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NYT hard Sudoku 23 Feb 2021: Solution for Beginners

NYT hard Sudoku 23 Feb 2021 Solution for Beginners

NYT hard Sudoku 23rd February, 2021: Easy to understand Solution in details

In this solution to NYT hard Sudoku Feb 23, 2021, each step is explained for even a beginner to understand. Hard Sudoku not any more beyond reach.

How to find the valid single digit for empty cells and achieving breakthroughs by Sudoku techniques shown.

First part is on meaning of a few terms used and the more frequently used four Sudoku techniques to solve a hard Sudoku puzzle without much effort.

Solution is the second part.

To skip the section on Sudoku techniques, click here.


Terminologies used in explaining solution to the NYT hard Sudoku

Row column labels and names for 9 cell squares

The 9 by 9 Sudoku game has 81 cells in 9 rows and 9 columns.

Rows are labelled R1 to R9 and columns C1 to C9.

81 cells are further divided into 9 groups of 9 square cells. These are either referred as top - left, middle, right; bottom - left, middle, right and left middle, central middle and middle right.

These 9 square cell groups are also referred by their specific labels joining the labels for the three rows and three columns. For example the specific label for the left middle 9 cell square will be R4R5R6-C1C2C3.

Convention followed is to mention row label suffixed by the column label. For example the top leftmost cell is referred as R1C1.

Sudoku rules

9 by 9 Sudoku has three conditions for a digit to occupy a cell,

  1. No digit can be repeated in any row. Every row must have the 9 digits 1 to 9 without repetition.
  2. No digit can be repeated in any column. Every column must have the 9 digits from 1 to 9 without repetition.
  3. Each of the 9 numbers of 9 cell squares must be filled up by exactly 9 digits from 1 to 9 without repeating any digit.

Sudoku techniques to make a breakthrough using specific patterns in the digits

Row column scan for a digit

This is the simplest for identifying a cell where only one digit CAN be placed. No other cell in the 9 cell square, the row or the column containing this specific cell would have this VALID digit except this specific cell that you have found.

For example, if digit 5 appears in R1 (in top left 9 cell square), R2 (in top right 9 cell square), and C4, C5, the only valid cell where 5 can be placed in R3 and top middle 9 cell square will be R3C6.

Possible digit subset (DS) in a cell

For every empty cell there will be a subset of the 9 digits each of which will be a potential candidate for the cell. This is the possible digit subset or DS for a cell.

In the NYT hard Sudoku Feb 22, 2021, DS for R2C9 is [3,8] as these two digits do not appear in the parents: top right 9 cell square, the parent row R2 and the parent column C8. In the 9 cell square, DS is [1,3,8,9]. Cell R2C9 is in two more affecting areas, the row R2 that has DS [2,5,7,9] and the column with DS [1,4]. These two together has 1, 9 extra to what the 9 cell square has. These two cannot appear in R2C9. Its DS reduces from [1,3,8,9] to [3,8]

Digit subset analysis or DSA

When no valid cell by row column scan is visible, try to find a cell for which DS has a single digit. The cell will then have a valid digit and complexity of the game will reduce a little.

As a strategy, DS is evaluated when promise of a simplifying pattern is expected. DS is generally evaluated in heavily filled up zones to keep its length short. This speeds up solution and reduces labor.

This technique is still easy and is the savior oftentimes.

Cycle of digits

If 2 digits, say [4,6], appear as a lone pair in 2 cells (without any other digits possible in the two cells) of a 9 cell square, a column or a row, a Cycle of 2 digits is formed. These two digits have to be placed in either of these two cells in the Cycle and in no other cell of the parent zones 9 cell square, the column or the row containing the lock. (a ZONE is a 9 cell square, a row or a column).

The two digits can then be eliminated from all the DSs of the parent zones of the Cycle.

A Cycle of 3 digits in 3 cells are encountered often. 4 cell Cycles can also be useful if it can be located.

Cycles are valuable patterns to identify.

Single digit lock - Conditions for single digit lock - how to identify it

Two conditions for single digit lock pattern,

  1. the digit can be placed in only two or three cells of a column or a row, AND,
  2. the locking cells must also be in SAME 9 cell square.
  3. the locked digit should not be present as a single cell candidate in both the adjacent two 9 cell squares through which the locked column or row passes.

The following shows and example of single digit lock of 5 in cells R7C1 and R9C1.

sudoku single digit lock: what it is and what it does

How a single digit lock is formed

Look at columns C1, C2 and C3 in the bottom left 9 cell square R7R8R9-C1C2C3. Out of 3 empty cells, the cell R7C3 is debarred for placing digit 5 as column C3 has a 5 and it lights up the cell for digit 5.

5 can appear only in two cells in column C1, R7C1 and R9C1 and in no other cell in the 9 cell square or the column C1.

It is locked inside these two cells in C1 and 9 cell parent square.

How a Sudoku single digit lock is used - What it does

The locked digit 5 eliminates itself from the DSs of the other two empty cells R5C1 and R6C1 and a new Cycle (2,3) is created in C1.

Focus again on the bottom left 9 cell square. With Cycle (2,3) in C1, another Cycle (5,9) is formed in the two cells of the 9 cell square. As a result, digit 1 becomes the only digit left and cell R7C3 only cell left for it.

Still more happens. With 1 in C3 now, digit 9 now must occupy the cell R6C3.

These two single digit candidates obtained by the single digit lock of 5 affects other cells and breaks the bottleneck.

As a strategy, always form a single digit lock as soon as it is discovered.

It is a powerful digit pattern. Even if its effect is not immediate, it should have positive results later.


The NYT hard Sudoku Feb 22, 2021

nytimes hard sudoku 23rd february 2021

Solution to the NYT hard Sudoku Feb 23, 2021 for beginners

Row column scan for 4 in top middle 9 cell square → 4 in R1, R3, C5, C6 → R2C4 4.

Row column scan for 4 in bottom left 9 cell square → 4 in R7, C1 → R9C2 4.

Row column scan for 4 in bottom right 9 cell square → 4 in R7, R9, C8 → R8C7 4.

Row column scan for 4 in left middle 9 cell square → 4 in R4, R6, C1, C2 → R5C3 4.

Row column scan for 8 in left middle 9 cell square → 8 in R1, C1, C3 → R3C2 8 → R2C9 8.

Row column scan for 8 in central middle 9 cell square → 8 in R4, C4, C5 → R5C6 8.

DS for R2, R2C5[1,3,6] and [1,6] in C5 → R2C5 3.

Cycle (1,5,7) formed in R3 as 3 in C3, C4 and top middle 9 cell square → R3C9 3, the only digit left in the row. Digits left in the top right 9 cell square [1,9] and 1 in C9 → R1C9 9 → R1C7 1.

By digit subset analysis or DSA on R8C8 → R8C8 7 as the only possible digit to place.

DSA on R4C7 → R4C7 5 as R4, C5 together has rest 8 digits.

With no more row column scan, DSA, Cycle or single digit lock available for breaking through the deadlock, DS enumeration for some of the empty cells done.

Further analysis next stage.

Results shown below.

NYT hard Sudoku 23 Feb 2021 Solution for Beginners Stage 1

DS enumeration started from smallest [1,6] in R2.

Identified that [1,6] in R2C1 is a smaller portion of DS [1,5,6,9] in bottom left 9 cell square as well as its three empty cells in column C1 sharing [1,6].

With four digits in four cells a Cycle will be created.

Indeed the Cycle [1,5,6,9] is created in cells R2C1, R7C1, R8C1, R9C1.

This four cell Cycle eliminates all these digits from rest of the three empty cell DSs in C1 and creates a second Cycle [23,7].

But its major effect is not creation of the second Cycle, but locking digit 9 in single column C1 in bottom left square. The effect is same as a scan for digit 9 in left middle square in C1.

With 9 in C2 and R4, only the cell R6C3 is left for 9 → R2C3 9.

This is the most important first breakthrough.

How the breakthrough achieved is shown below at this stage so that you can check its logic.

Check this carefully while we move on to next stage.

Results till now shown below.

NYT hard Sudoku 22 Feb 2021: Solution for Beginners Stage 2

Even after the breakthrough of single digit candidate 9, finding more hits is not easy.

But now if you examine the sets of DSs, you will find a number of patterns to use for positive results.

First our already known single digit clock on 9 in C1.

At this moment more important is though the single digit lock of digit 5 in C1 in the same three cells R7C1, R8C1, R9C1.

With 5 in R4, this lock of 5 in C1 creates a second lock on 5 in R5C2, R6C2.

Otherwise it is visible but it may be overlooked. This has been the target all along.

By this single digit lock on 5, DS [3,5] in R1C2 reduces to single digit candidate 3 and this is the real effect of the breakthrough.

R1C2 3 is the breakthrough that should open the gates for a number of single digit candidates.

Any single digit candidate obtained from a two digit DS is important, especially in the early stage of the game.

Again the results are shown and we'll move on to next stage.

Results shown below.

NYT hard Sudoku 23 Feb 2021: Solution for Beginners Stage 3

Due to lock on 5 in left middle 9 cell square cells of C2, R1C2 3R1C1 7.

Scan 7 in C1, C2 → R4C3 7 → R4C5 2 → R4C6 1 → R2C6 6 → R2C1 1.

It creates a Cycle (5,6,9) in C1 cells of bottom left 9 cell square → R7C3 1 → R3C3 5 → R1C3 6.

With R2C6 6, 1 in R3C4 became single digit candidate in the top middle square → R3C4 1 → R3C5 7, being the lone digit.

Even at this late stage, a single digit lock on 7 is formed in R5C4, R6C4.

With 7 in R7, R8, and effectively in C4 → R9C6 7 → R8C6 5 by elimination of 2,3 in R8 from column DS [2,3,5] → R1C5 2 → R1C4 5 → R7C6 3 → R8C5 9 → R8C1 6 → R5C5 5.

Rest next stage. Results shown.

NYT hard Sudoku 23 Feb 2021: Solution for Beginners Stage 4

With 5 in R5C5 → R5C2 1 → R6C2 5.

With 9 in R6C3 → R6C4 7 → R5C4 9 → R5C9 7 being the lone digit → R6C8 1 being the lone digit → R6C7 8 being the lone digit → R6C9 2 → R6C1 3 → R5C1 2.

R5C8 3 → R5C7 6 → R7C7 9 → R9C7 3 → R7C1 5 → R9C1 9 → R9C8 2 → R7C8 8 → R9C4 6 → R7C4 2 → R9C9 5 → R7C9 6.

END.

Final fully filled Sudoku shown. Check for no digit repeated in a row, column or a 9 cell square.

nytimes hard sudoku 23rd february 2021 beginners solution stage 5 final

The hard Sudoku gave away easy single digit candidates at the start but then it raised a block that forced use of possible digit enumeration for a number of cells.

The DSs enumerated for selected cells formed a not so common 4 digit Cycle for breakthrough.

And the puzzle resisted total crash down quite late in the game.

It is a fine hard Sudoku puzzle.

If you are still reading this and not so experienced in solving hard Sudoku puzzles, just leave aside the solution and solve the puzzle yourself.

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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