Hi folks,
Now that
A188 has been and gone, I'd like to share a move with you here that was used by Ed in his walkthrough (
here), and which was on my intended solving path when I posted the puzzle. As a basis, I'll take the grid state that would have been reached after Ed's step 8, but deferring his steps 4 and 5:
At this point, we can retrospectively apply Ed's steps 4 and 5:
Ed wrote:
4. "45" on r89: 1 outie r7c1 + 2 = r8c9 = [24/35] = {[3]/[4]..}
5. "45" on r12: 2 innies r2c19 = 7 = [43/52/61] ([34] blocked by step 4.
5a. r2c1 = (456); r2c9 = (123)
5b. no 1 in r3c9
Here, Ed was making clever use of the fact that there were fixed relationships between the digits in
R2C1 and
R2C9, and between
R7C1 and
R8C9. But the interesting thing is that it would have been possible to delete the 4 from
R2C9 even if this had not been the case! To see why, notice that (using innie/outie difference on
R1289) the three innies (
R2C19+R8C9, marked in green) sum to the single outie (
R7C1, marked in yellow) +
9. We can therefore safely eliminate 4 from
R2C9, because if
R2C9 were 4, this would force R28C9 to the naked pair {45}, thus summing to nine, requiring
R2C1 and
R7C1 to be equal, which is impossible because they are peers of each other.
Note that we can make a formal solving technique (IOU4) for this based on the following requirements, which could also be put to good use in more complicated scenarios.
IOU4- There are three innies and one outie (or vice-versa, of course).
- The three innies sum to the outie + D.
- One of the innies can "see" the outie.
- The other two ("satellite") innies can also see each other (but are not required to see the outie).
- One of the two satellite innies is a bivalue cell containing the candidates {xy}, where x + y = D.
- If the above conditions are met, we can eliminate {xy} from the other satellite innie.
Incidentally, you may also be interested to know my intended follow-up move to this, which I don't think anybody spotted:
Note that the yellow cell,
R2C6, can see all of the 4 dark green cells at
R1C7+R2C789. Therefore, these 4 cells cannot contain all three of {123}, because that would leave no possibilities for
R2C6. Consequently, one of {123} must go in
R3C9, the only other possible place in N3. This constrains the 5(2) cage at
R23C9 to {23}, which effectively cracks the puzzle.