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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
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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
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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
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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
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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 relation6
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
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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
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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
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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
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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?
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2201290142
117
1000
11
1000
7
1000
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BABABA
Principle of inclusion-exclusion
• Consider union of n sets, where n is a positive integer
• Let n=3
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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|21
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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
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n
k
k
k
n
0
0)1(
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
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4321
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434232413121
43214321
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