I think it can be explained. So far I have looked through all and the only thing I cannot explain is the 45 degree angle on the third I from right in I9-. I can't treat this in my method so I'm going to consider that to be a 90 degree caused by 2 lone pairs.
If you think of the I3- chain, the middle I has negative formal charge. It has 3 lone pairs so the shape it keeps it pretty much linear. If we want the chain to have another I, we would then have two I atoms with negative formal charge. If we want a 5-I chain, this will incur -3 negative formal charge in total, and so on.
Then we have bridges. So here the angle changes by 90 degrees because there are 2 lone pairs, or 45 degrees because there is 1 lone pair. So we deduce that I atoms which link 2 chains (chains described above as linear) are +1 and those linking 3 chains are +2 etc.
Given this and the coefficients on the species, we can write the following general statements:
(Symbolism: v[I] is the coefficient of I in the species; v[C,x] is the number of chains of length x I atoms in the species; v[B,y] is the number of bridging I atoms which connect y chains; z is the valency of the species)
v[I] = 3*v[C,3]+4*v[C,4]-v[B,2]-2*v[B,3]-3*v[B,4]
z=-v[C,3]-2*v[C,4]+v[B,2]+2*v[B,3]+3*v[B,4]
Now obviously this is not a perfectly determined system, but you can use it to make good headway towards the solutions, particularly keeping in mind that for Lewis structures it is better to have less formal charges, so the v[C,4] and v[B,4] will be minimized as much as possible in favour of v[C,3] and v[B,2] or v[B,3] respectively.