How to solve NYT Hard Sudoku March 3, 2021 by advanced Sudoku techniques
The NYT hard Sudoku 3rd March 2021 solved step by step quickly and easily, applying advanced Sudoku techniques. All breakthroughs adequately explained.
The NYT hard Sudoku 3rd March 2021
Before going through the solution solve the puzzle first.
Step by step solution to the NYT hard Sudoku 3rd March, 2021 Stage 1: Breakthrough by Single digit lock, Parallel digit scan, Double digit scan, Cycles and DSA technique
Strategy adopted: aggressive breakthroughs by advanced Sudoku techniques.
First valid cells by row column scan: R2C3 1 by row column scan for 1 in R1, R3 and R4C4 4 by row column scan for 4 in R5, C5, C6.
Followed by R4C2 6 by scan for 6 in R6, C1, C3 -- R5C8 6 by scan in R4, R6.
Still scanning for digit 6, a single lock on 6 is formed in R7C6, R9C6 by cross scan for 6 in R8, C4.
This gives valid cell R3C5 6 by scan for 6 in R2, C4 and lock on 6 in C6.
Followed by a special breakthrough in R8C5 3 by parallel scan for 3 in empty cells of C5 eliminating R4C5, R5C5, R6C5 because of 3 in central middle major square and eliminating R2C5 by 3 in R2.
A breakthrough R8C9 4 by DSA reduction of [3,5,6,8] from possible digit subset of [3,4,5,6,8] in C9.
Now consider a double digit scan for [1,9] in both C7 and C9 but not in right middle major square -- forms Cycle (1,9) in R4C8, R6C8 at one stroke.
This reduces the DS in all the empty cells in the major square and breakthrough R6C7 4 by scan for 4 in R4, C9 -- R4C7 7 by scan for 7 in C9 -- R4C9 3 by scan for 3 in R6 -- R6C9 5 by exception in right middle major square.
Reduced DS in C7 [2,3,5,6] further reduced by [3,5,6] in intersecting row R8 to produce R8C7 2.
Before closing this stage we'll form two power-packed Cycles in C2.
First notice that both row R9 and top left major square having the digits [1,9], all empty cells of C2 except R6C3 and R7C3 are eliminated for [1,9] thus forming the Cycle (1,9) in C2 -- Cycle (2,7,8) formed in C2 by the remaining three digits for the three cells R1C3, R3C3 and R9C3.
These form a major breakthrough. We'll see effects next stage.
Results of the actions taken shown.
Solution to the NYT hard Sudoku, 3rd Match, 2021 Stage 2: Cycles, DSA technique and Parallel digit scan
Breakthrough Cycle: [1,9] in R6C2 joins with [1,9] in R6C8 to form a Cycle of (1,9) in R6.
This gives us the first breakthrough in R6C5 8 by DSA reduction of 7 in C5 from DS [7,8] in R6 -- R6C1 7 by exception in C6.
Second breakthrough is in R2C8 8 by parallel scan elimination of R2C4 and R2C5 for 8 by 8 in C4 and C5 leaving only R2C8 for 8.
A Cycle (2,5) formed in R2.
With the major breakthrough in R2C8 8, R1C9 6 by reduction of 8 -- R8C8 8 by reduction of 6.
R7C7 6 by scan for 6 in C8 -- R9C6 6 by reduction of single digit lock partner on 6 in R7C6.
R8C8 7 by reduction of [3,5] from DS [3,5,7] in C8 -- Cycle (1,9) lock partner on 6 in R7C6. R8C8 7 by reduction of [3,5] from DS [3,5,7] in C8 -- Cycle (1,9).
Next a critical breakthrough by parallel digit scan for 4 on empty cells of R1: R1C8 4 by elimination of R1C2 by 4 in C2, elimination of R1C4, R1C6 by 4 in top middle major square and elimination of R1C7 and R1C9 by 4 in C7 to leave single cell R1C8 for 4 -- R1C8 4.
In total, as many as four out of five empty cells are eliminated to get the breakthrough by parallel scan. The alternative would have been to form the Cycle (2,3,7,8) of rest four digits in rest four cells in R1 by DSA on each.
That would have been a more time-consuming alternative.
Cycle (3,5) in C8 reduces [3,5] to result in R3C8 2.
With 6 in R1C9, R1C7 3 by reduction -- R3C7 5 by scan for 5 in R1 -- R1C7 3 by exception on top right major square.
More valid cells are visible but we won't disturb this status for ease of understanding and close the stage here.
Game status shown below.
Solution to the NYT hard Sudoku 3rd March Stage 3: Important breakthrough by Parallel digit scan
With 3 in R1C7, R1C4 7 by reduction -- R1C6 8 by reduction -- R1C2 2 by reduction -- DS [3,9] in R3C4, R3C6 -- R3C6 9 by reduction by 3 in C6 -- R3C4 3 by exception in top middle major square.
With 9 in R3C6, R8C6 1 by reduction -- R8C1 9 by reduction -- R7C2 1 by reduction -- R6C2 9, R7C4 9 by reduction.
R5C6 7 by scan for 7 in R4, C5 -- R5C4 1 by scan for 1 in C5, C6 With 9 in R6C2, R6C8 1 by reduction -- R4C8 9 by reduction.
Possible digit subset in C2 reduced to [7,8] -- R9C2 8 by 7 in R9 -- R3C2 7 by exception in C2.
An important breakthrough by parallel scan for 7 on empty cells of R7: R7C3 7 by elimination of R7C1 by 7 in C1, elimination of R7C6 by 7 in C6 and elimination of R7C8 by 7 in C8.
Game status shown below. No further challenges left.
Solution to the NYT hard Sudoku 3rd March: Final Stage 4: No more challenges left
R4C1 by scan for 1 in R5C3 -- R5C5 9 by scan for 9 in R4 -- R5C1 3 by reduction -- R7C1 2 by reduction -- R9C1 4 by reduction -- R9C3 3, R3C1 8 by reduction of 4 -- R3C3 4 by reduction of 8, R5C3 5 by reduction of 3 -- R4C3 8 by reduction.
R9C8 5 by reduction of 3 -- R9C4 2, R7C8 3 by reduction of 5 -- R7C6 5 by reduction of 2 -- R4C6 2 by exception in C6 -- R4C5 5 by reduction -- R2C5 2 by reduction -- R2C4 5 by reduction and exception in whole game.
Final solution below.
This 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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