stuck in this sudoku | |
LMI Essentials -> Solving Techniques | 159 posts • Page 6 of 7 • 1 2 3 4 5 6 7 |
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 | But I have not got some initial markings, which you have shown by deducing from the number clues, Eventually, I got them while solving. Yes .Some of the deductions are obtained as the solve proceeds. | ||
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 | 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. | ||
swaroop2011 |
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PR 2020 (Shading and Loops) Author Posts: 668 Location: India | Nice. | ||
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 . | ||
debmohanty |
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Location: India | kishy72 - 2015-11-13 2:29 AM I am wondering the source of the example . I picked it from my IPC2011 folder - so it had to be bit puzzle-y in nature. However, I'm not able to see this puzzle in any IPC2011 booklet, so this may be one of the rejects because "too-hard-to-start" | ||
swaroop2011 |
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PR 2020 (Shading and Loops) Author Posts: 668 Location: India | debmohanty - 2015-11-13 5:26 AM kishy72 - 2015-11-13 2:29 AM I am wondering the source of the example . I picked it from my IPC2011 folder - so it had to be bit puzzle-y in nature. However, I'm not able to see this puzzle in any IPC2011 booklet, so this may be one of the rejects because "too-hard-to-start" wow ! not ISC but IPC ?? Well anyways 2011 bangalore offline event was real fun. My first National championship :) . | ||
vopani |
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WSPC Organizer Posts: 739 Location: India | kishy72 - 2015-11-13 2:29 AM 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 . Nope. I would have guessed after 30-secs. It took me ~5-6mins to come up with this logic. Instead, I could've solved this by guessing multiple times in 5-6mins :-) | ||
RameshLMI |
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SM 2020 (Converse and Odd Even) Author Posts: 51 Location: India | Thanks Rohan. But I am unable to decipher this logical deduction, R5C1 + R5C2 = 8 R6C1 + R6C2 = 8 Can someone make this clear? | ||
vopani |
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WSPC Organizer Posts: 739 Location: India | RameshLMI - 2015-11-13 2:26 PM Thanks Rohan. But I am unable to decipher this logical deduction, R5C1 + R5C2 = 8 R6C1 + R6C2 = 8 Can someone make this clear? Look at Row4 and Row5. Two sums of 18 and 9 are given (which used 8 cells out of 12). So, the remaining 4 cells: R4C1 + R4C2 + R5C1 + R5C2 = 42 - 9 - 18 = 15 (to make the total of the two rows 42) Since R4C1 and R4C2 were found to be 3 & 4, we get R5C1 + R5C2 = 15 - 7 = 8. And using the 16-sum, R6C1 + R6C2 = 8. | ||
neerajmehrotra |
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Posts: 329 Location: India | I was lucky to have guessed a better start...could solve it easily then.... Its almost impossible for me decipher Rohan's Logic during a competition.... :) | ||
Fred76 |
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Diagonal Vision Author Posts: 337 Location: Switzerland | I thought that you created 2 group sum sudoku 6*6 and this one was too easy for the tournament | ||
ghirsch |
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Posts: 102 Location: United States | I've been working my way through Thomas Snyder's Art of Sudoku book, and I'm pretty stuck on this one. Can anyone help me out (preferably with just a step or two to get me on the right track)? I'm probably just missing something obvious but I can't seem to come up with any ideas for what to do. Edited by ghirsch 2015-11-15 4:51 AM | ||
swaroop2011 |
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PR 2020 (Shading and Loops) Author Posts: 668 Location: India | ghirsch - 2015-11-15 4:47 AM I've been working my way through Thomas Snyder's Art of Sudoku book, and I'm pretty stuck on this one. Can anyone help me out (preferably with just a step or two to get me on the right track)? I'm probably just missing something obvious but I can't seem to come up with any ideas for what to do. 9 at r7c7 due to 34 pair at r1c7 and r6c7 Edited by swaroop2011 2015-11-15 6:13 AM | ||
ghirsch |
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Posts: 102 Location: United States | Thanks Swaroop, that did the trick. It's a pretty tough deduction to find though, I guess I still need more practice. | ||
kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | Anti-diagonal from the Russian GP .I solved this particular sudoku for a long time and had to guess to finish it .Could someone tell me how to continue logically here ? Edited by kishy72 2016-04-05 1:39 PM | ||
rajeshk |
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WSPC Organizer Posts: 542 Location: India | kishy72 - 2016-04-05 1:38 PM Anti-diagonal from the Russian GP .I solved this particular sudoku for a long time and had to guess to finish it .Could someone tell me how to continue logically here ? Center has to be 1. In case we put 7 there then there will be no place left to put 7 in 6th Box. | ||
kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | The following sudoku is from CSOC 63 PB.I couldn't understand the English version of the rules given in the example image.I assumed it to be a clone sudoku and started solving like that in contest.However, it quickly broke after that. Can someone clarify what the rules imply ? (CP.png) (IB.png) Attachments ---------------- CP.png (61KB - 1 downloads) IB.png (100KB - 0 downloads) | ||
kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | Also, can the admin kindly specify alternate websites for uploading images here ? Earlier, I used to upload from tinypic which seems to have shut down now. The images posted above are screenshots and are too big and uncomfortable looking. | ||
xiao01wei |
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Posts: 6 Location: China | CSOC 63 P13's rule: if there is two same shape, for example, two vertical domino shape, there will be 4 numbers in these 2 shapes: A C B D and if A>B, and then the result is C>D at the same time if A<B, and then the result is C<D at the same time if C>D, and then the result is A>B at the same time if C<D, and then the result is A<B at the same time the same inequality between the four numbers in two shapes with corresponding position Edited by xiao01wei 2019-10-26 7:30 PM | ||
kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | xiao01wei - 2019-10-26 7:23 PM CSOC 63 P13's rule: if there is two same shape, for example, two vertical domino shape, there will be 4 numbers in these 2 shapes: A C B D and if A>B, and then the result is C>D at the same time if A if C>D, and then the result is A>B at the same time if C Thanks a lot Xiao Wei for the clarification ! I finished the sudoku now.It is certainly an interesting variant and I will look forward to seeing more of it in future. | ||
kishy72 |
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SM 2020 (Math) Author Posts: 419 Location: India | Little Killer from LMD portal by 'Realshaggy' This sudoku is pure evil.I tried everything that I usually encounter in Little Killer Sudokus from totalling clues,seeing min-max possibilities, clue interaction etc., but this sudoku just yielded nothing.I have been trying for the past week or so to complete this without success.It's kind of demotivating in a way that after so many years of solving sudoku, there are still some that I am unable to be complete. Kindly someone share the break-in for this ' psycho little killer '. (LK.jpg) Attachments ---------------- LK.jpg (45KB - 2 downloads) | ||
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