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#1: Kristen Vognild (kristen) on Jun 8, 2013

I had to guess a few times to solve this, and not I'm not sure what the figure at the lower left is supposed to be.

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Comment Suppressed:Click below to view spoilers

=) It's not super obvious.

There is probably a quicker way to solve this, but this is how I did it...

After very little line logic, simple edge logic with the 8 clue in C30 makes R1C30 and R24C30-R30C30 dots. Some more line logic can now be done (but not much).

Continuing with the 8 clue in C30 from the bottom, if you look how column 28 will be affected, you can make both R22C30 and R23C30 dots.

Back to simple edge logic, R15C30 - R21C30 can be made dots by using the 8 clue in C30. Now a lot more line logic can be done.

Next I did a two-way logic. Either R5C30 or R11C30 has to be black. In either case, this forces R5C29 to be black. More line logic.

If both R30C13 and R30C14 are dots, then R20C13 and R20C14 would have to be black. But this would make R20 invalid. So one of them must be black (but in this case both are black as it would satisfy the 2 clue). A little more line logic.

On to row 30. Edge logic will make R30C27-R30C29 dots. Next we know that either R30C22 or R30C23 has to be black. From this, we know that R28-R29C22 or R28-R29C23 must be black. This lets us dot R28C28, R28C29, and R29C27-R29C29. (And then R28C23 is black.)

Next I imagined that R28C22 was a dot. Looking 2 steps ahead, columns 24-26 would become invalid from this. So R28C22 is black. In the same way, if R28C21 was a dot, then 2 steps ahead, columns 24-25 would be invalid. So R28C21 is also black. (Another way, would be to see that no matter how you fill in the 4 clue in row 30, R28C21-R28C23 would always be black.)

Using edge logic with the 5 clue in row 27, R27C28 and R27C29 have to be dots. (And R27C23 becomes black.) Lots more line logic.

R14C29 cannot be black without making column 28 invalid. So it is a dot. Either R13C28 or R15C28 will be black from the 2 clue in column 28. This makes R12C28 and R14C28 dots.

Next I looked at row 15 on the right. If either the 3 or 4 clue started at R15C21, then row 14 would become invalid. So R15C21 is a dot. R15C22 is also a dot by the same logic. This makes R13C22 black.

Now look at the 2 clue in column 26. If it was at R12C26-R13C26, then looking 2 moves ahead, column 29 would be invalid as the 2 clue would not be able to be placed. So R12C26 is a dot. This makes R14C26 black. A little more line logic.

Next the other 3 clue in row 14 cannot be at R14C21-R14C23 without making row 13 invalid, so R14C21-R14C23 are dots. More line logic and now columns 17-30 are complete.

If the 8 clue in column 16 extends down to row 12, this will cause column 15 to be invalid. Since it cannot extend this far, then R4C16 has to be black. More line logic.

Using edge logic with the 4 clue in row 11, R11C1-R11C8 will be dots. More line logic.

The 4 clue in row 19 must either be at R19C6-R19C9 or R19C11-R19C14. If it is on the right, then row 18 would be invalid, so it must be on the right. More line logic.

Back to the 4 clue in row 11. It cannot extend to R11C9 without making row 12 invalid. So R11C9 is a dot.

Next lets imagine that the remaining 2 1 clues in column 10 have to go at R11C10-R12C10 and R14C10. This configuration will make column 9 invalid, so the 1 clue would have to be at R18C10.

Next with edge logic, we can make both R14C10 and R13C10 dots with the 2 clue in column 10.

The rest of the puzzle solves with line logic.

Joe - would you please review my steps when you get to chance to make sure I did not miss anything? I believe this technically falls under moderate lookahead as each step was no more that 2 steps ahead. Please mark it how you think it should be though.

hard -solved with a couple of reverts

Couldn't get it.

Don't understand the references. The image in the lower left doesn't make sense. I like the castle and paths. Very tough. Had to guess a lot.

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I enjoyed this puzzle. It can be solved with minimal look ahead by imagining where the 8 can be in the right column, then where the 4 can be on the bottom row, then where the 4 can be in row 11.

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