CpEHW1 - violation of the course rule. Note: Solutions must...

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1 Homework 1 CpE 360 – Computational Algorithms and Data Structures Instructor: Ashish Patel Due on September 26, 2006 All solutions must be (i) either delivered at the beginning of the class on September 19, 2006 (ii) or submitted per email at apatel7@stevens.edu by 12:30 pm September 19, 2006, (iii) or delivered to the instructor’s mailbox in the Burchard 212 by September 18, 2006. No late submission will be taken into consideration! Any homework submitted per email must contain full student’s name and description which homework has been submitted. The work you turn in must be your own personal work, composed and written by you. If you discuss a problem with a fellow student, you must mention this clearly in your homework (name and the fellow student before the solution of the problem). Your work will then be compared to the other student’s work to verify that your solution was written by you and reflect your own personal effort. If you don’t report it, it will be considered a
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Unformatted text preview: violation of the course rule. Note: Solutions must be readable (especially handwriting!!), concise, complete and must include pledge of the Honor system. Reading Assignment: Chapter 2 and Chapter 3 1. Explain the meaning of the following expressions: a. f(n) is ( ) 1 . b. f(n) is ( ) 1 . c. f(n) is ( ) 1 . 2. Prove or disprove each of the following conjectures: a. 2n + 7 = (1). b. 2 n+5 = (2 n ). c. 2 2n = (2 n ). d. f(n) = (g(n)) implies 2 f(n) = (2 g(n) ). e. f(n) = (f(n/2)). f. 2n + 7 = (n 2 ). 3. Indicate, for each pair of expressions (A, B) in the table below, whether A is , o, , , or of B. Assume that k 1, > 0, > 0, and c >1 are constant. Your answer should be in the form of the table with yes or no written in each box. 2 4. Rank the following functions by order of growth, that is, find an arrangement g 1 , g 2 , .g 15 of the function satisfying g 1 = g 2 , g 2 = g 3 , ., g 14 = g 15 ....
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CpEHW1 - violation of the course rule. Note: Solutions must...

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