Here is an alternative arrangement for 17 RSCs, using the same 3-level tree structure approach:
Assign TWO RSCs as the top level "roots" (X,Y) (or one can consider them as "grandpa & grandma"
), which are both linked to 5 other RSCs (A,B,C,D,E) - the middle level "moms". Each mom is linked to 2 other bottom level RSCs - "kids" (A1,A2,B1,B2,C1,C2,D1,D2,E1,E2).
Code:
_________X___Y_________
// // \ / \\ \\
A B C D E
/ \ / \ / \ / \ / \
A1 A2 B1 B2 C1 C2 D1 D2 E1 E2
As for additional links among "kids", each kid is linked to the same index kids of adjacent groups (A & E are considered adjacent) and different index kids of the groups 2 steps away. E.g. B2 is linked to A2,C2,D1,E1.
Verification:
Obviously the roots X,Y are connected to all other RSCs.
Each mom is directly linked to X,Y and its own kids. X or Y bridges it to all other moms. Its own kids bridge it to all other kids.
Each kid is directly linked to its own mom and 1 kids each from the 2 neighbour groups, and 1 kids each from the 2 distant groups. Its mom bridges it to X,Y and the other kid of the same group. The linked kids bridge it to other moms and all other kids (e.g. for B2, the linked kids A2,C2 bridge it to A,C,D2,E2 while D1,E1 bridge it to D,E,A1,C1).
Here is the SCM:
Code:
X Y A B C D E A1 A2 B1 B2 C1 C2 D1 D2 E1 E2
X . . 1 1 1 1 1 . . . . . . . . . .
Y . . 1 1 1 1 1 . . . . . . . . . .
A 1 1 . . . . . 1 1 . . . . . . . .
B 1 1 . . . . . . . 1 1 . . . . . .
C 1 1 . . . . . . . . . 1 1 . . . .
D 1 1 . . . . . . . . . . . 1 1 . .
E 1 1 . . . . . . . . . . . . . 1 1
A1 . . 1 . . . . . . 1 . . 1 . 1 1 .
A2 . . 1 . . . . . . . 1 1 . 1 . . 1
B1 . . . 1 . . . 1 . . . 1 . . 1 . 1
B2 . . . 1 . . . . 1 . . . 1 1 . 1 .
C1 . . . . 1 . . . 1 1 . . . 1 . . 1
C2 . . . . 1 . . 1 . . 1 . . . 1 1 .
D1 . . . . . 1 . . 1 . 1 1 . . . 1 .
D2 . . . . . 1 . 1 . 1 . . 1 . . . 1
E1 . . . . . . 1 1 . . 1 . 1 1 . . .
E2 . . . . . . 1 . 1 1 . 1 . . 1 . .
X Y A B C D E A1 A2 B1 B2 C1 C2 D1 D2 E1 E2
X 0 2 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2
Y 2 0 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2
A 1 1 0 2 2 2 2 1 1 2 2 2 2 2 2 2 2
B 1 1 2 0 2 2 2 2 2 1 1 2 2 2 2 2 2
C 1 1 2 2 0 2 2 2 2 2 2 1 1 2 2 2 2
D 1 1 2 2 2 0 2 2 2 2 2 2 2 1 1 2 2
E 1 1 2 2 2 2 0 2 2 2 2 2 2 2 2 1 1
A1 2 2 1 2 2 2 2 0 2 1 2 2 1 2 1 1 2
A2 2 2 1 2 2 2 2 2 0 2 1 1 2 1 2 2 1
B1 2 2 2 1 2 2 2 1 2 0 2 1 2 2 1 2 1
B2 2 2 2 1 2 2 2 2 1 2 0 2 1 1 2 1 2
C1 2 2 2 2 1 2 2 2 1 1 2 0 2 1 2 2 1
C2 2 2 2 2 1 2 2 1 2 2 1 2 0 2 1 1 2
D1 2 2 2 2 2 1 2 2 1 2 1 1 2 0 2 1 2
D2 2 2 2 2 2 1 2 1 2 1 2 2 1 2 0 2 1
E1 2 2 2 2 2 2 1 1 2 2 1 2 1 1 2 0 2
E2 2 2 2 2 2 2 1 2 1 1 2 1 2 2 1 2 0
Note that once again 5 of the nodes (A,B,C,D,E) have 4 direct links only, giving it the same amount of "link efficiency" as the previous arrangement.
Of course, if we remove one of the roots (X or Y), this automatically reduces to an alternative arrangement for 16 RSCs, and is more "link efficient" than the previous one, which uses up all 5 RPDs on each RSC.
And needless to say, if we add one extra root (Z), this becomes a very straight forward arrangement for 18 RSCs, which uses up all 5 RPDs on each RSC. However, I will later provide an alternative 18 RSCs scenario, which does not use the tree structure.
And come to think of it, from the first 17 RSCs arrangement (the one with 1 root, 4 moms and 12 kids) can also become an arrangement 18 RSCs by adding an additional root, and is more link efficient as the 2 roots have 4 direct links only. So it seems the possibility is endless once a certain structure is established. Just not for the (19,5) case though.