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#1: Jota (jota) on Apr 9, 2014 [SPOILER]

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i agree with jota, nice image rita :)

New version published by vitanulka.

Thanks :)

Very nice. Glad I didn't have to blow them up!

This is how I did it. It has not been rearranged for clarity. It's written as I found it.

After LL,

The 7 in r30 can go in only one spot using edge logic.

edge logic on the 3 in r7 makes c15,c21-26 white

edge logic on the 3 in c3 makes r25-26 white

edge logic on either 3 in c28 makes r10-11 and r22-26 white

edge logic on the bottom 2 in c27:
Note that it would both have to be under the two blue pixels in r19-20 and that it would have to have the 1 in c26 (or nothing) next to it. So that makes r22-26 white

edge logic 2 in r25 makes c5-6 white

Extended two-way logic on the 2 in c5: it can be in r26-27 or r29-30.
> If it's in r26-27, then the 3 in r26 extends to c7, and everything under the 3 in rows 28-30 is white.

> If it's in r29-30, then r29 is white from c6-30 and r30c6 is white

= Either way, r29c6-7 and r30c6 are white

Then line logic finishes the lower left.

edge logic on either 4 in c14 makes r8, r24-26 white

edge logic on the 6 in c14: It can be either within r9-17 or IN r21-26. If it were in r21-26, then the crossing 4 clue in r27 would give the 3 in c17 nowhere to go. So the blue pixels in c14r21-22 must be the 2 clue and c14r23+ must be white.

Then LL to finish

The extended two-way logic pushes the edge of moderate, but I believe it is still within as it is regular advanced logic + one step.

So - moderate lookahead, but barely.

Found to be solvable with moderate lookahead by infrapinklizzard.

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Nice puzzle....enjoyed the solve....thanks!

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