CSCI 125 & 161 / ENGR 144 Lecture 12

CSCI 125 & 161 / ENGR 144 Lecture 12 Martin van Bommel

Prime Numbers • Prime number is one whose only divisors are the number 1 and itself • Therefore, number is prime if it has two positive divisors • One way to test for prime is to count its divisors

Prime- First Try bool IsPrime(int n) { int i, divisors = 0; for (i=1; i<=n; i++) { if (n % i == 0) divisors++; } return (divisors == 2); }

Prime - Second Thought • Number is not prime if it has divisor other than 1 and itself • If number not divisible by 2, will not be divisible by any even number • Check for two, then only check odds • Only have to check up to square root of n

Prime - Second Try bool IsPrime(int n) { int i, limit; if (n == 2) return true; if (n % 2 == 0) return false; limit = sqrt(n) + 1; for (i = 3; i <= limit; i += 2) if (n % i == 0) return false; return true; }

Efficiency Trade-off • Recall implementations of IsPrime • Final version more efficient • Original is more readable and easier to prove correct • Principal concern must be correctness • Secondary factors are efficiency, clarity, and maintainability • No “best” algorithm from all perspectives

GCD • Greatest Common Divisor of two numbers • largest number that divides evenly into both • Function to determine GCD of two values int GCD(int x, int y); • e.g. • GCD(49, 35) = 7 • GCD(6, 18) = 6 • GCD(32, 33) = 1

Brute Force GCD int GCD(int x, int y) { int g = x; while (x % g != 0 || y % g != 0) { g--; } return g; }

Improved GCD int GCD(int x, int y) { int g; if (x < y) g = x; else g = y; while (x % g != 0 || y % g != 0) { g--; } return g; }

Problems with Brute Force • Poor choice for efficiency • e.g. GCD(10005, 10000) = 5 • Long running loop to find simple answer • Can’t count up! Why? • Other choices?

Euclid’s Algorithm for GCD 1. Divide x by y; call remainder r 2. If r is zero, answer is y. 3. If r is not zero, set x equal to old value of y, set y equal to r, repeat entire process • Difficult to prove correct

Euclid’s GCD int GCD(int x, int y) { int r = x % y; while (r != 0) { x = y; y = r; r = x % y; } return y; }

CSCI 125 & 161 / ENGR 144 Lecture 12