Volumetric argument in more dimensions

If we decide that our lattice is not constrained to planes, then we propose the following nodes with constant degree $M$ as building blocks.

Figure 11: Building blocks, lattice where we embed $G$ and $G_{sub}$.

We chose building blocks so that they generate homogeneous grids and with fixed length edges. Both constraints are necessary for the volumetric argument we want to use. Homogeneity is similar to the requirement for planar graphs in normal graph theory that edges do not cross. Edge-crossing would generate inhomogeneities with fixed length edges on 2D grids.

If $M$ is even, we naturally choose edges perpendicular to each other. Two counter-directed edges of the building block form an independent axis in the $d$ dimensional axis. Therefore, we set

$$d=\frac{M}{2}$$

If $M$ is odd, we might looked puzzled at the fractional dimension, a sort of in between dimension.

Furthermore, we know that the boundary of a subgraph has a dimension less (for a sphere $V = \frac{4\pi}{3} r^3>$, but $A_{surface} = 4 \pi r^2$). In general, if the subgraph is $d$-dimensional, the volume is proportional to $r^d$ and the boundary to $r^{d-1}$. Size of the random subgraph $G_{sub}$ with $n$ nodes is then:

$$V_{sub}=n \propto r^d.$$

The number of outgoing edges is proportional to the size of the boundary, $r^{d-1}$ namely.

$$e_{n} \propto r^{d-1}$$

We now start in the middle of our subgraph, and we subsequently add nodes until we reach the boundary. In the analogous mechanism of adding nodes means putting them into the bin, $i$, representing the height of the tree and the boundary is reached until all nodes are into the bin.

$$\sum_{1}^{l}{i^{d-1}}\propto N$$

We imagine a tiny lattice of nodes and we therefore turn summation into an integral:

$$\sum_{1}^{l}{i^{d-1}}=\int_{0}^{l}{i^{d-1}di}=\frac{i^{d}}{d}\mid_{0}^{l}\propto N$$

Thus:

$$\frac{l^{d}}{d}\propto N$$

and:

$$N \propto \frac{l^{\frac{M}{2}}}{\frac{M}{2}}$$

Solving for $l$ gives:

$$l \propto (N \frac{M}{2})^{\frac{2}{M}}.$$

Solving for $M$ requires a numerical approximation.

Resuming, we consider $G_{sub}$ embedded in $G$, both in a homogeneous $d$-dimensional grid composed by myriads of the above mentioned building blocks. We let $G_{sub}$ grow until it reaches the same size of $G$ through summation and we infer $l$.

Results are very similar to the Mandelbrot's formula

$$N \propto l^{d}$$

that relates mass ($N$) and linear extension $l$ with the exponent $d$ [25].