## NYTimes hard Sudoku puzzle 16th February, 2021 Solved in easy steps

The detailed solution of NYTimes hard Sudoku puzzle, 16th February, 2021 explained step by step in simple language bringing out how to use special patterns.

### The hard Sudoku puzzle NY Times, 16th February, 2021

It is a hard puzzle you'll like solving.

### Solution to the the hard Sudoku puzzle NY Times, 16th February, 2021

Use row-column scan, the simplest method to get a unique digit for a cell. Exhaust all possible finds by this method quickly.

But, for this hard Sudoku none of the 9 digits could give us our first simple digit placements by row-column scan.

This is expected in a hard Sudoku puzzle.

So take up the next essential job of **enumerating the digits possible for all empty cells.**

It's a tedious and boring task, but must be done WITHOUT ANY ERROR.

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

- Locate the region (junction of a row, column and 9 cell 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.
- Go on to complete enumeration for rest of empty cells in the promising region.
- Move on to the adjacent 9 cell square.
- After writing possible digits in a few empty cells,
**have a quick look to identify a Cycle or any other useful pattern formed.** - In short, enumerate possible digits for empty cells with minimum number of possible digits.
- Stop at 4 digit possibilities. Never use 5 digit possibilities. That's counter-productive.
- Finally, stop enumeration when a Cycle or any useful pattern is formed.
- 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 R2C5 gets (1,9) first.

Rest three cells of this promising 9 cell square filled up before we move on the adjacent 9 cell square on top right.

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 each empty cell

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 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,

$(2, 3, 5, 6, 7) \cup (2, 4, 7) \cup (2, 5, 6, 7, 8)=(2, 3, 4, 5, 6, 7, 8)$.

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 (1, 9) are valid in cell R2C5.

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

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

Three reasons why you should stop,

- Writing long digit sets takes more time and effort,
- Chance of getting a useful pattern from a 5 digit possibility in a cell is remote, and,
- 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.

When the three cells R7C1, R7C2 and R7C3 are enumerated we get out **first breakthrough.**

The

Cycle (1,5,8)is formed in these three cells.

Results shown below.

Digits (1,5,8) being locked up in these three cells, digit possibility (4,5,8,9) in R1C1 is simplified to (4,9). It cannot have 5 or 8. This forms a second Cycle (4,9) in R1C1, R1C6 in top row.

And this Cycle gives **first hit of the day** → **R1C4 3** when digits (4,9) are **removed** from (3,4,9).

Next form the **third Cycle of (1,9)** in R2C5 and R3C5 in column C5.

Look at the cell R7C5 down below. The Cycle of (1,9) in this column gives us second hit → **R7C5 8** followed by **R8C5 5**.

And these last two will decide digits 1, 8, and 5 respectively in R7C1, R8C1 and R9C1. We'll show this effect in next stage for convenience.

Results till now shown below.

**R7C1 1 **→ **R8C1 8 **→ **R9C1 5.**

The Cycle (1,9) further eliminates 9 from R1C6 → *R1C6 4 and R1C1 9 → R2C2 8 → R1C3 5 → R3C2 2 → R3C3 4.*

4 in row R6 → **R6C1 3** → **R4C1 4.**

4 in R4, R6 → **R5C9 4.**

3 in R6 → *3 is locked in R4C9.*

Results shown below.

By the **Cycle (1,3,9)** in R3C5, R3C7, R3C8 → **R3C9 5.**

3, 6 left in C5 cells R4C5, R5C5.

With 3 in *R4C9 → R4C5 6 → R5C5 3.*

Forgot to cancel 2 from (2,6) in *R9C2 → R9C2 6 →R5C2 5 because of Cycle (7,9) in R4C2 and R6C2.*

Results shown.

Cycle (2,7) in C3 → digits (1,6,8) left for three empty cells in C3 → **R4C3 8** → The empty cells in the central 9 cell square are surrounded by cells with single digits.

These are promising.

And that's right → *R5C4 1 → R5C6 8 → R5C3 6 → R6C3 1.*

Out of digits (5,7,9) left in in R6 → (7,9) in R6C4 creates a Cycle of (7,9) →** R6C6 5,**

Results shown now for better understanding.

Cycle (2,9) formed in R4C6, R7C6 → **R9C6 1.**

R9 and C9 together have the digits 1,2,3,4,5,6,7 and 9, only 8 missing → **R9C9 8 **→ *R6C9 6 → R1C9 7 → R7C9 9 → R8C9 1.*

Results shown below.

With 9 in R7 → *R7C6 2 → R4C6 9 → R4C2 7 → R6C2 9 → R6C4 7 → R4C4 2 → R8C4 9 → R9C4 4.*

2 in *R7C6 → R7C3 7 → R8C3 2.*

4 is the only digit left in R7 → R7C7 4 → R2C7 1 → R2C5 9 → R3C5 1 → R3C7 3 → R3C8 9 → R2C8 4.

Results shown. Next stage will see the end.

3 in R8, C7 → R9C8 3 → R9C7 2, the only missing digit in R9.

2 in R9C7 → R8C8 6 → R8C7 7.

6 in R8C8 → R1C8 8 → R1C7 6.

8 in R1C8 → R6C8 2 → R6C7 8.

Final solution below.

Identifying and using series of Cycles helped quick 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 belowshould 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.

#### Suggestion

For full enjoyment, avoid looking into any solution as well as the answer.

The joy of discoveries will then all be yours.