Hard Sudoku New York Times 21st February, 2021: Easy to understand Solution
New York Times hard Sudoku February 21, 2021: Easy Solution explains step by step how each valid cell is identified for a unique digit using digit patterns.
The New York Times hard Sudoku 21st February 2021
Before going through the solution solve the puzzle first.
Step by step solution to the New York Times hard Sudoku 21st February, 2021: Stage 1: Breakthroughs by DSA and Cycles
First valid cell by row column scan is R5C2 6 by row column scan for 6 in R4, R6, C3 -- R1C8 6 by row column scan for 6 in R3C9.
And that's all of the valid cells by row column scan.
Next take up valid cell identification by possible digit subset analysis or DSA in promising zone of a row, a column or a major square with good number of filled cells.
Accordingly, get the valid cell R2C1 3 by DSA reduction of [4,5,7] in C1 from possible digit subset in promising row R2 and hence in R2C1 [3,4,5,7].
Reduced digit subset for R2 is now [4,5,7].
With [4,5] in C9 and top right 9 cell major square together -- R2C9 7.
Again we get the next valid cell R6C7 7 by DSA: Possible digit subset in empty cells of R6 and hence in R6C7 [2,4,7,8,9] reduced by DS [2,4,8,9] in right middle major square and C7 together.
Possible digits in top right major square is [1,3,4,6].
With [4,6] in C9, Cycle (1,3) formed in C9 -- R3C8 4 by exception in top right major square.
R4C7 1 by row scan for 1 in R5, R6 and Cycle (1,3) in C9.
R5C8 3 by scan for 3 in R6 and Cycle (1,3) in C9 -- R5C4 7 by DS reduction of [8,9] from DS [7,8,9] in R5 -- R5C3 9 by reduction of 9 in C3 -- R5C5 8 by exception in R5.R3C6 7 by scan for 7 in R1, C4, C5.
R1C6 8 by scan for 8 in C4, C5.
Cycle (3,4,6) formed in C7 by exception of remaining three digits in three remaining cells of C7.
Results of the actions taken shown below.
Solution to the New York Times hard Sudoku, 20th February, 2021: Stage 2: Major breakthrough by DSA reduction technique
The most important breakthrough at this point is R8C3 1 by DSA reduction of [2,4,7,8] from possible digit subset DS [1,2,4,7,8] in C3.
R3C1 1 by scan for 1 in C2, C3 -- R3C9 3 by reduction -- R1C9 1 by reduction -- R3C2 9 by scan for 9 in R1, R2.
Cycle (2,4) formed in R1C2, R1C3 by DS reduction -- R1C3 5 by DS reduction of [2,4] -- Cycle (2,5) formed in R3C4, R3C5 by DS reduction of 4 from DS [2,4,5] -- R2C4 4 -- R2C2 5.
With 4 in R2C4, R6C4 2 by reduction -- R4C4 3 by reduction -- R3C4 5 by reduction -- R3C5 2 by reduction.
With 2 in R6C4, R6C1 8 by reduction -- R6C8 9 by reduction -- R6C4 4 by reduction -- R4C6 9 by reduction -- R4C9 8 by reduction.
Results of game status shown below.
Solution to the New York Times hard Sudoku 21st February: Final Stage 3: Easy valid cells with no more breakthroughs remaining
With 9 in R4C6, R7C6 3 by reduction -- R7C7 6 by reduction -- R7C4 1 by reduction -- R7C4 1 by reduction -- R9C4 6 by reduction.
R8C1 6 by scan for 6 in R7, R9, C2.
R8C2 3 by DSA reduction of [4,5,9] from DS [3,4,5,9] in R8 -- R8C7 4 by reduction -- R9C7 3 by reduction.
R4C1 2 by DSA reduction of 9 from DS [2,9] in C1 -- R9C1 9 by exception -- R9C6 4 by reduction.
R8C5 9 by scan for 9 in R9 -- R9C5 5 by exception in C5 -- R8C9 5 by reduction -- R9C9 2 by reduction. With 9 in R8C5, R8C9 5 by reduction -- R7C9 9 by reduction.
With 1 in R7C4, R7C8 8 by reduction -- R9C8 1 by reduction -- R7C2 2 by reduction -- R7C3 5 by exception in R7. With 2 in R7C2, R1C2 4 by reduction -- R4C3 4 by scan for 4 in C2 -- R4C2 7 by exception -- R9C3 7 by exception in C3 -- R9C2 8 by exception in the whole game.
Final solution below.
Solution of this hard Sudoku puzzle didn't need any of the advanced Sudoku digit patterns such as double digit scan, parallel digit scan, single digit lock or X wing. Forming unique digit in a cell and Cycles by possible digit subset analysis provided the main breakthroughs.
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