Running in pairs (and, more generally, in groups) can be more rewarding than running alone. Running buddies can motivate each other, share the mental load of maintaining pace, and provide competition and accountability.

The main problem with running buddies is that they can be hard to find. Not everyone is a runner, and runners vary in their abilities and training goals. Moreover, these abilities and goals are mostly unobservable by other runners searching for a buddy. This unobservable variation creates “matching frictions” that prevent runners from sorting into “optimal” pairs.

If prospective running buddies could observe each others’ abilities and training goals then they could form preferences over whom they want to be paired with.1 Runners could rank potential buddies from most to least prefered, and submit these rankings to a central match-maker (e.g., a team coach) whose task would be to partition the (assumedly even-sized) set of $$2n$$ runners into $$n$$ pairs. The socially optimal partition $$\mathcal{P}^*$$ would minimise the sum $$S(\mathcal{P})=\sum_{\{i,j\}\in\mathcal{P}}(x_{ij}+x_{ji}),$$ where $$x_{ij}$$ is the rank that runner $$i$$ assigns to potential buddy $$j$$. Minimising $$S(\mathcal{P})$$ would ensure that, on average, runners are paired with their most preferred buddies.

Let $$X=(x_{ij})$$ be the matrix of preference rankings and let $$Y=X+X^T$$. One way to find $$\mathcal{P^*}$$ would be to choose $$2n$$ entries of $$Y$$ such that (a) the sum of the chosen entries is minimised, and (b) each row and column of $$Y$$ contains exactly one chosen entry.2 This choice problem is equivalent to the (balanced) assignment problem, and can be solved using the Hungarian algorithm or via linear programming.

The socially optimal partition $$\mathcal{P^*}$$ of the set of runners into pairs may be “unstable:” there may exist two runners who would rather run with each other than with their assigned buddies.3 For example, suppose there is a set $$\{a, b, c, d\}$$ of four runners with preference rankings captured by the matrix $$X=\begin{bmatrix} & 1 & 2 & 3 \\ 2 & & 1 & 3 \\ 1 & 2 & & 3 \\ 1 & 2 & 3 & \end{bmatrix}.$$ The corresponding matrix $$Y=X+X^T$$ of bidirectional sums is $$Y=\begin{bmatrix} & 3 & 3 & \underline{4} \\ 3 & & \underline{3} & 5 \\ 3 & \underline{3} & & 6 \\ \underline{4} & 5 & 6 & \end{bmatrix},$$ where the underlined entries correspond to the socially optimal partition $$\mathcal{P}^*=\{\{a,d\},\{b,c\}\}$$ with $$S(\mathcal{P}^*)=14$$. This partition is unstable because runner $$a$$ would prefer to run with $$c$$ than $$d$$, and runner $$c$$ is indifferent between runners $$a$$ and $$b$$. Runners $$a$$ and $$c$$ would ignore the match-maker and become buddies, resulting in a socially inferior partition $$\mathcal{P}_*=\{\{a,c\},\{b,d\}\}$$ with $$S(\mathcal{P}_*)=16$$.

If the socially optimal partition of runners into pairs is unstable then the match-maker would need to prevent, or at least discourage, so-called “blocking pairs” from deviating from the optimum. For example, the match-maker could restrict runners’ access to training areas (e.g., running tracks and trails) so that no blocking pairs have concurrent access. However, such restrictions may be detrimental to runners’ training and camaraderie, and, consequently, reduce social welfare.

1. For example, if I wanted to improve my pace then I might prefer to run with someone slightly faster than me so that I can try to match their speed without them racing ahead of me. ↩︎

2. Equivalently, one could choose $$n$$ entries in the upper-right triangle of $$Y$$ that satisfy criteria (a) and (b). This works because $$Y$$ is symmetric. If $$y_{ij}$$ is chosen when minimising sums over $$Y$$ then $$y_{ji}$$ is chosen when minimising sums over $$Y^T$$. But $$Y=Y^T$$, so the sets of chosen entries when minimising over $$Y$$ and $$Y^T$$ must be equal. Thus, $$y_{ij}$$ and $$y_{ji}$$ must belong to both sets, and so the lower-left triangle can be ignored. ↩︎

3. On the other hand, I conjecture that every stable partition is socially optimal. ↩︎