1. Discrete Math For Computing II

2. Contact Information B. Prabhakaran Department of Computer Science University of Texas at Dallas Mail Station EC 31, PO Box 830688 Richardson, TX 75083 Email: praba@utdallas.edu Phone: 972 883 4680; Fax: 972 883 2349 URL: http://www.utdallas.edu/~praba/cs3305.html Office: ES 3.706 Office Hours: 1-2pm Tuesdays, Thursdays Other times by appointments through email Announcements: Made in class and on course web page. TA: TBA.

3. Example Applications Growth of bank account Initial Amount P 0 =$10,000 After n Years= P n Compound Interest I = 11% Soln: P n =P n-1 +(I/100)P n-1 =(1.11)P n-1 Using Iteration we get, P n =(1.11) n P 0 i.e P 30 =(1.11) 30 10,000=$228,922.97

4. Example Application Rabbits and Fibonacci Numbers Growth of rabbit population in which each rabbit yields 1 new one every period starting 2 periods after its birth. P n = P n−1 + P n−2 (Fibonacci relation)

5. Classic Tower of Hanoi Example Problem: Get all disks from peg 1 to peg 2. – Only move 1 disk at a time. – Never set a larger disk on a smaller one.

6. Hanoi Recurrence Relation Let H n = # moves for a stack of n disks. Optimal strategy: – Move top n−1 disks to spare peg. (H n−1 moves) – Move bottom disk. (1 move) – Move top n−1 to bottom disk. (H n−1 moves) Note: H n = 2H n−1 + 1

7. Why is H n = 2H n-1 + 1 – Only move 1 disk at a time. – Never set a larger disk on a smaller one.

8. Solving Tower of Hanoi RR H n = 2 H n−1 + 1 = 2 (2 H n−2 + 1) + 1 = 2 2 H n−2 + 2 + 1 = 2 2 (2 H n−3 + 1) + 2 + 1= 2 3 H n−3 + 2 2 + 2 + 1 … = 2 n−1 H 1 + 2 n−2 + … + 2 + 1 = 2 n−1 + 2 n−2 + … + 2 + 1(since H 1 =1) = 2 n − 1

9. §6.2: Solving Recurrences Definition: A linear homogeneous recurrence relation of degree k with constant coefficient is a recurrence relation of the form a n = c 1 a n−1 + … + c k a n−k, where the c i are all real, and c k ≠ 0. The solution is uniquely determined if k initial conditions a 0 …a k−1 are provided

10. §6.2: Solving Recurrences.. Linear?: Right hand side is sum of multiples of previous terms. Homogenous?: No terms that are NOT multiples of the a j s. Degree?: k-degree since previous k terms are used.

11. Solving with const. Coefficients Basic idea: Look for solutions of the form a n = r n, where r is a constant. This requires the characteristic equation: r n = c 1 r n−1 + … + c k r n−k, i.e., r k − c 1 r k−1 − … − c k = 0 (Dividing both sides by r n-k and subtracting right hand side from left). The solutions (characteristic roots) can yield an explicit formula for the sequence.

12. Solving …… Theorem1: Let c 1 and c 2 be real numbers. Suppose that r 2 − c 1 r − c 2 = 0 has two distinct roots r 1 and r 2. Then the sequence {a n } is a solution of the recurrence relation a n = c 1 a n−1 + c 2 a n−2 if and only if a n = α 1 r 1 n + α 2 r 2 n for n≥0, where α 1, α 2 are constants.

13. Example Solve the recurrence a n = a n−1 + 2a n−2 given the initial conditions a 0 = 2, a 1 = 7. An = rn  rn = r n-1 + 2rn-2  r2 = r +2 Solution: Use theorem 1 – c 1 = 1, c 2 = 2 – Characteristic equation: r 2 − r − 2 = 0 – Solutions: r = [−(−1)±((−1) 2 − 4·1·(−2)) 1/2 ] / 2·1 = (1±9 1/2 )/2 = (1±3)/2, so r = 2 or r = −1. – So a n = α 1 2 n + α 2 (−1) n.

14. Example Continued… To find α 1 and α 2, solve the equations for the initial conditions a 0 and a 1 : a 0 = 2 = α 1 2 0 + α 2 (−1) 0 a 1 = 7 = α 1 2 1 + α 2 (−1) 1 Simplifying, we have the pair of equations: 2 = α 1 + α 2 7 = 2α 1 − α 2 which we can solve easily by substitution: α 2 = 2−α 1 ; 7 = 2α 1 − (2−α 1 ) = 3α 1 − 2; 9 = 3α 1 ; α 1 = 3; α 2 = -1. Final answer:a n = 3·2 n − (−1) n