This is closely related to my previous post Double line notation - gluon propagator
I'm trying to understand the double-line vertices for the gluon in the case of a $U(N)$ gauge group. Normally, the vertex factors are given by: $$ \mathrm{Three} \ \to \ g f^{a}_{\ bc} \left[ h^{\mu \nu} ( k - p )^{\rho} + h^{\nu \rho} ( p - q )^{\nu} + h^{\rho \mu} ( q - k )^{\nu} \right] \\ \mathrm{Four}\ \ \to \ - i g^{2} \left[ f^{a}_{\ be}\ f^{c}_{\ de} \left( h^{\mu \rho} h^{\nu \sigma} - h^{\mu \sigma} h^{\nu \rho} \right) + f^{a}_{\ ce}\ f^{b}_{\ de} \left( h^{\mu \nu} h^{\rho \sigma} - h^{\mu \sigma} h^{\nu \rho} \right) + f^{a}_{\ de}\ f^{b}_{\ ce} \left( h^{\mu \nu} h^{\rho \sigma} - h^{\mu \rho} h^{\nu \sigma} \right) \right] $$
Since the gauge group is $U(N)$, if you write the group indices in the form $a = (\bar{j}, k)$, then the structure constants are going to be given by $$ f^{( \bar{m}, n )}_{ ( \bar{j},k )(\bar{p},q) } = i \big( \delta_{jq} \delta_{kn} \delta_{pm} - \delta_{jm} \delta_{kp} \delta_{qn} \big) $$ This inspires the use of the double line notation.
The way I understand it though, the three vertex factor now has TWO pieces with $\delta\delta\delta$'s - which means that I need to consider two double line diagrams to represent just one ordinary three gluon feynman diagram?
And futhermore the factors $ff$ yield six combinations of $\delta \delta \delta \delta$'s - which would mean I need 6 double line diagrams to represent just one ordinary four gluon feynman diagram?
I believe my understanding of this is correct (please call me out on this if its not).
My question is; how is this a simplification at all? Now instead of my original two diagrams, I've got 8 that I'm looking at? It seems to me that the double line notation is making the problem more convoluted rather than helping to simplify it? What is the point of this?