Composite checkerboard

[davar55]

Originally Posted by davieddy

If 25 swaps are needed to make the top half all even/odd, then
at least five(?) of the boundary pieces will be involved. By doing these
first, surely we can get the cross boundary sum associated with these
pieces to be composite, reducing the 10 boundary swaps
subsequently needed.

It's possible that all twenty-five swaps needed will be "behind the lines"
and not on the boundary (which may be evens and odds already).
In that case all 35 swaps may be needed to get the composite
configuration desired.
It's still open as to whether separating into even/odd halves
is in fact optimal.

[davieddy]

Originally Posted by davar55

It's possible that all twenty-five swaps needed will be "behind the lines"
and not on the boundary (which may be evens and odds already).

In that case you can elect to make the top half all odd instead of
all even (or vice versa), thereby having to swap all the boundary pieces
in the first 25 swaps..

[davieddy]

Originally Posted by Wacky

Go on and swap it. In its present location, it will need to be swapped with the "correct" replacement anyway. That swap can use the token in the current position just as well.

Yes. Since all the even counterparts are distinct, so is
99-<even counterpart> mod 100. So there is no contention for
the same number along the odd boundary.

[davieddy]

Yes I made a booboo.
Maybe 199 is prime or maybe it isn't.
But if you dont want to talk to me,
I don't want to talk to you.

David