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

New York Times Sudoku Hard February 15th, 2021 Solved

New York Times Sudoku Hard 15th February, 2021 Solved in quick easy steps

Solution to New York Times Sudoku hard, 15th February, 2021 explains step by step how to use Sudoku hard techniques to achieve quick breakthroughs.

The New York Times Sudoku Hard, 15th February, 2021

New York Times Sudoku Hard February 15th, 2021

It is a fairly hard Sudoku puzzle. To solve quickly you need to concentrate.

Solution to the New York Times Sudoku Hard, 15th February, 2021: Stage 1: Early breakthroughs by Single digit lock and parallel scan

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 number of times and select a promising 9 cell major square for row column scan on its empty cells. Identify the rows and columns with the digit that intersect in a cell in the the 9 square major square.

With this basic concept in place start row column scan for the lowest digit 1 and continue the process till a digit is finished or ends in a block of no more success. Then start the process for the next higher digit.

For easier lower level Sudoku games this systematic method of row column scan should provide a smooth path to the final solution. But when you face a Sudoku hard, valid cell by row column scan may be few or non-existent to start with.

That's why though we start a Sudoku hard solution with systematic row column scan, we also look for other advanced digit patterns for early breakthroughs of valid digits.

In this game having no success with 1 or 2 by row column scan select digit 3 as promising and get a few easy placements.

R2C3 3 by row column scan for 3 in R1, R3, C2 -- R6C7 3 by scan for 3 in R4, R5, C9 -- R8C8 3 by scan in C7, C9.

No more easy valid digit 3 possible at this point.

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

An easy catch by opportunistic scan for 9: R9C9 9 by scan in R7, R8.

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

So look for useful digit patterns.

Identify Single digit lock on 6 in R7C7, R7C9 by scan for 6 in R8 -- this lock on 6 participates in next valid cell R9C2 6 by scan for 6 in R7 by the lock, 6 in R8 and 6 in C1.


A note on single digit lock on 6

An effective single digit lock is formed invariably by scan for a digit in a single column or row or more frequently by cross-scan over a row and a column.

In this case observe that by scan for 6 in R8, digit placement of digit 6 is restricted to only two cells in R8, R8C7 and R8C9. In the final solution, 6 can appear in only one of these two cells and in no other cell of the parent bottom right major square or the parent row R8. Reduction of instances of 6 from all such empty cells is the gain.

Alternately this lock on 6 acts as if 6 actually exists in R8 and that's why a lock on a digit can be used for a scan for the digit.

In this case we get the breakthrough of R9C2 6 by scan for 6 in R7, R8 and C1, 6 in R7 being actually a lock on digit 6.

A single digit lock is always a valuable asset for hard Sudoku simplification and whenever yoi identify such a possibility, just note it in your mind for future use at the right timer. Never let go of an effective single digit lock.


Next identify the advanced technique of Parallel scan for 8 on empty cells of R8: 8 in bottom left major square eliminates R8C1, R8C2, R8C3 for 8 and 8 in C9 eliminates R8C9 for 8 leaving the single cell R8C7 for 8 -- R8C7 8.

Note: For more details, you may click on the section link on "How the parallel digit scan works" above and return by clicking on browser back button.

Identify a second Single digit lock on 5 in R9C5, R9C6 by scan for 5 in R7, C4 -- R9C1 3 by possible digit DS elimination of 5 (because of single digit lock on 5 in R9) from possible digit subset [3,5] in R9C1 (because of [1,4] in C1). This is a breakthrough by single digit lock and DSA.

R7C4 3 by scan for 3 in R9, C5, C6.

Result of actions shown below.

New York Times Sudoku Hard February 15th, 2021 solution Stage 1

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


DSA technique - Enumerating digits possible for an empty cell: If number of digits possible is 1, we get a valid cell hit

After speed of solution, reducing labor being the main objective, we won't enumerate 4 or 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 major square. Unique set of existing 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 missing digits (2, 8) are valid for the cell R2C5.

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

We stop in a 9 cell major square when the rest of the empty cells would all have digit possibilities 4, 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.

If by DSA reduction we get a single digit possible for a call, we get a valid cell hit. This is a powerful way to get a valid cell. For example, R5C5 2 by DSA reduction of [1,5] in the parent top middle major square itself from the possible digit DS of [1,2,5] in C5.

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


Solution to the New York Times Sudoku Hard, 15th February, 2021: Stage 2: Breakthrough by DSA technique and Cycles

Cycle (1,4,5) formed in R9 in cells R9C4, R9C5, R9C6 -- DS for two empty cells of bottom middle major square [7,8] -- R7C5 8 by reduction of 7 by 7 in C5 -- R7C6 7 by exception in bottom middle major square as the only possible digit left for the only truly empty cell.

R7C1 2 by DS reduction of [1,4,6] in C1 from possible digit subset DS [1,2,4,6] in four empty cells of R7.

R3C5 6 by DSA reduction of [1,2,5] from possible digit subset DS [1,2,5,6] in empty cells of C5 and hence in R3C5 -- R2C5 2 by DSA reduction of [1,5] from DS [1,2,5] in C5.

Note: To know more on DSA reduction technique, click on the above internal link and return (after going through it) by clicking on browser back button.

R1C7 2 by scan for 2 in R2, R3, C8 -- R1C8 6 by scan for 6 in R2, R3.

Cycle (4,7) in R1C4, R1C6 in top middle major square and in R1 by reduction of [8,9] in R1 from possible digit subset [4,7,8,9] in top middle major square -- R1C3 5 by exception in R1 as the only cell left for 5.

Game status at this point shown below.

New York Times Sudoku Hard February 15th, 2021 Solution Stage 2

Solution to the New York Times Sudoku Hard, 15th February, 2021: Stage 3: Critical Breakthrough by DSA reduction caused by a Cycle

With 7 in R7C6, R1C6 4 by reduction -- R9C6 5 by reduction -- R9C5 1 by reduction -- R9C4 4 by reduction -- R6C5 5 by reduction and exception in C5 -- R1C4 7 by reduction and exception in R1.

Single digit lock on 4 in R2C2, R3C2 by scan for 4 in C1 -- DS in R7C2 [1,4] is reduced by 4 by the lock on 4 in C2 resulting in R7C2 1 -- R8C9 1 by scan for 1 in R7 -- R8C3 4 by scan for 4 in C1, lock on 4 in C2.

Leftover three digits in C3 form a Cycle (1,6,7) in R4C3, R5C3, R6C3. This will result in a series of valid cells and is a major breakthrough.


A note on Cycle (1,6,7)

By possible digit analysis we have the DSs in R4C3, R5C3 and R6C3 as [1,6], [1,6,7] and [1,6,7]. Three digits appear in in the column in only these three cells and nowhere else. We can say with 100% certainty that these three digits will be placed in the final solution in only these three cells and nowhere else.

It is as if the digits Cycle between the cells: if R4C3 1, 6 and 7 can appear in either of the cells R5C3, R6C3. Alternately, if R4C3 6, digut [1,7] will appear in eithe of the cells R5C3, R6C3.

Because of this locked status of the digits in the cells involved in the Cycle, all occurrrences of these locked digits in zones containing the Cycle are reduced.

For example, in this case, reduction of 7 from DS of R6C2 gives us a valid cell hit as R6C2 2. This is a major breakthrough and we'll see its effects in the next stage.


Results shown below.

New York Times Sudoku Hard February 15th, 2021 Solution Stage 3

Solution to the New York Times Sudoku Hard, 15th February, 2021: Stage 4: No more breakthroughs needed - All are easy valid cells

R6C2 2 by reduction of 2 by the Cycle (1,6,7) in left middle major square -- R6C6 9 by reduction -- R3C6 8 by reduction -- R2C4 9 by reduction -- R3C1 7 by reduction -- R8C1 5 by reduction -- R8C2 7 by reduction.

R2C2 8 by scan for 8 in R3 -- R3C2 4 by exception in top left major square.

R5C2 5 by exception in C2 -- R6C9 6 by DSA reduction of [1,7] in C9 from DS [1,6,7] in three empty cells of R6 -- R6C4 1 by reduction of 7 from reduced DS [1,7] in R6 -- R6C3 7 by exception in R6 and by reduction.

With 6 in R6C9, R7C9 4 by reduction -- R7C7 6 by reduction and exception in C7.

R4C6 2 by exception in C6.

R5C9 2 by scan for 2 in R4, C7, C8.

R4C9 9 by parallel scan for 9 on empty cells of C9: 9 in R2 eliminates R2C9 for 9 leaving R5C9 for 9 -- R2C9 5 by reduction -- R4C7 5 by scan for 5 in R5, C8.

With 9 in R4C9, R4C1 8 by reduction of 9 from possible digit subset of [8,9] in C1 -- R5C1 9 by exception in C1.

R5C4 8 by scan for 8 in R4 -- R4C4 6 by exception in C4 -- R4C3 1 by reduction -- R5C3 6 by exception in C3 -- R4C8 4 by exception in R4. R2C8 7 by DSA reduction of 4 in C8 from DS [4,7] -- R2C7 4 by exception in R2 -- R3C8 9 by scan for 9 in C7 -- R3C7 1 by exception in R3 -- R5C7 7 by exception in C7 -- R5C8 1 by exception in R5.

Final solution below.

New York Times Sudoku Hard February 15th, 2021 Solution Stage 4 final


Advanced Sudoku technique for finding a valid cell : Parallel scan for a digit on empty cells of a row or column

Parallel scan Sudoku technique

We have detected the possibility of getting a valid cell R2C7 8. But how?

This is the quick method of parallel scan for digit 8 on empty cells of the row R8.

Observe how three empty cells of R8C1, R8C2, R8C3 are disallowed for digit 8 by the presence of digit 8 in the bottom left major square.

With this, two more empty cells R8C7 and R8C9 are left for 8 in R8.

But the cell R8C9 is again disallowed for digit 8 by the presence of 8 in C9 and the only cell left in R8 for digit 8 is the cell R8C7.

This is how we get a breakthrough valid cell R8C7 8 by parallel scan.

This is similar ro normal row column scan which isĀ  carried out on the empty cells of a major square. Instead, the parallel digit scan is carried out on the empty cells of a row or a column.

The other important property of a parallel digit scan is, it will always be associated with a resultant Cycle in rest of the cells after you get the valid cell.

For example, in this case, after you place 8 in R8C7, it will be easy for you to identify a Cycle of (1,4,5,7) in rest four empty cells without specifically evaluating the possible digit subsets in these cells.

You may perhaps appreciate that, if you have proceeded to form first the possible digit subsets in these four cells and then identified the formation of the Cycle resulting in the valid cell R2C7 8, it would have taken much longer.

That's why for speed, we prefer to identify parallel digit scan possibilities.

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 with only 23 filled cells is fairly 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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