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I having a problem with predicate which creates list of list for example
?-listoflist([p v q, p, r], R).
R=[ [p,q],[p],[r] ]
So far i have:
:- op(500, xfy, v).
listoflist([],[]):-!.
listoflist([H], [[H]]):-!
listoflist([H|T], [Result]):-
change_to_list(H,Tmp),
listoflist(T, [Tmp|Result])..
change_to_list(X v Y, [X|List]):-
change_to_list(Y,List),!.
change_to_list(X,[X]).
For operator declarations, always look what Prolog already has and fit your own operators into it:
?- current_op(Pri,Fix,\/).
Pri = 500, Fix = yfx.
A left-associative operator makes much more sense here. Maybe you can reuse this, instead of defining your own? And in case you want your own, take the very same priorities.
:- op(500, yfx, v).
operands(Op) --> [Op], {functor(Op,Op,0)}.
operands(L v R) --> operands(L), operands(R).
expr_operands(Expr, Ops) :-
phrase(operands(Expr), Ops).
?- maplist(expr_operands, [p v q, p, r], R).
R = [[p,q],[p],[r]].
Your listoflists/2 should be simpler and doesn't need cuts:
listoflists([], []).
listoflists([X|Xs], [Y|Ys]) :-
change_to_list(X, Y),
listoflists(Xs, Ys).
The other predicate, change_to_list/2 seems fine to me.
Notice how you collect the results in the head of the clause, not in the recursive call!
I am trying to break a word into different syllables in Prolog according to 2 different rules ..
rule 1: vowel-consonant-vowel (break word after second vowel)
rule 2: vowel-consonant-consonant-vowel (break word between the 2
consonant) , for example, calculator = cal-cula-tor ..
I already have the following code in Prolog, however, it only analyzes the first 3 or 4 letters of the word ..
I need it to process and analyze the entire word.
vowel(a).
vowel(e).
vowel(i).
vowel(o).
vowel(u).
consonant(L):- not(vowel(L)).
syllable(W, S, RW):-
atom_chars(W, [V1, C, V2|Tail]),
vowel(V1),
consonant(C),
vowel(V2),
!,
atomic_list_concat([V1, C, V2], S),
atomic_list_concat(Tail, RW).
syllable(W, S, RW):-
atom_chars(W, [V1, C, C2, V2|Tail]),
vowel(V1),
consonant(C),
consonant(C2),
vowel(V2),
!,
atomic_list_concat([V1, C, C2, V2], S),
atomic_list_concat(Tail, RW).
syllable(W, W, _).
break(W, B):-
syllable(W, B, ''), !.
break(W, B):-
syllable(W, S, RW),
break(RW, B2),
atomic_list_concat([S, '-', B2], B).
First, a setting that makes it much more convenient to specify lists of characters, and which I recommend you use in your code if you process text a lot:
:- set_prolog_flag(double_quotes, chars).
Second, the data, represented in such a way that the definitions can be used in all directions:
vowel(a). vowel(e). vowel(i). vowel(o). vowel(u).
consonant(C) :- maplist(dif(C), [a,e,i,o,u]).
For example:
?- consonant(C).
dif(C, u),
dif(C, o),
dif(C, i),
dif(C, e),
dif(C, a).
whereas the version you posted incorrectly says that there is no consonant:
?- consonant(C).
false.
The rules you outline are readily described in Prolog:
% rule 1: vowel-consonant-vowel (break after second vowel)
rule([V1,C,V2|Rest], Bs0, Bs, Rest) :-
vowel(V1), consonant(C), vowel(V2),
reverse([V2,C,V1|Bs0], Bs).
% rule 2: vowel-consonant-consonant-vowel (break between the consonants)
rule([V1,C1,C2,V2|Rest], Bs0, Bs, [C2,V2|Rest]) :-
vowel(V1), consonant(C1), consonant(C2), vowel(V2),
reverse([C1,V1|Bs0], Bs).
% alternative: no break at this position
rule([L|Ls], Bs0, Bs, Rest) :-
rule(Ls, [L|Bs0], Bs, Rest).
Exercise: Why am I writing [V2,C,V1|_] instead of [V1,C,V2|...] in the call of reverse/2?
Now, it only remains to describe the list of resulting syllables. This is easy with dcg notation:
word_breaks([]) --> [].
word_breaks([L|Ls]) --> [Bs],
{ rule([L|Ls], [], Bs, Rest) },
word_breaks(Rest).
word_breaks([L|Ls]) --> [[L|Ls]].
Now the point: Since this program is completely pure and does not incorrectly commit prematurely, we can use it to show that there are also other admissible hyphenations:
?- phrase(word_breaks("calculator"), Hs).
Hs = [[c, a, l], [c, u, l, a], [t, o, r]] ;
Hs = [[c, a, l], [c, u, l, a, t, o], [r]] ;
Hs = [[c, a, l], [c, u, l, a, t, o, r]] ;
Hs = [[c, a, l, c, u, l, a], [t, o, r]] ;
Hs = [[c, a, l, c, u, l, a, t, o], [r]] ;
Hs = [[c, a, l, c, u, l, a, t, o, r]].
In Prolog, it is good practice to retain the generality of your code so that you can readily observe alternative solutions. See logical-purity.
I guess its time for a DCG push back solution. The push back is used in the second rule of break//1. It is to reflect that we look at four characters but only consume two characters:
vowel(a). vowel(e). vowel(i). vowel(o). vowel(u).
consonant(C) :- \+ vowel(C).
break([V1,C,V2]) -->
[V1,C,V2],
{vowel(V1), consonant(C), vowel(V2)}.
break([V1,C1]), [C2,V2] -->
[V1,C1,C2,V2],
{vowel(V1), consonant(C1), consonant(C2), vowel(V2)}.
syllables([L|R]) --> break(L), !, syllables(R).
syllables([[C|L]|R]) --> [C], syllables([L|R]).
syllables([[]]) --> [].
So the overall solution doesn't need some extra predicates such as append/3 or reverse/2. We have also placed a cut to prune the search, which can be done because of the character catchall in the second rule of syllables//1.
Here are some example runs:
Jekejeke Prolog 2, Laufzeitbibliothek 1.1.6
(c) 1985-2016, XLOG Technologies GmbH, Schweiz
?- set_prolog_flag(double_quotes, chars).
Ja
?- phrase(syllables(R), "calculator").
R = [[c,a,l],[c,u,l,a],[t,o,r]] ;
Nein
?- phrase(syllables(R), "kitchensink").
R = [[k,i,t,c,h,e,n],[s,i,n,k]] ;
Nein
P.S.: In some older draft standards this DCG technique was
called "right-hand-context", and instead of the verb "push
back", the verb "prefixing" was used. In a newer draft standard
this is called "semicontext", and instead of the verb "push back",
the verb "restoring" is used.
https://www.complang.tuwien.ac.at/ulrich/iso-prolog/dcgs/dcgsdraft-2015-11-10.pdf
I think you could write it more simply.Here is my implementation:
syllable( Input, Final_Word):-
atom_chars( Input, Char_list),
(split(Char_list, Word)-> atom_chars( Final_Word, Word);
Final_Word=Input).
split([],[]).
split([X,Y,Z|T],[X,Y,Z,'-'|T1]):-
vowel(X),vowel(Z),
atom_chars( Input, T),
syllable(Input,T2),
atom_chars( T2, T1).
split([X,Y,Z,W|T],[X,Y,'-',Z|T1]):-
vowel(X),\+vowel(Y),\+vowel(Z),vowel(W),
atom_chars( Input, [W|T]),
syllable(Input,T2),
atom_chars( T2, T1).
split([X|T],[X|T1]):- \+vowel(X),split(T,T1).
split/2 splits the word adding '-' where it could be added following the above rules you stated and returns a list to syllable. atom_chars/2 transforms the list to a word. If the word couldn't be split then the output is the input.
Example:
?- syllable(calculator,L).
L = 'calcu-lato-r'.
I'm don't understand why you wrote 'calculator = cal-cula-tor ' since it doesn't follows the rules stated, since "cal" is not vowel-constant-vowel but constant-vowel-constant and same for the rest of thr word...
Second day I am trying to make something like this work:
?- sm([1,2,3,4], [3,4,5,6], X).
X = [4,6,8,10].
I have something like this for now:
sm([], []).
sm([Head1|Rest1], [Head2|Rest2], R) :-
ResultElem is Head1 + Head2,
append([ResultElem], R, R1),
sm(Rest1, Rest2, R1).
I get only:
Singleton variables: [X]
false
Why does it not working and how this can finally be overcome?
The problem is that your recursive sm/3 does not have a matching base, because you wrote sm/2 by mistake:
sm([], [], []). /* You forgot the third pair of [] brackets */
As far as the recursive clause goes, it is easier to put the result in the header of the rule, and it lets you avoid using append:
sm([Head1|Rest1], [Head2|Rest2], [ResultElem|R]) :-
ResultElem is Head1 + Head2,
sm(Rest1, Rest2, R).
you can use maplist :
add(X,Y,Z) :-
Z is X+Y.
my_sum(L1, L2, L) :-
maplist(add, L1,L2,L).
With SWI-Prolog and module lambda, you can write
:- use_module(library(lambda)).
my_sum(L1,L2,L) :-
maplist(\X^Y^Z^(Z is X+Y), L1, L2, L).
Finally, with module clpfd, you can have:
:- use_module(library(lambda)).
:- use_module(library(clpfd)).
my_sum(L1,L2,L) :-
maplist(\X^Y^Z^(Z #= X+Y), L1, L2, L).
and we have :
?- my_sum([1,A, 9] ,[B,5,6], [5,7,C]).
A = 2,
B = 4,
C = 15.
I am new to Prolog and when I query
sortedUnion([1,1,1,2,3,4,4,5], [0,1,3,3,6,7], [0,1,2,3,4,5,6,7]).
I get an error
Exception: (7) unite([_G114, _G162, _G201, _G231, _G243], [_G249, _G297, _G336, _G357, _G369], [0, 1, 2, 3, 4, 5, 6, 7]) ?
So I am hoping someone will be able to tell me where my code is mistaken and why it is wrong?
%undup(L, U) holds precisely when U can be obtained from L by eliminating repeating occurrences of the same element
undup([], []).
undup([X|Xs], [_|B]) :- remove(X,Xs,K), undup(K, B).
remove(_,[],[]).
remove(Y,[Y|T],D) :- remove(Y,T,D).
remove(Y,[S|T],[S|R]) :- not(Y = S), remove(Y,T,R).
%sortedUnion(L1,L2,U) holds when U contains exactly one instance of each element
%of L1 and L2
sortedunion([H|T], [S|R], [F|B]) :- undup([H|T], N), undup([S|R], M), unite(N,M,[F|B]).
unite([], [], []).
unite([X], [], [X]).
unite([], [X], [X]).
unite([H|T], [S|R], [X|Xs]) :- S=H, X is S, unite(T, R, Xs).
unite([H|T], [S|R], [X|Xs]) :- H<S, X is H, unite(T, [S|R], Xs).
unite([H|T], [S|R], [X|Xs]) :- S<H, X is S, unite([H|T], R, Xs).
An advice first: try to keep your code as simple as possible. Your code can reduce to this (that surely works)
sortedunion(A, B, S) :-
append(A, B, C),
sort(C, S).
but of course it's instructive to attempt to solve by yourself. Anyway, try to avoid useless complications.
sortedunion(A, B, S) :-
undup(A, N),
undup(B, M),
unite(N, M, S).
it's equivalent to your code, just simpler, because A = [H|T] and so on.
Then test undup/2:
1 ?- undup([1,1,1,2,3,4,4,5],L).
L = [_G2760, _G2808, _G2847, _G2877, _G2889] ;
false.
Clearly, not what you expect. The culprit should that anon var. Indeed, this works:
undup([], []).
undup([X|Xs], [X|B]) :- remove(X,Xs,K), undup(K, B).
2 ?- undup([1,1,1,2,3,4,4,5],L).
L = [1, 2, 3, 4, 5] ;
false.
Now, unite/3. First of all, is/2 is abused. It introduces arithmetic, then plain unification suffices here: X = S.
Then the base cases are hardcoded to work where lists' length differs at most by 1. Again, simpler code should work better:
unite([], [], []).
unite( X, [], X).
unite([], X, X).
...
Also, note the first clause is useless, being already covered by (both) second and third clauses.
I am trying to write a program in Prolog to find a Latin Square of size N.
I have this right now:
delete(X, [X|T], T).
delete(X, [H|T], [H|S]) :-
delete(X, T, S).
permutation([], []).
permutation([H|T], R) :-
permutation(T, X),
delete(H, R, X).
latinSqaure([_]).
latinSquare([A,B|T], N) :-
permutation(A,B),
isSafe(A,B),
latinSquare([B|T]).
isSafe([], []).
isSafe([H1|T1], [H2|T2]) :-
H1 =\= H2,
isSafe(T1, T2).
using SWI-Prolog library:
:- module(latin_square, [latin_square/2]).
:- use_module(library(clpfd), [transpose/2]).
latin_square(N, S) :-
numlist(1, N, Row),
length(Rows, N),
maplist(copy_term(Row), Rows),
maplist(permutation, Rows, S),
transpose(S, T),
maplist(valid, T).
valid([X|T]) :-
memberchk(X, T), !, fail.
valid([_|T]) :- valid(T).
valid([_]).
test:
?- aggregate(count,S^latin_square(4,S),C).
C = 576.
edit your code, once corrected removing typos, it's a verifier, not a generator, but (as noted by ssBarBee in a deleted comment), it's flawed by missing test on not adjacent rows.
Here the corrected code
delete(X, [X|T], T).
delete(X, [H|T], [H|S]) :-
delete(X, T, S).
permutation([], []).
permutation([H|T], R):-
permutation(T, X),
delete(H, R, X).
latinSquare([_]).
latinSquare([A,B|T]) :-
permutation(A,B),
isSafe(A,B),
latinSquare([B|T]).
isSafe([], []).
isSafe([H1|T1], [H2|T2]) :-
H1 =\= H2,
isSafe(T1, T2).
and some test
?- latinSquare([[1,2,3],[2,3,1],[3,2,1]]).
false.
?- latinSquare([[1,2,3],[2,3,1],[3,1,2]]).
true .
?- latinSquare([[1,2,3],[2,3,1],[1,2,3]]).
true .
note the last test it's wrong, should give false instead.
Like #CapelliC, I recommend using CLP(FD) constraints for this, which are available in all serious Prolog systems.
In fact, consider using constraints more pervasively, to benefit from constraint propagation.
For example:
:- use_module(library(clpfd)).
latin_square(N, Rows, Vs) :-
length(Rows, N),
maplist(same_length(Rows), Rows),
maplist(all_distinct, Rows),
transpose(Rows, Cols),
maplist(all_distinct, Cols),
append(Rows, Vs),
Vs ins 1..N.
Example, counting all solutions for N = 4:
?- findall(., (latin_square(4,_,Vs),labeling([ff],Vs)), Ls), length(Ls, L).
L = 576,
Ls = [...].
The CLP(FD) version is much faster than the other version.
Notice that it is good practice to separate the core relation from the actual search with labeling/2. This lets you quickly see that the core relation terminates also for larger N:
?- latin_square(20, _, _), false.
false.
Thus, we directly see that this terminates, hence this plus any subsequent search with labeling/2 is guaranteed to find all solutions.
I have better solution, #CapelliC code takes very long time for squares with N length higher than 5.
:- use_module(library(clpfd)).
make_square(0,_,[]) :- !.
make_square(I,N,[Row|Rest]) :-
length(Row,N),
I1 is I - 1,
make_square(I1,N,Rest).
all_different_in_row([]) :- !.
all_different_in_row([Row|Rest]) :-
all_different(Row),
all_different_in_row(Rest).
all_different_in_column(Square) :-
transpose(Square,TSquare),
all_different_in_row(TSquare).
all_different_in_column1([[]|_]) :- !.
all_different_in_column1(Square) :-
maplist(column,Square,Column,Rest),
all_different(Column),
all_different_in_column1(Rest).
latin_square(N,Square) :-
make_square(N,N,Square),
append(Square,AllVars),
AllVars ins 1..N,
all_different_in_row(Square),
all_different_in_column(Square),
labeling([ff],AllVars).