双方完全相同. 胜, 负还是和? 假设红先.
╔╤╤士将╤╤╤╗
╟┼炮┼士兵┼┼╢
╟┼┼┼炮┼┼┼╢
╟┼┼┼┼┼┼┼卒
╟┴┴┴┴┴┴┴╢
╟┬┬┬┬┬┬┬╢
╟┼┼┼┼┼┼┼兵
╟┼┼┼炮┼┼┼╢
╟┼炮┼仕卒┼┼╢
╚╧╧仕帅╧╧╧╝
Red wins, first move, cannon 7 up 1. |
sean9991® 如果我拿黑炮堵住你,你又如何? |
是不是和棋?好像没有谁能赢啊? |
红先胜！如黑中跑堵，折红边跑上至距红边跑一格处。黑被动，输！ |
黑中炮当然动不了啦(红炮在中间,动了就死了)..炮3--9?你怎么办? |
This problem is equivalent to removing matches from several stacks. You can remove any positive number of matches from one stack at any time. Whoever gets the last match wins the game. |
cba® 有没有答案啊? 如果红炮移动,黑炮可以放底线将军. |
红胜 最快情形： 兵一进一, 炮五进四 炮七进六，卒六平七 帅五平四，炮七二退一 炮七进一，炮五退一 炮五进一，炮五退一 炮五进一，炮五退一 炮五进一，炮五退一 炮五进一，卒九进一 兵一进一 （转自：顶顶华闻 www.TopChineseNews |
It's equivalent to 3 piles of stones (or matches): 1,4,6. While "1" (the pawns) can't be moving backward, the "4" and "6" (the cannons) can be increased in a certain situation (by moving backward) to a certain upper bound. However, I believe, the principle still applies. Therefore, "cannon 7 up 1" is the critical move, the 3 piles now become 1, 4, 5. If the black cannon comes forward, it will either become 1,4,0 or 1,0,5. The next move red has to make is turn it to 1,1,0 or 1,0,1. Black is end played, whether black moving pawn forward or the the cannon forward by 1 step or moving the other cannon backward won't help. |
if red pawn 1 up 1, black will cannon 3 up 2. Remember the objective of this game is to have the last move so that four cannons are touched and have one spot between the pawns. |
This is a famous ending named "双炮禁双炮". As I said before, it's equivalent to a game of removing matches (or stones) from several stackes. The rule is that two players take turns removing any positive number of matches from a single stack and whoever gets the last match wins. To win this game, represent the number of matches of each stack in binary format, and try to maintain even number of one's in every digit. To be specific, as in this game, there are 4 and 6 spots between 2 cannons and 2 spot between the pawns. Our objective is to have the last move so that the four cannons are touches, and have 1 spot left between the pawns. So we can think of 3 stackes of matches with 1, 4, and 6 matches, respectively. Represent them as binary number, we have 1, 100, and 110. So the first move is reduce 6 to 5, and make it 101. Now we have 1, 100, and 101. Note here we have even number of 1's in every digit. |
红炮七进一 以下变化是黑方最顽强的抵抗: 炮3退1 炮七进一 炮3进2 炮五进二 炮3进1 兵一进一 炮3进1 炮五进一 炮3进1 炮五进一 至此四个炮和两个兵已无法纵向移动, 红方每步应对不可有差错, 轮到黑走 炮3平5 仕五进四 炮5平3 仕六进五 炮3平5 帅五平六 炮5平3 帅六进一 红方调整双仕, 使黑方一将一歇求和的企图失败, 红胜. 以上参考答案为本人心得, 欢迎讨论. 请看----看给出了类似棋局, 多了一对兵. 由于两对兵是见合的, 所以感觉难度降低了. |
红炮3进8，底线将军，你怎么办？ |
对象棋平时小有爱好,这是我请教我们公司一位象棋高人(有多高?反正是狠~~狠厉害的那种...),他给我解的. """"""当然红先胜，还没听说红先输的残局，呵呵。 看看这样下对不对？炮七进五，黑方有三种应法： 1、黑方“卒9进1”，红方就“炮七再进一”步步逼死黑方； 2、黑方“炮3进1”，红方就“兵一进一”，然后黑只能退炮被红逼死； 3、黑方“炮3退1”，红方就“炮七进一”，黑方又只能选1与2中的一种下法，当然也是死路一条。 该残局的关键在于第一步进炮不要一下子逼死黑方，而有意留下一点余地，正所谓“得饶人处且饶人”。呵呵，胡诌的。""""""" |
黑方没有一将一歇求和的可能, 红可速胜 炮3平5! 仕五进六 炮5平3! 后仕进五 炮3平5! 仕五退六 炮5平3! 后仕进五 黑方长将作负 |
思路对了, 别忘了五路中炮之间有四空. 红方若炮七进五, 黑方炮5进4, 黑胜 |
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