prolog

问题 I'm trying to test if a list of brackets is valid. My code: checkbrackets([]). checkbrackets(['('|T]):- T = [')'|List], checkbrackets(List). checkbrackets(['('|T]):- T = ['('|List], append(Rest,[')'],T), checkbrackets(Rest). My code works for ['(', '(', ')', '(', '(', ')', ')', ')'] but it fails for ['(', '(', ')', ')', '(', ')'] . What am I doing wrong? Is it possible to write such a test without additional arguments like counters? 回答1: For the sake of completeness, here is a solution

问题 I'm trying to test if a list of brackets is valid. My code: checkbrackets([]). checkbrackets(['('|T]):- T = [')'|List], checkbrackets(List). checkbrackets(['('|T]):- T = ['('|List], append(Rest,[')'],T), checkbrackets(Rest). My code works for ['(', '(', ')', '(', '(', ')', ')', ')'] but it fails for ['(', '(', ')', ')', '(', ')'] . What am I doing wrong? Is it possible to write such a test without additional arguments like counters? 回答1: For the sake of completeness, here is a solution

问题 I have a csv file example.csv which contains two columns with header var1 and var2. I want to populate an initially empty Prolog knowledge base file import.pl with repeated facts, while each row of example.csv is treated same: fact(A1, A2). fact(B1, B2). fact(C1, C2). How can I code this in SWI-Prolog ? EDIT, based on answer from @Shevliaskovic : :- use_module(library(csv)). import:- csv_read_file('example.csv', Data, [functor(fact), separator(0';)]), maplist(assert, Data). When import. is

问题 I have a csv file example.csv which contains two columns with header var1 and var2. I want to populate an initially empty Prolog knowledge base file import.pl with repeated facts, while each row of example.csv is treated same: fact(A1, A2). fact(B1, B2). fact(C1, C2). How can I code this in SWI-Prolog ? EDIT, based on answer from @Shevliaskovic : :- use_module(library(csv)). import:- csv_read_file('example.csv', Data, [functor(fact), separator(0';)]), maplist(assert, Data). When import. is

问题 I have a csv file example.csv which contains two columns with header var1 and var2. I want to populate an initially empty Prolog knowledge base file import.pl with repeated facts, while each row of example.csv is treated same: fact(A1, A2). fact(B1, B2). fact(C1, C2). How can I code this in SWI-Prolog ? EDIT, based on answer from @Shevliaskovic : :- use_module(library(csv)). import:- csv_read_file('example.csv', Data, [functor(fact), separator(0';)]), maplist(assert, Data). When import. is

问题 In ISO Prolog unification is defined only for those cases that are NSTO (not subject to occurs-check). The idea behind is to cover those cases of unifications that are mostly used in programs and that are actually supported by all Prolog systems. More specifically, ISO/IEC 13211-1:1995 reads: 7.3.3 Subject to occurs-check (STO) and not subject to occurs-check (NSTO) A set of equations (or two terms) is "subject to occurs- check" (STO) iff there exists a way to proceed through the steps of the

问题 In ISO Prolog unification is defined only for those cases that are NSTO (not subject to occurs-check). The idea behind is to cover those cases of unifications that are mostly used in programs and that are actually supported by all Prolog systems. More specifically, ISO/IEC 13211-1:1995 reads: 7.3.3 Subject to occurs-check (STO) and not subject to occurs-check (NSTO) A set of equations (or two terms) is "subject to occurs- check" (STO) iff there exists a way to proceed through the steps of the

问题 In ISO Prolog unification is defined only for those cases that are NSTO (not subject to occurs-check). The idea behind is to cover those cases of unifications that are mostly used in programs and that are actually supported by all Prolog systems. More specifically, ISO/IEC 13211-1:1995 reads: 7.3.3 Subject to occurs-check (STO) and not subject to occurs-check (NSTO) A set of equations (or two terms) is "subject to occurs- check" (STO) iff there exists a way to proceed through the steps of the

问题 Got this strange behaviour. I was running these test cases: s1 :- Q=[[lambda,symbol(_3026),[cons,[quote,_3434], [quote,_3514]]],[quote,_3206]], P=[_3434|_3514], freeze(_3434, (write(foo), nl)), unify_with_occurs_check(P, Q). s2 :- Q=[[lambda,symbol(_3026),[cons,[quote,_3434], [quote,_3514]]],[quote,_3206]], P=[_3434|_3514], freeze(_3434, (write(foo), nl)), freeze(_3514, (write(bar), nl)), unify_with_occurs_check(P, Q). Now I get these results, where the outcome of s2 is wrong. The outcome is

问题 Got this strange behaviour. I was running these test cases: s1 :- Q=[[lambda,symbol(_3026),[cons,[quote,_3434], [quote,_3514]]],[quote,_3206]], P=[_3434|_3514], freeze(_3434, (write(foo), nl)), unify_with_occurs_check(P, Q). s2 :- Q=[[lambda,symbol(_3026),[cons,[quote,_3434], [quote,_3514]]],[quote,_3206]], P=[_3434|_3514], freeze(_3434, (write(foo), nl)), freeze(_3514, (write(bar), nl)), unify_with_occurs_check(P, Q). Now I get these results, where the outcome of s2 is wrong. The outcome is