Here is an example (
A 153 V2) for which a large difference between innies and outies can be used fruitfully.
Using 45 rule for c89, one obtains : r7c8+r8c8 = 8 + r1c9+r2c9+r3c9 (1)
In fact, the innies-outies difference is so large that digits at r78c8 cannot be equal to digits at r123c9. For instance, we can prove that r7c8 <> r1c9 (the following can be generalized for the other cells of the Innies- outies)
From (1) , we deduce : r7c8 - r1c9 = (8 - r8c8) + (r2c9+r3c9) (2)
On the one hand, 8 - r8c8 >= -1 since r8c8 <= 9
On the other hand, r2c9+r3c9 >= 1+2=3 since r2c9 and r3c9 see each other.
From (2), we deduce : r7c8 - r1c9 >= -1 + 3 = 2 -> r7c8 <> r1c9
Concerning the above puzzle, we get :
1) digit at r7c8 is locked for c9 at cage 14(2) : r7c8=(5689)
2) digit at r8c8 is locked for c9 at r456c9 : no 1,2,3.
We can go further :
3)r67c9 <> 7. Like this : if one of r67c9 was 7, the other one and r7c8 would total 14 with the same combination as 14(2) at
n9 since r7c8 is locked at 14(2), and there would be a clash between 14(2) and 21(3)
4) No 7 at 21(3) at n6 : combination {489}
The puzzle is not cracked yet, but everything becomes smoother from here...