[ Help | Earliest Comments | Latest Comments ][ List All Subjects of Discussion | Create New Subject of Discussion ][ List Earliest Comments Only For Pages | Games | Rated Pages | Rated Games | Subjects of Discussion ]Single Comment Bipartite Chess. Small chess variant with two phases. (5x7, Cells: 43) [All Comments] [Add Comment or Rating]Nicholas Kuschinski wrote on 2003-04-08 UTCPoor ★This game is bogus. I'm going to begin with the more trivial problems, and will move on to write a mathematical proof (that can be understood by just about anyone, you don't have to be a mathematician) that stage 1 is entirely pointless because black wins no matter how well or badly the game is played through this stage. <br> <br> The orthogonal power piece is obviously far more powerful than the diagonal one, and assures that the winner of stage one will have an easy win at the end if he picks it. I'm not sure stage one is necessarily conclusive, and would like to see you prove that the two players will not get locked someplace in the middle, and be unable to advance. I actually doubt very mucht that stage one must necessarily have such a conclusion at all. Even if it is conclusive, the difference in the power pieces will assure that the winner of stage one will win the game, and stage one is hardly complicated enough to be called chess. <br> <br> <br> Assuming for the sake of argument that stage 1 actualy DOES end, it gets even worse: Stage one will always be won by the player who moves first. Here is a mathematical proof of the fact: <br> All pieces move exactly one rank forwards on each turn, hence, in order to transfer one piece from the second rank onto the last rank, it will take exactly 7 moves. For a piece on the first rank to reach the last rank it takes exactly 8 moves. For a piece on the first rank to reach the penultimate rank it takes exactly 7 moves. For a piece on the second rank to reach the penultimate rank, it takes exactly 6 moves. Assume all pieces on the second rank but one move to the last rank, all of the moves made by these pieces will total 7x4=28 moves. The remaining piece on the second rank takes another 6 moves to reach the penultimate rank, and the pieces that start on the first rank each take seven moves, so their move total is 28. The total number of moves is 28+28+6=62. Now assume that the pieces in the second rank move to the last or penultimate rank in some other pattern than the one described above. For every piece in the second rank that moves 6 moves to the penultimate rank, there is a piece on the first rank which must take 8 moves to reach the last one for stage 1 to be complete. 6+8=14=7+7, so it doesn't make any difference how many pieces from which rank end up where, stage 1 will always take exactly 62 moves. Since the moves go in alternating order, the person who plays first will always complete 62 moves before his opponent, and will win no matter what happens. QED <br> It doesn't even matter how the well or badly the players play stage 1! assuming that it is even possible to complete this stage, it will take exactly 62 moves for the first player to move (black, in this case) to win stage 1. Hence, this game is completely bogus and pointless.