Minimum Spanning Tree

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Lecture 12

• Minimum Spanning Tree

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Motivating Example Point to Multipoint
Communication

• Single source, Multiple Destinations
• Broadcast
• All nodes in the network are destinations
• Multicast
• Some nodes in the network are destinations
• Only one copy of the information travels along
  common edges
• Message replication along forking points only.

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Spanning Tree

• We consider undirected graphs here.
• A tree is a connected graph without a cycle
• A spanning tree is a tree which has all vertices
  of the graph
• There may be multiple spanning trees
• We need to choose the minimum weight tree for
  broadcast.

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Blue edge spanning tree is the minimum weight
spanning tree
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Properties of a Tree

• A tree of V vertices has V-1 edges
• There exists a unique path between any two
  vertices of a tree.
• Adding any edge to a tree creates a unique cycle.
  Breaking any edge on this cycle restores a tree.

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Minimum Spanning Tree Construction

• We maintain a set of edges A, which is initially
  empty.
• Edges are added to A one at a time such that
  finally A becomes a minimum spanning tree.
• Edges are never removed from A.
• So safe edges must be added to A, i.e., at
  any stage A must be a part of a spanning tree.

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Safe Edge Addition

• Consider a cut in a graph (a cut consists of 2
  sets which partition the vertex set).

A,B,C D,E,F constitute a cut.

• A cut respects a set of edges I if no edge in the
  set crosses the cut.

Let I edge AB, edge EF.Cut A,B,C, D,E,F
respects set I

• A minimum weight edge crossing a cut is denoted
  a light weight edge in the cut

Edge CD is a light weight edge in the example cut.
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• Let A be a subset of a MST (minimum weight
  spanning tree).
• Let (S, V S) be any cut that respects A
• Let edge (u, v) be a light edge crossing the cut
• Then A ? (u, v) is subset of a MST

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Assume that all the edge weights are distinct.
Let no MST containing A contain edge (u, v).
Consider a MST T containing A
Add the edge (u, v) to T
Since (u, v) is not in T, T ? (u, v) contains a
cycle, and (u, v) is in the cycle.
Edge (u, v) are in the opposite sides of the cut.
Since any cycle must cross the cut even number of
times, there exists at least one other edge (x,
y) crossing the cut. Clearly, w(x, y) gt w(u, v).
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The edge (x, y) is not in A because (x, y)
crosses the cut, and the cut respects A.
Removing (x, y) from the cycle, breaks the cycle
and hence creates a spanning tree, T, s.t. T
T ? (u, v) (x, y)
w(T) w(T) w(u, v) w(x, y) ?
w(T) (as w(u, v) lt w(x, y))
This contradicts the fact that T is a MST.
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• So we always find a cut that respects A,
• And add a light edge across the cut to A.
• Kruskals and Prims algorithms find the cut
  differently.

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Kruskals Algorithm

• A ?
• For each vertex u in V, Create_Set(u)
• Sort E in increasing order by weight w
• For each edge (u,v) in the sorted list
• If Set(u)? Set(v)
• Add (u,v) to A
• Union Set(u) and Set(v)
• Return A

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Complexity Analysis

• The operations create set, testing whether set(u)
  set(v), union operations can be done in log V
  operations, using Union find data structure.
• Step 1 can be done in Vlog V
• Step 2 can be done in Elog E
• Step 3 can be done in Elog V
• Overall complexity is O(Vlog V Elog E Elog V)
  or O((V E)log V)

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Prims Algorithm

• Maintains a set of vertices S already in the
  spanning tree.
• Initially, S consists of one vertex r, selected
  arbitrarily.
• For every vertex u in V S, maintain the weight
  of the lightest edge between u and any vertex in
  S.
• If there is no edge between u and S, then this
  weight associated with u is infinity.

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• Add the vertex with the least weight to S.
• This is in effect adding the light weight edge
  crossing the cut S and V-S.
• Whenever a vertex is added to S, the weights of
  all its neighbors are reduced, if necessary.

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Pseudo-Code
S ?
For each u in V, keyu ?
Keyr 0
Predr NULL
While V ? S u Extract_Min(V-S) For each (v
in Adj(u)) if (v not in S) key(v)
min(w(u, v), key(v)) and pred(v) u
Add u in S
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For each v in Adju. can be done in E
complexity
Rest of the loop can be done in V2 complexity
So, overall O(V2)
Using heaps we can solve in O((V E)logV)
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Example