bat999 wrote:With this puzzle I know that "If r1c2 is not 7 then r1c1 is 1" is true because I can make it work with pencil and paper (pencilmarks on my pencilmarks).
But it's not a valid AIC so it's wrong to post the solution expressed as an AIC.
Got it.
Many people post network solutions in a chain-style notation that looks like an AIC. But, they don't claim it's an AIC. Your solution would have been okay if you hadn't placed "AIC:" at the beginning.
Now, this leads to the use of asterisk (*) in chains to denote where network information is being remembered/applied. Your chain, after some compacting, would look like:
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(7=12)r1c12 - (2=5)r3c2 - (5=149)r4c2,r6c12 - (9*)r5c1 =
(9-3)r5c6 = (3)r8c6 - (3=1)r8c8 - (1=3)r7c8 - (*93=256)r357c1 -
(2=1)r1c1 => -1 r1c2
Note, it could also be trimmed to get a different elimination:
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(2=5)r3c2 - (5=149)r4c2,r6c12 - (9*)r5c1 = (9-3)r5c6 = (3)r8c6 - (3)r8c8 = (3)r7c8 - (*93=256)r357c1 => -2 r1c1
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