<?xml version="1.0" encoding="UTF-8" standalone="yes"?><oembed><version><![CDATA[1.0]]></version><provider_name><![CDATA[Chaos at the Sky]]></provider_name><provider_url><![CDATA[https://chaosatthesky.wordpress.com]]></provider_url><author_name><![CDATA[chaotic_iak]]></author_name><author_url><![CDATA[https://chaosatthesky.wordpress.com/author/chaoticiak/]]></author_url><title><![CDATA[The Genius, by A Skymin&#8217;s Mind&nbsp;#7]]></title><type><![CDATA[link]]></type><html><![CDATA[<h1>Monorail</h1>
<p>Used as Death Match of Season 3, Round 10 and Season 4, Round 6. (Discussion of S4E6 in particular will come later.)</p>
<p><b>Rules</b></p>
<p>The objective is to be the one that completes a train track, or to be the one that declares it&#8217;s impossible (correctly).</p>
<p>There are 18 square tiles used in the game. Two of them show straight tracks, the station tiles, which are already placed on the grid; the remaining 16 tiles can be made to show a straight track or a bend track, which are to be placed. Players place tiles so that sides match (a track leading to a non-track side is not permitted; think of tile-based games (such as <a href="https://en.wikipedia.org/wiki/Dominoes">Dominoes</a> or <a href="https://en.wikipedia.org/wiki/Carcassonne_%28board_game%29">Carcassonne</a>, where sides of tiles must match). A player may place one to three tiles at once, but all of them must be played in a straight line contiguously (think of <a href="https://en.wikipedia.org/wiki/Scrabble">Scrabble</a>, where you cannot place your tiles separated, only that it&#8217;s even more strict; the tiles must be contiguous, not separated by already existing tiles). However, the track itself doesn&#8217;t need to be connected; as long as the tiles placed are connected, it&#8217;s fine.</p>
<p>The person who places the last tile, completing the track using all tiles already placed, wins. However, a player can also declare that it&#8217;s impossible to complete the track, in which the opponent must either complete it (for a win to the non-challenger) or resign (for a win to the challenger).</p>
<p>Also see <a href="https://www.facebook.com/video.php?v=893305020682343">this video from the Facebook page</a>, which is unfortunately in Korean, but gives illustrations.</p>
<p><!--more--></p>
<p><b>Discussion</b></p>
<p>I wrote up <a href="https://chaosatthesky.wordpress.com/2015/01/02/intro-to-cgt-1/">an introduction to combinatorial game theory</a>, which is used in this discussion, if you&#8217;re not familiar with it before.</p>
<p>First, since in all cases moves are available to both players, this is an impartial game and so <a href="https://en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem">Sprague-Grundy theorem</a> declares that it has a <a href="https://en.wikipedia.org/wiki/Nimber">Grundy value</a> (aka nim-value, nimber). This makes it possible to analyze the game by finding nimbers.</p>
<p>Sprague-Grundy theorem requires the game to have the normal play convention (player that cannot move loses). For the purpose of this game, it&#8217;s easy to decide whether a track can still be completed or not; if it&#8217;s not, the current player can declare so for a victory. It&#8217;s easier to interpret this as &#8220;making a move that makes the track impossible to complete is an illegal move&#8221;, so that in case there is an impossible track, the previous player must have made an illegal move; if they indeed didn&#8217;t have any legal move, they have lost by that point for having no legal move. This makes the game to satisfy the normal play convention.</p>
<p>Now, this should be solvable by bruteforcing the way, but I&#8217;m too lazy to code or let enough computer-hours to do the job. Instead, I will provide one move that I&#8217;ve analyzed, which is losing for the first player but the forced win is so deep that there are only two winning moves, both hard to find, and any other move (about 100+ moves exist in any position) loses. The station is marked in orange and the move I suggest is marked in green; the only two winning moves are those in red (both playing one tile).</p>
<div data-shortcode="caption" id="attachment_1611" style="width: 123px" class="wp-caption aligncenter"><a href="https://chaosatthesky.files.wordpress.com/2014/12/monorail-hard-move.png"><img loading="lazy" aria-describedby="caption-attachment-1611" data-attachment-id="1611" data-permalink="https://chaosatthesky.wordpress.com/2014/12/08/the-genius-by-a-skymins-mind-7/monorail-hard-move/" data-orig-file="https://chaosatthesky.files.wordpress.com/2014/12/monorail-hard-move.png" data-orig-size="147,195" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;,&quot;orientation&quot;:&quot;0&quot;}" data-image-title="Monorail &#8211; Hard move" data-image-description="" data-image-caption="&lt;p&gt;Monorail &#8211; A hard move to counter&lt;/p&gt;
" data-medium-file="https://chaosatthesky.files.wordpress.com/2014/12/monorail-hard-move.png?w=147" data-large-file="https://chaosatthesky.files.wordpress.com/2014/12/monorail-hard-move.png?w=147" class="size-thumbnail wp-image-1611" src="https://chaosatthesky.files.wordpress.com/2014/12/monorail-hard-move.png?w=113&#038;h=150" alt="Monorail - A hard move to counter" width="113" height="150" srcset="https://chaosatthesky.files.wordpress.com/2014/12/monorail-hard-move.png?w=113&amp;h=150 113w, https://chaosatthesky.files.wordpress.com/2014/12/monorail-hard-move.png 147w" sizes="(max-width: 113px) 100vw, 113px" /></a><p id="caption-attachment-1611" class="wp-caption-text">Monorail &#8211; A hard move to counter</p></div>
<p>The first reason is that the <em>only possible track</em> involving these tiles is the following:</p>
<div data-shortcode="caption" id="attachment_1612" style="width: 130px" class="wp-caption aligncenter"><a href="https://chaosatthesky.files.wordpress.com/2014/12/monorail-only-completion.png"><img loading="lazy" aria-describedby="caption-attachment-1612" data-attachment-id="1612" data-permalink="https://chaosatthesky.wordpress.com/2014/12/08/the-genius-by-a-skymins-mind-7/monorail-only-completion/" data-orig-file="https://chaosatthesky.files.wordpress.com/2014/12/monorail-only-completion.png" data-orig-size="195,243" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;,&quot;orientation&quot;:&quot;0&quot;}" data-image-title="Monorail &#8211; Only completion" data-image-description="" data-image-caption="&lt;p&gt;Monorail &#8211; Only way to complete&lt;/p&gt;
" data-medium-file="https://chaosatthesky.files.wordpress.com/2014/12/monorail-only-completion.png?w=195" data-large-file="https://chaosatthesky.files.wordpress.com/2014/12/monorail-only-completion.png?w=195" class="size-thumbnail wp-image-1612" src="https://chaosatthesky.files.wordpress.com/2014/12/monorail-only-completion.png?w=120&#038;h=150" alt="Monorail - Only way to complete" width="120" height="150" srcset="https://chaosatthesky.files.wordpress.com/2014/12/monorail-only-completion.png?w=120&amp;h=150 120w, https://chaosatthesky.files.wordpress.com/2014/12/monorail-only-completion.png 195w" sizes="(max-width: 120px) 100vw, 120px" /></a><p id="caption-attachment-1612" class="wp-caption-text">Monorail &#8211; Only way to complete</p></div>
<p>This can be proven, but it&#8217;s a good puzzle nevertheless. (Aka I&#8217;m too lazy; you go figure it out yourself.)</p>
<p>Now, with all orientation of tiles fixed, the game suddenly becomes a completely new face: You may put one to three tiles in a contiguous straight line over the red tiles (still with connectivity to the rest of the board), where the last person to place a tile wins. This is way simpler to analyze, especially armed with Sprague-Grundy theorem. We can prove that the left L-shaped part has nimber <img src="https://s0.wp.com/latex.php?latex=%2A6&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" srcset="https://s0.wp.com/latex.php?latex=%2A6&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002 1x, https://s0.wp.com/latex.php?latex=%2A6&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002&#038;zoom=4.5 4x" alt="*6" class="latex" />, and the right d-shaped part has nimber <img src="https://s0.wp.com/latex.php?latex=%2A7&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" srcset="https://s0.wp.com/latex.php?latex=%2A7&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002 1x, https://s0.wp.com/latex.php?latex=%2A7&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002&#038;zoom=4.5 4x" alt="*7" class="latex" />. Thus, by using <a href="https://en.wikipedia.org/wiki/Nim">Nim</a> strategy, the second player must play in the right part to reduce its nimber to <img src="https://s0.wp.com/latex.php?latex=%2A6&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" srcset="https://s0.wp.com/latex.php?latex=%2A6&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002 1x, https://s0.wp.com/latex.php?latex=%2A6&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002&#038;zoom=4.5 4x" alt="*6" class="latex" />. As it turns out, only the two moves above have this property; the rest doesn&#8217;t make the nimber <img src="https://s0.wp.com/latex.php?latex=%2A6&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" srcset="https://s0.wp.com/latex.php?latex=%2A6&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002 1x, https://s0.wp.com/latex.php?latex=%2A6&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002&#038;zoom=4.5 4x" alt="*6" class="latex" />, and thus the first player wins by moving on the left part appropriately. (Or if the second player plays in the left part, the first player can play in the right part to equalize the nimbers.)</p>
<p>With human players (and hence possible to make mistakes, especially those as deep as above), that move is a strong one even if losing. I haven&#8217;t found a winning first player move—indeed, I&#8217;m not even sure whether this is first-player or second-player win—but first player has a major advantage as they can force a major part of the track, if not completely, so that the second player must think carefully. On the other hand, once the track is completely determined, the game becomes simpler to analyze as above, so one must be careful of this if the opponent is aware of the strategy.</p>
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