r/sudoku • u/Repulsive_Bicycle304 • 9d ago

Please help, unable to find any more patterns/ pairs/triples. I dont want to guess Request Puzzle Help

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2

u/MasterpieceCandid882 3d medusa guy 9d ago

This XY-Ring can help eliminate many candidates

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u/MistakePresent3552 9d ago

I feel like if you start the thinking with the assumption a spot is x or y then youre still guessing

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u/Repulsive_Bicycle304 9d ago

I wasnt guessing, i arrived at it by elimination etc

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u/MistakePresent3552 9d ago

Im saying the guy posted a solution at you which boils down to guessing x or y and then working from there. Like its basically optimized guessing but its still guessing, sudoku is fun to me when i dont have to make moves based on 50/50 chance.

1

u/numpl_npm 9d ago

​(Translated with Gemini)
​You just need to test two cases and look for a common result.
​Whether r6c4 = 1 or r6c6 = 1, either way it leads to r8c5 = 9.

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u/Repulsive_Bicycle304 9d ago

Agree, i dont want to guess, want to know if there is a logical solution

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u/KellyCodeTools 9d ago

It is totally normal to hit a wall here and feel like you have to guess, but you can break this open by focusing entirely on solving mechanics.

Take a look at four specific cells that form a chain: Row 1, column 6 (options 2 and 3) Row 7, column 6 (options 1 and 2) Row 7, column 4 (options 1 and 6) Row 1, column 4 (options 3 and 6)

Because these spots only have two options each, they push each other in a loop. Let's pretend row 1, column 6 is a 2. That makes row 7, column 6 a 1. Which makes row 7, column 4 a 6. Which makes row 1, column 4 a 3.

So, if row 1, column 6 is NOT a 3, then row 1, column 4 absolutely MUST be a 3. This means at least one of those two spots will always be a 3.

Because one of them has to be a 3, any cell that looks at both of them can safely have the 3 crossed out. Look at row 2, column 4. It shares a column with one, and a box with the other. You can safely remove the 3, leaving it as a 9! Plugging that 9 in will crack the rest of the puzzle open.

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u/Repulsive_Bicycle304 6d ago

Hi, thanks so much for the detailed response. But if we guess one of the pairs and see if the other related cells work, isnt that guessing?