CSE115/ENGR160 Discrete Mathematics
04/19/12
Ming-Hsuan Yang
UC Merced
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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(n1))-3(n-2)=3n=an.
• Thus, {an} where an=3n is a solution for the
recurrence relation
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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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8.2 Solving linear recurrence
relations
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From mathematical induction
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Linear homogenous recurrence
relations with constant coefficients
characteristic equation
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Theorem 1
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Example
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Fibonacci numbers
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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?
| A  B || A |  | B |  | A  B |
1000  1000  1000 






 7   11   7 11 
 142  90  12  220
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Principle of inclusion-exclusion
• Consider union of n sets, where n is a positive
integer
• Let n=3
| A  B  C || A |  | B |  | C |  | A  B |  | B  C |  | C  A |  | A  B  C |
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Principle of inclusion-exclusion
• Let A1, A2, …, An be finite sets. Then
| A1  A2    An |
| A |  | A  A
1i  n
i
1i , j  n
i
j
|
n 1
|
A

A

A
|



(

1
)
| A1  A2    An |
 i j k
1i , j , k  n
• 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|
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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)
n
k n
• Recall  (1)  k   0 , C(r,0)-C(r,1)+C(r,2)-C(r,3)-…+(-1)rC(r,r)=0
k 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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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
| A1  A2  A3  A4 || A1 |  | A2 |  | A3 |  | A4 |
 | A1  A2 |  | A1  A3 |  | A1  A4 |  | A2  A3 |  | A2  A4 |  | A3  A4 |
 | A1  A2  A3 |  | A1  A2  A4 |  | A1  A3  A4 |  | A2  A3  A4 |
 | A1  A2  A3  A4 |
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Example
• For the union of 4 sets, there are 15 different
terms, one for each nonempty subset of {A1,
A2, A3, A4}
| A1  A2  A3  A4 || A1 |  | A2 |  | A3 |  | A4 |
 | A1  A2 |  | A1  A3 |  | A1  A4 |  | A2  A3 |  | A2  A4 |  | A3  A4 |
 | A1  A2  A3 |  | A1  A2  A4 |  | A1  A3  A4 |  | A2  A3  A4 |
 | A1  A2  A3  A4 |
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