1. Discrete Mathematics Recurrence Relations Chapter 5 R. Johnsonbaugh
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
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: First few terms:
f1 = 1, f2 = 2
Recursive formula:
f n+1 = f n-1 + f n for n > 3
First few terms: n 1 2 3 4 5 6 7 8 9 10 11 fn 13 21 34 55 89
144

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 Catalan numbers are generated by the formula:
Belgian mathematician,
Catalan numbers are generated by the formula:
Cn = C(2n,n) / (n+1) for n > 0
The first few Catalan numbers are:             n 1 2 3 4 5 6 7 8 9 10 11 Cn 14 42
132
429
1430
4862
16796
58786

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 condition a0 = 1
Solution: an = 2n

16. More on the iteration method
Example: Deer Population 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 : 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
for n = 0, 1, 2, 3,…

19. More on linear homogeneous recurrence relations
Theorem : 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 : The merge sort algorithm is (n lg n) in the worst case.