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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Tue Jul 25, 2017 11:12 pm Post subject: |
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| Ajò Dimonios wrote: | Hi JC
Perhaps the logic you described to me is also included in the following AIC, which first eliminates 1 in R5C456 and consequently also eliminates 5 in R8C7 and R6C4.
AIC
(1)R5C9=(1)R9C9-(1)R7C78=(1-5)R7C5=(5)R8C4-(5=1)R6C4=>-1R5C456=>R6C4=1;-5R6C4;R6C7=5 and -5R8C7=>solution stte.
Ciao a tutti
Paolo |
Paolo,
Couldn't the chain be shortened by eliminating one more 1, as shown below, with the change to the term in red.
(1)R5C9=(1)R9C9-(1)R7C78=(1-5)R7C5=(5)R8C4-(5=1)R6C4=>-1R5C456,r6c7 stte |
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Ajò Dimonios
Joined: 01 May 2017
Posts: 339
Location: Sassari Italy
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| Posted: Wed Jul 26, 2017 6:12 am Post subject: |
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Hi Marty
Yes, of course Marty. I wrote this chain so long to point out that the AIC first eliminates 1 in R5C456 ,R6C7 and after 5 in R6C4 and R8C7 as JC indicated by another method that starts with the same constraints.
Ciao a tutti
Paolo |
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