The New York Times publishes three puzzle challenges every [] It a Sudoku hard rich with learning potential. R5C5 7 by DSA reduction of [1,5,8] from DS [1,5,7,8] in C5 -- R6C4 4 by DS reduction of [3,8,9] in C4 from DS [3,4,8,9] in R6. Nevertheless, a distinction between hard, medium and easy Sudoku categories can be made. Unique set of existing digits in all these three together are. R1C6 9 by reduction -- R9C6 1 by reduction -- R9C4 9 by reduction R2C4 6 by DS reduction of [2,5] from DS [2,5,6] in C4 -- R3C4 5 by scan for 5 in R1 -- R1C4 2 by exception in C4. This Sudoku hard is an especially challenging puzzle rich with Sudoku digit patterns. The possible digit subset in the three empty cells is [1,4,8].
How to Solve New York Times Sudoku Hard February 20, 2021 - SureSolv For full enjoyment, avoid looking into any solution as well as the answer. 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. Step by step solution to the New York Times Sudoku Hard, 16th February, 2021: Stage 1: Breakthroughs by DSA technique and Cycles. Use Snyder until it fails. About New York Times Games. In the final solution, 6 can appear in only one of these two cells and in no other cell of the parent bottom right major square or the parent row R8.
Wordle today: Here's the answer, hints for November 3 This breakthrough gives us two more valid cells, R3C7 8 by reduction from DS [4,8] -- R1C7 4 by reduction and exception in top right major square. With 1 in R7C1, R7C7 5 by reduction -- R9C7 1 by reduction -- R9C4 5 by exception in R9 -- R7C4 9 by exception in whole game. That's why for speed, we prefer to identify parallel digit scan possibilities. Play the Daily New York Times Crossword puzzle edited by Will Shortz online.
New York Times Sudoku Hard 16 February, 2021: Quick Solution - SureSolv But by DSA we get the valid cell, R8C2 9 by reduction of [1,5,7] from bottom left major square DS [1,5,7,9] -- R8C3 1 by reduction of [5,7] in R8 from DS [1,5,7] in bottom left major square.
Instead, focus has been to aggressively create a breakthrough at the earliest as well as taking the opportunity of a valid cell by the simplest technique of row column scan. With 5 in R1C1, R7C1 1 by reduction -- R8C1 2 by reduction -- R8C3 6 by reduction -- R8C2 5 by reduction -- R8C5 1 by reduction -- R3C5 5 by reduction -- R9C4 8 by reduction -- R9C2 3 by reduction -- R9C3 9 by reduction -- R7C3 8 by reduction -- R4C3 7 by reduction -- R4C2 8 by reduction. How debt-for-climate swaps can help solve low-income countries' fiscal and environmental challenges at the same time R2C2 8 by scan for 8 in R3 -- R3C2 4 by exception in top left major square. This is a major breakthrough and we'll see its effects in the next stage. With 9 in R8C2, R8C9 4 by reduction -- R8C9 4 by reduction -- R9C9 9 by reduction. R3C2 4 by scan for 4 in R2, C1, C3 -- R3C3 1 by scan for 1 in R2, C1 -- R6C3 8 by reduction -- R6C2 1 by reduction -- R2C3 7 by reduction -- R2C8 6 by reduction -- R2C2 8 by reduction and exception -- R1C8 4 by reduction -- R1C7 5 by reduction -- R1C1 6 by reduction and exception in R1 -- R7C1 7 by reduction -- R3C1 5 by reduction. Hint #2 . For easier lower level Sudoku games this systematic method of row column scan should provide a smooth path to the final solution. An easy catch by opportunistic scan for 9: R9C9 9 by scan in R7, R8. Analyze in which cell digit 5 can appear in R2. We'll select relatively more promising cells in a row, a column or a 9 cell square (each a zone) to evaluate the smallest length of possible digits in the cells first.
Printable Sudoku New York Times Hard - Lyana Printable Sudoku Alternately this lock on 6 acts as if 6 actually exists in R8 and that's why a lock on a digit can be used for a scan for the digit. Then start the process for the next higher digit. With 1 in R1C6, R3C5 2 by reduction -- R1C5 8 by reduction -- R2C6 4 by reduction -- R9C6 8 by reduction caused by the Cycle (3,9) formed in R6C6, R8C6.
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. R9C8 2 by scan for 2 in C7 -- R9C3 5 by reduction -- R8C3 2 by reduction -- R9C4 4 by reduction -- R8C4 5 by reduction and exception. That means, these two digits light up or affect the intersecting cell R6C1 and reduce its possible digit subset to single digit 3. We can say with 100% certainty that these three digits will be placed in the final solution in only these three cells and nowhere else. With six digits filled in the left middle major square, it is a very promising zone to look for more valid cell hits. Single digit lock on 4 in R2C2, R3C2 by scan for 4 in C1 -- DS in R7C2 [1,4] is reduced by 4 by the lock on 4 in C2 resulting in R7C2 1 -- R8C9 1 by scan for 1 in R7 -- R8C3 4 by scan for 4 in C1, lock on 4 in C2. 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. So we'll next start evaluating 2 or 3 digit possible digit subsets DSs in . This is a breakthrough by single digit lock and DSA. With 4 in R1C8, R5C8 9 by reduction -- R5C7 4 by reduction -- R3C7 9 by exception in C7 -- R3C8 9 by exception in whole game. It's a pleasant surprise to identify that a Cycle of [1,2,5,6] is formed in the four cell DSs in R7C1, R8C1, R8C2 and R8C3. Solve New York Times Sudoku hard February 20, 2021 in quick steps. Next valid cell R5C6 6 will be shown in next stage. Now in addition, a third possible digit subset of [2,6] is formed in R8C3 by DSA reduction of [1,5] from possible digit subset [1,2,5,6] in R8. There is no metric or measurement to decide for sure that a Sudoku puzzle is hard, or not so hard. More Sudoku hard puzzles you may like to solve and learn how to solve. Before going through the solution solve the puzzle first. The online Sudoku NYT puzzle in this game must be solved using logic. To skip the section, click here. As a strategy we always start row column scan from digit 1 and continue till digit 9. Solution to New York Times Sudoku hard, 15th February, 2021 explains step by step how to use Sudoku hard techniques to achieve quick breakthroughs. The elimination of four empty cells for 5 in row R2 is shown by red arrows in the Stage 2 solution figure below. Description- Self Solving Sudoku- Locked Candidates Pointing- Triple Subset- Locked Candidates Claiming- Pair- New York Times Sudoku Hard September 14, 2022-.
NY Times solving times : r/sudoku - reddit New York Times Sudoku Hard 17 February, 2021: Quick Solution - SureSolv $(1, 3, 6) \cup (3, 4, 7, 9) \cup (1, 3, 5)=(1, 3, 4, 5, 6, 7, 9)$. JUMBLE. The updated list of Solutions to level 3, level 4 and NYTimes Sudoku hard puzzle games: How to solve Sudoku hard puzzle games full list (includes very hard Sudoku). DS in R8C2 [2,5,6] by reduction of 1 in C2 from DS [1,2,5,6] in R8 and DS in R8C3 [2,6] by reduction of [1,5] in C3 from DS [1,2,5,6] in R8. Now do a parallel scan for 1 in empty cells of column C6 to get the valid cell R1C6 1. The joy of discoveries will then all be yours. An interesting property of a valid cell by parallel digit scan is, there will always be an associated Cycle after you put the valid digit in its identified cell. With 6 in R6C9, R7C9 4 by reduction -- R7C7 6 by reduction and exception in C7. An often asked questions is, "What makes a Sudoku puzzle hard?". By the way, Sudoku hard solution techniques are included with many of the solutions. . The New York Times regularly publishes a number-based puzzle game called NYT Sudoku. About New York Times Games. For full enjoyment, avoid looking into any solution as well as the answer. After speed of solution, reducing labor being the main objective, we won't enumerate 4 or 5-digit possibilities or enumerate possible digits for ALL the cells. Short length easy possible digit subsets are evaluated for nearly all empty cells without any easy success. With 7 in R4C3, R1C3 3 by reduction -- R1C2 7 by reduction. You may perhaps appreciate that, if you have proceeded to form first the possible digit subsets in these four cells and then identified the formation of the Cycle resulting in the valid cell R2C7 8, it would have taken much longer. And no more for 2. R5C4 8 by scan for 8 in R4 -- R4C4 6 by exception in C4 -- R4C3 1 by reduction -- R5C3 6 by exception in C3 -- R4C8 4 by exception in R4. With 5 in C2, C6, C8, C9 and in C3 by single digit lock, digit 5 can be placed in R2 only in the single cell R2C5. A possible digit subset DS [2,6] already exists in R6C3 as a part of easily formed digit subset in two empty cells of R6. 3 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. Results of the actions taken shown below. Both digits 1 and 8 appear together in column C1 but not in any cell of the left major square. Well not disturb the Cycle of (6,8) for better understanding of what happened at this stage. The updated list of Solutions to level 3, level 4 and NYTimes Sudoku hard puzzle games: How to solve Sudoku hard puzzle games full list (includes very hard Sudoku). So only the two missing digits (2, 8) are valid for the cell R2C5.
Sudoku - New York Times Number Puzzles- The New York Times Next breakthroughs by forming Cycle (8,9) in R7C, R7C9 by DSA reduction in both cases by [5,7] from possible digit subset [5,7,8,9] in four empty cells of R7.
How to Solve New York Times Sudoku Hard October 21, 2022 Never let go of an effective single digit lock. The joy of discoveries will then all be yours.
It Takes a Lot of Elephant Brains to Solve This Mystery - The New York 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. To start with, R4C9 1 by DSA reduction of [4,9] in R4 from reduced DS [1,4,9] in C9 -- R8C8 1 by single digit lock partner reduction of 1 in R8C9. Since the launch of The Crossword in 1942, The Times has captivated solvers by providing engaging word and logic games.
How to Solve New York Times Sudoku Hard February 15, 2021 - SureSolv By the way, Sudoku hard solution techniques are included with many of the solutions. To be able to solve a Sudoku puzzle in a short time you should know how to use the pencil to trace the board. Step by step solution to the NY Times Sudoku Hard 4th March, 2021 Stage 1: Breakthrough by Single digit lock, Parallel digit scan, DSA and Cycles. Possible digit subset in central middle major square reduced to [1,4,5,7] -- R5C2 5 by DSA reduction of [2,6] in R5 from [2,5,6] possible digit subset in C2. 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. Note: To know more on DSA reduction technique, click on the above internal link and return (after going through it) by clicking on browser back button. There is no metric or measurement to decide for sure that a Sudoku puzzle is hard, or not so hard. SuDoku Archive for October 2012. An effective single digit lock is formed invariably by scan for a digit in a single column or row or more frequently by cross-scan over a row and a column. This eliminates the cell R6C1 for any of the two digits. In medium or easy Sudoku puzzles, number of filled up cells will be more. Wordle players can use these five hints to solve puzzle #500. Once a few cells are filled in a zone, enumerating the rest gets easier. You get a valid cell breakthrough in R2C5 5.
Solving a puzzle with a pen is a very different beast than using software that can fill and track pencil marks, highlight all cells with a given candidate and support non destructive cell highlights for chains. The NY Times Sudoku Hard 4th March 2021 solved step by step quickly and easily by advanced Sudoku techniques with breakthroughs explained adequately. It reduces the possible digit subset DS in right middle major square to [4,5,7,9] -- breakthrough valid cell R6C8 5 by reduction of [4,7,9] (with [4,7] in C8 and 9 in C6) from DS [4.5.7.9] -- followed by R7C8 3 by DSA reduction of [1,5,7] from DS [1,3,5,7] in C8 -- R7C9 7 by reduction of [1,5] from DS [1,5,7]. In this case for example, after you put 5 in R2C5, a Cycle of remaining four digits [4,6,7,8] is formed. 9 in R8C2, R8C9 4 by reduction -- R9C9 9 by reduction R9C9... Any easy success 6,8 ) for better understanding of What happened at this stage speed, we to. 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