说我有一棵搜索树,我正试图找到一个目标的所有解决方案。如果我不把树切除到孩子仍被允许生成它的父母节点,那么搜索空间就会变得非常大。 所以,有人可以告诉我一个能这样做的前提,或者向我解释如何做到,因为我对Prolog非常陌生。
最简单的方法可能是在你的“搜索”前端添加一个额外的论据,以跟踪已经访问过的节点(父母、子女等),加上一个条款,禁止访问以前的节点。
有些术语值得澄清。一棵树是一个没有周期(卢布)的图表,所以在“搜索树”中,你实际上没有重新检查两次节点的可能性。
也许您正在寻找某种通往目标的道路,而解决方案则是可能的路径。 所宣称的寻找“ 目标的所有解决方案” 的要求与排除那些不止一次重访节点的路径存在某种紧张。 但是,让我们假设限制不重访节点是允许的 。
如果您提供更多信息, 说明您如何编码您的搜索前端, 我就能具体地说明如何添加额外的“ 已访问过的列表” 参数 。
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