Puzzle Description:

#1: Norma Dee (norm0908) on Jan 4, 2014

A very nice little horse. Hope it came assembled!

nice thales :)

Very nice rocking horse and a fun puzzle

Found this to require deep lookahead. The Undo button got a lot of clicks while I tried to edge-logic the top row, which is where the lookahead was needed. After that it solved with line logic.

Even early in the puzzle a lot of edge logic was required.

very nice Thales.:)

Fun solve, and great image!

Awesome!

Found to be solvable with deep lookahead by infrapinklizzard.

after six pixels of line logic,

edge logic on the 3 in r1 makes c1-2, c8-11 and c18-20 white.

so much for vanilla and even chocolate logic. Now we need some neapolitan logic.

Look at the 12 in r19. Nearly every clue it crosses is a 2, and each is greater than 1. Now look at the rows above and below it. Two 5s and a 6. Since none are as long as the 12, we're going to need psuedo-smile logic. (It's not "real" smile logic which is a very specific set of circumstances.)

Because this pseudo-smile logic doesn't really tell us where to start placing clues, we must think a bit.

The 6 in r20 cannot start at the edge because we know it must be part of a pseudo-smile. In fact, it cannot be in c1-5, nor in c16-20 for that reason. So r20c1-5 & c16-20 must be white. Then LL



Then the least lookahead I found was:

Look at the left 5 in r18: If it were to go in c1-5, then
> the 12 in r19 would also have to start in c1 (because of the white in c1r17).
> that would force the left 4 in r17 into c5-8
= and the crossing clues would then conflict with r16.

So c1r18 is white. That gets a surprising amount of LL, But Wait! There's More!

The same logic can be used on the other end of the line, mirrored. So c20,r18 is also white. And more LL, but not as impressively much.


Now we can use plain old edge logic again. If the 12 were to go left into c4, it would conflict with the 4 in r17. So r18c2-4 must be white. Then LL solves the bottom half.


Internal edge logic on the 4 in c2: if it were to go in r6-9, then
> c2r5 would be white
> the 2 in r6 would force c3r6 to be black which would
> go up into r5.
= Which would conflict with r5's clues - the 5 would not fit.

So it can't go all the way down and c2r9 must be white.

Then LL

Two-way logic on the 5 in r5:
> if it goes right into c6, then c6r8 has to be white and r8c12 would be black,
> if it goes left into c1, then c1r8 has to be white and r8c12 would be black.
= either way, the 9 in r8 will be forced to the right at least one space, so c12r8 must be black. Then LL

Edge logic on the 3 in r1 now forces c12-15 to be white. Then LL

Then things get really sticky. I tried to do something with the fact that the 5 in r9 has to be next to the 9 in r8, but it fits at both ends of the 9's range.

There are a few edge-logics that don't go anywhere: the right 2 in r2 cannot be in c13; the 2 in c1 cannot be in r7 (it cannot be in either r6-7 nor r7-8).

Picture "logic" seems to imply that the 2s in r1-4 and c1-4 make a zigzag, and the 3 in r1 probably spans the 1. However, there is no easy way to eliminate the 3 in c3-5 nor in c5-7. (Remember, never rely on picture "logic"! The picture may not end up as you imagined, and it's really nothing but a guess. At best use it as a place to start looking for real logic.)

Okay, some deep lookahead with the 5 in r8. Via a roundabout way.

Look at the 2 in r7.
If it were to go into c8-9, then
> c9r9 would be white, and the 2s in c10-12 would be forced down into r9
> That would force the 5 in r9 into c10-14.
= A quick check rules out c10-14 (it would force the 9 to be ten long because of the pixel in c4). So the 2 cannot be here.

So that leaves only two possibilities for the black pixel in r7: Either

(A) the 2 is in c7-8; then
> c7r9 would be white and the 5 would have to be within c8-13 (remember 10-14 is bad).
> which would make c9-12 black

(B) or the pixel is the 1 and so
> r7c7&9 are white. forcing the 2s in those columns down into r9.
> that would pin the 5 to within c6-11, making c7-10 black.

== Either way r9c9-10 are black.

And that helps us not very much.

We really need to rule out the two pixels in r2c14-15. So what would happen if we say they are black?
> That would make the rest of those columns white.
> The 9 would be forced into c4-12.
> The important thing to note is that the 9 will need companions in the adjacent rows in all but possibly c5 (where the crossing clue is either a 1 or a 3)
= including the 2 in c13, there is no way to arrange the clues in r7&9 to fit this. (and this was deep lookahead.)

So the 2 in r2 cannot be in c14-15 and they must be white.

Then LL to finish.

Bleah. This was an ugly solve.

Thank you for the extensive analysis, Joe!

Goto next topic

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