How to solve NYT Sudoku Hard February 27, 2021 by advanced Sudoku techniques
How to solve NYT Sudoku hard February 27, 2021 explains each breakthrough using advanced Sudoku techniques. It's a quick and easy solution to a hard Sudoku.
The NYT Sudoku Hard, 27th February, 2021
Before going through the solution solve the puzzle first.
Step by step solution to the NYT Sudoku Hard 27th February, 2021 Stage 1: Breakthroughs by DSA and Cycles
First valid cell hit by row column scan is for 5: R1C4 5 by scan for 5 in R2, R3, C6 -- R8C5 5 by scan for 5 in R7, C4.
No more valid digit or even single digit lock by scan for any digit possible at this point of the game.
So we take up valid cell by possible digit analysis or DSA technique in promising cells and get immediate success: first R1C3 6 by reduction of digit subset [3,8,9] in R1 and C3 combined set of digits, from possible digit subset DS [3,6,8,9] in top left major square.
With 6 in R1C3 DS in top left major square and in R1C2 reduced to [3,8,9] -- R1C2 9 by reduction of [3,8] in R1 from the DS [3,8,9].
Again it's a situation of no easy breakthroughs.
This is the time to start enumerating the possible digit subsets of length preferably 2 digits or occasionally 3 digits in promising zones and cells.
It doesn't take long to identify a breakthrough by forming a Cycle (4,8) in R3C9 (by reduction of [1,5,9] from DS of [1,4,5,8,9] in C9) and also [4,8] in R7C9 (by reduction of [1,5,9] in R7 from DS [1,4,5,8,9] in C9).
This break immediately produced a series of valid cells: R2C9 1 by reduction of [8] from DS [1,8] because of the Cycle [4,8] in C9.
With 1 in R2C5, R2C6 6 by reduction of 1 -- R2C4 3 by reduction of [1,6] -- R2C3 8 by reduction of 3 -- R3C2 3 by reduction of 8. R1C6 1 by scan for 1 in R3.
A Cycle of (2,4) formed in R1 as well as in top right major square where it will reduce 4 in R3C9 [4,8] to create a hit. But we won't disturb the breakthrough Cycle now.
Results of the actions taken shown.
Solution to the NYT Sudoku Hard, 27th February, 2021 Stage 2: Breakthrough by Parallel digit scan and Cycle
As explained, first we'll show R3C9 8 by reduction of 4 by the Cycle (2,4) in top right major square -- R3C8 6 as the only digit left for the cell.
Also R7C9 4 by reduction of 8.
Now will use parallel digit scan for 3 on empty cells of R7: R7C7 3 as 3 in bottom middle square eliminates three cells in R7, R7C4, R7C5, R7C6 and 3 in R3C2 eliminates R7C2 for 3 leaving the single cell R7C7 for 3.
In the same way all three cells in R7 and bottom middle major square are disallowed by 8 in the major square leaving R7C2 8. This is the second case of applying parallel digit scan.
As a result Cycle (2,6,7) formed in R7C4, R7C5, R7C6 -- Cycle (4,9) formed in the remaining two cells R8C4, R9C4 in bottom middle major square.
Note: In the Cycle of (2,6,7) formed in three cells, each of the three digits appear in the set of possible digits at least twice and makes the digits self-sufficiently cycling between the three cells. This debars appearance of any of the three digits outside the Cycle in R7.
This second Cycle produces the breakthrough in R3C4 2 by reduction of [4,9] -- R3C5 9 by reduction -- R3C6 4 by exception -- R5C6 3 by reduction. By this last valid digit, the Cycle (2,6,7) in R7 would be broken up. To preserve the Cycle we'll close the stage here.
Game status shown below.
Solution to the NYT Sudoku Hard 25th February Stage 3: Breakthroughs by Single digit lock
As explained, with 7 in R5C6, R7C6 2 by reduction -- R6C6 9 by exception -- R7C5 6 by reduction -- R7C4 7 by exception and reduction.
With 6 in R7C5, R5C5 3 -- R4C5 2 by exception and reduction.
With 9 in R6C6, R6C9 5 by reduction -- R9C9 9 by reduction -- R9C4 4 by reduction -- R8C4 9 by reduction.
R5C2 5 by scan for 5 in R4, R6, C3 -- R9C8 5 by scan for 5 in R8, C7.
A single digit lock on 6 formed in R6C1, R6C2 by scan for 6 in R4, C3 and this lock gives us a breakthrough in R5C4 6 by scan for 6 in R4, and lock on 6 in R6.
Note on effective single digit lock on 6: An effective sigle digit lock can be discovered only by scanning a row and a column cross scan for the digit or by scanning one column or one row for the digit. Digit 6 is locked in R6 as well as in left middle major square in the two cells so that it debars all other empty cells in both R6 and the left middle major square for 6. It's a powerful asset to have.
Cycle (1,8) formed by the leftout 2 digits in central middle major square and Cycle (1,4,8) formed in R5 with the leftout 3 digits in R5.
DS in R7C7 [1,8] by reduction of [3,4,7,9] from DS [1,3,4,8,7,9] in right middle major square.
[1,8] in R7C7 further forms a Cycle of [1,4,8] with the other two cells R5C7, R5C8 in left middle major square -- DS in the major square reduces to [3,7,9] -- R4C7 9 by scan for 9 in C8.
Lastly observe the critical breakthrough to be caused by the single digit lock on 4 in R5C7 R5C8.
The lock on 4 reduces 4 from the DS [1,4] in R4C3 -- breakthrough R5C3 1.
Game status shown below. No further challenges left.
Solution to the NYT Sudoku Hard 27th February: Final Stage 4: No more challenges
First valid digit, R6C7 1 by scan for 1 in R5, C8 -- R6C4 8 by reduction -- R4C4 1 by reduction. Breakthrough in R4C2 7 by DSA reduction of [3,4,8] from DS [3,4,7,8] in R4 -- R4C8 3 by reduction -- R6C8 7 by reduction.
DS in leftover two cells R4C1, R4C3 of R4 is [4,8] -- R4C3 4 by reduction of 8 by 8 in C3 -- R4C1 8 by exception in R4.
R7C2 1 by scan for 1 in R9, C1, C3.
R9C3 7 by scan for 7 in R8, C1, C2.
R9C1 3 by scan for 3 in R8, C2 -- R6C1 6 by reduction -- R6C2 2 by reduction -- R9C2 6 by exception in C2 -- R8C1 4 by exception in C1 -- R8C3 2 by exception in bottom left major square -- R6C3 3 by reduction and exception in C3 -- With 2 in R8C3, R8C8 8 by reduction -- R5C8 4 by reduction -- R5C7 8 by reduction.
With 4 in R5C8, R1C8 2 by reduction -- R1C7 4 by reduction.
With 2 in R8C3, R8C7 6 by reduction -- R9C7 2 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.
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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