Puzzle Description:

Answer in first spoiler.

#1: Brian Bellis (mootpoint) on May 11, 2014 [SPOILER]

Beer nuts and bolts

The top right is really tricky. If you notice the 1,3,2 2,1 1,2 2,3,1 pattern appear twice you can use symmetry to solve. Maybe some of you can find a better way to solve it.

I look forward to your thoughts.

idk, B...I think it's kinda stretching it. A pattern is easy enough to see, but not as easy to decipher where to start without Trial&Error or deep l-a. interested to hear what Joe, David, and that ilk has to say. one could safely place some el and ex el dots, but I don't think too much more.

Well, I saw at the first sight there will be a two-way beginning. Patterns were clear.
Choose one way until the first error ... (or will be finished) :)

Found to be solvable with deep lookahead by infrapinklizzard.

This is on the very edge of guessing. I did find some deep lookahead, but it's very deep. It was not easy to keep in my head as I looked ahead.

Note: I used symmetry in the solve of this, but not in a prohibited way:

If clues are symmetrical, you know that if the solve on one side works, the solve on the other will work the same way.
Also, you know that if the blocks are placed asymmetrically and it still solves, that there are multiple solutions.

For Purist Approvedâ„¢ solving, what you cannot do with symmetry is placing clues with symmetry without disproving the other possible placements.

Here's the deep lookahead:

After line logic, you have a block of unsolved blankness from c9 to c20 and r1 to r9.

===
All the horizontal clues are greater than 1.
Therefore, the 1 in c20 will have the 3 in c19 next to it.

If the 1 in c20 is in r1, then the 3 in c19 will force a block of two in c18. This completes one of the two 2 clues. The other would then force a block of two in c17, which is a conflict.

Therefore c20r1 has to be white. The same can be done to make c20r9 white.

However, we can also generalize this. Look at what we did. Why was there a conflict in c17? Because there was a free block of two in c18. Why was there a free block of two? Because the 3 in c19 only took up one of the two 2 clues in c18.

So the 3 in c19 must cross BOTH the 2 clues in c18. This means that it cannot be in r1 or r9 (or the top or bottom 2 respectively would not fit). So c19r1 & c19r9 are white.

Note that the 1 in c20 must be in the middle of the 3 in c19 to provide the "space" between the two 2s n c18. So it cannot be in r2 or r8 and they must be white.

But also: the 1 cannot be in a row whose clue is greater than 2, or it would not provide that "space". So c20r5 is also white.

Now more deep lookahead:

The 3 in c19 can be in one of four positions now: c2-4, c3-5, c5-7, and c6-8.

If the c19's 3 were in r3-5, then note what the 3 in r5 would do. It would force the 2 in c18 down into r6. There it would run into a conflict as it would have to be part of a 2, but it would have a white on either side.

So the c19's 3 cannot be in c3-5 and thus the 1 in c20 cannot be in r4 and it must be white.

The same can be done on the 3's position in r5-7 to disprove it. So c20r6 must be white. But, also, the only positions left for the 3 in c19 are r2-4 and 6-8. So c19r5 is also white.

Now everything that's been done (from the === to here) can be done mirrored on the 3 in c10.

That will give us eight pixels from line logic.

Unfortunately that still leaves us with some more deep lookahead to look forward to. (Of course, there's a 50% chance if you want to guess now, but that's not how I roll, G.)

Look at c13 & c16. Each has only 2s for clues. So each pixel in r5 must go either up or down one more row.

Look at r4 and r6. Each also only has 2s. Therefore we know that both c13 and c16 cannot go in the same direction or the 2s would crash into each other. So one must go up and one must go down.

If c13 goes up and c16 down, then their crossing 2 would both have to be part of the 3 clue in the column next to them.

If c13 goes down and c16 up, then their crossing 2s could either be both 1s or ONE could be part of a 3 (but not both).

Since the remaining clues are symmetrical, the only way that only one could be off kilter is if the puzzle has multiple solutions. So let's concentrate on that one and try to eliminate it. This is pretty deep, tho.

So let's assume c13 goes down and c16 goes up. Then assume 2 in r6 will be part of the 3 in c14. That 3 will have to be in r5-7. (It's prevented from going into r4 by the 2 in r4 coming from the other side.) This puts the 1 in c14 in r9.

Now in c15, the 3 is forced into the bottom (available) row as well (r7-9 to be exact). This then soaks up a 2 in r8, forcing it into c16, which then gets forced down by the white after the 2 in r7 (like a zig-zag). All that means the 3 in r9 cannot go left into c13.
Whew.
Now THAT means that the second 2 in c13 must be within r1-3, and since all the clues in c14 are taken, both lines will be forced left into c12, creating a conflict.

So we've disproven that one of the 2s in center crosses a 3 while the other crosses a 1. We still have two more cases: they both cross 3s or they both cross 1s.

Just looking at it with experience suggests that crossing the 3s is more likely, so I'll try disproving the 1s first.

So, if we assume c13 goes down and c16 goes up and then the 2s in r4&6 each cross the 1s in the next column:

That would force the 3 in c14 to r1-3 (because the white before the 2 on the other side in r4 forces it up).
As all of the horizontal clues are greater than 1 and c15 must be white above the 1 we've placed in r4, all three rows must continue into c13, a conflict.

That wasn't so bad (comparatively), and we now know that the 2s in c13 and 16 must go up and down respectively.

That gives us another twenty pixels with line logic.

Then edge logic on the 3 in r1 rules out c16-18.

And line logic to finish.

oh, that's all.....lol

*snap*

good one brian :)

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