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#1: Joe (infrapinklizzard) on Jan 26, 2013 [SPOILER]

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One black pixel can be placed with line logic. However, all the blue and red can be placed with color logic. Then after a little line logic, there will be a total of five black pixels placed.

Then edge logic on the 8 in c25 can place it entirely (anywhere but r9-16 will make impossible blocks in c24).

Line logic makes an arc in the upper right. Then edge logic on the 5 in r1. A quick glance makes it seem like it could go in either c6-10 or c14-18, but the left one is ruled out because it would cause two more blocks in r2. There's only one more block available in r2 - one of the 3s having been taken up in the arc. So r1c14-18 is black.

Line logic completes the arc up over the top and adds a few pixels in c1-3. Plain edge logic on the 4 in c1 rules out c29-30. Deeper edge logic can rule out a few more. Imagine the 4 in r25-28. R1c28 and 27 must be white otherwise you end up with a block of three in c3 (in r25-27).

Then line logic fills quite a bit in the lower left. Looking around for more edge logic, the 6 in r6 looks likely, but examining it only rules out c9. Not enough to place any black.

Try edge logic on the 4 in c22. You can rule out r7&8. Not enough to place any of the 4, but it narrows r8 enough to place a pixel. That little thing brings on a whole lotta line logic.

Now make sure to put in dots that the purple arrow misses. For example, look at the black in r28c7. The solver doesn't know which clue it belongs to, so it leaves it alone. However, we know it must be either the first 1 or the second 1 in r28 - either way it's a 1, so r28c8 must be white.

Also, both in c9 and c13, two whites can be placed around r27 as it must be a 1 in each column. Then a little line logic - but then stuck again.

But wait, there's another white space that can be found with non-purple line logic. Look at the black pixel in c17r22. If we count off the remaining clues in c17 (1|1|2|1), we can see that it must be one of three clues. It must be one of the top two ones, or the first pixel of the 2. In any of those cases it would have a white space directly above it. So c17r21 is white.

That allows more line logic. Then some more non-purple LL -
= in c13, the black pixel in r21 must be a 1 as the 2 is fixed by the pixel in r27.
= C16r26 is a 1 as there are only 1s left in c16.

There's one more that is not necessary but a good mental exercise:
= C14 requires a tiny bit more thinking - the pixel in c14r21 must be a 1. Why? because there is a pixel below it (c14r26). While that pixel doesn't have to be the 2, it does means that any pixel above it can definitely not be the bottom clue - which is the only non-1 clue.

Then line logic to finish.

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Interesting to solve, Thanks Joe.

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Thanks, Kurt & Jota (and Adam, I guess?)

Kurt, that is a good shortcut for an advanced solver, but the deep edge logic needed for 6 in r6 is at least a three-step look-ahead, (place 6, extend four clues, dot end of clues, compare to r8). Too far for a simple explanation and for "moderate look-ahead."

ah....well, it's good you give exercises like you did though. helps people think and see things a lil differently.... can't believe this isn't getting rated better... it's fantastic!

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Found to be solvable with moderate lookahead by infrapinklizzard.

Seven years after the fact I found this among my saved puzzles. The solve was tough, with a fantastic image to make it well worth the challenge. Belated thanks, Joe!

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