Tough Sudoku for 1/July/2010

Today's puzzle has some rather interesting moves available immediately following the standard easy stuff - of which there is but little!

I should have a blog page up and running regarding at least the beginning of today's puzzle shortly.

A link to the new blog page will appear at the top of the right column of this page, directly under "Check out the Sudoku Blog".

Oh! What a pretty color!

time for Steve, rwm, sotir to lock and load.
I'll be surprised if it takes less than seven gnarly steps.

maen all!...so that's what munchkins wear in their hats...the aces killed me on this one.

Have a great day/night, everyone, and celebrate well tomorrow, fellow Canuks!

There is a backdoor of a5=4 in this puzzle (basic techniques only after that to UP=81) - but I can find no exclusions to justify this. Best regards, Neil

Hmm. The only dent in the armor of this problem I can make (after basic UP 27) is the following.

The 6 of the top left box goes into either c8 or b9.
The two options share as a common quick consequence d5=2. The logic goes thusly:

If c8=6, then b4=g6=6, h6=g2=f5=1, so d5=2.

If b9=6, then e8=6, d8=1, so d5=2.

This takes us to UP 28, but for now I don't see how to go on...

A slight change from the path I blogged in the last step: AAIC's used in steps 4,5,6.
1) Start 22 UP 26 LC(9)ab2 => g2<>9; NT(126)def5 => ai5,f6,ef4<>126UP 27
2) (83=1)hi6 - g6 = g2 - f2 = HP(16-62)f59 = FSF(2)cfi237i9- (2)g7 = (389)g78h8 => g46i89<>8, h9<>3; UP 29
3) (6)e8 = c8 - c6 = g6 - (6=2)g4 - (4=2)h4 - i5 = a5 - a8 = HP(46)ce8 => e8<>1358 UP 31; HP(12)eh1
4) (8)a1 = d1 - f3 = WWing(24)f3,h4,(2)eh1 - (4)h3 = h4 - i5 = (4)a5 => a1<>4; UP 32
5) (8)f3 = WWing(24)f3,h4,(2)eh1 - (4)h3 = (4-2)h4 = (2-1)h1 = e1 - f2 = HP(16)f59 => f9<>8
6) XYZ Chain (3*)b9 = [(2)i9 = i23 - (NP12=3)h1g2 - (3*)b2 = (3*-9)b7 = (9-2)g7 = (2)hi9] - (2=6)f9] => b9<>6; UP 81

Acronym guide:
LC - Locked Candidate
NT - Naked Triple
HP - Hidden Pair
FSF - Finned Swordfish
WWing - W Wing: (2)cells (ab), strong link (b) between them => a = a in the cells.
XYZ Chain: A kraken in which one of the (3) items considered directly forbids that which an AIC shows is forbidden by the other two choices.
Notes: in step 6 it the right-most bracket that appears in the chain is a typo.

1)Start at 22,SSTS to UP =27
2)Loop:(6)b4=b9-f9=(6-1)f5=f2-g2=(1-6)g6=(6)g4,
=>e9<>6,f5=16,e2<>1,g6=16,c4< ;>6
3)(2=1)d5-d8=(1-6)e8=(6)e5,=>e5<>2.UP=28
4)Kraken row (2)dehi9,=>i89<>8.UP=30
(2)e9-e1=h1-h4=(2-8)g4=(8)i6
||
(2-6)f9=(6-1)f5=f2-g2=(1-6)g6=(6 -8)g4=(8)i6
||
(2)h9-h4=(2-8)g4=(8)i6
||
(2-5)i9=(5)i8
5)Kraken row:(4)ace8,=>e8<>1.UP=31
(4)a8-(4=3)a5-(3=6)c6-c8=(8)e8
||
(4-6)c8=(6)e8
||
(4)e8
6) (4=2)h4-(2=1)h1-g2=(1-6)g6=(6-2)g4=(2)h4,=>h29<>2
7)(2)i9=g7-g4=(2-4)h4=(4-7)h3=(7)h9,=> ;i9<>7.UP=32

I'm trying to find a short way to finish.
I appreciate the help from others.

Horrendously –possibly illegally –long -gnarly? -first step, but it does it (I think?):

1. STs to 27.

2. G6=6 or ghi6=138. If ghi6=138, i5=4, f6=5, g4=c6=e8=f5=6, f2=h1=1, h6=3, h3=4, d8=1, d5=e1=2, h8=9, e2=4, a2=9, a8=c4=4, b4=5, b2=3, g2=2, and there are no 2s in row 7.
So g6=6. UPs to 36.

3. i6=3 or 8. If i6=3, g4=f6=8 (gh8=39), g7=2, i5=4, h4=2 (bh1=34), a1=8, h3=4 (to avoid AUR gh38=39), h1=3, b1=4 (de1=12 and f23=45) and all candidates are eliminated from f7.
So i6=8 and UPS go to 81.

Alfred has pointed out a mistake, so, although the first step may be invalidly long, I may as well make it technically correct:

After e1=2, I should have included f3=8 & a1=8, which is needed to force the a8=4.

Regards to all. It's nice to see all the new and old names.

sotir: it takes about 4 more kraken. no short path exists outside of Steve's proprietary methods.

SE=8.4

Done