SteveG48 wrote:8r5c1 = r2c1 - 8r2c6 = (HP(7&8))r23c6 - (7|8=2&3)r3c49 - (2|3=9)r3c3 - (9=6&8)r13c2 => -8 r2c1,r5c2 ; stte
I like your latest chain as it is very ingenious, but (8)r2c6 is a member of both terms connected by the red link. When (8)r2c6 is false two strong links are being followed a) (8=7)r2c6 and (8)r2c6 = (8)r3c6 to force (7,8)r2c6,r3c6. Now the next weak link (78)r23c6 - (7|8)r3c49 is dependent on that distribution because if the digits were swapped over it wouldn't be valid.
Ah! you say but they aren't swapped over! To which I stand on my high horse and tell you that every link in a Eureka AIC chain must be independent and stand on its own, but your chain is remembering what went on before.
Fortunately there is a way out of this by using a 'split' node.
(8)r5c1 = (8)r2c1 - (8)r2c6 =
(7,8)HP:r2c6,r3c6 - (7|8=2&3)HP:r3c49 - (2|3=9)r3c3 - (9=6&8)HP:r13c2 => r2c1,r5c2 <>8; stte
The Boolean term (7,8)HP:r2c6,r3c6 is true when cells occupied by the digits match according to their listing orders and false otherwise. I have no problems with this because I believe that within reason we should be able to use any True/False condition we want in a chain provided it's clear what's needed to make it true. Others may differ however as split notes amount to concealed branching. That's true, but the branching involved will be no worse than what we accept in an ALS.
Note now that (8)r2c6 is no longer a possibility in two successive terms as it was before, which makes the chain far more digestible. Note too that the chain can be read backwards.
DPB.
TAGdpbSplitNodes