prolog-toplevel

I'm using SWI-Prolog on Windows and am getting the following error: 14 ?- parent(X, Y) :- child(Y, X). ERROR: toplevel: Undefined procedure: (:-)/2 (DWIM could not correct) I'm not entirely sure what's going on, as this worked last week and I am just starting to learn Prolog. The FAQ says it all: http://www.swi-prolog.org/FAQ/ToplevelMode.html You need to create a file and write your program with rules there. The top level command line will only allow you to issue queries. You can try it this way 1 ?- assert(a(A,B):-A=B). true. 2 ?- a(B,c). B = c. 来源： https://stackoverflow.com/questions

I'm new to Prolog. I'm just trying simple examples to learn. I have this .pl file with these lines: parent(pam,bob). parent(tom,bob). parent(tom,lio). parent(bob,ann). parent(bob,pat). parent(pat,jim). After consulting and testing, it only shows the first answer. For example: 5 ?- parent(X,Y). X = pam, Y = bob . Isn't it supposed to give all the combinations that satisfy the relation parent ? Do anyone have idea what the problem is? don't hit enter after your first results shows, use spacebar instead [Enter] stops execution even if the backtracking is not completed yet [Spacebar] or [;]

?- length(L,25). L = [_G245, _G248, _G251, _G254, _G257, _G260, _G263, _G266, _G 269|...]. If I use write(L) following the length predicate then the interpreter prints the list twice, one expanded and the other not. There is a limit on the depth to prevent too long output. You can change it with set_prolog_flag/1. ?- length(L, 25). L = [_G257, _G260, _G263, _G266, _G269, _G272, _G275, _G278, _G281|...]. ?- current_prolog_flag(toplevel_print_options, V). V = [quoted(true), portray(true), max_depth(10), priority(699)]. ?- set_prolog_flag(toplevel_print_options, [quoted(true), portray(true), max

My SWI-Prolog knowledge base contains the following two facts: f(a,b). f(a,c). Now if I pose the query ?- f(a,c). true. But ?- f(a,b). true ; false. Why is f(a,b) both true and false? This also happens when there are three facts in the KB. If I append f(a,d). to the KB, then f(a,d) is true (only), but f(a,b) and f(a,c) are both true and false. What's going on, and what can I do so that Prolog answers (only) true to these queries? (Note: this answer is somewhat of a guess) Consider how Prolog determines whether f(a,c) is true or not. It checks the first rule, f(a,b) , and doesn't find a match,

In normal situation, we can use ";" to show the next answer if there is one. But if I do this, it shows me error: char_code/2: Cannot represent due to 'character_code' In stead of ";" , I use "shift + ;" , and prolog gives me a prompt Unknown action: : (h for help) Action? then if I input ";" , the designable answers will be shown one by one. What is the problem? Use swipl-win.exe instead of swipl.exe can solve this problem.(I tried 6.x and 7.x) By the way, swipl works well under MacOS. 来源： https://stackoverflow.com/questions/23133168/after-the-first-answer-prolog-shows-the-error-char-code-2

Here is what I have understood about Prolog variables. A single underscore stands for anonymous variable, which is like a new variable each time it occurs. A variable name starting with underscore like _W is not an anonymous variable. Or, the variable names generated inside Prolog, like _G189, is not considered anonymous: ?- append([1,2],X,Y). X = _G189 Y = [1, 2|_G189] Could you please help me understand? By the way, I got the above example from some tutorials, but when I run it in SWI-Prolog version 6, I get the following: ?- append([1,2],X,Y). Y = [1, 2|X]. Thanking you. Variables The

I come up against this all the time, and I'm never sure which way to attack it. Below are two methods for processing some season facts. What I'm trying to work out is whether to use method 1 or 2, and what are the pros and cons of each, especially large amounts of facts. methodone seems wasteful since the facts are available, why bother building a list of them (especially a large list). This must have memory implications too if the list is large enough ? And it doesn't take advantage of Prolog's natural backtracking feature. methodtwo takes advantage of backtracking to do the recursion for me,

I am trying to get a feel for Prolog programming by going through Ulle Endriss' lecture notes . When my solution to an exercise does not behave as expected, I find it difficult to give a good explanation. I think this has to do with my shaky understanding of the way Prolog evaluates expressions. Exercise 2.6 on page 20 calls for a recursive implementation of a predicate last1 which behaves like the built-in predicate last . My attempt is as follows: last1([_ | Rest], Last) :- last1(Rest, Last). last1([Last], Last). It gives the correct answer, but for lists with more than one element, I have