Convergence of edge relaxations

Dependencies:

  1. Edge relaxations

Let $G = (V, E)$ be a weighted graph. Let $s$ be the source vertex from which we wish to find the shortest path to every vertex.

While relaxations are iteratively applied to $G$, the following properties hold:

The path relaxation property implies that if we intelligently choose the order of relaxations, we can make $d$ converge to $\delta$ if there is a shortest path to every vertex (A shortest path will not exist for every vertex if, for example, there is a cycle with only negative-weight edges).

Proof

Convergence

Since $u$ is a predecessor of $v$ in a shortest path to $v$, $\delta(v) = \delta(u) + w(u, v)$.

After relaxing $(u, v)$, \[ d(v) \le d(u) + w(u, v) = \delta(u) + w(u, v) = \delta(v) \]

By the upper-bound property, $d(v) \ge \delta(v)$. Therefore, $d(v) = \delta(v)$.

Path-relaxation property

Can be easily proved by using induction over length of $p$ and using the convergence property. The base case $|p| = 0$ is true, since even before any relaxations, $\delta(s) = d(s) = 0$. By the upper-bound property and monotonicity of $d$, $\delta(s) = d(s)$ forever after.

Dependency for:

  1. Dijkstra's algorithm

Info:

Transitive dependencies:

  1. /sets-and-relations/equivalence-classes
  2. /sets-and-relations/relation-composition-is-associative
  3. /sets-and-relations/equivalence-relation
  4. Graph
  5. Path in Graph
  6. Triangle inequality of shortest paths
  7. Edge relaxations