Hi

Bill,

With the details you give, our solutions appear to be closer than I thought. We use the same central part, but we deal with it in different ways.

Smythe Dakota wrote:denis_berthier wrote: the central puzzle that remains after cutting out .... the above NW and SE sub-puzzles, ....

That's just using the singularities. Once you notice they're there, you know those little pieces will solve by themselves, otherwise the overall solution wouldn't be unique.

Actually, uniqueness of the global puzzle doesn't imply uniqueness for a sub-puzzle (even for one defined by a 1-cut).

Well, that depends on how you define a sub-puzzle. But when I say it can be solved independently, I mean that, after assigning proper sums to the parts of the segments it includes (and that have been cut into two parts), it can be solved. For the NW puzzle, for instance, this means assigning sum 5 to the trivial 1-cell horizontal sector starting at the right of r5c4 (the remaining two cells have been cut out).

The reason why uniqueness is not guaranteed in this case (I mean in the general case, not in the NW special 1-cell example) is that the constraints that should be inherited by the global segments are (momentarily) forgotten (the simplest example is the two small 2x2 NE and SW sub-puzzles).

I don't mean that 1-cuts must always be used this way. Most of the time, it's more efficient not to isolate a sub-puzzle from the global one.

But in the present case, the central sub-puzzle can indeed be solved independently.

From my experience of trying to solve independent sub-puzzles, it is sometimes easier to solve the largest one than the smallest - always for the same reason that part of the constraints are momentarily forgotten.

Smythe Dakota wrote:denis_berthier wrote: .... the small 2x2 NE and SW sub-puzzles, ....

I don't really see the point of this. Those 2x2 sub-puzzles don't solve by themselves. (The NE one will solve with just a little outside help, from r1c12+r2c12+r3c12+r4c12 only.)

My idea was to have the smallest possible central part.

It doesn't matter if the two small ones can't be solved independently. Once the NW, SE and central ones are solved, the rest is trivial.