denis_berthier wrote:Could you be more explicit on the following two chains in your solution and what they eliminate?

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r5c2-7(-r5c4)-r5c789=7=hidden pair r46c7-7-r2c7=7=hidden pair r2c89=5=r2c5-5-r4c5-9-r5c4-4-r5c7-1-r5c8, no value for r5c9

hidden pair in c2

r8c2=8

3 is locked in c3b7

r2c7-7(-r7c7)-naked triple r456c7-1-r7c7-2(-r7c5)-r7c1-1(-r4c1)-r7c5-5-r4c5-9(-r5c4)-r4c1-8-r4c7-1-r5c7-4-r5c4-7, no place for 7 in b6

The first chain eliminates 7 from r5c2. It is completely identical to ttt's first move but written in a different way. (I could have started it r5c236-7-...)

The second chain eliminates 7 from r2c7. However, ttt describes a different chain for his second move which eliminates different candidates but with similar effects. I admire greatly the simplicity of his loop compared to mine.

Being completely unfamiliar with nrczt-notation, I describe my second step like this:

7r2c7 creates a triple of 148 in c7 => r7c7=2 => r7c1=1 and r7c5=5 => r4c5=9 (=>r5c4<>9) => r4c1=8 => r4c7=1 => r5c7=4 => r5c4=7 which crashes the puzzle because there is no place for 7 in b6.