數獨又卡關了 - 數獨

By John
at 2020-08-02T10:10

All Comments

By Mary
at 2020-08-07T01:09

1. C1R678/C6R67/C9R78 HAS 6, C234R678 NO 6

By Caitlin
at 2020-08-09T03:51

2. C1R4 C3R6 HAS 39, C3R4=5

By Zanna
at 2020-08-13T00:16

1. 考慮R1和R8的4, R1C3(4,6)=>R8C2(4,5,6)不能是6。因為
會變成非唯一解

By Odelette
at 2020-08-16T09:01

2. R1C3(4,6),R8C2(4,5) and R9C3(5,6)=>R1C3=4 or R8C2=
4 => R8C3 不是 4

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