NYT hard Sudoku Feb 20, 2021 - Simple Solution for Beginners
Beginner's guide to solving NYT hard Sudoku Feb 20, 2021 explains the solution in simple language. How to make the breakthroughs in steps shown clearly.
The solution is in simple language with step by step details.
NYT hard Sudoku, Feb 20, 2021
You should target to solve this hard Sudoku puzzle in quick time.
Solution to NYT hard Sudoku, Feb 20, 2021
Identified single lock digit of 1 in R4C8, R5C8 while doing row-column scan for 1 → scan 1 in R6, C9.
To skip detailed explanation of how to identify and use single digit lock pattern click here.
Conditions for single digit lock - how to identify it
Two conditions for single digit lock pattern,
- the digit can be placed in only two or three cells of a column or a row, AND,
- the locking cells must also be in SAME 9 cell square.
- 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.
Digit 1 not only can appear ONLY IN TWO CELLS R4C8, R5C8 of column C8, the cells are in the SAME 9 cell square R4R5R6-C7C8C9. This digit 1 lock satisfies the third condition as well because the right bottom adjacent 9 cell square doesn't have locked digit 1 as a pre-filled digit.
Even if the third condition is not satisfied, single digit lock may be present in a 9 cell square, but that cannot be used for any positive outcome.
Positive effects of single digit lock
Two positive effects of a single digit lock pattern,
- the locked digit CANNOT APPEAR IN ANY OTHER CELL OF THE 9 CELL SQUARE, and,
- also CANNOT APPEAR IN ANY OTHER CELL OF THE LOCKED COLUMN or ROW.
This reduces the number of possible digits in the corresponding cells, generally resulting in a breakthrough.
While on row column scan for 8, the second single lock on digit 8 in same two cells R4C8, R5C8.
This lock will disallow 8 in R1C8, R2C8 and lock on 1 will disallow 1 in the single cell R7C8.
Digits possible in R7C8 → (3,5): the possible 1 in this cell is no longer possible because of the single digit lock of 1.
More important is elimination of digit 1 in cell R8C8 by the single digit lock of 1.
Digits possible in R7C7 → (1,5,7); in R7C9 → (3,7), in R9C7 → (1,5,7), in R9C9 → (2,3,7).
A Cycle of (1,2,5,7) created in R7C7, R7C8, R7C9 and R9C7 → R9C9 2.
This is the first result of the breakthrough.
Reason of the Cycle: With 3 or 5 in cell R7C8, Cycles (1,5) or (1,7) will be in R7C7, R9C7 and 7 or 3 in R7C9 respectively.
These are the only valid combinations of these digits. These cannot then appear outside these four cells, and the Cycle is formed. With 1 in R8C8 the Cycle won't have formed as well as there won't have been any breakthrough.
We'll show more simplification next stage. Results below.
By single digit locks, digits 1, 8 can appear in 9 cell square in only two cells R4C8 and R5C8.
These two form a Cycle of (1,8). No other digits can appear in these two cells.
This makes 4 as the only digit that can be placed in R6C9 → R6C9 4.
This is the second result of the breakthrough caused by two single digit locks and subsequent Cycle.
Next (5,7,9) are the digits left in R4C8, R5C8 and R6C7.
Also 7 in C8 and 9 in R6 → R6C8 5 → R6C7 7 → R5C9 9 → R3C9 3 → R7C9 7.
A Cycle of (1,5) formed in R7C7, R9C7 → R7C8 3 → R2C8 2 → R1C8 9.
Only 8 is possible in R2C1 → R2C1 8 → R2C7 6 → R2C4 3.
A new cycle is formed in R6C3, R6C4 → R6C5 3.
Lots of positive outcome of the breakthrough.
We'll see more in next stage.
Results till now shown below.
Row column scan on 3: R7, C4, C5 → R8C6 3.
This creates a new Cycle (2,6) in R6C3, R8C3 → R3C3 4 → R3C7 8 → R1C7 4.
Row column scan on 7 → 7 in C4, C6 → R5C5 7.
6 in C1, R9 and Cycle (2,6) in C3 creates single digit lock on 6 in cells R8C2, R8C3.
Digits possible in R8C5 are (1,5) → Cycle (1,5) formed in C5 in cells R3C5, R8C5 → R9C5 8 → R1C6 6.
This is the second breakthrough.
Row column scan on 8 → 8 in R3, C5, C6 → R1C4 8.
8 in R8, R9, C1 → R7C3 8.
More simplification in next stage.
Results shown below.
Cycle (2,4,6) in left middle 9 cell square → Cycle (7,8) in R4C2, R4C3 → R4C8 1 → R5C8 8 → R4C6 5 as the only digit left in the cell.
→ R1C6 2.
Cycle (3,5,7) form in R1 and top left 9 cell square → R3C2 2 → R5C2 6 → R8C3 6 → R6C3 2 → R6C4 6.
In R4, 4 is the only possible digit → R4C4 4.
Rest in the next final stage.
In R7 because of Cycle (1,5), the only digit left for the two empty cells are (6,9).
With 6 in C4 → R7C4 9 → R7C6 6.
Only digit 1 possible in R5C6 → R5C6 1 → R3C6 9.
Only digit 2 possible in R5C4 → R5C4 2 → R3C1 4.
With 8 in R7C3 → R4C3 7 → R4C2 8.
With 7 in R4C3 → R1C3 3 → Only digit possible in R8C2 5 → R8C1 2 → R7C1 1 → R7C7 5 → R9C7 1.
With 5 in R8C2 → R8C5 1.
With 5 in left bottom 9 cell square, only digit possible in R9C1 7 → R1C1 5 → R1C2 7.
With 1 in R8C5 → R3C5 5 → R3C4 1.
As the only digits left in the cells, R9C2 3, R9C3 9, R9C4 5.
Final solution shown below.
Repeated use of multiple single digit locks right from the beginning results in quick breakthroughs.
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.
For full enjoyment, avoid looking into any solution as well as the answer.
The joy of discoveries will then all be yours.