am_lect_10 - x:= a y:= b while y ≠ r:= x mod y x:= y y:=...

This preview shows pages 1–4. Sign up to view the full content.

Quiz 1 Start time: 9:00am End time: 9:20am

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
GCD Recall that gcd ( a , b ) is the greatest common divisor of a and b For any non-zero integer n , gcd ( n , 0) = n gcd (0, 0) is undefined If a Z , b Z + and a = bq + r then gcd ( a , b ) = gcd ( r , b ) Proof : Suppose gcd ( a , b ) = k Then n Z , a = kn and m Z , b = km So kn = ( km ) q + r and hence r = k ( n mq ) Thus k is a common divisor of b and r Suppose gcd ( b , r ) = t > k Since t b and t r then t ( bq + r ) So t is a common divisor of a and b , a contradiction (since t > k ) See 1pm lecture notes for a different proof
Euclidean algorithm for GCD Corollary : For positive integers a and b , gcd ( a , b ) = gcd ( b , a mod b ) A 2,300 year old algorithm (pseudocode): procedure gcd(a, b: positive integers)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x := a y := b while y ≠ r := x mod y x := y y := r return x Example : gcd (105, 252) x y r = x mod y 105 252 105 252 105 42 105 42 21 42 21 21 = 21 Recursive algorithm for GCD Recursion is a technique for reducing a big problem into one or more similar, smaller , problems — with a base case to handle the simplest problems procedure gcd(a, b: positive integers) r := a mod b if (r = 0) return b else return gcd(b, r) Example : gcd (105, 252) a b r = a mod b 105 252 105 252 105 42 105 42 21 42 21 21 = 21...
View Full Document

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern