Quick and easy solution to NYT Sudoku Hard February 25, 2021 by advanced Sudoku techniques
Solution to the NYT Sudoku Hard February 25 2021 achieved in four easy stages by breakthroughs using advanced techniques of single digit lock and more.
The NYT Sudoku Hard, 25th February, 2021
Before going through the solution solve the puzzle first.
Step by step solution to the NYT Sudoku Hard 25th February, 2021 Stage 1: Breakthroughs by two Single digit locks and DSA
Solution process starts with an opportunistic scan success in R4C5 8 by row column scan for 8 in R4, R5, C6.
Followed by another opportunistic scan in a promising zone: R6C8 2 by scan for 2 in C7, C9 -- R6C7 3 by scan for 3 in R4, R5.
Single digit lock in R3C5, R3C6 by cross scan for 6 in R2, C4 -- breakthrough R1C9 6 by scan for 6 in R2, lock on 6 in R3, 6 in C7 and C8. The single digit lock on 6 participates in this case in an usual scan for 6 to produce the breakthrough.
It is followed by R3C9 8 by scan for 8 in C7, C8 -- R1C3 8 by scan for 8 in R2, R3, C1 -- R8C8 7 by scan for 7 in R7, R9, C9.
A second single digit lock on 5 formed in R2C7, R3C7 by cross scan for 5 in R1, C8 -- R9C7 1 by DSA reduction [5,7,9] from DS [1,5,7,9] in C7. The lock in the case contributes to the breakthrough by DS reduction.
This is followed by R1C7 7 by DSA reduction of [5,9] in R1 from DS [5,7,9] in C7 -- R1C4 1 by DSA reduction of [3,4] from DS [1,3,4] in R1.
R7C8 9 by scan for 9 in C9 by Cycle (1,9).
A big breakthrough in R7C4 8 by DSA reduction of [1,3,4] in C4 and  in R7 from possible digit subset [1,2,3,4,8] in bottom middle major square
It is followed by R9C2 8 by scan for 8 in R7, R8, C3 -- R9C4 2 by DSA reduction of [3,4] in C4 and  in R9 from reduced DS [1,2,3,4] in bottom middle major square -- R5C5 2 by scan for 2 in R6, C4, C6.
Results of the actions taken shown.
Solution to the NYT Sudoku Hard, 25th February, 2021 Stage 2: Breakthrough by Single digit lock, Cycle and Parallel digit scan
Breakthrough single digit lock on 1 formed in R4C6, R5C6 by cross scan for 1 in R6, C4 -- R8C5 1 by scan for 1 in R9 and lock on 1 in C6.
Use of Single digit lock:
In the single digit lock, the digit 1 can appear in only the two cells.
That debars all other empty cells in lock column C6. Thus the lock on 1 can be used for a breakthrough either taking part in a scan like a normal case of 1 in C6, or it reduces a digit 1 in a DS of a cell in C6 to create the breakthrough.
In our game, breakthrough by the lock is by taking part in a simple row column scan only.
Okay returning to the solution, again we look for advanced digit patterns.
And sure enough we have one in R3C4 7 by parallel digit scan for 7 on empty cells of R2:
Digit 7 in C1, C5, C7, C8 eliminates four of the five empty cells on R2 leaving the R2C3 for 7. It is a major breakthrough obtained quickly.
Next breakthrough is achieved first by forming a Cycle (3,4) by DSA and then by reduction effect of the Cycle.
DS in R7C2 [3,4] by DSA reduction of [1,5,6] in C2 from DS [1,3,4,5,6] in R7.
This forms a breakthrough Cycle of (3,4) with R7C6 in R7 -- R7C9 5 by reduction of [3,4] -- R9C3 5 by scan in R7, R8 -- R5C1 5 by scan for 5 in R4, C2, C3.
Having been used to identifying advanced digit patterns, next breakthrough identified again by a parallel digit scan,
R7C1 6 by parallel digit scan for 6 on empty cells of C1.
6 in top left major square eliminates R1C1, R2C1, R3C1 for 6 and 6 in R4 eliminates R4C1 for 6 leaving the single cell R7C1 for 6.
R7C3 1 by exception in R7.
Game status shown below.
Solution to the NYT Sudoku Hard 25th February Stage 3: Breakthrough by Parallel digit scan
First breakthrough at this stage we'll achieve by parallel digit scan:
Out of four empty cells of R5, 6 in C2, C4 and C9 eliminates three for 6 leaving the single cell R5C6 for 6
This digit 6 in R5C6 gives us further valid cells:
R5C6 6 -- R3C5 6 by single digit lock reduction for 6 -- R4C6 1 by single digit lock partner reduction for 1 -- R4C9 9 by reduction of 1 -- R5C9 1 by reduction of 9 -- R4C1 4 by reduction of 9 -- R1C1 3 by reduction -- R1C8 4 by exception in R1.
Game status shown below. No further challenges left.
Solution to the NYT Sudoku Hard 25th February: Final Stage 4: DSA reduction, simple reduction and exceptions only
R6C6 4 by DSA reduction of [3,7] from DS [3,4,7] in C6 .
R7C6 3 by reduction -- R9C5 4 by reduction -- R7C2 4 by exception in R7. With 4 in R6C6, R6C5 5 by reduction -- R6C4 9 by reduction -- R6C3 6 by exception in C6 -- With 3 in R7C6, R3C6 7 by reduction -- R3C3 5 by reduction -- R2C5 3 by reduction -- R9C5 4 by reduction -- R9C9 3 by reduction -- R8C9 4 by exception in C9.
With 5 in R3C4, R3C7 9 by reduction -- R2C7 5 by reduction.
R2C1 9 by scan for 9 in C3 -- R3C1 1 by exception in C1 -- R2C8 1 by exception in R2 -- R3C8 3 by exception in C8.
R3C3 4 by scan for 4 in C2 -- R3C2 2 by exception in R3.
R4C3 2 by scan for 2 in C2 -- R8C3 9 by exception in C3 -- R8C2 3 by exception in bottom left major square -- R5C4 7 by exception in C4 -- R5C2 9 by exception in R5 -- R4C2 7 by exception in whole game.
This Sudoku game is a hard Sudoku puzzle that needed multiple breakthroughs by advanced Sudoku techniques - more than once Single digit lock and Parallel digit scan provided the main breakthroughs.
Focus on early breakthroughs and avoidance of unnecessary enumeration of possible digit subsets in empty cells speeds up the solution.
Though the number of filled cells 25 in this Sudoku hard is a little high for a hard Sudoku game, the digit arrangements in this Sudoku puzzle made sure that enough challenges are posed to the solver.
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