SteveG48 wrote:- Code: Select all
*--------------------------------------------------*
| 4 TU13 T26 | 7 T23 5 | 6-1 8 9 |
| 5 9 38 | 6 38 1 | 2 7 4 |
| 168 78-1 267 | 9 28 4 | 16 3 5 |
*----------------+----------------+----------------|
| 18 5 189 | 3 79 6 | 4 2 17 |
| 36 U37 S67 | 4 1 2 | 9 5 8 |
| 2 4-1 149 | 5 79 8 | 3 6 17 |
*----------------+----------------+----------------|
| 13 6 13 | 8 4 7 | 5 9 2 |
| 9 48 48 | 2 5 3 | 7 1 6 |
| 7 2 5 | 1 6 9 | 8 4 3 |
*--------------------------------------------------*
Stem [67]r5c3
Petal T: [1236]r1c235 (locked by stem 6)
Petal U: [137]r15c2 (locked by stem 7) => -1 r1c7,r36c2 ; stte
Bat, Steve's original post was not AIC (Eureka).
Notated in AIC it would be something like
(123=6)r1c235-(6=7)r5c3-(7=13)r15c2
So either 123 is in r1c235, implying 1r1c2 (and 3r1c5, 2r1c3), or 13 in r15c2, implying 1r1c2 (and 3r5c2) too.
[Added]
This is the long version of the same:
(1=3)r2c1-(3=2)r1c5-(2=6)r1c3-(6=7)r5c3-(7=3)r5c2-(3=1)r1c2
My suggested ABX notation:
A=r5c2, B=r5c3, X=r1c2
A<>7: r5c2=3->r1c2=1
B<>7: r5c3=6->r1c3=2->r1c5=3->r1c2=1
=>r1c2=1
If you inverse the first line (A=>B is equivalent to (not B) => (not A)), you get
r1c2<>1 => r5c2<>3 => A=7, this implies
B<>7 => r5c3=6->r1c3=2->r1c5=3->r1c2=1
Written as AIC:
(1=3)r1c2-(3=7)r5c2-(7=6)r5c3-(6=2)r1c3-(2=3)r1c5-(3=1)r2c1
So you get the same, just read from right to left.
So we have the same logic in 4 notations.