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How to Solve New York Times Hard Sudoku Puzzle February 16, 2021

Hard Sudoku NYtimes 16 February, 2021: Step by Step Easy Solution

New York Times Sudoku hard puzzle 16th February, 2021 Solved in easy steps

New York Times Sudoku Hard, 16 February, 2021 solved in four stages explaining all breakthroughs and how those are achieved by advanced Sudoku techniques.

The New York Times Sudoku Hard, 16th February, 2021

New York Times Sudoku Hard, 16th February 2021

Try solving the puzzle first.

Step by step solution to the New York Times Sudoku Hard, 16th February, 2021: Stage 1: Breakthroughs by DSA technique and Cycles

None of the digits could produce any success of valid cell by row column scan. This is the hallmark of a truly Sudoku hard puzzle game.

So we'll next start evaluating 2 or 3 digit possible digit subsets DSs in promising cells with many existing digits in the three interacting zones.

Just by superficial assessment of promise we land a success in the form of Cycle (1,5,8) in R7C1, R8C1, R9C1 by DSA.

Explanation on how Cycle (1,5,8) is formed

DS in promising cell R7C1 is [1,8] by reduction of [2,6,7] contributed by C1, [3,5] contributed by R7 and [4,9] contributed by bottom left major square. Altogether seven existing digit subset [2,3,4,5,6,7,9] affects or lights up the cell R7C1 so that only either of the two missing digits [1,8] can be placed in it.

[1,8] is the possible digit subset or DS in R7C1.

Difference of this cell with the other two cells, R8C1, R9C1 is the absences of digit 5 that makes the DSs for these two cells [1,5,8]. This is how the critical Cycle (1,5,8) is formed in R7C1, R8C1 and R9C1.

As a result, these three digits are locked in these three cells in the column C1 and parent bottom left major square and will disallow any of these three digits in the two parent zones, C1 and bottom left major square.

By this reduction of DSs of the affected zones, a Cycle reduces overall uncertainty in the game and often helps to produce a breakthrough.

Time saving strategy in enumeration of possible digits in empty cells of a Sudoku hard puzzle

  1. Locate the region (junction of a row, column and 9 cell major square) visually where maximum number of unique digits are filled up. These are the regions in which to start the tedious job. Number of possible digits in a cell in such a region should be minimum.
  2. Go on to complete enumeration for rest of empty cells in the promising region.
  3. Move on to the adjacent 9 cell major square.
  4. After writing possible digits in a few empty cells, have a quick look to identify a Cycle or any other useful pattern formed.
  5. In short, enumerate possible digits for empty cells with minimum number of possible digits.
  6. Stop at 4 digit possibilities. Never use 5 digit possibilities. That's counter-productive.
  7. Finally, stop enumeration when a Cycle or any useful pattern is formed.
  8. Use the positive results from the just discovered pattern as much as possible. Then start enumeration of possible digits in empty cells again.

By this strategy cell we get the Cycle (1,5,8) in C1.

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

How to enumerate digits possible in an empty cell by DSA technique

Reducing labor being an important objective, don't enumerate 5-digit possibilities or enumerate possible digits for ALL the cells.

Select relatively more promising cells in a row, a column or a 9 cell major 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 R9C4. It is in three zones or areas - row R9, column C4 and bottom middle major square. Unique set of digits in all these three together are,

$(7, 9) \cup (1, 3, 5, 6, 8) \cup (3, 5, 6, 7, 8)=(1, 3, 5, 6, 7, 8, 9)$.

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

So only the two digits (2, 4) are valid in cell R9C4. This is the possible digit subset or DS in the cell.

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

Stop in a 9 cell major square when the rest of the empty cells would all have possible digits 5 or more.

Three reasons why you should 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.

Strategy for quick enumeration,

Digit possibilities for empty cells are enumerated ONLY WHEN NECESSARY.

For example, if you get a useful pattern and a hit after enumerating possible digits for just a few cells, continue exploring the hit to get further hits leaving enumeration behind for the time being.

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

Using the reductions caused by this Cycle next we form the DS [4,9] in R1C1 and another [4,9] in R1C6 in the promising zone top middle major square with as many as 5 digits filled.

This second Cycle (4,9) gives us our first breakthrough. Let's see how.

Sensing our first valid cell breakthrough we push on to evaluate the DS of R1C4 as [3,4,9] by reduction of 1 in R1 from DS [1,3,4,9] in top middle major square -- R1C4 3 by reduction of [4,9] by Cycle (4,9).

Form now the breakthrough Cycle (1,9) in R2C5, R3C5 by reduction of 4 in C5 from DS [1,4,9] in the top middle major square. Note that though we know that R1C6 has to have digit 4, we are not destroying the Cycle (4,9) for ease of understanding.

Okay, then what does the Cycle (1,9) do?

It reduces the DS in C5 to [3,5,6,8] and creates the second breakthrough in R7C5 8 by reduction of [3,5,6] from DS [3,5,6,8].

This is followed by R8C5 5 by reduction of [3,6] from now reduced DS [3,5,6] in C5.

Rest of the two digits [3,6] form yet another Cycle (3,6) in cells R5C5, R5C6 that are in C5 as well as in the parent central middle major square. As a result we get the next valid cell R5C4 1 by possible digit analysis DSA on the cell.

Evaluating a few short length 2 digit DSs in promising cells we close the stage here. For ease of understanding we have also preserved the Cycle (1,5,8) which would give us three valid cells.

Results shown below.

New York Times Sudoku Hard, 16th February 2021 Stage 1 Solution

Solution to the New York Times Sudoku Hard, 16th February, 2021: Stage 2: Breakthrough by Cycles and DSA

First thing to do at this stage is to break the Cycle (4,9} in R1 by R1C6 4 by scan for 4 in C5 -- R9C4 4 by scan for 4 in R8, C6 -- R1C1 9 by reduction -- R2C2 8 by reduction -- R1C3 5 by reduction.

R3C3 4 by scan for 4 in C2 -- R3C2 2 by exception in top left major square.

R3C9 5 by scan for 5 in R1, R2, C7, C8.

Now we'll break the Cycle (1,5,8) by reducing 8 from DS [1,8] in R7C1 -- R7C1 1 -- R8C1 8 by reduction -- R9C1 5 by reduction -- R9C2 6 by DS reduction of [2,7] from DS [2,6,7] in bottom left major square.

R6C1 3 by reduction of 4 in R6 from reduced DS [3,4] in C1 -- R4C1 4 by exception in C1.

R6C3 1 by scan for 1 in R4, R5, C2.

Cycle (2,7) formed in R7C3, R8C3 in C3 -- Cycle (6,8) formed in R4C3, R5C3 in C3 as well as in parent left middle major square.

This reduces the possible digits DS in the rest of the three cells in left middle major square to [5,7,9] -- R5C2 5 by DS reduction of [7,9] in R5 from reduced DS [5,7,9] in C2.

New York Times Sudoku Hard, 16th February 2021 Stage 2 Solution

Solution to the New York Times Sudoku Hard 16th February: Stage 3: Breakthrough by Single digit lock

With 5 in R5C2, R5C6 8 by reduction of 5 -- R5C3 6 by reduction of 8 -- R5C5 3 by reduction of 6 -- R5C9 4 by exception in R5 -- R4C3 8 by reduction -- R4C5 6 by reduction -- R4C9 3 by reduction.

R6C6 5 by scan for 5 in R4, C4.

Cycle (2,9) formed in C6 cells R4C6 and R7C6 reducing 2 in DS [1,2] in R9C6 -- R9C6 1.

R7C7 4 by scan for 4 in R8, R9, C9.

Even at this late stage breakthrough by single digit lock on 1 in R2C7, R3C7 -- R8C9 1 by scan for 1 by R7, R9, lock on 1 in C7 and 1 in C8.

See how the allowed cells for 1 are reduced to a single cell by the scan indicated by arrows.

How a Single Digit Lock is formed and its effect

An effective single digit lock is formed by scan for the digit on a row, column or usually by a cross-scan of row and column. In such a single digit lock on 1 in this game, the digit 1 can appear in the final solution only in either of these two cells in column C7.

Digit 1 is thus locked in C7 as well as parent top right major square thus disallowing digit 1 in any other empty cell of these two parent zones C7 and top right major square.

A single digit lock creates a breakthrough either by reducing a critical digit from the DS of a cell or as in this case, by participating in a row column scan for the digit.

Breakthrough in this game by the single digit lock is R8C8 1 by scan for 1 in lock in C7, 1 in C8 and 1 in R7, R9.

Results shown below.

New York Times Sudoku Hard, 16th February 2021 Stage 3 Solution

Solution to the New York Times Sudoku Hard, 16th February 2021: Stage 4: Final solution: Single remaining breakthrough by DSA technique

Possible digit subset in C9 and R9C9 [6,7,8,9] reduced by [6,7,9] in R9 to give the first breakthrough at this stage by DSA -- R9C9 8 -- R6C9 6 by reduction -- R1C9 7 by reduction -- R7C9 9 by exception in C9 as well as by reduction.

With 9 in R7C9, R7C6 2 by reduction -- R8C4 9 by reduction -- R7C3 7 by reduction -- R8C3 2 by reduction.

With 2 in R7C6, R4C6 9 by reduction -- R4C2 7 by reduction -- R6C2 9 by reduction.

With 7 in R4C2, R4C4 2 by reduction of 7 -- R6C4 7 by reduction.

R8C7 7 by scan for 7 in R9, C8 -- R8C8 6 by exception in R8.

With 6 in R8C8, R1C7 6 by scan for 6 in R2, R3, C8 -- R1C8 8 by exception in R1.

R2C8 4 by scan for 4 in R3, C7 -- R2C7 1 by reduction -- R2C5 9 by reduction -- R3C5 1 by reduction and exception in top middle major square.

With 1 in R3C5, R3C7 3 by reduction -- R3C8 9 by reduction and exception in R3.

R6C7 8 by scan for 8 in C8 -- R6C8 2 by exception in R6 -- R9C8 3 by exception in C8 --R9 C7 2 by exception in the whole game.

Final solution below.

New York Times Sudoku Hard, 16th February 2021 Stage 4 Final Solution

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 just-so hard with as low as 23 filled cells.


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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By the way, Sudoku hard solution techniques are included with many of the solutions.

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