Puzzle Description:

Oh my, oh my!

#1: Kim (kjh) on Jun 22, 2013

Very cute, Tom. Please note, this is exactly what I would say - and be sure that I would divert my eyes as well! Thank you!

yrqw

Loved the solve!

Found to be solvable with moderate lookahead by infrapinklizzard.

After line logic,

Pseudo-edge logic on the 11 in c15. I say pseudo, because we're not looking at anything near the 11. Look at c6. the 3 is fixed by the pixels in c6r5-6. That means the top 1 must be somewhere within r2-4.

Now look at the 11 again. If it were to go all the way up into r2, the 1s in r2-4 would mean there's nowhere for the 1 in c6 to go. So the 11 cannot go all the way up and c15r2 must be white.

That gets a couple pixels with line logic. But then we're stuck again.

Do the same thing again, but using c5 as the comparison. If the 11 were to go up into r3, then c5 wouldn't have room for its 1. So c15r3 must be white. (This could actually have been done first, but it would have been much harder to explain.)

Then LL to finish.

thx Joe

Is that a baby fallen out of a high chair?
Oh my!

Alternate solution to Joe's instructive one: I was able to do enough line logic that c14 had r5 and r9 as black. I had placed c4r1. Then r15c2 or r15c3 being black would make r15c4 black and r14c4 white. But then there would not be space for the 2.

So that places r14c15 and r14c14, placing r4c14. So r5c14 is part of the 2. Then I got the rest by line logic.

?

Looks like someone praying at the porcelain god.

Goto next topic

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