a) s (b) -0.02 (c) 2/7
c
Let x = 17 - 13. Which is the second biggest value? (a) 0.2 (b) x (c) 0.3
c
Let f = -23 - -23.5. What is the third smallest value in 3/4, f, 0.09?
3/4
Let l be (-3)/(0 + 12/(-8)). Which is the second biggest value? (a) -1/5 (b) l (c) 3
b
Suppose 4*n + 2*w + 12 = 0, 3*n + 0*n + 18 = 3*w. Let k be 10/(-35)*(-2)/n. Which is the second biggest value? (a) k (b) 4 (c) 0.4
c
Let f(m) = m**3 - 6*m**2 + 5*m + 7. Let l be f(5). Let u be (-6)/(-21)*l/9. Which is the third smallest value? (a) -0.06 (b) -3/8 (c) u
c
Let j be (27/12 - 2)*2. What is the biggest value in 1/5, -0.3, j?
j
Let v = 1.07 + -0.07. Which is the second smallest value? (a) 0.3 (b) v (c) 3/5
c
Let l be 59/84 - (-8)/(-28). Let g = -1/6 + l. What is the second biggest value in 2/7, -0.4, g?
g
Let z = -0.45 + 0.05. Let p = -105.5 - -105. Which is the third smallest value? (a) z (b) 2 (c) p
b
Let m be (-11)/3 - (-3)/(-9). Let u = 104 - 99. Which is the second smallest value? (a) u (b) m (c) -0.2
c
Let h = 12.8 + -13. Let r = 2036987/7560 + -64/945. Let t = -269 + r. What is the smallest value in h, t, -4?
-4
Let i be 2/(-12) + 146/12. Let y = -8 + i. Which is the second biggest value? (a) 1 (b) 5 (c) y
c
Let h(q) = -q**2 - 6*q + 4. Let b be h(-6). Let g(v) = 3*v**2 - 1. Let t be g(1). Let y = t - 5. What is the second smallest value in y, 0.4, b?
0.4
Suppose 18 = 31*i - 33*i. Which is the smallest value? (a) -1/3 (b) i (c) 4
b
Suppose -6*y + 4*y = 6. Which is the smallest value? (a) 4 (b) 0.5 (c) y
c
Suppose 4*g - 6*g = 0, -2*y = -3*g. Let v = -7 - -6.9. Let t = -0.3 - v. Which is the smallest value? (a) t (b) y (c) 3
a
Let y be (-2*(0 + 1))/(24 - 25). Which is the second smallest value? (a) 7 (b) y (c) -2/11
b
Let q(j) = j**3 + 3*j**2 + j + 2 - 3*j - 2*j**2. Let v be q(-3). Let a(o) = -o - 6. Let g be a(v). What is the smallest value in -5, -0.3, g?
-5
Let f = -87 + 86.9. What is the smallest value in 0.2, f, -4, 2?
-4
Let d = 12017/12 - 1002. Let k = d - -1/4. Which is the biggest value? (a) -5 (b) k (c) -0.5
b
Let t = 0 - -0.4. Which is the smallest value? (a) t (b) -4 (c) 4
b
Let x(g) = -2*g**2 + 2*g**2 - 15 + 14 + |
Dr. J.D. Watts House
The Dr. J.D. Watts House is a historic house located at 205 West Choctaw Street in Dumas, Arkansas. It is a well preserved local example of a transitional Queen Anne/Colonial Revival residence.
Description and history
The 1-1/2 story timber-framed house was built around 1909 by a Mr. Williams, and was purchased by Dr. James David Watts in 1918, when he moved to the area. It has a hip roof, with cross-gable dormers on the sides and rear, and a large projecting gable-end dormer centered on the front facade. This dormer features a Palladian window, with the surrounding walls covered in diamond-cut and fish-scale shingles. The gable itself is decorated with jigsaw-cut boards. There is a single story porch, supported by Tuscan columns, that wraps around both sides of the house. The front entry is flanked by sidelight windows and pilasters supporting an entablature.
The house was listed on the National Register of Historic Places on December 9, 1994.
See also
National Register of Historic Places listings in Desha County, Arkansas
References
Category:Houses on the National Register of Historic Places in Arkansas
Category:Queen Anne architecture in Arkansas
Category:Colonial Revival architecture in Arkansas
Category:Houses in Desha County, Arkansas
Category:Houses completed in 1909
Category:National Register of Historic Places in Desha County, Arkansas |
Let f(z) be the second derivative of 0 - 15*z - 11/6*z**3 + 1/2*z**2. What is the derivative of f(d) wrt d?
-11
Find the second derivative of -47*x**2 - 32 - 15*x**2 + 32 + 3*x**2 + 40*x wrt x.
-118
Let i(w) be the first derivative of 0*w - 5*w**2 - 1/2*w**4 + 0*w**3 + 16. What is the second derivative of i(u) wrt u?
-12*u
Let g(s) = -s + 35. Let d be g(13). What is the second derivative of -2*b - 43*b**2 + 22 - d - 12*b wrt b?
-86
Let w(x) = -x**2 + 3. Let b(f) = -3*f**2 - 9*f + 247. Let a(c) = -b(c) + 6*w(c). What is the derivative of a(d) wrt d?
-6*d + 9
Let r(n) = -n + 7. Let o be r(3). Suppose -3*l + o + 5 = 0. Differentiate 2*c**l - 8*c**4 - 2*c**3 - 11 - 5*c**4 wrt c.
-52*c**3
Let t(n) = -47*n**3 + 2*n + 8*n**2 + 96 - 96. Let s(m) = -48*m**3 + 8*m**2 + 3*m. Let f(l) = 2*s(l) - 3*t(l). Find the third derivative of f(b) wrt b.
270
Let k(f) be the second derivative of 2/5*f**6 + 11/6*f**3 + 5*f + 0*f**2 + 0*f**5 + 0*f**4 + 0. What is the second derivative of k(t) wrt t?
144*t**2
Let c be ((-5)/2)/(1/(-2)). What is the second derivative of -4*v**5 + v + 0*v**5 + 7*v**5 - 2*v**c wrt v?
20*v**3
Suppose -4*d + 0*d - 4*v = -24, d - 4*v = 21. What is the derivative of 29*h + d - 44*h + 28*h wrt h?
13
Find the third derivative of 24*t**2 + 1211*t**5 - 1377*t**5 - t + t wrt t.
-9960*t**2
Let t(f) be the first derivative of -24/5*f**5 + 43 + 0*f**4 + 7*f**3 + 0*f + 0*f**2. What is the third derivative of t(h) wrt h?
-576*h
Let p(u) be the second derivative of -u**6/10 - 6*u**5/5 + 7*u**3/3 - u**2/2 - 76*u + 2. What is the second derivative of p(q) wrt q?
-36*q**2 - 144*q
Let q(x) = 112*x**2 - 42*x + 4. Let f(h) = 338*h**2 - 126*h + 11. Let r(t) = -3*f(t) + 8*q(t). Find the second derivative of r(l) wrt l.
-236
Let d(v) = 3*v + 5. Let b be d(3). Find the second derivative of x + 6*x - 28*x**2 + b*x**2 + 4*x**2 wrt x.
-20
Find the second derivative of 3*g - 1 + 4*g**4 + 4 + 5*g**4 + 19*g**4 wrt g.
336*g**2
Let b(x) be the first derivative of x**5/5 + 72*x - 74. Differentiate b(k) wrt k.
4*k**3
Let x(l) be the second derivative of 7*l**6/10 + l**5/20 + 767*l**4/12 + 88*l. Find the third derivative of x(n) wrt n.
504*n + 6
Let i(b) = b**4 - b**3 + b**2 - b + 2. Let s(u) = -30*u**4 + 4*u**3 - 4*u**2 + 6*u - 105. Let j(h) = -4*i(h) - s(h). What is the derivative of j(v) wrt v?
104*v**3 - 2
Find the second derivative of -4*i**2 + i + 50 + 76 + 89 + 39*i**2 + 83*i**2 wrt i.
236
Let x(b) = -b**3 - 152*b**2 - 755. Let u(k) = 3*k**3 + 454*k**2 + 2263. Let c(t) = 6*u(t) + 17*x(t). Find the first derivative of c(v) wrt v.
3*v**2 + 280*v
Let p(x) = -4*x**2 - 3*x + 7. Let m(c) = -7*c**2 - 5*c + 14. Suppose 3*t + 3 - 18 = 0. Let z(i) = t*p(i) - 3*m(i). Differentiate z(d) wrt d.
2*d
Let c(h) be the second derivative of h**8/14 - h**5/10 - 43*h**4/6 + 2*h + 15. Find the third derivative of c(o) wrt o.
480*o**3 - 12
Let c(i) be the second derivative of 3*i**8/28 + i**6/30 + 40*i**4/3 - 138*i. Find the third derivative of c(f) wrt f.
720*f**3 + 24*f
Let a = 28 - -47. What is the second derivative of -a*r - 12*r**4 + 151*r - 49*r wrt r?
-144*r**2
Let v(h) be the first derivative of 0*h**4 + 12/7*h**7 + 0*h**2 + 0*h**6 + h**3 + 3 + 0*h**5 + 0*h. What is the third derivative of v(c) wrt c?
1440*c**3
Suppose -m - 2*m = -3. Suppose 4*a = 3*i + 1, 5*a - m = i + 3*i. Differentiate 0*v**3 + v**3 - i - 4*v**3 with respect to v.
-9*v**2
Let n(q) = -27*q**2 + 2*q - 13. Let m(u) = u. Let p be (3/(-1))/(1/2 + 1). Let o(s) = p*m(s) + n(s). What is the first derivative of o(t) wrt t?
-54*t
Let q(f) = -453*f**4 + 4*f**3 + 2*f**2 + 455. Let c(v) = 927290*v**4 - 8190*v**3 - 4095*v**2 - 931385. Let d(u) = -2*c(u) - 4095*q(u). Differentiate d(n) wrt n.
1820*n**3
Let k(s) be the second derivative of 103*s**5/20 - s**4/4 - s**3/2 - 8*s + 1. Find the third derivative of k(g) wrt g.
618
Suppose 0 = p + 2*r, 0 = -5*r - 5. What is the derivative of u**2 + 0*u**p + 3941 - 3986 + 8*u wrt u?
2*u + 8
What is the second derivative of -74*w**3 - 29*w**3 + 76*w - 27*w wrt w?
-618*w
Let i(h) be the first derivative of 3*h**4/2 - h**3 + 37*h**2/2 + 40. What is the second derivative of i(m) wrt m?
36*m - 6
Let q(w) = -377*w**5 - 2*w**3 - 62*w - 1. Let l(j) = 1884*j**5 + 11*j**3 + 310*j + 4. Let r(c) = -2*l(c) - 11*q(c). What is the second derivative of r(z) wrt z?
7580*z**3
Let b(u) = 10*u**2 - 3*u - 405. Let x(l) = -20*l**2 + 6*l + 813. Let f(t) = -11*b(t) - 6*x(t). What is the derivative of f(y) wrt y?
20*y - 3
Let z(k) be the first derivative of -k**6/12 - k**5/30 + 10*k**3 - 32. Let m(x) be the third derivative of z(x). What is the second derivative of m(y) wrt y?
-60
Find the third derivative of 15*i**5 + i**2 - 47*i**2 + 2*i**4 + 24*i**5 - 48*i**2 wrt i.
2340*i**2 + 48*i
Let l(b) be the first derivative of 29*b**2/2 + 33*b + 10. Let i(c) = 43*c + 50. Let z(f) = -5*i(f) + 7*l(f). Find the first derivative of z(v) wrt v.
-12
Find the second derivative of -1534*m**5 - 25*m - 3*m - 8*m + 1541*m**5 - 30*m - m**4 wrt m.
140*m**3 - 12*m**2
Let b(c) = -5*c + 29. Let a(f) = 24*f - 147. Let p(h) = -6*a(h) - 33*b(h). Differentiate p(m) wrt m.
21
Let n = -20 - -27. What is the first derivative of 7*d + 22 + n*d - 21 wrt d?
14
Let u(d) = 47*d - 349. Let k(j) = 95*j - 686. Let t(q) = 3*k(q) - 5*u(q). Differentiate t(r) wrt r.
50
Let p = -255 + 165. Let a(f) = 198*f**3 + 45*f + 243. Let l(v) = 9*v**3 + 2*v + 11. Let x(y) = p*l(y) + 4*a(y). Differentiate x(t) with respect to t.
-54*t**2
Let w(t) = -t + 21 - 36 - 24. Let v(b) = 37. Let s(h) = 4*v(h) + 3*w(h). Differentiate s(o) with respect to o.
-3
Let y(r) be the third derivative of -71*r**7/70 + 239*r**5/60 - r**2 - 140. Find the third derivative of y(v) wrt v.
-5112*v
Let b(p) be the second derivative of 41*p**6/30 - p**3/3 + 9*p**2/2 + 442*p. Find the second derivative of b(t) wrt t.
492*t**2
What is the third derivative of -6*j**3 + 7*j**3 - 228*j**2 + 123*j**2 + 2*j**4 + 13*j**4 wrt j?
360*j + 6
Let d(k) = 53*k**3 - 5*k**2 - 3*k + 31. Let y(x) = 54*x**3 - 6*x**2 - 4*x + 30. Let v(o) = -4*d(o) + 3*y(o). Differentiate v(i) with respect to i.
-150*i**2 + 4*i
Let r(n) be the third derivative of 15*n**8/56 - 5*n**4/24 + n**3/6 + 75*n**2. Find the second derivative of r(t) wrt t.
1800*t**3
Find the first derivative of -18 - 180 - 413*g - 137 + 6 wrt g.
-413
Let y(m) be the first derivative of 7*m**5/5 - 10*m**3/3 - 134*m - 60. Find the first derivative of y(i) wrt i.
28*i**3 - 20*i
Suppose -7 = 4*v - 5*v + 2*p, 4*v = 4*p + 20. Find the second derivative of w - 4*w**2 + v*w**2 - 3*w - 5*w**2 wrt w.
-12
Let r(s) = 5*s**3 + 1 + 13*s**2 - 15*s**2 - 2*s**3 + 2*s**3. Let z be r(1). Differentiate -z*g - 1 + g**2 + 4*g wrt g.
2*g
Let l(s) be the first derivative of s**3/3 - 5*s**2/2 + 2*s + 2. Let f be l(5). Differentiate -3 - q + f*q - 3*q wrt q.
-2
What is the second derivative of 163*g**4 - 454596*g**2 + 454596*g**2 - 387*g wrt g?
1956*g**2
Suppose -3*k - 125 = -734. Find the second derivative of 12*d + 104*d**4 - k*d**4 + 86*d**4 wrt d.
-156*d**2
Differentiate -8532*d**3 - 83*d**4 + 8532*d**3 + 69 with respect to d.
-332*d**3
Let q(d) = 146*d**2 - 176. Let u(c) = -4378*c**2 + 5280. Let p(z) = 121*q(z) + 4*u(z). What is the first derivative of p(s) wrt s?
308*s
Find the third derivative of o**6 + 275*o**2 - 409*o**2 - 282*o**4 + 473*o**4 wrt o.
120*o**3 + 4584*o
Suppose 2*j = 7*j - 15. Suppose 0*i + b - 48 = -j*i, -3*b = 0. Differentiate -11 + 8*o**2 - i*o**2 + 4 wrt o.
-16*o
Find the first derivative of 60*g + 102*g - 210 + 8*g wrt g.
170
Differentiate 126 + 51*n**2 + 9*n + 7*n + n**3 - 33*n + 17*n wrt n.
3*n**2 + 102*n
Suppose -2 = 4*v - 26. Let k(w) = -w**3 + 7*w**2 - 6*w + 5. Let r be k(v). What is the second derivative of -3*o**5 - 5*o - 2*o**r + 9*o + 3*o**5 wrt o?
-40*o**3
Let i(p) be the first derivative of p - 7 + p - 87*p**2 + 71*p**2. What is the derivative of i(a) wrt a?
-32
Let t be (0/(-2))/(-17 + 24). Let a(s) be the first derivative of -1/2*s**4 + s**3 + t*s + 0*s**2 - 3. Find the third derivative of a(k) wrt k.
-12
Let s(t) be the second derivative of -t**6/3 - 3*t**5/10 - 8*t**2 - 17*t. Let q(x) = x**4 - x**3 - 1. Let p(m) = 6*q(m) - s(m). Differentiate p(n) wrt n.
64*n**3
Let n(p) be the second derivative of -17*p**6/6 + 17*p**3/6 - p**2 - 5*p |
Q:
Differentiability of $a:K\to \mathcal{B}(E,F)$?
Assume that $K\subseteq \mathbb{R}$ is an open subset and $E,F$ are Banach spaces.
Denote $\mathcal{B}(E,F)$ to be collection of all bounded linear operators from $E$ into $F.$
Endow $\mathcal{B}(E,F)$ with the strong operator topology.
Question: Suppose that $a:K\to \mathcal{B}(E,F)$ is a bounded linear operator.
To show that $a$ is differentiable at some $t_0\in K,$ what should we show?
I think we can show the following:
Fix $t_0\in K.$
For $e\in E,$ if $a(\cdot)e:K\to F$ is differentiable at $t_0,$ then we can conclude that $a$ is differentiable at $t_0.$
Am I right?
A:
You're right.
$$\begin{eqnarray} &a(t)\text{ is differentiable at }t=t_0& \\\iff& \text{strong limit } \lim_{h\to 0} \frac{a(t_0+h)-a(t_0)}{h}&\text{ exists in }\mathcal{B}(E,F). \\\iff&
\forall e \in E, \;\lim_{h\to 0} \left(\frac{a(t_0+h)-a(t_0)}{h}\right)(e)&\text{ exists in }F.\\
\iff&\forall e\in E,\; \lim_{h\to 0} \left(\frac{a(t_0+h)e-a(t_0)e}{h}\right)&\text{ exists in }F.\\
\iff&\forall e\in E,\; t\mapsto a(t)e\text{ is differentiable at }&t=t_0.
\end{eqnarray}$$
|
package openpgp
import (
"bytes"
"crypto"
"strings"
"testing"
"time"
"golang.org/x/crypto/openpgp/errors"
"golang.org/x/crypto/openpgp/packet"
)
func TestKeyExpiry(t *testing.T) {
kring, err := ReadKeyRing(readerFromHex(expiringKeyHex))
if err != nil {
t.Fatal(err)
}
entity := kring[0]
const timeFormat = "2006-01-02"
time1, _ := time.Parse(timeFormat, "2013-07-01")
// The expiringKeyHex key is structured as:
//
// pub 1024R/5E237D8C created: 2013-07-01 expires: 2013-07-31 usage: SC
// sub 1024R/1ABB25A0 created: 2013-07-01 23:11:07 +0200 CEST expires: 2013-07-08 usage: E
// sub 1024R/96A672F5 created: 2013-07-01 23:11:23 +0200 CEST expires: 2013-07-31 usage: E
//
// So this should select the newest, non-expired encryption key.
key, _ := entity.encryptionKey(time1)
if id, expected := key.PublicKey.KeyIdShortString(), "96A672F5"; id != expected {
t.Errorf("Expected key %s at time %s, but got key %s", expected, time1.Format(timeFormat), id)
}
// Once the first encryption subkey has expired, the second should be
// selected.
time2, _ := time.Parse(timeFormat, "2013-07-09")
key, _ = entity.encryptionKey(time2)
if id, expected := key.PublicKey.KeyIdShortString(), "96A672F5"; id != expected {
t.Errorf("Expected key %s at time %s, but got key %s", expected, time2.Format(timeFormat), id)
}
// Once all the keys have expired, nothing should be returned.
time3, _ := time.Parse(timeFormat, "2013-08-01")
if key, ok := entity.encryptionKey(time3); ok {
t.Errorf("Expected no key at time %s, but got key %s", time3.Format(timeFormat), key.PublicKey.KeyIdShortString())
}
}
func TestMissingCrossSignature(t *testing.T) {
// This public key has a signing subkey, but the subkey does not
// contain a cross-signature.
keys, err := ReadArmoredKeyRing(bytes.NewBufferString(missingCrossSignatureKey))
if len(keys) != 0 {
t.Errorf("Accepted key with missing cross signature")
}
if err == nil {
t.Fatal("Failed to detect error in keyring with missing cross signature")
}
structural, ok := err.(errors.StructuralError)
if !ok {
t.Fatalf("Unexpected class of error: %T. Wanted StructuralError", err)
}
const expectedMsg = "signing subkey is missing cross-signature"
if !strings.Contains(string(structural), expectedMsg) {
t.Fatalf("Unexpected error: %q. Expected it to contain %q", err, expectedMsg)
}
}
func TestInvalidCrossSignature(t *testing.T) {
// This public key has a signing subkey, and the subkey has an
// embedded cross-signature. However, the cross-signature does
// not correctly validate over the primary and subkey.
keys, err := ReadArmoredKeyRing(bytes.NewBufferString(invalidCrossSignatureKey))
if len(keys) != 0 {
t.Errorf("Accepted key with invalid cross signature")
}
if err == nil {
t.Fatal("Failed to detect error in keyring with an invalid cross signature")
}
structural, ok := err.(errors.StructuralError)
if !ok {
t.Fatalf("Unexpected class of error: %T. Wanted StructuralError", err)
}
const expectedMsg = "subkey signature invalid"
if !strings.Contains(string(structural), expectedMsg) {
t.Fatalf("Unexpected error: %q. Expected it to contain %q", err, expectedMsg)
}
}
func TestGoodCrossSignature(t *testing.T) {
// This public key has a signing subkey, and the subkey has an
// embedded cross-signature which correctly validates over the
// primary and subkey.
keys, err := ReadArmoredKeyRing(bytes.NewBufferString(goodCrossSignatureKey))
if err != nil {
t.Fatal(err)
}
if len(keys) != 1 {
t.Errorf("Failed to accept key with good cross signature, %d", len(keys))
}
if len(keys[0].Subkeys) != 1 {
t.Errorf("Failed to accept good subkey, %d", len(keys[0].Subkeys))
}
}
func TestRevokedUserID(t *testing.T) {
// This key contains 2 UIDs, one of which is revoked:
// [ultimate] (1) Golang Gopher <no-reply@golang.com>
// [ revoked] (2) Golang Gopher <revoked@golang.com>
keys, err := ReadArmoredKeyRing(bytes.NewBufferString(revokedUserIDKey))
if err != nil {
t.Fatal(err)
}
if len(keys) != 1 {
t.Fatal("Failed to read key with a revoked user id")
}
var identities []*Identity
for _, identity := range keys[0].Identities {
identities = append(identities, identity)
}
if numIdentities, numExpected := len(identities), 1; numIdentities != numExpected {
t.Errorf("obtained %d identities, expected %d", numIdentities, numExpected)
}
if identityName, expectedName := identities[0].Name, "Golang Gopher <no-reply@golang.com>"; identityName != expectedName {
t.Errorf("obtained identity %s expected %s", identityName, expectedName)
}
}
// TestExternallyRevokableKey attempts to load and parse a key with a third party revocation permission.
func TestExternallyRevocableKey(t *testing.T) {
kring, err := ReadKeyRing(readerFromHex(subkeyUsageHex))
if err != nil {
t.Fatal(err)
}
// The 0xA42704B92866382A key can be revoked by 0xBE3893CB843D0FE70C
// according to this signature that appears within the key:
// :signature packet: algo 1, keyid A42704B92866382A
// version 4, created 1396409682, md5len 0, sigclass 0x1f
// digest algo 2, begin of digest a9 84
// hashed subpkt 2 len 4 (sig created 2014-04-02)
// hashed subpkt 12 len 22 (revocation key: c=80 a=1 f=CE094AA433F7040BB2DDF0BE3893CB843D0FE70C)
// hashed subpkt 7 len 1 (not revocable)
// subpkt 16 len 8 (issuer key ID A42704B92866382A)
// data: [1024 bits]
id := uint64(0xA42704B92866382A)
keys := kring.KeysById(id)
if len(keys) != 1 {
t.Errorf("Expected to find key id %X, but got %d matches", id, len(keys))
}
}
func TestKeyRevocation(t *testing.T) {
kring, err := ReadKeyRing(readerFromHex(revokedKeyHex))
if err != nil {
t.Fatal(err)
}
// revokedKeyHex contains these keys:
// pub 1024R/9A34F7C0 2014-03-25 [revoked: 2014-03-25]
// sub 1024R/1BA3CD60 2014-03-25 [revoked: 2014-03-25]
ids := []uint64{0xA401D9F09A34F7C0, 0x5CD3BE0A1BA3CD60}
for _, id := range ids {
keys := kring.KeysById(id)
if len(keys) != 1 {
t.Errorf("Expected KeysById to find revoked key %X, but got %d matches", id, len(keys))
}
keys = kring.KeysByIdUsage(id, 0)
if len(keys) != 0 {
t.Errorf("Expected KeysByIdUsage to filter out revoked key %X, but got %d matches", id, len(keys))
}
}
}
func TestKeyWithRevokedSubKey(t *testing.T) {
// This key contains a revoked sub key:
// pub rsa1024/0x4CBD826C39074E38 2018-06-14 [SC]
// Key fingerprint = 3F95 169F 3FFA 7D3F 2B47 6F0C 4CBD 826C 3907 4E38
// uid Golang Gopher <no-reply@golang.com>
// sub rsa1024/0x945DB1AF61D85727 2018-06-14 [S] [revoked: 2018-06-14]
keys, err := ReadArmoredKeyRing(bytes.NewBufferString(keyWithSubKey))
if err != nil {
t.Fatal(err)
}
if len(keys) != 1 {
t.Fatal("Failed to read key with a sub key")
}
identity := keys[0].Identities["Golang Gopher <no-reply@golang.com>"]
// Test for an issue where Subkey Binding Signatures (RFC 4880 5.2.1) were added to the identity
// preceding the Subkey Packet if the Subkey Packet was followed by more than one signature.
// For example, the current key has the following layout:
// PUBKEY UID SELFSIG SUBKEY REV SELFSIG
// The last SELFSIG would be added to the UID's signatures. This is wrong.
if numIdentitySigs, numExpected := len(identity.Signatures), 0; numIdentitySigs != numExpected {
t.Fatalf("got %d identity signatures, expected %d", numIdentitySigs, numExpected)
}
if numSubKeys, numExpected := len(keys[0].Subkeys), 1; numSubKeys != numExpected {
t.Fatalf("got %d subkeys, expected %d", numSubKeys, numExpected)
}
subKey := keys[0].Subkeys[0]
if subKey.Sig == nil {
t.Fatalf("subkey signature is nil")
}
}
func TestSubkeyRevocation(t *testing.T) {
kring, err := ReadKeyRing(readerFromHex(revokedSubkeyHex))
if err != nil {
t.Fatal(err)
}
// revokedSubkeyHex contains these keys:
// pub 1024R/4EF7E4BECCDE97F0 2014-03-25
// sub 1024R/D63636E2B96AE423 2014-03-25
// sub 1024D/DBCE4EE19529437F 2014-03-25
// sub 1024R/677815E371C2FD23 2014-03-25 [revoked: 2014-03-25]
validKeys := []uint64{0x4EF7E4BECCDE97F0, 0xD63636E2B96AE423, 0xDBCE4EE19529437F}
revokedKey := uint64(0x677815E371C2FD23)
for _, id := range validKeys {
keys := kring.KeysById(id)
if len(keys) != 1 {
t.Errorf("Expected KeysById to find key %X, but got %d matches", id, len(keys))
}
keys = kring.KeysByIdUsage(id, 0)
if len(keys) != 1 {
t.Errorf("Expected KeysByIdUsage to find key %X, but got %d matches", id, len(keys))
}
}
keys := kring.KeysById(revokedKey)
if len(keys) != 1 {
t.Errorf("Expected KeysById to find key %X, but got %d matches", revokedKey, len(keys))
}
keys = kring.KeysByIdUsage(revokedKey, 0)
if len(keys) != 0 {
t.Errorf("Expected KeysByIdUsage to filter out revoked key %X, but got %d matches", revokedKey, len(keys))
}
}
func TestKeyWithSubKeyAndBadSelfSigOrder(t *testing.T) {
// This key was altered so that the self signatures following the
// subkey are in a sub-optimal order.
//
// Note: Should someone have to create a similar key again, look into
// gpgsplit, gpg --dearmor, and gpg --enarmor.
//
// The packet ordering is the following:
// PUBKEY UID UIDSELFSIG SUBKEY SELFSIG1 SELFSIG2
//
// Where:
// SELFSIG1 expires on 2018-06-14 and was created first
// SELFSIG2 does not expire and was created after SELFSIG1
//
// Test for RFC 4880 5.2.3.3:
// > An implementation that encounters multiple self-signatures on the
// > same object may resolve the ambiguity in any way it sees fit, but it
// > is RECOMMENDED that priority be given to the most recent self-
// > signature.
//
// This means that we should keep SELFSIG2.
keys, err := ReadArmoredKeyRing(bytes.NewBufferString(keyWithSubKeyAndBadSelfSigOrder))
if err != nil {
t.Fatal(err)
}
if len(keys) != 1 {
t.Fatal("Failed to read key with a sub key and a bad selfsig packet order")
}
key := keys[0]
if numKeys, expected := len(key.Subkeys), 1; numKeys != expected {
t.Fatalf("Read %d subkeys, expected %d", numKeys, expected)
}
subKey := key.Subkeys[0]
if lifetime := subKey.Sig.KeyLifetimeSecs; lifetime != nil {
t.Errorf("The signature has a key lifetime (%d), but it should be nil", *lifetime)
}
}
func TestKeyUsage(t *testing.T) {
kring, err := ReadKeyRing(readerFromHex(subkeyUsageHex))
if err != nil {
t.Fatal(err)
}
// subkeyUsageHex contains these keys:
// pub 1024R/2866382A created: 2014-04-01 expires: never usage: SC
// sub 1024R/936C9153 created: 2014-04-01 expires: never usage: E
// sub 1024R/64D5F5BB created: 2014-04-02 expires: never usage: E
// sub 1024D/BC0BA992 created: 2014-04-02 expires: never usage: S
certifiers := []uint64{0xA42704B92866382A}
signers := []uint64{0xA42704B92866382A, 0x42CE2C64BC0BA992}
encrypters := []uint64{0x09C0C7D9936C9153, 0xC104E98664D5F5BB}
for _, id := range certifiers {
keys := kring.KeysByIdUsage(id, packet.KeyFlagCertify)
if len(keys) == 1 {
if keys[0].PublicKey.KeyId != id {
t.Errorf("Expected to find certifier key id %X, but got %X", id, keys[0].PublicKey.KeyId)
}
} else {
t.Errorf("Expected one match for certifier key id %X, but got %d matches", id, len(keys))
}
}
for _, id := range signers {
keys := kring.KeysByIdUsage(id, packet.KeyFlagSign)
if len(keys) == 1 {
if keys[0].PublicKey.KeyId != id {
t.Errorf("Expected to find signing key id %X, but got %X", id, keys[0].PublicKey.KeyId)
}
} else {
t.Errorf("Expected one match for signing key id %X, but got %d matches", id, len(keys))
}
// This keyring contains no encryption keys that are also good for signing.
keys = kring.KeysByIdUsage(id, packet.KeyFlagEncryptStorage|packet.KeyFlagEncryptCommunications)
if len(keys) != 0 {
t.Errorf("Unexpected match for encryption key id %X", id)
}
}
for _, id := range encrypters {
keys := kring.KeysByIdUsage(id, packet.KeyFlagEncryptStorage|packet.KeyFlagEncryptCommunications)
if len(keys) == 1 {
if keys[0].PublicKey.KeyId != id {
t.Errorf("Expected to find encryption key id %X, but got %X", id, keys[0].PublicKey.KeyId)
}
} else {
t.Errorf("Expected one match for encryption key id %X, but got %d matches", id, len(keys))
}
// This keyring contains no encryption keys that are also good for signing.
keys = kring.KeysByIdUsage(id, packet.KeyFlagSign)
if len(keys) != 0 {
t.Errorf("Unexpected match for signing key id %X", id)
}
}
}
func TestIdVerification(t *testing.T) {
kring, err := ReadKeyRing(readerFromHex(testKeys1And2PrivateHex))
if err != nil {
t.Fatal(err)
}
if err := kring[1].PrivateKey.Decrypt([]byte("passphrase")); err != nil {
t.Fatal(err)
}
const identity = "Test Key 1 (RSA)"
if err := kring[0].SignIdentity(identity, kring[1], nil); err != nil {
t.Fatal(err)
}
ident, ok := kring[0].Identities[identity]
if !ok {
t.Fatal("identity missing from key after signing")
}
checked := false
for _, sig := range ident.Signatures {
if sig.IssuerKeyId == nil || *sig.IssuerKeyId != kring[1].PrimaryKey.KeyId {
continue
}
if err := kring[1].PrimaryKey.VerifyUserIdSignature(identity, kring[0].PrimaryKey, sig); err != nil {
t.Fatalf("error verifying new identity signature: %s", err)
}
checked = true
break
}
if !checked {
t.Fatal("didn't find identity signature in Entity")
}
}
func TestNewEntityWithPreferredHash(t *testing.T) {
c := &packet.Config{
DefaultHash: crypto.SHA256,
}
entity, err := NewEntity("Golang Gopher", "Test Key", "no-reply@golang.com", c)
if err != nil {
t.Fatal(err)
}
for _, identity := range entity.Identities {
if len(identity.SelfSignature.PreferredHash) == 0 {
t.Fatal("didn't find a preferred hash in self signature")
}
ph := hashToHashId(c.DefaultHash)
if identity.SelfSignature.PreferredHash[0] != ph {
t.Fatalf("Expected preferred hash to be %d, got %d", ph, identity.SelfSignature.PreferredHash[0])
}
}
}
func TestNewEntityWithoutPreferredHash(t *testing.T) {
entity, err := NewEntity("Golang Gopher", "Test Key", "no-reply@golang.com", nil)
if err != nil {
t.Fatal(err)
}
for _, identity := range entity.Identities {
if len(identity.SelfSignature.PreferredHash) != 0 {
t.Fatalf("Expected preferred hash to be empty but got length %d", len(identity.SelfSignature.PreferredHash))
}
}
}
func TestNewEntityCorrectName(t *testing.T) {
entity, err := NewEntity("Golang Gopher", "Test Key", "no-reply@golang.com", nil)
if err != nil {
t.Fatal(err)
}
if len(entity.Identities) != 1 {
t.Fatalf("len(entity.Identities) = %d, want 1", len(entity.Identities))
}
var got string
for _, i := range entity.Identities {
got = i.Name
}
want := "Golang Gopher (Test Key) <no-reply@golang.com>"
if got != want {
t.Fatalf("Identity.Name = %q, want %q", got, want)
}
}
func TestNewEntityWithPreferredSymmetric(t *testing.T) {
c := &packet.Config{
DefaultCipher: packet.CipherAES256,
}
entity, err := NewEntity("Golang Gopher", "Test Key", "no-reply@golang.com", c)
if err != nil {
t.Fatal(err)
}
for _, identity := range entity.Identities {
if len(identity.SelfSignature.PreferredSymmetric) == 0 {
t.Fatal("didn't find a preferred cipher in self signature")
}
if identity.SelfSignature.PreferredSymmetric[0] != uint8(c.DefaultCipher) {
t.Fatalf("Expected preferred cipher to be %d, got %d", uint8(c.DefaultCipher), identity.SelfSignature.PreferredSymmetric[0])
}
}
}
func TestNewEntityWithoutPreferredSymmetric(t *testing.T) {
entity, err := NewEntity("Golang Gopher", "Test Key", "no-reply@golang.com", nil)
if err != nil {
t.Fatal(err)
}
for _, identity := range entity.Identities {
if len(identity.SelfSignature.PreferredSymmetric) != 0 {
t.Fatalf("Expected preferred cipher to be empty but got length %d", len(identity.SelfSignature.PreferredSymmetric))
}
}
}
func TestNewEntityPublicSerialization(t *testing.T) {
entity, err := NewEntity("Golang Gopher", "Test Key", "no-reply@golang.com", nil)
if err != nil {
t.Fatal(err)
}
serializedEntity := bytes.NewBuffer(nil)
entity.Serialize(serializedEntity)
_, err = ReadEntity(packet.NewReader(bytes.NewBuffer(serializedEntity.Bytes())))
if err != nil {
t.Fatal(err)
}
}
|
Manisha
Manisha is the Hindu goddess of the mind. When used in this context, it symbolizes intelligence and desire. Manisha = Mann + Iccha according to Sandhi Vicched in Hindi, which means my wish.
Category:Hindu goddesses
The word Manisha also represents the sacred Sanskrit of Bundhu-khaka. This represents all well-being in the body of a gungru showing enlightenment and happiness.
The Hindu upanishad also describes the theory of Manisha. |
Na+/K+ ATPase activity in mouse lung fibroblasts and HeLa S3 cells during and after hyperthermia.
The ouabain-sensitive ATP-hydrolysing activity, representing the Na+/K+ ATPase capacity, of isolated membranes and whole cells during and after hyperthermia treatments was investigated. In isolated membranes no heat damage after treatments up to 46 degrees C during 45 min or up to 6 h at 44 degrees C could be detected. The ATP hydrolysing activity of Na+/K+ ATPase seems not to be impaired by direct heat attack in the range of commonly used hyperthermic temperatures (39-46 degrees C). Heat effects on the ATP hydrolysing activity of Na+/K+ ATPase of whole mouse fibroblasts could only be detected after heat doses (greater than 40 min at 44 degrees C) necessary to yield over 99 per cent dead cells. Potassium influx, measured with 86RB+ as the K+ tracer, was initially enhanced during incubation at 44 degrees C proportionally with the enhancement of the ATP-hydrolysing activity after raising the temperature. Replacement of non-lethally (10 min at 44 degrees C) and lethally (40 min at 44 degrees C) treated mouse fibroblasts to 37 degrees C showed complete reversibility of the enhanced activity at 44 degrees C to the control level at 37 degrees C. For comparison, the ATP-hydrolysing activity of Na+/K+ ATPase of HeLa S3 cells growing as monolayer was also tested. The activity after heat treatments up to 60 min at 44 degrees C was also found to be unchanged in these experiments. No indication of irreversible damage to the ATP-hydrolysing capacity of mouse fibroblasts and HeLa S3 cells, or K+ pumping activity of mouse fibroblasts by heat treatments up to 40 min at 44 degrees C was found. |
Laborers work in a construction site in the Israeli settlement of Ramat Givat Zeev in the occupied-West Bank November 19, 2019. (Reuters)
Asharq Al-Awsat
Israel's defense minister on Wednesday said he aimed to boost the number of Jewish settlers in the occupied West Bank to one million within a decade, from around 400,000 at present.
Naftali Bennett, a hawk who draws much of his support from settlers, is leading his New Right party to elections in March.
He was speaking at a Jerusalem congress on Washington's November policy shift stating that it no longer considers Israeli settlements illegal, alongside Prime Minister Benjamin Netanyahu and US ambassador David Friedman
"Our aim is that within a decade a million Israeli citizens will live in Judaea and Samaria," Bennett said, using the biblical term for the Israeli-occupied West Bank.
Friedman, who is also Jewish and a strong supporter of settlements, disputed the use of the term "occupied".
"We are not occupiers in our homeland, we are not occupiers in our own land, we are not like the Belgians in the Congo," he said.
Both men's comments drew swift condemnation from the Palestinian Authority.
The PA foreign ministry described them as "racist" and "reflecting the Jewish colonial nature of the deal of the century" -- a reference to US President Donald Trump's so-far undisclosed peace plan.
The ministry added that Friedman and Bennett's statements were "official confessions of their involvement in the crime of settlement and the confiscation of Palestinian land".
On November 18, US Secretary of State Mike Pompeo said the United States no longer considers Israeli settlements to be "inconsistent with international law".
Previously US policy was based, at least in theory, on a legal opinion issued by the State Department in 1978 which said that establishing settlements in Palestinian territories captured a decade earlier by Israel went against international law.
The Fourth Geneva Convention on the laws of war explicitly forbids moving civilians into occupied territories.
The about-turn brought stiff international and Palestinian criticism.
The United Nations and European Union said the decision would not change the reality that the settlements were illegal, while the Arab League condemned Washington's unilateral move.
More than 600,000 Israelis live in settlements in east Jerusalem and the West Bank, alongside more than three million Palestinians.
Israel seized control of the territories, seen as pivotal parts of any future Palestinian state, in the 1967 Six-Day War.
Shortly after taking up his post in November Bennett announced a plan to double the number of settlers in the flashpoint West Bank city of Hebron. |
About Biomass
Biomass is a quite general term for material derived from growing plants or from animal manure. Bioenergy refers to the technical systems through which biomass is produced or collected, converted and used as an energy source. A wide variety of conversion routescan be distinguished that produce a variety of energy carriers either in a solid, liquid or gaseous form. These energy carriers address all types of energy markets: heat, electricity and transportation.
Bioenergy already provides the majority of renewable energy worldwide and is considered to have the potential to provide a large fraction of world energy demand over the next century. At the same time, if biomass systems are managed properly, bioenergy will contribute to meet the requirement of reducing carbon emissions.
Bioenergy represented a 68% of the total gross inland consumption of renewables in 2011. Solid biomass and renewable waste are the main source of bioenergy accounting for 115Mtoe in 2011, an 8,4% of the total final energy consumption in Europe. It plays an essential role in countries such as Estonia, Latvia, Finalnd and Sweden whre the consumption is above 25%
Estimated demand for biomass for energy in the EU27 countries based on national renewable energy projections and reported conversion efficiencies (Bentsen and Felby, Biotechnology for Biofuels, 2012)
Heat consumption from solid biomass (2011): 64,9 MTOW
Electricity produced from solid biomass (2011): 72,8TWh
Primary energy production from solid biomass (2011): 78,8 MTOE
The combined European (EU-27) consumption of biofuels represented 13,615,000 ktoe (tonnes of oil equivalent) in 2013 (EUR-ObservER, 2014), with a strong prominence of biodiesel which represents 79% of European biofuel consumption and 19,9% in the case of bioethanol. The orverall figures represent a growth of 28% between 2008 and 2013. Furthermore, the production of biodiesel reached 91,879,000 tonnes of oil equivalent in 2012 experiencing a 28% growth in the last five years, while bioethanol reached 20,359,000 and had a 32% growth in the same period. |
No. 19 Gym’Backs travel to SECs Saturday
BY ANDRES FOCIL
FAYETTEVILLE — The No. 19 University of Arkansas gymnastics team begins post-season action this weekend traveling to the Southeastern Conference Championship in Birmingham, Ala., Saturday. Arkansas is making its fourth appearance in four seasons at the SEC Championship and returns to Birmingham after a two-year hiatus in Duluth, Ga., for the 2004 and 2005 championships. "This is a big meet this weekend," said co-head coach Mark Cook. "It’s hard to believe we’ve had 10 meets already this year. We know it will be a battle at SECs as it always is but we feel confident and prepared heading into it." The Gym’Backs are no strangers to the tough competition that awaits them in Birmingham. Headlining the field is defending national champion and top-ranked Georgia. They are joined by No. 3 Florida, No. 5 Alabama, No. 8 LSU, No. 14 Auburn, No. 18 Kentucky and No. 19 Arkansas in what has become one of the toughest conference championships in the country. The Gym’Backs have continued to improve during their first three-plus seasons of competition and picked up narrow wins over Auburn and Kentucky at home this year for the second consecutive season. But both of those SEC foes have continued to score well all season and Arkansas will have it’s hands full this weekend. Another new feature (besides the location) that the Gym’Backs face this weekend is the arena setup. This weekend’s competition takes place with the gymnastics equipment on an elevated platform in the center of the arena. The setup is similar to the national championship structure and is familiar to the gymnasts from their club days. "I’d say about half of our team has seen this setup before," said co-head coach Rene Cook. "It’s a little different look and, yes, we do have some experience with it but it’s a different environment. We’ve competed well in the last five meets though and we’re going in well prepared and on a high note. I can’t wait to see what we can do." In addition to the goal of a top finish at the SEC Championship, Arkansas enters the meet looking for a high road score in the hopes of moving up into the top 18 in the country next week. The Gym’Backs, currently 19th in the country, are shooting for one of the coveted 18 seeded spots with the regionals just around the corner. "We’re in a really good spot right now," said Rene Cook, "and moving up is well within our grasp." Arkansas returns from SEC action to prepare for the 2006 NCAA South Central Regional Championship, a meet the Gym’Backs host for the first time in school history April 8 in Barnhill Arena. Tickets for the NCAA Regional are currently on sale at the Lady Razorback Ticket Office. Please call 57-LBACK or log on to LADYBACKS.COM for more information. |
---
layout: base
title: 'Statistics of CCONJ in UD_Russian-Taiga'
udver: '2'
---
## Treebank Statistics: UD_Russian-Taiga: POS Tags: `CCONJ`
There are 18 `CCONJ` lemmas (0%), 18 `CCONJ` types (0%) and 2680 `CCONJ` tokens (4%).
Out of 17 observed tags, the rank of `CCONJ` is: 16 in number of lemmas, 17 in number of types and 8 in number of tokens.
The 10 most frequent `CCONJ` lemmas: <em>и, а, но, или, да, ни, то, либо, зато, иль</em>
The 10 most frequent `CCONJ` types: <em>и, а, но, или, да, ни, то, либо, зато, иль</em>
The 10 most frequent ambiguous lemmas: <em>и</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 1738, <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 159, <tt><a href="ru_taiga-pos-X.html">X</a></tt> 1), <em>а</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 434, <tt><a href="ru_taiga-pos-NOUN.html">NOUN</a></tt> 2, <tt><a href="ru_taiga-pos-INTJ.html">INTJ</a></tt> 1, <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 1), <em>да</em> (<tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 53, <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 43), <em>ни</em> (<tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 42, <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 39), <em>то</em> (<tt><a href="ru_taiga-pos-PRON.html">PRON</a></tt> 152, <tt><a href="ru_taiga-pos-SCONJ.html">SCONJ</a></tt> 40, <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 26, <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 15, <tt><a href="ru_taiga-pos-DET.html">DET</a></tt> 1), <em>либо</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 9, <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 1), <em>+</em> (<tt><a href="ru_taiga-pos-SYM.html">SYM</a></tt> 9, <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 4), <em>причем</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 4, <tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> 2), <em>иначе</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 3, <tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> 2), <em>однако</em> (<tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> 36, <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 3)
The 10 most frequent ambiguous types: <em>и</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 1492, <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 156, <tt><a href="ru_taiga-pos-X.html">X</a></tt> 1), <em>а</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 257, <tt><a href="ru_taiga-pos-ADP.html">ADP</a></tt> 1, <tt><a href="ru_taiga-pos-INTJ.html">INTJ</a></tt> 1, <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 1), <em>да</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 31, <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 25), <em>ни</em> (<tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 41, <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 34), <em>то</em> (<tt><a href="ru_taiga-pos-PRON.html">PRON</a></tt> 41, <tt><a href="ru_taiga-pos-SCONJ.html">SCONJ</a></tt> 36, <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 25, <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 14, <tt><a href="ru_taiga-pos-DET.html">DET</a></tt> 8), <em>либо</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 9, <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 1), <em>+</em> (<tt><a href="ru_taiga-pos-SYM.html">SYM</a></tt> 9, <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 4), <em>причем</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 3, <tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> 2), <em>иначе</em> (<tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 3, <tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> 2), <em>однако</em> (<tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> 11, <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 2)
* <em>и</em>
* <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 1492: <em>Мы собираем свой #Топ100 - что увидеть <b>и</b> попробовать во Вьетнаме .</em>
* <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 156: <em>@screened-200 спасибо <b>и</b> Вам за прекрасный отдых 👌🏻🌸</em>
* <tt><a href="ru_taiga-pos-X.html">X</a></tt> 1: <em>@xxxxxx конечно , у Вас же цель сократить явку , уменьшив количество голосов у <b>и</b></em>
* <em>а</em>
* <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 257: <em>Добрый вечер , вид изумительный , <b>а</b> что это за номер ?</em>
* <tt><a href="ru_taiga-pos-ADP.html">ADP</a></tt> 1: <em>Даже не знаю , как относиться <b>а</b> этому .</em>
* <tt><a href="ru_taiga-pos-INTJ.html">INTJ</a></tt> 1: <em>Ведь в троллейбус с тобой нельзя ... <b>а</b> , Петрович ?</em>
* <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 1: <em><b>а</b> да ладно что плести врачу осталось уползти</em>
* <em>да</em>
* <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 31: <em>@screened-57 тоже вариант , <b>да</b> ) от солнца они очень усердно прячутся 🙈</em>
* <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 25: <em>@screened-172 <b>да</b> !</em>
* <em>ни</em>
* <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 41: <em>я тоже зимой <b>ни</b> азу не путешествовала ))) и не хочу</em>
* <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 34: <em>@Zhirinovskiy <b>ни</b> чего он не решает просто воду наливает</em>
* <em>то</em>
* <tt><a href="ru_taiga-pos-PRON.html">PRON</a></tt> 41: <em>А <b>то</b> , что культуры у нас нет , это да !!!</em>
* <tt><a href="ru_taiga-pos-SCONJ.html">SCONJ</a></tt> 36: <em>Кстати , вот если у детей разница в 5 - 6 лет , <b>то</b> вообще легко !</em>
* <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 25: <em>Да , наконец - <b>то</b> увидим солнце 🌤</em>
* <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 14: <em>Напротив сидел парень , <b>то</b> поднимал глаза на меня , <b>то</b> опускал .</em>
* <tt><a href="ru_taiga-pos-DET.html">DET</a></tt> 8: <em>Каждый год он отмечается в одно и <b>то</b> же время — 19 января .</em>
* <em>либо</em>
* <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 9: <em>5⃣ И главное - запретить увеличение каких <b>либо</b> налогов . ⠀</em>
* <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> 1: <em>И результат самый большой танец в Питере из когда <b>либо</b> организованных !</em>
* <em>+</em>
* <tt><a href="ru_taiga-pos-SYM.html">SYM</a></tt> 9: <em>@xxxxxx английский , общагу <b>+</b> рус и математика , надо ещё какой то</em>
* <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 4: <em>Полный комплект , готовый к полетам <b>+</b> дополнительная батарея .</em>
* <em>причем</em>
* <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 3: <em>" Необходимо очистить систему от мздоимцев , которые место своей работы превратили в бизнес , <b>причем</b> , доходный " , — заявил глава правительства .</em>
* <tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> 2: <em>люди умирают не в 90 лет , не в авариях и не <b>причем</b> тут грипп .</em>
* <em>иначе</em>
* <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 3: <em>Никогда не изменяйте своим принципам — принципам свободы и духовности ... <b>иначе</b> вы погибнете ...</em>
* <tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> 2: <em>3 . Еще 30 секунд — на подтверждение аккаунта ( <b>иначе</b> ничего не получится )</em>
* <em>однако</em>
* <tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> 11: <em>я дурак но <b>однако</b> где кудесник где психолог божества</em>
* <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> 2: <em>но <b>однако</b> шли года шёл туман и ерунда</em>
## Morphology
The form / lemma ratio of `CCONJ` is 1.000000 (the average of all parts of speech is 1.613758).
The 1st highest number of forms (1) was observed with the lemma “+”: <em>+</em>.
The 2nd highest number of forms (1) was observed with the lemma “А”: <em>А</em>.
The 3rd highest number of forms (1) was observed with the lemma “а”: <em>а</em>.
`CCONJ` occurs with 2 features: <tt><a href="ru_taiga-feat-Polarity.html">Polarity</a></tt> (31; 1% instances), <tt><a href="ru_taiga-feat-Foreign.html">Foreign</a></tt> (2; 0% instances)
`CCONJ` occurs with 2 feature-value pairs: `Foreign=Yes`, `Polarity=Neg`
`CCONJ` occurs with 3 feature combinations.
The most frequent feature combination is `_` (2647 tokens).
Examples: <em>и, а, но, или, да, то, либо, ни, зато, иль</em>
## Relations
`CCONJ` nodes are attached to their parents using 12 different relations: <tt><a href="ru_taiga-dep-cc.html">cc</a></tt> (2623; 98% instances), <tt><a href="ru_taiga-dep-advmod.html">advmod</a></tt> (28; 1% instances), <tt><a href="ru_taiga-dep-fixed.html">fixed</a></tt> (8; 0% instances), <tt><a href="ru_taiga-dep-conj.html">conj</a></tt> (6; 0% instances), <tt><a href="ru_taiga-dep-orphan.html">orphan</a></tt> (5; 0% instances), <tt><a href="ru_taiga-dep-parataxis.html">parataxis</a></tt> (3; 0% instances), <tt><a href="ru_taiga-dep-root.html">root</a></tt> (2; 0% instances), <tt><a href="ru_taiga-dep-appos.html">appos</a></tt> (1; 0% instances), <tt><a href="ru_taiga-dep-case.html">case</a></tt> (1; 0% instances), <tt><a href="ru_taiga-dep-discourse.html">discourse</a></tt> (1; 0% instances), <tt><a href="ru_taiga-dep-goeswith.html">goeswith</a></tt> (1; 0% instances), <tt><a href="ru_taiga-dep-obj.html">obj</a></tt> (1; 0% instances)
Parents of `CCONJ` nodes belong to 17 different parts of speech: <tt><a href="ru_taiga-pos-VERB.html">VERB</a></tt> (1270; 47% instances), <tt><a href="ru_taiga-pos-NOUN.html">NOUN</a></tt> (839; 31% instances), <tt><a href="ru_taiga-pos-ADJ.html">ADJ</a></tt> (212; 8% instances), <tt><a href="ru_taiga-pos-PROPN.html">PROPN</a></tt> (112; 4% instances), <tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> (111; 4% instances), <tt><a href="ru_taiga-pos-PRON.html">PRON</a></tt> (70; 3% instances), <tt><a href="ru_taiga-pos-DET.html">DET</a></tt> (19; 1% instances), <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> (17; 1% instances), <tt><a href="ru_taiga-pos-NUM.html">NUM</a></tt> (15; 1% instances), <tt><a href="ru_taiga-pos-INTJ.html">INTJ</a></tt> (3; 0% instances), <tt><a href="ru_taiga-pos-AUX.html">AUX</a></tt> (2; 0% instances), <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> (2; 0% instances), (2; 0% instances), <tt><a href="ru_taiga-pos-SCONJ.html">SCONJ</a></tt> (2; 0% instances), <tt><a href="ru_taiga-pos-SYM.html">SYM</a></tt> (2; 0% instances), <tt><a href="ru_taiga-pos-ADP.html">ADP</a></tt> (1; 0% instances), <tt><a href="ru_taiga-pos-X.html">X</a></tt> (1; 0% instances)
2643 (99%) `CCONJ` nodes are leaves.
29 (1%) `CCONJ` nodes have one child.
5 (0%) `CCONJ` nodes have two children.
3 (0%) `CCONJ` nodes have three or more children.
The highest child degree of a `CCONJ` node is 5.
Children of `CCONJ` nodes are attached using 9 different relations: <tt><a href="ru_taiga-dep-fixed.html">fixed</a></tt> (27; 52% instances), <tt><a href="ru_taiga-dep-punct.html">punct</a></tt> (15; 29% instances), <tt><a href="ru_taiga-dep-det.html">det</a></tt> (2; 4% instances), <tt><a href="ru_taiga-dep-goeswith.html">goeswith</a></tt> (2; 4% instances), <tt><a href="ru_taiga-dep-parataxis.html">parataxis</a></tt> (2; 4% instances), <tt><a href="ru_taiga-dep-cc.html">cc</a></tt> (1; 2% instances), <tt><a href="ru_taiga-dep-conj.html">conj</a></tt> (1; 2% instances), <tt><a href="ru_taiga-dep-nsubj.html">nsubj</a></tt> (1; 2% instances), <tt><a href="ru_taiga-dep-nummod-gov.html">nummod:gov</a></tt> (1; 2% instances)
Children of `CCONJ` nodes belong to 13 different parts of speech: <tt><a href="ru_taiga-pos-PART.html">PART</a></tt> (18; 35% instances), <tt><a href="ru_taiga-pos-PUNCT.html">PUNCT</a></tt> (15; 29% instances), <tt><a href="ru_taiga-pos-ADV.html">ADV</a></tt> (4; 8% instances), <tt><a href="ru_taiga-pos-DET.html">DET</a></tt> (3; 6% instances), <tt><a href="ru_taiga-pos-PRON.html">PRON</a></tt> (3; 6% instances), <tt><a href="ru_taiga-pos-CCONJ.html">CCONJ</a></tt> (2; 4% instances), <tt><a href="ru_taiga-pos-ADJ.html">ADJ</a></tt> (1; 2% instances), <tt><a href="ru_taiga-pos-INTJ.html">INTJ</a></tt> (1; 2% instances), <tt><a href="ru_taiga-pos-NOUN.html">NOUN</a></tt> (1; 2% instances), <tt><a href="ru_taiga-pos-NUM.html">NUM</a></tt> (1; 2% instances), <tt><a href="ru_taiga-pos-SCONJ.html">SCONJ</a></tt> (1; 2% instances), <tt><a href="ru_taiga-pos-SYM.html">SYM</a></tt> (1; 2% instances), <tt><a href="ru_taiga-pos-VERB.html">VERB</a></tt> (1; 2% instances)
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