# Dominance

Topics: Big 12 Conference, The A-Team, Kansas Pages: 6 (1521 words) Published: April 2, 2013
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Dominance Matrices
Text Reference: Section 2.1, p. 114 The purpose of this set of exercises is to apply matrices and their powers to questions concerning various forms of competition between individuals and groups. In Section 2.1 powers of a square matrix are deﬁned: ¡¡¡

ßÞ

To see one area in which matrix powers are useful, consider the following situation. A football conference consists of 5 teams: Teams A, B, C, D, and E. Each team plays the other 4 once, and the results this season were: A B C D E defeated defeated defeated defeated defeated C and D A, C, and E D B A, C, and D

The question is how to rank these teams. Teams B and E both won 3 games while losing 1. Which team is more powerful? How shall the tie between these teams be broken? One way is to note that B defeated E, and thus should be more powerful. In this case B has one-step dominance over E. There is a problem with this method of tie-breaking: in larger conferences each team may not play every other team. Another way to handle the question of tie-breaking is to investigate two-step dominance. For example, A has two-step dominance over B because A defeated D and D defeated B. Notice that one team can have two-step dominance over another in multiple ways: team B has two-step dominance over team C in two ways: B defeated A who defeated C, and B defeated E who defeated C. Question: 1. Find all of the two-step dominances in the example above, keeping track of the number of ways each dominance occurs. Luckily, there is an easier way to calculate two-step dominances. Create a ¢ matrix by ﬁrst associating each team with a corresponding row and column of the matrix: team A corrsponds to row 1 and column 1, team B to row and column 2, etc. Place a 1 at the ´ µ entry of the matrix if the team corresponding to row defeated the team corresponding to column . Otherwise a 0 is recorded. The example generates the following dominance matrix:

¾

¼ ½ ¼ ¼ ½

¼ ¼ ¼ ½ ¼

½ ½ ¼ ¼ ½

½ ¼ ½ ¼ ½

¼ ½ ¼ ¼ ¼

¿

1

Thus each win for team A, for example, is encoded in the matrix as a 1 in the appropriate place in the ﬁrst row of the matrix . If the entries in the ﬁrst row of the matrix are added, the total number of wins (or one-step dominances) which Team A has will result. To compute two-step dominances, consider the matrix

¾

¼ ½ ¼ ½ ¼

½ ¼ ½ ¼ ½

¼ ¾ ¼ ½ ½ µ

½ ¿ ¼ ¼ ¾

¼ ¼ ¼ ½ ¼

¿

¾

By the row-column rule (p. 103 of the text), the ´
½ ½ · ¾ ¾ ·

entry in
·

¾

is

¿ ¿ ·

Each term in this sum is either a 0 or a 1, since the terms in are either 0’s or 1’s. If a term of ½, then ½ and ½. In terms of the athletic competition, ½ and the form Ø team defeated the Ø team and the Ø team defeated the Ø team. ½ implies that the Thus there is one way in which the Ø team has two-step dominance over the Ø team. Adding up will thus count how many two-step dominances the Ø team has over all terms of the form the Ø team. Question: 2. By comparing the entries in ¾ with your answer to Question 1, conﬁrm that the ´ µ entry in ¾ is the number of ways in which the team corrsponding to row has two-step dominance over the team corresponding to column . Now the entries in each row of ¾ may be added to see how many two-step dominances each team has. Team A has 2, team B has 6, team C has 1, team D has 3, and team E has 4. So since team B has 6 two-step dominances and team E has only 4, two-step dominances can be used to break their tie and declare B the champion. Alternatively, one could count both one-step and two-step dominances by computing

¾

¼ ¾ ¼ ½ ½

½ ¼ ½ ½ ½

½ ¿ ¼ ½ ¾

¾ ¿ ½ ¼ ¿

¼ ½ ¼ ½ ¼

¿

·

¾

then adding the entries in each row of this matrix. Some authors in the ﬁelds of sociology and psychology call this sum the power of the team associated with row . The powers of the example teams are: A B C D E 2...