Consider an undirected graph with two distinguished nodes $u\neq v$. How hard is it to find an $u-v$ path, such that its length is as close to the average $u-v$ path length as possible?

Formally, for a path $P$ let $\ell(P)$ denote its length (number of edges on $P$, but a weighted version may also be considered). Let $A$ denote the average length of simple $u-v$ paths: $$A= \frac{1}{N}\sum \ell(P)$$ where $N$ is the number of simple $u-v$ paths and the summation is taken over all such paths. Let us call a simple $u-v$ path $P_0$ typical if $|\ell(P_0)-A|$ is minimum.

Question: What is the complexity of finding such a typical path? Is anything known about this problem?

  • 1
    $\begingroup$ Does anybody know the complexity of sampling a simple $u$-$v$ path uniformly at random? That could be useful for approximating the unweighted case. $\endgroup$
    – Neal Young
    Oct 22, 2022 at 10:59

1 Answer 1


Lemma 1. The weighted problem is NP-hard by reduction from Partition.

Lemma 2. The unweighted problem is NP-hard by reduction from Hamiltonian Path.

Proof of Lemma 1. Given a Partition instance $(x_1, \ldots, x_n)$, construct the multigraph $G=(V,E)$ with $V=[n+1]$ and, for each $i\in [n]$, two copies of edge $(i, i+1)$, one with weight $x_i$ and the other with weight zero. Then ask for the path from $1$ to $n+1$ whose weight is as close to average over paths from $1$ to $n+1$.

By linearity of expectation, the average weight is $\sum_{i=1}^n x_i/2$, so there is a path with average weight if and only if the Partition instance is feasible.

(If desired, the multigraph can easily be converted to an equivalent graph by splitting each edge $(i, i+1)$ of weight zero into two zero-weight edges.) $~~~\Box$

Proof sketch for Lemma 2. Given a Hamiltonian Path instance $G=(V,E)$ with source $s$ and sink $t$, construct the following multigraph $G'$. Let $n=|V|$.

First, add a new, long "super-path" from $s$ to $t$ as follows. Fix some $p, q, k, \ell$ to be determined later. Add $k$ new vertices $a_1, a_2, \ldots, a_k$, with $p$ new multi-edges $(a_i, a_{i+1})$ between each consecutive pair. Add edges $(s, a_1)$ and $(a_k, t)$. This addition adds $p^k$ paths of length $k$ from $s$ to $t$.

Now add another (separate) super-path to add $q^\ell$ paths of length $\ell$.

Choose $p, q, k, \ell$ so that the number of added paths is much larger than $n!$, so that the added paths determine (up to lower-order terms) the average path length. Choose $k$ and $\ell$ with $k < n-1 \ll \ell$, so that the average path length is larger than $n$, and closer to $n$ than to $\ell$. (Details below.)

Then the typical path will be a Hamiltonian path from $s$ to $t$ in the original graph, if there is one.

Here are the details for choosing $p, q, k, \ell$. Choose $p=n^{30}$, $q=n^2$, $k=n/3$, and $\ell=5n$. Then the addition adds $p^k = n^{10n}$ paths of length $k=n/3$, and $q^\ell = n^{10n}$ paths of length $\ell= 5n$. The average length of the added paths is then $(1/3 + 5)n/2 = 8n/3$. If there is a Hamiltonian $s$-$t$ path in the original graph, its length will be about $5n/3$ shorter than the average. The longer added paths, of length $5n$, are about $7n/3$ longer than the average, so they are not better. (Note that there are at most $n! \ll n^{10n}$ paths in the original graph, of length at most $n$, so they affect the average path length in the final graph by only lower-order terms.)

If a multigraph is not allowed, the construction can be adjusted appropriately by splitting each multi-edge as usual, then taking account how this affects $k$ and $\ell$. $~~~\Box$

  • $\begingroup$ This can surely be extended to show hardness of approximating even the unweighted case, by reducing from the longest-path problem (which is hard to approximate) instead of from hamiltonian path. $\endgroup$
    – Neal Young
    Oct 22, 2022 at 19:31

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