Let's see the effect of this Cycle (1,8). Copyright: Atanu Chaudhuri and respective Authors. 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. That's why for speed, we prefer to identify parallel digit scan possibilities. Move around the grid frequently. In the process we may get valid cells, but otherwise our lookout is for identifying for a single digit lock by cross scan. Because of 8 in central middle major square, R5C6 6 by reduction of 8 -- R8C6 8 by reduction of 6 -- R8C5 6 by reduction of 8. This is similar ro normal row column scan which is carried out on the empty cells of a major square. 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 20th February, 2021: Stage 1: Breakthroughs by Double digit scan, DSA technique and Cycles First success by row-column scan: R8C8 4 because of 4 in R7, R9 -- R9C9 2 by scan for 2 in R7, C7. DS in C1 [1,2,5,7] reduced by [2,7] in R7 to form DS [1,5] in R7C1, reduced by 7 in R8 to form DS [125] in R8C1 and reduced by 2 in R9 to form DS [1,5,7] in R9C1. These are rough figures drawn from experience. If you get stuck, we're here to help with handy hints, as we are every day. 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. This breakthrough DS of [6,8] joins with [6,8] in R8C6 to form breakthrough Cycle [6,8] in C6 reducing DS in C6 to [1,2,9] and causing the breakthrough in R6C6 2 by reduction of [1,9] from C6 DS [1,2,9] -- R6C3 4 by reduction of [2,8] -- R6C4 8 by exception in R6 -- R1C2 4 by scan for 4 in R3, C1, C3. Every such easy fill at this stage is a bonus. With 6 in R6C9, R7C9 4 by reduction -- R7C7 6 by reduction and exception in C7. 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. An often asked questions is, "What makes a Sudoku puzzle hard?". This Sudoku hard is an especially challenging puzzle rich with Sudoku digit patterns. R2C9 2 by DSA reduction of [1,4,9] in R2 from DS [1,2,4,9] in C9. Alternately, if R4C3 6, digut [1,7] will appear in eithe of the cells R5C3, R6C3. Reduced DS in two empty cells in C8 is [2,9] -- R2C8 2 by reduction of 9 by R2 -- R2C1 8 by reduction -- R2C7 6 by reduction -- R2C4 3 by exception in R2. Bur that would have been time-taking. It a Sudoku hard rich with learning potential. As a strategy we always start row column scan from digit 1 and continue till digit 9. R2C8 7 by DSA reduction of 4 in C8 from DS [4,7] -- R2C7 4 by exception in R2 -- R3C8 9 by scan for 9 in C7 -- R3C7 1 by exception in R3 -- R5C7 7 by exception in C7 -- R5C8 1 by exception in R5. R3C5 6 by DSA reduction of [1,2,5] from possible digit subset DS [1,2,5,6] in empty cells of C5 and hence in R3C5 -- R2C5 2 by DSA reduction of [1,5] from DS [1,2,5] in C5. These puzzles are But how? It can also be helpful to mark the columns and rows with pencil marks to work through elimination methods. Copyright: Atanu Chaudhuri and respective Authors. About New York Times Games. The other important property of a parallel digit scan is, it will always be associated with a resultant Cycle in rest of the cells after you get the valid cell. 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. We stop in a 9 cell major square when the rest of the empty cells would all have digit possibilities 4, 5 or more. 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. Description- Self Solving Sudoku- Slice and Dice- Locked Candidates Pointing- Triple Subset- Pair- Locked Candidates Claiming- New York Times Sudoku Hard October 21, 2022- Repeat from 19 January 2019- Watch as a Sudoku Classic puzzle is clearly solved step by step.- Stop video, solve the next square, play to check- Each square is solved in order of easy to hard.- Thanks for watching- Share if you like- A more important issue is reminding us to:- Observe the 6 foot rule- Wash our hands- Cough into our elbows- Avoid touching our faces- Wear a better mask- Be glad to get the jabs- Lets keep everyone safe- Cheers#RSOUDREUNSUDOKU#ResolviendoSudoku#SudokuNYT#NYTimes#Sudoku#Puzzle#Puzzles#Solutions 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. Before going through the solution solve the puzzle first. R2C2 8 by scan for 8 in R3 -- R3C2 4 by exception in top left major square. Look for placing any other possible digit by row-column scan. Before going through the solution solve the puzzle first. That means, these two digits light up or affect the intersecting cell R6C1 and reduce its possible digit subset to single digit 3. 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. So only the two missing digits (2, 8) are valid for the cell R2C5. Because of the series of early breakthroughs by advanced digit patterns, the Sudoku hard could be solved rather easily. The . 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. 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 Cycles First valid cell by row column scan is for 5: R4C2 5 by scan for 5 in R5, R6. Elimination of four cells of column C6 for digit 1 by parallel scan is shown by blue arrows in Stage 2 solution figure below. Advanced Sudoku technique of double digit 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. This is a breakthrough by single digit lock and DSA. Most important breakthrough at this stage finally has been provided by the single digit lock on 6 in R8C2, R8C3. If you are alert for detecting a parallel scan opportunity, breakthrough would be much quicker. About New York Times Games. Learn more about how to play The New York Times' word games and logic puzzles: Wordle, Spelling Bee, Letterboxed, Tiles, Vertex, and Sudoku. Anyone can read what you share. Next R1C8 7 by scan for 7 in R2, R3, C9 and that's all by row column scan at this point. 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. 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. Wordle players can use these five hints to solve puzzle #500. You could have identified the formation of this Cycle first to get the breakthrough. 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. The New York Times Sudoku Hard, 17th February, 2021 Before going through the solution solve the puzzle first. Consider 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. R5C2 5 by exception in C2 -- R6C9 6 by DSA reduction of [1,7] in C9 from DS [1,6,7] in three empty cells of R6 -- R6C4 1 by reduction of 7 from reduced DS [1,7] in R6 -- R6C3 7 by exception in R6 and by reduction. This is how we get a breakthrough valid cell R8C7 8 by parallel scan. Unique set of existing digits in all these three together are. Since the launch of The Crossword in 1942, The Times has captivated solvers by providing engaging word and logic games. Hint #1: The word contains three vowels (madness isn't it) one of which, in a different word structure, could be a consonant. 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. About New York Times Games. 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. Cycle reduces possible digit subset for R7C4, R7C6 to [5,7] and with [7] in C6 we get the breakthrough R7C5 5 by reduction of 7 followed by R7C4 7 by exception. One of the best ways is to personalize them using QR codes. 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. R6C5 3 by scan for 3 in R4, R5, C4 -- R8C6 3 by scan in R7, C4, C5. 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. 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. With R7C5 9, R5C5 1 by reduction -- R5C4 8 by reduction -- R4C4 2 by reduction -- R6C4 6 by reduction -- R6C5 3 by reduction -- R6C6 9 by reduction -- R8C6 3 by reduction. As a byproduct the leftover two digits [1,8] form a Cycle of (1,8) in R6C2 and R6C3. 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. Play the Daily New York Times Crossword puzzle edited by Will Shortz online. Never lose an opportunity for a valid cell hit by row column scan. In this case for example, after you put 5 in R2C5, a Cycle of remaining four digits [4,6,7,8] is formed. Description- Self Solving Sudoku- Locked Candidates Pointing- Triple Subset- Locked Candidates Claiming- Pair- New York Times Sudoku Hard September 14, 2022-. In this case we get the breakthrough of R9C2 6 by scan for 6 in R7, R8 and C1, 6 in R7 being actually a lock on digit 6. By possible digit analysis we have the DSs in R4C3, R5C3 and R6C3 as [1,6], [1,6,7] and [1,6,7]. Solve New York Times Sudoku hard February 20, 2021 in quick steps. Hint #2 . If you cannot logically place a value in a cell then do not place that value in a cell. For example, in this case, reduction of 7 from DS of R6C2 gives us a valid cell hit as R6C2 2. We don't bother about other digits appearing in these two cells. 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). Observe how the other four empty cells of C6 have been disallowed for 1 by scan for 1 in R2 (disallowing R2C6 for 1), Cycle (1,8) in R6 (disallowing 1 in R6C6), single digit lock in R8 (disallowing R8C6 for 1) and 1 in R9 (disallowing R9C6 for 1). Results of the actions taken shown below. A single digit lock is always a valuable asset for hard Sudoku simplification and whenever yoi identify such a possibility, just note it in your mind for future use at the right timer. Be VERY careful about your notes. There is no metric or measurement to decide for sure that a Sudoku puzzle is hard, or not so hard. Sunday Los Angeles Times crossword Sunday New York Times crossword Sunday Premier crossword SUDOKU. $(1, 3, 6) \cup (3, 4, 7, 9) \cup (1, 3, 5)=(1, 3, 4, 5, 6, 7, 9)$. With 6 in R1C5, R1C6 2 by reduction -- R1C1 5 by reduction -- R1C4 8 by reduction. If yes, then we have provided all the NY Times Sudoku Answers. Unnecessary enumeration of possible digit subsets in empty cells is avoided to speed up the solution. If a square is incorrect, it will be marked as such. Instead, the parallel digit scan is carried out on the empty cells of a row or a column. 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. First valid cell by row column scan is for 5: R4C2 5 by scan for 5 in R5, R6. Strategy adopted: start with row column scan coupled with aggressive breakthroughs by advanced Sudoku techniques. 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. This is a breakthrough that otherwise couldn't have been achieved. This is shown by "5" in the two cells. We decide to continue to form DSs in empty cells of the promising zone bottom left major square. JUMBLE. To be quick select the digit that appears maximum number of times and select a promising 9 cell major square for row column scan on its empty cells. In easy ones it can be 36 or more. Because of this locked status of the digits in the cells involved in the Cycle, all occurrrences of these locked digits in zones containing the Cycle are reduced. 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. By the way, Sudoku hard solution techniques are included with many of the solutions. Chose between Classic, Monster, Kids and Squiggly, and easy, medium, hard and very hard skill levels We "live-solve" the New York Times "Hard" Sudoku on 31 Jan 19. 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. 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. 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]. Abdlomax 3 yr. ago In 2014, we introduced The Mini Crossword followed by Spelling Bee, Letter Boxed, Tiles and Vertex. Enjoy also learning how to solve Sudoku hard in easy steps. First success by row-column scan: R8C8 4 because of 4 in R7, R9 -- R9C9 2 by scan for 2 in R7, C7. Three digits appear in in the column in only these three cells and nowhere else. 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.
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how to solve new york times sudoku