@ 2016-02-12 8:45 AM (#20963 - in reply to #20961) (#20963) Top | ||||||||||||||||
Guest | Guest posted @ 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. | |||||||||||||||
@ 2016-02-12 8:59 AM (#20964 - in reply to #20963) (#20964) Top | ||||||||||||||||
Posts: 774 Country : India | rakesh_rai posted @ 2016-02-12 8:59 AM 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) ? | |||||||||||||||
@ 2016-02-12 9:01 AM (#20965 - in reply to #20964) (#20965) Top | ||||||||||||||||
Posts: 774 Country : India | rakesh_rai posted @ 2016-02-12 9:01 AM >>>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. | |||||||||||||||
@ 2016-02-12 11:11 AM (#20966 - in reply to #20965) (#20966) Top | ||||||||||||||||
harmeets | harmeets posted @ 2016-02-12 11:11 AM >>>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. | |||||||||||||||
@ 2016-02-12 4:03 PM (#20967 - in reply to #20869) (#20967) Top | ||||||||||||||||
Posts: 739 Country : India | vopani posted @ 2016-02-12 4:03 PM
Really nice sudokus! I was lost on both the Average sudokus... but otherwise it went smooth. QuadMax 9x9 and X-Frame 9x9 were my favourites. Thanks Harmeet and Rakesh. | |||||||||||||||
@ 2016-02-12 4:12 PM (#20968 - in reply to #20869) (#20968) Top | ||||||||||||||||
An LMI player | An LMI player posted @ 2016-02-12 4:12 PM
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@ 2016-02-12 5:29 PM (#20969 - in reply to #20869) (#20969) Top | ||||||||||||||||
Posts: 12 Country : Poland | marred posted @ 2016-02-12 5:29 PM
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@ 2016-02-12 7:49 PM (#20970 - in reply to #20869) (#20970) Top | ||||||||||||||||
An LMI player | An LMI player posted @ 2016-02-12 7:49 PM
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@ 2016-02-12 10:13 PM (#20971 - in reply to #20869) (#20971) Top | ||||||||||||||||
An LMI player | An LMI player posted @ 2016-02-12 10:13 PM
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@ 2016-02-12 10:40 PM (#20972 - in reply to #20869) (#20972) Top | ||||||||||||||||
Posts: 30 Country : Poland | margareta456 posted @ 2016-02-12 10:40 PM
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@ 2016-02-12 10:46 PM (#20973 - in reply to #20869) (#20973) Top | ||||||||||||||||
Posts: 44 Country : Thailand | MrLiang posted @ 2016-02-12 10:46 PM
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@ 2016-02-13 12:40 AM (#20975 - in reply to #20869) (#20975) Top | ||||||||||||||||
Posts: 45 Country : India | Swagatam posted @ 2016-02-13 12:40 AM
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@ 2016-02-13 3:15 AM (#20976 - in reply to #20869) (#20976) Top | ||||||||||||||||
Posts: 24 Country : Poland | wgryciuk posted @ 2016-02-13 3:15 AM
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@ 2016-02-13 3:20 AM (#20977 - in reply to #20869) (#20977) Top | ||||||||||||||||
Posts: 63 Country : United Kingdom | David McNeill posted @ 2016-02-13 3:20 AM
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@ 2016-02-13 3:23 PM (#20979 - in reply to #20869) (#20979) Top | ||||||||||||||||
Posts: 51 Country : India | RameshLMI posted @ 2016-02-13 3:23 PM
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@ 2016-02-13 5:14 PM (#20980 - in reply to #20869) (#20980) Top | ||||||||||||||||
Posts: 22 Country : India | Vishal posted @ 2016-02-13 5:14 PM
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@ 2016-02-13 6:29 PM (#20981 - in reply to #20869) (#20981) Top | ||||||||||||||||
Posts: 139 Country : Estonia | TiiT posted @ 2016-02-13 6:29 PM
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@ 2016-02-13 10:05 PM (#20984 - in reply to #20869) (#20984) Top | ||||||||||||||||
Posts: 102 Country : United States | ghirsch posted @ 2016-02-13 10:05 PM
Thanks for another great test Rakesh and Harmeet. All of the puzzles I solved were very nice, but my favorites were QuadMax 2 and X Sum Frame 2 (I really liked the deductions for X in this one). | |||||||||||||||
@ 2016-02-13 11:14 PM (#20985 - in reply to #20869) (#20985) Top | ||||||||||||||||
Posts: 52 Country : India | gaurav.kjain posted @ 2016-02-13 11:14 PM Really nice sudokus....Liked construction of all of them. Hats off, Rakesh and Harmeet. Kropki seems to be a no starter for me logically, Rakesh/Harmeet, please provide me THE logical starting point. Technically for me 2015-16 SM is ended as, I will be co-authoring last round of SM along with Amit. All the rounds were really great and I enjoyed them a lot. Thanks for the authors and organizers for all great SM rounds throughout the year. | |||||||||||||||
@ 2016-02-13 11:54 PM (#20986 - in reply to #20985) (#20986) Top | ||||||||||||||||
Posts: 774 Country : India | rakesh_rai posted @ 2016-02-13 11:54 PM gaurav.kjain - 2016-02-13 11:14 PM I guess you are referring to the 9x9. Kropki seems to be a no starter for me logically, Rakesh/Harmeet, please provide me THE logical starting point. (1) One starting point is R1C3/C4 -> one of them has to be 3. You can find out easily that the 3 must go in R1C4. You can then fill out all cells in row 1. R1C1 cannot be 5 (look at the long sequence in col 1) so it will be 1. (2) R5C1 is even (count odd/even in box4). Using this, we can complete col 1, as there is only one place for the number 2. After this, it should be simpler. | |||||||||||||||
@ 2016-02-14 12:20 AM (#20988 - in reply to #20869) (#20988) Top | ||||||||||||||||
Posts: 152 Country : United Kingdom | detuned posted @ 2016-02-14 12:20 AM
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@ 2016-02-14 1:16 AM (#20989 - in reply to #20869) (#20989) Top | ||||||||||||||||
Posts: 419 Country : India | kishy72 posted @ 2016-02-14 1:16 AM
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@ 2016-02-14 1:51 AM (#20990 - in reply to #20869) (#20990) Top | ||||||||||||||||
An LMI player | An LMI player posted @ 2016-02-14 1:51 AM
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@ 2016-02-14 2:05 AM (#20991 - in reply to #20869) (#20991) Top | ||||||||||||||||
Posts: 172 Country : ITALY | forcolin posted @ 2016-02-14 2:05 AM The puzzles were all right but the point distribution was rather poor. IMHO for the big kropki 16 points were faaaaaar too many, while the big average with only 13 points was underrated. | |||||||||||||||
@ 2016-02-14 11:44 AM (#20992 - in reply to #20986) (#20992) Top | ||||||||||||||||
Posts: 52 Country : India | gaurav.kjain posted @ 2016-02-14 11:44 AM rakesh_rai - 2016-02-13 11:54 PM gaurav.kjain - 2016-02-13 11:14 PM I guess you are referring to the 9x9. Kropki seems to be a no starter for me logically, Rakesh/Harmeet, please provide me THE logical starting point. (1) One starting point is R1C3/C4 -> one of them has to be 3. You can find out easily that the 3 must go in R1C4. You can then fill out all cells in row 1. R1C1 cannot be 5 (look at the long sequence in col 1) so it will be 1. (2) R5C1 is even (count odd/even in box4). Using this, we can complete col 1, as there is only one place for the number 2. After this, it should be simpler. Thanks a lot Rakesh, I really forgot to use Even/Odd marking , it makes it so easy which was looking a non starter to me , I remember Prasanna also mentioned it earlier. |