prolog-dif

问题 Is there value in Prolog that is not equal to itself? I write answer to some question about min of tree and this answer also says that if tree is empty min is null. Sounds good idea first but now when I think it sounds like bad idea. It is kinda OK if null <> null , no problem. But in Prolog I see null is just atom so.... ?- null = null. true. ?- null == null. true. ?- dif(null, null). false. How can I make some term in Prolog that always say: ?- dif(Something, Something). true. But if it is

问题 I'm having some trouble removing values from a list in prolog. I have a list of colors and I want to add a list of colors to it and keep all the values that have no duplicate and remove the rest. [green, red, blue, purple, yellow, brown, orange, black, purple] so purple appears twice in this list and I want to remove both of them. This is the list I want to be returned. [green, red, blue, yellow, brown, orange, black] I currently have this to remove all the duplicates but I can't get both

问题 I'm having some trouble removing values from a list in prolog. I have a list of colors and I want to add a list of colors to it and keep all the values that have no duplicate and remove the rest. [green, red, blue, purple, yellow, brown, orange, black, purple] so purple appears twice in this list and I want to remove both of them. This is the list I want to be returned. [green, red, blue, yellow, brown, orange, black] I currently have this to remove all the duplicates but I can't get both

问题 I'm reading the "7 Languages in 7 Days"-book, and have reached the Prolog chapter. As a learning exercises I'm trying to solve some textual logic puzzles. The puzzle goes as follow: Five sisters all have their birthday in a different month and each on a different day of the week. Using the clues below, determine the month and day of the week each sister's birthday falls. Paula was born in March but not on Saturday. Abigail's birthday was not on Friday or Wednesday. The girl whose birthday is

问题 I want to access list permutation and pass it as argument to other functions. This is the permutation code: takeout(X,[X|R],R). takeout(X,[F|R],[F|S]) :- takeout(X,R,S), write(S). perm([X|Y],Z) :- perm(Y,W), takeout(X,Z,W). perm([],[]). 回答1: To start with, let's redefine your predicates so they don't do any unnecessary I/O: takeout(X,[X|R],R). takeout(X,[F |R],[F|S]) :- takeout(X,R,S). perm([X|Y],Z) :- perm(Y,W), takeout(X,Z,W). perm([],[]). Now you have what could be considered a "pure"

问题 I need to filter the list [#,d,e,#,f,g] such that I get the output as [[d,e],[f,g]] , I am stuck while creating a new list every time I encounter '#' is there a way to do this? I tried the code below, filterL([],List) :-[]. filterL([Head|Tail],X) :- ( Head \='#'-> append(X,Head,List), filterL(Tail,List) ; filterL(Tail,X) ). 回答1: Your problem is not very well defined. Are empty sequences allowed or not? Shall [#] be related to [[],[]] (there is an empty sequence before and after) or [] ? You

问题 I have a prolog assignment. I need to look at the first item in a list, see if its following items are the same until they are not and separate the lists by the first item and its duplicates. e.g if my list was a,a,a,b,c it would separate it into first: a,a,a. second: b,c. My current solution works except that the final matching item comes into the second list, not the first. I can't seem to think of a way to get it to appear in the first list instead. grab([],[],[]). grab([A,A|L],[A|L2],Rest

问题 (This is NOT a coursework question. Just my own personal learning.) I'm trying to do an exercise in Prolog to delete elements from a list. Here's my code : deleteall([],X,[]). deleteall([H|T],X,Result) :- H==X, deleteall(T,X,Result). deleteall([H|T],X,[H|Result]) :- deleteall(T,X,Result). When I test it, I first get a good answer (ie. with all the Xs removed.) But then the backtracking offers me all the other variants of the list with some or none of the instances of X removed. Why should

问题 I'm having some trouble understanding why my code in prolog does something based on the order I put my rules in. Here is my database: parent(tom, bob). parent(tom, liz). parent(mary, bob). parent(mary, liz). male(tom). male(bob). female(mary). female(liz). And here are the rules: %difference(X, Y) ==> Predicate to check if two people X and Y are not the same person. difference(X, Y) :- \==(X, Y). father(X, Y) :- male(X), parent(X, Y), difference(X, Y). mother(X, Y) :- female(X), parent(X, Y),

问题 Write a predicate allDistinct/1 whose parameter is a list (of symbols) and which succeeds if all symbols in the list are different. notin(A,[]). notin(A,[B|C]) :- A\=B, notin(A,C). allDistinct([]). allDistinct([_]). allDistinct([A|B]) :- notin(A,B), allDistinct(B). 回答1: Following up on the previous sketch by @whd we can proceed like this. Based on iwhen/2 we can succinctly define distinct/1 like so: :- use_module(library(lists), [same_length/2]). distinct(Es) :- iwhen(ground(Es), (sort(Es,Fs)