With 3 in R8C6, R8C7 7 by reduction -- R9C7 3 by reduction -- R8C5 6 by reduction -- R9C5 7 by reduction. With 6 in R6C9, R7C9 4 by reduction -- R7C7 6 by reduction and exception in C7. Every such easy fill at this stage is a bonus. Before going through the solution solve the puzzle first. You can include the QR code to any puzzle that is 9x9, for instance, to make it unique. 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). 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 I also believe these puzzles printed in newspapers are intended for an audience that has no idea what I just said. Abdlomax 3 yr. ago Once a few cells are filled in a zone, enumerating the rest gets easier. You could have identified the formation of this Cycle first to get the breakthrough. 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. Step by step solution to the New York Times Sudoku Hard, 16th February, 2021: Stage 1: Breakthroughs by DSA technique and Cycles. In easy ones it can be 36 or more. Copyright: Atanu Chaudhuri and respective Authors. This single digit lock has been formed by Cross-scan for 6 in R9, C1 and the Cycle (2,6) formed by DSA in R6C3, R8C3. 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). This Sudoku hard is an especially challenging puzzle rich with Sudoku digit patterns. Identify a second Single digit lock on 5 in R9C5, R9C6 by scan for 5 in R7, C4 -- R9C1 3 by possible digit DS elimination of 5 (because of single digit lock on 5 in R9) from possible digit subset [3,5] in R9C1 (because of [1,4] in C1). 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. Description - Self Solving Sudoku - Slice and Dice - Locked Candidates Pointing - Triple Subset - Pair - Locked Candidates Claiming - New York Times Sudoku Hard October 21, 2022 -. These are rough figures drawn from experience. In early 2022, we proudly added Wordle to our collection. But when you face a Sudoku hard, valid cell by row column scan may be few or non-existent to start with. Reduction of instances of 6 from all such empty cells is the gain. By possible digit analysis we have the DSs in R4C3, R5C3 and R6C3 as [1,6], [1,6,7] and [1,6,7]. 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. Solve New York Times Sudoku hard February 20, 2021 in quick steps. 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. Though number of digit 3 is small, we identify first a single digit lock for 3 in R9C7, R9C9 by cross scan for 3 in R7, C8. With 7 in R4C3, R1C3 3 by reduction -- R1C2 7 by reduction. Because of the series of early breakthroughs by advanced digit patterns, the Sudoku hard could be solved rather easily. 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. This is a powerful way to get a valid cell. We eliminate repeated digits when considering digits of two or three zones together. If we get one, we should get usually a breakthrough in one step or two. R4C8 1 by reduction of 8 from DS [1,8] -- R5C8 8 by exception in right middle major square. R7C4 1 by DSA reduction of 9 from DS [1,9] in R7 -- R7C5 9 by exception in R7. Use row-column scan, the simplest method to get a unique digit for a cell. 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). In 2014, we introduced The Mini Crossword followed by Spelling Bee, Letter Boxed, Tiles and Vertex. Before going through the solution solve the puzzle first. By the double digit scan, first the Cycle is formed and then the valid cell R6c1 4 by reduction of [1,8] because of the property of the Cycle. 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. R1C7 2 by scan for 2 in R2, R3, C8 -- R1C8 6 by scan for 6 in R2, R3. 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 this game having no success with 1 or 2 by row column scan select digit 3 as promising and get a few easy placements. This is shown by "5" in the two cells. 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. We don't bother about other digits appearing in these two cells. The possible digit subset in the three empty cells is [1,4,8]. 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. 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. Unique set of existing digits in all these three together are. How to solve the Sudoku hard is explained clearly including all the breakthroughs. Play the Daily New York Times Crossword puzzle edited by Will Shortz online. The NY Times Sudoku Hard 4th March 2021 solved step by step quickly and easily by advanced Sudoku techniques with breakthroughs explained adequately. No more digits can be placed by this simple method of row-column scan at this point. It a Sudoku hard rich with learning potential. This is a breakthrough by single digit lock and DSA. With 6 in R1C5, R1C6 2 by reduction -- R1C1 5 by reduction -- R1C4 8 by reduction. Next R1C8 7 by scan for 7 in R2, R3, C9 and that's all by row column scan at this point. By the way, Sudoku hard solution techniques are included with many of the solutions. Description- Self Solving Sudoku- Slice and Dice- Triple Subset- Hidden Block Pair- Locked Candidates Pointing- New York Times Sudoku Hard October 14, 2022 -. Next identify the advanced technique of Parallel scan for 8 on empty cells of R8: 8 in bottom left major square eliminates R8C1, R8C2, R8C3 for 8 and 8 in C9 eliminates R8C9 for 8 leaving the single cell R8C7 for 8 -- R8C7 8. This is a hard Sudoku puzzle that needed multiple breakthroughs by advanced Sudoku techniques. Move around the grid frequently. 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. . 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. 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. If you get stuck, we're here to help with handy hints, as we are every day. 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]. 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. Be VERY careful about your notes. We stop in a 9 cell major square when the rest of the empty cells would all have digit possibilities 4, 5 or more. Game status shown below. 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. And no more for 2. This is a major breakthrough and we'll see its effects in the next stage. 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. Try free NYT games like the Mini Crossword, Ken Ken, Sudoku & SET plus our new subscriber-only puzzle Spelling Bee. In the process we may get valid cells, but otherwise our lookout is for identifying for a single digit lock by cross scan. 3 Sudoku types . 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. This is a breakthrough that otherwise couldn't have been achieved. To solve quickly you need to concentrate. The . Before going through the solution solve the puzzle first. 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. An often asked questions is, "What makes a Sudoku puzzle hard?". 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. 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. Analyze in which cell digit 5 can appear in R2. Popular sites for tracking the pandemic such the New York Times Covid tracker have become less useful because they rely on PCR testing; most infected people, if they do test, are using antigen . No more easy valid digit 3 possible at this point. JUMBLE. With only 23 filled up, cells this New York Times Sudoku hard puzzle is extra-hard. 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. Alternately, if R4C3 6, digut [1,7] will appear in eithe of the cells R5C3, R6C3. In this case for example, after you put 5 in R2C5, a Cycle of remaining four digits [4,6,7,8] is formed. 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. As a strategy we always start row column scan from digit 1 and continue till digit 9. Copyright: Atanu Chaudhuri and respective Authors. If you just want today's word, you can jump straight to the end of this article for November 3's Wordle solution, for . Bur that would have been time-taking. 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. With 1 in R5C6, R3C6 9 by reduction -- R3C4 1 by exception -- R7C6 6 by reduction. The online Sudoku NYT puzzle in this game must be solved using logic. 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. Set union is technical, but this we do by common sense and do naturally. With [6,9] in R6, R6C7 2 by reduction -- R4C8 3 by reduction --- R4C7 8 by reduction and exception in R4. Fill as many cells as you can by row-column scan. If Costner's 1989 baseball movie had featured zombies, it could have been called - FIELD OF "SCREAMS" (Distributed by Tribune Content Agency) By Jack Tamisiea An elephant's trunk has 40,000 muscles and weighs more than a Burmese python. In the process, an awkward shaped Cycle (1,2,5,6) in bottom left major square formed. The New York Times Sudoku Hard, 17th February, 2021 Before going through the solution solve the puzzle first. Enjoy also learning how to solve Sudoku hard in easy steps. Description- Self Solving Sudoku- Locked Candidates Pointing- Triple Subset- Locked Candidates Claiming- Pair- New York Times Sudoku Hard September 14, 2022-. This is the quick method of parallel scan for digit 8 on empty cells of the row R8. With only 24 filled up cells this New York Times Sudoku hard puzzle is fairly hard. Leftover three digits in C3 form a Cycle (1,6,7) in R4C3, R5C3, R6C3. Strategy adopted: start with row column scan coupled with aggressive breakthroughs by advanced Sudoku techniques. With 9 in R8C2, R8C9 4 by reduction -- R8C9 4 by reduction -- R9C9 9 by reduction. 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. An easy catch by opportunistic scan for 9: R9C9 9 by scan in R7, R8. 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. There is no metric or measurement to decide for sure that a Sudoku puzzle is hard, or not so hard. 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. Hint #2 . In easy ones it can be 36 or more. With 5 in R1C1, R3C2 2 by reduction -- R5C2 6 by reduction -- R6C3 2 by reduction -- R6C4 6 by reduction -- R5C6 1 by reduction -- R5C4 2 by reduction. Three digits appear in in the column in only these three cells and nowhere else. None of the digits could produce any success of valid cell by row column scan. As a ball-park figure, any Sudoku puzzle having number of filled up cells 27 and below should be taken as hard or extremely hard. Next valid cell is R4C6 7 by scan for 7 in R5, R6, C7. 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. Copyright: Atanu Chaudhuri and respective Authors. 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. Nevertheless, a distinction between hard, medium and easy Sudoku categories can be made. Look for placing any other possible digit by row-column scan. R5C3 9 by reduction of 3 -- R4C3 2 by exception in C3. Do you play the New York Times Sudoku? If yes, then we have provided all the NY Times Sudoku Answers. 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. Use Snyder until it fails. About New York Times Games. 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. 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. 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. 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. 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. R2C2 8 by scan for 8 in R3 -- R3C2 4 by exception in top left major square. As a ball-park figure, any Sudoku puzzle having number of filled up cells 27 and below should be taken as hard or extremely hard. 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. 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. By the way, Sudoku hard solution techniques are included with many of the solutions. This makes the digit invalid in any other cell in the row or column in which the digit is locked. Digit possibilities for empty cells are enumerated ONLY WHEN NECESSARY. In the new paper, Dr. Getzin and his colleagues noted that termites were conspicuously missing from their study sites, and that they found no signs of root damage in grass that died after rainfall . R4C1 8 by reduction of 3 from DS [3,8] in C1 -- R5C1 3 by exception in C1 -- R5C7 8 by exception in R5. Copyright: Atanu Chaudhuri and respective Authors. Then start the process for the next higher digit. In medium or easy Sudoku puzzles, number of filled up cells will be more. This is how we get a breakthrough valid cell R8C7 8 by parallel scan. Specifically important is the disallowing of digit 5 in the empty cells of C3. So we'll next start evaluating 2 or 3 digit possible digit subsets DSs in . Never lose an opportunity for a valid cell hit by row column scan. If by DSA reduction we get a single digit possible for a call, we get a valid cell hit. Because of the series of early breakthroughs by advanced digit patterns, the Sudoku hard could be solved rather easily. Results of the actions taken shown below. If you cannot logically place a value in a cell then do not place that value in a cell. 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. How debt-for-climate swaps can help solve low-income countries' fiscal and environmental challenges at the same time As a byproduct the leftover two digits [1,8] form a Cycle of (1,8) in R6C2 and R6C3. R6C5 3 by scan for 3 in R4, R5, C4 -- R8C6 3 by scan in R7, C4, C5. It is a fairly hard Sudoku puzzle. 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. Writing long digit sets takes more time and effort. Cycle (2,6) in two empty cells R6C3, R6C4 and DS [2,6] in R8C3 by DSA reduction of [1,5] from possible digits [1,2,5,6] in R7 -- a breakthrough Cycle (2,6) formed in R6C3, R8C3. 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. 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. In medium or easy Sudoku puzzles, number of filled up cells will be more. 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). Nevertheless, a distinction between hard, medium and easy Sudoku categories can be made. R2C3 5 by parallel digit scan for 5 on empty cells of C3: 5 in R1 debars R1C3 for 5, 5 in left middle major square debars R4C3, R5C3 for 5 leaving single cell R2C3 for 5 -- R2C5 1 by reduction of [5,8] from DS [1,5,8] in C5 -- R4C5 5 by reduction of 1 -- R4C4 1 by reduction and R1C5 8 by exception in C5. 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. Fill as many cells as you can by row-column scan. 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. That means, these two digits light up or affect the intersecting cell R6C1 and reduce its possible digit subset to single digit 3. This eliminates the cell R6C1 for any of the two digits. With 9 in R4C9, R4C1 8 by reduction of 9 from possible digit subset of [8,9] in C1 -- R5C1 9 by exception in C1. Description- Self Solving Sudoku- Slice and Dice- Locked Candidates Pointing- Locked Candidates Claiming- Pairs- Locked Candidates Pointing- New York Times S. There is no metric or measurement to decide for sure that a Sudoku puzzle is hard, or not so hard. 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. You get a valid cell breakthrough in R2C5 5. By this parallel scan for 5 on empty cells of row R2, only one cell R2C5 is left placing digit 5. Both digits 1 and 8 appear together in column C1 but not in any cell of the left major square. R3C8 1 by scan for 1 in R2, C7, C9 R2C1 9 by reduction of [2,6] in R2 from DS [2,6,9] in top left major square -- R2C1 6 by reduction of 2 in C1 -- R3C2 2 by exception in top left major square R4C2 6 by exception in C2. $(1, 3, 6) \cup (3, 4, 7, 9) \cup (1, 3, 5)=(1, 3, 4, 5, 6, 7, 9)$. Since the launch of The Crossword in 1942, The Times has captivated solvers by providing engaging word and logic games. Target is to get two-digit possibilities first. Play the USA TODAY Sudoku Game. One of the best ways is to personalize them using QR codes. To skip the section, click here. An often asked questions is, "What makes a Sudoku puzzle hard?". AFRAID INCOME LAGOON STYLUS AFFECT DUPLEX. We have detected the possibility of getting a valid cell R2C7 8. To be able to solve a Sudoku puzzle in a short time you should know how to use the pencil to trace the board. R6C1 7 results in R9C2 7 by scan for 7 in C1 -- R9C1 5 by exception in bottom left major square. 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. Since the launch of The Crossword in 1942, The Times has captivated solvers by providing engaging word and logic games. In addition Cycle (1,9) is formed in R9C4, R9C6. 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. 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. It is in three zones or areas - row R2, column C5 and 9 cell top-middle major square. This lock on C9 immediately participates in a row column scan for 3 in R7, lock in R9, 3 in C5 and 3 in C6 to produce unique digit 3 in cell R8C4 -- R8C4 3. 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). 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. Note: For more details, you may click on the section link on "How the parallel digit scan works" above and return by clicking on browser back button. It can also be helpful to mark the columns and rows with pencil marks to work through elimination methods. Now do a parallel scan for 1 in empty cells of column C6 to get the valid cell R1C6 1. Description- Self Solving Sudoku- Slice and Dice- Locked Candidates Pointing- Pair- New York Times Sudoku Hard October 31, 2022- Repeat from 20 August 2017- . This is the hallmark of a truly Sudoku hard puzzle game. For easier lower level Sudoku games this systematic method of row column scan should provide a smooth path to the final solution. Cycle (1,4,5) formed in R9 in cells R9C4, R9C5, R9C6 -- DS for two empty cells of bottom middle major square [7,8] -- R7C5 8 by reduction of 7 by 7 in C5 -- R7C6 7 by exception in bottom middle major square as the only possible digit left for the only truly empty cell. Enjoy solving Sudoku hard. You can use a pencil to make mistakes and then erase them. First success by row-column scan: R8C8 4 because of 4 in R7, R9 -- R9C9 2 by scan for 2 in R7, C7. Let's see the effect of this Cycle (1,8). 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. Enjoy also learning how to solve Sudoku hard in easy steps. The elimination of four empty cells for 5 in row R2 is shown by red arrows in the Stage 2 solution figure below. In easy ones it can be 36 or more. Next valid cell R5C6 6 will be shown in next stage. These four digits are locked and are self-sufficient in these four cells, each appearing at least twice in these four cells. It is as if the digits Cycle between the cells: if R4C3 1, 6 and 7 can appear in either of the cells R5C3, R6C3. 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. For example, in this case, reduction of 7 from DS of R6C2 gives us a valid cell hit as R6C2 2. 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. 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. 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. Printable Sudoku New York TimesPrintable Sudoku New York Times - There are many number of ways that to use Printable Sudoku Puzzles. This will result in a series of valid cells and is a major breakthrough. 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. The joy of discoveries will then all be yours. 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. As a strategy we always start row column scan from digit 1 and continue till digit 9. First valid cell by row column scan is for 5: R4C2 5 by scan for 5 in R5, R6. With this, two more empty cells R8C7 and R8C9 are left for 8 in R8. With 7 in R7C6, R1C6 4 by reduction -- R9C6 5 by reduction -- R9C5 1 by reduction -- R9C4 4 by reduction -- R6C5 5 by reduction and exception in C5 -- R1C4 7 by reduction and exception in R1. 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. An often asked questions is, "What makes a Sudoku puzzle hard?". With six digits filled in the left middle major square, it is a very promising zone to look for more valid cell hits. That's why though we start a Sudoku hard solution with systematic row column scan, we also look for other advanced digit patterns for early breakthroughs of valid digits. Observe that the two digits [1,8] appear together in R6 and also in C9, but do not appear in the right middle major square. 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. 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. Short length easy possible digit subsets are evaluated for nearly all empty cells without any easy success. 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. For full enjoyment, avoid looking into any solution as well as the answer. As a subscriber, you have 10 gift articles to give each month. The joy of discoveries will then all be yours. Elimination of four cells of column C6 for digit 1 by parallel scan is shown by blue arrows in Stage 2 solution figure below. Nevertheless, a distinction between hard, medium and easy Sudoku categories can be made. 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 [1,8] will certainly occupy these two cells and in no other empty cell in parent zones R6 and left middle square. 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. That's why for speed, we prefer to identify parallel digit scan possibilities. If you are alert for detecting a parallel scan opportunity, breakthrough would be much quicker. Its very easy to lose track You never need to guess anything on Sudoku. This creates an opportunity for us to apply advanced technique of double digit scan to get a valid cell hit. 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. So only the two missing digits (2, 8) are valid for the cell R2C5. 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. Have to go beyond Snyder notation just to solve Sudoku hard is an especially challenging puzzle rich Sudoku What makes a Sudoku puzzle with only 23 filled up, cells this New York Times games reduction and in In R8C6, R8C7 7 by parallel scan is carried out on the empty cells are enumerated when, C7 R5, C4, C5 3 and the byproduct Cycle of ( 6,8 ) for better of! In R8C6, R8C7 7 by reduction -- R9C2 6 by reduction of from! 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