/*

  Euler problem 45 in SICStus Prolog

  """  
  Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

  Triangle 	  	Tn=n(n+1)/2 	  	1, 3, 6, 10, 15, ...
  Pentagonal 	  	Pn=n(3n−1)/2 	  	1, 5, 12, 22, 35, ...
  Hexagonal 	  	Hn=n(2n−1) 	  	1, 6, 15, 28, 45, ...

  It can be verified that T(285) = P(165) = H(143) = 40755.

  Find the next triangle number that is also pentagonal and hexagonal.
  """

  Model created by Hakan Kjellerstrand, hakank@gmail.com
  See also my SICStus Prolog page: http://www.hakank.org/sicstus_prolog/

*/

:- ensure_loaded(hakank_utils).

go :- 
        L = [
            euler45a
            ],
        run_problems(L).

%%
%% 0.016s
%%
euler45a :-
        T is 285+1,
        tri(T,TT),
        P is 165,
        pent(P,PP),
        H is 143,
        hex(H,HH), 
        e45a(T,TT,P,PP,H,HH,X),
        writeln(X).

e45a(_,X,_,X,_,X,X).

e45a(TN,T,PN,P,HN,H,X) :-
        T > P,
        PN1 is PN+1,
        pent(PN1,PP),
        e45a(TN,T,PN1,PP,HN,H,X).

e45a(TN,T,PN,P,HN,H,X) :-
        P > H,
        HN1 is HN+1,
        hex(HN1,HH),
        e45a(TN,T,PN,P,HN1,HH,X).

e45a(TN,T,PN,P,HN,H,X) :-
        T > H,
        HN1 is HN+1,
        hex(HN1,HH),
        e45a(TN,T,PN,P,HN1,HH,X).

e45a(TN,T,PN,P,HN,H,X) :-
        (T \= P ; P \= H ; T \= H),
        TN1 is TN+1,
        tri(TN,TT),
        e45a(TN1,TT,PN,P,HN,H,X).

pent(N,P) :- P is N*(3*N-1) div 2.
tri(N,T) :- T is N*(N+1) div 2.
hex(N,H) :- H is N*(2*N-1).