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	<title>kleene &amp;laquo; WordPress.com Tag Feed</title>
	<link>http://en.wordpress.com/tag/kleene/</link>
	<description>Feed of posts on WordPress.com tagged "kleene"</description>
	<pubDate>Sat, 26 Dec 2009 20:02:15 +0000</pubDate>

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<title><![CDATA[Arithmetic Hierarchy and P=NP]]></title>
<link>http://rjlipton.wordpress.com/2009/05/27/arithmetic-hierarchy-and-pnp/</link>
<pubDate>Wed, 27 May 2009 22:48:50 +0000</pubDate>
<dc:creator>rjlipton</dc:creator>
<guid>http://rjlipton.wordpress.com/2009/05/27/arithmetic-hierarchy-and-pnp/</guid>
<description><![CDATA[The complexity of open problems via the arithmetic hierarchy Stephen Kleene is a famous logician who]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p> <font color="”#0066cc?"><br />
<em> The complexity of open problems via the arithmetic hierarchy </em><br />
<font color="”#000000?"></p>
<p><img src="http://rjlipton.wordpress.com/files/2009/05/images7.jpeg" alt="images" title="images" width="92" height="125" class="alignright size-full wp-image-2468" /></p>
<p>
Stephen Kleene is a famous logician who got his PhD under Alonzo Church at Princeton University. Kleene has many important concepts named after him: the Kleene hierarchy, the Kleene star, Kleene&#8217;s recursion theorem and the Kleene fixed-point theorem. His 1952 book, <em>Introduction to Meta-mathematics</em>, is a classic; more on his book in a moment, which while a classic has an unusual feature.</p>
<p>
Today, I want to talk about the Kleene Hierarchy, which is now usually called the <a href="http://en.wikipedia.org/wiki/Arithmetical_hierarchy">Arithmetic Hierarchy</a> (AH). We can use the AH to classify <em>problems,</em> in a sense, that I will shortly make precise. In this classification, the P=NP question comes out with high marks, which is both cool&#8211;we have a hard open problem&#8211;and is uncool, since we have a hard open problem.<br />
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<p>
I had the honor of meeting Kleene at his retirement conference, which was held in June 1978 in Madison, Wisconsin. That summer I was also privileged to sit in the office of Barkley Rosser, who was away on leave. Somehow I thought a lot about logic that summer.</p>
<p>
Kleene had many terrific PhD students: two I know well are Robert Constable and Richard Vesley. Richard, who is a wonderful teacher, taught me first order logic and recursion theory, and helped me solve my first &#8220;open&#8221; problem. But, that is another story, for another day.</p>
<p>
There is a barber shop story about Kleene, which I cannot validate with certainty; and there is a retirement story that I can validate&#8211;since I saw it first hand.</p>
<p>
While working for Church, Stephen was trying to show that the lambda calculus could compute anything &#8220;computable&#8221;. Unfortunately, Church and Kleene could not see how to compute even the following function:
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+x+%5Crightarrow+x%2B1.%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle x \rightarrow x+1.&amp;fg=000000' title='\displaystyle x \rightarrow x+1.&amp;fg=000000' class='latex' /></p>
<p> Yes, even the lowly successor function was not obviously computable in their calculus. The story goes that one day Stephen was getting his hair cut, when he realized how to encode the successor function into the lambda calculus. It is claimed, he jumped out of the barber chair, with half a haircut, and ran to tell Church. Not quite Archimedes running naked through the streets of Syracuse, shouting &#8220;Eureka,&#8221; but close. I think the story, true or not, says more about the nature of the lambda calculus as a basis of computing than anything else, but that is my take.</p>
<p>
At Kleene&#8217;s retirement conference the final talk was given by the great man himself. After a long and well deserved introduction Kleene stepped up to the stage, and placed his first overhead slide onto the projector. Before he could say a word there was a roar of laughter throughout the packed auditorium. He ignored the laughter and proceeded with his talk. The reason for the laughter was that the slide looked like a page from his famous book. The slide looked like this:
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++1.+%5Cdots+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  1. \dots &amp;fg=000000' title='\displaystyle  1. \dots &amp;fg=000000' class='latex' /></p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Cvdots+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \vdots &amp;fg=000000' title='\displaystyle  \vdots &amp;fg=000000' class='latex' /></p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Cquad+1.1+%5Cdots%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \quad 1.1 \dots&amp;fg=000000' title='\displaystyle  \quad 1.1 \dots&amp;fg=000000' class='latex' /></p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Cvdots+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \vdots &amp;fg=000000' title='\displaystyle  \vdots &amp;fg=000000' class='latex' /></p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Cquad+%5Cquad+1.1.1+%5Cdots+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \quad \quad 1.1.1 \dots &amp;fg=000000' title='\displaystyle  \quad \quad 1.1.1 \dots &amp;fg=000000' class='latex' /></p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Cquad+%5Cvdots+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \quad \vdots &amp;fg=000000' title='\displaystyle  \quad \vdots &amp;fg=000000' class='latex' /></p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Cquad+%5Cvdots%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \quad \vdots&amp;fg=000000' title='\displaystyle  \quad \vdots&amp;fg=000000' class='latex' /></p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++2.+%5Cdots+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  2. \dots &amp;fg=000000' title='\displaystyle  2. \dots &amp;fg=000000' class='latex' /></p>
<p> <em>Meta-mathematics</em> his book, is known for its detailed numbering scheme that only a logician could love. Phrases like  	&#8220;by <img src='http://l.wordpress.com/latex.php?latex=%7B2.4.1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{2.4.1}&amp;fg=000000' title='{2.4.1}&amp;fg=000000' class='latex' /> we see that <img src='http://l.wordpress.com/latex.php?latex=%7B2.7.4%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{2.7.4}&amp;fg=000000' title='{2.7.4}&amp;fg=000000' class='latex' /> is true&#8221;  abound. We all thought that Kleene was making fun of his own style, and would soon switch into story mode. The audience stopped laughing as soon as it dawned on everyone that this was no joke. Kleene was going to read and present a very detailed argument on something. Too bad. I was hoping to hear stories about his advisor, or his results, or his students&#8211;his life. But that was not to be: we were treated to an hour of a detailed talk on some exotic topic in logic.</p>
<p>
<p><b> Arithmetic Hierarchy </b></p>
<p><p>
The AH classifies all formulas according to their quantifier complexity. Thus, a sentence is in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B2%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{2}&amp;fg=000000' title='\Pi_{2}&amp;fg=000000' class='latex' /> provided it is equivalent to,
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Cforall+x+%5Cexists+y+%5Cpsi%28x%2Cy%29+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \forall x \exists y \psi(x,y) &amp;fg=000000' title='\displaystyle  \forall x \exists y \psi(x,y) &amp;fg=000000' class='latex' /></p>
<p> and <img src='http://l.wordpress.com/latex.php?latex=%7B%5Cpsi%28x%2Cy%29%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\psi(x,y)}&amp;fg=000000' title='{\psi(x,y)}&amp;fg=000000' class='latex' /> is quantifier free. </p>
<p>
For example, Fermat&#8217;s Last Theorem (FLT) is in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{1}&amp;fg=000000' title='\Pi_{1}&amp;fg=000000' class='latex' />. This follows since we can write it in the following way.
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Cforall+x%2Cy%2Cz%2Cn%26%2362%3B2+%5Cquad+%28x%5E%7Bn%7D%2By%5E%7Bn%7D+%3D+z%5E%7Bn%7D+%5Crightarrow+xyz%3D0%29+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \forall x,y,z,n&gt;2 \quad (x^{n}+y^{n} = z^{n} \rightarrow xyz=0) &amp;fg=000000' title='\displaystyle  \forall x,y,z,n&gt;2 \quad (x^{n}+y^{n} = z^{n} \rightarrow xyz=0) &amp;fg=000000' class='latex' /></p>
<p>
In a reversal of history, if you know the polynomial hierarchy, then you know AH: just remove the restrictions on the quantifiers. There was a time when many great ideas of complexity theory could be found by taking a concept from the area of recursion theory and &#8220;just&#8221; making the concept feasible: diagonalization, complete sets, various notions of reducibility, oracles, and more. That was not always how they were actually discovered, but many of the key notions were first defined in recursion theory. There may be other concepts still lurking there <img src='http://l.wordpress.com/latex.php?latex=%7B%5Cdots%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\dots}&amp;fg=000000' title='{\dots}&amp;fg=000000' class='latex' /></p>
<p>
<p><b> Classification of Open Problems </b></p>
<p><p>
We saw how we could write the FLT in AH. In a natural way we can classify any open problem of arithmetic by writing it as a formal statement in AH. Here are some other problems&#8211;some open and some known&#8211;with respect to their AH complexity. This idea of viewing a problem&#8217;s complexity via the AH is an old idea, but one that I find useful. I hope you will agree.</p>
<p>
<img src='http://l.wordpress.com/latex.php?latex=%7B%5Cbullet%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\bullet}&amp;fg=000000' title='{\bullet}&amp;fg=000000' class='latex' /> <em>The Four Color Theorem is in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{1}&amp;fg=000000' title='\Pi_{1}&amp;fg=000000' class='latex' />.</em> This is easy to see, since it states that for every planar graph, there is a <img src='http://l.wordpress.com/latex.php?latex=%7B4%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{4}&amp;fg=000000' title='{4}&amp;fg=000000' class='latex' />-coloring of the faces.</p>
<p>
<img src='http://l.wordpress.com/latex.php?latex=%7B%5Cbullet%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\bullet}&amp;fg=000000' title='{\bullet}&amp;fg=000000' class='latex' /> <em>The Finite Simple Group Classification is in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{1}&amp;fg=000000' title='\Pi_{1}&amp;fg=000000' class='latex' />.</em> This is the famous &#8220;conjecture&#8221;, that is now a theorem, that all finite simple groups are either one of the <img src='http://l.wordpress.com/latex.php?latex=%7B26%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{26}&amp;fg=000000' title='{26}&amp;fg=000000' class='latex' /> so-called sporadic simple groups, or come from the known infinite families of finite simple groups.</p>
<p>
These first two raise a annoying point. If a theorem is proved, then it can be written as anything. This is another way of saying that <img src='http://l.wordpress.com/latex.php?latex=%7B1%3D1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{1=1}&amp;fg=000000' title='{1=1}&amp;fg=000000' class='latex' /> is equivalent to the Four Color Theorem. What I mean above is: what is the best that you can do <em>without</em> using the proof of the theorem. Textbook exercises have this problem all the time: &#8220;without using X prove Y.&#8221; I have the same problem here.</p>
<p>
<img src='http://l.wordpress.com/latex.php?latex=%7B%5Cbullet%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\bullet}&amp;fg=000000' title='{\bullet}&amp;fg=000000' class='latex' /> <em>The Jacobian Conjecture is in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{1}&amp;fg=000000' title='\Pi_{1}&amp;fg=000000' class='latex' />.</em> This is, one of my favorite non-complexity conjectures. I will discuss it another time. The fact that it is of this form is not trivial, since it concerns the behavior of polynomials with complex coefficients. It requires a lemma to re-state it as a statement about integers.</p>
<p>
<img src='http://l.wordpress.com/latex.php?latex=%7B%5Cbullet%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\bullet}&amp;fg=000000' title='{\bullet}&amp;fg=000000' class='latex' /> <em>The Riemann Hypothesis is in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{1}&amp;fg=000000' title='\Pi_{1}&amp;fg=000000' class='latex' />.</em> The fact that this is in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{1}&amp;fg=000000' title='\Pi_{1}&amp;fg=000000' class='latex' /> is plausible, since it states, of course, that <em>all</em> the non-trivial roots of the <img src='http://l.wordpress.com/latex.php?latex=%7B%5Czeta%28%5Csigma%2Bit%29%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\zeta(\sigma+it)}&amp;fg=000000' title='{\zeta(\sigma+it)}&amp;fg=000000' class='latex' /> function have <img src='http://l.wordpress.com/latex.php?latex=%7B%5Csigma%3D1%2F2%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\sigma=1/2}&amp;fg=000000' title='{\sigma=1/2}&amp;fg=000000' class='latex' />. However, the obvious way to encode this runs into many technical issues about the locations of the zeroes of the <img src='http://l.wordpress.com/latex.php?latex=%7B%5Czeta%28%5Csigma%2Bit%29%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\zeta(\sigma+it)}&amp;fg=000000' title='{\zeta(\sigma+it)}&amp;fg=000000' class='latex' /> function. However, Jeff Lagarias to the rescue. He has proved the following wonderful <a href="http://arxiv.org/pdf/math/0008177v2">result:</a> Let <img src='http://l.wordpress.com/latex.php?latex=%7BH_%7Bn%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{H_{n}}&amp;fg=000000' title='{H_{n}}&amp;fg=000000' class='latex' /> be equal to the <img src='http://l.wordpress.com/latex.php?latex=%7Bn%5E%7Bth%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n^{th}}&amp;fg=000000' title='{n^{th}}&amp;fg=000000' class='latex' /> harmonic number,
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++H_n+%3D+%5Csum_%7Bi%3D1%7D%5E%7Bn%7D+%5Cfrac%7B1%7D%7Bi%7D.+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  H_n = \sum_{i=1}^{n} \frac{1}{i}. &amp;fg=000000' title='\displaystyle  H_n = \sum_{i=1}^{n} \frac{1}{i}. &amp;fg=000000' class='latex' /></p>
<p> Consider, the statement (E), that for all <img src='http://l.wordpress.com/latex.php?latex=%7Bn+%5Cge+1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n \ge 1}&amp;fg=000000' title='{n \ge 1}&amp;fg=000000' class='latex' />,
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Csum_%7Bd+%26%23124%3B+n%7D+d+%5Cle+H_%7Bn%7D+%2B+%5Cexp%28H_%7Bn%7D%29%5Clog%28H_%7Bn%7D%29+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \sum_{d &#124; n} d \le H_{n} + \exp(H_{n})\log(H_{n}) &amp;fg=000000' title='\displaystyle  \sum_{d &#124; n} d \le H_{n} + \exp(H_{n})\log(H_{n}) &amp;fg=000000' class='latex' /></p>
<p> with equality only for <img src='http://l.wordpress.com/latex.php?latex=%7Bn+%3D+1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n = 1}&amp;fg=000000' title='{n = 1}&amp;fg=000000' class='latex' />. Lagarias&#8217; result is that the statement E is equivalent to the Riemann Hypothesis. This will show that the Riemann Hypothesis is in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{1}&amp;fg=000000' title='\Pi_{1}&amp;fg=000000' class='latex' />.</p>
<p>
Even with Jeff&#8217;s result there is a small technical point. We must show that it is computable to test whether or not
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++a+%5Cle+%5Cexp%28b%29%5Clog%28c%29+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  a \le \exp(b)\log(c) &amp;fg=000000' title='\displaystyle  a \le \exp(b)\log(c) &amp;fg=000000' class='latex' /></p>
<p> for rational numbers <img src='http://l.wordpress.com/latex.php?latex=%7Ba%2Cb%2Cc%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{a,b,c}&amp;fg=000000' title='{a,b,c}&amp;fg=000000' class='latex' />. By clearing the denominator of <img src='http://l.wordpress.com/latex.php?latex=%7Ba%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{a}&amp;fg=000000' title='{a}&amp;fg=000000' class='latex' /> and using the elementary fact that <img src='http://l.wordpress.com/latex.php?latex=%7B+r+%5Clog%28c%29+%3D+%5Clog%28c%5E%7Br%7D%29%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ r \log(c) = \log(c^{r})}&amp;fg=000000' title='{ r \log(c) = \log(c^{r})}&amp;fg=000000' class='latex' />, we can show that we need only be able to test
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++a+%5Cle+%5Cexp%28b%29%5Clog%28c%29+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  a \le \exp(b)\log(c) &amp;fg=000000' title='\displaystyle  a \le \exp(b)\log(c) &amp;fg=000000' class='latex' /></p>
<p> for natural numbers. This is easy: compute the right-hand side until the error is below <img src='http://l.wordpress.com/latex.php?latex=%7B1%2F3%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{1/3}&amp;fg=000000' title='{1/3}&amp;fg=000000' class='latex' />. Then, since <img src='http://l.wordpress.com/latex.php?latex=%7Ba%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{a}&amp;fg=000000' title='{a}&amp;fg=000000' class='latex' /> is a whole number, there can be no mistake.</p>
<p><p>
<img src='http://l.wordpress.com/latex.php?latex=%7B%5Cbullet%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\bullet}&amp;fg=000000' title='{\bullet}&amp;fg=000000' class='latex' /> <em>The Twin Prime Conjecture is in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B2%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{2}&amp;fg=000000' title='\Pi_{2}&amp;fg=000000' class='latex' />.</em> This is the famous conjecture that there are an infinite number of primes <img src='http://l.wordpress.com/latex.php?latex=p+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p &amp;fg=000000' title='p &amp;fg=000000' class='latex' /> so that <img src='http://l.wordpress.com/latex.php?latex=p%2B2+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p+2 &amp;fg=000000' title='p+2 &amp;fg=000000' class='latex' /> is also prime. Such a prime is, of course, called a <em>twin prime.</em> The obvious way to say this is: for all <img src='http://l.wordpress.com/latex.php?latex=x+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x &amp;fg=000000' title='x &amp;fg=000000' class='latex' />, there is a twin prime <img src='http://l.wordpress.com/latex.php?latex=p+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p &amp;fg=000000' title='p &amp;fg=000000' class='latex' /> that is greater than <img src='http://l.wordpress.com/latex.php?latex=x+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x &amp;fg=000000' title='x &amp;fg=000000' class='latex' /></p>
<p><b> Where is P=NP in the AH? </b></p>
<p><p>
The next question is where does P=NP lie with respect to AH? I do not know if either one of P=NP or P<img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cneq%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\neq} }&amp;fg=000000' title='{ {\neq} }&amp;fg=000000' class='latex' />NP are in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B1%7D+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{1} &amp;fg=000000' title='\Pi_{1} &amp;fg=000000' class='latex' />. I will explain the best that I know, and I hope that someone can improve these simple bounds&#8211;perhaps they have already? Of course, as I pointed out earlier, if the P=NP question is resolved, then better bounds must follow.</p>
<p>
First, lets look at how to encode P <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cneq%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\neq} }&amp;fg=000000' title='{ {\neq} }&amp;fg=000000' class='latex' /> NP. We need the standard idea of adding a &#8220;clock&#8221; to a Turing machine. Let <img src='http://l.wordpress.com/latex.php?latex=%7B%5Bx%2Cy%5D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{[x,y]}&amp;fg=000000' title='{[x,y]}&amp;fg=000000' class='latex' /> be a pairing function on the natural numbers: that is <img src='http://l.wordpress.com/latex.php?latex=%7Bx%2Cy+%5Crightarrow+%5Bx%2Cy%5D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{x,y \rightarrow [x,y]}&amp;fg=000000' title='{x,y \rightarrow [x,y]}&amp;fg=000000' class='latex' /> is one-to-one and is easy to compute. Define,
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++M_%7B%5Bx%2Cc%5D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  M_{[x,c]}&amp;fg=000000' title='\displaystyle  M_{[x,c]}&amp;fg=000000' class='latex' /></p>
<p> as the deterministic Turing machine that operates as follows on an input <img src='http://l.wordpress.com/latex.php?latex=%7By%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{y}&amp;fg=000000' title='{y}&amp;fg=000000' class='latex' />. The machine treats <img src='http://l.wordpress.com/latex.php?latex=%7Bx%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{x}&amp;fg=000000' title='{x}&amp;fg=000000' class='latex' /> as a deterministic program, and simulates <img src='http://l.wordpress.com/latex.php?latex=%7Bx%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{x}&amp;fg=000000' title='{x}&amp;fg=000000' class='latex' /> on input <img src='http://l.wordpress.com/latex.php?latex=%7By%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{y}&amp;fg=000000' title='{y}&amp;fg=000000' class='latex' />. At the same time the machine runs a counter that stops its execution after <img src='http://l.wordpress.com/latex.php?latex=%7B%26%23124%3By%26%23124%3B%5E%7Bc%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{&#124;y&#124;^{c}}&amp;fg=000000' title='{&#124;y&#124;^{c}}&amp;fg=000000' class='latex' /> steps. If the machine accepts before the counter stops, then it accepts; otherwise, it rejects. </p>
<p>
Then, </p>
<blockquote><p><b>Theorem: </b> <em> P <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cneq%7D+%7D%26%2338%3Bbg%3De8e8e8%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\neq} }&amp;bg=e8e8e8&amp;fg=000000' title='{ {\neq} }&amp;bg=e8e8e8&amp;fg=000000' class='latex' /> NP is a <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B2%7D+%26%2338%3Bbg%3De8e8e8%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{2} &amp;bg=e8e8e8&amp;fg=000000' title='\Pi_{2} &amp;bg=e8e8e8&amp;fg=000000' class='latex' /> sentence. 	</em></p></blockquote>
<p> <em>Proof:</em>  Define <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cpsi%28x%2Cc%2Cy%29%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\psi(x,c,y)} }&amp;fg=000000' title='{ {\psi(x,c,y)} }&amp;fg=000000' class='latex' /> to denote that
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++M_%7B%5Bx%2Cc%5D%7D%28y%29+%3D+z%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  M_{[x,c]}(y) = z&amp;fg=000000' title='\displaystyle  M_{[x,c]}(y) = z&amp;fg=000000' class='latex' /></p>
<p> and either (i) <img src='http://l.wordpress.com/latex.php?latex=%7By%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{y}&amp;fg=000000' title='{y}&amp;fg=000000' class='latex' /> is in <img src='http://l.wordpress.com/latex.php?latex=%7B%7B%5Cmathsf%7BSAT%7D%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{{\mathsf{SAT}}}&amp;fg=000000' title='{{\mathsf{SAT}}}&amp;fg=000000' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=%7Bz%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{z}&amp;fg=000000' title='{z}&amp;fg=000000' class='latex' /> is &#8220;reject&#8221;, or (ii) <img src='http://l.wordpress.com/latex.php?latex=%7By%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{y}&amp;fg=000000' title='{y}&amp;fg=000000' class='latex' /> is not in <img src='http://l.wordpress.com/latex.php?latex=%7B%7B%5Cmathsf%7BSAT%7D%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{{\mathsf{SAT}}}&amp;fg=000000' title='{{\mathsf{SAT}}}&amp;fg=000000' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=%7Bz%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{z}&amp;fg=000000' title='{z}&amp;fg=000000' class='latex' /> is &#8220;accept.&#8221; I claim that P <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cneq%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\neq} }&amp;fg=000000' title='{ {\neq} }&amp;fg=000000' class='latex' /> NP is equivalent to
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%09%5Cforall+x%2Cc+%5C%2C+%5Cexists+y+%5C%2C+%5Cpsi%28x%2Cc%2Cy%29.+%5C+%5C+%5C+%5C+%5C+%281%29%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  	\forall x,c \, \exists y \, \psi(x,c,y). \ \ \ \ \ (1)&amp;fg=000000' title='\displaystyle  	\forall x,c \, \exists y \, \psi(x,c,y). \ \ \ \ \ (1)&amp;fg=000000' class='latex' /></p>
<p>
Suppose that P <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cneq%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\neq} }&amp;fg=000000' title='{ {\neq} }&amp;fg=000000' class='latex' /> NP is true. Then, I claim that (1) is true. Suppose that it is false, then it follows that there are <img src='http://l.wordpress.com/latex.php?latex=%7Bx%2Cc%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{x,c}&amp;fg=000000' title='{x,c}&amp;fg=000000' class='latex' /> so that <img src='http://l.wordpress.com/latex.php?latex=%7B+M_%7B%5Bx%2Cc%5D%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ M_{[x,c]}}&amp;fg=000000' title='{ M_{[x,c]}}&amp;fg=000000' class='latex' /> correctly computes <img src='http://l.wordpress.com/latex.php?latex=%7B%7B%5Cmathsf%7BSAT%7D%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{{\mathsf{SAT}}}&amp;fg=000000' title='{{\mathsf{SAT}}}&amp;fg=000000' class='latex' />. This is a contradiction.</p>
<p>
Next suppose that P=NP. Then, I claim that (1) is false. Let <img src='http://l.wordpress.com/latex.php?latex=%7Bx%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{x}&amp;fg=000000' title='{x}&amp;fg=000000' class='latex' /> be the program that runs in polynomial time and computes <img src='http://l.wordpress.com/latex.php?latex=%7B%7B%5Cmathsf%7BSAT%7D%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{{\mathsf{SAT}}}&amp;fg=000000' title='{{\mathsf{SAT}}}&amp;fg=000000' class='latex' /> correctly. Then, this is again a contradiction. <img src='http://l.wordpress.com/latex.php?latex=%5CBox%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Box&amp;fg=000000' title='\Box&amp;fg=000000' class='latex' /></p>
<p>
Second, now let us look at how to encode P=NP. It is easy to see that it is a <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5CSigma_%7B2%7D%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\Sigma_{2}} }&amp;fg=000000' title='{ {\Sigma_{2}} }&amp;fg=000000' class='latex' /> sentence, since it is the negation of the first sentence, P <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cneq%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\neq} }&amp;fg=000000' title='{ {\neq} }&amp;fg=000000' class='latex' /> NP. Thus, P=NP is:
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Cexists+x%2Cc+%5C%2C+%5Cforall+y+%5C%2C+%5Cneg%5Cpsi%28x%2Cc%2Cy%29+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \exists x,c \, \forall y \, \neg\psi(x,c,y) &amp;fg=000000' title='\displaystyle  \exists x,c \, \forall y \, \neg\psi(x,c,y) &amp;fg=000000' class='latex' /></p>
<p>
A question that I wonder about is : can we do better? Can we encode, for example, P <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cneq%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\neq} }&amp;fg=000000' title='{ {\neq} }&amp;fg=000000' class='latex' /> NP, as a <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B1%7D+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{1} &amp;fg=000000' title='\Pi_{1} &amp;fg=000000' class='latex' /> sentence?</p>
<p>
<p><b> A Curious Property of <img src='http://l.wordpress.com/latex.php?latex=%7B%5Cmathsf%7BSAT%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\mathsf{SAT}}&amp;fg=000000' title='{\mathsf{SAT}}&amp;fg=000000' class='latex' /> </b></p>
<p><p>
We have shown that P <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cneq%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\neq} }&amp;fg=000000' title='{ {\neq} }&amp;fg=000000' class='latex' /> NP is a <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B2%7D+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{2} &amp;fg=000000' title='\Pi_{2} &amp;fg=000000' class='latex' /> sentence. Every such sentence can be viewed as defining a Skolem function. Recall if
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%5Cforall+x+%5C%2C+%5Cexists+y+%5C%2C+A%28x%2Cy%29+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  \forall x \, \exists y \, A(x,y) &amp;fg=000000' title='\displaystyle  \forall x \, \exists y \, A(x,y) &amp;fg=000000' class='latex' /></p>
<p>is true, then there is a function <img src='http://l.wordpress.com/latex.php?latex=f+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f &amp;fg=000000' title='f &amp;fg=000000' class='latex' />, called the  <em>Skolem function</em>, so that for all <img src='http://l.wordpress.com/latex.php?latex=x+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x &amp;fg=000000' title='x &amp;fg=000000' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=A%28x%2Cf%28x%29%29+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A(x,f(x)) &amp;fg=000000' title='A(x,f(x)) &amp;fg=000000' class='latex' />.</p>
<p>
Let <img src='http://l.wordpress.com/latex.php?latex=%7Bf%28i%2Cc%29%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{f(i,c)}&amp;fg=000000' title='{f(i,c)}&amp;fg=000000' class='latex' /> be the smallest natural number <img src='http://l.wordpress.com/latex.php?latex=%7By%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{y}&amp;fg=000000' title='{y}&amp;fg=000000' class='latex' /> so that <img src='http://l.wordpress.com/latex.php?latex=%7BM_%7B%5Bi%2Cc%5D%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{M_{[i,c]}}&amp;fg=000000' title='{M_{[i,c]}}&amp;fg=000000' class='latex' /> makes a mistake on the input <img src='http://l.wordpress.com/latex.php?latex=%7By%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{y}&amp;fg=000000' title='{y}&amp;fg=000000' class='latex' />. Then, if P <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cneq%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\neq} }&amp;fg=000000' title='{ {\neq} }&amp;fg=000000' class='latex' /> NP is true, the function <img src='http://l.wordpress.com/latex.php?latex=%7Bf%28i%2Cc%29%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{f(i,c)}&amp;fg=000000' title='{f(i,c)}&amp;fg=000000' class='latex' /> is always defined.</p>
<p>
The curious property of this function is this. If <img src='http://l.wordpress.com/latex.php?latex=%7Bf%28i%2Cc%29%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{f(i,c)}&amp;fg=000000' title='{f(i,c)}&amp;fg=000000' class='latex' /> grows &#8220;fast&#8221; for an infinite number of inputs, then while P <img src='http://l.wordpress.com/latex.php?latex=%7B+%7B%5Cneq%7D+%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{ {\neq} }&amp;fg=000000' title='{ {\neq} }&amp;fg=000000' class='latex' /> NP, my <a href="http://rjlipton.wordpress.com/2009/02/12/a-nightmare-about-sat/">nightmare</a> happens. In this case, there are an infinite number of <img src='http://l.wordpress.com/latex.php?latex=%7Bn%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n}&amp;fg=000000' title='{n}&amp;fg=000000' class='latex' /> so that <img src='http://l.wordpress.com/latex.php?latex=%7B%7B%5Cmathsf%7BSAT%7D%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{{\mathsf{SAT}}}&amp;fg=000000' title='{{\mathsf{SAT}}}&amp;fg=000000' class='latex' /> for inputs of length <img src='http://l.wordpress.com/latex.php?latex=%7Bn%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n}&amp;fg=000000' title='{n}&amp;fg=000000' class='latex' /> has circuit size <img src='http://l.wordpress.com/latex.php?latex=%7Bn%5E%7BO%28%5Clog+n%29%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n^{O(\log n)}}&amp;fg=000000' title='{n^{O(\log n)}}&amp;fg=000000' class='latex' />. </p>
<blockquote><p><b>Theorem: </b> <em> Suppose that there are infinite number of <img src='http://l.wordpress.com/latex.php?latex=i+%26%2338%3Bbg%3De8e8e8%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='i &amp;bg=e8e8e8&amp;fg=000000' title='i &amp;bg=e8e8e8&amp;fg=000000' class='latex' /> for which there exists a <img src='http://l.wordpress.com/latex.php?latex=c+%26%2338%3Bbg%3De8e8e8%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c &amp;bg=e8e8e8&amp;fg=000000' title='c &amp;bg=e8e8e8&amp;fg=000000' class='latex' /> so that 	</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+f%28i%2Cc%29+%26%2362%3B+2%5E%7B2%5E%7B%26%23124%3Bi%26%23124%3B%2Bc%7D%7D%26%2338%3Bbg%3De8e8e8%26%2338%3Bfg%3D000000.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle f(i,c) &gt; 2^{2^{&#124;i&#124;+c}}&amp;bg=e8e8e8&amp;fg=000000.' title='\displaystyle f(i,c) &gt; 2^{2^{&#124;i&#124;+c}}&amp;bg=e8e8e8&amp;fg=000000.' class='latex' /></p>
<p>Then, for infinitely many <img src='http://l.wordpress.com/latex.php?latex=%7Bn%7D%26%2338%3Bbg%3De8e8e8%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n}&amp;bg=e8e8e8&amp;fg=000000' title='{n}&amp;bg=e8e8e8&amp;fg=000000' class='latex' />,  <img src='http://l.wordpress.com/latex.php?latex=%7B%5Cmathsf%7BSAT%7D%7D%26%2338%3Bbg%3De8e8e8%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\mathsf{SAT}}&amp;bg=e8e8e8&amp;fg=000000' title='{\mathsf{SAT}}&amp;bg=e8e8e8&amp;fg=000000' class='latex' /> has circuit size <img src='http://l.wordpress.com/latex.php?latex=%7Bn%5E%7BO%28%5Clog+n%29%7D%7D%26%2338%3Bbg%3De8e8e8%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n^{O(\log n)}}&amp;bg=e8e8e8&amp;fg=000000' title='{n^{O(\log n)}}&amp;bg=e8e8e8&amp;fg=000000' class='latex' />. </em></p></blockquote>
<p><p>
<em>Proof:</em>  Let <img src='http://l.wordpress.com/latex.php?latex=%7Bi%26%2362%3B1%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{i&gt;1}&amp;fg=000000' title='{i&gt;1}&amp;fg=000000' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=%7Bc%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{c}&amp;fg=000000' title='{c}&amp;fg=000000' class='latex' /> be so that
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+f%28i%2Cc%29+%26%2362%3B+2%5E%7B2%5E%7B%26%23124%3Bi%26%23124%3B%2Bc%7D%7D.%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle f(i,c) &gt; 2^{2^{&#124;i&#124;+c}}.&amp;fg=000000' title='\displaystyle f(i,c) &gt; 2^{2^{&#124;i&#124;+c}}.&amp;fg=000000' class='latex' /></p>
<p> Define <img src='http://l.wordpress.com/latex.php?latex=%7Bn+%3D+2%5E%7B%26%23124%3Bi%26%23124%3B%2Bc-1%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n = 2^{&#124;i&#124;+c-1}}&amp;fg=000000' title='{n = 2^{&#124;i&#124;+c-1}}&amp;fg=000000' class='latex' />. Note, that <img src='http://l.wordpress.com/latex.php?latex=%7Bc%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{c}&amp;fg=000000' title='{c}&amp;fg=000000' class='latex' /> is at most <img src='http://l.wordpress.com/latex.php?latex=%7B%5Clog+n%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\log n}&amp;fg=000000' title='{\log n}&amp;fg=000000' class='latex' />. Then, <img src='http://l.wordpress.com/latex.php?latex=%7BM_%7B%5Bi%2Cc%5D%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{M_{[i,c]}}&amp;fg=000000' title='{M_{[i,c]}}&amp;fg=000000' class='latex' /> on all <img src='http://l.wordpress.com/latex.php?latex=%7By%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{y}&amp;fg=000000' title='{y}&amp;fg=000000' class='latex' /> of length <img src='http://l.wordpress.com/latex.php?latex=%7Bn%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n}&amp;fg=000000' title='{n}&amp;fg=000000' class='latex' /> is correct, since
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++y+%5Cle+2%5E%7Bn%7D+%3D+2%5E%7B2%5E%7B%26%23124%3Bi%26%23124%3B%2Bc-1%7D%7D+%26%2360%3B+f%28i%2Cc%29.%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  y \le 2^{n} = 2^{2^{&#124;i&#124;+c-1}} &lt; f(i,c).&amp;fg=000000' title='\displaystyle  y \le 2^{n} = 2^{2^{&#124;i&#124;+c-1}} &lt; f(i,c).&amp;fg=000000' class='latex' /></p>
<p> The size of the circuit that simulates this Turing machine on inputs of length <img src='http://l.wordpress.com/latex.php?latex=%7Bn%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n}&amp;fg=000000' title='{n}&amp;fg=000000' class='latex' /> is polynomial in <img src='http://l.wordpress.com/latex.php?latex=%7B%26%23124%3Bi%26%23124%3B%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{&#124;i&#124;}&amp;fg=000000' title='{&#124;i&#124;}&amp;fg=000000' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%7Bn%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{n}&amp;fg=000000' title='{n}&amp;fg=000000' class='latex' />, and the running time of the machine. The machine, by definition, runs in time
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle++%26%23124%3By%26%23124%3B%5E%7Bc%7D+%5Cle+n%5E%7Bc%7D+%5Cle+n%5E%7B%5Clog+n%7D.%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle  &#124;y&#124;^{c} \le n^{c} \le n^{\log n}.&amp;fg=000000' title='\displaystyle  &#124;y&#124;^{c} \le n^{c} \le n^{\log n}.&amp;fg=000000' class='latex' /></p>
<p>	 Thus, the theorem is proved. <img src='http://l.wordpress.com/latex.php?latex=%5CBox%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Box&amp;fg=000000' title='\Box&amp;fg=000000' class='latex' /></p>
<p><b> Open Problems </b></p>
<p>
Before I state some open problems, I want to thank Subrahmanyam Kalyanasundaram and Ken Regan for many helpful comments on this post. Subrahmanyam has helped me many times, and is terrific at helping improve the posts.</p>
<p>Ken pointed out several interesting things that I will have to leave for a future post. The main issues he raised are: First, there is previous work on the &#8220;curious&#8221; property done by Shai Ben-David in an apparently unpublished 1992 <a href="http://www.cs.uwaterloo.ca/~shai/pub.html#onthep">technical report.</a> Further, Ken also points out that any NP-complete language can be used in the place of <img src='http://l.wordpress.com/latex.php?latex=%7B%5Cmathsf%7BSAT%7D%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\mathsf{SAT}}&amp;fg=000000' title='{\mathsf{SAT}}&amp;fg=000000' class='latex' />.</p>
<p>
Second, that the placement of P=NP into AH has been thought about before&#8211;I am not surprised. Ken mentions that the logician Alex Wilkie repeated a rumor to the effect that P=NP was in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_1+%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_1 &amp;fg=000000' title='\Pi_1 &amp;fg=000000' class='latex' /> in a <a href="http://en.wikipedia.org/wiki/Public_house">pub</a> <del datetime="2009-05-28T11:04:41+00:00">one of his publications</del> in the mid-1980s, but nothing more ever came of it. Then, Ken recalls, that Tomoyuki Yamakami and others claimed this again around 2002-2003, but their claim was later withdrawn. </p>
<p>
Finally, one of the reasons that a <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B2%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{2}&amp;fg=000000' title='\Pi_{2}&amp;fg=000000' class='latex' /> sentence <img src='http://l.wordpress.com/latex.php?latex=%7B%5Cforall+x+%5C%2C+%5Cexists+y+%5C%2C+A%28x%2Cy%29%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\forall x \, \exists y \, A(x,y)}&amp;fg=000000' title='{\forall x \, \exists y \, A(x,y)}&amp;fg=000000' class='latex' /> can be <em>unprovable</em> in Peano Arithmetic is that the Skolem function for this sentence may grow too fast. This happens, for example, for the famous <a href="http://en.wikipedia.org/wiki/Paris-Harrington_theorem">Paris-Harrington</a> version of the Ramsey Theorem. So what the above theorem says is this: if P <img src='http://l.wordpress.com/latex.php?latex=%7B%5Cneq%7D%26%2338%3Bfg%3D000000&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='{\neq}&amp;fg=000000' title='{\neq}&amp;fg=000000' class='latex' /> NP is unprovable in Peano because the Skolem function grows too fast, then my nightmare about P=NP happens. This is the core idea of the paper of Ben-David. More on his interesting work in a future post. </p>
<p>
So here are the open questions: What is the complexity of other problems? What is the complexity of your favorite problem? Is P=NP in <img src='http://l.wordpress.com/latex.php?latex=%5CPi_%7B1%7D%26%2338%3Bfg%3D000000%3F&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Pi_{1}&amp;fg=000000?' title='\Pi_{1}&amp;fg=000000?' class='latex' /> </p>
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<title><![CDATA[Da importância do simbolismo]]></title>
<link>http://linguagemobjeto.wordpress.com/2009/02/26/da-importancia-do-simbolismo/</link>
<pubDate>Thu, 26 Feb 2009 23:55:43 +0000</pubDate>
<dc:creator>Cleber</dc:creator>
<guid>http://linguagemobjeto.wordpress.com/2009/02/26/da-importancia-do-simbolismo/</guid>
<description><![CDATA[Anyone who doubts the advantages of symbols (in their proper place) is invited to solve the equation]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><blockquote>
<p style="text-align:justify;">Anyone who doubts the advantages of symbols (in their proper place) is invited to solve the equation x2 + 3x &#8211; 2 = 0 by completing the square (as taught in high school), but doing all the work in words. We start him off by stating the equation in words: The square of the unknown, increased by three times the unknown, and diminished by two, is equal to zero.</p>
<p style="text-align:justify;">Anyone who doubts that apt choices of mathematical symbolism have played a major role in the modern development of mathematics and science is invited to multiply 416 by 144, but doing all the manipulations in Roman numerals. His problem is thus to multiply CDXVI by CXLIV.</p>
</blockquote>
<p style="text-align:justify;">Kleene, <a href="http://www.amazon.com/Mathematical-Logic-Stephen-Cole-Kleene/dp/0486425339/ref=pd_bbs_sr_1?ie=UTF8&#38;s=books&#38;qid=1235692500&#38;sr=8-1">&#8216;Mathematical Logic&#8217;</a>, p. 5.</p>
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<item>
<title><![CDATA[... und nun hat sie es doch wieder erwischt]]></title>
<link>http://2exilberliner.wordpress.com/2008/02/17/und-nun-hat-sie-es-doch-wieder-erwischt/</link>
<pubDate>Sun, 17 Feb 2008 11:41:47 +0000</pubDate>
<dc:creator>2exilberliner</dc:creator>
<guid>http://2exilberliner.wordpress.com/2008/02/17/und-nun-hat-sie-es-doch-wieder-erwischt/</guid>
<description><![CDATA[Nach drei Tagen Fieber bis zu 40°C sind wir heute Morgen nun doch mit unserer kleinen Maus zum Notar]]></description>
<content:encoded><![CDATA[Nach drei Tagen Fieber bis zu 40°C sind wir heute Morgen nun doch mit unserer kleinen Maus zum Notar]]></content:encoded>
</item>
<item>
<title><![CDATA[neue Frisur, neuer Job und ein bisschen Urlaub ist alles was ich mir wünsche]]></title>
<link>http://2exilberliner.wordpress.com/2008/02/12/ne-neue-frisur-neuer-job-und-ein-bisschen-urlaub-mussen-her/</link>
<pubDate>Tue, 12 Feb 2008 19:17:06 +0000</pubDate>
<dc:creator>2exilberliner</dc:creator>
<guid>http://2exilberliner.wordpress.com/2008/02/12/ne-neue-frisur-neuer-job-und-ein-bisschen-urlaub-mussen-her/</guid>
<description><![CDATA[&#8230; und was soll ich sagen ich habe mich von meinen inzwischen doch sehr sehr langen Haaren getr]]></description>
<content:encoded><![CDATA[&#8230; und was soll ich sagen ich habe mich von meinen inzwischen doch sehr sehr langen Haaren getr]]></content:encoded>
</item>
<item>
<title><![CDATA[unsere kleene]]></title>
<link>http://2exilberliner.wordpress.com/2008/01/11/unsere-kleene/</link>
<pubDate>Fri, 11 Jan 2008 20:09:43 +0000</pubDate>
<dc:creator>2exilberliner</dc:creator>
<guid>http://2exilberliner.wordpress.com/2008/01/11/unsere-kleene/</guid>
<description><![CDATA[]]></description>
<content:encoded><![CDATA[]]></content:encoded>
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<title><![CDATA[Lamas und Krokodile]]></title>
<link>http://2exilberliner.wordpress.com/2008/01/05/lamas-und-krokodile/</link>
<pubDate>Sat, 05 Jan 2008 23:31:16 +0000</pubDate>
<dc:creator>Marc</dc:creator>
<guid>http://2exilberliner.wordpress.com/2008/01/05/lamas-und-krokodile/</guid>
<description><![CDATA[Was gibt es neues aus unserem Hause zu berichten? Nun ja, Paula hat einen neuen Zeitvertreib gefunde]]></description>
<content:encoded><![CDATA[Was gibt es neues aus unserem Hause zu berichten? Nun ja, Paula hat einen neuen Zeitvertreib gefunde]]></content:encoded>
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