@ 2015-09-26 2:42 PM (#19465 - in reply to #19454) (#19465) Top | |
Posts: 52 Country : India | gaurav.kjain posted @ 2015-09-26 2:42 PM 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 |
@ 2015-09-26 6:38 PM (#19468 - in reply to #19465) (#19468) Top | |
Posts: 417 Country : India | kishy72 posted @ 2015-09-26 6:38 PM 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. |
@ 2015-11-12 9:15 PM (#19907 - in reply to #8611) (#19907) Top | |
Posts: 417 Country : India | kishy72 posted @ 2015-11-12 9:15 PM 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 |
@ 2015-11-12 10:51 PM (#19909 - in reply to #8611) (#19909) Top | |
Posts: 739 Country : India | vopani posted @ 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. |
@ 2015-11-13 12:26 AM (#19911 - in reply to #19909) (#19911) Top | |
Posts: 668 Country : India | swaroop2011 posted @ 2015-11-13 12:26 AM Nice. |
@ 2015-11-13 2:29 AM (#19912 - in reply to #19909) (#19912) Top | |
Posts: 417 Country : India | kishy72 posted @ 2015-11-13 2:29 AM 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 . |
@ 2015-11-13 5:26 AM (#19913 - in reply to #19912) (#19913) Top | |
Country : India | debmohanty posted @ 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" |
@ 2015-11-13 7:43 AM (#19914 - in reply to #19913) (#19914) Top | |
Posts: 668 Country : India | swaroop2011 posted @ 2015-11-13 7:43 AM 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 :) . |
@ 2015-11-13 9:21 AM (#19915 - in reply to #19912) (#19915) Top | |
Posts: 739 Country : India | vopani posted @ 2015-11-13 9:21 AM 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 :-) |
@ 2015-11-13 2:26 PM (#19917 - in reply to #19909) (#19917) Top | |
Posts: 51 Country : India | RameshLMI posted @ 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? |
@ 2015-11-13 2:33 PM (#19918 - in reply to #19917) (#19918) Top | |
Posts: 739 Country : India | vopani posted @ 2015-11-13 2:33 PM 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. |
@ 2015-11-13 11:29 PM (#19919 - in reply to #8611) (#19919) Top | |
Posts: 328 Country : India | neerajmehrotra posted @ 2015-11-13 11:29 PM 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.... :) |
@ 2015-11-13 11:57 PM (#19920 - in reply to #8611) (#19920) Top | |
Posts: 337 Country : Switzerland | Fred76 posted @ 2015-11-13 11:57 PM I thought that you created 2 group sum sudoku 6*6 and this one was too easy for the tournament |
@ 2015-11-15 4:47 AM (#19946 - in reply to #8611) (#19946) Top | |
Posts: 102 Country : United States | ghirsch posted @ 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. Edited by ghirsch 2015-11-15 4:51 AM |
@ 2015-11-15 6:13 AM (#19947 - in reply to #19946) (#19947) Top | |
Posts: 668 Country : India | swaroop2011 posted @ 2015-11-15 6:13 AM 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 |
@ 2015-11-15 6:39 AM (#19948 - in reply to #8611) (#19948) Top | |
Posts: 102 Country : United States | ghirsch posted @ 2015-11-15 6:39 AM Thanks Swaroop, that did the trick. It's a pretty tough deduction to find though, I guess I still need more practice. |
@ 2016-04-05 1:38 PM (#21433 - in reply to #8611) (#21433) Top | |
Posts: 417 Country : India | kishy72 posted @ 2016-04-05 1:38 PM |
@ 2016-04-05 6:00 PM (#21436 - in reply to #21433) (#21436) Top | |
Posts: 542 Country : India | rajeshk posted @ 2016-04-05 6:00 PM |
@ 2019-10-25 1:25 PM (#27343 - in reply to #8611) (#27343) Top | |
Posts: 417 Country : India | kishy72 posted @ 2019-10-25 1:25 PM 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) |
@ 2019-10-25 1:40 PM (#27344 - in reply to #8611) (#27344) Top | |
Posts: 417 Country : India | kishy72 posted @ 2019-10-25 1:40 PM 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. |
@ 2019-10-26 7:23 PM (#27348 - in reply to #8611) (#27348) Top | |
Posts: 6 Country : China | xiao01wei posted @ 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<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 |
@ 2019-10-26 11:38 PM (#27350 - in reply to #27348) (#27350) Top | |
Posts: 417 Country : India | kishy72 posted @ 2019-10-26 11:38 PM 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. |
@ 2019-11-23 12:36 PM (#27362 - in reply to #8611) (#27362) Top | |
Posts: 417 Country : India | kishy72 posted @ 2019-11-23 12:36 PM 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) |
@ 2019-11-23 6:15 PM (#27363 - in reply to #8611) (#27363) Top | |
Posts: 25 Country : United Kingdom | Puzzle_Maestro posted @ 2019-11-23 6:15 PM If you add up the clues on the top and bottom rows, you get 369. Adding up the clues in the left and right gives 327. The difference between the two is 42. Observe that the arrows on the left and right point to 6 cells which are not pointed to by the arrows on top and bottom (with minimum sum 3*(1+2)=9), and similarly the arrows on top and bottom point to 6 cells not pointed to by arrows on the left and right (maximum sum 3*(8+9)=51). The difference between them must be 42, and since the maximum difference between them is 42, we can establish a few 12 and 89 pairs. |
@ 2019-11-25 8:52 AM (#27364 - in reply to #27363) (#27364) Top | |
Posts: 417 Country : India | kishy72 posted @ 2019-11-25 8:52 AM Puzzle_Maestro - 2019-11-23 6:15 PM If you add up the clues on the top and bottom rows, you get 369. Adding up the clues in the left and right gives 327. The difference between the two is 42. Observe that the arrows on the left and right point to 6 cells which are not pointed to by the arrows on top and bottom (with minimum sum 3*(1+2)=9), and similarly the arrows on top and bottom point to 6 cells not pointed to by arrows on the left and right (maximum sum 3*(8+9)=51). The difference between them must be 42, and since the maximum difference between them is 42, we can establish a few 12 and 89 pairs. Thanks a lot ! I could complete it with your pointer. |