## Looking for help

All about puzzles in newspapers, magazines, and books

### Not really

Not really
when I am looking at R6C2 and R7C2 I know that 3,6 is impossible, the 4,6 I removed immediately and looked for the 1,9 which is not that tough.

Regarding rubylips argument the other candidates are not relevant here
he is saying r5c1=4 =>r5c1<>8 => r4c1=8 => r6c2=3 => r8c2=6 => r8c1=4
and that a contradiction so r5c1<>4
bennys

Posts: 156
Joined: 28 September 2005

bennys and rubylips thanks for your patient replies -

The chain is perfectly clear - I had a problem understanding the original notation.

bennys - how does this combination {39} {29} {289} {689} lead to

Code: Select all
`R5C2 cant be 3 because of C (the only 3 in C is in R5)`
?
emm

Posts: 987
Joined: 02 July 2005

If you have locked set (its locked because the 6 is removed)and a candidate appear only in one square it force to get that value.(R5C7=3)
bennys

Posts: 156
Joined: 28 September 2005

Of course, r6c2=6. Thanks.
emm

Posts: 987
Joined: 02 July 2005

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