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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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4/5 divided by 1/3 plus 1/5 minus 3/5
r-ruslan [8.4K]

Answer:

  \frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}=-12

Step-by-step explanation:

Considering the expression

\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}

Solution Steps:

\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}

as

\mathrm{Combine\:the\:fractions\:}\frac{1}{5}-\frac{3}{5}:\quad -\frac{2}{5}

so

=\frac{\frac{4}{5}}{\frac{1}{3}-\frac{2}{5}}    

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}

=\frac{4}{5\left(\frac{1}{3}-\frac{2}{5}\right)}

join  \frac{1}{3}-\frac{2}{5}:\quad -\frac{1}{15}

so

=\frac{4}{5\left(-\frac{1}{15}\right)}

\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a

=\frac{4}{-5\cdot \frac{1}{15}}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{-b}=-\frac{a}{b}

=-\frac{4}{5\cdot \frac{1}{15}}

\mathrm{Multiply\:}5\cdot \frac{1}{15}\::\quad \frac{1}{3}

so

=-\frac{4}{\frac{1}{3}}

\mathrm{Simplify}\:\frac{4}{\frac{1}{3}}:\quad \frac{12}{1}

so

=-\frac{12}{1}

\mathrm{Apply\:rule}\:\frac{a}{1}=a

=-12

Therefore

                  \frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}=-12

4 0
3 years ago
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