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#1: David Bouldin (dbouldin) on Oct 30, 2018

i should've tracked my solve because it ended up being challenging but fun! it's definitely solvable with moderate lookahead. i'll resolve it if anybody needs some clues. thanks for the puzzle!

I made some guesses to get started, but this was surprisingly solvable!

I solved it, but I'd call it deep lookahead. Anyone got an easier hint than this?

After LL, no edge had any simple EL involving a single clue.

This is a sort of counting edge logic, feels like deep lookahead to me:
C20 has a total of nine black, C19 only totals eight, and only two of the rightmost clues (in all rows) are 1: R6 and R13.
So, 3 possibilities:
#1: C20 clues intersect neither R6 nor R13, making *nine* blacks in C19. Contradicts the total of eight in C19.
#2: C20 clues intersect either R6 or R13, making *all* eight blacks in C19. Possible...
#3: C20 clues intersect both R6 and R13, making seven blacks in C19, so one black in C19 will have white to its right in C20. Possible...

Consider scenario #2 again: Since C20 makes *all* eight blacks in C19, C19 must be formed by taking C20 and whiting out one square. So C19 could be:
white out a 1 clue: [4,1,1,2]? nope.
white out part of the 2 clue: [4,1,1,1,1]? nope.
white out part of the 4 clue: [3,1,1,1,2]? nope. [2,1,1,1,1,2]? nope. [1,2,1,1,1,2]? nope.

So scenario #2 is impossible, leaving #3.
Now we know that C20 R6 and R13 are black. But we can take this a little farther...

This is a little tricker: C20 only makes seven blacks in C19, so after whiting out two blacks from C20, we get to add one to form C19.
It helps narrow things down that R6 and R13 are too far apart to both impact the same C20 clue (biggest is 4), and too close to impact both the 4 and the 2. So from C20 we can white out a 1 clue and part of the 2 clue or 4 clue, or white out two 2 clues:
white out part of 2 clue: [4,1,1,1]
white out part of 4 clue: [3,1,1,2] or [2,1,1,1,2] or [1,2,1,1,2]
white out two 1 clues: [4,1,2]
Now adding one black could either simply add one to a single clue, or could join two adjacent clues, adding one (ex. [1,1]+1 -> [2,1] or [1,2] or [3]). So, trying to reach [4,1,3] from those five options:
[4,1,1,1]: must get a rightmost 3, which can be done by joining the two rightmost 1 clues.
[3,1,1,2]: can get a leftmost 4 or a rightmost 3, but not both. Impossible.
[2,1,1,1,2]: five clues cannot become 3 with only a single join. Impossible.
[1,2,1,1,2]: can get a leftmost 4 or a rightmost 3, but not both. Impossible.
[4,1,2]: must get a rightmost 3, which can be done by adding to the 2 clue.

So, C20 R6 is a 1 clue and C20 R13 is either a 1 clue or a 2 clue. Phew!

LL to finish.

this time didn't seem familiar, so i'm not sure that this was how i did it the first time, but here is what i just did after LL:

- EL: 4 in R20 is restricted to R12-20, but no matter where in that range, C17R19 is black

- LL

- EL: 4 in C20 is restricted to R1-10, and no matter where in that range, will form part of the 4 clue in C17, dotting C17R10-14

-LL to finish

Thanks, David! Your way is much easier.

Here's my way
After LL,
Look at r20 vs r19. In order for r20 to fit, two of its 1s must cross vertical 1s. Therefore r20c7 & c13 must be black with white on either side of them. (Note that r20c4 cannot be one of the 1s)
LL to finish

Found to be solvable with moderate lookahead by infrapinklizzard.

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especially one with a GMT hand :)

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