Sudoku Puzzles

Looks like a perfect pastel, Wendy!

Love the colours!

Gorge-eous!

1.UP=27 (good chains, but tough to find)
2.(5=2)f6-(2=349)cfi8-(3)e8=HT(136)e79f9-(5)e79=e12 => f19<>5, e7<>2, UP=37
3.(579=3)d79e8-(3=49)ci8 => f8<>9, UP=47
4.(9)a2=d2-f1=f9-(1=579)f9d79-(7)e8=b8-(379=8)a129 => a19<>9 UP=81.

Sure are a lot of Georges in Australia.

29:13

to you all from foggy Portland, Oregon.

I found this difficult, and it took me a while to write out the solution. I apologize for the heavy use of krakens. Comments and criticisms are welcome!
========================
UP=27.
(i) More...

my first two steps can be expressed in a single step using farpointer's strategy:
(5=2)f6-(2=349)cfi(8)-(3)e8=HT(136)e79f9-(39)d79e8=(5)d79=>d45<>5,f19<>5, e7<>2.

Thanks for the comment, Neil.

gorgeous picture, ugly puzzle

looks like heavy use of reverse engineering Jiminoregon - a 20 link chain is way too long and never necessary. Two of them in a four chain kraken?? life is too short to verify these - especially when combined with overuse of forward weak link instantiation, all those (mn=xyz) links look highly More...

Yes, I know I was focused on the possible HP at h79. It really blinded me to other possibilities. I have to learn how to screen out these "shortcuts" :-(

two: try not to improvise eureka notation
what is (!n=n)c supposed to mean??
not clear at all where 'kraken' iii proves anything - what happened to (58)aeh9 - quite a muddle!

I thought "!n" meant "not n".
(5)e9 knocked out by (5)e2;
(8)h9 knocked out by (8) g7.
But you are right - "muddle" says it.

how is (1) applied to (156=37)e5 - parity wrong.
(1)x == ...--(?1...)y or
(1)x -- ...--(1...)y you can't have it both ways.

(1=9)f1-(9=2)f8-(2=6)e7-(69=1)f9

For this one I managed to get to UP 27 and when I followed the f19<>5 and e7<>2 as suggested in farpointer's step 2 I got to UP47 without the need for f8<>9. Don't know if I did something wrong or if you missed something.

1)Start at 22,basic techniques to UP=27
2)(2=9)f8-d79=(49-7)d12=d79-e8=(7)b8,=>b8<>2.UP=32
3)(57=9)d79-(9=2)f8-(2=5)f6,=> ;d45<>5.UP=47
4)(9):a2=d2-f1=f9,=>a9<>9
5)(57=8)ah9-(8=3)g7-(3=4)g2-(4=9)d2-f1=(9-1 )f9=(1)e9,=>e9<>37.UP=81

5) To read (37=8)ah9..

~20 min (probably a fortunate error)

This meant back to "if..then" for me. The reason is apparently that I suck at translating into AICs such pieces of logic, where a deduction depends simultaneously on several consequences of the contrapositive assumption. Unfortunately I don't have the time to study this much at this More...

26:21, hi all.

15:02

you're quite right Ants - don't know what I missed?
1.(5=2)f6-(2=349)cfi8-(3)e8=HT(136)e79f9-(5)e79=e12 => f19<>5, e7<>2, UP=47
2.(9)a2=d2-f1=f9-(1=579)f9d79-(7)e8=b8-(379=8)a129 => a19<>9 UP=81.

1. Given UP=22. UPs at e5, f5, pairs 39 at bc4, UPs at c6, c3 and i4. UP=27.
2. If f6=2, then f8=9, pairs 49 at d12, pairs 15 at d45, d9=7, d7=2=b8, b7=7, which leaves top right block withou 7. So f6<>2, but d5=2. UP=37.
3. If f9=6, then b7=6, b8=2=e7, pairs 47 at d12, pairs 59 at d79, More...

Sorry for the spelling error of 'leaves' in step 3.

SE=8.3