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#1: Tom O'Connell (sensei69) on May 27, 2011 [HINT]

again, i used smile logic at the top

I really do not see a good way to solve this one without looking more than two moves ahead.

Tom - that doesn't fit the definition of smile logic.

hmmmmm your the judge, gator

i could not solve this logically

I guessed on the 5 in C1, then was able to work out the rest (I think).

Comment Suppressed:Click below to view spoilers

- LL, EL the 3 in R2 and the 5 in C1 and more LL...

- no matter how you slide the 8, either C8R4 or C8R11 will be black, making C10R11 a dot no matter what.

- realizing that wherever the 2 in C10 goes it will trigger the 2 in C9, there will always be dots in C9R5, C9R8 and C9R13

- LL

- imagine the other 2 in C2 being in R11&12. That would dot C3R9, violating R10 on the other side, so dot C2R11

- LL to end

That's the best way that I found to solve it with the least look-ahead...that last step before LL. I think it's relatively easy to eyeball (the paper/pencil test), but I still think it qualifies as more than two moves to get there.

I thoroughly enjoyed the puzzle though!

"- no matter how you slide the 8, either C8R4 or C8R11 will be black, making C10R11 a dot no matter what." I can't agree with this. If column 10 had clues "2 2", then I could.

wow, what was i thinking? ..i'm going to blame it on being up way past my bedtime. i'll relook at it. thanks for the catch gator.

how about this (assume LL start/between-each/end):

- EL the 3 in R2 and the 5 in C1

- realizing that wherever the 2 in C10 goes it will trigger the 2 in C9, there will always be dots in C9R5, C9R8 and C9R13

- internal edge logic on the 8 (if it starts in R13 or R12, it violates C10 with too many dots)

- EL the bottom 2 in C2

and here is the new look-ahead move:

- if you imagine C4R4 is black, that would push the bottom 2 in C2 into the lower slot (via C1), violating C4...so it's a dot.

Thanks, y'all.

I looked at David's (#10) comments since this is not a logical solve. I don't get what he's saying in the last bit with how the last step violates Column 4.

However, I do get this. Blacking R4C4 and thus forcing the 2s to R8-9 and R12-13 (what it's down to by this point) forces the 6 all the way into the last column. This winds up with a pair of 2s in R12, which is a violation. We don't need the step in R4C4 for this; once you reach that step, black C2, R12-13, and you'll find the violation. He winds up being correct on R4C4 becoming a dot, but it's not the crux of that step.

I needed a form of counting logic to finish up, however. By blacking C3, R12-13 after doing some line logic, you wind up having three squares that have to be false to verify C3. This causes a 6 in C1, a violation. So I dotted those and moved to the finish. I would argue that guessing is required, however.

Found to be solvable with deep lookahead by infrapinklizzard.

Yet another emendment to an emendment:

In #10 David is right up to the last step. However, the bottom 2 in r2 is already placed. So what he said is indeed wrong. But what he meant is not. What he meant is the (only) 2 in c3. So:

If r4c4 were black:
> r4c1-2 would be white, pushing the 5 in c1 all the way down.
> the 5 would then fill in the 2s in r8 & r9 so c3r8-9 would be white
> which would force the 2 in c3 into r12-13, triggering the 2 in r12
> which would make another block of 1 in c4.
= That would make three blocks of 1 in c4, so our original premise must be wrong and r4c4 must therefore be white.

Show: Spoilers

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