Floyd-Warshall

Floyd-Warshall

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Abstract View

Algorithmic problem: All pairs shortest paths

Prerequisites: None

Type of algorithm: loop

Auxilliary data:

1. A constant representing the number of node: [math]\displaystyle{ n = |V| }[/math]
2. A real-valued [math]\displaystyle{ (n \times n) }[/math] matrix [math]\displaystyle{ M }[/math]
3. An ordering of all nodes, that is, [math]\displaystyle{ V = \{u_1, \ldots ,u_n\} }[/math]

Notation: On this page, [math]\displaystyle{ M^i }[/math] refers to the contents of [math]\displaystyle{ M }[/math] after [math]\displaystyle{ i\geq 0 }[/math] iterations.

Invariant: After [math]\displaystyle{ i \ge 0 }[/math] iterations, for [math]\displaystyle{ v,w \in V }[/math], [math]\displaystyle{ M(v,w) }[/math] is the length of a shortest [math]\displaystyle{ (v,w) }[/math]-path subject to the constraint that all internal nodes of the path belong to [math]\displaystyle{ \{u_1, \ldots , u_i\} }[/math].

Variant: [math]\displaystyle{ i }[/math] increases by [math]\displaystyle{ 1 }[/math].

Break condition: It is [math]\displaystyle{ i=n }[/math]

Induction basis

Abstract view: For all [math]\displaystyle{ v,w \in V }[/math] we set

1. [math]\displaystyle{ M(v,v):=0, \forall v \in V }[/math]
2. [math]\displaystyle{ M(v,w):=l(v,w), (v,w) \in A }[/math]
3. [math]\displaystyle{ M(v,w):= +\infty }[/math], otherwise

Implementation: Obvious.

Proof: Nothing to show.

Induction step

Abstract view: Bring [math]\displaystyle{ v_i }[/math] into the game.

Implementation: For all [math]\displaystyle{ v,w\in V }[/math], set [math]\displaystyle{ M^{i+1}(v,w):=\min\{M^i(v,w),M^i(v,u_i)+M^i(u_i,w)\} }[/math].

Correctness: Let [math]\displaystyle{ p }[/math] denote a shoortest [math]\displaystyle{ (v,w }[/math]-path subject to the constraint that all internal nodes are taken from [math]\displaystyle{ \{u_1, \ldots, u_n \} }[/math].

1. If [math]\displaystyle{ p }[/math] does not contain [math]\displaystyle{ u_i }[/math], [math]\displaystyle{ p }[/math] is even a shortest [math]\displaystyle{ (v,w) }[/math]-path such that all internal nodes are taken from [math]\displaystyle{ \{u_1, \ldots , u_{i-1} \} }[/math]. In this case, the induction hypothesis implies that the value of [math]\displaystyle{ M(v,w) }[/math] immediately before the i-th iteration equals the length of [math]\displaystyle{ p }[/math]. Clearly, the i-th iteration does not change the value of [math]\displaystyle{ M(v,w) }[/math] in this case.
2. On the other hand, suppose [math]\displaystyle{ p }[/math] does contain [math]\displaystyle{ u_i }[/math]. Due to the prefix property, the segment of [math]\displaystyle{ p }[/math] from [math]\displaystyle{ v }[/math] to [math]\displaystyle{ u_i }[/math] and from [math]\displaystyle{ u_i }[/math] to [math]\displaystyle{ w }[/math] are a shortest [math]\displaystyle{ (v,u_i) }[/math]-path and a shortest [math]\displaystyle{ (u_i,w) }[/math]-path, respectively, subject to the constraint that all internal nodes are from [math]\displaystyle{ \{u_1, \ldots , u_{i-1} \} }[/math]. The induction hypothesis implies that tehese lengths equal the values [math]\displaystyle{ M(v,u_i) }[/math] and [math]\displaystyle{ M(u_i, w) }[/math], respectively.

Complexity

Statement: [math]\displaystyle{ \mathcal{O}(n^3) }[/math]

Proof: The overall loop has [math]\displaystyle{ n }[/math] iterations. In each iteration, we update all [math]\displaystyle{ n^2 }[/math] matrix entities. Each update requires a constant number of steps.