*4.4. The Number of Spanning Trees*

The number of spanning trees has a very important role in various networks, and it can be computed using the product of nonzero eigenvalues of the Laplacian matrix [16];

$$N\_{ST}(G\_t) \;=\; \frac{\prod\_{k=2}^{N\_t} \nu\_i}{N\_t} = \frac{A\_t}{N\_t} = \frac{\prod\_{i=0}^{t-1} \prod\_{j=0}^{n-1} \nu\_{i,j}}{N\_t}, \quad (i,j) \neq (0,0)$$

The number of spanning trees for the categorical path-path product network:

$$N\_{ST}(G\_t) \quad = \frac{\prod\_{i=0}^{t-1} \prod\_{j=0}^{n-1} \left( (2 - 2\cos\frac{j\pi}{n})d\_{i+1} + d\_{j+1}(2 - 2\cos\frac{j\pi}{t}) - (2 - 2\cos\frac{j\pi}{n})(2 - 2\cos\frac{j\pi}{t}) \right)}{nt}$$

The number of spanning trees for the categorical cycle-path product network:

$$N\_{ST}(\mathcal{G}\_{\mathbf{f}}) \quad = \frac{\prod\_{i=0}^{t-1} \prod\_{j=0}^{n-1} ((2 - 2\cos\frac{2j\pi}{n})d\_{i+1} + d\_{j+1}(2 - 2\cos\frac{i\pi}{l}) - (2 - 2\cos\frac{2j\pi}{n})(2 - 2\cos\frac{i\pi}{l}))}{n\mathbf{f}}$$

The number of spanning trees for the categorical cycle-cycle product network:

$$N\_{ST}(\mathcal{G}\_t) \quad = \frac{\prod\_{i=0}^{t+1} \prod\_{j=0}^{n-1} ((2 - 2\cos\frac{2j\pi}{n})d\_{i+1} + d\_{j+1}(2 - 2\cos\frac{2i\pi}{l+2}) - (2 - 2\cos\frac{2j\pi}{n})(2 - 2\cos\frac{2i\pi}{l+2}))}{n(t+2)}$$
