failure-slice

问题 I would like to use DCGs as a generator. As of now, the syntax is s-->a,b. a-->[]. a-->a,c. c-->[t1]. c-->[t2]. b-->[t3]. b-->[t4]. I would like to generate all s where the length of a is < someNumber . Using ?- phrase(a,X),length(X,Y),Y<4. i can get all a with less than 4 items. However, when all combinations are exhausted, the system (SWI-Prolog 6.2.5) seems to stall. Sometimes ago, a similar question was asked here. However, being new to Prolog i am not able to make it work with the

来源： https://stackoverflow.com/questions/63371755/why-does-my-prolog-rule-get-stuck-in-infinite-recursion

来源： https://stackoverflow.com/questions/63371755/why-does-my-prolog-rule-get-stuck-in-infinite-recursion

来源： https://stackoverflow.com/questions/55208901/why-do-i-get-a-stack-limit-exceeded-error-when-defining-a-predicate-that-convert

问题 I am trying to run the following program in Prolog. mama_mia1(A,M,LI,HI,LO,HO,AA) :- p1(A,M,LI,HI,LO,HO,PROGS), reverse(PROGS,PROG), atom_chars(AA,PROG), !. p1(_,_,LO,LO,LO,_,[]). p1(_,_,HO,HO,_,HO,[]). p1(_,_,LO,HO,LO,HO,[]). p1(_,_,X,LO,LO,HO,[]) :- X>LO,X<HO. p1(_,_,X,HO,LO,HO,[]) :- X>LO,X<HO. p1(_,_,LO,Y,LO,HO,[]) :- Y>LO,Y<HO. p1(_,_,HO,Y,LO,HO,[]) :- Y>LO,Y<HO. p1(_,_,X,Y,LO,HO,[]) :- X>LO,X<HO,Y>LO,Y<HO. p1(A,M,X,Y,LO,HO,PROG) :- ( (X1 is X+A, H1 is HO+1, X1<H1, Y1 is Y+A, Y1<H1 ) ->

问题 I am trying to run the following program in Prolog. mama_mia1(A,M,LI,HI,LO,HO,AA) :- p1(A,M,LI,HI,LO,HO,PROGS), reverse(PROGS,PROG), atom_chars(AA,PROG), !. p1(_,_,LO,LO,LO,_,[]). p1(_,_,HO,HO,_,HO,[]). p1(_,_,LO,HO,LO,HO,[]). p1(_,_,X,LO,LO,HO,[]) :- X>LO,X<HO. p1(_,_,X,HO,LO,HO,[]) :- X>LO,X<HO. p1(_,_,LO,Y,LO,HO,[]) :- Y>LO,Y<HO. p1(_,_,HO,Y,LO,HO,[]) :- Y>LO,Y<HO. p1(_,_,X,Y,LO,HO,[]) :- X>LO,X<HO,Y>LO,Y<HO. p1(A,M,X,Y,LO,HO,PROG) :- ( (X1 is X+A, H1 is HO+1, X1<H1, Y1 is Y+A, Y1<H1 ) ->

问题 I'm trying to write reversible relations in "pure" Prolog (no is , cut, or similar stuff. Yes it's homework), and I must admit I don't have a clue how. I don't see any process to create such a thing. We are given "unpure" but reversible arithmetic relations (add,mult,equal,less,...) which we must use to create those relations. Right now I'm trying to understand how to create reversible functions by creating the relation tree(List,Tree) which is true if List is the list of the leaves of the

问题 likes(tom,jerry). likes(mary,john). likes(mary,mary). likes(tom,mouse). likes(jerry,jerry). likes(jerry,cheese). likes(mary,fruit). likes(john,book). likes(mary,book). likes(tom,john). likes(john,X):-likes(X,john), X\=john. Hi there, above is a very simple prolog file, with some facts and only one rule: John likes anyone who likes him. But after loading this file and ask Prolog the following query: likes(john,X). The program crashes. The reason is somehow prolog gets stuck at likes(john,john)

问题 This question came up while answering another question on StackOverflow on (generalizing a bit) generating all sequences formed out of a finite set of elements with no duplicate occurrences. As Boris rightly indicated in the comments, there are many existing solutions to this problem. However, I am interested in a solution that does not use an accumulator (i.e., a list of already picked elements against which a newly selected element is to be compared) but that uses dif/2 statements instead.

问题 I have to find if two members of a list are adjacent. The restriction is to use the append/3 predicate. So far I've done the below, it works if it's true, otherwise I get no answer, it's like it runs eternally. adjacent(X,Y,L):- append(L1,[X,Y],T1),append(T1,T2,L). 回答1: To see that your program will loop, it suffices to consider the following failure-slice: adjacent(X,Y,L):- append(L1,[X,Y],T1), false , append(T1,T2,L) . If this program will loop, then the original program will loop too. It