Hw3s - ENGRI 1101 Engineering Applications of OR Fall 2008...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ENGRI 1101 Engineering Applications of OR Fall 2008 Homework 3 Homework 3 Solutions 1. (a) Each sequence corresponds to a node of the graph, so node 1=0000000101, node 2=1100111111, node 3=1100111110, node 4=0000001001, and node 5=0000111111. There is an edge between every pair of nodes, and the cost edge { i,j } is the number of bits that need to be changed to convert sequence i into sequence j (or vice versa). They are given in the following table: Node 1 2 3 4 5 1 6 7 2 4 2 1 6 2 3 7 3 4 4 5 The graph is depicted here: 5 2 4 3 1 6 7 2 4 1 6 2 7 3 4 (b) The minimum spanning tree calculated using Kruskal’s algorithm is presented below. Note that there are several edges of the same cost in the tree, so, depending on how you broke the ties between these edges, you may have ended up with a different tree. The cost of you minimum spanning tree should be the same, though. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
5 4 3 2 1 4 4 7 3 2 6 1 7 2 5 4 3 2 1 4 4 7 3 2 6 1 7 2 5 4 3 2 1 4 4 7 3 2 6 1 7 2 5 4 3 2 1 4 4 7 3 2 6 1 7 2 Edges were considered in the following order:
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/12/2009 for the course ENGRD 2700 taught by Professor Staff during the Spring '05 term at Cornell University (Engineering School).

Page1 / 4

Hw3s - ENGRI 1101 Engineering Applications of OR Fall 2008...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online