The CCC in the previous post worked because R8C456 was between 1 and 9 greater than R89C5. In this example from
Assassin 187 V4, that situation doesn't exist but instead there's the interesting situation that the hidden cage R8C456 has the same total as the 3-cell cage at R8C6. This can lead to an interesting
variant of CCC which doesn't give any immediate eliminations but can still be very useful, as can be seen below.
This is the position after the Prelims, then
45 rule on R9 3 outies R8C456 = 21 = {489/579} (cannot be {678} because no 6,7,8 in R8C5) -> R8C5 = {45}, R8C46 = {789}, 9 locked for R8 and N8, clean-up: no 4,5 in R9C5
and 45 rule on N9 2 innies R7C9 + R9C7 = 14 = {59/68}
Here R8C456 = 21 = {489/579}, 21(3) cage at R8C6 = {489/579/678} and because R8C6 is the crossover
and R9C6 "sees" R8C456 it follows that R8C456 and the 21(3) cage at R8C6 must have different combinations. This is clearly useful for cages which only have two or three combinations, less so when there are more possible combinations.
In this particular position we can make more progress, because there's no 7 in R9C7, which does lead to candidate eliminations.
R8C456 and 21(3) cage at R8C6 = {489/579/678} cannot have the same combination because they share cell R8C6 and R9C6 “sees” R8C456 -> at least one of the combinations must contain 7 (and no 7 in R9C7) -> must have 7 in R8C46 + R9C6, locked for N8
and that's not the end of it because there's no 4 in R9C7
Because 7 only in R8C46 + R9C6, either R8C456 or 21(3) cage at R8C6 must be {489} -> must have 4 in R8C5 + R9C6, locked for N9