I'm trying to learn Prolog. I looked at this script:
:- use_module(library(clpfd)).
puzzle([S,E,N,D] + [M,O,R,E] = [M,O,N,E,Y]) :-
Vars = [S,E,N,D,M,O,R,Y],
Vars ins 0..9,
all_different(Vars),
S*1000 + E*100 + N*10 + D + M*1000 + O*100 + R*10 + E #= M*10000 + O*1000 + N*100 + E*10 + Y,
M #\= 0,
S #\= 0.
Source: https://github.com/Anniepoo/prolog-examples/blob/master/sendmoremoney.pl
I run it like so and get some output:
$ swipl -q -s sendmoremoney.pl
?- puzzle(X).
X = ([9, _G2009, _G2012, _G2015]+[1, 0, _G2024, _G2009]=[1, 0, _G2012, _G2009, _G2042]),
_G2009 in 4..7,
all_different([9, _G2009, _G2012, _G2015, 1, 0, _G2024, _G2042]),
91*_G2009+_G2015+10*_G2024#=90*_G2012+_G2042,
_G2012 in 5..8,
_G2015 in 2..8,
_G2024 in 2..8,
_G2042 in 2..8.
It looks like that is giving me a range of possible values for each letter. But how can I get a single solution where each letter is assigned to one of the possible values? Seems like a very basic question, but I can't figure it out.
Ok, looks like this wasn't so much a Prolog question as a clpfd question.
?- puzzle(As + Bs = Cs), label(As).
As = [9, 5, 6, 7],
Bs = [1, 0, 8, 5],
Cs = [1, 0, 6, 5, 2] ;
false.
Found my answer here: http://www.swi-prolog.org/man/clpfd.html
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