A Difficult Wacky Sudoku - SS#8

Question

^{An entry in Fortnightly Topic Challenge #47: "Wacky Sudokus"}

Other puzzles in this series

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Welcome to the eighth puzzle in this series! For more information about the series, see the first puzzle and the introduction. Enjoy!

This is the first of what I would say are the two hardest puzzles in this series, so good luck!!

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What's a good wacky sudoku without some small coloured dots amirite?

Google Sheets Link (Doesn't contain the dots, just the numbers)

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RULES:

• Normal Sudoku rules apply
• There are coloured circles place in the corner of some cells. These are near the centre of four cells
  • If the dot is red, the number in the cell of the dot is greater (and not equal to) than the other three numbers in the four
  • If the dot is blue, the number in the cell of the dot is smaller (and not equal to) than the other three numbers in the four

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As I'm expecting this to be a bit harder, if possible please give a detailed explanation of how you solved and the logic used! :)

Answer 1

COMPLETED GRID

REASONING

Using the dots, we get some quick deductions:

R2C2 is 3, since it must be less than 4, and cannot be 1 or 2. R8C7 must be 8, since Sudoku rules require one of R7-9C7 to be 8, and R7 is blocked and R9 cannot be 8 since it is not less than 5. Staying here, R9C7 must be 2, since it must be less than 5, but cannot be 1, 3 or 4. In this square, Sudoku rules force R8C8 to be 4.

R6C5 must be 9, since it is greater than 8, and the same logic fills R5C7. Looking at R2C8, we see it must be less than 5, since it must be less than its three neighbors, and one of those can be at most 5. Since R2C8 cannot 1, 3 or 4, it must be 2. Finally, in the upper left square, either R3C1 or R3C3 is a 1 by Sudoku rules, but 3 is greater than 1 so it must be R3C1. Sudoku forces R8C3 and R9C4 to be 1. This means one of R5C9 or R6C9 must be 1, but it cannot be R6C9 since it is less than R5C9. This completes the 1s. The grid thus far:

Not stuck in the middle:

Sudoku rules force R8C5 to be 2, and then R9C6 to be 7...we then get R4C5 to be 7 for free. The 9 in R9 must then be in R9C9. Up to the middle square, the only remaining values to place are 2, 4, 5, and 8, which forces R4C6 to be 2 and R4C6 to be 8, making R5-6 in C4 4 and 5 in some order. Combined with R7-8 in C4 being 6 and 9 forces R1C4 to be 8. We now know R1-3 in C6 must be 3, 6 and 9 in some order, forcing R1C6 to be 3, and we can also fix R2C5=4 and R3C5=5. This also forces R2C9 to be 5 by Sudoku rules. The grid thus far:

Finishing up:

R5C2 is forced to be 7 because it must be greater than 6, and 8 and 9 are blocked. Continuing in row 5, the remaining values to fill are 2, 3, 4, and 5. Using the blue dot in R4C4 to note that R5C3-4 must be 4 and 5, we can fill this row. The rest of the grid fills in with Sudoku logic.