The MiniZinc constraint solver allows to express cardinality constraints very easily using the built-in sum()
function:
% This predicate is true, iff 2 of the array
% elements are true
predicate exactly_two_sum(array[int] of var bool: x) =
(sum(x) == 2);
The cardinality constraint is met, if and only if the number if true elements in the array of Boolean variables is as specified. Booleans are automatically mapped to integer values 0
and 1
to compute the sum.
I implemented my own cardinality constraint predicate as set of counter slices:
% This predicate is true, iff 2 of the array
% elements are true
predicate exactly_two_serial(array[int] of var bool: x) =
let
{
int: lb = min(index_set(x));
int: ub = max(index_set(x));
int: len = length(x);
}
in
if len < 2 then
false
else if len == 2 then
x[lb] /\ x[ub]
else
(
let
{
% 1-of-3 counter is modelled as a set of slices
% with 3 outputs each
array[lb+1..ub-1] of var bool: t0;
array[lb+1..ub-1] of var bool: t1;
array[lb+1..ub-1] of var bool: t2;
}
in
% first two slices are hard-coded
(t0[lb+1] == not(x[lb] \/ x[lb+1])) /\
(t1[lb+1] == (x[lb] != x[lb+1])) /\
(t2[lb+1] == (x[lb] /\ x[lb+1])) /\
% remaining slices are regular
forall(i in lb+2..ub-1)
(
(t0[i] == t0[i-1] /\ not x[i]) /\
(t1[i] == (t0[i-1] /\ x[i]) \/ (t1[i-1] /\ not x[i])) /\
(t2[i] == (t1[i-1] /\ x[i]) \/ (t2[i-1] /\ not x[i]))
) /\
% output 2 of final slice must be true to fulfil predicate
((t1[ub-1] /\ x[ub]) \/ (t2[ub-1] /\ not x[ub]))
)
endif endif;
This implementation is using a parallel encoding with fewer lines/variables between the slices:
% This predicate is true, iff 2 of the array
% elements are true
predicate exactly_two_parallel(array[int] of var bool: x) =
let
{
int: lb = min(index_set(x));
int: ub = max(index_set(x));
int: len = length(x);
}
in
if len < 2 then
false
else if len == 2 then
x[lb] /\ x[ub]
else
(
let
{
% counter is modelled as a set of slices
% with 2 outputs each
% Encoding:
% 0 0 : 0 x true
% 0 1 : 1 x true
% 1 0 : 2 x true
% 1 1 : more than 2 x true
array[lb+1..ub] of var bool: t0;
array[lb+1..ub] of var bool: t1;
}
in
% first two slices are hard-coded
(t1[lb+1] == (x[lb] /\ x[lb+1])) /\
(t0[lb+1] == not t1[lb+1]) /\
% remaining slices are regular
forall(i in lb+2..ub)
(
(t0[i] == (t0[i-1] != x[i]) \/ (t0[i-1] /\ t1[i-1])) /\
(t1[i] == t1[i-1] \/ (t0[i-1] /\ x[i]))
) /\
% output of final slice must be 1 0 to fulfil predicate
(t1[ub] /\ not t0[ub])
)
endif endif;
Question:
Does it make sense to use home-grown cardinality predicates? Or is the MiniZinc implementation of
sum()
beyond all doubts in terms of solution speed?
Update:
I am using Gecode as solver back-end.