How to Solve the New York Times Sudoku Hard 22nd February, 2021 with step by step explanation
Solve New York Times Sudoku Hard February 22, 2021 in quick steps. How to solve the Sudoku hard is explained clearly including all the breakthroughs.
The New York Times Sudoku Hard, 22nd February, 2021
Before going through the solution solve the puzzle first.
Step by step solution to the New York Times Sudoku Hard 22nd February, 2021: Stage 1: Breakthroughs by DSA technique and Cycles
First few valid cells by row column scam: R7C9 1 by row column scan for 1 in R8, R9, C7 -- R7C6 7 by opportunistic scan for 7 in C4, C5 -- R6C6 3 in promising neighborhood by row column scan for 3 in R4, R5, C5.
R5C6 8 by DSA reduction of [2,9] from DS [2,8,9] in C6.
Note on DSA reduction of valid cell R5C6 8: Possible digit subset in promising zone of C6 with 6 filled digits is the three missing digits [2,8,9]. With 9 in R5 and 2 in parent central middle major square for cell R5C6, the effective possible digit subset is reduced from [2,8,9] by [2,9] to the single digit 8. This is the unique naked digit that has to be placed in R5C6.
Remaining two digits in C6 with just 2 empty cells left, Cycle (2,9) formed in R2C6, R3C6 in C6 and parent top middle major square.
Note on Cycle (2,9): The Cycle in C6 won't have any further effect in C6, but being in the parent top middle major square, it reduces all possible digits in the top middle major square by [2,6]. This is because, the two digits [2,6] are locked in the two cells of the Cycle and cannot appear outside the Cycle in the empty cells of parent zones of the Cycle.
Next valid cells are by DSA: R1C5 5 by DSA reduction of [2,3,9] from possible digit subset [2,3,5,9] in R1 and hence in R1C5 -- R1C3 2 by DS reduction of [3,9] from reduced DS [2,3,9] in R1 -- R1C8 3 by reduction of 9 -- R1C2 9 by exception in R1.
R4C1 7 by row column scan for 7 in R5, R6, C2. Don't let go of any opportunity for a valid cell by an easy row column scan.
Last in this stage, R2C8 2 by DSA reduction of [7,9] from possible digit subset [2,7,9] in top right major square.
Results of the actions taken shown below.
Solution to the New York Times Sudoku Hard, 22nd February, 2021: Stage 2: Breakthrough by Single digit lock
With R2C8 2, R2C6 9 by reduction -- R3C6 2 by reduction -- R2C9 7 by reduction -- R3C9 9 by reduction.
R3C3 7 by row column scan for 7 in R2, C1, C2.
R3C1 1 by row column scan for 1 in R2, C2 -- R6C3 1 by scan for 1 in C1, C2 -- R4C5 1 by scan for 1 in R6, C4 -- R5C8 1 by scan for 1 in R4, R6, C7. Digit 1 fully filled.
R6C9 6 by scan for 6 in R4, C7.
Single digit lock on 5 formed in R4C8, R4C9 by scan for 5 in C7 -- R5C4 5 by scan for 5 in lock on 5 in R4, 5 in C5 -- R5C2 2 by DSA reduction of [4,6] in C1 from DS [2,4,6] in R5 -- R5C4 4 by reduction -- R5C2 6 by exception in R5. R9C2 7 by scan for 7 in R7, R8, C8, C9.
Note on Single digit lock: 5 in C7 blocks R6C7 for 5 and leaves only the two consecutive cells R4C8, R4C9 for 5. Digit 5 can appear in only these two cells in the final solution. It means digit 5 is locked and these two cells and thus diallows appearance of 5 in any empty cell of the parent zone R4. This ios the way a single digit lock is formed and takes part in finding a valid cell by reduction of 5 in R4 or by scan for 5 in R4.
Last few valid cells are by row column scan. Time to time you should try for this easiest way to get valid cells.
R8C7 3 by scan for 3 in R7, C8, C9 -- R9C1 3 by scan for 3 in R7, R8, C3.
R8C1 9 by scan for 9 in R7, C3.
R8C9 4 by scan for 4 in R9, C8.
Results of game status shown below.
Solution to the New York Times Sudoku Hard 22nd February: Stage 3: Breakthrough by Parallel digit scan, Single digit lock and DSA reduction
Breakthrough by Parallel digit scan for 2 on empty cells of C2: With 2 in left middle major square, R4C2, R6C2 disallowed for 2 and with 2 in top left major square, cells R2C2, R3C2 are disallowed for 2 leaving the single cell R7C2 for 2 -- R7C2 2.
How Parallel digit scan carried out and what is its effect
Instead of scanning for a digit on empty cells of a major square, in parallel digit scan empty cells of a row (or column) are scanned in parallel by presence of the digit in columns (or rows).
We do this scan only to get a single valid cell left for the digit in the scanned row (or column), rest eliminated by the scan.
In this game, 2 in top left major square eliminates two empty cells R2C2, R3C2 for 2 and 2 in left middle major square eliminates the two other empty cells R4C2, R6C2 for 2 in C2.
This leaves the single cell R7C2 for 2 and gives us a good breakthrough.
Special property of a parallel digit scan is the byproduct of a Cycle in the row or column scan. Here, we have the Cycle (3,4,5,8) formed in C2 by rest of the four digits.
You could have first formed the DSs by tedious possible digit analysis of these four cells to form the Cycle and as result obtained the valid cell R7C2 2.
We always prefer parallel digit scan because of its speed of resolution, only if we can identify it.
R7C7 8 by DSA reduction of [4,5,6] in C7 from possible digits DS [4,5,6,8] in R7.
R7C4 6 by DS reduction of [4,5] in bottom middle major square from reduced DS [4,5,6] in R7 -- R7C1 5 by reduction of 4 -- R7C3 4 by exception in R7.
R6C1 8 by exception in C1 -- R4C2 4 by reduction of 5 by single digit lock on 5 (the lock on 5 is still working) -- R6C2 5 by exception in left middle major square -- R4C4 9 by reduction -- R6C5 4 by reduction.
Game status shown below.
Solution to the New York Times Sudoku Hard, 22nd February 2021: Stage 4: Final solution: With game wide open, all valid cells obtained easily
R6C7 9 by exception in R7 -- R4C7 2 by exception in C7 -- R4C8 8 by scan for 8 in C9 -- R4C9 5 by exception.
With 8 in R4C8, R8C8 6 by reduction -- R9C8 5 by exception in C8 -- R9C9 2 by exception in C9. R2C3 5 by scan for 5 in C2.
R3C5 6 by scan for 6 in R2, C4 -- R9C3 6 by scan for 6 in R8 -- R8C3 8 by exception in C3.
R8C5 2 by exception in R8.
R9C5 9 by scan for 9 in C4 -- R9C4 8 by exception in R9 -- R3C4 3 by reduction -- R2C4 4 by exception in C4 -- R2C5 8 by exception in C5 -- R2C2 3 by exception in C2.
R3C2 8 by exception in whole game.
Final solution below.
This Sudoku game is a fairly hard Sudoku puzzle needing two of the advanced Sudoku techniques - Single digit lock and Parallel digit scan.
Focus on early breakthroughs speeds up the solution.
A question often asked, "What makes a Sudoku puzzle hard?"
There is no metric or measurement parameter to decide for sure that a Sudoku puzzle is hard, or not so hard.
Nevertheless, a distinction between hard, medium and easy Sudoku categories can be made.
As a ball-park figure, any Sudoku puzzle having number of filled up cells 27 and below should be taken as hard or extremely hard.
In medium or easy Sudoku puzzles, number of filled up cells will be more. In easy ones it can be 36 or more.
These are rough figures drawn from experience.
With only 24 filled up cells this New York Times Sudoku hard puzzle is fairly hard.
For full enjoyment, avoid looking into any solution as well as the answer.
The joy of discoveries will then all be yours.
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How to solve Sudoku hard puzzle games full list (includes very hard Sudoku).
Enjoy solving Sudoku hard.
By the way, Sudoku hard solution techniques are included with many of the solutions.
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