NYTimes hard Sudoku 17 February, 2021 - Solution | Suresolv

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How to Solve New York Times Hard Sudoku Puzzle February 17, 2021

nytimes hard sudoku 17th february 2021 solution

NYTimes hard Sudoku puzzle 17th February, 2021 Solved in easy steps

The detailed solution of NYTimes hard Sudoku puzzle, 17th February, 2021 explained step by step in simple language bringing out how to use special patterns.

The hard Sudoku puzzle NY Times, 17th February, 2021

nytimes hard sudoku 17th february 2021 puzzle

It is an especially hard puzzle that will challenge you.

Solution to the hard Sudoku puzzle NY Times, 17th February, 2021

Simplest method of row-column scan for digit 3 in 9 cell square R1R2R3-C1C2C3:

3 in R1, R3, C2, C3 → R2C1 3.

Getting 3 in rest four 9 cell squares not possible at this stage.

But 5 is easy to place in R4C2 by scan of two rows R5, R6 → R4C2 5.

No more hits by row column scan is possible.

Next method to get unique digits in a cell is a by-product of the process of enumeration of digits possible in a cell.

When the digit possible turns out to be a lone digit, it becomes a hit.

By possible digit analysis on cell R6C1, the digits in its three parent zones: row R6, column C1 and 9 cell square R4R5R6-C1C2C3 are,

1, 2, 3, 5, 6, 7, 8, 9.

Only 4 is missing.

We get the minor breakthrough → R6C1 4.

The technique in this method of possible digit analysis is to,

  • Look for maximum number of unique filled digits for a cell (its row, column and 9 cell square combined).
  • Identify a promising cell and form the set of unique filled digits in its parent row, column and 9 cell square.

When no more easy hits by row-column scan is possible, this is the next best action take as you have only to gain from this action.

Even if you don't get a unique possible digit, you'll get a small set of possible digits in the cell. Small sets of possible digits in empty cells are valuable assets for future use.

The 9 cell square of interest now has six filled up cells.

Three empty cells can have the three remaining digits (1,7,8).

As R6C9 has digit 7, you get a Cycle (1,8) in R6C2, R6C3.

And 7 in remaining cell of the 9 cell square R5C2 → R5C2 7.

The new filled digit 7 creates another hit by row-column scan → R4C6 7.

Results shown.

nytimes hard sudoku 17th february 2021 solution stage 1

The third important pattern of single digit lock

In a single digit lock, a lone digit is locked in two or three 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.

With digit 5 in R4C2, and R7C9, 5 can appear only in two cells R8C3 and R9C3 of 9 cell square R7R8R9-C1C2C3.

Digit 5 is locked in C3. This is shown by "…5…" in the two cells.

We don't bother about other digits appearing in these two cells. Important is - digit 5 can appear in only these two cells.

Fourth important pattern of parallel scan of multiple rows or columns

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 HIT.

R2C5 5.

This is a breakthrough.

Out of 9 cells in row R2, two cells were already occupied and five parallel columns intersecting R2 had 5. Only the sixth empty cell R2C5 could accept digit 5 giving us a breakthrough hit.

This is powerful pattern to discover and use.

Results till now shown below.

nytimes hard sudoku 17th february 2021 solution stage 2

Continue to look for single digit lock and discover the lock of digit 1 in R8C8, R8C9 by 1 in R9 and C7.

Identify another lock of digit 4 in R1C6, R2C6 to form possible digits (3,9) in R8C6. (3,9) in R6C6 is the motivation for this discovery.

The two creates Cycle (3,9) in C6 and gives us an unexpected hit → R9C6 8.

Let's go over to next stage showing results achieved till now.

Results shown below.

nytimes hard sudoku 17th february 2021 solution stage 3

8 in C6 gives R2C6 4 → R1C6 1 → R3C5 2 → R1C5 8.

Form digit possibilities in the empty cells in promising zones of central and right middle 9 cell square.

With no positive results change focus to top right square.

The most promising cell R2C9 2 as the only digit left in its three affecting zones R2, C9 and 9 cell square R1R2R3-C7C8C9.

Row-column scan for 2 in R2, R3, C1 → R1C2 2.

(4,9) in R4 cancels these two digits in possible digits (1,4,9) in R4C9 → R4C9 1 → R8C8 1.

Close here to start fresh in next stage. Results till now shown.

Results shown.

nytimes hard sudoku 17th february 2021 solution stage 4

Digit 3 locked in R6C5, R6C6 by 3 in R5, C4 → (2,9) in R6C7 with no immediate effect.

With digit 4 locked in R3C2 and R3C3 it reduces the possible digits in R3C8 to (5,7,9) and R3C8 to (7,9).

Cycle (4,6,7,9) in four cells R1C8, R2C8, R3C8 and R5C8 creates a two digit lock of digits (6,7) in R1C8, R2C8 and R3C8. This eliminates (6,7) from possible digits in R1C7 and R3C7.

Also down below, because of Cycle (4,9) in R8C9 and R9C9, possible digits (6,7) in R7C7 creates a Cycle of (6,7) with already existing (6,7) in R7C1.

We'll see the simplifying effects of these in next stage.

Results till now shown below.

nytimes hard sudoku 17th february 2021 solution stage 5

The Cycle (6,7) created series of single digit candidates:

R7C3 4 → R7C4 1 → R7C5 9 → R5C5 1 → R5C4 8 → R4C4 2 → R6C4 6 → R6C5 3 → R6C6 9 → R8C6 3 → R6C7 2.

Results shown below.

nytimes hard sudoku 17th february 2021 solution stage 6

Further by 2 in R4C4 → R4C8 3 → R9C8 2.

Row column scan on 3: 3 in R8, C8, C9 → R9C7 3 → R4C7 8.

Enumerating the possible digits in rest of the empty cells created the important Cycle of (6,7) in R8C5 and R8C7.

R8C3 2 as it is the only place in the 9 cell square in which it can appear.

R8C2 9 as R7C3 4 → R8C9 4 → R9C9 9 → R9C2 6 → R7C1 7 → R9C3 5.

R9C4 4 → R9C5 7 → R8C5 6 → R8C4 5 → R8C7 7 → R7C7 6.

Rest in the next final stage.

Results shown.

nytimes hard sudoku 17th february 2021 solution stage 7

With 6 in R9C2 → R2C2 8 → R2C3 7 → R3C3 1 → R3C2 4 → R3C1 5 → R3C7 9 → R3C8 7 → R2C8 6.

With 9 in R3C7 → R5C7 4 → R5C8 9.

With 1 in R3C3 → R6C3 8 → R3C2 1.

With 5 in R3C1 → R1C1 6 → R1C8 4 → R1C7 5.

Final solution below.

nytimes hard sudoku 17th february 2021 solution stage 8 final

This Sudoku refused to break down to a late stage. It is an especially challenging puzzle rich with Sudoku digit patterns.

End note

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 NY Times Sudoku puzzle is extra-hard.

Though it relented and provided a breakthrough time to time, just after a breakthrough the puzzle posed a new hurdle to cross.

It a hard Sudoku rich with learning potential.

Suggestion

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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