I want to freeze my goal until some variable, for example list, is unbounded, right now I have
sieve(N,L) :-
freeze(Aux,sieve(N,L,[],Aux)),
numlist(2,N,Aux).
sieve(N,L,R,[H|T]) :-
freeze(X, X mod H =\= 0 ; X == H),
findall(X,select(X,T,_),P),
sieve(N,L,[H|R],P).
sieve(_,L,L,[]).
But it stop after some operations and waits forever. Could someone tell me how to correct this?
Ok i found out solution i had to change recursive call in sieve so now i call it in freeze predicate.
as requested
i found clue here Lazy lists in Prolog?
sieve(N,L) :-
sieve(L,Strumien,[]),
numlist(2,N,X),
X = Strumien.
sieve(L,Strumien,X) :-
freeze(Strumien,
( Strumien =[H|T],
filter(H,T,Z),
sieve(L,Z,[H|X])
)).
sieve(L,[],L).
filter(H,S,X) :-
filter(H,S,X,[]).
filter(_,[],X,X).
filter(H,S,X,Y) :-
freeze(S,S =[H1|T]),
( H1 mod H =\= 0 ->
append(Y,[H1],Y2),
filter(H,T,X,Y2)
;
filter(H,T,X,Y)
).
I'm trying to figure out a way to check if two lists are equal regardless of their order of elements.
My first attempt was:
areq([],[]).
areq([],[_|_]).
areq([H1|T1], L):- member(H1, L), areq(T1, L).
However, this only checks if all elements of the list on the left exist in the list on the right; meaning areq([1,2,3],[1,2,3,4]) => true. At this point, I need to find a way to be able to test thing in a bi-directional sense. My second attempt was the following:
areq([],[]).
areq([],[_|_]).
areq([H1|T1], L):- member(H1, L), areq(T1, L), append([H1], T1, U), areq(U, L).
Where I would try to rebuild the lest on the left and swap lists in the end; but this failed miserably.
My sense of recursion is extremely poor and simply don't know how to improve it, especially with Prolog. Any hints or suggestions would be appreciated at this point.
As a starting point, let's take the second implementation of equal_elements/2 by #CapelliC:
equal_elements([], []).
equal_elements([X|Xs], Ys) :-
select(X, Ys, Zs),
equal_elements(Xs, Zs).
Above implementation leaves useless choicepoints for queries like this one:
?- equal_elements([1,2,3],[3,2,1]).
true ; % succeeds, but leaves choicepoint
false.
What could we do? We could fix the efficiency issue by using
selectchk/3 instead of
select/3, but by doing so we would lose logical-purity! Can we do better?
We can!
Introducing selectd/3, a logically pure predicate that combines the determinism of selectchk/3 and the purity of select/3. selectd/3 is based on
if_/3 and (=)/3:
selectd(E,[A|As],Bs1) :-
if_(A = E, As = Bs1,
(Bs1 = [A|Bs], selectd(E,As,Bs))).
selectd/3 can be used a drop-in replacement for select/3, so putting it to use is easy!
equal_elementsB([], []).
equal_elementsB([X|Xs], Ys) :-
selectd(X, Ys, Zs),
equal_elementsB(Xs, Zs).
Let's see it in action!
?- equal_elementsB([1,2,3],[3,2,1]).
true. % succeeds deterministically
?- equal_elementsB([1,2,3],[A,B,C]), C=3,B=2,A=1.
A = 1, B = 2, C = 3 ; % still logically pure
false.
Edit 2015-05-14
The OP wasn't specific if the predicate
should enforce that items occur on both sides with
the same multiplicities.
equal_elementsB/2 does it like that, as shown by these two queries:
?- equal_elementsB([1,2,3,2,3],[3,3,2,1,2]).
true.
?- equal_elementsB([1,2,3,2,3],[3,3,2,1,2,3]).
false.
If we wanted the second query to succeed, we could relax the definition in a logically pure way by using meta-predicate
tfilter/3 and
reified inequality dif/3:
equal_elementsC([],[]).
equal_elementsC([X|Xs],Ys2) :-
selectd(X,Ys2,Ys1),
tfilter(dif(X),Ys1,Ys0),
tfilter(dif(X),Xs ,Xs0),
equal_elementsC(Xs0,Ys0).
Let's run two queries like the ones above, this time using equal_elementsC/2:
?- equal_elementsC([1,2,3,2,3],[3,3,2,1,2]).
true.
?- equal_elementsC([1,2,3,2,3],[3,3,2,1,2,3]).
true.
Edit 2015-05-17
As it is, equal_elementsB/2 does not universally terminate in cases like the following:
?- equal_elementsB([],Xs), false. % terminates universally
false.
?- equal_elementsB([_],Xs), false. % gives a single answer, but ...
%%% wait forever % ... does not terminate universally
If we flip the first and second argument, however, we get termination!
?- equal_elementsB(Xs,[]), false. % terminates universally
false.
?- equal_elementsB(Xs,[_]), false. % terminates universally
false.
Inspired by an answer given by #AmiTavory, we can improve the implementation of equal_elementsB/2 by "sharpening" the solution set like so:
equal_elementsBB(Xs,Ys) :-
same_length(Xs,Ys),
equal_elementsB(Xs,Ys).
To check if non-termination is gone, we put queries using both predicates head to head:
?- equal_elementsB([_],Xs), false.
%%% wait forever % does not terminate universally
?- equal_elementsBB([_],Xs), false.
false. % terminates universally
Note that the same "trick" does not work with equal_elementsC/2,
because of the size of solution set is infinite (for all but the most trivial instances of interest).
A simple solution using the sort/2 ISO standard built-in predicate, assuming that neither list contains duplicated elements:
equal_elements(List1, List2) :-
sort(List1, Sorted1),
sort(List2, Sorted2),
Sorted1 == Sorted2.
Some sample queries:
| ?- equal_elements([1,2,3],[1,2,3,4]).
no
| ?- equal_elements([1,2,3],[3,1,2]).
yes
| ?- equal_elements([a(X),a(Y),a(Z)],[a(1),a(2),a(3)]).
no
| ?- equal_elements([a(X),a(Y),a(Z)],[a(Z),a(X),a(Y)]).
yes
In Prolog you often can do exactly what you say
areq([],_).
areq([H1|T1], L):- member(H1, L), areq(T1, L).
bi_areq(L1, L2) :- areq(L1, L2), areq(L2, L1).
Rename if necessary.
a compact form:
member_(Ys, X) :- member(X, Ys).
equal_elements(Xs, Xs) :- maplist(member_(Ys), Xs).
but, using member/2 seems inefficient, and leave space to ambiguity about duplicates (on both sides). Instead, I would use select/3
?- [user].
equal_elements([], []).
equal_elements([X|Xs], Ys) :-
select(X, Ys, Zs),
equal_elements(Xs, Zs).
^D here
1 ?- equal_elements(X, [1,2,3]).
X = [1, 2, 3] ;
X = [1, 3, 2] ;
X = [2, 1, 3] ;
X = [2, 3, 1] ;
X = [3, 1, 2] ;
X = [3, 2, 1] ;
false.
2 ?- equal_elements([1,2,3,3], [1,2,3]).
false.
or, better,
equal_elements(Xs, Ys) :- permutation(Xs, Ys).
The other answers are all elegant (way above my own Prolog level), but it struck me that the question stated
efficient for the regular uses.
The accepted answer is O(max(|A| log(|A|), |B|log(|B|)), irrespective of whether the lists are equal (up to permutation) or not.
At the very least, it would pay to check the lengths before bothering to sort, which would decrease the runtime to something linear in the lengths of the lists in the case where they are not of equal length.
Expanding this, it is not difficult to modify the solution so that its runtime is effectively linear in the general case where the lists are not equal (up to permutation), using random digests.
Suppose we define
digest(L, D) :- digest(L, 1, D).
digest([], D, D) :- !.
digest([H|T], Acc, D) :-
term_hash(H, TH),
NewAcc is mod(Acc * TH, 1610612741),
digest(T, NewAcc, D).
This is the Prolog version of the mathematical function Prod_i h(a_i) | p, where h is the hash, and p is a prime. It effectively maps each list to a random (in the hashing sense) value in the range 0, ...., p - 1 (in the above, p is the large prime 1610612741).
We can now check if two lists have the same digest:
same_digests(A, B) :-
digest(A, DA),
digest(B, DB),
DA =:= DB.
If two lists have different digests, they cannot be equal. If two lists have the same digest, then there is a tiny chance that they are unequal, but this still needs to be checked. For this case I shamelessly stole Paulo Moura's excellent answer.
The final code is this:
equal_elements(A, B) :-
same_digests(A, B),
sort(A, SortedA),
sort(B, SortedB),
SortedA == SortedB.
same_digests(A, B) :-
digest(A, DA),
digest(B, DB),
DA =:= DB.
digest(L, D) :- digest(L, 1, D).
digest([], D, D) :- !.
digest([H|T], Acc, D) :-
term_hash(H, TH),
NewAcc is mod(Acc * TH, 1610612741),
digest(T, NewAcc, D).
One possibility, inspired on qsort:
split(_,[],[],[],[]) :- !.
split(X,[H|Q],S,E,G) :-
compare(R,X,H),
split(R,X,[H|Q],S,E,G).
split(<,X,[H|Q],[H|S],E,G) :-
split(X,Q,S,E,G).
split(=,X,[X|Q],S,[X|E],G) :-
split(X,Q,S,E,G).
split(>,X,[H|Q],S,E,[H|G]) :-
split(X,Q,S,E,G).
cmp([],[]).
cmp([H|Q],L2) :-
split(H,Q,S1,E1,G1),
split(H,L2,S2,[H|E1],G2),
cmp(S1,S2),
cmp(G1,G2).
A simple solution using cut.
areq(A,A):-!.
areq([A|B],[C|D]):-areq(A,C,D,E),areq(B,E).
areq(A,A,B,B):-!.
areq(A,B,[C|D],[B|E]):-areq(A,C,D,E).
Some sample queries:
?- areq([],[]).
true.
?- areq([1],[]).
false.
?- areq([],[1]).
false.
?- areq([1,2,3],[3,2,1]).
true.
?- areq([1,1,2,2],[2,1,2,1]).
true.
As a prolog newbie, I wanted to try to implement a binary sudoku solver.(Code is below, swi-prolog). Binary sudoku is explained here : https://cstheory.stackexchange.com/questions/16982/how-hard-is-binary-sudoku-puzzle
However, when performing the following query :
binarySudoku([[1,0],[0,1]]). I get "true."
binarySudoku([[1,_],[_,_]]). I get "false."
Now obviously it shouldn't return false as there is a solution... Why is this happening/how can I fix this?
:-use_module(library(clpfd)).
validRow(Row) :-
Row ins 0..1,
length(Row,L),
sum(Row,#=,L/2).
matrixNth(Matr,X,Y,El) :-
nth1(Y,Matr,CurRow),
nth1(X,CurRow,El).
allDifferent([]).
allDifferent([X|Y]) :-
not(member(X,Y)),
allDifferent(Y).
invalid(Rows,X,Y) :-
AboveY is Y-1,
BelowY is Y+1,
matrixNth(Rows,X,Y,1),
matrixNth(Rows,X,AboveY,1),
matrixNth(Rows,X,BelowY,1).
invalid(Rows,X,Y) :-
LeftX is X-1,
RightX is X+1,
matrixNth(Rows,X,Y,1),
matrixNth(Rows,LeftX,Y,1),
matrixNth(Rows,RightX,Y,1).
binarySudoku(Rows) :-
length(Rows,Height),
transpose(Rows,Cols),
length(Cols,Height),
maplist(validRow,Rows),
foreach(between(1,Height,X),foreach(between(1,Height,Y),not(invalid(Rows,X,Y)))),
allDifferent(Rows).
Instead of (\+)/1, which is not logically sound in this case, use the pure constraint dif/2: Your code works as expected of you change the line not(member(X,Y)) to:
maplist(dif(X), Y)
Example query (notice that I also use a_more_readable_naming_convention insteadOfMixingTheCases):
?- binary_sudoku([[1,A],[B,C]]), label([A,B,C]).
A = B, B = 0,
C = 1 ;
false.
+1 for using CLP(FD), which is a good fit for this task.
I need to create a Prolog predicate for power of 2, with the natural numbers.
Natural numbers are: 0, s(0), s(s(0)) ans so on..
For example:
?- pow2(s(0),P).
P = s(s(0));
false.
?- pow2(P,s(s(0))).
P = s(0);
false.
This is my code:
times2(X,Y) :-
add(X,X,Y).
pow2(0,s(0)).
pow2(s(N),Y) :-
pow2(N,Z),
times2(Z,Y).
And it works perfectly with the first example, but enters an infinite loop in the second..
How can I fix this?
Here is a version that terminates for the first or second argument being bound:
pow2(E,X) :-
pow2(E,X,X).
pow2(0,s(0),s(_)).
pow2(s(N),Y,s(B)) :-
pow2(N,Z,B),
add(Z,Z,Y).
You can determine its termination conditions with cTI.
So, how did I come up with that solution? The idea was to find a way how the second argument might determine the size of the first argument. The key idea being that for all i ∈ N: 2i > i.
So I added a further argument to express this relation. Maybe you can strengthen it a bit further?
And here is the reason why the original program does not terminate. I give the reason as a failure-slice. See tag for more details and other examples.
?- pow2(P,s(s(0))), false.
pow2(0,s(0)) :- false.
pow2(s(N),Y) :-
pow2(N,Z), false,
times2(Z,Y).
It is this tiny fragment which is the source for non-termination! Look at Z which is a fresh new variable! To fix the problem, this fragment has to be modified somehow.
And here is the reason why #Keeper's solution does not terminate for pow2(s(0),s(N)).
?- pow2(s(0),s(N)), false.
add(0,Z,Z) :- false.
add(s(X),Y,s(Z)) :-
add(X,Y,Z), false.
times2(X,Y) :-
add(X,X,Y), false.
pow2(0,s(0)) :- false.
pow2(s(N),Y) :- false,
var(Y),
pow2(N,Z),
times2(Z,Y).
pow2(s(N),Y) :-
nonvar(Y),
times2(Z,Y), false,
pow2(N,Z).
This happends because the of evaluation order of pow2.
If you switch the order of pow2, you'll have the first example stuck in infinite loop..
So you can check first if Y is a var or nonvar.
Like so:
times2(X,Y) :-
add(X,X,Y).
pow2(0,s(0)).
pow2(s(N),Y) :-
var(Y),
pow2(N,Z),
times2(Z,Y).
pow2(s(N),Y) :-
nonvar(Y),
times2(Z,Y),
pow2(N,Z).