Sudoku Puzzles

No worries. It's just water over the dam.

1.UP=30 (kraken knockout)
2.Kraken Row(8)abe7=> (4)a1=(8)a7, a1<>8; UP=81
(8-6)b7=(6-3)b9=(3-4)i9=i2-h1=a1
||
(8-4)e7=h7-h1=a1.

sotir: your Kraken AUR yesterday is impressive, a type I haven't yet considered.

Watch your footing - it looks pretty slippery to me!

1)Start at 22,SSTS to UP=30
2)(8)b3=(8-4)a1=h1-h7=(4-8)e7=(8)e8,=>b8<>8
3)(8)b3=(8-4)a1=h1-i2=(4-3)i9=(3-6)b9=( 6)b7,=>b7<>8.UP=81

Thanks farpointer.I have learned that from Aran.
See Aran's comment for Extreme #200.

Hi sotir,
very nice on using AUR to attack yesterday tough puzzle. I'm happy to see you on improving your skills to solve tough puzzles.

~40 min

23:03

Hi ttt,
Thanks for your very kind words.I always try to learn from you and appreciate your help.
Best Regards,Sotir

Looks refreshing!

Hi farpointer.This time you have selected a long path, to fit in with the Kraken logic. After 8(b7), 8(a1) is the immediate outcome for column a. You selected a long path, via i9 and i2 to reach the desired result: a1<>8. Shouldn't you really say that b7<>8, because it leads to a contradiction at a1 ?
Regards, Alfred.

Enjoy your day/night, everybody!

Alfred: true enough that (8)b7 produces UP (8)a1, but that proves nothing but (8)a7 OR (8)a1 if (8)b7 - almost a truism. For the kraken logic to work here (~x)a1 after (8)b7 AND (~y)a1 after (8)e7 must be shown. Then either a7=8 OR a1=x OR a1=y has been shown, QED a1<>8.

-in the above x, y <> 8.

-excuse the notation (x)a1 from (8)b7 and (y)a1 from (8)e7 must be shown - x, y <> 8.
(8)a7=[AAIC(8)be7: (4)a1=...=(8)b7=e7-...=(4)a1].
the above AIC prohibits (8)a1.

in Steve's notation:
SIS(8)abe7:
(8)a7*=[(4)a1=h1-i2=(4-3)i9=(3-6)b9=(6-8*)b7=(8*-4)e7=h7-h1=a1] => a1<>8.

IMO, the kraken format is easier to read and understand.

Alfred: if one could have written:
(8)a7=b7...=(4)a1, the kraken wouldn't be needed.
but we only have (8)a7=(8)be7.

Alfred: obviously one can easily prove b7<>8 with sotir's second chain - I used the kraken as to capture the logic of both chains in one step. Normally I wouldn't do this, but as a one step KO it seems reasonable (plus- counting sets - it was the minimum.

Thanks for the detailed response,farpointer,but I still think it's not a good categorization of a Kraken proof, because one of the three options does not lead to the same clear result at a1.

Alfred: the result is crystal clear - requiring krakens and AIC chains to all be perfect loops ending on the same bool node would be extremely limiting aside from ignoring all wraps or continuous networks which are very rich in their exclusion sets. (in general, any bool which is weak with all More...

Be interesting what Mr Kraken would have to say on this one. Hope he is not already rolling in his grave!

Mr Kraken?? does he speak from the grave you have dug him? He rolls - but with mirth at the flimsy chains you propose.

My go for today:
1) Basic techniques to UP=30.
2) (8)b3=b8-e8=(8-4)e7=h7-i9=i2-h1=(4-8)a1=(8)g1=>i3<>8,UP=81.

Neil, you missed (8)b7.
Regards,Sotir

Neil,you can use kraken col.: (8)b378,=>a1<>8.UP=81
(8)b3
||
(8-6)b7=(6-3)b9=(3-4)i9=i2-h1=(4)a1
||
(8)b8-e8=(8-4)e7 =h7-h1=(4)a1

Sotir,
Thanks for your notes above - useful. However, I still am not an expert on AICs, so those are the ones that I look for first.
Best regards, Neil

SE=7.3