CS 223 Purdue

Assembler 80186 and higher TRANSFER Name Comment MOV Move (copy) XCHG Exchange STC CLC CMC STD CLD STI CLI PUSH PUSHF PUSHA POP POPF POPA CBW CWD CWDE Set Carry Clear Carry Complement Carry Set Direction Clear Direction Set Interrupt Clear Interrupt

Solutions of Homework 5: Big O, Important Note: log n Q1. 1. 2. 3. 4. Q.2 i. 6n2 - 2n 6n2 = O(n2) . . . . . . . . . . . . . . . (1) 6n2 - 2n 6n2 - 2n2 = 4n2 = (n2 ) . . . . (2) Form (1) and (2): 6n2 - 2n log n = (n2 ). 2 ii. log6nn+1 6n2 n3 = O

Solutions of Homework 6: Programming Project The following is the complete C program that implements the given function with the required modifications. #include <math.h> #include <stdio.h> #include <stdlib.h> int count; double fun(double n){ count+

Solutions of HW#1: Basic Logic Q.1 Make truth tables for the following statement: (p q) (q p) Answer p T T F F q T F T F pq T F T T qp T T F T (p q) (q p) T T F T (p q) (p q) Answer p T T F F q T F T F pq T F T T p q T T T F (p q) (p

[50] Homework 7: Basic Number Theory [10] Use the Euclidean algorithm to find (a) gcd(1529, 14039), (b) gcd(1111, 11111). [10] Compute 61531 mod 713. [10] Prove that 937 is an inverse of 13 modulo 2436. [10] Solve 4x = 5 mod 9. [10] Encrypt the mess

Chapter 18: Introduction to Assurance Overview Why assurance? Trust and assurance Life cycle and assurance June 1, 2004 Computer Security: Art and Science Slide #18-1 Overview Trust Problems from lack of assurance Types of assurance Lif

Solution of Homework 6: Basic Number Theory Q1. Use the Euclidean algorithm to find (a) gcd(1529, 14039), (b) gcd(1111, 11111). Answer Note that in each iteration the larger value is replaced by the smaller value in the previous iteration and the sma

Solutions of Homework #4: Proof Techniques Q1. Show that 5 5 is irrational. Answer Proof by contradiction: Assume that 5 5 = p is in its simplest form, i.e., both p q and q do not have a common divisor and therefore the fraction p cannot be q simpl

Solution of Homework 9: Discrete Probability Q1. Show that if A and B are independent events, then A and B are also independent events. Answer P (A)P (B) = 1 - P (A) 1 - P (B) = 1 - P (A) - P (B) + P (A)P (B) = 1 - P (A) + P (B) - P (A)P (B) =