Dijkstra

Dijkstra

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Dijstra's algorithm is a graph algortihm solving the single-source shortest-paths problem.

Requirements

• directed Graph [math]\displaystyle{ G=(V,E) }[/math]
• weight function [math]\displaystyle{ w\colon E\to\mathbb R }[/math]
• [math]\displaystyle{ \forall(u,v)\in E\colon w(u,v)\geq 0 }[/math]
• source [math]\displaystyle{ s\in V }[/math]
• Datastruct [math]\displaystyle{ S }[/math]
• Datastruct [math]\displaystyle{ Q }[/math]

General information

Algorithmic problem: Single source shortest paths

Prerequisities: For all [math]\displaystyle{ \alpha \in A }[/math], it is [math]\displaystyle{ l(a) \geq 0 }[/math].

Type of algortihm: loop

Auxiliary data:

1. A temporary distance value [math]\displaystyle{ \delta(v) \in \R }[/math] for each node [math]\displaystyle{ v \in V }[/math]. At termination, it is [math]\displaystyle{ \delta(v) = \Delta(v) }[/math] for all [math]\displaystyle{ v \in V }[/math].
2. A bounded priority queue [math]\displaystyle{ Q }[/math] of size [math]\displaystyle{ |V|-1 }[/math], which contains nodes from [math]\displaystyle{ V }[/math] and takes their [math]\displaystyle{ \delta }[/math]-values as keys. The node with minimal key is returned.

Abstract view

Invariant: After [math]\displaystyle{ i \geq 0 }[/math] iterations:

1. [math]\displaystyle{ Q }[/math] contains all nodes of [math]\displaystyle{ V }[/math] except for [math]\displaystyle{ s }[/math] and [math]\displaystyle{ i }[/math] further nodes.
2. For the nodes [math]\displaystyle{ v \in V }[/math] not in [math]\displaystyle{ Q }[/math], it is [math]\displaystyle{ \delta(v) = \Delta(v) }[/math].
3. For the nodes [math]\displaystyle{ v \in V }[/math] in [math]\displaystyle{ Q }[/math], [math]\displaystyle{ \delta(v) }[/math] is the length of a shortes [math]\displaystyle{ (s,v) }[/math]-path that solely contains nodes not in [math]\displaystyle{ Q }[/math] (except for [math]\displaystyle{ v }[/math] itself, of course). As usual, this means [math]\displaystyle{ \delta(v) = +\infty }[/math] if there is no such path.

In particular, it is [math]\displaystyle{ \delta(v) \geq \Delta(v) }[/math] for each node [math]\displaystyle{ v \in V }[/math].

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

Break condition:

• [math]\displaystyle{ Q = \empty }[/math], which means that all nodes are reachable from [math]\displaystyle{ s }[/math] and have been processed;
• otherwise, if [math]\displaystyle{ \delta(v) = +\infty }[/math] for the noext node [math]\displaystyle{ v }[/math] of [math]\displaystyle{ Q }[/math], which means that [math]\displaystyle{ \delta(v) = + \infty }[/math] for every node in [math]\displaystyle{ Q }[/math] and hence none of the nodes in [math]\displaystyle{ Q }[/math] is reachable from [math]\displaystyle{ s }[/math].

Induction basis

Abstract view:

1. All nodes of [math]\displaystyle{ V }[/math] except for [math]\displaystyle{ s }[/math] have to be members of [math]\displaystyle{ Q }[/math].
2. Root [math]\displaystyle{ s }[/math] must have correct distance [math]\displaystyle{ \Delta(s) = 0 }[/math].
3. The other nodes must meet Invariant #3.

Implementation:

1. [math]\displaystyle{ \delta(s) := 0 }[/math];
2. For all [math]\displaystyle{ a = (s,v) \in A }[/math], set [math]\displaystyle{ \delta(v) := l(a) }[/math].
3. For all [math]\displaystyle{ v \in V\setminus\{s\} }[/math] with [math]\displaystyle{ (s,v) \notin A }[/math] set [math]\displaystyle{ \delta(v) := +\infty }[/math].
4. Insert all nodes in [math]\displaystyle{ V\setminus\{s\} }[/math] into [math]\displaystyle{ Q }[/math].

Proof:

Obvious.

Induction step

Abstract view:

One node is extracted from [math]\displaystyle{ Q }[/math], and the distance values of the endnodes of its outgoing arcs are updated to meet Invariant #3.

Implementation:

1. Extract the next node [math]\displaystyle{ v }[/math] from [math]\displaystyle{ Q }[/math].
2. For each outgoing arc [math]\displaystyle{ a = (v,w) \in A }[/math] such that [math]\displaystyle{ w \in Q }[/math], set [math]\displaystyle{ \delta(w) := min\{\delta(w), \delta(v) + l(a)\} }[/math].

Correctness:

The first invariant is trivially maintained. So we will focus on the second and third invariants.
Consider the moment immediately before the [math]\displaystyle{ i }[/math]-iteration. The core of the proof is to show that [math]\displaystyle{ \delta(v) = \Delta(v) }[/math] held at that moment (and hence forever). For a contradiction, suppose there is an [math]\displaystyle{ (s,v) }[/math]-path [math]\displaystyle{ p }[/math] whose length is strictly less than [math]\displaystyle{ \delta(v) }[/math]. The third invariant implies that [math]\displaystyle{ p }[/math] has nodes in [math]\displaystyle{ Q }[/math] besides [math]\displaystyle{ v }[/math]. Let [math]\displaystyle{ u }[/math] denote the first such node on [math]\displaystyle{ p }[/math]. Then none of the nodes on the subpath of [math]\displaystyle{ p }[/math] from [math]\displaystyle{ s }[/math] to [math]\displaystyle{ u }[/math] belonged to [math]\displaystyle{ Q }[/math] at that moment (except for [math]\displaystyle{ u }[/math] itself, of course). Due to the prefix property, this subpath is a shortest [math]\displaystyle{ (s,u) }[/math]-path, so Invariant #3 implies [math]\displaystyle{ \delta(u) = \Delta(u) }[/math]. On the other hand, the specific choice of [math]\displaystyle{ v }[/math] implies [math]\displaystyle{ \delta(u) \geq \delta(v) }[/math]. In summary, the subpath of [math]\displaystyle{ p }[/math] from [math]\displaystyle{ s }[/math] to [math]\displaystyle{ w }[/math] is not shorter than [math]\displaystyle{ \delta(v) }[/math]. To obtain [math]\displaystyle{ l(p) \lt \delta(v) }[/math], the subpath of [math]\displaystyle{ p }[/math] from [math]\displaystyle{ u }[/math] to [math]\displaystyle{ v }[/math] must have negative total length, which contradicts the prerequisite [math]\displaystyle{ l \geq 0 }[/math].
Now we know [math]\displaystyle{ \delta(v) = \Delta(v) }[/math] Therefore, [math]\displaystyle{ v }[/math] fulfills the statement of the second invariant after the [math]\displaystyle{ i }[/math]-th iteration. Of course, the second invariant is not violated by any other node either, and is thus maintained.
For a node [math]\displaystyle{ w \in V }[/math] let [math]\displaystyle{ P_i (w) }[/math] denote the set of all [math]\displaystyle{ (s,w) }[/math]-paths such that no node (except for [math]\displaystyle{ w }[/math] itself) is in [math]\displaystyle{ Q }[/math] immediately after the [math]\displaystyle{ i }[/math]-th iteration. In particular, the statement of the third invariant means that [math]\displaystyle{ \delta(w) }[/math] is the length of a shortest path in [math]\displaystyle{ P_i (w) }[/math]. Tthe induction hypothesis says that [math]\displaystyle{ \delta(w) }[/math] is the minimum length of all [math]\displaystyle{ (s,w) }[/math]-paths in [math]\displaystyle{ P_{i-1} (w) }[/math]. The difference [math]\displaystyle{ P_i (w) \setminus P_{i-1} (w) }[/math] consists of all [math]\displaystyle{ (s,w) }[/math]-paths such that [math]\displaystyle{ (v,w) }[/math] is the last arc and all nodes except for [math]\displaystyle{ v }[/math] and [math]\displaystyle{ w }[/math] belong to [math]\displaystyle{ V_{i-1} }[/math]. Therefore, the second step of the iteration incorporates [math]\displaystyle{ P_i (w) \setminus P_{i-1} (w) }[/math] in the computation of [math]\displaystyle{ \delta(w) }[/math] appropriately to maintain Invaraint #3.

Pseudocode

DIJKSTRA(G,w,s)

DIJKSTRA(G,w,s)
1 INITIALIZE-SINGLE-SOURCE(G,s) 
2      S = ∅
3      Q = G.V
4      while Q ≠ ∅
5            u = EXTRACT-MIN(Q)
6            S = S ∪ {u} 
7            for each vertex v ∈ G.Adj[u]
8                 RELAX(u,v,w)

RELAX(u,v,w)

RELAX(u,v,w)
1 if v.d > u.d + w(u,v)
2       v.d = u.d + w(u,v)
3       v.π = u

INITIALIZE-SINGLE-SOURCE(G,s)

INITIALIZE-SINGLE-SOURCE(G,s) 
1 for each vertex v ∈ G.V
2         v.d = ∞
3         v.π = NIL
4 s.d = 0