@ 2015-09-08 4:33 PM (#19275 - in reply to #8611) (#19275) Top | |
Posts: 1801 Country : India | prasanna16391 posted @ 2015-09-08 4:33 PM 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) |
@ 2015-09-08 6:25 PM (#19279 - in reply to #19275) (#19279) Top | |
Posts: 419 Country : India | kishy72 posted @ 2015-09-08 6:25 PM prasanna16391 - 2015-09-08 4:33 PM What I did was, 1. Mark odds/evens. For reasons unknown,I fail to use this simple yet highly powerful notation.Now,I clearly understand where I am missing the trick in this variant of sudoku.Often it is the consecutive sudoku that tends to trip me up a lot in any test.Just like the in/out rule of an irregular,I am beginning to comprehend that this notation of odd/even is indispensable when it comes to consecutive and possibly a lot other variants. Can't be in R1C7 or R2C7 because both are 3rd/4th cells of a 6-cell series I never spotted this 6-cell series and I never would have if it hadn't been for the explanation you gave.This is mainly due to the fact that the series bends back on itself.I was under the impression that it leads to additional possibilities rather than a straight 6-cell consecutive chain.I am curious to know whether you spotted it right away?If yes,what made you sure that it is a series of straight 6 consecutive numbers ? and thanks a ton for the lovely explanation and for making me a better solver right from this moment ! Kishore |
@ 2015-09-08 8:24 PM (#19280 - in reply to #8611) (#19280) Top | |
Posts: 1801 Country : India | prasanna16391 posted @ 2015-09-08 8:24 PM Because of the converse rule, a 'T' shape of consecutive bars will always be a series of 4. The series in question clearly extends on the T since the R2C7 continuation to R2C8 cannot be the same as R2C6. I also notice a series of 8 in the bottom left btw, but didn't use that till later. The series of 8 is a bit more difficult to see, but note that R7C3 is the intersection of two T-shapes, so if it is the same series going outward from there then the same digit will be in R6C2 and R8C3. |
@ 2015-09-08 8:52 PM (#19281 - in reply to #19280) (#19281) Top | |
Posts: 419 Country : India | kishy72 posted @ 2015-09-08 8:52 PM Thank you.That makes it clear! |
@ 2015-09-16 7:43 PM (#19405 - in reply to #8611) (#19405) Top | |
Posts: 10 Country : India | abhi265645 posted @ 2015-09-16 7:43 PM Not been able to move after this... Please help. TIA<a href="http://tinypic.com?ref=32zjnma" target="_blank"><img src="http://i60.tinypic.com/32zjnma.jpg" border="0" alt="Image and video hosting by TinyPic"></a> |
@ 2015-09-16 8:32 PM (#19407 - in reply to #8611) (#19407) Top | |
Posts: 668 Country : India | swaroop2011 posted @ 2015-09-16 8:32 PM I think last row, 1 will come at C3. |
@ 2015-09-17 11:58 AM (#19411 - in reply to #19407) (#19411) Top | |
Posts: 10 Country : India | abhi265645 posted @ 2015-09-17 11:58 AM Thank you @Swaroop. |
@ 2015-09-24 6:59 PM (#19448 - in reply to #8611) (#19448) Top | |
Posts: 419 Country : India | kishy72 posted @ 2015-09-24 6:59 PM |
@ 2015-09-24 10:29 PM (#19449 - in reply to #19448) (#19449) Top | |
Posts: 52 Country : India | gaurav.kjain posted @ 2015-09-24 10:29 PM 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 |
@ 2015-09-24 10:51 PM (#19450 - in reply to #19449) (#19450) Top | |
Posts: 419 Country : India | kishy72 posted @ 2015-09-24 10:51 PM Thanks!!Was difficult to spot.I could get only the A group .Now I see what I missed. Kishore |
@ 2015-09-25 11:39 AM (#19451 - in reply to #19450) (#19451) Top | |
Posts: 52 Country : India | gaurav.kjain posted @ 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. imag |
@ 2015-09-25 12:32 PM (#19452 - in reply to #19451) (#19452) Top | |
Posts: 419 Country : India | kishy72 posted @ 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. |
@ 2015-09-25 12:42 PM (#19453 - in reply to #19452) (#19453) Top | |
Posts: 52 Country : India | gaurav.kjain posted @ 2015-09-25 12:42 PM 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. |
@ 2015-09-25 1:31 PM (#19454 - in reply to #19453) (#19454) Top | |
Posts: 419 Country : India | kishy72 posted @ 2015-09-25 1:31 PM 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 |
@ 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: 419 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: 419 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: 419 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. |