clpfd

I have been given an exercise to solve the zebra puzzle using a constraint solver of my choice, and I tried it using the Prolog clpfd library . I am aware that there are other more idiomatic ways to solve this problem in Prolog, but this question is specifically about the clpfd package! So the specific variation of the puzzle (given that there are many of them) I'm trying to solve is this one: There are five houses The Englishman lives in the red house The Swedish own a dog The Danish likes to drink tea The green house is left to the white house The owner of the green house drinks coffee The

How to define a as a integer/float number ? I want to find the results of a+b+c+d=10 where a,b,c,d is integer and >=0 . Here is a simple, modern, pure Prolog, non-CLP-library solution: range(X):- member(X,[0,1,2,3,4,5,6,7,8,9,10]). ten(A,B,C,D):- range(A), range(B), range(C), range(D), 10 =:= A + B + C + D. with SWI-Prolog you can use CLP(FD) library 1 ?- use_module(library(clpfd)). % library(error) compiled into error 0.00 sec, 9,764 bytes % library(clpfd) compiled into clpfd 0.05 sec, 227,496 bytes true. 2 ?- Vars=[A,B,C,D],Vars ins 0..10,sum(Vars,#=,10),label(Vars). Vars = [0, 0, 0, 10], A

I can make lists of ascending integer like so: ?- findall(L,between(1,5,L),List). I know I can also generate values using: ?- length(_,X). But I don't think I can use this in a findall, as things like the following loop: ?- findall(X,(length(_,X),X<6),Xs). I can also generate a list using clpfd . :- use_module(library(clpfd)). list_to_n(N,List) :- length(List,N), List ins 1..N, all_different(List), once(label(List)). or list_to_n2(N,List) :- length(List,N), List ins 1..N, chain(List,#<), label(List). The last method seems best to me as it is the most declarative, and does not use once/1 or

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 binary tree Tree . To achieve such a thing I'm trying to create the tree_size(Tree,N) relation which is

Can someone provide a simple example of channelling constraints? Channelling constraints are used to combine viewpoints of a constraint problem. Handbook of Constraint Programming gives a good explanation of how it works and why it can be useful: The search variables can be the variables of one of the viewpoints, say X1 (this is discussed further below). As search proceeds, propagating the constraints C1 removes values from the domains of the variables in X1. The channelling constraints may then allow values to be removed from the domains of the variables in X2. Propagating these value

The term fib(N,F) is true when F is the N th Fibonacci number. The following Prolog code is generally working for me: :-use_module(library(clpfd)). fib(0,0). fib(1,1). fib(N,F) :- N #> 1, N #=< F + 1, F #>= N - 1, F #> 0, N1 #= N - 1, N2 #= N - 2, F1 #=< F, F2 #=< F, F #= F1 + F2, fib(N1,F1), fib(N2,F2). When executing this query (in SICStus Prolog), the first (and correct) match is found for N (rather instantly): | ?- fib(X,377). X = 14 ? When proceeding (by entering ";") to see if there are any further matches (which is impossible by definition), it takes an enormous amount of time (compared