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Daily Sudoku Answer 



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Feb 19 - Super Hard
Puzzle Copyright © Kevin Stone

Share link – www.brainbashers.com/s201429



Reasoning 



R4C4 can only be <2>

R4C6 can only be <8>

R7C3 can only be <2>

R7C7 can only be <4>

R4C2 can only be <5>

R6C4 can only be <9>

R6C6 can only be <6>

R2C4 can only be <7>

R5C5 can only be <3>

R7C1 can only be <1>

R2C6 can only be <9>

R8C4 can only be <1>

R2C5 can only be <2>

R8C6 can only be <7>

R4C8 can only be <4>

R6C2 can only be <2>

R6C8 can only be <5>

R5C2 can only be <9>

R7C9 can only be <8>

R2C8 can only be <8>

R1C5 can only be <6>

R5C1 can only be <7>

R1C2 is the only square in row 1 that can be <8>

R2C2 is the only square in row 2 that can be <1>

R9C9 is the only square in row 9 that can be <1>

R9C8 is the only square in row 9 that can be <7>

R1C8 can only be <2>

R5C8 can only be <6>

R5C9 can only be <2>

R8C8 can only be <3>

R3C1 is the only square in row 3 that can be <2>

R9C1 is the only square in row 9 that can be <3>

Squares R8C2 and R9C2 in block 7 form a simple naked pair. These 2 squares both contain the 2 possibilities <46>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the block.

R8C1 - removing <4> from <459> leaving <59>

Squares R2C1 and R2C9 in row 2 and R8C1 and R8C9 in row 8 form a Simple X-Wing pattern on possibility <5>. All other instances of this possibility in columns 1 and 9 can be removed.

R1C1 - removing <5> from <459> leaving <49>

R1C9 - removing <5> from <457> leaving <47>

The puzzle can be reduced to a Bivalue Universal Grave (BUG) pattern, by making this reduction:

R1C3=<57>

These are called the BUG possibilities. In a BUG pattern, in each row, column and block, each unsolved possibility appears exactly twice. Such a pattern either has 0 or 2 solutions, so it cannot be part of a valid Sudoku

When a puzzle contains a BUG, and only one square in the puzzle has more than 2 possibilities, the only way to kill the BUG is to remove both of the BUG possibilities from the square, thus solving it

R1C3 - removing <57> from <579> leaving <9>

R1C1 can only be <4>

R1C7 can only be <5>

R3C3 can only be <7>

R9C3 can only be <5>

R9C7 can only be <6>

R2C9 can only be <4>

R2C1 can only be <5>

R1C9 can only be <7>

R3C9 can only be <6>

R3C7 can only be <9>

R8C9 can only be <5>

R8C1 can only be <9>

R9C2 can only be <4>

R8C5 can only be <4>

R8C2 can only be <6>

R9C5 can only be <9>



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