/*

Langford's number problem in ECLiPSe.

Langford's number problem (CSP lib problem 24)
http://www.csplib.org/Problems/prob024/
"""
Arrange 2 sets of positive integers 1..k to a sequence,
such that, following the first occurence of an integer i,
each subsequent occurrence of i, appears i+1 indices later
than the last.
For example, for k=4, a solution would be 41312432
"""

* John E. Miller: Langford's Problem
http://www.lclark.edu/~miller/langford.html

* Encyclopedia of Integer Sequences for the number of solutions for each k
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=014552

Also, see the following models:
* MiniZinc: http://www.hakank.org/minizinc/langford2.mzn
* Comet   : http://www.hakank.org/comet/langford.co
* Gecode/R: http://www.hakank.org/gecode_r/langford.rb

Model created by Hakan Kjellerstrand, hakank@gmail.com

*/

% Licenced under CC-BY-4.0 : http://creativecommons.org/licenses/by/4.0/

:-lib(ic).
:-lib(ic_global).
:-lib(ic_search).
%:-lib(branch_and_bound).
:-lib(listut).
:-lib(util). % for time
%:-lib(propia).

go :-
K :: 2..8,
indomain(K),
writeln(K),
Selection = first_fail,
Choice = indomain_min,
writeln([selection:Selection, choice:Choice]),
time(findall([K,Backtracks,Solution,Position], langford(K,Solution,Position,Selection,Choice,Backtracks),
L)),
length(L,Len),
writeln(L),
writeln(len:Len),
nl,fail.

% For a specific K, check all possible variants of Selection and
% Choice methods.
go2 :-
K = 8,
selection(Selections),
choice(Choices),
writeln(K),
( foreach(Selection,Selections), param(Choices,K) do
( foreach(Choice, Choices),
param(K,Selection) do
writeln([selection:Selection, choice:Choice]),
time(findall(Backtracks,
langford(K,_Solution,_Position,Selection,Choice,Backtracks),
L)),
length(L,Len),
writeln(backtracks:L),
writeln(len:Len),
nl,
flush(output)
)
).

selection([input_order,first_fail, anti_first_fail, smallest,largest,
occurrence,most_constrained,max_regret]).
choice([indomain,indomain_min,indomain_max,indomain_middle,
indomain_median,indomain_split, indomain_random,
indomain_interval]).

langford(K, Solution, Position) :-
langford(K, Solution, Position,first_fail,indomain,_Backtracks).

langford(K, Solution, Position,Selection,Choice,Backtracks) :-
K2 is 2*K,
length(Position, K2),
Position :: 1..K2,

length(Solution,K2),
Solution :: 1..K,

ic_global:alldifferent(Position),

%  symmetry breaking
nth1(1,Solution,Solution1),
nth1(K2,Solution,SolutionK2),
Solution1 #< SolutionK2,

( for(I,1,K), param(Position,Solution,K) do
IK is I+K,
nth1(IK, Position, PositionIK),
nth1(I, Position, PositionI),
I1 is I+1,
PositionIK #= PositionI + I1,
nth1(PositionI,Solution,SolutionPositionI),
SolutionPositionI #= I,
nth1(PositionIK,Solution,SolutionPositionIK),
SolutionPositionIK #= I
),

term_variables([Position,Solution], Vars),
search(Vars,0,Selection,Choice,complete,[backtrack(Backtracks)]).