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  1. PO
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    <p>Probably I'm missing something, but just by inspection, it looks like this sequence of steps should work? :</p> <ul> <li>start with "weight <code>(2,0,-1)</code> and one <code>red</code> play"</li> <li>take a <code>red</code> play, piece <code>C</code>, which "adds weight <code>(2,0,-1)</code> and two <code>red</code> plays", leaving you with weight <code>(4,0,-2)</code> and two <code>red</code> plays.</li> <li>take a <code>red</code> play, piece <code>A</code>, which "adds weight <code>(0,-2,-1)</code>", leaving you with weight <code>(4,-2,-3)</code> and one <code>red</code> play.</li> <li>take a <code>red</code> play, piece <code>D</code>, which "adds weight <code>(0,2,1)</code> and two <code>blue</code> plays", leaving you with weight <code>(4,0,-2)</code> and two <code>blue</code> plays.</li> <li>take a <code>blue</code> play, piece <code>A</code>, which "adds weight <code>(-2,0,1)</code>", leaving you with weight <code>(2,0,-1)</code> and one <code>blue</code> play.</li> <li>take a <code>blue</code> play, piece <code>A</code>, which "adds weight <code>(-2,0,1)</code>", leaving you with weight <code>(0,0,0)</code> and no plays.</li> </ul> <p>A bit more schematically:</p> <pre><code>move weight plays ------ --------- ------- (2,0,-1) red red C (4,0,-2) red x2 red A (4,-2,-3) red red D (4,0,-2) blue x2 blue A (2,0,-1) blue blue A (0,0,0) - </code></pre> <p>. . . no?</p> <hr> <p><strong>Edited to add:</strong></p> <p><em>How I Found This:</em></p> <p>Since this question has garnered a lot of interest, maybe I should explain how I found the above solution. It was basically luck; I happened upon two key observations:</p> <ul> <li><code>red A</code> and <code>red D</code> cancel each other out weight-wise (the former adds <code>(0,-2,-1)</code>, the latter adds <code>(0,2,1)</code>), and add a total of two <code>blue</code> plays (both from <code>red D</code>) and no <code>red</code> plays; so if you play one right after the other, you "convert" two <code>red</code> plays into two <code>blue</code> plays.</li> <li><code>blue A</code> cancels out the initial weight (it adds <code>(2,0,-1)</code>) and adds no plays, so the entire problem can be solved by "converting" one <code>red</code> play into one <code>blue</code> play.</li> </ul> <p>That gave me a good start. I started with <code>red C</code> so as to get the two <code>red</code> plays that I could "convert" to two <code>blue</code> plays, and I immediately saw that <code>red C</code> was also the "opposite" of <code>blue A</code> weight-wise, so could also be canceled with a <code>blue A</code>. In my head, it all seemed to cancel out perfectly; then I wrote it down to make sure.</p> <p><em>Proof That It's Minimal:</em></p> <p>Also, while I didn't bother to reason through it at the time, I can also prove that this is a "minimal" winning sequence for that starting position &mdash; by which I mean, that if a sequence starts with "weight <code>(2,0,-1)</code> and one <code>red</code> play" and ends with weight <code>(0,0,0)</code> and no plays, then the sequence must contain at least five moves. To do so, assume that we have a sequence that meets this condition and has <em>fewer</em> than five moves. Then:</p> <ol> <li>Since the first component of the weight starts out positive, and no <code>red</code> play decreases it, the sequence will require at least one <code>blue</code> play, which means that it will necessarily include <code>red D</code> at least once (since that's the only way to acquire a <code>blue</code> play if you don't start out with one).</li> <li>Since the sequence includes <code>red D</code> at least once, and <code>red D</code> gives us two <code>blue</code> plays, the sequence will necessarily include <code>blue</code> at least twice (since the game doesn't end until no plays remain).</li> <li>Since the sequence includes <code>red D</code> at least once, and <code>red D</code> adds 2 to the second component of the weight, it will necessarily either contain <code>red A</code> at least once, or else contain <code>red B</code> at least twice (since there is no other way to decrease the second component of the weight by at least 2). But clearly, if it contains <code>red D</code> at least once, <code>blue</code> at least twice, and <code>red B</code> at least twice, then it will contain at least five moves, which is forbidden by assumption; so it must use the <code>red A</code> approach.</li> <li>So, the sequence contains <code>red D</code> at least once, <code>blue</code> at least twice, and <code>red A</code> at least once. And by assumption, it contains fewer than five moves. So it must consist of <em>exactly</em> one <code>red D</code>, two <code>blue</code>s, one <code>red A</code>, and zero other moves.</li> <li>The starting position gives us only one <code>red</code> play, and the sequence includes exactly two <code>red</code> plays. Therefore, the sequence must contain a move that yields exactly one <code>red</code> play. But the only possible move that could do that is <code>red B</code>, which the sequence doesn't contain.</li> </ol> <p>Therefore, no such sequence is possible.</p> <p><em>The Other Starting Position:</em></p> <p>I can also prove, using similar sorts of arguments, that any solution starting with the "weight <code>(3,1,-1)</code> and two <code>red</code> plays" option must also contain at least five moves. One such five-move solution is:</p> <pre><code>move weight plays ------ --------- ------- (3,1,-1) red x2 red A (3,-1,-2) red red B (4,-2,-3) red red D (4,0,-2) blue x2 blue A (2,0,-1) blue blue A (0,0,0) - </code></pre>
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    1. POFind an algorithm to balance this game
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    1. COThank you! The only problem now is that this answer doesn't work when I plug it back into the original problem that I'm working on. I'll have to see what's wrong there, but this looks like an answer to me. Thanks again!!
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    2. CO@Daniel: You're welcome!
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