Complete graph with n vertices. Walk from vertex u to vertex v of length k. I don't understand how the number of walks between the two of length k is $n^{k-1}$

I've tried this formula on an example where n =5 and k = 2 and I can't even find 5 walks. Any help understanding this formula and example would be greatly appreciated. Thanks.


1 Answer 1


I think there may be a mistake in your statement, unless you allow a walk to stay in the same vertex on a step, instead of crossing an edge (but that's not the usual definition of walk), or unless you consider each vertex in a complete graph to have a loop (but again, that's not the usual definition).

I think you could do a proof by induction on $k$, distinguishing the cases where $u = v$ and $u \neq v$. Denote $A(k)$ the number of walks of length $k$ from $u$ to $v$ if we suppose $u = v$ and $B(k)$ if we suppose $u \neq v$.

  • the base cases are as follow:

    • $A(1) = 0$ (there is no edge from $u$ to itself); $A(2) = n-1$ (a $2$-walk is $u-w-u$, with $w\neq u$);
    • $B(1) = 1$ (there is exactly one edge from $u$ to $v$); $B(2) = n-2$ (a $2$-walk is $u-w-v$ with $w\neq u$ and $w\neq v$);
  • the induction case can be described as such: a walk of length $k+1$ from $u$ to $v$ is $u-w\sim v$, where $w\sim v$ is a walk of length $k$ from $w$ to $v$.

    • if $u = v$, then $w\neq u$ and $w\neq v$, and there are $n-1$ choices for $w$. That means that $A(k+1) = (n-1) B(k)$;
    • if $u \neq v$, then $w\neq u$, but $w$ could be equal to $v$. Distinguishing the cases $v = w$ and $v\neq w$, we get $B(k+1) = A(k) + (n-2)B(k)$.

Combining the two, we get $B(k+1) = (n-1)B(k-1) + (n-2)B(k)$. This is a constant-recursive sequence of order $2$.

Solving this, we get an expression $B(k) = \alpha(-1)^k + \beta(n-1)^k$. Since $B(1) = 1$ and $B(2) = n-2$, that means that $\alpha = -\frac{1}n$ and $\beta = \frac1n$. We conclude that: $$B(k) = \frac1n((n-1)^k - (-1)^k)\qquad A(k) = \frac1n((n-1)^k+(-1)^k(n-1))$$ This is indeed an integer.

For example, the number of walks of length $3$ in a complete graph of order $5$ from the vertex $1$ to the vertex $3$ is $\frac15(4^3-(-1)^3) = \frac{65}5 = 13$. Indeed, those walks are: $$1013, 1023, 1043, 1203, 1213, 1243, 1303, 1313, 1323, 1343, 1403, 1413, 1423$$


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