How to Solve the New York Times Sudoku Hard 20th February, 2021 with step by step explanation
Solve New York Times Sudoku hard February 20, 2021 in quick steps. How to solve the Sudoku hard is explained clearly including all the breakthroughs.
The New York Times Sudoku Hard, 20th February, 2021
Before going through the solution solve the puzzle first.
Step by step solution to the New York Times Sudoku Hard 20th February, 2021: Stage 1: Breakthroughs by Double digit scan, DSA technique and Cycles
First success by row-column scan: R8C8 4 because of 4 in R7, R9 -- R9C9 2 by scan for 2 in R7, C7.
Advanced Sudoku technique of double digit scan
Observe that the two digits [1,8] appear together in R6 and also in C9, but do not appear in the right middle major square.
This creates an opportunity for us to apply advanced technique of double digit scan to get a breakthrough Cycle (1,8).
Instead of scanning the promising empty cells of a major square FOR A SINGLE DIGIT, in a double digit scan, the promising empty cells of the right middle major square are scanned for TWO DIGITS (1,8) TOGETHER -- Cycle (1,8) formed in R4C8, R5C8 in C8 and parent right middle major square.
This is how double digit scan works.
Let's see the effect of this Cycle (1,8).
It reduces the possible digit subset DS in right middle major square to [4,5,7,9] -- breakthrough valid cell R6C8 5 by reduction of [4,7,9] (with [4,7] in C8 and 9 in C6) from DS [188.8.131.52] -- followed by R7C8 3 by DSA reduction of [1,5,7] from DS [1,3,5,7] in C8 -- R7C9 7 by reduction of [1,5] from DS [1,5,7].
Other valid cells by DSA and Cycles
R6C9 4 by DSA reduction of [7,9] from DS [4,7,9] -- R6C7 7 by reduction -- R5C9 9 by exception in right middle major square -- R3C9 3 by exception in C9.
Reduced DS in two empty cells in C8 is [2,9] -- R2C8 2 by reduction of 9 by R2 -- R2C1 8 by reduction -- R2C7 6 by reduction -- R2C4 3 by exception in R2.
R6C5 3 by scan for 3 in R4, R5, C4 -- R8C6 3 by scan in R7, C4, C5.
Cycle (2,6) in two empty cells R6C3, R6C4 and DS [2,6] in R8C3 by DSA reduction of [1,5] from possible digits [1,2,5,6] in R7 -- a breakthrough Cycle (2,6) formed in R6C3, R8C3.
Breakthrough R3C3 4 by DSA reduction of [3,7,8,9] from possible digit subset of [3,4,7,8,9] -- R3C7 8 by reduction from DS [4,8] -- R1C7 4 by reduction and exception in top right major square.
Explanation on how Cycle (2,6) is formed in C2 and how it results in a breakthrough
A possible digit subset DS [2,6] already exists in R6C3 as a part of easily formed digit subset in two empty cells of R6.
Now in addition, a third possible digit subset of [2,6] is formed in R8C3 by DSA reduction of [1,5] from possible digit subset [1,2,5,6] in R8.
[2,6] in two cells of C3 forms a Cycle in which the two digits get locked and thus disallows appearance of 2 and 6 in any other empty cell of C3.
Because of this effect of limiting the digits inside the Cycle, the possible digit subset in the column C3 is reduced to [3,4,7,8,9] and this creates the breakthrough in R3C3 4 by DSA reduction of [3,7,8,9] (with [8,9] in parent top left major square and [3,7] in second parent R3) from DS [3,4,7,8,9] in C3.
This breakthrough gives us two more valid cells, R3C7 8 by reduction from DS [4,8] -- R1C7 4 by reduction and exception in top right major square.
Results of the actions taken shown below.
Solution to the New York Times Sudoku Hard, 20th February, 2021: Stage 2: A major breakthrough by a Cycle in only one parent major square
R5C1 4 by scan for 4 in C2, C3.
DS in both R4C2 and R4C3 [7,8] by reduction of [2,6] in R4 from possible digit subset [2,6,7,8] in left middle major square. This creates Cycle (7,8) in R4.
R4C8 1 by reduction of 8 from DS [1,8] -- R5C8 8 by exception in right middle major square.
R4C4 4 by scan for 4 in R5, R6, C6 -- R4C6 5 by exception in R4 -- R5C5 7 by scan for 7 in C4, C6.
With much of the empty cells filled up with valid digits, short length possible digits are evaluated conveniently in promising zones by possible digit analysis technique.
In the process, an awkward shaped Cycle (1,2,5,6) in bottom left major square formed. This reduces [1,5] from DS [1,5,7] to produce a major breakthrough of R9C1 7. We'll close here and explore the rest later.
How the Cycle (1,2,5,6) is formed in only the bottom left major square
DS in C1 [1,2,5,7] reduced by [2,7] in R7 to form DS [1,5] in R7C1, reduced by 7 in R8 to form DS  in R8C1 and reduced by 2 in R9 to form DS [1,5,7] in R9C1.
We decide to continue to form DSs in empty cells of the promising zone bottom left major square.
DS in R8C2 [2,5,6] by reduction of 1 in C2 from DS [1,2,5,6] in R8 and DS in R8C3 [2,6] by reduction of [1,5] in C3 from DS [1,2,5,6] in R8.
It's a pleasant surprise to identify that a Cycle of [1,2,5,6] is formed in the four cell DSs in R7C1, R8C1, R8C2 and R8C3.
These four digits are locked and are self-sufficient in these four cells, each appearing at least twice in these four cells.
The specialty of this Cycle is - it belongs to only the single parent of A MAJOR SQUARE and to no other parent row or column.
This Cycle immediately causes the breakthrough of R9C1 7 by reduction of [1,5] from DS of [1,5,7] in R9C1 because of the property of locking the four digits inside the Cycled cells only.
Results of game status shown below.
Solution to the New York Times Sudoku Hard 20th February: Stage 3: Breakthrough by single digit lock
So the last breakthrough has been R9C1 7 by reduction of [1,5] from DS [1,5,7] because of the Cycle (1,2,5,6) in bottom left major square.
Short length easy possible digit subsets are evaluated for nearly all empty cells without any easy success.
Most important breakthrough at this stage finally has been provided by the single digit lock on 6 in R8C2, R8C3.
This single digit lock has been formed by Cross-scan for 6 in R9, C1 and the Cycle (2,6) formed by DSA in R6C3, R8C3.
Effect of this critically important single digit lock is reduction of 6 in possible digit subset [1,5,6] to [1,5] in R8C5.
This helps to form a new Cycle (1,5,8) in R3C5, R8C5, R9C5 -- R1C5 6 by reduction.
This is the big breakthrough.
With 6 in R1C5, R1C6 2 by reduction -- R1C1 5 by reduction -- R1C4 8 by reduction.
To preserve the breakthrough digit patterns the stage is closed here with status shown below.
Solution to the New York Times Sudoku Hard, 20th February 2021: Stage 4: Final solution: With game wide open, all valid cells obtained easily
With 5 in R1C1, R3C2 2 by reduction -- R5C2 6 by reduction -- R6C3 2 by reduction -- R6C4 6 by reduction -- R5C6 1 by reduction -- R5C4 2 by reduction.
With 5 in R1C1, R7C1 1 by reduction -- R8C1 2 by reduction -- R8C3 6 by reduction -- R8C2 5 by reduction -- R8C5 1 by reduction -- R3C5 5 by reduction -- R9C4 8 by reduction -- R9C2 3 by reduction -- R9C3 9 by reduction -- R7C3 8 by reduction -- R4C3 7 by reduction -- R4C2 8 by reduction.
With 7 in R4C3, R1C3 3 by reduction -- R1C2 7 by reduction.
With 1 in R5C6, R3C6 9 by reduction -- R3C4 1 by exception -- R7C6 6 by reduction.
With 1 in R7C1, R7C7 5 by reduction -- R9C7 1 by reduction -- R9C4 5 by exception in R9 -- R7C4 9 by exception in whole game.
Final solution below.
This Sudoku hard is a challenging puzzle rich with Sudoku digit patterns. Because of the series of early breakthroughs by advanced digit patterns, the Sudoku hard could be solved rather easily.
An often asked questions is, "What makes a Sudoku puzzle hard?"
There is no metric or measurement 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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