The New Sudoku Players' Forum

[Jeff]

Please help.

[jimbob]

From the position you posted I would look at 'column' 8. There are only 2 places for a 4.

If the 4 is in row 5 then r5c7=1
If the 4 is in row 6 then r6c7=1, r6c6=5

Either way, 1 can be eliminated from other cells in column 7. This leaves r4c7 as being 2 or 5, so the 8 cage (is that the right term in a kakuro?) in column 7 must be 1,5,2.

Can you progress from that point?

[udosuk]

Okay, since r4c7 has to be 2|5, r4c8 cannot be 6.
And if r6c8=6, r6c6=3 => r6c7=1 => r5c7=2 => r5c8=3, and we must have 9s in both r28c8... Therefore r6c8 cannot be 6.
Which leaves r8c8=6, r8c5=1 and r7c5=7, and also r2c8=9...

And the rest would flow from there...

[Jean-Christophe]

Another way:
Since R7C45 <> {15} -> R7C6 <> 2
-> R6C6 <> 5
Since R6C6 <> {15} -> R6C8 <> 4
-> R5C8 = 4 (hidden single in C8)
R5C7 = 1, R46C7 = [25], R4C68 = [71]
...

[Jeff]

Good moves, thank you very much.