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MATH/CS 011 Final Exam Study Guide $9.99
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MATH/CS 011 Final Exam Study Guide
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This is a comprehensive study guide for MATH/CS 011 at the University of California, Riverside based on the class taught by Professor "Joe" for Winter 2023. Includes definitions and all major topics covered by the course, as well as practice questions and answers from the final exam.

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  • August 18, 2024
  • 14
  • 2023/2024
  • Exam (elaborations)
  • Questions & answers

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MATH/CS 011: Study Guide
Type Study Guide

Class CS 011

Date @March 18, 2024 8:00 AM

Quarter Winter



MATH/CS 011: Study Guide
UCR’s Math/CS 011 - Introduction to Discrete Structures

Professor Kejia Zhu (Joe) [Winter 2024 Final Exam: Study Guide]




1. Logic and Propositions
1.1 Logic Analysis

⚔️ Knights & Knaves
Problem: On an island, you encounter two inhabitants, A and B.



MATH/CS 011: Study Guide 1

, A says, "B is a Knave."
B says, "A is a Knight."
Solution:

1. If A were a Knight, then A's statement "B is a Knave" would be true. This
implies that B cannot tell the truth. Given that B's statement is "A is a Knight,"
which in this case would be true, it creates a contradiction if B were indeed a
Knave, as Knaves cannot tell the truth.

2. If A were a Knave, then A's statement "B is a Knave" would be a lie, meaning B
must be a Knight. Since B claims that "A is a Knight," and if B were a Knight
(hence, always telling the truth), this would again create a contradiction
because we assumed A is a Knave.

To solve this paradox, analyze the logic more carefully:

If A's statement about B being a Knave were true, then A would be a Knight
(since Knights tell the truth). However, for B to say "A is a Knight" would also
be true, which cannot happen if B were a Knave, as Knaves cannot tell the
truth (a contradiction).

The only scenario that does not result in a logical contradiction is if A is a
Knave making a false statement about B, and B is a Knight, correctly
identifying A's status.

Therefore, the correct conclusion is that A is a Knave and B is a Knight , as this is
the only configuration that does not result in a contradiction.


1.2 Logical Symbols & Propositions
Symbols: ∃(exists), ∀(for all), →(implies).




MATH/CS 011: Study Guide 2

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