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puzzles/project_euler/pe/p149.html
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puzzles/project_euler/pe/p149.html
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<!DOCTYPE html>
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<html lang="en">
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<head>
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<meta charset="utf-8" />
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<meta name="author" content="Colin Hughes" />
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<title>Problem 149 - Project Euler</title>
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<h2>Searching for a maximum-sum subsequence.</h2><div class="info" style="cursor:help;width:200px;margin-bottom:10px;"><h3>Problem 149</h3><span style="width:300px;color:#666;">Published on Friday, 13th April 2007, 10:00 pm; Solved by 2350</span></div>
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<div class="problem_content" role="problem">
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<p>Looking at the table below, it is easy to verify that the maximum possible sum of adjacent numbers in any direction (horizontal, vertical, diagonal or anti-diagonal) <span style="white-space:nowrap">is 16 (= 8 + 7 + 1).</span></p>
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<div style="text-align:center">
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<table border="1" cellpadding="6" cellspacing="0" style="margin:auto"><tbody align="right"><tr><td><img src='images/symbol_minus.gif' width='9' height='3' alt='−' border='0' style='vertical-align:middle;' />2</td ><td >5</td><td>3</td><td>2</td></tr>
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<tr><td>9</td><td><img src='images/symbol_minus.gif' width='9' height='3' alt='−' border='0' style='vertical-align:middle;' />6</td><td>5</td><td>1</td></tr>
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<tr><td>3</td><td>2</td><td>7</td><td>3</td></tr>
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<tr><td><img src='images/symbol_minus.gif' width='9' height='3' alt='−' border='0' style='vertical-align:middle;' />1</td><td>8</td><td><img src='images/symbol_minus.gif' width='9' height='3' alt='−' border='0' style='vertical-align:middle;' />4</td><td> 8</td></tr></tbody></table>
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</div>
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<p>Now, let us repeat the search, but on a much larger scale:</p>
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<p>First, generate four million pseudo-random numbers using a specific form of what is known as a "Lagged Fibonacci Generator":</p>
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<p>For 1 <img src='images/symbol_le.gif' width='10' height='12' alt='≤' border='0' style='vertical-align:middle;' /> <i>k</i> <img src='images/symbol_le.gif' width='10' height='12' alt='≤' border='0' style='vertical-align:middle;' /> 55, <i>s</i><sub><i>k</i></sub> = [100003 <img src='images/symbol_minus.gif' width='9' height='3' alt='−' border='0' style='vertical-align:middle;' /> 200003<i>k</i> + 300007<i>k</i><sup>3</sup>] (modulo 1000000) <img src='images/symbol_minus.gif' width='9' height='3' alt='−' border='0' style='vertical-align:middle;' /> 500000.<br />
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For 56 <img src='images/symbol_le.gif' width='10' height='12' alt='≤' border='0' style='vertical-align:middle;' /> <i>k</i> <img src='images/symbol_le.gif' width='10' height='12' alt='≤' border='0' style='vertical-align:middle;' /> 4000000, <i>s</i><sub><i>k</i></sub> = [<i>s</i><sub><i>k<img src='images/symbol_minus.gif' width='9' height='3' alt='−' border='0' style='vertical-align:middle;' />24</i></sub> + <i>s</i><sub><i>k<img src='images/symbol_minus.gif' width='9' height='3' alt='−' border='0' style='vertical-align:middle;' />55</i></sub> + 1000000] (modulo 1000000) <img src='images/symbol_minus.gif' width='9' height='3' alt='−' border='0' style='vertical-align:middle;' /> 500000.</p>
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<p>Thus, <i>s</i><sub>10</sub> = <img src='images/symbol_minus.gif' width='9' height='3' alt='−' border='0' style='vertical-align:middle;' />393027 and <i>s</i><sub>100</sub> = 86613.</p>
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<p>The terms of <i>s</i> are then arranged in a 2000<img src='images/symbol_times.gif' width='9' height='9' alt='×' border='0' style='vertical-align:middle;' />2000 table, using the first 2000 numbers to fill the first row (sequentially), the next 2000 numbers to fill the second row, and so on.</p>
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<p>Finally, find the greatest sum of (any number of) adjacent entries in any direction (horizontal, vertical, diagonal or anti-diagonal).</p>
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</div><br />
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