R9C7 4 by reduction of 6 from DS [4,6] from R9 -- R9C8 6 by exception in R9 -- R2C8 4 by exception in C8 -- R2C7 3 by exception in whole game. Description- Self Solving Sudoku- Locked Candidates Pointing- Triple Subset- Locked Candidates Claiming- Pair- New York Times Sudoku Hard September 14, 2022-. Effect of this critically important single digit lock is reduction of 6 in possible digit subset [1,5,6] to [1,5] in R8C5. Short length easy possible digit subsets are evaluated for nearly all empty cells without any easy success. This Cycle immediately causes the breakthrough of R9C1 7 by reduction of [1,5] from DS of [1,5,7] in R9C1 because of the property of locking the four digits inside the Cycled cells only. For example, in this case, after you place 8 in R8C7, it will be easy for you to identify a Cycle of (1,4,5,7) in rest four empty cells without specifically evaluating the possible digit subsets in these cells. With 6 in R9C5, R8C2 9 by reduction -- R9C2 6 by reduction. Enjoy also learning how to solve Sudoku hard in easy steps. It is in three zones or areas - row R2, column C5 and 9 cell top-middle major square. Specifically important is the disallowing of digit 5 in the empty cells of C3. Next R1C8 7 by scan for 7 in R2, R3, C9 and that's all by row column scan at this point. With 6 in R1C5, R1C6 2 by reduction -- R1C1 5 by reduction -- R1C4 8 by reduction. To be precise, this is what happens in the double digit scan for [1,8] together on the empty cells of the left middle major square. SuDoku Archive for October 2012. It can also be helpful to mark the columns and rows with pencil marks to work through elimination methods. 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. Three digits appear in in the column in only these three cells and nowhere else. Printable Sudoku New York TimesPrintable Sudoku New York Times - There are many number of ways that to use Printable Sudoku Puzzles. 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. Chance of getting a useful pattern from a 5 digit possibility in a cell is remote, and, Such long list of digits CLUTTERS the view and. With [6,9] in R6, R6C7 2 by reduction -- R4C8 3 by reduction --- R4C7 8 by reduction and exception in R4. Writing long digit sets takes more time and effort. Turning attention first on promising zone C6 we get the breakthrough DS [6,8] in R5C6 by DSA reduction of [1,2,9] from column C6 DS [1,2,6,8,9]. Game status shown below. This is a major breakthrough and we'll see its effects in the next stage. You can use a pencil to make mistakes and then erase them. Sunday Los Angeles Times crossword Sunday New York Times crossword Sunday Premier crossword SUDOKU. These four digits are locked and are self-sufficient in these four cells, each appearing at least twice in these four cells. Wordle players can use these five hints to solve puzzle #500. 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. R1C3 3 by DS reduction of [1,6] from DS [1,3,6] in R1 -- R1C9 6 by reduction -- R1C1 1 by exception in R1. Solution to the New York Times Sudoku Hard, 15th February, 2021: Stage 1: Early breakthroughs by Single digit lock and parallel scan Use row-column scan, the simplest method to get a unique digit for a cell. Because of this effect of limiting the digits inside the Cycle, the possible digit subset in the column C3 is reduced to [3,4,7,8,9] and this creates the breakthrough in R3C3 4 by DSA reduction of [3,7,8,9] (with [8,9] in parent top left major square and [3,7] in second parent R3) from DS [3,4,7,8,9] in C3. 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. R2C9 2 by DSA reduction of [1,4,9] in R2 from DS [1,2,4,9] in C9. We'll end this stage by an important breakthrough in R6C1 7 by parallel digit scan for 7 on empty cells of R6: 7 in C3 debars cell R6C3, 7 in C4 debars 7 in R6C4 and 7 in C6 debars R6C6 for 7 leaving single cell R6C1 for 7. Elimination of four cells of column C6 for digit 1 by parallel scan is shown by blue arrows in Stage 2 solution figure below. Use row-column scan, the simplest method to get a unique digit for a cell. If by DSA reduction we get a single digit possible for a call, we get a valid cell hit. 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. Analyze in which cell digit 5 can appear in R2. But the cell R8C9 is again disallowed for digit 8 by the presence of 8 in C9 and the only cell left in R8 for digit 8 is the cell R8C7. Use Snyder until it fails. To solve quickly you need to concentrate. These puzzles are We decide to continue to form DSs in empty cells of the promising zone bottom left major square. Enjoy also learning how to solve Sudoku hard in easy steps. Advanced Sudoku technique of double digit scan 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). 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. Next valid cell is R4C6 7 by scan for 7 in R5, R6, C7. It a Sudoku hard rich with learning potential. So the last breakthrough has been R9C1 7 by reduction of [1,5] from DS [1,5,7] because of the Cycle (1,2,5,6) in bottom left major square. 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. This is a hard Sudoku puzzle that needed multiple breakthroughs by advanced Sudoku techniques. So only the two missing digits (2, 8) are valid for the cell R2C5. None of the digits could produce any success of valid cell by row column scan. This creates an opportunity for us to apply advanced technique of double digit scan to get a breakthrough Cycle (1,8). $(1, 3, 6) \cup (3, 4, 7, 9) \cup (1, 3, 5)=(1, 3, 4, 5, 6, 7, 9)$. If you just want today's word, you can jump straight to the end of this article for November 3's Wordle solution, for . No more easy valid digit 3 possible at this point. You get a valid cell breakthrough in R2C5 5. [2,6] 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. This makes the digit invalid in any other cell in the row or column in which the digit is locked. About New York Times Games. Solution to New York Times Sudoku hard, 15th February, 2021 explains step by step how to use Sudoku hard techniques to achieve quick breakthroughs. In medium or easy Sudoku puzzles, number of filled up cells will be more. This is shown by "5" in the two cells. Since the launch of The Crossword in 1942, The Times has captivated solvers by providing engaging word and logic games. This creates an opportunity for us to apply advanced technique of double digit scan to get a valid cell hit. R4C9 9 by parallel scan for 9 on empty cells of C9: 9 in R2 eliminates R2C9 for 9 leaving R5C9 for 9 -- R2C9 5 by reduction -- R4C7 5 by scan for 5 in R5, C8. We'll close here and explore the rest later. For full enjoyment, avoid looking into any solution as well as the answer. The New York Times Sudoku Hard, 17th February, 2021 Before going through the solution solve the puzzle first. Copyright: Atanu Chaudhuri and respective Authors. Description- Self Solving Sudoku- Slice and Dice- Locked Candidates Pointing- Pair- New York Times Sudoku Hard October 31, 2022- Repeat from 20 August 2017- . The Hint button will highlight the next logical square to solve that is empty or incorrect. This is the quick method of parallel scan for digit 8 on empty cells of the row R8. Once a few cells are filled in a zone, enumerating the rest gets easier. This creates Cycle (7,8) in R4. In this case observe that by scan for 6 in R8, digit placement of digit 6 is restricted to only two cells in R8, R8C7 and R8C9. Step by step solution to the New York Times Sudoku Hard, 16th February, 2021: Stage 1: Breakthroughs by DSA technique and Cycles. That means, these two digits light up or affect the intersecting cell R6C1 and reduce its possible digit subset to single digit 3. R6C2 2 by reduction of 2 by the Cycle (1,6,7) in left middle major square -- R6C6 9 by reduction -- R3C6 8 by reduction -- R2C4 9 by reduction -- R3C1 7 by reduction -- R8C1 5 by reduction -- R8C2 7 by reduction. Before going through the solution solve the puzzle first. This is the hallmark of a truly Sudoku hard puzzle game. With 6 in R6C9, R7C9 4 by reduction -- R7C7 6 by reduction and exception in C7. Digit possibilities for empty cells are enumerated ONLY WHEN NECESSARY. This will result in a series of valid cells and is a major breakthrough. R4C7 7 by scan for 7 in C8, C9 -- R3C7 9 by reduction of 8 from DS [8,9] in R3 -- R3C9 8 by exception in R3 -- R7C9 9 by reduction -- R7C8 8 by reduction -- R4C9 4 by reduction of 9 from DS [4,9] in R4 -- R4C8 9 by exception -- R9C9 3 by exception in C9. Select a Word Game to learn more: Wordle. Take cell R2C5. You could have identified the formation of this Cycle first to get the breakthrough. Identify the rows and columns with the digit that intersect in a cell in the the 9 square major square. By the way, Sudoku hard solution techniques are included with many of the solutions. R7C1 2 by DS reduction of [1,4,6] in C1 from possible digit subset DS [1,2,4,6] in four empty cells of R7. 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. Identify Single digit lock on 6 in R7C7, R7C9 by scan for 6 in R8 -- this lock on 6 participates in next valid cell R9C2 6 by scan for 6 in R7 by the lock, 6 in R8 and 6 in C1. Okay, continuing to identify more possible single digit locks and breakthroughs, with focused intent it doesn't take much to identify a second single digit lock this time on 1 in cells R8C8, R8C9 by scan for 1 in R9, C7. In this game having no success with 1 or 2 by row column scan select digit 3 as promising and get a few easy placements. 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. 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. Reduction of instances of 6 from all such empty cells is the gain. But when you face a Sudoku hard, valid cell by row column scan may be few or non-existent to start with. Before going through the solution solve the puzzle first. DS in both R4C2 and R4C3 [7,8] by reduction of [2,6] in R4 from possible digit subset [2,6,7,8] in left middle major square. . Breakthrough R3C3 4 by DSA reduction of [3,7,8,9] from possible digit subset of [3,4,7,8,9] -- R3C7 8 by reduction from DS [4,8] -- R1C7 4 by reduction and exception in top right major square. Copyright: Atanu Chaudhuri and respective Authors. R6C9 4 by DSA reduction of [7,9] from DS [4,7,9] -- R6C7 7 by reduction -- R5C9 9 by exception in right middle major square -- R3C9 3 by exception in C9. Otherwise we know that the single digit lock might become useful later and we ignore it for the present. 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). R2C3 3 by row column scan for 3 in R1, R3, C2 -- R6C7 3 by scan for 3 in R4, R5, C9 -- R8C8 3 by scan in C7, C9. Observe how three empty cells of R8C1, R8C2, R8C3 are disallowed for digit 8 by the presence of digit 8 in the bottom left major square. Enjoy also learning how to solve Sudoku hard in easy steps. Never lose an opportunity for a valid cell hit by row column scan. Try free NYT games like the Mini Crossword, Ken Ken, Sudoku & SET plus our new subscriber-only puzzle Spelling Bee. But how? There is no metric or measurement to decide for sure that a Sudoku puzzle is hard, or not so hard. In this case for example, after you put 5 in R2C5, a Cycle of remaining four digits [4,6,7,8] is formed. It is a fairly hard Sudoku puzzle. The joy of discoveries will then all be yours. 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). By the way, Sudoku hard solution techniques are included with many of the solutions. Its very easy to lose track You never need to guess anything on Sudoku. In this process of parallel digit scan, presence of digit 5 in four columns C2, C6, C8 and C9 in parallel affected or lighted up four empty cells of the scanned row R2 and thus made these four cells invalid for placing digit 5. Abdlomax 3 yr. ago Starting the solution process, first valid cell success by row column is for 2: R8C7 2 by scan for 2 in R9, C8, C9 -- R7C1 2 by scan for 2 in R8, R9. With six digits filled in the left middle major square, it is a very promising zone to look for more valid cell hits. R6C1 7 results in R9C2 7 by scan for 7 in C1 -- R9C1 5 by exception in bottom left major square. 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. As a strategy we always start row column scan from digit 1 and continue till digit 9. Section below explains in a bit of detail the process of enumerating digits possible in a cell. 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