re-scaling a ratio to reflect a change in value

Please assume you are replying to somebody with limited knowledge in mathematics. I am drawing a blank.

I made up a formula which quantifies how similar two graphs are G and G' (Conceptual Graphs to be precise), scaled between 0 and 1, with 0 being completely dissimilar, and 1 being identical, where nodes can belong to either a set of concepts C or a set of relations R (but not both):

$$C_{G} \sim C'_{G'} = \frac{C(G \cap G')}{ \mid C_{G} \mid + \mid C_{G'} \mid }.$$

$$R_{G} \sim R'_{G'} = \frac{R(G \cap G')}{ \mid R_{G} \mid + \mid R_{G'} \mid }.$$

$$E_{G} \sim E'_{G'} = \frac{E(G \cap G')}{ \mid E_{G} \mid + \mid E_{G'} \mid }.$$

By obtaining the ratio of identical concepts, relations and Edges, I can calculate the normalised average graph similarity as shown below:

$$G \sim G' = \frac{ (C_{G} \sim C'_{G'}) + (R_{G} \sim R'_{G'}) + (E_{G} \sim E'_{G'})}{3}.$$

This works, but it is biased towards nodes (concepts C and relations R), since the ratio of nodes to edges is 2:1.

Now assume I have some graph metric, e.g., 0.79 (or 79%) and I want to adjust it, so that the graph ratio for nodes and edges is 1:1, so that the change in the ratio reflects the change in the metric. I know that since the ratio is 2:1, thus 0.33 of 0.79 is attributed to edges, hence edges provide 0.26 of the total.

1. How can I adjust the formula so that the ratio of Edges and Nodes is 1:1?
2. Is it possible to adjust the actual value (e.g., 0.79) to reflect the ratio change, while keeping it normalised between 0 and 1?

I don't know what exactly you're doing, but it sounds like you want to use some kind of weighted average. Given three quantities $x,y,z$, their weighted average with respective weights $a,b,c$ would be $\dfrac{ax+by+zc}{a+b+c}$. Often it is convenient (and always possible) to choose weights whose sum is $1$, so that the denominator vanishes, but that's not important. In your case, if you want to give $x$ twice as much weight as $y$, and say for example give $y$ and $z$ the same weight, then it would be $\dfrac{2x+y+z}{4}$.