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Submitted by: Gath

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Saw several of those on the east side of Lake Superior.
20/Dec/14 1:05 AM
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Fitting that we had Easter Island as a Hard, and this as a Tough.
20/Dec/14 2:43 AM
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1. Unique possibilities to 29.
2. Whether d7=1,d3=5;OR d7=7,e8=5;e2=2.UP81.
20/Dec/14 3:43 AM
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Q-10: Symmetry=None, Start=25, SSTS=32, SG=11.5, SE=8.9
300041900096037000000800050060000003000008002504900000000700200400000801007010030

easy but site puzzle is again trivial, Q-10 is 1 AIC + 4 patterns = 2.25 * points
20/Dec/14 4:35 AM
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Q-11: Symmetry=AntiDiagonal, Start=25, SSTS=30, SG=11.5, SE=8.5
000020001000009400008100520006000000900063000050080072087000010004008900100006008

bonus round, kraken KO with patterns (1 chaining step) **
20/Dec/14 5:25 AM
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#29 8@df1=>b1<>8
#30 46@gi2=>gi2<>12=>1@ac2=>[ab1<>1&a3<>1]
1?@c2=>e2=2=>e4=9=& GT;c4={}=>c2<>1
#31 48@df1=>df1<>5
5?@h3=>e2=5=>e8=7=>d7=1=>d3={}=>h3<>5

#32 More...
20/Dec/14 7:48 AM
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Long path to Q-11
Standard techniques UP 30
1. (6)a3=i3-i7=ab7 =>-6a9
2. (1=3)g6-(3=2)g3-f3=(2)f6 =>-1f6 UP 33
3. Kraken row (7)aei8 =>-2bc1
||(7-2)a8=a23
||(7)e8-f79=(7-2)f6=f3-d2=(2)ab2
||(7)i8-i2=(7-2)d2=(2)ab2
4.Kraken column (4)b9765 More...
20/Dec/14 10:25 AM
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Q-10,SE=8.9
SSTS
1)(8):c6=c9-i9=i4,=>b4<>8
2)(4=14)d156-(3)d5=(3-6)f4=f7-(6=2)d9,=>d1 <>2
3)(2=6)e4-d4=d7-(6=2)d9,=>d6<>2.UP=81
20/Dec/14 6:40 PM
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Sorry,
The solution is not valid.
20/Dec/14 6:47 PM
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Path to Q-10, with AICs
Standard techniques UP 32
1. Kite (8)h8=a8-c9=c6 =>-8h6
2. SC (3)b4=f4-f23=d2 =>-3b2
3. (1)h4=(1-2)b4=(28-9)ac6=(9)h6 =>-1h6
4. (4)f6=f13-(4=2)d1-(2*=6)d9-hi9=(6-7)i7=a7-(7=582)b129-(2)b4=(28)ac6 =>-2d6,-2f6
//(4)f6==(28)ac6-(9)ac6=h6 =>-4h6 More...
21/Dec/14 10:14 AM
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Q-10

NT(236)d5.ef4=*[(124)d156=*7e4-(7=64)g41-(4=2)d1]-(2=6)d9-(6=278)hi9.h8-(7=258)b129 :=> -2b4; UP44
XYWing(1-35)b34.f3-3f4; UP81
21/Dec/14 10:36 AM
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Hi JC. I am not used to juggling so many large groups of cells, so I could not follow your first step. However, here is another way to show b4<>2:
Whether b4=3;OR e4=3=d2;and now whether d9=6=i7,i9=7,b129=258;OR d9=2,c2=9 to avoid 69UR @ ef27,b2=2 to avoid 59UR @ fi13; and again b2<>2.

BUT, after ac6=2, efg6=457, I could only find one more UP, d6=1. How did you reach UP44?
21/Dec/14 9:14 PM
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Should read:

Whether b4=3;OR f4=3=d2.....etc.
21/Dec/14 9:17 PM
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Another typo, right at the end, should read:

....again b4<>2.
21/Dec/14 9:22 PM
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Hi Alfred,

Here is a detailed solution of Q-10 :

#1. Basics
5 Singles
1ac6-1a8; XWing(3Ccd)-3b2; i13=59
2 Singles
6def2-6df13; 7g456-7i4; XWing(8Cci)-8b4.h6

The structure of the B/B-Plot shows that one can best enumerate the solutions of the puzzle from More...
22/Dec/14 12:47 AM
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Thanks a lot JC. I will work through that soon.
d2<>6, because I am testing for f4=3=d2 at that stage.

Best Regards, Alfred.
22/Dec/14 3:31 AM
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Hi again Alfred,

OK, I see now. Very well done! It is even possible to simplify your step like this :

i9=7 -> b129=258 [3b2 being removed by XWing(3bf4,3d25)]
or
i7=7 -> b2=2 or b2=5 -> d2=6 to avoid DP[(59)fi13 + (69)ef27], d5=3=b4

Conclusion : More...
22/Dec/14 4:42 AM
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Another way to prove -2b4, after the simple colourings
// Kite (8)h8=a8-c9=c6 =>-8h6
// SC (3)b4=f4-f23=d2 =>-3b2
Proof of -2b4 with a double kraken
1a. Kraken cell (267)e4 =>SIS (6)e4==[(2)e4==6d9]
||(2)e4
||(6)e4
||(7)e4-(7=6)g4-(6=4)g1-(4=2)d1-(2=6)d9
1. Kraken More...
22/Dec/14 8:59 AM
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I follow it now, JC, thanks. It looks like d9=26 is the useful cell in each of the methods. When d9=6,b129 becomes 258;and when d9=2,d5ef4 becomes 236. Even cenoman's Krakens make a lot of reference to d9.
I still only managed to reach UP 43 instead of UP44, but never mind, your final XY Wing reached UP81.
Best Regards, Alfred.
22/Dec/14 1:47 PM
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