Download %% FILE kakuro

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FILE kakuro.pl
SYSTEM LOGRES
CREATED TA-070911
REVISED TA-070912
%% This program describes a small Prolog program
%% for solving a Kakuro inspired problem
%% Together with it follows a set of useful list
manipulation predicates
:-prolog_flag(single_var_warnings,_,off).
%% Problem
/*
Make a Prolog predicate
kakuro(L,S)}
that tests whether L is a list of different
integers between 1 and 9, and S is the
sum of the integers in L.
Example:
kakuro([1,3],4) is true
kakuro([1,3],5) is false
kakuro([1,3,0],4) is false
kakuro([1,1,3],4) is false
For that purpose, we need a Prolog predicate
listsum(L,S)
where L is a list of integers , and
sum of the numbers in L.
S is the
Examples:
listsum([1,3,6],Z) is true with Z=10
listsum([],Z)
is true with Z=0
We also need a Prolog predicate
alldifferent(L)
where L is a list of integers,
and all the elements of X are different.
Example
alldifferent([]) is true
alldifferent([3,2,1]) is true
alldifferent([3,2,1,3]) is false
*/
%% Main programs
kakuro(L,S) :alldigits(L),
alldifferent(L),
listsum(L,S).
alldigits([]).
alldigits([X|Y]) :digit1(X),
alldigits(Y).
alldifferent([]).
alldifferent([X|Y]):alldifferent(Y),
notmember(X,Y).
listsum([],0).
listsum([X|Y],S):listsum(Y,S1),
S is S1 + X.
%% Sub programs
digit1(X) :member(X,[1,2,3,4,5,6,7,8,9]).
%% List predicates
car([Head|Tail],Head).
cdr([Head|Tail],Tail).
cons(Head,Tail,[Head|Tail]).
member(X,[X|_]).
member(X,[U|V]):member(X,V).
append([],X,X).
append([U|V],Y,[U|Z]):append(V,Y,Z).
last(Elt,[Elt]).
last(Elt,[_|Xs]):last(Elt,Xs).
listlength([],0). %% length/2 is predefined
listlength([X|Y],N) :listlength(Y,N1),
N is N1 + 1.
notmember(X,[]).
notmember(X,[U|V]):differ(X,U),
notmember(X,V).
/*
notmember(X,L) :-
%% Alternative
\+ member(X,L). %% Negation as failure
%% (\+ means not provable).
*/
%% Utility
ans(X) :write(X),nl,fail.
differ(X,Y) :- X \== Y.
%%%%%%%%%%%%%%%%%%%%%%%%%%
/*
% Example Run.
% sicstus
?-[kakuro].
?-kakuro([1,3],4).
yes
?- kakuro([1,3],5).
no
?-kakuro([1,3,0],4).
no
?-kakuro([1,1,3],4).
no
?- kakuro([1,2,3],H).
H = 6 ?
yes
%% Inverting Input/Output is possible here
| ?- kakuro([X,Y,Z],7).
X = 1,
Y = 2,
Z = 4 ?
X = 1,
Y = 4,
Z = 2 ?
X = 2,
Y = 1,
Z = 4 ?
X = 2,
Y = 4,
Z = 1 ?
X = 4,
Y = 1,
Z = 2 ?
X = 4,
Y = 2,
Z = 1 ?
no
| ?*/
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