Puzzle Description Suppressed:Click below to view spoilers

#1: derby (Derby) on Oct 14, 2018 [HINT] [SPOILER]

Comment Suppressed:Click below to view spoilers

Comment Suppressed:Click below to view spoilers

Initial line & color logic completes 93% with only black remaining.

C7: EL places black in R22. LL.

R23: EL narrows the 4 down to 3 possible positions: C26-29, C27-30, and C30-33. Any of these makes the black in R22C30 part of the last 2 in that row. R22 C32-34 are white. LL.

R17: EL marks C12 black. LL.

R17: The remaining 2 is either in C22-23 or it places a black in C26. If it's in the latter position, it moves the 3 in R16 into C23. Either way, R17C23 is black.

R18: The black in C12 must be part of a 2. Whichever way it goes, R18 C9-10 are white.

C14: The last black 1 can either go in R22 or R23. Either way, R23C13 is white.

R23: We now know the blacks in C15-16 must be part of the first 3. If the second 3 is placed in C24-26, it will make R22 invalid. R23C25 can't be part of the 4 from our previous EL and it can't be the second 1 either, so it must be white. Minimal LL.

R23: Whichever of the remaining positions for the second 3 is used, the 3 in R22 goes into C24-26. LL.

C27: Whether the first black 1 goes into R17 or R18, R18C26 is white.

R17: If the remaining 2 is placed in C25-26, it places whites in R19-20 C26 and a black in R20C25. The blacks that would now be in R19-20 C25 would both need to go into C24, which is invalid. Therefore, R17C25 is white. LL.

R19: If the 3rd black 2 is placed in C22-23, it places black in R19C26, which would make the black in R20C22 part of a 2 in that row. This would place a black into R20C23, which would make C23 invalid. Therefore R19C22 must be white. LL.

R18: If the 3rd 2 goes into C22-23, it makes R17 invalid, so it must go into C18-19 instead. LL.

R20: If the remaining 2 is placed within C17-19, it makes R19 invalid, so these squares are white. LL.

R19: No matter how the remaining 2 is placed, it will start the remaining 2 in R20, and R20C13 is white.

Deep lookahead R18: If the remaining black is in C11, it places black in R15C11 and completes that column. A black would need to be placed in R17C13, which in turn places a black in R22C10. This leaves R21C13 for the remaining black in C13. There would then need to be a black in R23C10, which makes C7 invalid. Therefore R18C13 must be black instead. LL.

R19: No matter how the remaining 2 is placed, R20C10 is black.

LL to finish.

Found to be solvable with deep lookahead by wombatilim.

Show: Spoilers

Goto next topic

You must register and log in to be able to participate in this discussion.