mini-kanren cond-a cond-u 和 cond-e 有什么区别? [英] mini-kanren what is the difference between cond-a cond-u and cond-e?

问题描述

我尝试在 clojure 中使用 mini-kanren 的实现.但是一直在努力理解 cond-a cond-u 和 cond-e 之间的区别.我似乎对 cond-e 很清楚，但对 cond-a 和 cond-u 的理解却很糟糕.

I have tried to use an implementation of mini-kanren in clojure. But been struggling to understand the difference between cond-a cond-u and cond-e. I seem to be quite clear about cond-e but understanding of cond-a and cond-u is quiet bad.

cond-e 设定一组目标，然后尝试每个目标……即尝试所有成功的分支.

cond-e takes a set of goals and then tries each of them... i.e tries all the branches that succeed.

相反，cond-a 和 cond-u 提交到第一个谓词成功的分支，并且 cond-a 在那里返回所有可能的成功统一.虽然 cond-u 只返回第一个成功的统一......但是当我使用它时，这种理解似乎是不正确的.有人能帮我澄清一下这个理解吗.

cond-a and cond-u on the contrary commit to the branch whose first predicate succeeds and cond-a returns all the possible successful unifications there after. while cond-u returns only the first successfull unification ... but however it seems like this understanding is not correct when I use it. Can somebody help me clarify this understanding.

推荐答案

cond-e 给出所有成功分支的所有答案

cond-e gives every answer from all branches that succeed

cond-i 给出每个答案，但将它们交错.

cond-i gives every answer but interleaves them.

cond-a 只给出第一个成功分支的答案

cond-a only gives the answers from the first successful branch

cond-u 只给出第一个成功分支的第一个答案

cond-u only gives the first answer from the first successful branch

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