New York Times Sudoku hard puzzle 17th February, 2021 Solved in easy steps
New York Times Sudoku Hard 17 February, 2021 solved in four stages that explain all breakthroughs and how those are achieved by advanced Sudoku techniques.
The New York Times Sudoku Hard, 17th February, 2021
Before going through the solution solve the puzzle first.
Step by step solution to the New York Times Sudoku Hard 17th February, 2021: Stage 1: Breakthroughs by Double digit scan, Parallel digit scan and and Cycles
First valid cell by row column scan is for 5: R4C2 5 by scan for 5 in R5, R6.
It is followed immediately by a second valid cell for 7 in the promising neighborhood of left middle major square: R5C2 7 by scan for 7 in R6.
Advanced Sudoku technique of double digit scan
This creates an opportunity for us to apply advanced technique of double digit scan to get a valid cell hit.
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 a major square are scanned for TWO DIGITS TOGETHER.
As [1,8] appear together in C1, and also as none of these two digits are placed yet in left middle major square, we get Cycle (1,8) in cells R6C2, R6C3 by a double digit scan of [1,8] in C1.
This eliminates the cell R6C1 for any of the two digits.
Primarily, with only two cells left for the two digits, a breakthrough Cycle of (1,8) is formed in a single strike and it leaves only the cell R6C1 for digit 4 by exception in left middle major square.
This is our breakthrough.
Advanced Sudoku technique of parallel digit scan for 3
Still analyzing this promising neighborhood we detect the possibility of the next breakthrough of R2C1 3 by parallel scan for 3 on empty cells of C1.
3 in bottom left major square eliminates R7C1 for 3 and 3 in R1 and R3, eliminate the other two empty cells R1C1 and R3C1 for 3. This leaves the single cell R2C1 for 3 and the byproduct Cycle of (5,6,7) in the other three empty cells of C1.
Next valid cell is R4C6 7 by scan for 7 in R5, R6, C7.
Never lose an opportunity for a valid cell hit by row column scan.
Explanation on how Cycle (1,8) is formed and how it results in a breakthrough
With six digits filled in the left middle major square, it is a very promising zone to look for more valid cell hits.
The possible digit subset in the three empty cells is [1,4,8].
Both digits 1 and 8 appear together in column C1 but not in any cell of the left major square. That means, these two digits light up or affect the intersecting cell R6C1 and reduce its possible digit subset to single digit 3.
As a byproduct the leftover two digits [1,8] form a Cycle of (1,8) in R6C2 and R6C3.
We are not certain at this point which of the two digits will finally occupy which of the two cells, but taking the two cells together, we are certain that digits [1,8] will certainly occupy these two cells and in no other empty cell in parent zones R6 and left middle square.
To be precise, this is what happens in the double digit scan for [1,8] together on the empty cells of the left middle major square. By the double digit scan, first the Cycle is formed and then the valid cell R6c1 4 by reduction of [1,8] because of the property of the Cycle.
Results of the actions taken shown below.
Solution to the New York Times Sudoku Hard, 17th February, 2021: Stage 2: Breakthroughs by Single digit lock and Parallel digit scan
We'll now show you a special valid cell breakthrough by a combination of advanced patterns of single digit lock and parallel scan.
The important pattern of single digit lock
In a single digit lock, a lone digit is locked in two cells in a 9 cell square as well as in a row or column. This makes the digit invalid in any other cell in the row or column in which the digit is locked.
Consider the row column cross-scan: With digit 5 in R4C2, and R7C9, 5 can appear only in two cells R8C3 and R9C3 of bottom left major square Thus the digit 5 is locked in C3 as well as in the parent major square.
This is shown by "…5…" in the two cells. We don't bother about other digits appearing in these two cells. We are only concerned with the most important fact that digit 5 can appear in only these two cells and NOWHERE ELSE IN THE TWO PARENT ZONES.
Specifically important is the disallowing of digit 5 in the empty cells of C3.
This prompts us to look for a valid cell caused by this single digit lock and thus discover the difficult to identify R2C5 5 by parallel digit scan for 5 on empty cells of R2.
Parallel digit scan for a single digit on the empty cells of a row (or column)
Analyze in which cell digit 5 can appear in R2.
With 5 in C2, C6, C8, C9 and in C3 by single digit lock, digit 5 can be placed in R2 only in the single cell R2C5.
You get a valid cell breakthrough in R2C5 5.
This is a difficult breakthrough.
In this process of parallel digit scan, presence of digit 5 in four columns C2, C6, C8 and C9 in parallel affected or lighted up four empty cells of the scanned row R2 and thus made these four cells invalid for placing digit 5. By this parallel scan for 5 on empty cells of row R2, only one cell R2C5 is left placing digit 5.
Thus you get the breakthrough R2C5 5.
The elimination of four empty cells for 5 in row R2 is shown by red arrows in the Stage 2 solution figure below.
An interesting property of a valid cell by parallel digit scan is, there will always be an associated Cycle after you put the valid digit in its identified cell. In this case for example, after you put 5 in R2C5, a Cycle of remaining four digits [4,6,7,8] is formed. You could have identified the formation of this Cycle first to get the breakthrough. Bur that would have been time-taking.
If you are alert for detecting a parallel scan opportunity, breakthrough would be much quicker.
Okay, continuing to identify more possible single digit locks and breakthroughs, with focused intent it doesn't take much to identify a second single digit lock this time on 1 in cells R8C8, R8C9 by scan for 1 in R9, C7.
Now do a parallel scan for 1 in empty cells of column C6 to get the valid cell R1C6 1.
Observe how the other four empty cells of C6 have been disallowed for 1 by scan for 1 in R2 (disallowing R2C6 for 1), Cycle (1,8) in R6 (disallowing 1 in R6C6), single digit lock in R8 (disallowing R8C6 for 1) and 1 in R9 (disallowing R9C6 for 1).
Elimination of four cells of column C6 for digit 1 by parallel scan is shown by blue arrows in Stage 2 solution figure below.
With 1 in R1C6, R3C5 2 by reduction -- R1C5 8 by reduction -- R2C6 4 by reduction -- R9C6 8 by reduction caused by the Cycle (3,9) formed in R6C6, R8C6. R2C9 2 by DSA reduction of [1,4,9] in R2 from DS [1,2,4,9] in C9.
Results of game status shown below.
Solution to the New York Times Sudoku Hard 17th February Stage 3: Routine breakthroughs by Cycles and by DSA
To start with, R4C9 1 by DSA reduction of [4,9] in R4 from reduced DS [1,4,9] in C9 -- R8C8 1 by single digit lock partner reduction of 1 in R8C9.
R1C2 2 by scan for 2 in R2, R3, C1.
Because of the Cycle (6,7) in R7, reduced possible digit subset in three empty cells of R7 is [1,4,9] -- breakthrough R7C3 4 by DSA reduction of [1,9] from DS [1,4,9].
R7C4 1 by DSA reduction of 9 from DS [1,9] in R7 -- R7C5 9 by exception in R7.
With R7C5 9, R5C5 1 by reduction -- R5C4 8 by reduction -- R4C4 2 by reduction -- R6C4 6 by reduction -- R6C5 3 by reduction -- R6C6 9 by reduction -- R8C6 3 by reduction.
With [6,9] in R6, R6C7 2 by reduction -- R4C8 3 by reduction --- R4C7 8 by reduction and exception in R4.
Results shown below.
Solution to the New York Times Sudoku Hard, 17th February 2021: Stage 4: Final solution: No more breakthrough needed
R3C2 4 by scan for 4 in R2, C1, C3 -- R3C3 1 by scan for 1 in R2, C1 -- R6C3 8 by reduction -- R6C2 1 by reduction -- R2C3 7 by reduction -- R2C8 6 by reduction -- R2C2 8 by reduction and exception -- R1C8 4 by reduction -- R1C7 5 by reduction -- R1C1 6 by reduction and exception in R1 -- R7C1 7 by reduction -- R3C1 5 by reduction.
With 7 in R7C1, R7C7 6 by reduction.
R9C8 2 by scan for 2 in C7 -- R9C3 5 by reduction -- R8C3 2 by reduction -- R9C4 4 by reduction -- R8C4 5 by reduction and exception.
With 3 in R8C6, R8C7 7 by reduction -- R9C7 3 by reduction -- R8C5 6 by reduction -- R9C5 7 by reduction. With 6 in R9C5, R8C2 9 by reduction -- R9C2 6 by reduction.
With 9 in R8C2, R8C9 4 by reduction -- R8C9 4 by reduction -- R9C9 9 by reduction.
With 4 in R1C8, R5C8 9 by reduction -- R5C7 4 by reduction -- R3C7 9 by exception in C7 -- R3C8 9 by exception in whole game.
Final solution below.
This Sudoku hard is an especially 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 23 filled up, cells this New York Times Sudoku hard puzzle is extra-hard.
It a Sudoku hard rich with learning potential.
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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