盲公顶棍有点怪

sdndsx

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[DhtmlXQ_adddate]2019-04-20 13:42:50[/DhtmlXQ_adddate]
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[DhtmlXQ_title]盲公顶棒（三路对头兵卒）01[/DhtmlXQ_title]
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[DhtmlXQ_remark]小巷弯弯[/DhtmlXQ_remark]
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[DhtmlXQ_title]盲公顶棒（三路对头兵卒）02[/DhtmlXQ_title]
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[DhtmlXQ_open]A00 非常见开局[/DhtmlXQ_open]
[DhtmlXQ_result]红胜[/DhtmlXQ_result]
[DhtmlXQ_remark]小巷弯弯[/DhtmlXQ_remark]
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[DhtmlXQ_title]盲公顶棒（三路对头兵卒）03[/DhtmlXQ_title]
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[DhtmlXQ_open]A00 非常见开局[/DhtmlXQ_open]
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sdndsx

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[DhtmlXQ_adddate]2019-04-20 13:42:50[/DhtmlXQ_adddate]
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[DhtmlXQ_title]盲公顶棒（三路对头兵卒）01[/DhtmlXQ_title]
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[DhtmlXQ_remark]小巷弯弯[/DhtmlXQ_remark]
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[DhtmlXQ_comment7]如果你之前都看懂了的话，前3回合就不用解释了吧。第四回合是个妙着，两对炮的距离分别是2，1，黑方要想把距离变成1，1的话就要走炮5进1，可是再一看，这个点刚好被红炮控制住了不能走。[/DhtmlXQ_comment7]
[DhtmlXQ_comment15]这里又故技重施，黑方仍然苦于不能走炮5进1，只能退炮[/DhtmlXQ_comment15]
[DhtmlXQ_comment23]胜。【小结】：||||这个例局的精髓就在于第4回合和第8回合，巧妙控制，让黑方无法进中炮，而被迫退边炮，一直退到无路可退为止。||那么怎样达成这个条件呢？仔细观察就会发现，需要红中炮到黑7卒的距离等于黑边炮到黑7卒的距离。在压制的前提下，如果红方能让中炮抢到这个位置就能赢，否则就是和棋（例局3）。||如果红中炮到黑7卒的距离小于黑边炮到黑7卒的距离，红方可以通过主动退中炮的方法来调整（第6回合就是这样）。但是反过来，如果红中炮到黑7卒的距离大于黑边炮到黑7卒的距离，就不好调整了，所以红方要想办法尽量的拉大黑边炮到黑7卒的距离，缩小红中炮到黑7卒的距离。具体点说就是红方绝对不要进三兵，那么黑方早晚要先进7卒。（当然如果初始局面红方三兵就已经动过了，那就算了。）||||要点：压制边炮，推进中炮，迫使对方先进7卒，让红中炮到黑7卒的距离等于黑边炮到黑7卒的距离。||知道了这个要点之后，炮的位置，兵卒的位置全都围绕这一点进行推算，就可以知道能不能赢了。[/DhtmlXQ_comment23]
[DhtmlXQ_refer]http%3A//www.hgchess.com/%0D%0Ahttp%3A//www.hgchess.com/bbs/forum.php%3Fmod%3Dviewthread%26tid%3D55961%26extra%3D[/DhtmlXQ_refer]
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sdndsx

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[DhtmlXQ_adddate]2019-05-03 21:58:11[/DhtmlXQ_adddate]
[DhtmlXQ_editdate]2019-05-03 21:58:11[/DhtmlXQ_editdate]
[DhtmlXQ_title]盲公顶棒（三路对头兵卒）02[/DhtmlXQ_title]
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[DhtmlXQ_comment0]收集自网上的散局||zqhuang[/DhtmlXQ_comment0]
[DhtmlXQ_comment1]柳暗指出这样红胜[/DhtmlXQ_comment1]
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sdndsx

谢谢各位老师的解拆研究！能否找出规律来呢？