UMBC CMSC203 Discrete Structures, Section 06, Spring 2016

Homework 9, Due Thursday, 04/14

Instructions: In the following questions you are asked to use proof by induction. Your proof must not simply be a sequence of equations, even if the statement you are proving is arithmetic in nature. Clearly indicate using complete English sentences: 1) what you are allowed to assume from the induction hypothesis, 2) what you need to show to establish the induction step, and 3) which steps of the proof uses the induction hypothesis.

Responses that do not include well-written English sentences that clearly explain your proof will receive a grade of no more than 50%.

1. Fibonacci Numbers. [Adapted from Rosen 5/e]
   The Fibonacci numbers f₀,  f₁,  f₂, ... are defined recursively by f₀ = 0,  f₁ = 1,  f₂ = 1 and for all n ≥ 3,

   f_n = f_{n − 1} + f_{n − 2} .

   Use structural induction to show that

   ( f₁ )² + ( f₂ )² + ⋅⋅⋅ + ( f_n )² = f_n f_n+1

2. String Reversal. [Adapted from Rosen 5/e]
   In this problem, we consider only strings over {0, 1} of length greater than or equal to 0. For example, the string 01001 is a string of length 5. We use λ to denote the empty string. For a string w, we can define the reversal of the string, written w^R, as follows:
   • λ^R = λ
   • for a string x, ( 0 x )^R = x^R 0.
   • for a string x, ( 1 x )^R = x^R 1.
   For example, we can unwind the recursive definition of 01001^R by:

   01001^R = 1001^R0 = 001^R10 = 01^R010 = 1^R0010 = λ^R10010 = λ10010 = 10010.

   Use structural induction to prove that for all strings w and x, (w x)^R = x^R w^R.

   Note: For this problem, you may assume without proof that string concatenation is associative. So, given 3 strings x, y and z,  ( x y ) z = x ( y z ).

3. Tournament Graphs. [Adapted from Rosen 5/e]
   Recall that a tournament graph G is a directed graph (without self loops) such that for any two distinct vertices u and v in G, exactly one of (u, v) or (v, u) is an edge in G. (Here, u and v being distinct means u ≠ v.)

   You can think of the vertices of a tournament graph as teams participating in a round-robin tournament (where each team plays every other team once) in a sport that does not allow tie games. Then, for any two teams u and v, either u defeats v or v defeats u (but not both). If u defeats v, then the edge (u, v) is placed in the graph. Otherwise, v defeats u and (v, u) is placed in the graph.

   In this problem, you are asked to prove by induction that every tournament graph G has a "winner". Here, a winner is a vertex w that defeats every vertex (other than w) either directly or indirectly. We say that w defeats x directly if (w, x) is an edge in G. Vertex w defeats vertex x indirectly if there exists a vertex u such that (w, u) and (u, x) are edges in G (that is, w defeats u which in turn defeats x).

   Note: a tournament graph can have more than one winner.

   Hint: for a vertex v think about the set of vertices that v defeats directly and a second set of vertices that v defeats indirectly but not directly.