How to solve NYT Sudoku Hard February 28, 2021 by advanced Sudoku techniques
How to solve NYT Sudoku hard February 28, 2021 explains each breakthrough using advanced Sudoku techniques. It's a quick and easy solution to a hard Sudoku.
The NYT Sudoku Hard, 28th February, 2021
Before going through the solution solve the puzzle first.
Step by step solution to the NYT Sudoku Hard 28th February, 2021 Stage 1: Breakthrough by Parallel digit scan and Double digit scan
First valid cell R9C3 8 by row column scan for 8 in R7, R8, C2 -- R3C1 8 by scan in C2, C3.
Next breakthrough is R4C1 1 by parallel scan for 1 empty cells of C1 eliminating R6C1 by 1 in R6 and eliminating R8C1 by 1 in R8. This gives us the next valid cell R5C6 1 by scan for 1 in R4, R6, C4.
R2C9 9 by scan for 9 in R1, C8.
Analyzing the possible digits in neighborhood of top right major square identify the formation of a Cycle (1,2) in R1C7, R1C9 by double digit scan for [1,2] in C8.
Breakthrough by double digit scan, Cycle and parallel digit scan:
First double digit cross scan for [4,6] in R4 and C7 but not in right middle major square forms the Cycle (4,6) in R5C9, R6C9 in C9 at one go.
Check and verify.
As a result, a Cycle of (1,2,3,5) formed in rest of the cells in C9.
Now we are ready for the breakthrough of R5C2 5 by parallel scan for 5 on empty cells of R5: cell eliminations for 5 in R5 are - R5C3 because of 5 in C3, R5C4 because of 5 in C4, R5C7 because of 5 in C7 and R5C9 because of 5 in Cycle (1,2,3,5) in C9. The only cell left in R5 for 5 is - R5C2 5.
Results of the actions taken shown.
Solution to the NYT Sudoku Hard, 28th February, 2021 Stage 2: Breakthrough by a combination of DSA and Cycles
With the last breakthrough of R5C2 5, R6C1 3 by reduction -- R8C1 5 by reduction.
R8C9 3 by reduction of DS [1,2,5] from Cycle (1,2,3,5) in C9 and in R8C9.
Now we'll have a special breakthrough by a combination of DSA and Cycles.
As [3,8] are locked in the Cycle (3,4,8) in C8, DS in R4C8 is [5,7] reduced by [2,3,8] from DS [2,3,5,7,8] in right middle major square.
DS in R4C9 is [2,5] reduced by [3,7,8] from DS [2,3,5,7,8] and DS in R6C7 is [2,5,7] reduced by [3,8] from [2,3,5,7,8].
These three DSs thus form a Cycle of (2,5,7) in right middle major major square reducing [7] in R5C7 to R5C7 8 -- R5C4 6 by reduction -- R5C9 4 by reduction -- R5C3 7 by reduction -- R6C9 6 by reduction -- R4C7 3 by exception in right middle major square.
R4C4 8 by scan for 8 in R6, C5.
R6C3 4 by scan for 4 in R4 -- R4C2 9 by exception in left middle major square.
R1C2 7 by scan for 7 in R3, C3.
Game status shown below.
Solution to the NYT Sudoku Hard 28th February Stage 3: Breakthrough by DSA
First valid cell is by scan: R3C2 4 by scan for 4 in C3 -- R3C8 3 by reduction.
Cycle (1,3) in R1C3, R2C3 as the two digits left for the two cells left in the top left major square.
Because of this Cycle in C3, a second Cycle of (2,9) is formed in R7C3, R8C3 in C3 and in bottom left major square this time.
It reduces the possible digit subset DS for the rest three cells to [1,3.6] -- breakthrough R8C2 6 by reduction of [1,3] from DS [1,3,6] -- R8C8 7 by reduction -- R4C8 5 by reduction -- R4C9 2 by reduction -- R6C7 7 by reduction.
With 2 in R4C9, R1C9 1 by reduction -- R9C9 5, R1C7 2, R1C3 3 all by reduction of 1 -- R2C3 1 by reduction.
R4C5 7 by exception in R4.
R3C5 1 by scan for 1 in R1, R2, C4, C6.
A Cycle of (2,9) is formed in R3C4, R3C6 as well as in R3C4, R6C4.
Game status shown below. No further challenges left.
Solution to the NYT Sudoku Hard 28th February: Final Stage 4: No more challenges left
R2C5 4 by DSA reduction of [3,8] from DS [3,4,8] in three cells of R2 -- R2C8 8 by reduction, R2C6 3 by exception in R2 -- R1C8 4 by reduction -- R7C8 6 by exception.
R1C6 8 by scan for 8 in C5, R3 -- R1C5 6 by exception in R1.
R9C3 6 by scan for 3 in R7, R8, C4, C5.
R7C6 7 by scan for 7 in R8, C4, C5 -- R7C2 3 by reduction -- R9C2 7 by reduction -- R6C6 5 by parallel digit scan for 5 on empty cells of C6 eliminating R8C6 for 5 by 5 in R8, eliminating R3C6 by 5 in R3 leaving the lone cell R8C6 5.
R7C5 5 by scan for 5 in R9, C4, C6.
With Cycle (2,9) in C4, DS in leftout two cells is [3,4] -- R7C4 3 by reduction of 4 in R7 -- R7C2 1 by reduction -- R9C2 3 by reduction -- R7C7 9 by reduction of 2 in C7 from DS [2,9] in R7 -- R7C3 2 by reduction -- R8C3 9 by reduction -- R8C6 2 by reduction and exception in R8.
With 2 in R8C6, R3C6 9 by reduction -- R3C4 2 by reduction -- R6C4 9 by reduction -- R6C5 2 by reduction -- R9C4 4 by exception in C4 -- R9C5 9 by exception in C5 -- R9C7 1 by exception in whole game.
Final solution below.
This Sudoku game is a hard Sudoku puzzle that needed multiple breakthroughs by advanced Sudoku techniques.
Unnecessary enumeration of possible digit subsets in empty cells is avoided to speed up the solution.
Instead, focus has been to create a breakthrough at the earliest.
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