[FIXED] What is the relaxation condition in graph theory

Issue

I'm trying to understand the main concepts of graph theory and the algorithms within it. Most algorithms seem to contain a "Relaxation Condition" I'm unsure about what this is.

Could some one explain it to me please.

An example of this is dijkstras algorithm, here is the pseudo-code.

 1  function Dijkstra(Graph, source):
 2      for each vertex v in Graph:           // Initializations
 3          dist[v] := infinity               // Unknown distance function from source to v
 4          previous[v] := undefined          // Previous node in optimal path from source
 5      dist[source] := 0                     // Distance from source to source
 6      Q := the set of all nodes in Graph
    // All nodes in the graph are unoptimized - thus are in Q
 7      while Q is not empty:                 // The main loop
 8          u := vertex in Q with smallest dist[]
 9          if dist[u] = infinity:
 10              break                         // all remaining vertices are inaccessible from source
 11          remove u from Q
 12          for each neighbor v of u:         // where v has not yet been removed from Q.
 13              alt := dist[u] + dist_between(u, v)
 14              if alt < dist[v]:             // Relax (u,v,a)
 15                  dist[v] := alt
 16                  previous[v] := u
 17      return dist[]

Thanks

Solution

Relaxation step:

• You have two nodes, u and v
• For every node, you have a tentative distance from the source node (for all nodes except for the source, it starts at positive infinity and it only decreases up to reaching its minimum).

The relaxation step basically is asking this:

• I already know that I can reach v with some path of distance dist[v]. Could I improve on this by going to v through u instead? (where the distance of the latter would be dist[u] + weight(u, v))

Graphically:

s ~~~~~~~> v
 \         ^
  \        |
   \~~~~~> u

You know some path s~>v which has distance dist[v], and you know some path s~>u which has distance dist[u]. If dist[u] + weight(u, v) < dist[v], then the path s~>u->v is shorter than s~>v, so you'd better use that one!

(I write a~>b to mean a path of any length from a to b, while a->b I mean a direct single edge from a to b).

You may also want to check this lecture: http://ocw.mit.edu/OcwWeb/Electrical-Engineering-and-Computer-Science/6-046JFall-2005/VideoLectures/detail/embed17.htm

Answered By - Dimitris Andreou

Answer Checked By - Candace Johnson (PHPFixing Volunteer)