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Submitted by: vicky

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I peaked into the future and found this puzzle to be very interesting. I have written a new blog page conveying some of that interest. You should find a link at the top of the upper right column on this page that will take you to the new blog page. The link should appear today under "check out the sudoku blog". Hopefully, some of you find it interesting!
23/Feb/11 9:50 PM
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SE=8.9
24/Feb/11 12:11 AM
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Looks a lot like our "burning bush" in autumn.
24/Feb/11 12:48 AM
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Gorgeous Autumn color!
(We've got this planted on the north side of our house. It does not turn this beautiful there.)
24/Feb/11 12:55 AM
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3 aic + kraken is the pedestrian path.
24/Feb/11 2:06 AM
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1.SST to UP24 - not optimized
2.aur(78)ce46, FXW(3)di28, FSF(9)abi158 > -7i5, -9i79, -3i9
3.(4=6)g1-{gi3=(6-8)a3=i3-(8=246)egi1} > -24a13b1 -4i3f1 -8i2 -2d1
4.AALS(34567)g134 > -9a78 More...
24/Feb/11 2:21 AM
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-now I'll look at the blog... maybe learn something
24/Feb/11 2:24 AM
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-maybe step 3 can be depth 3 loop (-4i3 needed?)
then solution would have no chain fragment greater than depth 3 (with step 6 as WWing, 2 aic + kraken - max chain fragment depth 3)
24/Feb/11 3:12 AM
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haven't looked, but may be possible w/o aals
24/Feb/11 3:22 AM
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1. Unique possibilities to 24.
2. a5=7;to avoid the 78 unresolvable rectangle at ce46.UP27.
3. If b1=9=c6,c4=8=e6,b2=5=d3=f4,d1=1,f4=5,g4=6(to avoid the 26UR at bh45);68UR at ah13.So a1=9.UP30.
4. If f4=5,f2=4,f8=1,e79=49,e3=3,g3=4,c3=2=e1;i2 has no candidate.So f4=6.UP49.
5. Locked 9s at ce79,h17=46.UP81.
24/Feb/11 3:28 AM
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I got UP to 35 on this one - very unusual for me to get more than picked by the usuals so wonder if I did it by chance or missed something
24/Feb/11 1:10 PM
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Beautiful Bev!!!
24/Feb/11 2:54 PM
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Just for fun, a path to finish without using uniqueness or Krakens:

1) Start 22 UP 24 Some SSTS including coloring => i9<>3
2) FSF(9)abi158,g8 => i9<>9
3) (NT246=3)h456 - h89 = HP(37-79)gi8 = gi7 - e7 = (9-3)e9 = e3 - (3=NP46)g13 => g46<>46
4) (7)a5 = i5 More...
24/Feb/11 11:25 PM
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Kobold,
I am curious about the current concept of "depth".
If I understand it correctly, the max depth per chain fragment in your solution is 3.
Can one thus arbitrarily break apart an AIC that is not Krakenized to adjust the max depth per chain fragment?
For example, any AIC More...
24/Feb/11 11:38 PM
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Ants, after you deduce a1=9, you reach UP35. (Not UP 30 as I stated in step 3.)
24/Feb/11 11:58 PM
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Steve .. I wasn't really intending any serious metric, and I agree such are difficult to agree on. Simply a loose measure of how many sets the aic uses, for the kraken the depth is of course cumulative but one could argue the look-ahead is equal to that of (max) aic leg taken alone.
25/Feb/11 1:20 AM
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Another longer path w/o krakens:
1)SSTS to UP=24
2)(3):i2=d2-e3=e9,=>i9<>3
3)(9):i5=ab5-c6=c79-a8=gi8,=>i79<>9
4)AUR(78 )ce46,=>c46<>7.UP=27
5)(5)d3=a3-(5=248)b2c3a2-(8)df2=(8-1)d1=(1)d3,=>ab1a3<>24,d More...
25/Feb/11 2:32 AM
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Sorry for messy notation. Still new to sudoku.

1. SSTS to UP=24. Notice pair 78 e4=e6=(78).
2. If i5=7 than c4=c6=(78) that can't be. So a5=7.UP 27
3. if c6=9 than c4=8, g8=7, g7=9, g4=(46) and g3=3. SO f1=f3=f7=(24). So c6 !=9. UP34
4. If b2=2 than i1=d4=2, e1=4, f2=5 => b4=h4=(46) and f4=6. So b2 = 4. UP81.
02/Mar/11 6:49 AM
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