1. NC -> 7(2) = {16} or {25}
-> Of 7(2)n1 and 7(2)c4 - one of them is {25} and the other is {16}

2. If 7(2)c6 has different values from 7(2)c4 -> one each of (16) pairs with one each of (25)
(SS) But this would require 9(2)n9 (paired with 7(2)n1) to have values from the set (1256) which is impossible.
-> 7(2)c6 has the same values as 7(2)c4

3. Trying 7(2)c4 = {16} puts 8(2)c4 = {35}, 7(2)c6 = {16}, 8(2)c6 = {35}, 7(2)n1 = {25}
Puts {16} as a pair
Also puts (35) in n5 in r456c5
(NC) puts 4 in n5 in r5c46
(NC) puts r46c5 = {35} -> (35) is a pair
But this would require 9(2)n9 to have a 3 or 5. (SS)
But it cannot be {45} (NC) or {36} since (SS) 6 pairs with a 1 and there is no 1 or 6 in 7(2)n1
-> 7(2)c4 = {25}

4. -> 7(2)n1 = {16}, 8(2)c4 = {17}, 7(2)c6 = {25}, 8(2)c6 = [17]
-> (25) is a pair

5. NC -> 13(2)n4 from {58} or {49}
But cannot be {58} since that would require 10(2)n6 to have a 2. (SS)
-> 13(2)n4 = {49}
-> 10(2)n6 cannot have a 2 (since 2 would require a paired 5 in 13(2)n4)
-> 10(2)n6 = {37}
(NC) -> 14(2)n6 = {59}
(SS) -> 9(2)n4 has a 2 or a 5 -> (NC) 9(2)n4 = {27}
-> (79) is a pair
-> (34) is a pair

6. (79) is a pair -> r46c5 = [79]
-> HS 1 in c5 -> r5c5 = 1 (unpaired)
-> Last remaining pair = (68)

“I commented on the players forum that I could not find an NC puzzle that was fully Semi-symmetric (i.e. at least on number is paired with itself in at least one position pair) all I could find were asymmetric ones. Wecoc took this as a challenge and found a set of them, posting a nice vanilla puzzle.

From his puzzle I made a killer which I re-post here. I have solved it a couple of ways but find my solutions mucky, can anyone post a neat solution?”

Prelims, including effect of NC

a) R23C6 = {17/26/35}
b) R3C12 = {16/25} (cannot be {34})
c) R34C4 = {16/25} (cannot be {34})
d) R4C78 = {59/68}
e) R5C23 = {49/58} (cannot be {67})
f) R5C78 = {19/28/37/46}
g) R6C23 = {18/27/36} (cannot be {45})
h) R67C6 = {16/25} (cannot be {34})
i) R78C4 = {17/26/35}
j) R7C89 = {18/27/36} (cannot be {45})

NC only used as stated.

1a. R23C6 = {17/35} (cannot be {26} which clashes with R67C6)
1b. Killer pair 1,5 in R23C6 and R67C6, locked for C6
1c. R78C4 = {17/35} (cannot be {26} which clashes with R34C4)
1d. Killer pair 1,5 in R34C4 and R78C4, locked for C4

2a. R3C12 = {16/25} must have different combination from R34C4
2b. R3C12 = {16/25} corresponds with R7C89 = {18/27/36} -> {16} cannot correspond with {25} -> R34C4 must have the same combination as R67C6
[Possibly slightly simplified; see also wellbeback’s start.]
2c. R4C4 cannot be the same as R6C6 -> R34C4 and R67C6 must have the same vertical order

3a. R4C78 cannot be {68} because R5C78 = {19} (cannot be {37}, NC), R5C23 = {58}, R6C23 = {27/36}, R4C78 corresponds with R6C23, R5C23 corresponds with R5C78 but 8 cannot correspond both with one of 1,9 and one of 2,3,6,7
3b. R4C78 = {59}, locked for R4 and N6, clean-up: no 2 in R3C4, no 1 in R6C78
3c. R34C4 = {16}/[52] -> R67C6 = {16}[52] (steps 2b and 2c), no 2 in R6C6, no 5 in R7C6

4. R3C12 corresponds with R7C89, R4C78 corresponds with R6C23, R6C23 and R7C89 are both {18/27/36} neither containing 5, R4C78 = {59} corresponds with one of the combinations in R6C23
-> R3C12 = {16} (cannot be {25} because {25} and {59} cannot correspond with the same combination), locked for R3 and N1
4a. R3C45 = [52] -> R67C6 = [52] (step 2c), R78C4 = {17}, locked for C4 and N8, R23C6 = [17], clean-up: no 7 in R7C89
4b. R4C4 corresponds with R6C6 -> 2,5 paired
4c. R4C78 = {59} corresponds with R6C23 -> R6C23 = {27}, locked for R6 and N4
4d. 7 paired with 9
4e. R5C23 corresponds with R5C78, R5C23 contains one of 5,9 -> R5C78 must contain one of 2,7 but {58} cannot correspond with {28} since they are in the same row -> R5C23 = {49}, locked for R5 and N4, R5C78 = {37}, locked for R5 and N6
4f. R6C6 = 5 -> R5C46 = [68] (NC) -> R5C5 = 1, R5C19 = [52], R5C23 = [94] (NC), R5C78 = [37] (NC)
4g. 3,4 paired, 6,8 paired
4h. R4C5 = 7 (hidden single in N5)
4i. 1 can only correspond with itself
4j. R3C12 corresponds with R7C89, R3C12 = {16} -> R7C89 = {18} (cannot be {36} because 3 is paired with 4), locked for R7 and N9 -> R78C4 = [71]

[The rest will be very dependent on NC and corresponding cells.]
5a. R5C1 = 5 -> no 6 in R46C1 (NC)
5b. 6 in N4 only in R4C23, locked for R4
5c. R5C9 = 2 -> no 1 in R46C9 (NC)
5d. R46C1 correspond with R46C9, no 1 in R46C9 -> no 1 in R46C1
5e. Naked pair {38} in R46C1, locked for C1 and N4
5f. R46C1 corresponds with R46C9, 3 in R46C1 -> R46C9 must contain 4, locked for C9 and N6
5g. R5C8 = 7 -> R6C8 = 1 (NC), R7C89 = [81]
5h. R3C1 corresponds with R7C9 -> R3C12 = [16], R4C23 = [16]
5i. R1C7 = 1 (hidden single in N3)
5j. R9C3 = 1 (hidden single in N7)
5k. R4C5 corresponds with R6C5, R4C5 = 7 -> R6C5 = 9
5l. R6C6 = 5 -> R6C7 = 8 (NC), R46C9 = [46], R4C6 = 3, R46C1 = [83], R6C4 = 4, R4C78 = [59] (NC), R6C23 = [72] (NC)
5m. R3C4 = 5 -> no 4 in R3C5 (NC)
5n. R4C7 = 5 -> no 4 in R3C7 (NC)
5o. R3C8 = 4 (hidden single in R3)
5p. R6C3 = 2 -> no 3 in R7C3 (NC)
5q. R7C6 = 2 -> no 3 in R7C5 (NC)
5r. R7C2 = 3 (hidden single in R7)
5s. R7C8 = 8 -> no 9 in R7C7 (NC)
5t. 9 in R7 only in R7C13, locked for N7
5u. R3C3 corresponds with R7C7, R7C7 = {46} -> R3C3 = {38}
5v. 9 in R3 only in R3C79, locked for N3

6a. R12C4 = {38/39} (cannot be {89}, NC), 3 locked for C4 and N2
6b. R3C6 = 7 -> R3C5 = 2 (NC), R3C7 = 9
6c. R3C8 = 4 -> R3C9 = 8 (NC)
6d. R3C3 corresponds with R7C7, R3C3 = 3 -> R7C7 = 4
6e. R3C5 corresponds with R7C5, R3C5 = 2 -> R7C5 = 5, R7C13 = [69]
6f. R89C7 = {26/27} (cannot be {67}, NC), 2 locked for C7 and N9
6g. R89C8 = {35/36} (cannot be {56}, NC), 3 locked for C8 and N9
6h. R1C7 = 1 -> no 2 in R1C8 (NC)
6i. R2C8 = 2 (hidden single in N3)
6j. R7C2 = 3 -> no 2,4 in R8C2 (NC)
6k. R2C8 corresponds with R8C2, R2C8 = 2 -> R8C2 = 5
6l. R2C1 corresponds with R8C9, R8C9 = {79} -> R2C1 = {79}
6m. R2C1 = {79} -> R2C2 = 4 (NC)
6n. R2C2 corresponds with R8C8, R2C2 = 4 -> R8C8 = 3
6o. R7C5 = 5 -> R28C5 = [68] (NC), R2C7 = 7
6p. R2C2 = 4 -> R2C3 = 8 (NC)
6q. R8C8 = 3 -> R8C7 = 6 (NC)

and the rest is naked singles, without using corresponding pairs.