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 stuck in this sudoku | |
| LMI Essentials -> Solving Techniques | 159 posts • Page 5 of 7 • 1 2 3 4 5 6 7 |
prasanna16391 |
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2021 World Sudoku+Puzzle Convention Organizer Posts: 1809    Location: India | |||
| vopani |
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WSPC Organizer Posts: 739  Location: India | |||
| prasanna16391 |
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2021 World Sudoku+Puzzle Convention Organizer Posts: 1809 Location: India | |||
| gaurav.kjain |
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SM 2020 (Irregular and Outside) Author Posts: 52 Location: India | |||
| kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | A Consecutive Sudoku from the US Sudoku Team Qualification round by Wei-Hwa Huang.I greatly enjoyed this very delightful set ! All the sudokus had a unique solve path and were beautifully constructed.I was more or less sprinting through the test when I hit this one and spent around 27-28 mins on this single sudoku.I adopted a very mechanical approach and found that R2C6 was the start in my case.I eliminated digits one by one in R2C6 which was very tiresome.Surely, there must be an alternative easier solve than the one mentioned.If someone has a better start,please let me know.....  Kishore Edited by kishy72 2015-09-08 12:03 PM | ||
| prasanna16391 |
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2021 World Sudoku+Puzzle Convention Organizer Posts: 1809 Location: India | What I did was, 1. Mark odds/evens. In 2nd box, there are three distinct consecutive pairs, which would take one odd and one even each, meaning two odds and one even go in the series of 3. The series of 3 cannot have 7 and so cannot have 9 either and so must be 123 or 345. Either way, 3 is part of it, and there must be groups of 6-7 and 8-9 in that box elsewhere. 2. Then, note that R3C8 to R3C7 is a series of 6 and that R1C7 = R2C6. 3. Where can 8 be in the 3rd box? Can't be in R1C8 because that is odd (using C8). Can't be in R1C7 or R2C7 because both are 3rd/4th cells of a 6-cell series (considering R1C7=R2C6. 8 cannot be in R2C8 either because that causes a 5-6 pair in 2nd box which isn't allowed going by step 1. So, 8 is at an extreme of the 6-cell series. 4. The series has to be 8-7-6-5-4-3 or 3-4-5-6-7-8. Either way, R1C7, R2C7 will form a 5-6 pair, which means 7 cannot be in R2C9 or R3C9 because 6 is taken. So, 7 must be in R2C7 and the series of 6 is done. (Consecutive.png) Attachments ----------------  Consecutive.png (14KB - 6 downloads) | ||
| kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | |||
| prasanna16391 |
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2021 World Sudoku+Puzzle Convention Organizer Posts: 1809 Location: India | |||
| kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | |||
| abhi265645 |
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| swaroop2011 |
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PR 2020 (Shading and Loops) Author Posts: 668 Location: India | |||
| abhi265645 |
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| kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | 'Flower' from Riad's contest.I have managed to complete the other 8 except this.Strangely,most of the puzzles to me seemed to be a contest in itself and took more than an hour and close to a 2.Can someone post the continuation here?  Edited by kishy72 2015-09-24 7:00 PM | ||
| gaurav.kjain |
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SM 2020 (Irregular and Outside) Author Posts: 52 Location: India | Some cells are divided into GROUP A and B and possible Group positions are marked. 'A' group 2 cells should be same as 'A' middle cell group and 'B' group cells in Box2 should become part of B group cells in marked positions. B connections are shown, so because of that B group cells as shown connected, with violet colour should be same  kishy72 - 2015-09-24 6:59 PM 'Flower' from Riad's contest.I have managed to complete the other 8 except this.Strangely,most of the puzzles to me seemed to be a contest in itself and took more than an hour and close to a 2.Can someone post the continuation here? | ||
| kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | |||
| gaurav.kjain |
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SM 2020 (Irregular and Outside) Author Posts: 52 Location: India | Cross number sudoku, Either I didnot understand the rules or there is serious error in Cross Numbers marking, round represent cross number marking, you can check column 5 [1 1 1] is not satisfied. Someone please help.  imag | ||
| kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | gaurav.kjain - 2015-09-25 11:39 AM Cross number sudoku, Either I didnot understand the rules or there is serious error in Cross Numbers marking, round represent cross number marking, you can check column 5 [1 1 1] is not satisfied. Someone please help. This is how the crossword goes.See where you have gone wrong.  | ||
| gaurav.kjain |
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SM 2020 (Irregular and Outside) Author Posts: 52 Location: India | Ohh this R6c6/7 and R7C6/7 creating 2x2 which I was thinking no 2x2 is possible in cross number, This was my confusion and this was prohibiting me from putting any more markings. Thanks Kishore Please give me some starting steps for wall sudoku. kishy72 - 2015-09-25 12:32 PM gaurav.kjain - 2015-09-25 11:39 AM Cross number sudoku, Either I didnot understand the rules or there is serious error in Cross Numbers marking, round represent cross number marking, you can check column 5 [1 1 1] is not satisfied. Someone please help. This is how the crossword goes.See where you have gone wrong. | ||
| kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | Please give me some starting steps for wall sudoku. This one is hard to explain.I myself kind of stumbled to the solution after a long time.I will run you though the basic starting steps. Remember that in this sudoku,there are 2 "golden" rules. Rule 1 :: If there is no wall segment between 2 cells, those 2 cells are consecutive in value. Rule 2 :: Each digit(unless it is a 1 or 12) will have connection to 2 digits which is consecutive to it .This means that if you get connection to 2 cells ,you can draw wall segments on the other 2 borders and vice versa. Digits 1 and 12 will have wall segments on 3 border grid lines. * The zero clue indicates that there can be no wall segments on that line .Hence I marked 'x''s on that line .This means cells R23/C1,R23/C2,R23/C3.......R23/C12 are consecutive in value. *Also I have marked red lines on some borders .These are obtained by deduction from the given clues.For instance,reason out how the (37)clue at the top can be satisfied. *Look at R4C4 .It cannot be a 1 or 12.There are 2 red wall segments that are obtained by deduction from the clues.So from what I stated above in Rule 2 ,you can draw connections to 2 cells.  Give it a start and get back if you are stuck ! Kishore Edited by kishy72 2015-09-25 1:33 PM | ||
| gaurav.kjain |
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SM 2020 (Irregular and Outside) Author Posts: 52 Location: India | Finally I am done with THE Maze Monster. Kishore your initial steps and explanation helped a lot. Thanks for wonderful explanation.  But I have not got some initial markings, which you have shown by deducing from the number clues, Eventually, I got them while solving. Can you give me reason of markings highlighted in BLUE.  Edited by gaurav.kjain 2015-09-26 2:50 PM | ||
| kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | |||
| kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | After an indefinite hiatus,I am posting here in the forum again.At this instance, a Group sum sudoku from the IB of SM - 4 I tried this one a lot and got a lot of pencil marks going but not one concrete digit.I think it would be futile to post the one I had pencil marks on as it yielded nothing.Hence I am posting the image as is..... Can someone please tell me how to put one digit without much effort ?  Edited by kishy72 2015-11-12 9:16 PM | ||
| vopani |
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WSPC Organizer Posts: 739 Location: India | |||
| swaroop2011 |
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PR 2020 (Shading and Loops) Author Posts: 668 Location: India | |||
| kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | Rohan Rao - 2015-11-12 10:51 PM Not sure if this is the best start, but here's what I did. It took me a while though. R5C1 & R6C1 = 5 & 6 R5C2 + R6C2 = 5 R5C3 + R6C3 = 5 So, R2C3 + R3C3 = 5 OR 6 OR 7 It cannot be 7 because there is a 7-sum-pair below it (R3C3 + R4C3). If it is 6, then R2C2 & R3C2 = 4 & 6 and R5C2 & R6C2 = 1 & 4 which is not possible (Basically, you can't fill up the 16-sum and 10-sum of column2-column3 together). So, R2C3 + R3C3 = 5 R2C2 + R3C2 = 11 = 5 & 6 R2C1 + R3C1 = 3 = 1 & 2 R1C3 + R4C3 = 11 = 5 & 6 R3C1 & R3C3 = 1 & 2 R4C1 & R4C2 = 3 & 4 So, R5C1 + R5C2 = 8 R6C1 + R6C2 = 8 Hence, R5C2 & R6C2 = 2 & 3 R4C2 = 4 ! From there, it should get solved. Thanks Rohan ! Excellent chain of logical thinking .But seriously,were all those meant to deduced during a test and if anyone would wait for logic to put in a digit ?  I am wondering the source of the example . | ||
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