First INSANE September 27 anniversary

12 Comments

1. Nate
   Posted September 27, 2006 at 11:29 pm | Permalink

   For those of us who are navigationally challenged, can someone post a link or instructions on how I can download the famous SEPT 27, 2005 puzzle? I go to the archives and nothing happens, thanx in advance, N8

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2. udosuk
   Posted September 28, 2006 at 6:44 am | Permalink

   Nate, the link is under “Categories” from the right column menus. You have to click to the 2nd page to download the puzzle and the solution. 💡

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3. Fernando Alves de Castro
   Posted September 30, 2006 at 1:35 pm | Permalink

   I would like to try to solve the ‘anniversary puzzle’ but, probably, I don’t go to do it because for me doesn’t make sense to have ‘trial and error’ technique to solve this kind of puzzles. A puzzle can be very hard only with logic techniques. In the past, if I’m not wrong, you said in this site that ‘trial and error’ technique would never used in your puzzles. With this technique I don’t know how many attempts I need to use to find out a solution. I’m very sorry but I don’t go to loose my time.

   Best regards,

   FAC

   P.S. – Did you use the ‘trial and error’ in the last “insane’s” posted? For the puzzle of Sept 04, 06 I didn’t reach a soulution yet.

   Reply
4. HK James
   Posted October 1, 2006 at 5:52 am | Permalink

   It was a challenging puzzle, but I never used trial and error. I was always able to use logic to eliminate pencil marks. Sometimes it involved looking at possibilities in several dimensions.

   I started on the left side, geting about 5 numbers placed and relatively simple pencil marks. The right side went much the same way. Then I slowly filled in pencil marks for center, and placing all remaining numbers.

   Maybe I was lucky, but I actually did this puzzle faster than the easier puzzle posted on 29 Sep.

   Reply
5. djape
   Posted October 1, 2006 at 10:57 am | Permalink

   Fernando, this puzzle was posted to mark the anniversary of the “most difficult” puzzle posted on this site. One year ago, I only started making my Killer Sudoku puzzles and couldn’t really decide on how difficult I make them. Now I can. But the puzzle of Sep 27, 2005 was a success because it brought people to this site.

   “Trial&error” is a very dubious technique in Killer Sudoku puzzles. You should look into the forum to see how other folks cope with puzzles of this difficulty.

   For Sep 4, 2006 puzzle you should read this thread from the forum:
   http://www.djapedjape.com/sudoku/forum/viewtopic.php?t=468

   Regards,

   Dj.

   Reply
6. Bob
   Posted October 2, 2006 at 4:33 am | Permalink

   After your challenge “don’t be surprised if you can’t place a single number” it was mandatory to give this a go. I managed 14 cells so far. and will need to take a break before my eyes bleed.

   Because so many of your fans jump immediately to your Daily Killer page [it’s my homepage], I wonder if they are missing this gem. Surprisingly, it doesn’t [yet] appear in the forum.

   THANK YOU DJ, and Happy Anniversary!!

   Reply
7. Vihari
   Posted October 6, 2006 at 8:00 am | Permalink

   Apropos the September 26, 2006 killer INSANE – I reached a solution that is different from the one posted. My view has always been that there should be one unique solution to a puzzle; is this condition not necessary?

   Reply
8. Vihari
   Posted October 6, 2006 at 8:03 am | Permalink

   Apologies…I think there was an error in my sheet..

   Reply
9. Sandra
   Posted October 9, 2006 at 1:37 pm | Permalink

   I agree with HK James that this puzzle was easier than some of the others. I started with most of the right and the left side. The centre squares followed relatively easily.
   The key to the puzzle, for me, was to solve the ’11’ box; the choice was 2,9 or 3,8.

   Reply
10. JC
    Posted October 9, 2006 at 10:42 pm | Permalink

    I enjoyed the puzzle and for me (with my wife) it did take longer than usual. I achieved 14 unique numbers before having to assume one of a duplet and prove an inconsistency (though it took me almost to the end before it appeared). After that one quick dead end branch before the end solved it. Didn’t do much else today though! Start left and right and the middle sorts itself out. Regards, JC

    Reply
11. JC
    Posted October 9, 2006 at 10:59 pm | Permalink

    A tip for anyone interested. Once the puzzle is complex and the paper and pencil is messy use Word. I created a 9×9 grid with each cell divided into a thin, small box for options (eg point size 8) and a larger box for the solution. When you have a long branch to try put a marker (*, colour, etc) against the assumption and continue. If the branch is a dead end then use Ctrl Z repeatedly until you arrive back at the mark. Hope it helps. JC

    Reply
12. udosuk
    Posted November 15, 2006 at 1:31 am | Permalink

    Complete Walkthrough for Insane (7425379) (27/09/2006)

    (NB: most prepositions are denoted as @,
    HS=Hidden Single, NP=Naked Pair, NT=Naked Triple)

    Stage 1: Opening from the wings

    Innies @ n4 => r6c123=23={689} (NT)
    Innies @ n6 => r4c789=20 cannot have {12}
    Innies @ c89 => r46c8=6=[42|51]
    If r46c8=[42], r56c7=12-2=10={19|37}, and we cannot make r4c789=20={479}
    => r46c8=[51] => 14/2 @ r89c8={68} (NP)
    Outies @ c9 => r57c8=9={27} (NP) => r123c8={349} (NT)
    7/2 @ r7c89=[25] => r5c8=7 => r56c9=13-7=6={24} (NP) => r56c7=12-1=11=[83]
    r34c7=18-5=13=[76], r4c9=9, 7/2 @ r12c7={25}, 10/2 @ r89c9={37}
    Innies @ c12 => r46c2=11=[29|38]
    Outies @ c1 => r35c2=5={14} ({23} clashes with r4c2) => 5/2 @ r3c12={14} (NPs)
    21/3 @ r67 => r7c3 cannot be 1
    12/3 @ n4 & r4c2=2|3 => r45c3=12-(2|3)=10|9 cannot have 1 ({89} unavailable)
    => 1 @ c3 locked @ r89c3, 7/2 @ r89c3={16} (NP)
    HS @ n4 => r6c1=6, r6c23={89}, r7c3=4 (21/3)
    12/3 @ n4 comes from {2357} must be {237} => r4c3=7, r4c2+r5c3={23} (NP)
    HS @ n4 => r5c1=5, r4c1+r5c2={14}
    6 @ c2 locked @ r12c2, 13/2 @ r12c2={67} (NP)
    Innies @ n1 => r123c3=16 => r123c3=r789c2=16={5(29|38 )}
    r12c1=r46c2=r56c3=11={29|38}, r789c1=24-6=18={7(29|38 )}

    Stage 2: Advancing towards the middle

    r789c7={149}=14 & 19/4 @ r78 => r7c56=r9c7+5=6|9|14
    {245} unavailable => r7c56=9|14 => r9c7=4|9 => r78c7={1(9|4)}
    If r78c7={19}=10, r7c56=19-10=9={36}
    If r78c7={14}=5, r7c56=19-5=14={68}
    => r7c56={6(3|8 )} with 6 locked
    If r9c7=4, r9c56=15-4=11={29} ({38} clashes with r7c56)
    If r9c7=9, r9c56=15-9=6={15|24}
    => 15/3 @ r9c567={9(15|24)} with 9 locked
    19/4 @ r23 comes from {235689}={23(59|68 )}, r3c8=3|9, r3c9=6|8
    If r2c3=2, r3c345={3(59|68 )}, must clash with one of r3c89
    If r2c3=5, r3c345={239} clashes with r3c8
    => r2c3={389} => r13c3 cannot have 9 (r123c3=16 cannot be made)
    2 @ 19/4 locked @ r3c345, not in r3c6
    If r3c6=3|9, r3c68 would form an NP of {39}
    Then r3c345 must have both 5 and 6|8, cannot make {23(59|68 )} => r3c6=5|6|8
    r23c3 is part of {23(59|68 )}, cannot be [85] => r1c3 cannot be 3, =2|5|8
    r1c45 cannot have 9, otherwise 16/3 @ r1c345={259} would clash with r1c7
    r1c45 cannot have 1, otherwise 16/3 @ r1c345={178} would clash with r1c29
    9/3 @ r456c4 cannot have {789} => 9 @ r5/n5 locked @ r5c56

    Stage 3: The critical move that cracks it

    Suppose r2c3=9, then r3c345 must have a 5 => r3c6={68}
    Also, r2c456 and r3c45 cannot have 9 => r1c6=9 (HS @ n2) => r5c5=9 (HS @ n5)
    => r9c7=9 (HS @ r9) => r7c56=14={68} => r37c6 forms an NP of {68}
    This would force 17/3 @ r456c6 to come from {123457}, impossible!
    Therefore r2c3 cannot be 9!

    Stage 4: Cleaning up

    HS @ c3 => r56c3=[29], r467c2=[389], r7c7=1, r56c9=[42], r35c2=[41], r34c1=[14]
    19/3 @ r456c5 & r6c5=4|5|7 => r4c5 cannot be 1|2, must be 8
    => 19/3 @ r456c5=[865], 9/3 @ r456c4=[234], 17/3 @ r456c6=[197]
    => r7c56=[36], r8c7=9 (19/4), 15/3 @ r9c567=[924], r3c5=2
    16/3 @ r1c345 comes from {45678}, must be {457}
    => 16/3 @ r1c345=[574]… And the rest are naked singles…

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