Finding the longest contiguous sublist in Prolog

Question

I\'m a beginner in Prolog and this is my question:

I have a sorted list of integers without duplicates i.e. [1,2,3,11,12,13,14,21,22,23,24,25]

I

Answer 1

First, we define z_nonsucc_t/3 based on clpfd and bool01_t/2:

:- use_module(library(clpfd)).

z_nonsucc_t(X,Y,T) :-
   Y #\= X+1 #<==> B,
   bool01_t(B,T).

To split an integer list into consecutive runs, we use splitlistIfAdj/3 like this:

?- splitlistIfAdj(z_nonsucc_t,[1,2,3,11,12,13,14,21,22,23,24,25],Pss).   
Pss = [[1,2,3],[11,12,13,14],[21,22,23,24,25]].

Next, we define meta-predicate max_of_by/3 based on if_/3, (#>)/3, and (#>=)/3:

max_of_by(X,[E|Es],P_2) :-
   call(P_2,E,V),
   max_of_by_aux(Es,P_2,V,[E],Rs),
   reverse(Rs,Xs),
   member(X,Xs).

max_of_by_aux([]    , _ ,_ ,Bs ,Bs).
max_of_by_aux([E|Es],P_2,V0,Bs0,Bs) :-
   call(P_2,E,V),
   if_(V #>  V0, Bs1=[], Bs1=Bs0),
   if_(V #>= V0, (V1 = V , Bs2 = [E|Bs1]),
                 (V1 = V0, Bs2 =    Bs1 )),
   max_of_by_aux(Es,P_2,V1,Bs2,Bs).

To get the longest list(s), we use meta-predicate max_of_by/3 with length/2 like so:

?- max_of_by(Xs,[[1,2,3],[11,12,13,14],[21,22,23,24,25]],length).
Xs = [21,22,23,24,25].

Note that max_of_by/3 may succeed more than once in tie cases:

?- max_of_by(Xs,[[1,2,3],[11,12,13,14,15],[21,22,23,24,25]],length).
  Xs = [11,12,13,14,15]
; Xs = [21,22,23,24,25].

Put it all together in predicate maxCont/2:

maxCont(Zs,Xs) :-
   splitlistIfAdj(z_nonsucc_t,Zs,Pss),
   max_of_by(Xs,Pss,length).

Sample queries:

?- maxCont([1,2,3,11,12,13,14,   21,22,23,24,25],Xs).
Xs = [21,22,23,24,25].

?- maxCont([1,2,3,11,12,13,14,15,21,22,23,24,25],Xs).
  Xs = [11,12,13,14,15]
; Xs = [21,22,23,24,25].