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To see the invariant, first note that <math>v</math> has distance <math>k</math>. So, for every node <math>w</math> such that <math>(v,w)\in A</math>, the distance of <math>w</math> is at most <math>k+1</math>. We have to show: The distance of <math>w</math> is at least <math>k</math>, and if there are nodes with distance <math>k+1</math> in <math>Q</math>, the distance of <math>w</math> is <math>k+1</math>.
To see the invariant, first note that <math>v</math> has distance <math>k</math>. So, for every node <math>w</math> such that <math>(v,w)\in A</math>, the distance of <math>w</math> is at most <math>k+1</math>. We have to show: The distance of <math>w</math> is at least <math>k</math>, and if there are nodes with distance <math>k+1</math> in <math>Q</math>, the distance of <math>w</math> is <math>k+1</math>.
First suppose for a contradiction that the disance of <math>w</math> is less than <math>k</math>.


== Correctness ==
== Correctness ==

Revision as of 10:52, 10 October 2014


General information

Algorithmic problem: Graph traversal

Type of algorithm: loop.

Abstract view

Definition: On this page, the distance of a node [math]\displaystyle{ v\in V }[/math] is the minimal number of arcs on a path from the start node [math]\displaystyle{ s }[/math] to [math]\displaystyle{ v }[/math].

Specific characteristic: The nodes are finished in the order of increasing distance (which is not unique, in general).

Auxiliary data: A FIFO queue [math]\displaystyle{ Q }[/math] whose elements are nodes in [math]\displaystyle{ V }[/math].

Invariant: Before and after each iteration:

  1. There is a current distance [math]\displaystyle{ k\in\mathbb{N} }[/math].
  2. Let [math]\displaystyle{ n }[/math] denote the current size of [math]\displaystyle{ Q }[/math]. There is [math]\displaystyle{ \ell\in\{1,\ldots,n\} }[/math] such that the first [math]\displaystyle{ \ell }[/math] elements of [math]\displaystyle{ Q }[/math] have distance [math]\displaystyle{ k }[/math] and the last [math]\displaystyle{ n-\ell }[/math] elements have distance [math]\displaystyle{ k+1 }[/math].

Variant: One node is removed from [math]\displaystyle{ Q }[/math].

Break condition: [math]\displaystyle{ Q=\emptyset }[/math].

Induction basis

Abstract view: The start node is seen, no other node is seen. The start node is the only element of [math]\displaystyle{ Q }[/math]. The output sequence is empty.

Implementation: Obvious.

Proof: Obvious.

Induction step

Abstract view:

  1. Extract the first element [math]\displaystyle{ v }[/math] from [math]\displaystyle{ Q }[/math].
  2. For each outgoing arc [math]\displaystyle{ (v,w) }[/math] of [math]\displaystyle{ v }[/math] such that [math]\displaystyle{ w }[/math] is not yet seen:
    1. Label [math]\displaystyle{ w }[/math] as seen.
    2. Append [math]\displaystyle{ w }[/math] to [math]\displaystyle{ Q }[/math].
    3. Add [math]\displaystyle{ w }[/math] to [math]\displaystyle{ Q }[/math].

Implementation: Obvious.

Proof: The variant is obviously fulfilled.

To see the invariant, first note that [math]\displaystyle{ v }[/math] has distance [math]\displaystyle{ k }[/math]. So, for every node [math]\displaystyle{ w }[/math] such that [math]\displaystyle{ (v,w)\in A }[/math], the distance of [math]\displaystyle{ w }[/math] is at most [math]\displaystyle{ k+1 }[/math]. We have to show: The distance of [math]\displaystyle{ w }[/math] is at least [math]\displaystyle{ k }[/math], and if there are nodes with distance [math]\displaystyle{ k+1 }[/math] in [math]\displaystyle{ Q }[/math], the distance of [math]\displaystyle{ w }[/math] is [math]\displaystyle{ k+1 }[/math].

First suppose for a contradiction that the disance of [math]\displaystyle{ w }[/math] is less than [math]\displaystyle{ k }[/math].

Correctness

It is easy to see that each operation of the algorithm is well defined. Due to the variant, the loop terminates after a finite number of steps.


Complexity

Statement: The asymptotic complexity is in [math]\displaystyle{ \Theta(|V|+|A|) }[/math] in the best and worst case.

Proof:


Pseudocode

BFS(G,s)
 1  for each vertex uG.V - {s}
 2       u.color = WHITE
 3       u.d = ∞
 4       u.π = NIL
 5  s.color = GRAY
 6  s.d = 0
 7  s.π = NIL
 8  Q = Ø
 9  ENQUE(Q, s)
10  while Q ≠ Ø
11      u = DEQUEUE(Q)
12      for each vG.Adj[u]
13              if v.color == WHITE
14                       v.color = GRAY
15                       v.d = u.d + 1
16                       v.π = u
17                       ENQUEUE(Q, v)
18      u.color = BLACK