Prolog Cryptarithmetic Puzzle - prolog

Question

I have been asked to solve a Cryptarithmetic Puzzle using Prolog:
GIVE
* ME
------
MONEY
The above is the puzzle, I cannot figure out where is the problem, the result always returns false. Plus I am not allowed to use any library in SWI-Prolog.
solve(Z) :-
assign(Z,[0,1,2,3,4,5,6,7,8,9]),
check(Z).
find( VAL , G,I,V,E ) :- VAL is G * 1000 + I * 100 + V * 10 + E.
find2(VALR, M,E ) :- VALR is M * 10 + E.
find3(VALA, M,O,N,E,Y) :- VALA is M * 10000 + O * 1000 + N * 100 + E * 10 + Y.
check(Z) :-
G #>= 1,
M #>= 1,
find( VAL, G,I,V,E),
find2(VALR, M,E),
find3(VALA, M,O,N,E,Y),
VAL * VALR =:= VALA.
assign(Z,L) :-
permute(L,Z).
/* permute is similar to all_different in swi-prolog */
addany(X,K,[X|K]).
addany(X,[F|K],[F|L1]) :-
addany(X,K,L1).
permute([],[]).
permute([X|K],P) :-
permute(K,L1),
addany(X,L1,P).
Sample query:
?- solve([G,I,V,E,M,O,N,Y]).
false. % fails unexpectedly

Answer 1

Let's get right to the heart of the matter!
Every permutation of
[0,1,2,3,4,5,6,7,8,9] is a list of length 10.
[G,I,V,E,M,O,N,Y] is a list of length 8.
No permutation of [0,1,2,3,4,5,6,7,8,9] can be unified with [G,I,V,E,M,O,N,Y].
As a quick-fix, adapt the definition of check/1 like this:
check([G,I,V,E,M,O,N,Y,_,_]) :-
find( VAL, G,I,V,E),
G >= 1,
find2(VALR, M,E),
M >= 1,
find3(VALA, M,O,N,E,Y),
VAL * VALR =:= VALA.
Then, run the following "fixed" query:
?- Expr = ([G,I,V,E]*[M,E] = [M,O,N,E,Y]),
Zs = [G,I,V,E,M,O,N,Y,_,_],
time(solve(Zs)).
% 24,641,436 inferences, 7.692 CPU in 7.709 seconds (100% CPU, 3203506 Lips)
Expr = ([1,0,7,2] * [9,2] = [9,8,6,2,4]),
Zs = [1,0,7,2,9,8,6,4,3,5] ;
% 7,355 inferences, 0.007 CPU in 0.007 seconds (100% CPU, 1058235 Lips)
Expr = ([1,0,7,2] * [9,2] = [9,8,6,2,4]), % redundant
Zs = [1,0,7,2,9,8,6,4,5,3] ;
% 6,169,314 inferences, 1.935 CPU in 1.939 seconds (100% CPU, 3188312 Lips)
Expr = ([1,0,9,2] * [7,2] = [7,8,6,2,4]),
Zs = [1,0,9,2,7,8,6,4,3,5] ;
% 7,355 inferences, 0.005 CPU in 0.005 seconds (99% CPU, 1360603 Lips)
Expr = ([1,0,9,2] * [7,2] = [7,8,6,2,4]), % redundant
Zs = [1,0,9,2,7,8,6,4,5,3] ;
% 6,234,555 inferences, 1.955 CPU in 1.959 seconds (100% CPU, 3189462 Lips)
false.
Here's another way to solve the problem:
First, use clpfd!
:- use_module(library(clpfd)).
Second, (re-)use code presented earlier in my answer
to the related question Faster implementation of verbal arithmetic in Prolog:
?- Expr = ([G,I,V,E] * [M,E] #= [M,O,N,E,Y]),
Zs = [G,I,V,E,M,O,N,Y],
crypt_arith_(Expr,Zs),
time(labeling([],Zs)).
% 397,472 inferences, 0.088 CPU in 0.088 seconds (100% CPU, 4521899 Lips)
Expr = ([1,0,7,2] * [9,2] #= [9,8,6,2,4]), Zs = [1,0,7,2,9,8,6,4] ;
% 128,982 inferences, 0.037 CPU in 0.037 seconds (100% CPU, 3502788 Lips)
Expr = ([1,0,9,2] * [7,2] #= [7,8,6,2,4]), Zs = [1,0,9,2,7,8,6,4] ;
% 77,809 inferences, 0.028 CPU in 0.028 seconds (100% CPU, 2771783 Lips)
false.
No redundant solutions. Orders of magnitude faster than "generate & test". clpfd rocks!

Answer 2

The following article by Eric Weisstein and Ed Pegg will be useful. It offers several solutions for a similar problem in Mathematica.
Using a very brute-force approach, there are two solutions: 1072 * 92 = 98624 and 1092 * 72 = 78624. The code that I used:
In[16]:= Cases[
Permutations[
Range[0, 9], {5}], {g_, i_, v_, e_, m_} /; g > 0 && m > 0 :>
With[{dig = IntegerDigits[(g*10^3 + i*10^2 + v*10 + e) (10 m + e)]},
Join[{g, i, v, e, m}, dig[[{2, 3, 5}]]] /;
And[Length[dig] == 5, Unequal ## dig, dig[[{1, 4}]] == {m, e},
Intersection[dig[[{2, 3, 5}]], {g, i, v, e, m}] === {} ]
]]
Out[16]= {{1, 0, 7, 2, 9, 8, 6, 4}, {1, 0, 9, 2, 7, 8, 6, 4}}