CSE115/ENGR160 Discrete Mathematics 04/19/12

1. CSE115/ENGR160 Discrete Mathematics04/19/12 Ming-Hsuan Yang UC Merced

2. 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=2an-1 where n is a positive integer with a0=5

3. 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., a0, a1, …, an-1, for all integers n with n≥n0 where n0 is a nonnegative integer • A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation

4. 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

5. Example • Let {an} be a sequence that satisfies the recurrence relation an=an-1 – an-2 for n=2, 3, 4, … and suppose that a0=3 and a1=5, what are a2 and a3? • Using the recurrence relation, a2=a1-a0=5-3=2 and a3=a2-a1=2-5=-3

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

7. 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+0.11Pn-1=(1.11)Pn-1

8. Modeling with recurrence relations • The initial condition P0=10,000, thus • P1=(1.11)P0 • P2=(1.11)P1=(1.11)2P0 • P3=(1.11)P2=(1.11)3P0 • … • Pn=(1.11)Pn-1=(1.11)nP0 • We can use mathematical induction to establish its validity

9. Modeling with recurrence relations • We can use mathematical induction to establish its validity • Assume Pn=(1.11)n10,000. Then from the recurrence relation and the induction hypothesis • Pn+1=(1.11)Pn=(1.11)(1.11)n10,000=(1.11)n+110,000 • n=30, P30=(1.11)3010,000=228,922.97

10. 8.2 Solving linear recurrence relations

11. From mathematical induction

12. Linear homogenous recurrence relations with constant coefficients characteristic equation

13. Theorem 1

14. Example

15. Fibonacci numbers

18. 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

19. 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?

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

21. Principle of inclusion-exclusion • Let A1, A2, …, 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 A1, A2, …, An where 1≤r≤n • This element is counted C(r,1) times by ∑|Ai|

22. 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+1C(r,r) • Recall , C(r,0)-C(r,1)+C(r,2)-C(r,3)-…+(-1)rC(r,r)=0 • Thus, C(r,1)-C(r,2)+C(r,3)-…+(-1)r+1C(r,r)=C(r,0)=1 • Thus, this element a is counted exactly once by the right hand side

23. 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 2n-1 terms in the formula • Example: 15 terms

24. Example • For the union of 4 sets, there are 15 different terms, one for each nonempty subset of {A1, A2, A3, A4}