I m 写A,打心游戏,经过一些测试,I ve发现,使用MTD(f)法算法(一系列零窗口搜索)比自己使用甲型六氯环己烷更快。
MTD(f)算法在以下网址作了很好的描述:http://people.csail.mit.edu/plaat/mtdf.html。
我面临的问题是,对于MTD(f)搜索中的每一份通行证(针对每一方)而言,我并没有重复我所储存的以往任何职位,尽管该链接上的文字表明,我应当(事实上,在校对之间的表格加快算法)。
我的问题是,当我把立场和价值储存在我的转表上时,我也储存其有效的甲型和乙型价值。 因此,与不同的树木(以及因此的甲型和乙型)的第二次穿过树,可能重复任何信息。 难道这是人们期望的,还是我在此缺乏一些基本的东西?
例如,如果对甲型=3 beta=4来说,我们就取得了7项成果(显然是一种截断),那么我是否在表格中储存该表对甲型=3的β=6有效? Or beta=7?
你们的问题在于如何在甲型塔旁边使用一个转站表的概念理解。 这是一个巨大的问题,我也谈到这个问题,在审视我之后,我发现。
基本上,由于截断发生,结果只是约束性的,而不是真正的微型最高值,因此,你无法对所有甲型六氯环己烷结果一视同仁。 事实证明,使用约束器将始终给予你同样的最好的下一个国家,但可能没有确切的分数。 当你把国家从截肢中储存起来时,你需要把它当作一个约束,并试图在下一个通行证上加以改进。 这常常会多次评估同一节点,但会不断改进所需的实际分数。
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