by **tso** » Wed Aug 17, 2005 10:25 pm

Jeff,

The partial puzzle doesn't have enough information as posted. Even when considering the cells with 3 candidates, it has 3 solutions, two of which have different numbers in each cell, so no forcing chain or any other proof is possible.

Also,

1) r1c9 -- the 4 doesn't connect to anything so this loop in the upper right corner cannot form a forcing chain. (If r1c9=4, the rest of the loop can exist in two states.) These five cells are a dead end and can be ignored.

2) Exactly one of r89c3 must be 9

Exactly one of r79c6 must be 9

Therefore, these four cells are in a binary group. The line from r9c3-r9c6 is changed from RED to BLUE

Last edited by

tso on Wed Aug 17, 2005 7:06 pm, edited 1 time in total.