## New York Times Sudoku hard 23rd February, 2021 Easy Solution to a tough Sudoku

New York Times Sudoku Hard February 23, 2021 Easy Solution explains step by step how the multiple breakthroughs are achieved by advanced Sudoku techniques.

### The New York Times Sudoku Hard, 23rd February, 2021

Before going through the solution solve the puzzle first.

### Step by step solution to the New York Times Sudoku Hard 23rd February, 2021: Stage 1: Breakthroughs by DSA technique and Parallel digit scan

A good number of initial valid cells obtained by row column scan.

**R2C4 4** by row column scan for 4 in top middle major square: 4 in R1, R3, C5, C6 -- **R9C2 4** by row column scan for 4 in bottom left major square: 4 in R7, C1 -- **R8C7 4** by row column scan for 4 in bottom right major square: 4 in R7, R9, C8 - **R5C3 4** by row column scan for 4 in left middle 9 cell square: 4 in R4, R6, C1, C2.

Digit 4 fully filled.

Next valid cell by row column scan for 8: R3C2 8 by scan for 8 in left middle 9 cell square: 8 in R1, C1, C3 and **R2C9 8** by scan for 8 in R1, R3.

**R5C6 8** by row column scan for 8 in central middle major square: 8 in R4, C4, C5.

With no more valid cell by row column scan we'll use possible digit subset analysis or DSA technique now.

**R2C5 3** by DSA reduction of [1,6] in C5 from possible digit subset DS [1,3,6] in R2.

The **next breakthrough valid cell is R3C9 3 by parallel digit scan for 3.**

Let's explain how this happens.

**Parallel digit scan:** Empty cells of R3 is scanned in parallel simultaneously for digit 3 appearing in the intersecting columns. As a result, cells R3C3, R3C4 and R3C5 are prohibited for 3 by digit 3 in columns C3, C4 and C5.

So only cell left in R3 for 3 is R3C9 and we get the breakthrough **R3C9 3**. This is a quick trick to earn a valid cell.

Next valid cell is by an opportunistic scan: **R1C7 1** by scan for 1 in C9 -- **R1C9 9** by exception in top right major square.

And then the never-failing DSA:

**R8C8 7** by DSA reduction of [5,6,9] in C8 from DS [5,6,7,9] in R8.

**R4C7 5** by DSA reduction of [1,2,7] in C7 from DS [1,2,5,7] in R4.

Formation of a major breakthrough Cycle (1,5,6,9) is also shown, but this will be discussed in details next stage.

Results of the actions taken shown.

### Solution to the New York Times Sudoku Hard, 23rd February, 2021: Stage 2: Breakthrough by effective Single digit lock created by a Cycle

By DS enumeration the **breakthrough Cycle of (1,5,6,9)** formed in R2C1, R7C1, R8C1 and R9C1. This Cycle prohibits placement of any of these four digits in any other empty cell of column C1.

This is the property of a Cycle.

But **as digit 9 appears in only the three digit subsets of R7C1, R7C2, R7C3** you may appreciate the fact that **digit 9 must finally occur only in one of these three cells.** It means, even though 9 appears in the DS of R7C3, effectively this 9 has to be eliminated from the DS reducing it to [1,5,6].

The digit 9 in three cells of C1 and bottom left major square in this way acts as **a single digit lock on 9.**

The

lock disallows placement of 9 in any of the empty cells influenced by the Cycle, namely, empty cells of C1 and bottom left major square.

Reduction of 9 from the DS of R7C3 gives us the **major breakthrough in R6C3 9** as the only cell in C3 where 9 can appear, all others eliminated.

This breakthrough of R6C3 9 helps to give us the **next major breakthrough single digit lock on 5** in R5C2, R6C2 by single row scan for 5 in R4.

This is because the Cycle [1,5,6,9] prohibits digit 5 in R5C1, R6C1 as well.

This lock provides the **second hard-earned breakthrough in R1C2 3** by reduction of 5 from DS [3,5] in R1C2 by the single digit lock on 5 in C2 -- **R1C1 7** by reduction -- **Cycle (2,3)** created in C1.

Now by simple scan for 7 in C1 and C2, **R4C3 7**.

By 1 in C5, reduced DS [1,2] in C4 is further reduced to the valid cell **R4C5 2** and then **R4C6 1** by exception -- **R2C6 6** by reduction -- **R2C1 1** by reduction.

Two critical breakthroughs helped to give us so many valid cells.

Formation and effect of single digit lock on 5 are shown by red arrows in the game status figure shown below.

### Solution to the New York Times Sudoku Hard 23rd February: Stage 3: Breakthrough by Single digit lock again

With 1 in R2C1, **R3C3 5** by reduction -- **R1C3 6** by reduction -- **R7C3 1** by exception in C3.

With 5 in R3C3, **R3C5 7** by reduction -- **R3C4 1** by reduction and exception in R3.

A **single digit lock** is formed in R5C4, R6C4 by scan for 7 in C5 --** R9C6 7** by scan for 7 in R7, R8, lock on 7 in C4, 7 in C5.

R8C6 5 by DSA reduction of [2,3] in R8 from DS [2,3,5] in C6 -- **R1C6 1** by reduction -- **R1C4 5** by reduction -- **R7C6** 3 by reduction -- **R8C5 9** by reduction.

**R8C1 6** by reduction -- **R5C5 5** by reduction.

Formation and effect of breakthrough single digit lock are shown by red arrows in the game status figure shown below.

### Solution to the New York Times Sudoku Hard, 23rd February 2021: Stage 4: Final solution: All valid cells obtained easily

With 5 in R5C5, **R5C2 1** by reduction -- **R6C2 5** by reduction.

**R5C4 9** by scan for 9 in R6 -- **R6C4 7** by exception in central middle square.

**R6C8 1** by scan for 1 in R5, C7, C9 -- **R6C9 2** by DSA reduction of [3,8] in C9 fromDS [ 2,3,8] in R6 -- **R6C1 3** by reduction -- **R5C1 2** by reduction -- **R6C7 8** by exception in R6.

**R7C8 8** by scan for 8 in R9, C7, C9. R5C9 7 by scan for 7 in C7, C8.

**R5C7 6** by scan for 6 in C8 -- R5C8 3 by exception in C8.

**R9C8 2** by exception in C8.

With 2 in R9C8, **R9C4 6** by reduction -- **R7C4 2** by reduction.

**R7C9 6** by scan for 6 in R9, C7 -- **R9C9 5** by exception in C9 -- **R9C1 9** by reduction -- **R7C1 5** by reduction -- **R7C7 9** by exception in R7 -- **R9C7 3** by exception in whole game.

Final solution below.

This Sudoku game is a hard Sudoku puzzle that needed multiple breakthroughs by advanced Sudoku techniques - Single digit lock, Parallel digit scan and a special form of a Cycle parented to only a major square.

Focus on early breakthroughs and avoidance of unnecessary enumeration of possible digit subsets in empty cells speeds up the solution.

Though the number of filled cells 25 in this Sudoku hard is a little high for a hard Sudoku game, the digit arrangements in this Sudoku puzzle made sure that enough challenges are posed to the solver.

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