Puzzle Description:

Crosswalk aka zebra crossing

#1: Michael Eddy (meddy) on Jul 15, 2021

This was quite tough. I definitely needed moderate lookahead. I don't fully understand the distinction between moderate and deep lookahead so I can't say if deep lookahead is required.

I like the final image and enjoyed how it came together during the solve.

Oh please keep zebras away from the idiot drivers!
Or, perhaps more justly, keep idiot drivers away from zebras!

Nice!

They aren't normally called zebra stripes around these parts (the US). So, assuming this is a UK term, so you call them ZEB-ra stripes (as opposed to ZEE-bra)?

with a capital zed!

Zebra crossing or pedestrian crossing in Australia

EDIT: See comment 10 instead for an easier solve.

After the initial LL, EL on C30 makes R1-6,9-13,15C30 dots.

Similar logic can be used in C29 then C28 then C27 to make R1-6,9-13 in C29, C28, and C27 dots.

EL in R15 makes R15C28-29 dots.

EL in C1 makes R1-2C1 dots. Looking 2 moves ahead, R3C1 will also be a dot. Now we can use EL to make R6C1 a dot.

EL in R6 makes R6C2-4 dots.

If the 3 clue in R5 is in C1-3, then looking 2 moves ahead we will not be able to complete C2 as you could not place the 1 clue. So R5C1 is a dot (and then so is R4C1).

The last logic can be used to also make R10-13,15-18C1 dots.

Look at R13 on the left. If the 3 clue is placed at R13C2-4. Then R10C3 and R13C5 would be dots and R10C4 would be black. But that means R10C5 would also have to be black, but it cannot be with the dot at R13C5. So R13C2 is a dot and similarly R13C3 will be a dot.

Now imagine that 3 clue is at R13C4-6. The columns all have 4 clues so R10C4 - R13C6 would all be black. However, to complete the 4 clue in R12 would force a 1 clue in C3 or C7 which does not exist. So R13C4 is also a dot. LL.

EL in C5 makes R10-13C5 dots and then EL with C6 makes R10-13C6 dots. LL.

Look at the 3 clue in C3. Notice how this will affected C1. If I place it at R1-3C3, then the rest of the column will be dots. This would also make R22-25C1 dots leaving R19-21C1 open. But we need 4 cells open in order to place the 2 1 clues in C1. So R1C3 is a dot. We can continue on down C3 with this logic making R2-5,15-19C3 dots. LL.

EL in C4 makes R1-5,15C4 dots. EL in C5 makes R1-6,15C5 dots. EL in C6 makes R1-6,15C6 dots.

Look at the 2 clue in C1. The lowest it can go is R22-23C1. But it cannot be there with EL. Neither can it be at R21-22 looking a few moves ahead. So it can only be in R19-21 making R20C1 black. And fill in the dot.

EL in R15 makes R15C27 a dot as it would make R16 invalid.

Look at the 6 clue in R13. It has to be completed at R13C13 or R13C19. This will cause either R13C15 or R13C19 to be black. This constrains how the 4 4 clues can be completed in R15. They can only be in C11-22. So the rest of R15 is dots (R15C7-10,23-26).

LL makes R15C19 black, then R16C19 black.

If the 3 7 clues in C19 are completed in R1-11, this will make C14 invalid as you would not be able to complete the 5 clue. So the 7 clue must go lower. LL.

I have to stop here for now, but I'll come back later and hopefully finish this one.

Fun puzzle requiring some unusual logic to solve. I enjoyed it.

Found to be solvable with moderate lookahead by gator.

Continuing:

Either R15C15 or R15C18 has to be black; however, both of them cannot be. So one of the 14 clues in C15 or C18 will have to extend to R1. This will let us add quite a few dots in the upper section (R1,5C7-12,21-26 R2-4,6C7-10,23-26). LL to finish.

I solved the puzzle again with only using this last logic I found and that is all that you need. The rest is all line logic.

Very fun puzzle!

Very good illustration. Fun solve. No guessing.

Goto next topic

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