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How to Solve Hard Sudoku New York Times 4th March 2021

NY Times Sudoku Hard 4th March 2021 Quick Solution

How to solve NY Times Sudoku Hard March 4, 2021 by advanced Sudoku techniques

The NY Times Sudoku Hard 4th March 2021 solved step by step quickly and easily by advanced Sudoku techniques with breakthroughs explained adequately.

The NY Times Sudoku hard 4th March 2021

NYT hard Sudoku 4th March, 2021

Before going through the solution solve the puzzle first.

Step by step solution to the NY Times Sudoku Hard 4th March, 2021 Stage 1: Breakthrough by Single digit lock, Parallel digit scan, DSA and Cycles

Strategy adopted: start with row column scan coupled with aggressive breakthroughs by advanced Sudoku techniques.

As a strategy we always start row column scan from digit 1 and continue till digit 9. In the process we may get valid cells, but otherwise our lookout is for identifying for a single digit lock by cross scan.

If we get one, we should get usually a breakthrough in one step or two.

Otherwise we know that the single digit lock might become useful later and we ignore it for the present.

Starting the solution process, first valid cell success by row column is for 2: R8C7 2 by scan for 2 in R9, C8, C9 -- R7C1 2 by scan for 2 in R8, R9. And no more for 2.

Though number of digit 3 is small, we identify first a single digit lock for 3 in R9C7, R9C9 by cross scan for 3 in R7, C8.

This lock on C9 immediately participates in a row column scan for 3 in R7, lock in R9, 3 in C5 and 3 in C6 to produce unique digit 3 in cell R8C4 -- R8C4 3.

This is a breakthrough that otherwise couldn't have been achieved.

Next R1C8 7 by scan for 7 in R2, R3, C9 and that's all by row column scan at this point.

But by DSA we get the valid cell, R8C2 9 by reduction of [1,5,7] from bottom left major square DS [1,5,7,9] -- R8C3 1 by reduction of [5,7] in R8 from DS [1,5,7] in bottom left major square.

Next breakthroughs by forming Cycle (8,9) in R7C, R7C9 by DSA reduction in both cases by [5,7] from possible digit subset [5,7,8,9] in four empty cells of R7.

Cycle reduces possible digit subset for R7C4, R7C6 to [5,7] and with [7] in C6 we get the breakthrough R7C5 5 by reduction of 7 followed by R7C4 7 by exception.

In addition Cycle (1,9) is formed in R9C4, R9C6.

We'll end this stage by an important breakthrough in R6C1 7 by parallel digit scan for 7 on empty cells of R6: 7 in C3 debars cell R6C3, 7 in C4 debars 7 in R6C4 and 7 in C6 debars R6C6 for 7 leaving single cell R6C1 for 7.

Elimination of the cells in R6 for 7 by parallel scan are shown by colored arrows in the game status figure shown below.

R6C1 7 results in R9C2 7 by scan for 7 in C1 -- R9C1 5 by exception in bottom left major square.

Results of the actions taken shown.

Solution to the NY Times Sudoku hard, 4th Match, 2021 Stage 2: Breakthrough by force creation of Cycle by DSA

Turning attention first on promising zone C6 we get the breakthrough DS [6,8] in R5C6 by DSA reduction of [1,2,9] from column C6 DS [1,2,6,8,9].

This breakthrough DS of [6,8] joins with [6,8] in R8C6 to form breakthrough Cycle [6,8] in C6 reducing DS in C6 to [1,2,9] and causing the breakthrough in R6C6 2 by reduction of [1,9] from C6 DS [1,2,9] -- R6C3 4 by reduction of [2,8] -- R6C4 8 by exception in R6 -- R1C2 4 by scan for 4 in R3, C1, C3.

We’ll not disturb the Cycle of (6,8) for better understanding of what happened at this stage. Next valid cell R5C6 6 will be shown in next stage.

Game status shown below.

NY Times Sudoku hard 4th March, 2021 Solution Stage 2

Solution to the NY Times Sudoku hard 4th March Stage 3: Breakthrough by DSA and Parallel digit scan

Because of 8 in central middle major square, R5C6 6 by reduction of 8 -- R8C6 8 by reduction of 6 -- R8C5 6 by reduction of 8.

Possible digit subset in central middle major square reduced to [1,4,5,7] -- R5C2 5 by DSA reduction of [2,6] in R5 from [2,5,6] possible digit subset in C2.

R5C5 7 by DSA reduction of [1,5,8] from DS [1,5,7,8] in C5 -- R6C4 4 by DS reduction of [3,8,9] in C4 from DS [3,4,8,9] in R6.

R2C3 5 by parallel digit scan for 5 on empty cells of C3: 5 in R1 debars R1C3 for 5, 5 in left middle major square debars R4C3, R5C3 for 5 leaving single cell R2C3 for 5 -- R2C5 1 by reduction of [5,8] from DS [1,5,8] in C5 -- R4C5 5 by reduction of 1 -- R4C4 1 by reduction and R1C5 8 by exception in C5.

Game status shown below. No further challenges left.

Solution to the NYT hard Sudoku 3rd March: Final Stage 4: No more challenges left

R1C6 9 by reduction -- R9C6 1 by reduction -- R9C4 9 by reduction R2C4 6 by DS reduction of [2,5] from DS [2,5,6] in C4 -- R3C4 5 by scan for 5 in R1 -- R1C4 2 by exception in C4. R1C3 3 by DS reduction of [1,6] from DS [1,3,6] in R1 -- R1C9 6 by reduction -- R1C1 1 by exception in R1.

R5C3 9 by reduction of 3 -- R4C3 2 by exception in C3.

R3C8 1 by scan for 1 in R2, C7, C9 R2C1 9 by reduction of [2,6] in R2 from DS [2,6,9] in top left major square -- R2C1 6 by reduction of 2 in C1 -- R3C2 2 by exception in top left major square R4C2 6 by exception in C2.

R4C1 8 by reduction of 3 from DS [3,8] in C1 -- R5C1 3 by exception in C1 -- R5C7 8 by exception in R5.

R4C7 7 by scan for 7 in C8, C9 -- R3C7 9 by reduction of 8 from DS [8,9] in R3 -- R3C9 8 by exception in R3 -- R7C9 9 by reduction -- R7C8 8 by reduction -- R4C9 4 by reduction of 9 from DS [4,9] in R4 -- R4C8 9 by exception -- R9C9 3 by exception in C9.

R9C7 4 by reduction of 6 from DS [4,6] from R9 -- R9C8 6 by exception in R9 -- R2C8 4 by exception in C8 -- R2C7 3 by 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 aggressively create a breakthrough at the earliest as well as taking the opportunity of a valid cell by the simplest technique of row column scan.


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