The puzzle is from Andrew Stuart's Weekly Unsolvables #429 (from David Filmer)
The board position is after many moves in my solver in case your solver doesn't arrive at this position.
980700600750000090006000000640000000009600050000003000007900083005800960000020001
- Code: Select all
9 8 1234 7 1345 1245 6 124 t 24
7 5 1234 24 t 6 124 1234 9 8
124 b 12 b 6 1234 8 9 1237 12347 5
6 4 128 125 157 12578 1238 123 9 12
1238 123 9 6 14 1248 1248 5 7
5 7 128 124 9 3 1248 124 6 124
124 6 7 9 145 145 245 8 3
1234 123 5 8 1347 147 9 6 24
348 9 48 345 2 6 457 47 1 4
CLb CL1 CL2
(124)JE2:r3c12,r2c4,r1c8
I think this is a valid JE2 exocet with incompatible base candidates 12 in r3c12.
I further think that based on "04 Incompatible Base Pair Examples v2.odt", section 3 "Example 1 - JE2", since 12 are incompatible, I can solve r3c1 as a 4.
This is not the case since 4 is not the solution to r3c1 (it is a 2).
So, somewhere, my understanding of JEs and/or incompatible pairs is wrong.
Would someone be willing to point out my mistake(s)?
Thank you,
Tim