Fii, Prolog - rextester

Language: Layout:

:- use_module(library(clpfd)). %for transpose

%sort words by length then hand things off to puzsol predicate

puzzle_solution(Puz,Ws):-

sort_ws(Ws,Ss),

puzsol(Puz,Ss).

%finished when word list is empty

puzsol(_,[]).

%solve for one word then recursively for the rest of the words

puzsol(Puz,[W|Ws]):-

%try first the original then the transpose, each time naming the result P

(P=Puz;transpose(Puz,P)),

%get a horizontal line

member(M,P),

%get a slot in the line

slot(M,S),

%fill in the slot with this word

fill(S,W),

%solve for the rest of the words

puzsol(P,Ws).

%left edge is always a slot because there might not be a # before it. If it IS a # no need to check that because a word won't fit there anyeay

slot(L,L).

%a slot has anything to the left, a # A, then a var or a non-# B, and consists of B followed by the rest of the line

slot(L,[B|C]):-

append([_,[A,B],C],L),

%check first that A isn't a var so = will not bind the var to #

\+var(A), A='#',

%any var or non-# can start the slot

(var(B);B\='#').

%get length of the slot itself
slotlength([],0):- !.
slotlength([H|T],0):-
  \+var(H),H='#',!
slotlength([_|T],Le):-
  slotlength(T,Le2),
  Le is Le2+1.

%any empty slot can be filled with an empty word, allows words to run up to the rightmost edge

fill([],[]).

%if the slot is not empty but the word is empty, the very next thing after the word had better be a #

fill([H|_],[]):-

%check for non-var first so assignment to # is not made

\+var(H),H='#'.

%copy word into slot from head to tail

fill([H|T],[H|T2]):- fill(T,T2).

%sort words by length, long to short

sort_ws(Ws, Ws2):-

%assign a key of negative the length so that larger ones will come before smaller ones

findall(L-W,(member(W,Ws),length(W,L2),L is -L2),Lws),

%do the sort

keysort(Lws,Ss),

%strip off sort keys

findall(M,member(_-M,Ss),Ws2).

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