R. Johnsonbaugh Discrete Mathematics 5 th edition, 2001 Chapter 5 Recurrence Relations

1. R. Johnsonbaugh Discrete Mathematics 5th edition, 2001 Chapter 5 Recurrence Relations

2. 5.1 Introduction A recurrence relation • is an infinite sequence a1, a2, a3,…, an,… • in which the formula for the nth term an depends on one or more preceding terms, • with a finite set of start-up values or initial conditions

3. Examples of recurrence relations • Example 1: • Initial condition a0 = 1 • Recursive formula: a n = 1 + 2a n-1 for n > 2 • First few terms are: 1, 3, 7, 15, 31, 63, … • Example 2: • Initial conditions a0 = 1, a1 = 2 • Recursive formula: a n = 3(a n-1 + a n-2) for n > 2 • First few terms are: 1, 2, 9, 33, 126, 477, 1809, 6858, 26001,…

4. Fibonacci sequence • Initial conditions: • f1 = 1, f2 = 2 • Recursive formula: • f n+1 = f n-1 + f n for n > 3 • First few terms:

5. Compound interest • Given • P = initial amount (principal) • n = number of years • r = annual interest rate • A = amount of money at the end of n years At the end of: • 1 year: A = P + rP = P(1+r) • 2 years: A = P + rP(1+r) = P(1+r)2 • 3 years: A = P + rP(1+r)2 = P(1+r)3 … • Obtain the formula A = P (1 + r) n

6. Eugene Catalan • Belgian mathematician, 1814-1894 • Catalan numbers are generated by the formula: Cn = C(2n,n) / (n+1) for n > 0 • The first few Catalan numbers are:       

7. Catalan Numbers: applications • The number of ways in which a polygon with n+2 sides can be cut into n triangles • The number of ways in which parentheses can be placed in a sequence of numbers, to be multiplied two at a time • The number of rooted trivalent trees with n+1 nodes • The number of paths of length 2n through an n by n grid that do not rise above the main diagonal • The number of nonisomorphic binary trees with n vertices

8. Towers of Hanoi Start with three pegs numbered 1, 2 and 3 mounted on a board, n disks of different sizes with holes in their centers, placed in order of increasing size from top to bottom. • Object of the game: find the minimum number of moves needed to have all n disks stacked in the same order in peg number 3.

9. Rules of the game: Hanoi towers Start with all disks stacked in peg 1 with the smallest at the top and the largest at the bottom • Use peg number 2 for intermediate steps • Only a disk of smaller diameter can be placed on top of another disk

10. End of game: Hanoi towers • Game ends when all disks are stacked in peg number 3 in the same order they were stored at the start in peg number 1. • Verify that the minimum number of moves needed is the Catalan number C3 = 5. Start End

11. A problem in Economics • Demand equation: p = a - bq • Supply equation: p = kq • There is a time lag as supply reacts to changes in demand • Use discrete time intervals as n = 0, 1, 2, 3,… • Given the time delayed equations pn = a – bqn (demand) pn+1 = kqn+1 (supply) • The recurrence relation obtained is pn+1 = a – bpn /k

12. Economic cobweb with a stabilizing price

13. Ackermann’s function • Initial conditions: A(0,n) = n + 1, for n = 0, 1, 2, 3,… • Recurrence relations: A(m,0) = A(m – 1, 1), for m = 1, 2, 3,… A(m,n) = A(m -1, A(m, n -1)) for m = 1, 2, 3,… and n = 1, 2, 3,…

14. 5.2 Solving recurrence relations Two main methods: • Iteration • Method for linear homogeneous recurrence relations with constant coefficients

15. Method 1: Iteration • Problem: Given a recursive expression with initial conditions a0, a1 • try to express an without dependence on previous terms. Example: an = 2an-1 for n > 1, with initial conditiona0 = 1 • Solution: an = 2n

16. More on the iteration method Example: DeerPopulation growth • Deer population dn at time n • Initial condition: d0 = 1000 • Increase from time n-1 to time n is 10%. Therefore the recursive function is dn – dn-1 = 0.1dn-1  dn = 1.1dn-1 • Solution: dn = 1000(1.1)n

17. Method 2: Linear homogeneous recurrence relations Theorem 5.2.11: Given the second order linear homogeneous recurrence relation with constant coefficients an = c1an-1 + c2an-2 and initial conditions a0 = C0, a1 = C1 1. If S and T are solutions then U = bS + dT is also a solution for any real numbers b, d 2. If r is a root of t2 – c1t – c2 = 0, then the sequence {rn}, n = 0, 1, 2,… is also a solution

18. Case 1: Two different roots 3. If r1 and r2 (r1 r2) are solutions of the quadratic equation t2 – c1t – c2 = 0, then there exist constants b and d such that an = br1n + dr2n forn = 0, 1, 2, 3,…

19. More on linear homogeneous recurrence relations Theorem 5.2.14: Let an = c1an-1 + c2an-2 be a second order linear homogeneous recurrence relation with constant coefficients. • Let a0 = C0, a1 = C1 be the first two terms of the sequence satisfying the recurrence relation.

20. Case 2: One root of multiplicity 2 If r is a root of multiplicity 2 satisfying the equation t2 – c1t – c2 = 0, then: there exist constants b and d such that an = brn + dnrn for n = 0, 1, 2, 3,…

21. 5.3 Applications to the analysis of algorithms 1. Selection sorting a) Given a sequence of n terms ak, k = 1, 2,…, n to be arranged in increasing order b) Count the number of comparisons bn with initial condition b1 = 0 c) Obtain recursion relation bn = n – 1 + bn-1 for n = 1, 2, 3,… d) bn = n(n-1)/2 = (n2)

22. Binary search 2. Problem: Search for a value in an increasing sequence. Return the index of the value, or 0 if not found. • Initial condition a1 = 2 • Recurrence relation an = 1 + an/2 • Result: an = (lg n)

23. Merging two sequences 3. Problem: Combine two increasing sequences into a single increasing sequence (merge two sequences). • Theorem 5.3.7: To merge two sequences the sum of whose lengths is n, the number of comparisons required is n-1.

24. Merge sort 4. A recursive algorithm is used to sort a sequence into increasing order using the algorithm for merging two increasing sequences into one increasing sequence (merge sort). • Theorem 5.3.10: The merge sort algorithm is (n lg n) in the worst case.