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Nytimes Sudoku Hard Apr 4, 2024: Solution for Amateurs to Experts

NYT Hard Sudoku April 4, 2024: Easy to Understand Solution

NYT Hard Sudoku April 4, 2024 needs careful use of easy to understand Sudoku techniques based on fundamental three Sudoku rules.

First solve then learn from the solution. The puzzle and the solution should be enjoyable by both an amateur as well as an expert.

NYT Hard Sudoku April 4, 2024

This puzzle has 25 out of 81 cells filled with digits. With not so low filled to empty cell ratio (22 to 23, max 24 filled cells generally found to very hard), you may expect the Sudoku puzzle not to be very hard, but it is hard alright and needs careful solving.

NYT Hard Sudoku April 4, 2024

Solution: NYT Hard Sudoku April 4, 2024

Stage 1: Major breakthroughs

First the easy ones by simple row column digit scan: R8C9 5, R9C3 8, R8C7 8.

Naked single: R1C6 8.

Scan: R6C5 8, R2C8 8, R2C9 4, R3C9 2, R1C5 7.

Single digit lock on 1 in R9 eliminates 1 in R9C6: Cycle (6,9) in C6: Breakthrough R6C6 1.

Single digit lock on 2 in R2C1, R9C1 linked to second lock in R9C4, R7C5 linked to third lock on 2 in R7C5, R2C5.

The last cell in the third lock R2C5 shares the row with the first locked cell R2C1.

This chained single digit locks has a special property of elimination of the locked digit in the last cell of the last lock R2C5: Either of the first lock on 2 in R2C1 or R9C1 eliminates 2 in R2C5:

Critical breakthrough R2C5 1.

More next stage.

Results shown.

NYT Hard Sudoku April 4, 2024 Solution Stage 1

Stage 2: Effect of Chained single digit locks for 2: Critical breakthrough

Three single digit locks on 2 are linked in a closed chain:

First lock on 2: R2C1, R9C1 in C1.

Second lock on 2: R9C4, R7C5 in bottom middle major square. It shares row R9 with the first lock between cells R9C4, R9C1. Link between the first two locks is through the shared row R9.

Third lock on 2: R7C5, R2C5 in C5. At one end, R7C5 it is linked with the second lock. At the other end, it comes full circle and shares the row R2 with the first lock. between cells R2C5, R2C1.

Resolution: First lock possibilities:

  • 2 in R2C1 eliminates 2 in R2C5 in the third lock in C5.
  • 2 in R9C1 eliminates 2 in R9C5, results in 2 in R7C5: eliminates 2 in R2C5 in the third lock in C5.
  • For both the possible placements in the first lock in C1, 2 is eliminated in the last cell of the third lock in C5.
  • Critical breakthrough R2C5 1.

DS Reductions: R2C7 6, R1C7 3, R1C8 5, R1C4 2, R1C3 6, R7C5 2.

Parallel scan for 2 on C3: breakthrough R4C3 2.

Scan R8C4 1.

Cycles (3,4,6) and (5,9) in C4.

Reducing [3,4] because of Cycle (3,4) in R5: R5C4 6.

DS Reductions: R4C4 3, R9C4 4, R8C5 3, R8C2 6, R8C1 4, R5C5 4, R5C3 3, R3C3 5, R3C4 9, R2C4 5, R2C1 3.

Solution next stage.

Results shown.

NYT Hard Sudoku April 4, 2024 Solution Stage 2

Stage 3: All reductions

Rest reductions. Routine.

Final solution.

NYT Hard Sudoku April 4, 2024 Solution Stage 2 final

Sudoku Techniques: Based on the basic three Sudoku rules

Row column digit scan: Most basic: If a digit appears in a row and a column (or a second row) to eliminate all but one cell in the intersecting major square, the digit scanned must be placed in the single cell in the major square available for it.

DS reductions or possible digit subset reductions: The is used nearly at every step on the way to the solution. It specifically is useful for giving naked singles, hidden singles (both valid digit hits), or Cycles. DS reduction for breakthrough usually occurs when DS in one zone (say row) interacts with the existing common digits of a second intersecting zone (say another intersecting column) reducing the DS in the intersected cell to just 1. Example: DS {5,7,9] in Row R8 intersects with Column C8 containing [5,9] reducing DS of intersected cell to breakthrough R8C8 7.

Parallel digit scan: In parallel digit scan, a single digit appears in a number of rows (or columns) eliminating the cells of an intersecting column (or row) for occupancy of the digit scanned. This may leave a single cell in the affected column (or row) for the scanned digit providing a breakthrough.

Cycle: If the same set of 2 (3, 4 or 5) digits in different combinations appear in 2 (3, 4 or 5) cells of a row (or column or a major square), no other cell of the row (or column or major square) can have these Cycled digits. Example: A Cycle of (8,9) in two cells of a row debars any other cell of the row to have the digit 8 or 9.

Single digit lock: When a single digit appears in DSs of only two cells in a row (or column), the digit is locked in this row (or column) and its cells. No other cell in the affected row (or column) can host this locked digit. Usually, a single digit lock is sought for within a major square. This debars the cells of the major square from hosting the locked digit as well. For example: if digit 4 in R4 and R6 eliminates all cells of the central middle major square for 4 except R5C4 and R5C5, we get digit 4 lock in R5 and also in central middle major square. Digit 4 cannot appear in any other cell in R5 or the major square.

Single digit locks may occur also with same digit in three consecutive cells in a major square row (or column).

Rare is the single digit lock spread over more than one major square, but these may be of great value if a pair of such single digit locks happen to share two columns and two rows resulting in more valuable breakthrough digit pattern of X wing or still more powerful chained single digit locks.

Chained single digit locks: When three or more independent single digit locks on same digit are linked together in a chain by sharing rows, columns or common cells with the last two locks joined by a common cell, for both the values of the first lock, the locked digit in the last lock is eliminated.

Example: Three single digit locks on 2 form a chain: The first lock in R2C1, R9C1 in C1. Linked to the second lock in bottom middle major square cells R9C4, R7C5. Linked to the third lock in C5 cells R7C5, R2C5. Joins back to the first lock in sharing row R2 and forms a chain of three single digit locks on 2.

For 2 in both positions of the first lock, 2 in last cell R2C5 of the last lock is eliminated providing a breakthrough.


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