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Solutions to problem set #17 of cse 1400 and mth 2051 university courses, focusing on representing relations through graphs and adjacency matrices. The set includes rules for the rock-paper-scissors-lizard-spock game and instructions to construct adjacency matrices for the divides relation and congruence mod 5 relation.
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Typology: Exercises
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(a) Draw a graph which shows the rules of “Rock-Paper-Scissors- Lizard-Spock” Answer: Let R, P, Sc, L, Sp stand for Rock, Paper, Scissors, Lizard, Spock. The graph of rules is R
Sc L
Sp
(b) Construct an adjacency matrix which shows the rules of “Rock-Paper-Scissors-Lizard-Spock” Answer: Let R, P, Sc, L, Sp stand for Rock, Paper, Scissors, Lizard, Spock. The adjacency matrix is
R P Sc L Sp R 1 0 1 1 0 P 1 1 0 0 1 Sc 0 1 1 1 0 L 0 1 0 1 1 Sp 1 0 1 0 1 (c) Is the game a partial order? Why or why not? Answer: The answer is no, the game is not a partial order. I’ve made the game reflexive by my choice. By looking at the adja- cency matrix, you can determine that the game is antisymmet- ric. However, the game is not transitive, which you can see by looking at the graph. For example, Rock smashes Lizara and Lizard poisons Spock, but Rock does not beat Spock: The edges are directed. (d) Is the game an equivalence? Why or why not? Answer: The answer is no, the game is not an equivalence rela- tion. By looking at the adjacency matrix, you can determine that the game is not symmetric and the game is not transitive.
0 1 2 3 4 5 6 7 8 9 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 2 1 0 1 0 1 0 1 0 1 0 3 0 0 0 1 0 0 1 0 0 1 4 0 0 0 0 1 0 0 0 1 0 5 0 0 0 0 0 1 0 0 0 0 6 0 0 0 0 0 0 1 0 0 0 7 0 0 0 0 0 0 0 1 0 0 8 0 0 0 0 0 0 0 0 1 0 9 0 0 0 0 0 0 0 0 0 1