Puzzle Description:

But spilled wine...

#1: Teresa K (fasstar) on Nov 28, 2009 [HINT]

No guessing. A little edge logic here and there helped on the inside.

Found to be logically solvable by jan.

It took me a bit more than plain old fashioned edge logic to get the blue, but it is doable. Quite a nice puzzle, challenging and a good image.

Ditto #3. The black alone was fun enough to solve, without even doing the blue. Great job.

Thanks for the challenge :)

After initial line and color logic, all the red should be placed. In addition, there should be 7 black pixels placed (which adds up to *four* black 2 clues) and five blue pixels placed.

Edge logic on the 5 in c7 rules out r17 due to a conflict in c9. (The 2|2|3 clues in r15|16|17 would make a block of one in c9, conflicting with the 2 in that column.)

Now another round of edge logic on the 5 in c7 will rule out r14-16 due to a conflict in c8. (If the 5 extends below r13, it would create a block of one black in c8 conflicting with the 2 and 3 clues.)

Then line and color logic finishes the black.

We still only have five blue pixels placed. We must do several rounds of edge logic ruling out spaces before we can actually get around to placing any blue pixels.

Edge logic on the 3 in c1 will rule out r7&8 due to a conflict in c2. (It would create a block of one blue in c2 where there must be a block of 3.)

Then edge logic on the 3 in c2 will rule out r7&8 due to a conflict in c3. (It would create a block of one blue in c3 where there must be a block of 3.)


Now there's a bit of edge logic that's a bit harder to visualize since it's not on an edge.

But first, to help with the process, let's look at some line logic. Notice the 2|3 in c5. Also notice that r4 and r6 are already blue. Line logic on that column shows that those clues can only be in two positions. Either those two pixels are both in the 3, or they are one each in the two clues.

In other words, the three in that column can *only* be in r4-6 or r6-8.


OK, now we will do "edge logic" on the 5 in c4.
Just like edge logic, we will hypothetically place the 5 in r4-8. Now we see the ramifications:
- c5r8 must be white due to the 1 in r8
- c5r7 must be blue due to the 3 in r7
and that crashes the 3 in c5.
Therefore, the 5 in c4 cannot reach r8, and c5r8 is white.

Then line logic to finish.


Fun puzzle.

Goto next topic

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