pset1soln

pset1soln - EECS 310 Fall 2011 Instructor Nicole Immorlica...

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Unformatted text preview: EECS 310, Fall 2011 Instructor: Nicole Immorlica Problem Set #1 Due: Sept 27, 2011 1. (15 points) Graph coloring abstraction. Describe a problem (other than the two mentioned in lecture) that can be solved using a graph coloring abstraction. Give a (small) instance of the problem, convert it to a graph, describe a coloring of your graph, and explain the implied solution. Solution: Alice is planning the seating chart for a reception and she wants to know how many tables she is going to need. Some of the guests at the reception are enemies with some of the other guests. Alice wants to design a seating chart such that no two enemies are sitting together at the same table. The tables that she has are big enough to hold any number of guests, so the only constraint in the number of tables needed is to avoid having two enemies sit at the same table. How many tables does Alice need? To formulate this as a graph coloring problem, we will add represent each person as a node in the graph and place an edge between two nodes if they are enemies. Then in any coloring of the resulting graph, the color of a node determines which table that person will sit at. Then a minimum coloring of this graph is the minimum number of tables that Alice needs. For example, assume there are four guests at this party: Alice, Bob, Charlie, and Dave. Dave is enemies with Alice, Bob, and Charlie and there are no other enemy relationships. In the constructed graph, we have four nodes (one for each person) and three edges. The edges are (Alice, Dave), (Bob, Dave), and (Charlie, Dave). The solution to the coloring problem shows that we only need two tables to seat all the guests. Dave is at his own table and everybody else at the other table. 2. (15 points) Logic. (a) (5 points) The exclusive or operation is defined as: P ⊕ Q is true when exactly one of P or Q is true. Find an expression equivalent to P ⊕ Q using only not ( ¬ ), and ( ∧ ), and or ( ∨ ). (b) (5 points) The operators not ( ¬ ) and and ( ∧ ) are complete in the sense that any logical statement can be expressed using only these two operators. Find an expres- sion equivalent to P ⇐⇒ Q using only not ( ¬ ) and and ( ∧ ). (c) (5 points) Rewrite ¬ ( P → Q ) using only not ( ¬ ) and and ( ∧ ), distributing the negation throughout the parathenses (so that the symbol ¬ is followed directly by P or Q in your final statement). Solution: Part A We claim the following expression is equivalent to ⊕ : ( P ∨ Q ) ∧ ( ¬ P ∨ ¬ Q )) . The first expression guarantees that at least one of P or Q is true and the second guarantees that they are not both true. We can check our solution by writing the truth table for the expressions and noting that the columns corresponding to the two expressions are equivalent....
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pset1soln - EECS 310 Fall 2011 Instructor Nicole Immorlica...

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