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CIS 825 Lecture 9. Minimum Spanning tree construction Each node is a subtree/fragment by itself. Select the minimum outgoing edge of the fragment Send.

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Presentation on theme: "CIS 825 Lecture 9. Minimum Spanning tree construction Each node is a subtree/fragment by itself. Select the minimum outgoing edge of the fragment Send."— Presentation transcript:

1 CIS 825 Lecture 9

2 Minimum Spanning tree construction Each node is a subtree/fragment by itself. Select the minimum outgoing edge of the fragment Send message to connect to the fragment on the other side of this edge to form a larger fragment

3 Selecting minimum outgoing edge Each fragment has a leader/root Leader send FIND message to all nodes in the fragment Each node sends test message on outgoing edges, one at a time in increasing cost order, to find an outgoing edge Each node reports outgoing edges to the leader Leader then selects the minimum one Root is changed to the node with the minimum outgoing edge

4 Level Numbers Initially, fragments of size one are at level 0 When two fragments of size L merge, a new fragment of size L + 1 is created Level number is included in all messages When a node receives a message from a node with a higher level number X, it delays response until its fragment number becomes at least X.

5 When a node receives a message from a node with a higher level number X, it delays response until its fragment number becomes at least X

6

7 Fragment absorption If node x in fragment F at level L sends message to node y in fragment G at level L’ to combine such that L < L’, then if y is still finding outgoing edge then it absorbs F into G x

8 Message complexity A node is involved in at most one step of finding minimum outgoing edge per level number

9 Time complexity

10 Computing Functions Site i has x i. Compute F(x 1,...., x n )

11 Topology determination Is the topology of the network a tree? Is the topology of the network a ring?

12 Depth First Search

13 Initiator: not_visited = neighbor list select j from not_visited; remove j from not_visited; send visit() to j; : Any other site I - receive visit() from k visited i = true; parent = k; L: if (not_visited != {}) { select j from not_visited; remove j from not_visited; send visit() to j; } else send backtrack() to parent; -receive visit() from k remove k from not_visited; send ack() to k; -Receive backtrack() or ack() go to L:


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