Puzzle Description:

#1: Belita (belita) on Jan 21, 2023 [HINT] [SPOILER]

Nice pattern, but not solvable without educated guesses. I assumed a lot of symmetry because the title said it was a pattern.

Actually, you don't need to assume symmetry. If you compare the clues for C1-10, and the clues for C12-21, you'll find they're identical, thus confirming symmetry.

Once the symmetry has been confirmed, C11 can only be black where the horizontal clues are not an even quantity, in which case the center clue has to be centered on C11, which solves the whole column.

After this, you're really just solving C1-10, and mirroring it on C12-21. All of that can be done with LL and some moderate lookahead.

So, the question becomes whether using the knowledge of symmetry to solve C11 and to know that each side mirrors each other is considered to be logic, or guessing?

Where solvability ratings are concerned, symmetry counts as guessing. I'll see if I can find a logical means to solve this that doesn't require symmetry.

R1: Basic EL prevents any 3s from being placed in C5-7 or C15-17, though none of those individual squares can be eliminated. If we look 2-3 steps down, however, we can narrow things down a bit more: for example, we can't have a 3 in both C1-3 and C6-8, because we'd make C2 invalid with the next move, and C2-4 + C6-8 would be eliminated in 2 moves. The 2nd 3 therefore can't be placed any sooner than C7-9, which does allow us to mark R1C6 white. Working from the right, we can use the same logic to eliminate C16. This is enough to start each of the 3s in R1. LL.

R1: Regardless of how the first 3 is placed, R2C2 is black. Regardless of how the last 3 is placed, R2C20 is black. LL.

R1: If either of the middle two 3s goes into C11, it will make R2 invalid, so R1C11 is white. LL.

R1: Regardless of how the first 3 is placed, R3C4 is white. Regardless of how the last 3 is placed, R3C18 is white. LL.

R1: Whether the 2nd 3 goes to C7 or C10, R2C10 is black. Whether the 3rd 3 goes to C12 or C15, R2C12 is black. Minimal LL.

R7: If C2 or C20 is black, it will make R6 or R8 invalid, so these squares are white.

R17: If C1 or C21 is black, it will make R16 or R18 invalid, so these squares are white.

R20: LL puts the 3 in the C6-16 range. Any placement that does not use C11 will start a 3 in R19. If this is to the left of C11, placing the first 2 and 1 in R20 will make R19 invalid. If this is to the right of C11, placing the last 1 and 2 in R20 will make R19 invalid. Therefore, R20C11 must be black (and part of the 3). LL.

Deep lookahead R18: If the first 5 starts in C1, the first 3 in R19 would be in C3-5 and the first 2 in R20 would be in C4-5. The only way to fit the first 1 in R20 would be in C9, but this would also place the second 1 in R19, making the blacks in R19-20 C11 both the start of a 3. This makes C12 invalid. Therefore R18C1 must be white. The same logic can be used on the right to mark R18C21 white. LL.

I need a break from this one for a while. Currently at 15%, and I'm not at all sure this one can be done logically.

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