The problem is as follows: Write a predicate in Prolog most_common_sublist(L1,N,L2) that will find the sublist L2 with length N such that it is the most common sublist in L1.
//Example 1:
?- most_common_sublist([1,2,2,3,2,2,4,2,2,3],1,L).
L=[2];
//Example 2:
?- most_common_sublist([1,2,2,3,2,2,4,2,2,3],2,L).
L=[2,2];
//Example 3:
?- most_common_sublist([1,2,2,3,2,2,4,2,2,3],3,L).
L=[2,2,3];
My approach was to generate all the possible sublists of size N using the generator predicate, check which of those is the most common one in the list using the check predicate, and then just put that as my result.
The reason why I'm not using the built-in predicates for length and add is because I'm supposed to write my own.
My generator predicate works, it gives out the correct output.
?- generator([1,2,2,3,2,2,4,2,2,3],3,L).
L = [[1, 2, 2], [2, 2, 3], [2, 3, 2], [3, 2, 2], [2, 2, 4], [2, 4, 2], [4, 2|...], [2|...]] [write]
L = [[1, 2, 2], [2, 2, 3], [2, 3, 2], [3, 2, 2], [2, 2, 4], [2, 4, 2], [4, 2, 2], [2, 2, 3]]
I checked all my predicates and they all seem to work (at least for the test cases I'm using), the problem occurs with the check predicate. It seems to work fine until it gets to N>=P (when this is NOT true, works fine when it is true). I expect the program to go onto the next check predicate under it (the third check predicate) so that it stores Temp value in Result instead of the H value. For some reason it does not go to the third check predicate (I checked with debugger), instead it does something weird (I can't figure out what).
most_common_sublist(L,N,Result):-generator(L,N,LOP),check(LOP,_,Temp),add(Temp,[],Result).
add([],L,L).
add([X|L1],L2,[X|L3]):-add(L1,L2,L3).
length([],0).
length([X|O],N):-length(O,M),N is M+1.
sublist([H|_],1,[H]).
sublist([H|T],N,[H|LOP]):-M is N-1,sublist(T,M,LOP).
generator(L,N,[L]):-length(L,M),N=:=M.
generator([H|T],N,LOP):-sublist([H|T],N,PN),generator(T,N,LP),add([PN],LP,LOP).
check([],Z,K):-Z is 0,add([],[],K).
check([H|T],Hits,Result):-check_how_many(H,[H|T],N),check(T,P,_),N>=P,Hits is N,add(H,[],Result).
check([H|T],Hits,Result):-check_how_many(H,[H|T],N),check(T,P,Temp),Hits is P,add(Temp,[],Result).
check_how_many(X,[X],1).
check_how_many(_,[_],0).
check_how_many(Pattern,[H|T],Hits):-same(Pattern,H),check_how_many(Pattern,T,P),Hits is P+1.
check_how_many(Pattern,[_|T],Hits):-check_how_many(Pattern,T,P),Hits is P.
same([], []).
same([H1|R1], [H2|R2]):-
H1 = H2,
same(R1, R2).
Since I'm not familiar with your code I rewrote it with similar functionality. Lines followed by %here are my improvements (2 times used). For simplicity I used the inbuild predicates length/2 and append/3 instead of add/3. sublist/3 has a complete different code but same functionality, same/2 is not necessary at all. Most uses of you add/3 were not necessary as well as some equality statements.
most_common_sublist(L,N,Temp):-
generator(L,N,LOP),
check(LOP,_,Temp).
sublist(L,N,S):-
length(S,N),
append(S,_,L).
generator(L,N,[L]):-
length(L,N).
generator([H|T],N,LOP):-
sublist([H|T],N,PN),
generator(T,N,LP),
append([PN],LP,LOP).
check([],0,[]).
check([H|T],N,H):-
check_how_many(H,[H|T],N),
check(T,P,_),
N>=P.
check([H|T],P,Temp):-
check_how_many(H,[H|T],N),
check(T,P,Temp)
%here
, N=<P
.
check_how_many(X,[X],1).
check_how_many(_,[_],0).
check_how_many(H,[H|T],Hits):-
check_how_many(H,T,P),
Hits is P+1.
check_how_many(Pattern,[H|T],P):-
%here
Pattern \== H,
check_how_many(Pattern,T,P).
After giving up on tracing I just used the following call to debug after enabling long output (
?- set_prolog_flag(answer_write_options,[max_depth(100)]).
):
?- findall(Temp,check([[1, 2, 2], [2, 2, 1]],_,Temp),Out).
Initial output was
Out = [[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[2,2,1],[2,2,1],[],[],[2,2,1],[2,2,1],[],[]].
Which contains way to much empty lists. First fix (%here) was to set the condition N=<P for the last check/3 case. Until now it was possible to choose a P lower than N, which should be covered by the 2nd check/3 case. Output changed to
Out = [[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[2,2,1],[2,2,1],[2,2,1],[]].
Better, but still empty lists possible. A similar case happens in the last check_how_many/3 case: you have to state that H and Pattern are different, otherwise it would be possible for a fitting Pattern not to be counted. Lets check the output
Out = [[1,2,2],[1,2,2],[1,2,2],[2,2,1]].
Way better. Lets check another case:
?- findall(Temp,check([[1, 2, 2], [1, 2, 2], [2, 2, 1]],_,Temp),Out).
Out = [[1,2,2],[1,2,2],[1,2,2],[1,2,2]].
?- findall(Temp,check([[1, 2, 2], [2, 2, 2], [1, 2, 2]],_,Temp),Out).
Out = [[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[2,2,2],[2,2,2],[2,2,2],[1,2,2]].
Works... Almost.
So the problem seems to be check_how_many/3: alter
check_how_many(_,[_],0).
to
check_how_many(_,[],0).
and you should be fine.
?- findall(Temp,check([[1, 2, 2], [2, 2, 2], [1, 2, 2]],_,Temp),Out).
Out = [[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2],[1,2,2]].
Since it is way more fun to write the code yourself than to debug foreign code I'll add another answer with my attempt.
It is way more fun to code by yourself than to debug alien code. So here is my attempt. It works different than yours because I do not calculate possible subsets but work on the "leftover" list. I use the inbuild predicates length/2, append/3 and member/2 which are 3 lines each to write down.
% check how often 2.nd attribute List occurs in 1st attribute List.
countit([],_,Val,Val).
countit([H|In],Out,Past,Future):-
( append(Out,_,[H|In])
-> Present is Past+1,
countit(In,Out,Present,Future)
; countit(In,Out,Past,Future)
).
mostCommonSublist(In,N,Out):-
maxStartList(In,N,OutList,Max),
member((Max,Out),OutList).
% for every endlist calculate how often the first N elements appear within the endlist, track the max
maxStartList(In,N,[(1,In)],1):-
length(In,N),
!.
maxStartList([H|In],N,[(CntH,Curr)|MaxList],Max):-
length(Curr,N),
countit([H|In],Curr,0,CntH),
maxStartList(In,N,MaxList,CntIn),
Max is max(CntH , CntIn).
The main predicate mostCommonSublist/3 calls the predicate maxStartList/4 to get all sublists/countpairs. Afterwards it validates if the count of a sublist equals the maximum. This is neccessary to check for different answers with the same (maximum) count.
The maxStartList/4 drops elements from the inputlist and counts how often the start of the current list occurs within it. Also it keeps track of the maximum.
For the current inputlist the calculating predicate countit/4 is called. It calculated for a given inputlist (first argument) the number of occurences of a sublist (2nd argument).
My code actually uses a twist: The content of the sublist is not unified when calling countit/4 for the first time, just the sublist length is set. In the first recursion it will unify all entries with the start elements from the inputlist and count it. In the following recursion steps the sublist if fully known. Using an if-then-else (..->..;..) the two cases for remaining inputlist starts with the sublist or not, the predicate basically counts the occurences. Until the remaining inputlist has only N elements left (length(In,N)).
The calculated count/sublist pairs are stored in a list, the maximum is tracked as well.
After knowing all count/sublist pairs I finallize it all by stating that the count of an accepted sublist has to be equal to the maximum.
The nice thing is that there are no dublicate answers.
?- mostCommonSublist([1,2,2,3,2,2,4,2,2,3],3,L).
L = [2,2,3] ;
false.
?- mostCommonSublist([1,2,2,1,2,1,2,2,2,3],3,L).
L = [1,2,2] ;
L = [2,1,2] ;
false.
?- mostCommonSublist([1,2,2,1,2,1,2,2,2,1],2,L).
L = [1,2] ;
L = [2,2] ;
L = [2,1] ;
false.
I have a function which generates one list for each iteration.
(so there are many lists generated at the end of the query).
Is it possible to gather all the generated list into one big list outside of the predicate loop?
?- path(i,k,Path).
Path = [segmentTime(red,i,m,2),segmentTime(blue,m,j,4),segmentTime(blue,j,k,2)] ? ;
Path = [segmentTime(blue,i,m,3),segmentTime(blue,m,j,4),segmentTime(blue,j,k,2)] ? ;
Path = [segmentTime(purple,i,j,3),segmentTime(blue,j,k,2)] ? ;
into something like
AllPaths = [[segmentTime(red,i,m,2),segmentTime(blue,m,j,4),segmentTime(blue,j,k,2)],
[segmentTime(blue,i,m,3),segmentTime(blue,m,j,4),segmentTime(blue,j,k,2)],
[segmentTime(purple,i,j,3),segmentTime(blue,j,k,2)]]
*my code is complex so snippet code of an easy example of something like [1,2,3,4,5] will suffice.
You can use setof built-in predicate. I have linked how it works below.
path([1,2,3,4]).
path([1,2,3,5]).
path([1,2,3,6]).
pathAll(X):- setof(A,(path(A)),X).
OUTPUT
?- X = [[1, 2, 3, 4], [1, 2, 3, 5], [1, 2, 3, 6]]
You can rewrite your code as follows.
pathAll(X) :- setof(Path,(path(i,j,Path)) , X).
OUTPUT
?- pathAll(AllPaths).
AllPaths = [[segmentTime(red,i,m,2),segmentTime(blue,m,j,4),segmentTime(blue,j,k,2)],
[segmentTime(blue,i,m,3),segmentTime(blue,m,j,4),segmentTime(blue,j,k,2)],
[segmentTime(purple,i,j,3),segmentTime(blue,j,k,2)]]
Hope this helped you mate. If this helped you upvote is much appreciated
permutation([], []).
permutation(L, [X|Xs]) :- select(X, L, Rest), permutation(Rest, Xs).
If I type permutation([1,2,3],R), the first solution is "[1,2,3]" but how to get to the second one without using ";" or "fail". I need to use the 2nd solution like "[1,3,2]" or so in order compare it to another list.
What I mean is:
permutation([], []).
permutation(L, [X|Xs]) :- select(X, L, Rest), permutation(Rest, Xs).
go_perm(L,P) :-
L = P,
write(P),nl.
go_perm(L,P) :-
permutation(P,P2), % in this case i wanna get the next solution -.-
go_perm(L,P2).
If L = P then it finishes. Permutation of the first solution for "[1,2,3]" is "[1,2,3]". But that pulls me into stackoverflow because it runs into never-endless thing.
Perhaps you understand me. Thanks!
Assuming you want to loop over the solutions to print them
One standard way to accomplish this is to fail and backtrack, as in:
print_all_permutations(X)
:- permutation(X, Y), print(Y), nl, fail ; true.
Assuming you just want to check if a given solution is correct
You are already done. Just call the function with the reference list and the list you want to test:
permutation([1, 2, 3], [2, 1, 3]).
will return true, because [2, 1, 3] is a permutation of [1, 2, 3]. If the second argument is not a permutation, the goal will evaluate to false.
This is the magic of prolog: finding a solution, or checking if a given solution is correct, are the same thing.
In between: partial solution
The same reasoning still applies:
permutation([1, 2, 3], [2, X, 3]).
will display the only possible value for X.
Or, if you want the whole list to be the result:
X = [2, X, 3], permutation([1, 2, 3], X).
You need to look at various aggregate predicates. Here, findall would work nicely. you can invoke it:
ListIn=[1,2,3], findall(Perm, permutation(ListIn, Perm), Permutations).
This will call permutation on ListIn until it fails. Each Perm returned by permutation will be collected into the Permutations variable.
permutation is a predicate that succeeds when one list is a permutation of the other. You don't actually need to enumerate them; just write permutation([1, 2, 3], [2, 1, 3]) and Prolog will tell you "true".