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Converse - 12th-15th Feb - Sudoku Mahabharat & ISC Qualifier | |

LMI Tests -> Sudoku Mahabharat | 74 posts • Page 1 of 3 • 1 2 3 |

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Location: India | Link for rules and other details about Sudoku Mahabharat : http://logicmastersindia.com/SM/2015-16.asp Instructions for Converse - http://logicmastersindia.com/SM/201602/ | |||||||||||||||

smitra |
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Posts: 1 | how to participate in this ? | |||||||||||||||

prasanna16391 |
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PR 2020 (Evergreens) Author Posts: 1414 Location: India | Please go through the F.A.Q. here - http://logicmastersindia.com/forum/forums/thread-view.asp?tid=381 If you have any further queries, feel free to ask. | |||||||||||||||

debmohanty |
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Location: India | Since it coincides with TVC XIX, this will start a day earlier (i.e. on 12^{th}) than usual SM episodes. | |||||||||||||||

Administrator |
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Location: India | Instructions published. Link: http://logicmastersindia.com/SM/201602/ | |||||||||||||||

vopani |
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WSPC Organizer Posts: 725 Location: India | Can the duration of the test be added to the IB? I'm sure most of us know the timings and duration, but it was pointed out last time and there's a chance a newcomer 'might' miss it too. | |||||||||||||||

kishy72 |
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SM 2020 (Math) Author Posts: 395 Location: india | Just to clarify . Please correct me if I am wrong. In QuadMax 4Odd 4Even Sudoku ,the arrows provide 2 kinds of information. i)All 4 cells comprised by a single arrow are of the same parity . ii)The arrows point towards the cell with the highest digit.In case there are 2 digits with the highest value,they will be represented by arrows with double heads. Also ,apart from the above,I remember seeing Average in an earlier SM round (not this year) and finding it unpleasantly sticky.Can someone direct me straight to that round? I think that round was authored by Rohan or Rakesh ?! | |||||||||||||||

shera90 |
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Posts: 71 Location: India | In kropki, if no dots are there can it be supposed that neither the difference is 1 nor one one digit is half of other ? And one question more, can someone give me few starting clues as to how to solve average 6*6 and average 9*9 in IB because I am confused from where to start | |||||||||||||||

swaroop2011 |
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PR 2020 (Shading and Loops) Author Posts: 611 Location: india | kishy72 - 2016-02-07 3:07 AM Also ,apart from the above,I remember seeing Average in an earlier SM round (not this year) and finding it unpleasantly sticky.Can someone direct me straight to that round? I think that round was authored by Rohan or Rakesh ?! Yes , it was in last year's Converse SM round authored by Me :) Here is link: http://logicmastersindia.com/lmitests/?test=SM201503 | |||||||||||||||

swaroop2011 |
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PR 2020 (Shading and Loops) Author Posts: 611 Location: india | shera90 - 2016-02-07 3:35 AM And one question more, can someone give me few starting clues as to how to solve average 6*6 and average 9*9 in IB because I am confused from where to start For 6x6, look at 3rd row. There will be 1 at R3c6 and 6 at r4c1. This should give you start. Remember Extreme numbers can never be on the cell with average symbol. i.e. 1,6 for this example and 1,9 for 9x9 grid. I am posting one more post with practice links and tips. That will help you to get started. Edited by swaroop2011 2016-02-07 4:17 AM | |||||||||||||||

swaroop2011 |
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PR 2020 (Shading and Loops) Author Posts: 611 Location: india | This was from last year's Converse Round, authored by me, and i had shared some helpful links and tips that time. Hope that helps again. Follow the link: http://swaroopg92.blogspot.com/2015/03/sudoku-mahabharat-episode-7-... | |||||||||||||||

prasanna16391 |
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PR 2020 (Evergreens) Author Posts: 1414 Location: India | In addition to Swaroop's post of assembled links and tips, I thought I'd share the only site where I have prominently seen Quad Max 4 Even 4 Odd - http://www.argio-logic.net/ . If I remember right (I have been auto-logged in for years now), registration is done by clicking on the forum link. Once registered, clicking on the 'Sudoku' link near the top left of the home page link I shared should give a list of Sudoku variants where Quad Max 4 Even 4 Odd will be there. | |||||||||||||||

rakesh_rai |
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Mean Minis (2020) Author Posts: 771 Location: India | Rohan Rao - 2016-02-06 6:10 PM We will add to the IB in the next version when we also add the points table. Can the duration of the test be added to the IB? kishy72 - 2016-02-07 3:07 AM Precisely! There are three possible cases - (a) No arrows: The 4 cells do not all have the same parity, (b) One sided arrow: All 4 cells have same parity and the cell where arrow points has the highest value. (c) Two sided arrow: All 4 cells have same parity and both cells where the arrow points have the highest value. Just to clarify . Please correct me if I am wrong. In QuadMax 4Odd 4Even Sudoku ,the arrows provide 2 kinds of information. i)All 4 cells comprised by a single arrow are of the same parity . ii)The arrows point towards the cell with the highest digit.In case there are 2 digits with the highest value,they will be represented by arrows with double heads. shera90 - 2016-02-07 3:35 AM Absolutely!In kropki, if no dots are there can it be supposed that neither the difference is 1 nor one one digit is half of other ? | |||||||||||||||

Raphael |
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It is an interesting twist that the number of points that the X-sums frame sudoku is worth must be figured out during solving! ;) | ||||||||||||||||

vopani |
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WSPC Organizer Posts: 725 Location: India | For those who missed the Sudoku GP Round-2 last weekend, it might be useful to try out some of the variants. Though, not explicitly mentioned, many of them turned out to be 'Converse'-like variants. Good practise before the Converse round in SM. http://gp.worldpuzzle.org/content/sudoku-gp | |||||||||||||||

rakesh_rai |
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Mean Minis (2020) Author Posts: 771 Location: India | A partcipant asked this question. Sharing it here as it may be relevant for all. i hv a question regarding x sum frame in converse variations sudoku instruction booklet in the 6X6 grid m getting left with numbers 4 and 3 left in the second column of the first grid and same numbers in the parallel grid in 5th column. can you tell me where m getting it wrong. Since there is no X to the left of the first row, by converse rule, the sum of the first three numbers cannot be equal to X. (X=7 here) If row 1 column 2 has a 4 the sum of first three numbers in row 1 becomes 7. So this cell must contain 3. | |||||||||||||||

rajeshk |
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WSPC Organizer Posts: 521 Location: India | Here are some of the practice puzzles for the three variations appearing in this test Average Sudoku: http://www.funwithpuzzles.com/2015/11/average-sudoku-puzzles.html Consecutive Sudoku: http://www.funwithpuzzles.com/2015/06/non-consecutive-sudoku-puzzle... Kropki Sudoku: http://www.funwithpuzzles.com/2015/09/kropki-puzzles.html | |||||||||||||||

Administrator |
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Location: India | ## Points TableInstruction Booklet is re-uploaded with points table. Here it is for your quick reference. | |||||||||||||||

An LMI player |
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shera90 |
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Posts: 71 Location: India | Anybody give some clues on kropki sudoku 9*9 from IB. I want to know how to start as there are very few clues and I don't know where to start. And also please give some starting clues for x sum frame 9*9 from IB. Edited by shera90 2016-02-12 8:14 AM | |||||||||||||||

rakesh_rai |
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Mean Minis (2020) Author Posts: 771 Location: India | shera90 - 2016-02-12 7:44 AM Anybody give some clues on kropki sudoku 9*9 from IB. I want to know how to start as there are very few clues and I don't know where to start. Look at column 4. Because of the 2s in col5 and col6, a 2 must come in C4R1 or C4R2 -> this means C4R1 and C4R2 must have the numbers 2 and 3. Then C4R4 has possible options as 5,6,7. 5 and 7 are not possible because of converse rule. So it must have the number 6. C4R6 now has possible options 5 or 7. If it is 7 then C4R8 and C4R9 will be 4 and 5, which is not possible. So C4R6 must be 5. Now C4R9 must be 4 as we have a black circle. And C4R7 is 7. In general, try to start around black circles, as they are the most likely places to start. Then try to look for long consecutive sequences in a row/column/box. If you are stuck or there is less progress initially, some amount of pencilmarking around the marked circles can help get the breakthrough. After that make sure you do not forget classic rules AND converse rules. For example, if a cell has 3 and there are no circles then the neighbours cannot be 2,4, or 6. I hope you will be able to solve this one now. | |||||||||||||||

Guest |
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>>>And also please give some starting clues for x sum frame 9*9 from IB. IB X sum frame 9x9: C4R4 = 4 (singlet). Concentrate on C5... C5R4 and C5R6 must contain (2,6). So, R5C4 and R5C6 must contain (3,8). Also observe that, C3R4 = C7R6 = 8. | ||||||||||||||||

rakesh_rai |
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Mean Minis (2020) Author Posts: 771 Location: India | Guest - 2016-02-12 8:45 AM >>>And also please give some starting clues for x sum frame 9*9 from IB. IB X sum frame 9x9: C4R4 = 4 (singlet). Concentrate on C5... C5R4 and C5R6 must contain (2,6). So, R5C4 and R5C6 must contain (3,8). Also observe that, C3R4 = C7R6 = 8. Can you explain how you got (2,6) and (3,8) ? | |||||||||||||||

rakesh_rai |
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Mean Minis (2020) Author Posts: 771 Location: India | >>>And also please give some starting clues for x sum frame 9*9 from IB. The starting approach has to be to find the possibilities for X, without forgetting classic rules. For example, in the IB, if we look at column 6, the first three numbers can have three numbers from 3,5,8 and 9. So the possible values of X are 16, 17, 20 or 22. Now look at row 6. The possible values for first three numbers are 2,6,7,8 and 9. Using these numbers, we cannot make 20. So X cannot be 20. Also, if X=16, then we need to have 2,6 and 8 as first three numbers. Then we cannot place any number in R6C5. So X cannot be 16 either. The possible values for X are now 17 or 22. Similarly, you can deduct that X cannot be 17 as it leads to a contradiction in Boxes 4/5/6. Once X is identified as 22, there are only two ways to make 22 (589 or 679). You will be able to solve easily after that. | |||||||||||||||

harmeets |
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>>>Can you explain how you got (2,6) and (3,8) ? In column C5, the middle 3 numbers (in r4,r5,r6) must add up to 12. Two numbers are 4 and 2, so 3rd one must be 6. Why the middle numbers add to 12? Because in C5, 1st 3 numbers (r1,r2,r3) = (45 - X - 12), and last 3 numbers (r7,r8,r9) = X. So middle 3 numbers = (45 - (45 - X - 12) - X) = 12. | ||||||||||||||||

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