Puzzle Description Suppressed:Click below to view spoilers

#1: Heather M (AuntieH) on May 1, 2014

Lovely. Make mine a chai please. :)

Make mine iced :)

Couldn't get it. Tried twice.

Loved the solve, will wait for the experts. Thanks Rita.

Found to require some guessing by infrapinklizzard.

The bottom I got after some involved deep lookahead involving the 3 in r29.

The top, though seems to be a mess. After peeking, I thought maybe I could do something with the 2,1,2, in c15, but it doesn't have to be the 1; it could be a 2. (5up,7up,2down,3down,5up,2up; or 5up,7up,2down,3across,5up,1)

Also, it could go 5down,7across,2up,3down,5down,1 or many other positions. ugh.

So I'm calling guessing unless someone spots a way I missed.

==
Here's how I solved the bottom:

After initial line logic,

Edge logic on the 12 in r30 makes c25-30 white.

Deep edge logic on the 20 in r30 shows that it cannot go left past c11 without a conflict in r28, so that makes c18-22 black. Then LL

Edge logic on the 6 in c30 makes c2-7 white. Then LL

Edge logic on the 3 in r29 makes c23 and c27-28 white.

Now the 3 in r29 can be in one of two positions; it must be in either c10-12 or c24-26.

Some very deep lookahead can prove it's not in c10-12.

If it were, there'd be eight possibilities for those three columns in r30 (since the vertical clues are at least 2 in each case). We'll deal with them in binary order with 0 meaning the pixel in r30 is white, and 1 meaning it is black. (In all cases, all three pixels in r29 are black.)

000 - 1 in r27
001 - 2,2 or 5 in r28
010 - 1,12 in r30
011 - 2,2 or 5 in r28
100 - 1 in r27
101 - 1 in r27 & r28
110 - no immediate conflict
111 - 1 in r28

So as you can see, the only way that r29c10-12 can be the 3 is if the 2in c10 and the 3 in c11 go into r30, but the 2 in c12 does not.

This we can disprove in one more lookahead - if the 3 in r29 is in c10-12, that would force the 6 into c3-8. The 1 in c6 would then force the two currently black pixels in r28c7-8 to be the 2 clue in that row. That means that the two new black pixels in r28c11-12 would create a conflict as they could not be part of the 4 clue (because r28c10 would be white).


[[ Another way to do this lookahead would be to start with the assumption that the 3 is in c10-12. That would force the 6 to be in c3-8.

[[ The 1s in c4,5,6 combined with the 3 in c11 would force the two black pixels in c7-8 to be the 2 clue in r28. Then the 3 clue in c11 would force r28c9-13 black (because of the 4 in r28).

[[ That would cause two conflicts - it would make a block of 1 in r30 right next to the 12, and it would also cause a block of 1 in r27 (in c13).

So, whichever way you deep-lookahead it, the 3 cannot go in c10-12, and so must be in c24-26.

Then LL solves the bottom.

i feel like i should have tracked this one. it definitely made my head hurt. i made two look-ahead moves but neither seemed super complicated. one was with about 30 pixels left and the other with about 20...right in the steam. anyway, might have to redo this one and track my moves...when my head stops hurting.

For the bottom part some simple or more advanced EL was enough. It went pretty fast, no problems there. It sounds more complicated how Joe solved it. I guess it's matter of the starting point for building the contradictions.

For the upper part, the 12 on l8 and the succession of the 1-2-1-2 vertical clues gave its exact position, from were some internal edge logic solved it. So I would rate it some look-ahead. Not very deep, but not just simple eithe. Why does it still have a "?".

It's funny. I got to the very end but can't solve the last square. Not sure where I went wrong. I like the image and the challenge.

If you look above where it says solvability it says that guessing is required. So... guess away. :)

Found to be solvable with deep lookahead by infrapinklizzard.

Good catch on the 12 in r8, kibitzar. LL will then complete the upper rim of the cup.

However, the rest of the top does not seem to be so simple.

If the 5 in c13 is in r8-12, it will force the 7 in c14 up one more row and the crossing 2s will make a block of 3 in c15 where there can't be one. So c13r12 must be white.
ll
If the 5 in c13 is in r7-11 it will trigger the 2s in r9-11. This will not leave any room for the bottom 2 in c15. So c13r11 must be white.
ll
The 3 in c19 cannot be in r5-7 because of line logic. It cannot be in r6-8 or r7-9 because it would cause a block of three in c18. So, all-in-all it cannot be in r6 and c19r6 must be white.
ll (Note that line logic makes r11c19 white - no matter whether the pixel at r8c18 is part of the 3 or the 1 clue in c18.)
That's the extent of moderate lookahead I can find on the top.
For deep lookahead:
If the 7 in c14 were to reach up beyond r4 then the 2s in r4&5 would have to go right (because of the 1 in r3). That would force the 5 in c13 down into r10 which would trigger the 2 and make c14r10 black causing a conflict. So c14r32-3 must be white.

If the 7 in c14 was in r6-12, it would trigger the 2 in r11 which would trigger the 2 in c15. That would then force the 5 in c13 up one more row which would force r5c14 black creating a conflict. So r12c14 must be white.
ll

The left 2s in r9&10 cannot both go right without causing a conflict, so at least one must go left. No matter which goes left, r9c13 will be black.
ll
extended edge logic 3 c19: if it is in r8-10 then it will fix the 2 in c18. That will the force the 1 in c18 into r12. But the 1 in r19 would also have to be in r12 causing a conflict. So the 3 cannot be there and the black pixel in c19r8 must be the 1 clue.
ll
2-way logic on the 2 in c15: whether it goes up or down, c16r12 must be white.
ll
It occurs to me that the top would be much easier to solve if the unknown pixel at r12c18 were mistakenly made white. Unfortunately there's no good reason to do away with it.
deep lookahead:
if the 5 in c13 were to go up into r5, then r5c15 would be white forcing the 2 in that column up. Which then makes c16r3 white, forcing c16's upper 3 down and which forces the lower 3 down into r10. Which then forces the 2 in r10 left into c13 making a conflict with our original assumption. So the 5 cannot go up and c13r5 must be white.
LL to finish

I am happy to change the determination of a puzzle if a better solution is pointed out to me, but the solution must have specific steps.

Show: Spoilers

Goto next topic

You must register and log in to be able to participate in this discussion.