ronk wrote:But I don't see how -- in general -- you can say ...
RW wrote:- Code: Select all
ab abX
abY ab
where if abX=a => abY=a, without any strong links.
Why isn't the following a valid outcome?
- Code: Select all
b a
Y b
[edit: edited the original post to clarify, the diagram you quoted does not include the conditions that imply if abX=a => abY=a. I added the diagram to show the four UR cells, the next line added the condition abX=a => abY=a and the second diagram showed what this would look like in a grid.]
It is not valid because of the definition of the pattern I gave you. If abX=a => abY=a. If you put 'a' in abX and 'y' in abY, there would be a box without an 'a'. In this pattern (the 3 rows below the line are supposed to represent a box):
- Code: Select all
ab abX
-----------
abY bC ab
bD bE abF
bG bH abK
(C, D, E, F, G, H and K <> a)
candidate a appears only in column 3 and the cells that are part of the deadly pattern. If you put 'a' in abX, the only cell left for 'a' in the lower box is in abY.
ronk wrote:For this example, I see that
r6c3=3 -> r7c3<>3
r6c3=3 -> r8c3=7 -> r7c3<>7
r6c3=3 -> r6c3<>4 -> r8c1=5 -> r8c2=9 ->r7c2=1 -> r7c3<>1
To clarify my chain:
if r6c3=3 => r7c3<>3 => r8c3<>3 => r8c1=3 (=> r6c1=7 => r8c3=7 => UR: r6c3<>3)
As I said, this pattern is really common (allowed 5 eliminations in the mentioned puzzle) and it's also easy to spot. I would call it a UR because I can make the elimination without taking any steps outside the pattern (excluding the elimination steps, which I don't count as steps anyway).
Another version of this that I frequently use is:
- Code: Select all
ab ab
-----------
abY bC abX
bD bE abF
bG bH abK
eliminate b from abX
(if abX=b => ab(above)=a =>abY=a)
RW