Double Dobble - easy optimization


A quick optimization over the quest for Double Dobble arrangements.

In previous post Quest for Double Dobbles I described a possible approach to finding Double Dobbles, with all the limitations.

One observation that allowed me to reduce the search time was that it’s fine to always assume that the first index is $0$. Thanks to the rotation invariance, we can rotate any arrangement and make it so the first index is $0$.

It occurred to me that a similar reasoning can be also done to fix the following index to always be $1$. It works like this:

  • the difference $1$ MUST appear exactly twice in a valid arrangement;
  • hence, at least two indexes MUST be adjacent (modulo $N$);
  • if we apply a rotation over this arrangement to put the one on the left at index $0$, the other one goes into index $1$.

Hence, our iterator for possible combinations becomes this:

my $it = NestedLoops(
      [0], [1],
      map {
         my $end = $N - 1 - ($k - 1) + $_;
         sub { [($_ + 1) .. $end] },
      } (2 .. $k - 1)

which brings us down to ${N - 2} \choose {k - 2}$ possible combination to look through.

For values of $k$ that indeed have a solution, this is not an improvement. I mean, if there is a viable arrangement, the variant starting with $(0, 1, …)$ will be checked before all the others anyway.

The optimization comes handy when trying out values of $k$ that do not have a solution:

$ time perl 8

real	0m17.382s
user	0m17.040s
sys	0m0.096s

$ time perl 8

real	0m4.126s
user	0m4.088s
sys	0m0.008s

I suspect that there may be other low-hanging fruits to further enhance the search… who knows?!?

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