Puzzle Description Suppressed:Click below to view spoilers

#1: CB Paul (cbpaul) on May 15, 2021

Very tricky indeed! I'd love to read the description.

Initial line & color logic gets 62% complete with only black remaining.

R12: If the first black 3 extends to C31, it creates a contradiction in R17, so R12C31 is white. LL.
C41: IEL marks R8 white. LL.
R30: There is a black 2 that needs to fit into C1-6. The only way to avoid a contradiction in R31 is to have one of its squares in R30C3. LL.
R39: Same logic as the previous step places a black in R39C3. LL.
R30: Whether the first black 2 extends into C2 or C4, there is a 5 in that column, which would place a black in R32. Either of these marks R32C14 black, and R32 C1&C5-6 are white. LL.
R6: EL marks C47 white, and regardless of which remaining position the last black 3 uses, the black 3 in R7 goes into C46-48. LL.
C50: EL places the 6 in R15-20. LL.
R14: C44 is white to avoid a contradiction in R15. Minimal LL.
R14: Whichever way the last black 3 goes, it will complete R15-18. C44, C46, and C47 in these rows are white. R19C45 is also white.
R23: The last black 2 can either go into C48 or C50. If it goes into C48, it causes a contradiction at R26, so it must go into C50 instead. Minimal LL.
C49: EL places whites in R25-28 and R49-50. Minimal LL.
C48: EL marks R24-32 and R39-40 white.
C49: EL puts the 5 in R29-33, R33-37, or R34-38. Any of these positions will start the 3 in C50, placing it in R27-29, R31-33, or R38-40. The rest of the open squares in C50 can be marked white.

Deep lookahead R21: If C33 is black, it marks R23-24C33 black as well, and there is no way to place C34-35 in the remaining space without causing a contradiction. Regardless of whether the black is placed in R21 C34 or C35 instead, R22-24 C33 also get marked white, as does R25C35.

R25: whichever way the 2 goes, it marks R27-28 C39 white.

C48: If the 6 goes in the upper position, R19C47 is white. If it goes in the lower position, the 3 in C47 must be in R38-40. Either way, R19C47 is white.

C39: If R35 or R36 is black, it forces blacks into R30 C38 and C40, but these can't both be black. Therefore R35-36 C39 are white. LL.

At this point I'm 89% done and I need a break. I've had to backtrack on my progress 3 or 4 times already when I realized an earlier assumption hadn't been logically sound even though it turned out to be correct (hooray for taking notes). At this point I'm leaning toward "deep lookahead I haven't found yet" but it could still require guessing.

Picking up where I left off....

C33: If the black 4 starts in R25, the 5 in C34 will need to be in R29-33, R30-34, or R31-35, and the 4 in C35 will need to be in R21-24. If the 5 is in R29-33, the 2 in R29 makes C35 invalid. In the other two positions, the 2 in R33 makes C35 invalid. R25C33 must be white. Minimal LL.

C33: If the black 4 starts in R26, the 5 in C34 goes up to R21. The 4 in C34 is in R29-32. This doesn't leave enough room for both the 4 and the black 5 in C35, so R26C33 is white. EL can mark R27-28 C33 white as well.

Deep lookahead R36: There are 2 1s available for C34-40. One of those will be in either C38 or C40, but there can't be one in both of those columns (using either one will mark enough squares white for the other to not fit in these rows). This leaves one black for C34-35. If it's in C34, the 5 in C34 will be in R25-29, and the 4 in C35 will be in R21-14. If the remaining 1 in R36 is in C35, the 5 in C34 will be in R21-25 and the 4 in C35 will be in R26-29. Either way, R29C34 is black.

R25: Whether the 2 goes into C38 or C40, R29C39 is white. Minimal LL.

R29: If the 2 goes into C33, the 5 in C34 is in R25-29 and the 4 in C35 is in R21-24. If the 5 in C35 uses R33, it will make C34 invalid, so it must place black in R36-38. This does not leave enough room for the 4 in C34, so the 2 must go into C35 instead. Minimal LL.

C35: Whether the black in R29 goes into R28 or R30, R33C34 will be black. LL.

C43: EL marks R24 white.
C44: EL marks R38-40 white.
C45: EL marks R37-40 white.

Deep lookahead C50: If the 3 is placed in R27-29, the 6 in C48 will need to be in R33-38, which puts the 2 in R37 into C43-44. If the 3 in C50 is in either of the other two available positions, it places the 6 in C48 into R14-19, and R29C46 would need to be black. This would make R37C46 white, which once again places the 2 in R37 into C43-44.

LL to finish.

Found to have a unique solution by wombatilim.

Found to be solvable with deep lookahead by wombatilim.

Show: Spoilers

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