by **susume** » Tue Aug 05, 2008 12:38 pm

Thanks for the additional info and Gurth's proof (Im not sure it's a proof so much as a convincing argument, but in any case I'm convinced). In this particular puzzle, each equivalence set consists of 4 symmetrically placed candidates, each 90 degrees anti-clockwise from the last, not just 2, so I think you can get 4x as much information from each chain. Example: the three cells symmetrical to r1c7 are r3c1, r9c3, and r7c9. One set of candidates in them that must all be true together or all be false together are (4)r1c7, (1)r3c1, (2)r9c3, (3)r7c9. The set of digits 1234 are equivalent in every symmetrical set of cells in which one of them appears, and the set 6789 are equivalent in every symmetrical set of cells in which one of them appears.

I believe this allows substitution of any proposition in the equivalence set into a true proposition and it will still be true. Example: (4)r1c7=(4)r3c9 (biloc in block). The following must then also be true:

(4)r1c7=(1)r1c3

(4)r1c7=(2)r7c1

(4)r1c7=(3)r9c7

(1)r3c1=(4)r3c9

(2)r9c3=(4)r3c9

(3)r7c9=(4)r3c9

(please let me know if my notation is bad - still learning)

It may be unusual to have strong links on different digits in different cells, but the ones shown above still behave as regular strong links - i.e. exactly one of the pair must be true. I found a simple nice loop to accomplish the first elimination found by the SE solver.

Oops- Glyn I see you also stated 4-way symmetry while I was writing.... I chose anti-clockwise only because the clues in this puzzle are placed anti-clockwise, so it's easier for me to be sure I'm getting the right digits in the equivalence set.

susume