CSCI 125 & 161 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; }