Johnson's Algorithm

Johnson’s Algorithm is an algorithm that computes the length of the shortest path between every pair of vertices in a directed graph $G=⟨V,E⟩$ with edge weights and no negative cycles in

• $\mathcal{O}(|V{|}^3+|V||E|)$ time and $\mathcal{O}(|V{|}^2+|E|)$ space, or

• $\mathcal{O}((|V{|}^2+|V||E|)\log |E|)$ time and $\mathcal{O}(|V{|}^2+|E|)$ space.

Tip

This problem can also be solved by the Floyd-Warshall Algorithm in $\mathcal{O}(|V{|}^3+|E|)$ time and $\mathcal{O}(|V{|}^2)$ space.

Tip

This problem can also be solved by the Bellman-Ford Algorithm in $\mathcal{O}(|V{|}^2|E|)$ time and $\mathcal{O}(|V{|}^2)$ space.

Algorithm

Let ${\mathrm{dist}}_G(x,y)$ denote the length of the shortest path between $x$ and $y$ in graph $G$.

0. Let $s$ be a new vertex, and let $G^{\ast }=⟨V\cup \{s\},E\cup \{⟨s,v,0⟩:v\in V\}⟩$.

1. Apply the Bellman-Ford Algorithm to find ${\mathrm{dist}}_{G^{\ast }}(s,v)$ for each vertex $v$ in $V$.

Lemma

Let $h(v)={\mathrm{dist}}_{G^{\ast }}(s,v)$, then

$$\forall e\in E,w(e)+h(s(e))-h(t(e))\ge 0$$

Proof

• If $w(e)+h(s(e))-h(t(e))<0$, since $$w(e)+h(s(e))-h(t(e))<0⟺{\mathrm{dist}}_{G^{\ast }}(s,t(e))>{\mathrm{dist}}_{G^{\ast }}(s,s(e))+w(e)$$ it follows that, let $e_0{e}_1\dots e_{n-1}$ be a shortest path from $s$ to $s(e)$, then $e_0{e}_1\dots e_{n-1}e$ is a path from $s$ to $t(e)$ with a weight less than ${\mathrm{dist}}_{G^{\ast }}(s,t(e))$.

By contradiction, it follows that $w(e)+h(s(e))-h(t(e))\ge 0$.

3. Let $G^{\prime }=⟨V,\{⟨s(e),t(e),w(e)+h(s(e))-h(t(e))⟩:e\in E\}⟩$, then applying the lemma yields that $G^{\prime }$ is a directed graph with non-negative edge weights.
4. Apply Dijkstra’s Algorithm $|V|$ times to find ${\mathrm{dist}}_{G^{\prime }}(x,y)$ for each pair of vertices $x$ and $y$ in $V$.

Lemma

$$\forall x,y\in V,{\mathrm{dist}}_G(x,y)={\mathrm{dist}}_{G^{\prime }}(x,y)-h(x)+h(y)$$

Proof

Let $e_0{e}_1\dots e_{n-1}$ be a path from $x$ to $y$ in $G$, $e_0^{\prime }{e}_1^{\prime }\dots e_{n-1}^{\prime }$ be the corresponding path in $G^{\prime }$.

$$\begin{array}{rl}\sum _{i=0}^{n-1}w(e_i)& =-h(x)+\sum _{i=0}^{n-1}(w(e_i)+h(s(e))-h(t(e)))+h(y)\\ & =\sum _{i=0}^{n-1}w(e_i^{\prime })-h(x)+h(y)\end{array}$$

5. Apply the lemma to find ${\mathrm{dist}}_G$.

This algorithm solves the problem in

• $\mathcal{O}(|V{|}^3+|V||E|)$ time and $\mathcal{O}(|V{|}^2+|E|)$ space, or

• $\mathcal{O}((|V{|}^2+|V||E|)\log |E|)$ time and $\mathcal{O}(|V{|}^2+|E|)$ space.

std::vector<std::vector<int>> johnson(int n, int m, std::vector<int> u, std::vector<int> v, std::vector<int> w) {
	u.resize(m + n, n), v.resize(m + n), w.resize(m + n, 0);
	std::ranges::iota(v | std::views::drop(m), 0);
 
	auto h = bellman_ford(n + 1, m + n, u, v, w, n);
 
	u.resize(m), v.resize(m), w.resize(m);
	for (int i = 0; i < m; i++) {
		w[i] += h[u[i]] - h[v[i]];
	}
 
	std::vector dist(n, std::vector<int>(n));
	for (int i = 0; i < n; i++) {
		dist[i] = dijkstra(n, m, u, v, w, i);
		for (int j = 0; j < n; j++) {
			dist[i][j] -= h[i] - h[j];
		}
	}
	return dist;
}