The specialty of this Cycle is - it belongs to only the single parent of A MAJOR SQUARE and to no other parent row or column. These four digits are locked and are self-sufficient in these four cells, each appearing at least twice in these four cells. It's a pleasant surprise to identify that a Cycle of is formed in the four cell DSs in R7C1, R8C1, R8C2 and R8C3. We decide to continue to form DSs in empty cells of the promising zone bottom left major square.ĭS in R8C2 by reduction of 1 in C2 from DS in R8 and DS in R8C3 by reduction of in C3 from DS in R8. How the Cycle (1,2,5,6) is formed in only the bottom left major squareĭS in C1 reduced by in R7 to form DS in R7C1, reduced by 7 in R8 to form DS in R8C1 and reduced by 2 in R9 to form DS in R9C1. We'll close here and explore the rest later. This reduces from DS to produce a major breakthrough of R9C1 7. In the process, an awkward shaped Cycle (1,2,5,6) in bottom left major square formed. With much of the empty cells filled up with valid digits, short length possible digits are evaluated conveniently in promising zones by possible digit analysis technique. R4C4 4 by scan for 4 in R5, R6, C6 - R4C6 5 by exception in R4 - R5C5 7 by scan for 7 in C4, C6. R4C8 1 by reduction of 8 from DS - R5C8 8 by exception in right middle major square. Solution to the New York Times Sudoku Hard, 20th February, 2021: Stage 2: A major breakthrough by a Cycle in only one parent major squareĭS in both R4C2 and R4C3 by reduction of in R4 from possible digit subset in left middle major square. Results of the actions taken shown below. This breakthrough gives us two more valid cells, R3C7 8 by reduction from DS - R1C7 4 by reduction and exception in top right major square. in two cells of C3 forms a Cycle in which the two digits get locked and thus disallows appearance of 2 and 6 in any other empty cell of C3.īecause of this effect of limiting the digits inside the Cycle, the possible digit subset in the column C3 is reduced to and this creates the breakthrough in R3C3 4 by DSA reduction of (with in parent top left major square and in second parent R3) from DS in C3. Now in addition, a third possible digit subset of is formed in R8C3 by DSA reduction of from possible digit subset in R8. Explanation on how Cycle (2,6) is formed in C2 and how it results in a breakthroughĪ possible digit subset DS already exists in R6C3 as a part of easily formed digit subset in two empty cells of R6. R6C5 3 by scan for 3 in R4, R5, C4 - R8C6 3 by scan in R7, C4, C5.Ĭycle (2,6) in two empty cells R6C3, R6C4 and DS in R8C3 by DSA reduction of from possible digits in R7 - a breakthrough Cycle (2,6) formed in R6C3, R8C3.īreakthrough R3C3 4 by DSA reduction of from possible digit subset of - R3C7 8 by reduction from DS - R1C7 4 by reduction and exception in top right major square. Reduced DS in two empty cells in C8 is - R2C8 2 by reduction of 9 by R2 - R2C1 8 by reduction - R2C7 6 by reduction - R2C4 3 by exception in R2. R6C9 4 by DSA reduction of from DS - R6C7 7 by reduction - R5C9 9 by exception in right middle major square - R3C9 3 by exception in C9. It reduces the possible digit subset DS in right middle major square to - breakthrough valid cell R6C8 5 by reduction of (with in C8 and 9 in C6) from DS - followed by R7C8 3 by DSA reduction of from DS in C8 - R7C9 7 by reduction of from DS. Let's see the effect of this Cycle (1,8). Instead of scanning the promising empty cells of a major square FOR A SINGLE DIGIT, in a double digit scan, the promising empty cells of the right middle major square are scanned for TWO DIGITS (1,8) TOGETHER - Cycle (1,8) formed in R4C8, R5C8 in C8 and parent right middle major square. This creates an opportunity for us to apply advanced technique of double digit scan to get a breakthrough Cycle (1,8). Observe that the two digits appear together in R6 and also in C9, but do not appear in the right middle major square. Advanced Sudoku technique of double digit scan Step by step solution to the New York Times Sudoku Hard 20th February, 2021: Stage 1: Breakthroughs by Double digit scan, DSA technique and Cyclesįirst success by row-column scan: R8C8 4 because of 4 in R7, R9 - R9C9 2 by scan for 2 in R7, C7. The New York Times Sudoku Hard, 20th February, 2021īefore going through the solution solve the puzzle first. How to solve the Sudoku hard is explained clearly including all the breakthroughs. Solve New York Times Sudoku hard Februin quick steps. How to Solve the New York Times Sudoku Hard 20th February, 2021 with step by step explanation
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |