1. Joined
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    23 Apr '08 21:26


    How many shortest games?
  2. Standard memberSwissGambit
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    23 Apr '08 23:031 edit
    Originally posted by David113
    [fen] [/fen]

    How many shortest games?
    oops, hit post button by mistake...
  3. Standard memberSwissGambit
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    24 Apr '08 00:07
    My first stab at this yielded an answer of 75497472. That seems like far too many...
  4. Joined
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    24 Apr '08 01:41
    Originally posted by SwissGambit
    My first stab at this yielded an answer of 75497472. That seems like far too many...
    It is indeed wrong🙂
  5. Standard memberSwissGambit
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    24 Apr '08 07:43
    Originally posted by David113
    It is indeed wrong🙂
    This is more a math problem than a chess problem. I'll start by doing the 'chess' legwork, and maybe some math genius will fill in the rest.

    One of the SPGs leading to the diagram is:
    1. b4 a5 2. e4 a4 3. Bb2 Ra5 4. Ne2 Rf5 5. Ng3 e5 6. Be2 b5 7. O-O Qg5 8.
    Re1 Bb7 9. Nf1 Bd5 10. g3 Be6 11. Kg2 Bd6 12. Bd4 Nf6 13. Kh3 Kd8 14. Be3

    13.5 moves appears to be the shortest possible PG. For White, there are two main sequences. One is the b4-Bb2-Bd4-Be3 sequence, and the other is the e4-Ne2-Ng3-Be2-00-Re1-Nf1-g3-Kg2-Kh3 sequence. By filling in the shorter of the two sequences in various combinations on the 'scoresheet', the other sequence's moves are placed automatically, and thus do not generate any 'extra' combinations.

    more to come...
  6. Standard memberSwissGambit
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    25 Apr '08 00:32
    Originally posted by SwissGambit
    This is more a math problem than a chess problem. I'll start by doing the 'chess' legwork, and maybe some math genius will fill in the rest.

    One of the SPGs leading to the diagram is:
    1. b4 a5 2. e4 a4 3. Bb2 Ra5 4. Ne2 Rf5 5. Ng3 e5 6. Be2 b5 7. O-O Qg5 8.
    Re1 Bb7 9. Nf1 Bd5 10. g3 Be6 11. Kg2 Bd6 12. Bd4 Nf6 13. Kh3 Kd8 14. Be3

    13.5 moves appea ...[text shortened]... ced automatically, and thus do not generate any 'extra' combinations.

    more to come...
    So, to determine the number of white move orders...

    Label the sequence b4-Bb2-Bd4-Be3 Q1, Q2, Q3, Q4 respectively. Start with Q4 as move 14 and Q3 as move 13. This leaves 11 different places to put Q2. If Q2 is move 12, that leads to 11 places to put Q1 and thus 11 move orders. If Q2 is move 11, we get 10 move orders and so on. 13.Q3 and 14.Q4 leads to 1+2+3+4+5+6+7+8+9+10+11 = 66 move orders.

    With 12.Q3 and 14.Q4, there is one less place to stick Q2/Q1, so we only get 1+2+3+4+5+6+7+8+9+10 = 55 move orders.

    Each time I shift Q3 one move left [sooner in the game], I end up cutting a number off the addition stack. Spreadsheeting this, I get 66+55+45+36+28+21+15+10+6+3+1 = 286 move orders.

    Continued later...
  7. Joined
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    25 Apr '08 00:431 edit
    Originally posted by SwissGambit
    So, to determine the number of white move orders...

    Label the sequence b4-Bb2-Bd4-Be3 Q1, Q2, Q3, Q4 respectively. Start with Q4 as move 14 and Q3 as move 13. This leaves 11 different places to put Q2. If Q2 is move 12, that leads to 11 places to put Q1 and thus 11 move orders. If Q2 is move 11, we get 10 move orders and so on. 13.Q3 and 14.Q4 leads t ...[text shortened]... preadsheeting this, I get 66+55+45+36+28+21+15+10+6+3+1 = 286 move orders.

    Continued later...
    Edit: Your computation is too complicated. There is a simpler way.
  8. Standard memberSwissGambit
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    25 Apr '08 05:52
    Originally posted by David113
    Edit: Your computation is too complicated. There is a simpler way.
    Probably. But I'm hoping my tortured writhing inspires someone to produce a more elegant solution and put me out of my misery.
  9. SubscriberAThousandYoung
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    27 Apr '08 04:401 edit
    Two

    1. White resigns.
    1. Draw by agreement
  10. Standard memberSwissGambit
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    30 May '08 19:04
    OK, it's official: No one is going to solve this. Let's see the answer.
  11. Joined
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    31 May '08 17:23
    Originally posted by David113
    [fen]1n1k3r/2pp1ppp/3bbn2/1p2prq1/pP2P3/4B1PK/P1PPBP1P/RN1QRN2[/fen]

    How many shortest games?
    22.
  12. Joined
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    31 May '08 17:55
    Originally posted by curseknight
    22.
    Wrong🙁
  13. Standard memberSwissGambit
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    01 Jun '08 03:58
    Originally posted by David113
    Wrong🙁
    😠
  14. Joined
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    01 Jun '08 17:15
    Originally posted by SwissGambit
    😠
    I agree with SG there and in his earlier post asking for the solution.
  15. Standard memberSwissGambit
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    01 Jun '08 19:56
    The answer is 1 Million.

    http://www.combinatorics.org/Volume_11/PDF/v11i2a4.pdf

    Page 12, solution 8.
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