NYTimes hard Sudoku 15 February, 2021 - Solution | Suresolv

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How to Solve New York Times Sudoku Hard February 15, 2021

nytimes hard sudoku 15th february 2021 solution

NYTimes hard Sudoku puzzle 15th February, 2021 Solved in easy steps

The detailed solution to NYTimes hard Sudoku puzzle, 15th February, 2021 explained step by step in simple language bringing out how to achieve breakthrough.

The hard sudoku puzzle NY Times, 15th February, 2021

nytimes hard sudoku 15th february 2021 puzzle

It is a good hard Sudoku puzzle. You'll like solving it. I liked.

Solution to the hard Sudoku puzzle NY Times, 15th February, 2021

Use row-column scan, the simplest method to get a unique digit for a cell. Fill as many cells as you can by row-column scan. Every such easy fill at this stage is a bonus.

To be quick select the digit that appears maximum and select a promising 9 cell small square. Identify the rows and columns with the digit that intersect in the 9 square cell.

This is the basic preparation.

We select digit 3 as promising and get a few easy placements.

3 in R1, R3: 3 possible only in R2C2 or R2C3.

→ C2 has 3, R2C2 cannot have 3 → R2C3 3.

3 in R4, R5: 3 possible only in R6C7 or R6C9.

→ C9 has 3, R6C9 cannot have 3 → R6C7 3.

3 in C7, C9 → 3 possible only in R8C8.

No more easy 3 possible at this point.

Look for placing any other possible digit by row-column scan.

9 in R7, R8 → R9C7 9.

An easy catch.

No more digits can be placed by this simple method of row-column scan at this point.

Result of actions shown in stage 1 solution.

nytimes hard sudoku 15th february 2021 solution stage 1

In hard Sudoku puzzles, you may expect this situation at the start itself when simple hits by row-column scan is no longer possible. Good thing in this puzzle is, we have got four simple cell placements before this tedious stage.

Below section explains in a bit of detail the process of enumerating digits possible in a cell. To skip the section, click here.


Enumerating digits possible for each empty cell

After speed of solution, reducing labor being the main objective, we won't enumerate 5-digit possibilities or enumerate possible digits for ALL the cells.

We'll select relatively more promising cells in a row, a column or a 9 cell square (each a zone) to evaluate the smallest length of possible digits in the cells first. Target is to get two-digit possibilities first.

Take cell R2C5. It is in three zones or areas - row R2, column C5 and 9 cell top-middle square. Unique set of digits in all these three together are,

$(1, 3, 6) \cup (3, 4, 7, 9) \cup (1, 3, 5)=(1, 3, 4, 5, 6, 7, 9)$.

Set union is technical, but this we do by common sense and do naturally. We eliminate repeated digits when considering digits of two or three zones together.

So only the two digits (2, 8) are valid in cell R2C5.

Once a few cells are enumerated in a zone, enumerating the rest gets easier.

We stop in a 9 cell square when the rest of the empty cells would all have digit possibilities 5 or more.

Three reasons why we stop,

  1. Writing long digit sets takes more time and effort,
  2. Chance of getting a useful pattern from a 5 digit possibility in a cell is remote, and,
  3. Such long list of digits CLUTTERS the view and hinders fluid movement of the eye over the game while searching for a useful pattern. This is important.

It is a strategy born out of experience:

Digit possibilities for empty cells are enumerated ONLY WHEN NECESSARY.

For example, if we get a useful pattern and a hit after enumerating just a few cell digit possibilities, we continue exploring the hit to get further hits.

Reason is simple: With more number of cells filled up, cell digit possibilities get shorter and getting them easier.


Avoiding the effort of choosing a unique place for the next digit, we'll write the two-digit, three-digit and four-digit possibilities for each cell next.

We won't write the five-digit possibilities in a cell at present.

A cycle (1,3,4,5) formed in R9C1, R9C4, R9C5, R9C6.

The four digits are locked in these four cells and in no other cell in the row R9 any of these four digits can be placed.

The fifth empty cell in the row can have only digit 6.

We'll show this digit placed a bit later. In next game stage graphic below, only the cycle is shown.

nytimes hard sudoku 15th february 2021 solution stage 2

Place 6 in R9C2 6.

Continue analyzing the digit possibilities in the 9 cell square R7R8R9-c1C2C3.

Discover a second cycle (1,4,5,7) in R8C1, R8C2, R8C3, R8C9.

These digits cannot be placed in the fifth empty cell R8C7. It can have then only digit 8.

Results shown below.

nytimes hard sudoku 15th february 2021 solution stage 3

Place 8 in R8C7.

Eliminate every occurrence of 8 in its three zones (R8, C7 and 9 cell square R7R8R9-C7C8C9).

Identify a possible breakthrough in the interesting pattern of:

Digit set (1, 4) in R8C9 and its effects on digit sets (5,7), (1,4,5,7), (1,4,5,7) in R8C1, R8C2 and R8C3 respectively.

Consider the logic:

If R8C9 is 1, a cycle of (4,5,7) is formed in R8C1, R8C2 and R8C3. (5,7) cannot appear in any other cells in 9 cell square R7R8R9-C1C2C3.

If R8C9 is 4, again a cycle of (1,5,7) is formed in R8C1, R8C2 and R8C3 with same outcome.

It means, digits (5,7) are locked in R8C1, R8C2 and R8C3 in both the cases → (5,7) cannot appear in any other cells in this 9 cell square.

Eliminate these two digits in any cell digit possibility in the 9 cell square.

Elimination of 5 in R9C1 gives us the CRITICAL BREAKTHROUGH leaving single possible digit 3 in R9C1.

This is a Compound Cycle, a combination of two Cycles one inside the other. The Cycle of (1,4) is embedded inside the Cycle of (1,4,5,7).

Results of analysis and action shown in stage 4 graphic below.

nytimes hard sudoku 15th february 2021 solution stage 4

2 is placed in R7C1. It cannot have 3 anymore → R7C1 2.

Two new cycles (1,4,6), and (1,4,5) are formed in R7C2, R7C7, R7C9 and R9C4, R9C5, R9C6 respectively.

These results are shown next.

nytimes hard sudoku 15th february 2021 solution stage 5

8 is placed in R7C5. It cannot have 1 anymore → R7C5 8.

7 is placed in R7C6. It cannot have 8 anymore → R7C6 7.

3 is placed in R7C4 as the only remaining digit in the 9 cell square R7R8R9-C4C5C6 (digits 1, 4 and 5 are locked in the cycle (1, 4, 5) in cells R9C4, R9C5 and R9C6).

This placement of 3 doesn't have any impact on the rest of the puzzle, but the 8 in R7C5 has lots.

R2C5 gets 2 and R3C5 gets 6.

These results are shown next.

nytimes hard sudoku 15th february 2021 solution stage 6

By row-column scan for digit 2 on R2, R3 and C8, 2 is placed uniquely in R1C7 → R1C7 2. We get this positive result now because of recent placement of 2 in R2C5.

Always look for getting hits by this simple method.

Due to elimination of 2 and 6 in 9 cell square R1R2R3-C4C5C6, a cycle (4,7) is formed in R1C4 and R1C6.

We'll soon see how this new cycle results in multiple positive effects.

Results till now shown below.

nytimes hard sudoku 15th february 2021 solution stage 7

By elimination of digits 4 and 7 we get the unique placements,

9 in R2C4, 8 in R3C6, 5 in R1C3, and 6 in R1C8.

These results shown next.

nytimes hard sudoku 15th february 2021 solution stage 8

The cycle (4, 7) has done its bits. We'll now simplify these two cells to unique digit placements.

Down below 7 in R7C6 cancels out 7 in R1C7 leaving unique digit 4 → R1C6 4. This reduces (4,7) in R1C4 to unique digit 7 → R1C4 7.

Digit 4 in R1C6 starts a chain of hits in R9,

R9C6 5R9C5 1R9C4 4.

Digit 1 creates 5 in R6C5.

You could have made these hits earlier, but never mind, in Sudoku it happens.

More positive results obtained as,

R3C1 gets 7 as it cannot have 8 anymore → R3C2 gets 4 R2C2 gets 8.

R7C2 gets 1R8C3 gets 4R8C9 gets 1.

Results shown below.

nytimes hard sudoku 15th february 2021 solution stage 9

R8C1 gets 5 as it cannot have 7 anymore.

R8C2 gets 7, R6C2 gets 2, R5C2 gets 5 as the only remaining digit in C2.

With 9 in R9C7, R3C7 get 1 between possible digits (1, 9) → R3C8 gets 9.

With 2 in R6C2, R6C6 gets 9R4C6 gets 2.

Results shown below.

nytimes hard sudoku 15th february 2021 solution stage 10

R6C9 gets 6, R6C4 gets 1, R6C3 gets 7.

R7C9 gets 4, R7C7 gets 6, R2C9 gets 5.

By row-column scan for digit 5 on R5, C8 and C9, cell R4C6 gets digit 5.

As a result, cell R5C7 gets 7, R2C7 gets 4, R2C8 gets 7, R5C8 gets 1, R4C8 gets 4.

Results shown below.

nytimes hard sudoku 15th february 2021 solution stage 11

In column C9 digits 2 and 9 are left. As R4 has digit 2, digit 9 can only be placed in R4C9 → R4C9 9.

As a result, R4C1 gets 8, R5C1 gets 9, R5C9 gets 2, R4C4 gets 6, R5C4 gets 8, R4C3 gets 1 and the last cell R5C3 gets 6.

Final solution is shown below.

nytimes hard sudoku 15th february 2021 solution final stage 12

After the critical breakthrough when digit 3 is placed in R9C1 in Stage 4, rest didn't have any challenges left.

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.

This NY Times Sudoku puzzle is hard and well-balanced.

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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