Stuck, yet again. Hopefully I didn't make another stupid mistake like yesterday...
:( I suck
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i found just one step for now
by bennys » Mon Oct 03, 2005 5:46 am
r7c7 can't be 6 because then r8c7=2 and then both r5c8 and r9c8 have only 45 as candidats and that mean that r7c8=9 but it can't be because of r7c4
- bennys
- Posts: 156
- Joined: 28 September 2005
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