count successive occurrences of number in Prolog

Question

Hello I am trying to make a program in Prolog that given a list it counts the occurrences of each successive element in the list as follows:

count(1,[1,1,1,2,2,2         


        

Answer 1

Another solution (tail recursive) is this:

run_length_encode( Xs , Ys ) :- % to find the run length encoding of a list ,
  rle( Xs , 1 , Ys ) .          % - just invoke the helper

rle( []       , _ , []    ) .     % the run length encoding of the empty list is the empty list
rle( [A]      , N , [X:N] ) .     % A list of length 1 terminates the run: move the run length to the result
rle( [A,A|Xs] , N , Ys    ) :-    % otherwise, if the run is still going
  N1 is N+1 ,                     % - increment the count, 
  rle( [A|Xs] , N1 , Ys )         % - and recurse down
  .                               %
rle( [A,B|Xs] , N , [A:N|Ys] ) :- % otherwise, if the run has ended
  A \= B ,                        % - we have a break
  rle( [B|Xs] , 1 , Ys )          % - add the completed run length to the result and recurse down
  .                               %