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Aleonysh [2.5K]
2 years ago
12

Find the number of paths of length 4 between two different vertices in K5​

Mathematics
1 answer:
Mkey [24]2 years ago
4 0

K₅ is the 5-complete graph, with adjacency matrix

A = \begin{bmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{bmatrix}

The (i, j)-th entry of the matrix A⁴ gives the number of length-4 paths from vertex i to vertex j. Computing A⁴ isn't so bad:

A^2 =  \begin{bmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{bmatrix}  \begin{bmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{bmatrix} =  \begin{bmatrix} 4 & 3 & 3 & 3 & 3 \\ 3 & 4 & 3 & 3 & 3 \\ 3 & 3 & 4 & 3 & 3 \\ 3 & 3 & 3 & 4 & 3 \\ 3 & 3 & 3 & 3 & 4 \end{bmatrix}

A^4 = \begin{bmatrix} 4 & 3 & 3 & 3 & 3 \\ 3 & 4 & 3 & 3 & 3 \\ 3 & 3 & 4 & 3 & 3 \\ 3 & 3 & 3 & 4 & 3 \\ 3 & 3 & 3 & 3 & 4 \end{bmatrix} \begin{bmatrix} 4 & 3 & 3 & 3 & 3 \\ 3 & 4 & 3 & 3 & 3 \\ 3 & 3 & 4 & 3 & 3 \\ 3 & 3 & 3 & 4 & 3 \\ 3 & 3 & 3 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 52 & 51 & 51 & 51 & 51 \\ 51 & 52 & 51 & 51 & 51 \\ 51 & 51 & 52 & 51 & 51 \\ 51 & 51 & 51 & 52 & 51 \\ 51 & 51 & 51 & 51 & 52 \end{bmatrix}

We want the paths between two distinct vertices, so we ignore the entries on the diagonal and take the total of the non-diagonal entries, 20 • 51 = 1020.

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