Parallel digit scan for a single digit on the empty cells of a row (or column)Īnalyze in which cell digit 5 can appear in R2. This prompts us to look for a valid cell caused by this single digit lock and thus discover the difficult to identify R2C5 5 by parallel digit scan for 5 on empty cells of R2. Specifically important is the disallowing of digit 5 in the empty cells of C3. We are only concerned with the most important fact that digit 5 can appear in only these two cells and NOWHERE ELSE IN THE TWO PARENT ZONES. We don't bother about other digits appearing in these two cells. This makes the digit invalid in any other cell in the row or column in which the digit is locked.Ĭonsider the row column cross-scan: With digit 5 in R4C2, and R7C9, 5 can appear only in two cells R8C3 and R9C3 of bottom left major square Thus the digit 5 is locked in C3 as well as in the parent major square. In a single digit lock, a lone digit is locked in two cells in a 9 cell square as well as in a row or column.
The important pattern of single digit lock We'll now show you a special valid cell breakthrough by a combination of advanced patterns of single digit lock and parallel scan. Solution to the New York Times Sudoku Hard, 17th February, 2021: Stage 2: Breakthroughs by Single digit lock and Parallel digit scan Results of the actions taken shown below. By the double digit scan, first the Cycle is formed and then the valid cell R6c1 4 by reduction of because of the property of the Cycle. To be precise, this is what happens in the double digit scan for together on the empty cells of the left middle major square. We are not certain at this point which of the two digits will finally occupy which of the two cells, but taking the two cells together, we are certain that digits will certainly occupy these two cells and in no other empty cell in parent zones R6 and left middle square. That means, these two digits light up or affect the intersecting cell R6C1 and reduce its possible digit subset to single digit 3.Īs a byproduct the leftover two digits form a Cycle of (1,8) in R6C2 and R6C3. īoth digits 1 and 8 appear together in column C1 but not in any cell of the left major square. The possible digit subset in the three empty cells is. With six digits filled in the left middle major square, it is a very promising zone to look for more valid cell hits. Explanation on how Cycle (1,8) is formed and how it results in a breakthrough Never lose an opportunity for a valid cell hit by row column scan. Next valid cell is R4C6 7 by scan for 7 in R5, R6, C7. This leaves the single cell R2C1 for 3 and the byproduct Cycle of (5,6,7) in the other three empty cells of C1. Still analyzing this promising neighborhood we detect the possibility of the next breakthrough of R2C1 3 by parallel scan for 3 on empty cells of C1.ģ in bottom left major square eliminates R7C1 for 3 and 3 in R1 and R3, eliminate the other two empty cells R1C1 and R3C1 for 3.
Advanced Sudoku technique of parallel digit scan for 3 Primarily, with only two cells left for the two digits, a breakthrough Cycle of (1,8) is formed in a single strike and it leaves only the cell R6C1 for digit 4 by exception in left middle major square. This eliminates the cell R6C1 for any of the two digits. 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 a major square are scanned for TWO DIGITS TOGETHER.Īs appear together in C1, and also as none of these two digits are placed yet in left middle major square, we get Cycle (1,8) in cells R6C2, R6C3 by a double digit scan of in C1. This creates an opportunity for us to apply advanced technique of double digit scan to get a valid cell hit. Advanced Sudoku technique of double digit scan It is followed immediately by a second valid cell for 7 in the promising neighborhood of left middle major square: R5C2 7 by scan for 7 in R6. Step by step solution to the New York Times Sudoku Hard 17th February, 2021: Stage 1: Breakthroughs by Double digit scan, Parallel digit scan and and Cyclesįirst valid cell by row column scan is for 5: R4C2 5 by scan for 5 in R5, R6. The New York Times Sudoku Hard, 17th February, 2021īefore going through the solution solve the puzzle first. New York Times Sudoku Hard 17 February, 2021 solved in four stages that explain all breakthroughs and how those are achieved by advanced Sudoku techniques. New York Times Sudoku hard puzzle 17th February, 2021 Solved in easy steps
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