Sudoku Puzzles

Uluru and the Olgas were one of the highlights of our trip to Oz.

An interesting puzzle…
Start at 22, basic technique UP26
Look at (12389)bcdef8=(46)bcef8 and bivalue cell (46)a8
01: (12389)bcdef8=(46)abcef-(6)g8=(6)g5-(6=24)hi6-(2=8)d6-(8=3)d8 => ef7 & gh8<>3
02: (3)h7=(3)i7-(3)i5=(3-6)g5=(6)h6 => h7<>6, UP81

A nice week to all!!

maen all! do they let people climb or just look? nice simple benches...i put the design in my work project notes for this spring.

22:38

Hi ttt,
Excellent solution,step o1 is very interesting.
Thanks sharing with us.
Best Regards,Sotir

27:04

Here is a less technical (more amateurish?)interpretation:
1. Unique possibilities to 26.
2. If h7=6=g5=e6,i5=3,h4=5,i4=7,d6=8,d8=3;row 7 is devoid of 3.Hence quad 1239 at hi79;g9=8,locked 6s at gh8 and UPs to81.

A different view to the usual, glad you have seen them Hal, Olgas especially (now called by the Aboriginal name Kata tjuta which I understand to be 'many heads'), very mysterious place.

Hi All, My go for today, but same conclusion as ttt's, although a different method (which I am not certain that I've noted correctly.
1) Basic techniques to UP=26.
2) (6)h6*=(6)e6-(6=247)hi6i4-(247=3)i5-(3)i7=(3)gh8h7*
2a) (6)h6=(6-9)e6=(9-8)a6=(8)d6-(8=3)d8-(3)gh8**
2b) (3)gh8h7*-(3)gh8**=(3)h7=>h7<>6, UP=81.
Thanks ttt, I will study your method.

1.Given UP=22. Basic techniques UPs at c5,b4,b6,b1. UP=26.
2. 6 can be at ab8/b7. If 6 is at a or b 8, then h7=6, g5=6, i5=3, e6=6, d6=8, d8=3, which leaves no 3 in top right block. So 6 cannot be at ab8. Hence b7=6. UP=81.
(This is a 'non-technical' solution. I hope this is not the same as any of the earlier posted 'technical' solutions.)

Hi All,
I came back this one and see that:
1- Typo on my original post, I’m sorry about that – not included 2’s…:
(1389)bdef8=(46)abef-(6)g8=(6)g5-(6=24)hi6-(2=8)d6-(8=3)d8 => ef7<>3
2- There are some way to eliminate 3’s at ef7 & gh8 (as below…) but I like above way More...

SE=8.5

the puzzle is not as complex as above discussion might suggest - 2 aic's max depth 3 solve.

1.SST to UP26
2.(7=8)g8-(36')g58=i5-i7=(367')gh8h7 > -29h8 -28g8 UP28
3.(6=248)dhi6-(378=6)dgh8 > -6h7 UP81

Kobold,
A very clever solution, imo. Congratulations!
Your step 2 has both a typo, and an obscure usage of terminology (at least obscure to me).

I assume that ' is used to signify hidden sets.

Imo, given that usage, you meant:

(67')gh8 = (6-3)g5 = i5 - (3)i7 = (367')gh8h7

Hopefully that is a faithful representation of your meaning?