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Tough Sudoku for 16/August/2010

                 
                 
                 
                 
                 
                 
                 
                 
                 

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Submitted by: Gath

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1.UP=40 (I found it tough despite the initial UP)
2.loop (5=36)b56-(3)ac4=(3-5)d4=d5
=> a5<>5, b13.c46<>6, c6.a5<>3, d4<>24; UP=41
3.(3)b1=b56-ac4=d4-(3=2)d6-(2=9)c6-(9=35)a47 => a1<>3; UP=44
4.(2)e4=e1-(2=9)h1-i1=i6-c6=c4 => c4<>2; More...
16/Aug/10 12:44 AM
Is this a super highway Outback style?
16/Aug/10 1:16 AM
Looks like the border patrol are stuck out in the middle of nowhere!
16/Aug/10 2:33 AM
Hmm. I had to use a bit of "if..then" at one step.

1) UP 40.

2) parts A,B grew as an attempt to turn the logic of part C into an AIC
2A) (2)g3=d3-e1=e4 => g4<>2
2B) (6)i13=(189)i139-(189=36)i6-{NP36 in bi6}(3=2)d6-d3=(2)g3 => g3,i5,i6<>6
2C) If More...
16/Aug/10 2:39 AM
here's a kraken for you to glom Alfred, ends on 3 different nodes after UP=41 after step 2 above.
3.Kraken Row(3)abc1 => gh4<>9; UP=81
(3)a1-(35=9)a74
||
(3-4)b1=b3-d3=e1-e4=HP(468)fgh4
||
(3-6)c1=(6-9)i1=i6.

the 3 end nodes are (9)i6, (468)fgh4, (9)a4.
(9)gh4 makes all three false thus resulting in contradiction - NO 3's in row 1.
16/Aug/10 2:43 AM
Oops. My step 4A is rather clumsy. Doesn't need the Xwing, because (8)i6=f6, so
4A) (1=8)i9-i6=(8-1)f6=(1)gi6 =>i5<>1.
16/Aug/10 2:44 AM
1.UP=40 (kraken knockout!)
2.Kraken Cell(129)h1 => gi6=9; UP=81
(1)h1-a1=(1-9)a8=a4-gh4=gi6
||
(2)h1-e1=e4-c4=(2-9)c6=gi6
||
(9)h1-i1=i6-gh4=gi6.
16/Aug/10 3:21 AM
farpointer: That last kraken is very nice! At some point I also noticed that creating that 236 triple on row 6 would really get the puzzle moving, but couldn't see a way to do that. Well done!
16/Aug/10 4:18 AM
I am sure that there are simpler ways to do this, but here is my solution.
UP=41.
Kraken cell (4)b3=(9)hi1=>b1<>4
(4)b3=b1-(4=2)e1
||
(4)b3=(4-3)b1=b56-ac4=d4-(3=2)d6
||
(4)b3=(4-3 )b1=b56-ac4=(3-5)d4=ac4-(5=36)b56-(236=9)c6-i6=i1
||
(4)b3=(4-3)b1=b56-ac4=(3-5)d4=ac4-(5=36)b56-( 236=9)c6-(9=1)c8-a8=a1-(12=9)h1`
=>b1<>4.
UP=81.
16/Aug/10 4:32 AM
Jyrki: I had some difficulty parsing your 2B) as an AIC chain. Perhaps clearer notation is:

(6)i13=(189)i139-(189=36)bi6-(3=2)d6-d3=g3.

written this way it's not clear how the chain manages to forbid (6)i56. Which two nodes in the above prohibit (6)i5 and (6)i6 ?
16/Aug/10 4:40 AM
Jiminoregon: what kraken cell?
your four AIC chains (if correct) show:
1) (4)b3 == (2)e1
2) (4)b3 == (2)d6
3) (4)b3 == (9)i1
4) (4)b3 == (9)h1
the last two imply (4)b3 since (9)h1 and (9)i1 can't both be true at the same time if !(4)b3. The first two chains are irrelevant noise - and the conclusion has nothing to do with kraken anything!
16/Aug/10 5:10 AM
Jiminoregon: I haven't checked your very long chains, but the two chains 3) & 4) are actually one chain of the form a == b -- c == a => a.
Your solution then has a whopping 30 links!
16/Aug/10 5:28 AM
farpointer:

That chain itself forbids 6 only from g3, but that simultaneously locks 6 into i13, so (6)i56 are also prohibited. Sorry about being unclear.
16/Aug/10 5:29 AM
farpointer: the first two chains were used thus:
The first chain is needed to establish (4)b1=>h1<>2
The second chain is needed to establish c6<>2 under the same assumption.
In spite of your objections, I do think this argument is correct. And yes, it does seem to be some sort of nishio logic, not kraken.
16/Aug/10 5:30 AM
ok Jiminoregon. I tried reading those chains. You're using undocumented UP-style forward weak links to create in-chain strong links (explains (4)b3 == (2*)d6 anyway so you can say (36)b56--(2*36=9)c6 and (4)b3 == (2*)e1 for (1)a1--(12*=9)h1.
(hint - I've abandondoned the technique of UP-style More...
16/Aug/10 6:11 AM
note: my experience shows that most toughs can be solved using a series of very short AIC chains - max (and rarely at that of 5 sets).
16/Aug/10 6:17 AM
note: chains using UP-style forward weak links break the AIC logic over the span of the forwarded weak links. The chain loses symmetry and cannot be wrapped or read backwards. Sub-chain deductions cannot be freely made, but must jump over any spans used. The chains are very long and start to More...
16/Aug/10 6:56 AM
Jyrki: I apologize for being obtuse about (6)i56. - I caught the bug from Alfred! Anyway, I did manage to find a chain for your 2C) deduction:

(9=136)aci1-(3)b1=b56-ac4=d4-(3=2)d6-c6=c4-HT(359)acd4=(9)gh4.

complex - some things don't translate well into AIC chains.
16/Aug/10 7:23 AM
farpointer: I am also very impressed with your knockout Kraken solution. No complaints this time.

Regards, Alfred.
16/Aug/10 10:00 AM

farpointer,nice finding.We can use shorter:
2)Kraken cell:(129)h1,=>c6<>9.UP=81
(1)h1-a1=a8-(1=9)c8
||
(2)h1-e1=e4-c4=(2)c6
||
(9)h1-i1=(9) i6

Regards,Sotir
16/Aug/10 12:01 PM
I have this very same photo from last year.
16/Aug/10 12:28 PM
nice edit sotir! - always harder to see the elimination from three different end-points.
16/Aug/10 1:01 PM
19:29 - an easy start to 40, but took a little finding after that.
16/Aug/10 1:39 PM
farpointer: Thanks for working out the chain for my step 2C. Agreed - it doesn't translate well.

Hmm. Let's try Krakenese!

Kraken cell (469)g4 => i1=9
(9)g4-i6=(9)i1
||
(6)g4-(6=8)f4-f6=(8-9)i6=(9)i1
||
(4)g4-(4=2)e4-e1=(2-9)h1=(9)i1

A pale, distant cousin More...
16/Aug/10 5:19 PM
Another way with long chains: basic techniques UP40
01: (29)c46=(9)a4-(9=1)a8-(1=3)a1-b1=b56-ac4=d4-(3=2)d6-c6=c4
=> c4<>356, c6<>3 - extend chain at (3)b56 => (35)ad4, UP44
02: (9)i1=(9)i6-(9=2)c6-(2=3)d6-d4=a4-a7=c7-(3=6)c1 => i1<>6, UP81

The other viewing for step#1:
(2)c4=(2)c6-(2=3)d6-(X-wing:3’s)bi56=(3)b1-(3=1)a1-(1=9)a8-(9)a4=(29)c46
16/Aug/10 5:31 PM

SE=8.3
10/Sep/10 11:17 PM
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