UMBC CMSC441, Design & Analysis of Algorithms, Fall 2014


Homework 1

Due Thursday, September 4, 2014

Instructions:

In the following questions you are asked to provide a proof by induction. You must provide a proof by induction, written with complete English/Math sentences, even if there are other ways to prove the statement. Your proof cannot simply be a sequence of equations, even if the statement you are proving is numerical in nature.

Your proof must clearly indicate, with complete English/Math sentences:

  1. The statement to be proven
  2. The induction hypothesis and the induction parameter
  3. The base case(s) proven
  4. The statement you must prove to establish the induction step
  5. The place(s) in the proof of the induction step that use the induction hypothesis
  6. The justification(s) for the use(s) of the induction hypothesis in the proof of the induction step

Submissions that do not include well-written English/Math sentences that clearly explain your proof will receive a grade of less than 50%. Here is an example of a graded, well-written proof by induction: hw0.pdf.

  1. Series. Prove by induction on n, that for all n ≥ 1,
    12 + 22 + 32 + ··· + n2 = n ( n + 1 ) ( 2 n + 1 ) / 6
    Note: present your algebraic manipulations on only one side of the equation.

  2. Red & Black Checkers. You have a line of n checkers arranged left to right, where n ≥ 2. Each checker is either red or black. The leftmost checker is red and the rightmost checker is black. Prove by induction that when you examine the checkers from left to right, you must encounter a red checker followed immediately by a black checker.

  3. Regular Graphs. A k-regular graph is an undirected graph where every vertex has degree k. For example, a graph with 3 vertices connected in a triangle is a 2-regular graph, since each vertex has degree 2.

    Use induction to show that for every k ≥ 1, there exists a k-regular graph.

    Hint: Choose your induction hypothesis carefully.


Last Modified: 27 Aug 2014 21:49:18 EDT by Richard Chang
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