CSE 115ENGR 160 Discrete Mathematics 041912 MingHsuan Yang

CSE 115/ENGR 160 Discrete Mathematics 04/19/12 Ming-Hsuan Yang UC Merced 1

8. 1 Recurrence relations • Many counting problems can be solved with recurrence relations • Example: The number of bacteria doubles every 2 hours. If a colony begins with 5 bacteria, how many will be present in n hours? • Let an=2 an-1 where n is a positive integer with a 0=5 2

Recurrence relations • A recurrence relation for the sequence {an} is an equation that expresses an in terms of 1 or more of the previous terms of the sequence, i. e. , a 0, a 1, …, an-1, for all integers n with n≥n 0 where n 0 is a nonnegative integer • A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation 3

Recursion and recurrence • A recursive algorithm provides the solution of a problem of size n in terms of the solutions of one or more instances of the same problem of smaller size • When we analyze the complexity of a recursive algorithm, we obtain a recurrence relation that expresses the number of operations required to solve a problem of size n in terms of the number of operations required to solve the problem for one or more instance of smaller size 4

Example • Let {an} be a sequence that satisfies the recurrence relation an=an-1 – an-2 for n=2, 3, 4, … and suppose that a 0=3 and a 1=5, what are a 2 and a 3? • Using the recurrence relation, a 2=a 1 -a 0=5 -3=2 and a 3=a 2 -a 1=2 -5=-3 5

Example • Determine whether the sequence {an}, where an=3 n for every nonnegative integer n, is a solution of the recurrence relation an=2 an-1 – an-2 for n=2, 3, 4, … • Suppose an=3 n for every nonnegative integer n. Then for n≥ 2, we have 2 an-1 -an-2=2(3(n 1))-3(n-2)=3 n=an. • Thus, {an} where an=3 n is a solution for the recurrence relation 6

Modeling with recurrence relations • Compound interest: Suppose that a person deposits $10, 000 in a savings account at a bank yielding 11% per year with interest compounded annually. How much will it be in the account after 30 years? • Let Pn denote the amount in the account after n years. The amount after n years equals the amount in the amount after n-1 years plus interest for the n -th year, we see the sequence {Pn} has the recurrence relation Pn=Pn-1+0. 11 Pn-1=(1. 11)Pn-1 7

Modeling with recurrence relations • • The initial condition P 0=10, 000, thus P 1=(1. 11)P 0 P 2=(1. 11)P 1=(1. 11)2 P 0 P 3=(1. 11)P 2=(1. 11)3 P 0 … Pn=(1. 11)Pn-1=(1. 11)n. P 0 We can use mathematical induction to establish its validity 8

Modeling with recurrence relations • We can use mathematical induction to establish its validity • Assume Pn=(1. 11)n 10, 000. Then from the recurrence relation and the induction hypothesis • Pn+1=(1. 11)Pn=(1. 11)n 10, 000=(1. 11)n+110, 000 • n=30, P 30=(1. 11)3010, 000=228, 922. 97 9

8. 2 Solving linear recurrence relations • 10

From mathematical induction • 11

Linear homogenous recurrence relations with constant coefficients • characteristic equation 12

Theorem 1 • 13

Example • 14

Fibonacci numbers • 15

Recurrence relations • Play an important role in many aspects of algorithms and complexity • Can be used to – analyze the complexity of divide-and-conquer algorithms (e. g. , merge sort) – Solve dynamic programming problems (e. g. , scheduling tasks, shortest-path, hidden Markov model) – Fractal 18

8. 5 Inclusion-exclusion • The principle of inclusion-exclusion: For two sets A and B, the number of elements in the union is defined by |A⋃B|=|A|+|B|-|A⋂B| • Example: How many positive integers not exceeding 1000 are divisible by 7 or 11? 19

Principle of inclusion-exclusion • Consider union of n sets, where n is a positive integer • Let n=3 20

Principle of inclusion-exclusion • Let A 1, A 2, …, An be finite sets. Then • Proof: Prove it by showing that an element in the union is counted exactly once by the right-hand side of the equation • Suppose that a is a member of exactly r of the sets A 1, A 2, …, An where 1≤r≤n • This element is counted C(r, 1) times by ∑|Ai| 21

Principle of inclusion-exclusion • It is counted C(r, 2) times by ∑|Ai⋂ Aj | • In general, it is counted C(r, m) times by the summation involving m of the sets Ai. Thus, this element is counted exactly C(r, 1)-C(r, 2)+C(r, 3)-…+(-1)r+1 C(r, r) • Recall , C(r, 0)-C(r, 1)+C(r, 2)-C(r, 3)-…+(-1)r. C(r, r)=0 • Thus, C(r, 1)-C(r, 2)+C(r, 3)-…+(-1)r+1 C(r, r)=C(r, 0)=1 • Thus, this element a is counted exactly once by the right hand side 22

Principle of inclusion-exclusion • Gives a formula for the number of elements in the union of n sets for every positive integer n • There are terms in this formula for the number of elements in the intersection of every nonempty subset of the collection of the n sets. Hence there are 2 n-1 terms in the formula • Example: 15 terms 23

Example • For the union of 4 sets, there are 15 different terms, one for each nonempty subset of {A 1, A 2, A 3, A 4} 24