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ProofSample.html
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<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html >
<head><title>Proof Sample</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<meta name="generator" content="TeX4ht (http://www.tug.org/tex4ht/)">
<meta name="originator" content="TeX4ht (http://www.tug.org/tex4ht/)">
<!-- html -->
<meta name="src" content="ProofSample.tex">
<link rel="stylesheet" type="text/css" href="ProofSample.css">
</head><body
>
<div class="maketitle">
<h2 class="titleHead">
PROOF SAMPLE
</h2><div class="authors"><span class="author" >
<span
class="cmr-10">M.</span><span
class="cmr-10"> J.</span><span
class="cmr-10"> KLEIN</span>
</span></div>
<div class="submaketitle">
<div class="date" >
<!--l. 52--><p class="indent" > <span
class="cmti-12">Date</span>: November 21, 2017.</div></div></div>
<h3 class="sectionHead"><span class="titlemark">1. </span> <a
id="x1-10001"></a>Introduction</h3>
<!--l. 56--><p class="noindent" >This problem came up in the process of searching for lower bounds on the
number of cycles in a graph with certain properties.
<!--l. 58--><p class="indent" > I chose this proof as my sample because it’s my favorite short proof. The
final bit of reasoning is analogous to reducing the problem to a halting
problem and solving that halting problem.
<h3 class="sectionHead"><span class="titlemark">2. </span> <a
id="x1-20002"></a>Proof</h3>
<!--l. 61--><p class="noindent" >This is a proof of the lower bound of the pseudo-boolean function
<center class="math-display" >
<img
src="ProofSample0x.png" alt=" ( )
m∑ -1 ∑i 2
f(A ) := aj
i=1 j=1 " class="math-display" ></center>
given that <span
class="cmmi-12">A </span><span
class="cmsy-10x-x-120">∈ </span><span
class="msbm-10x-x-120">B</span><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup>, <span
class="cmmi-12">ε </span>= <span
class="cmex-10x-x-120">∑</span><sub>
<span
class="cmmi-8">i</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">i</span></sub> = constant, and <span
class="cmmi-12">m </span><span
class="cmsy-10x-x-120">≥ </span>2. In other
words, this is an optimization of the function <span
class="cmmi-12">f </span>: <span
class="msbm-10x-x-120">B</span><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup> <span
class="cmsy-10x-x-120">→ </span><span
class="msbm-10x-x-120">ℤ</span>, where the
binary input vector has a given Hamming Weight.
<div class="newtheorem">
<!--l. 66--><p class="noindent" ><span class="head">
<a
id="x1-2001r1"></a>
<span
class="cmbx-12">Lemma 2.1.</span> </span>
<center class="math-display" >
<img
src="ProofSample1x.png" alt="m∑ -1∑ i m∑ -1 m∑-1
aj = m ai - iai
i=1 j=1 i=1 i=1
" class="math-display" ></center>
</div>
<div class="proof">
<!--l. 71--><p class="indent" > <span class="head">
<span
class="cmti-12">Proof.</span> </span>
<center class="math-display" >
<img
src="ProofSample2x.png" alt="m∑ -1∑i
aj = (a1) + (a1 + a2) + (a1 + a2 + a3) + ...
i=1 j=1
" class="math-display" ></center> We have (<span
class="cmmi-12">m </span><span
class="cmsy-10x-x-120">- </span><span
class="cmmi-12">i</span>) copies of <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">i</span></sub> so <br
class="newline" />
<center class="math-display" >
<img
src="ProofSample3x.png" alt="m -1 i m- 1 m- 1 m -1
∑ ∑ ∑ ∑ ∑
aj = (m - i)ai = m ai - iai
i=1 j=1 i=1 i=1 i=1
" class="math-display" ></center> <span class="qed"><span
class="msam-10x-x-120">□</span></span>
</div>
<div class="newtheorem">
<!--l. 77--><p class="noindent" ><span class="head">
<a
id="x1-2002r2"></a>
<span
class="cmbx-12">Lemma 2.2.</span> </span>
<center class="math-display" >
<img
src="ProofSample4x.png" alt="m-1 i m-1 m- 1 m -1 m-1
∑ ∑ m2- ∑ m- ∑ 1∑ 1-∑ 2
aji = 2 ai - 2 ai + 2 iai - 2 i ai
i=1 j=1 i=1 i=1 i=1 i=1
" class="math-display" ></center>
</div>
<div class="proof">
<!--l. 86--><p class="indent" > <span class="head">
<span
class="cmti-12">Proof.</span> </span>
<table
class="align-star">
<tr><td
class="align-odd"><span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">i</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">j</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">i</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">j</span></sub><span
class="cmmi-12">i</span></td> <td
class="align-even"> = 1(<span
class="cmmi-12">a</span><sub><span
class="cmr-8">1</span></sub>) + 2(<span
class="cmmi-12">a</span><sub><span
class="cmr-8">1</span></sub> + <span
class="cmmi-12">a</span><sub><span
class="cmr-8">2</span></sub>) + 3(<span
class="cmmi-12">a</span><sub><span
class="cmr-8">1</span></sub> + <span
class="cmmi-12">a</span><sub><span
class="cmr-8">2</span></sub> + <span
class="cmmi-12">a</span><sub><span
class="cmr-8">3</span></sub>) + <span
class="cmmi-12">…</span></td> <td
class="align-label"></td> <td
class="align-label">
</td></tr><tr><td
class="align-odd"></td> <td
class="align-even"> = <span
class="cmmi-12">a</span><sub><span
class="cmr-8">1</span></sub> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">j</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">j </span>+ <span
class="cmmi-12">a</span><sub>
<span
class="cmr-8">2</span></sub> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">j</span><span
class="cmr-8">=2</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">j </span>+ <span
class="cmmi-12">a</span><sub>
<span
class="cmr-8">3</span></sub> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">j</span><span
class="cmr-8">=3</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">j </span>+ <span
class="cmmi-12">…</span></td> <td
class="align-label"></td> <td
class="align-label"></td></tr></table>
But <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">j</span><span
class="cmr-8">=</span><span
class="cmmi-8">i</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">j </span>= (<span
class="cmmi-12">m </span><span
class="cmsy-10x-x-120">- </span><span
class="cmmi-12">i</span>)(<span
class="cmmi-12">m </span>+ <span
class="cmmi-12">i </span><span
class="cmsy-10x-x-120">- </span>1)<span
class="cmmi-12">∕</span>2 so then we have:
<table
class="align-star">
<tr><td
class="align-odd"> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">i</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">j</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">i</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">j</span></sub><span
class="cmmi-12">i</span></td> <td
class="align-even"> = <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">i</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">i</span></sub>(<span
class="cmmi-12">m </span><span
class="cmsy-10x-x-120">- </span><span
class="cmmi-12">i</span>)(<span
class="cmmi-12">m </span>+ <span
class="cmmi-12">i </span><span
class="cmsy-10x-x-120">- </span>1)<span
class="cmmi-12">∕</span>2</td> <td
class="align-label"></td> <td
class="align-label">
</td></tr><tr><td
class="align-odd"></td> <td
class="align-even"> = <img
src="ProofSample5x.png" alt="1-
2" class="frac" align="middle"> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">i</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">i</span></sub>(<span
class="cmmi-12">m</span><sup><span
class="cmr-8">2</span></sup> <span
class="cmsy-10x-x-120">- </span><span
class="cmmi-12">m </span>+ <span
class="cmmi-12">i </span><span
class="cmsy-10x-x-120">- </span><span
class="cmmi-12">i</span><sup><span
class="cmr-8">2</span></sup>)</td> <td
class="align-label"></td> <td
class="align-label">
</td></tr><tr><td
class="align-odd"></td> <td
class="align-even"> = <img
src="ProofSample6x.png" alt="m2
---
2" class="frac" align="middle"> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">i</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">i</span></sub> <span
class="cmsy-10x-x-120">-</span><img
src="ProofSample7x.png" alt="m
--
2" class="frac" align="middle"> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">i</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">i</span></sub> + <img
src="ProofSample8x.png" alt="1
--
2" class="frac" align="middle"> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">i</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">ia</span><sub>
<span
class="cmmi-8">i</span></sub> <span
class="cmsy-10x-x-120">-</span><img
src="ProofSample9x.png" alt="1
--
2" class="frac" align="middle"> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">i</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">i</span><sup><span
class="cmr-8">2</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">i</span></sub></td> <td
class="align-label"></td> <td
class="align-label"></td></tr></table>
<span class="qed"><span
class="msam-10x-x-120">□</span></span>
</div>
<div class="newtheorem">
<!--l. 101--><p class="noindent" ><span class="head">
<a
id="x1-2003r3"></a>
<span
class="cmbx-12">Lemma 2.3.</span> </span>
<center class="math-display" >
<img
src="ProofSample10x.png" alt=" ( )2 ( )
∑m ∑i ∑m ∑m i-∑ 1
am+1-j = iai + 2jajai
i=1 j=1 i=1 i=2 j=1
" class="math-display" ></center>
</div>
<div class="proof">
<!--l. 106--><p class="indent" > <span class="head">
<span
class="cmti-12">Proof.</span> </span>Note that
<center class="math-display" >
<img
src="ProofSample11x.png" alt="( )
∑i 2 ∑i ∑i
am+1- j = a2m+1- j + 2 am+1-jam+1 -k
j=1 j=1 j<k
" class="math-display" ></center> But <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">m</span><span
class="cmr-8">+1</span><span
class="cmsy-8">-</span><span
class="cmmi-8">j</span></sub><sup><span
class="cmr-8">2</span></sup> = <span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">m</span><span
class="cmr-8">+1</span><span
class="cmsy-8">-</span><span
class="cmmi-8">j</span></sub>, because <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">m</span><span
class="cmr-8">+1</span><span
class="cmsy-8">-</span><span
class="cmmi-8">j</span></sub> <span
class="cmsy-10x-x-120">∈{</span>0<span
class="cmmi-12">, </span>1<span
class="cmsy-10x-x-120">} </span>so:
<center class="math-display" >
<img
src="ProofSample12x.png" alt="∑ i ∑i
a2 = a
m+1 -j m+1-j
j=1 j=1
" class="math-display" ></center> Thus
<center class="math-display" >
<img
src="ProofSample13x.png" alt=" m ( i )2 m ( i ) m ( i )
∑ ∑ ∑ ∑ ∑ ∑
am+1 -j = am+1 -j + 2 am+1 -jam+1- k
i=1 j=1 i=1 j=1 i=1 j<k
" class="math-display" ></center> But note that, because there are <span
class="cmmi-12">i </span>copies of <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">m</span><span
class="cmr-8">+1</span><span
class="cmsy-8">-</span><span
class="cmr-8">(</span><span
class="cmmi-8">m</span><span
class="cmr-8">+1</span><span
class="cmsy-8">-</span><span
class="cmmi-8">i</span><span
class="cmr-8">)</span></sub>,
<center class="math-display" >
<img
src="ProofSample14x.png" alt=" ( )
∑m ∑ i ∑m
am+1- j = iai
i=1 j=1 i=1
" class="math-display" ></center> Likewise, there are <span
class="cmmi-12">i </span>copies of each distinct pair of <span
class="cmmi-12">a</span>’s, so
<center class="math-display" >
<img
src="ProofSample15x.png" alt="∑m ( ∑i ) ∑m ( i∑-1 )
2 a a = 2ja a
m+1- j m+1-k j i
i=1 j<k i=1 j=2
" class="math-display" ></center> <span class="qed"><span
class="msam-10x-x-120">□</span></span>
</div>
<div class="newtheorem">
<!--l. 132--><p class="noindent" ><span class="head">
<a
id="x1-2004r1"></a>
<span
class="cmbx-12">Theorem 1.</span> </span>
<center class="math-display" >
<img
src="ProofSample16x.png" alt=" ( )2
1 m∑ -1 ∑i
--ε(ε + 1)(2 ε + 1) ≤ aj = f (A)
6 i=1 j=1
" class="math-display" ></center>
</div>
<div class="proof">
<!--l. 137--><p class="indent" > <span class="head">
<span
class="cmti-12">Proof.</span> </span>We begin by reversing the indices of the <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">j</span></sub>’s and applying
Lemma <a
href="#x1-2003r3">2.3<!--tex4ht:ref: lem12 --></a> to get
<center class="math-display" >
<img
src="ProofSample17x.png" alt="m∑-1( ∑ i )2 ∑m ∑m ( ∑i-1 )
a = ia + 2ja a
j i j i
i=1 j=1 i=1 i=2 j=1
" class="math-display" ></center> Recall that <span
class="cmmi-12">ε </span>= <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">i</span><span
class="cmr-8">=1</span></sub><sup><span
class="cmmi-8">m</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">i</span></sub>. We first consider the trivial cases of
<span
class="cmmi-12">ε </span>= 0<span
class="cmmi-12">, </span>1.
<!--l. 144--><p class="indent" > If <span
class="cmmi-12">ε </span>= 0, then all the <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">i</span></sub> = 0 and so the total is 0 and the theorem
holds.
<!--l. 146--><p class="indent" > If <span
class="cmmi-12">ε </span>= 1, then exactly one <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">i</span></sub><span
class="cmmi-12">≠</span>0. Let us call it <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">p</span></sub> for positive. Then the
total is equal to <span
class="cmmi-12">ka</span><sub><span
class="cmmi-8">k</span></sub>, because no pair of <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">i</span></sub>’s can be non-zero. Taking
the minimum <span
class="cmmi-12">k </span>= 1, we get the total is 1 and the theorem holds.
<!--l. 150--><p class="indent" > Note that if <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">z</span></sub> is positive, it adds exactly <span
class="cmmi-12">z </span>+ 2<span
class="cmmi-12">z</span> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">j</span><span
class="cmr-8">=</span><span
class="cmmi-8">z</span><span
class="cmr-8">+1</span></sub><sup><span
class="cmmi-8">m</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">j</span></sub> to the
total, independent of the rest.
<!--l. 152--><p class="indent" > Suppose that there exists some <span
class="cmmi-12">y </span>such that <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">y</span></sub> = 1 and for all <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">i</span></sub> = 0,
<span
class="cmmi-12">i < y</span>. Let <span
class="cmmi-12">y </span>be the minimum such index. The contribution of <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">y</span></sub> is
<span
class="cmmi-12">y </span>+ 2<span
class="cmmi-12">y</span> <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">j</span><span
class="cmr-8">=</span><span
class="cmmi-8">y</span><span
class="cmr-8">+1</span></sub><sup><span
class="cmmi-8">m</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">j</span></sub>. Similarly the contribution of <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">y</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sub>, if we swap the
values of <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">y</span></sub> and <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">y</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sub>, is (<span
class="cmmi-12">y </span><span
class="cmsy-10x-x-120">- </span>1) + 2(<span
class="cmmi-12">y </span><span
class="cmsy-10x-x-120">- </span>1) <span
class="cmex-10x-x-120">∑</span>
<sub><span
class="cmmi-8">j</span><span
class="cmr-8">=</span><span
class="cmmi-8">y</span></sub><sup><span
class="cmmi-8">m</span></sup><span
class="cmmi-12">a</span><sub>
<span
class="cmmi-8">j</span></sub>. However, if we
only swap these two values, the values, and thus the contributions of
those values, with index greater than <span
class="cmmi-12">y </span>are left unchanged. Therefore,
the contribution of <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">y</span><span
class="cmsy-8">-</span><span
class="cmr-8">1</span></sub>, if the values are swapped, is
<center class="math-display" >
<img
src="ProofSample18x.png" alt=" ∑m
(y - 1) + 2(y - 1) aj.
j=y+1
" class="math-display" ></center> As this operation of swapping the first positive value above all the
zero-values will always decrease the total, one may repeat this operation
as long as there are positive values with higher index than some zero
value. Therefore, the minimum total is exactly the assignment where
there is no non-positive assignment below a positive one, that is precisely
the assignment with all positive assignments at the lowest possible
indices.
<!--l. 158--><p class="indent" > Thus our minimum total, given that <span
class="cmmi-12">ε < i </span>implies that <span
class="cmmi-12">a</span><sub><span
class="cmmi-8">i</span></sub> = 0, is
<center class="math-display" >
<img
src="ProofSample19x.png" alt=" ( )
∑ε ∑ε ∑i- 1 1
i + 2jajai = -ε (ε + 1)(2ε + 1).
i=1 i=2 j=1 6
" class="math-display" ></center> <span class="qed"><span
class="msam-10x-x-120">□</span></span>
</div>
</body></html>