I\'m reading the paper Typed Logical Variables in Haskell, but I\'m failing to understand the details of the ultimate implementation. In particular, the backtracking state transformer introduced in se
I recently came across this code example in Mercury: append开发者_运维知识库(X,Y,Z) :- X == [],
Suppose I have a substitution S and list Xs, where each variable occurring in Xs also occurs in S. How would I find the list S(Xs), i.e., the list obtained by applying the substitution S to the list X
12 ?- 3+4*5 = X+Y. X = 3, Y = 4开发者_如何学编程*5. 13 ?- 3+4*5 = X*Y. false. 16 ?- 3*4+5 = X*Y. false.
EDIT: Solved. I was unware that enabling a language extension in the source file did not enable the language extension in GHCi. The solution was to :set FlexibleContexts in GHCi.
Well I know it might sound a bit strange but yes my question is: \"What is a unification algorithm\".
From an old final for my class: Here is some prolog code: mystery(1, 1). mystery(N, F) :- N1 is N-1, mystery(N1,F1),
I\'m developing (in Java), for fun, an applic开发者_如何学Cation which uses an unification algorithm.
Assume I have the following rules: unify(\'test\', \'this is a test\'). run :- write(\'Enter something: \'),
In prolog, we can do something like the following: myFunction a (a:xs) = ... This is, when the 1st argument of myFunction is the same as the first item of the list that\'s in the 2nd argument, this
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