This is the first prolog code I've written in 20 years. I used to be very into it but that was a long time ago.

surpassing_problem.prolog

% A brute force algorithm to solve the Surpassing Problem posed by Martin Rem in 1988
% 
% By definition, a surpasser of an element of an array is a greater element to the right, 
% so x[j] is a surpasser of x[i] if i < j and x[i] < x[j]. The surpasser count of an element 
% is the number of its surpassers. For example, the surpasser count of the string GENERATING 
% are given by 
% 
% G E N E R A T I N G
% 5 6 2 5 1 4 0 I 0 0
% The maximum surpasser count is 6. The first occurrence of the letter E has six surpassers, 
% namely N, R, T, I, N, and G. 

% ?- surpassingProblem('GENERATING', X).
% X = 6

surpassingProblem(Text, MaxSurpassor):-
  name(Text, CharList),
  surpassingProblemCharacters(CharList, SurpassorCounts),
  max(SurpassorCounts, MaxSurpassor).
  
surpassingProblemCharacters([], []).
surpassingProblemCharacters([FirstChar|CharList], [FirstSurpassors|RestSurpassors]) :-
  surpassor_count(FirstChar, CharList, FirstSurpassors),
  surpassingProblemCharacters(CharList, RestSurpassors).
  
surpassor_count(_, [], 0):- !.
surpassor_count(Element, [H|T], Count):-
  Element < H,
  !, % red cut
  surpassor_count(Element, T, CountOfRest),
  Count is CountOfRest + 1.
surpassor_count(Element, [_|T], Count):-
  surpassor_count(Element, T, Count).
  
max([x], x).
max([H|T], MaxTail) :-
  max(T, MaxTail),
  MaxTail > H,
  !. % red cut
max([H|T], H).