1. Application: Algorithms Lecture 19 Section 3.8 Tue, Feb 20, 2007

2. Algorithms An algorithm is a step-by-step procedure that is guaranteed to stop after a finite number of steps for all legitimate inputs.

3. Example: Max Algorithm Describe an algorithm for finding the maximum value in a list of numbers. Special cases: What if the list is empty? Others?

4. An Algorithmic Language An algorithmic language is much like a computer language. Its primary purpose is to express unambiguously the steps of an algorithm, including all decisions and special cases that may arise.

5. Assignment Statements An assignment statement is of the form where x is a variable and expr is an expression. Examples found := true count := count + 1 x := expr

6. Conditional Statements One-way conditional statements: Two-way conditional statements: if condition then sequence of statements if condition then sequence of statements else sequence of statements

7. Iterative Statements While loops: For loops: while condition sequence of statements end while for var := init expr to final expr sequence of statements next var

8. For Loops A for loop can always be rewritten as a while loop. var := init expr while var  final expr sequence of statements var := var + 1 end while

9. A Notation for Algorithms An algorithm will be organized into three parts: Input – List all input variables and any special assumptions about them. Algorithm body – Describe the step-by-step procedure. Output – List the output variables.

10. Example: Finding the Max Input: a, n a is a list of numbers. n is the size of the list, n  1. Algorithm body: Output: max max := a[1] for i := 2 to n if a[i] > max then max = a[i]

11. Tracing an Algorithm Write a trace table for the Max algorithm and the input a = {7, 3, 8, 6, 4}, n = 5. Iterationanmaxi 0{7, 3, 8, 6, 4}572 1 2 3 4

12. Tracing an Algorithm Write a trace table for the Max algorithm and the input a = {7, 3, 8, 6, 4}, n = 5. Iterationanmaxi 0{7, 3, 8, 6, 4}572 173 2 3 4

13. Tracing an Algorithm Write a trace table for the Max algorithm and the input a = {7, 3, 8, 6, 4}, n = 5. Iterationanmaxi 0{7, 3, 8, 6, 4}572 173 284 3 4

14. Tracing an Algorithm Write a trace table for the Max algorithm and the input a = {7, 3, 8, 6, 4}, n = 5. Iterationanmaxi 0{7, 3, 8, 6, 4}572 173 284 385 4

15. Tracing an Algorithm Write a trace table for the Max algorithm and the input a = {7, 3, 8, 6, 4}, n = 5. Iterationanmaxi 0{7, 3, 8, 6, 4}572 173 284 385 486

16. Greatest Common Divisors Let a and b be integers that are not both 0. The greatest common divisor of a and b, denoted gcd(a, b), is the unique positive integer d with the following properties: d | a and d | b. For every integer c, if c | a and c | b, then c | d.

17. Least Common Multiples Let a and b be nonzero integers. The least common multiple of a and b, denoted lcm(a, b), is the unique positive integer m with the following properties: a | m and b | m. For every integer c, if a | c and b | c, then m | c.

18. The High-school gcd Algorithm The high-school method is very inefficient. Factor each number into standard form. For each prime that appears in both factorizations, use it as a factor of the gcd along with the smaller of the two exponents.

19. Example Find the gcd of 81900 and 54810. We factor them as 81900 = 2 2  3 2  5 2  7 1  13 1  29 0 54810 = 2 1  3 3  5 1  7 1  13 0  29 1 Therefore, the gcd is 2  3 2  5  7 = 630

20. The Euclidean Algorithm Factoring is inefficient; therefore, this algorithm is inefficient. The run time of this algorithm is O(  10 d ), where d is the number of digits in the number. Euclid had a much better idea. The run time of the Euclidean Algorithm is O(d), where d is the number of digits in the number.

21. The Euclidean Algorithm Input: A, B (positive integers, not both 0)

22. The Euclidean Algorithm Algorithm body: Output: b a := A b := B if b = 0 then swap a and b r := a mod b while r > 0 a := b b := r r := a mod b end while

23. Example Apply the Euclidean Algorithm to A = 81900 and B = 54810. IterationABabr 08190054810819005481027090 1 2

24. Example Apply the Euclidean Algorithm to A = 81900 and B = 54810. IterationABabr 08190054810819005481027090 15481027090630 2

25. Example Apply the Euclidean Algorithm to A = 81900 and B = 54810. IterationABabr 08190054810819005481027090 15481027090630 2270906300

26. Least Common Multiples What is the efficient way to find lcm’s? What is the lcm of 81900 and 54810?

27. Proof of the Euclidean Algorithm Theorem: The Euclidean Algorithm terminates for all legitimate inputs A and B. Proof: We may assume that B > 0. After the first iteration of the while loop, 0  b < B since b is the remainder of A divided by B.

28. Proof of the Euclidean Algorithm Each iteration produces a nonnegative remainder that is smaller than the previous remainder. This cannot happen more than B times before the remainder is 0.

29. Proof of the Euclidean Algorithm Lemma 1: If b > 0, then gcd(b, 0) = b. Proof: b | b and b | 0. For all integers c, if c | 0 and c | b, then c | b. Therefore, b = gcd(b, 0).

30. Proof of the Euclidean Algorithm Lemma 2: If a and b are integers, with b  0, and q and r are integers such that a = qb + r then gcd(a, b) = gcd(b, r). Proof: Let d = gcd(b, r). Then d | b and d | r and any integer that divides b and r must also divide d.

31. Proof of the Euclidean Algorithm We must show that d | a and d | b and any integer that divides a and b must also divide d. We already know that d | b. Since a = qb + r, it follows that d | a. Let c be an integer such that c | a and c | b. Since r = a – qb, it follows that c | r and so c | d. Therefore, d = gcd(a, b).

32. Proof of the Euclidean Algorithm Theorem: The Euclidean Algorithm produces the gcd of A and B. Proof: After the final iteration of the while loop, r = 0. By Lemma 1, the output b is the gcd of b and r. By Lemma 2, that is equal to the gcd of “a” and “b” in the final iteration.

33. Proof of the Euclidean Algorithm But “a” and “b” on the last iteration were “b” and “r” on the previous iteration. Therefore, gcd(a, b) on the last iteration equals gcd(b, r) on the previous iteration, which equals gcd(a, b) on the previous iteration, and so on. Following this argument all the back to the first iteration, we see that the output is gcd(A, B).

34. Proof of the Euclidean Algorithm In the next chapter, we will study mathematical induction. At that point, we will be able to make this argument more rigorous.