I'm trying to create a query elimrow(_, [H|T], X) that deletes the nth row in a matrix array.
Sample:
?- elimrow(-3,[[1,2],[3,4],[5,6]], X). => X = [[1,2],[3,4],[5,6]]
?- elimrow(2,[[1,2],[3,4],[5,6]], X). => X = [[1,2],[5,6]]
So far I was able to create this:
elimrow(1, [H|_], H).
elimrow(I, [H|T], X) :-
I1 is I-1, elimrow(I1, T, X), delete(X, [H|T], B), writeln(B).
delete(A, [A|B], B).
delete(A, [B, C|D], [B|E]) :- delete(A, [C|D], E).
This is currently able to select the row which I want to delete. However the delete function isn't functioning fully as expected.
?- elimrow(2,[[1,2],[3,4],[5,6]],X).
[[1,2],[5,6]]
X = [3, 4]
It outputs the correct deleted array [[1,2], [5,6]], however it also outputs a X = [3,4]
I'm confused as to why there was a second output. (I only had one writeln(B)).
I also tried checking it with my first sample and it returned with false when it's not supposed to delete anything.
?- elimrow(-3, [[1,2],[3,4],[5,6]],X).
false.
Appreciate any help on this. Many thanks!
I think you make it too complicated. Your matrix is just a list of lists. So you can delete the I-th element of a list:
eliminate(_, [], []).
eliminate(0, [_|T], T).
eliminate(I, [H|T], [H|R]) :-
I > 0,
I1 is I-1,
eliminate(I1, T, R).
I am trying to get a set of elements from a list in prolog, such that a query:
get_elems([1, 2, 4, 10], [a, b, c, d, e], X).
yields:
X = [a, b, d]
I would like to implement it without using the built in predicate nth.
I have tried using the following, but it does not work:
minus_one([], []).
minus_one([X|Xs], [Y|Ys]) :- minus_one(Xs, Ys), Y is X-1.
get_elems([], _, []).
get_elems(_, [], []).
get_elems([1|Ns], [A|As], Z) :- get_elems(Ns, As, B), [A|B] = Z.
get_elems(Ns, [_|As], Z) :- minus_one(Ns, Bs), get_elems(Bs, As, Z).
Edit: The list of indices is guaranteed to be ascending, also I want to avoid implementing my own version of nth.
Give this a go:
get_elems(Xs,Ys,Zs) :- get_elems(Xs,1,Ys,Zs).
get_elems(Xs,_,Ys,[]) :- Xs = []; Ys = [].
get_elems([N|Xs],N,[H|Ys],[H|Zs]) :- !, N1 is N + 1, get_elems(Xs,N1,Ys,Zs).
get_elems(Xs,N,[_|Ys],Zs) :- N1 is N + 1, get_elems(Xs,N1,Ys,Zs).
This just keeps counting up and when the head of the second term is equal to the current index it peels off the head and makes it the head of the current output term. If it doesn't match it just discards the head and keeps going.
I'm searching for a compact predicate to swap sublists of fixed length within a larger list. For example, if sublists have size 3 then
[a,t,t,g,c,c]
becomes
[g,c,c,a,t,t]
I ended up with the following program:
dna_sub(A,B,X,Xe) :-
append(A1,_,A),
length(A1,Xe),
append(B1,B,A1),
length(B1,X).
dna_swap(A,B,X,Xe,Y,Ye) :-
length(A, Size),
dna_sub(A,Part1, 0, X),
dna_sub(A,Part2, X, Xe),
dna_sub(A,Part3, Xe, Y),
dna_sub(A,Part4, Y, Ye),
dna_sub(A,Part5, Ye, Size),
append(Part1, Part4, Tmp),
append(Tmp, Part3, Tmp2),
append(Tmp2, Part2, Tmp3),
append(Tmp3, Part5, B).
dna_swap(A,B) :-
length(A, Size),
Limit is Size - 3,
between(0,Limit, X),
Xe is X + 3,
Xs is Xe,
between(Xs, Size, Y),
Ye is Y + 3,
dna_swap(A,B,X,Xe,Y,Ye).
It seems to be working. For example, the following query:
dna_swap([t,a,g,t,g,c], L).
Obtains the correct answer in L.
Anyway, as you can see, it's very verbose. Is there a better way?
Edit
This seems to work a lot better:
dna_swap(A,B) :-
append(Left1, [X1,X2,X3|Right1], A),
append(Left2, [Y1,Y2,Y3|Right2], Right1),
append(Left1, [Y1,Y2,Y3|Left2], Tmp),
append(Tmp, [X1,X2,X3|Right2], B).
sublists(List,Count,A,B) :-
length(A,Count),
append(A,B,List).
swap(List,Count,SwappedList) :-
sublists(List,Count,A,B),
append(B,A,SwappedList).
Hope this is what you are looking for:
4 ?- swap([a,b,c,d],2,S).
S = [c, d, a, b].
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.
I'm trying to write a Prolog predicate (SWI) that would select N elements from a List, like this:
selectn(+N, ?Elems, ?List1, ?List2) is true when List1, with all Elems removed, results in List2.
selectn(N,Lps,L1s,[]) :- length(L1s,L), N >= L, permutation(L1s,Lps).
selectn(0,[],L1s,Lps) :- permutation(L1s,Lps).
selectn(N,[E|Es],L1s,L2s) :-
select(E,L1s,L0s),
N0 is N-1,
selectn(N0,Es,L0s,L2s).
My problem is that in some cases, I get duplicated results and I don't know how to avoid them:
?- findall(L,selectn(2,Es,[a,b,c],L),Ls),length(Ls,Solutions).
Ls = [[c], [b], [c], [a], [b], [a]],
Solutions = 6.
This is no homework, but if you want to help me as if it was, I'll be pleased as well.
this could answer your question (albeit I don't understand your first clause selectn/4, permutation is already done by 'nested' select/3)
selectn(0, [], Rest, Rest).
selectn(N, [A|B], C, Rest) :-
append(H, [A|T], C),
M is N-1,
selectn(M, B, T, S),
append(H, S, Rest).
yields
?- findall(L,selectn(2,Es,[a,b,c],L),Ls),length(Ls,Solutions).
Ls = [[c], [b], [a]],
Solutions = 3.