ch5Discrete Maths

7/28/2019 ch5Discrete Maths

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Recurrence Relations


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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 orinitial

conditions


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Examples of recurrence relations

Example 1: 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,


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Fibonacci sequence

Initial conditions:

f1

= 1, f2

= 2

Recursive formula:

fn+1 = fn-1 + fn for n > 3

First few terms:

n 1 2 3 4 5 6 7 8 9 10 11

fn 1 2 3 5 8 13 21 34 55 89 144


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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


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Eugene Catalan

Belgian mathematician, 1814-1894

Catalan numbers are generated by the

formula:

Cn = C(2n,n) / (n+1) for n > 0

The first few Catalan numbers are:

n 0 1 2 3 4 5 6 7 8 9 10 11

Cn 1 1 2 5 14 42 132 429 1430 4862 16796 58786


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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 twoat 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


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Towers of Hanoi

Start with three pegs numbered 1, 2 and 3mounted on a board, n disks of different sizeswith holes in their centers, placed in order ofincreasing size from top to bottom.

Object of the game: find the minimum numberof moves needed to have all n disks stackedin the same order in peg number 3.


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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


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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 movesneeded is the Catalan number C3 = 5.

Start End


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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 equationspn = a bqn (demand)

pn+1 = kqn+1 (supply)

The recurrence relation obtained ispn+1 = a bpn /k


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Economic cobweb

with a stabilizing price


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Ackermanns function

Initial conditions:

A(0,n) = n + 1, for n = 0, 1, 2, 3,

Recurrence relations:

A(m,0) = A(m1, 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,


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5.2 Solving recurrence relations

Two main methods:

Iteration Method for linear homogeneous recurrence

relations with constant coefficients


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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 conditiona0 = 1

Solution: an = 2n


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More on the iteration method

Example: DeerPopulation 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


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Method 2: Linear homogeneous

recurrence relations

Theorem 5.2.11: 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, d2. If r is a root of t2 c1t c2 = 0, then the

sequence {rn}, n = 0, 1, 2, is also a solution


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Case 1: Two different roots

3. If r1 and r2 (r1 r2) are solutions of the quadraticequation t2 c1t c2 = 0, then there exist

constants b and d such that

an = br1n + dr2

n

forn = 0, 1, 2, 3,


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More on linear

homogeneous recurrence relations

Theorem 5.2.14: Let an = c1an-1 + c2an-2 be a

second order linear homogeneous recurrencerelation with constant coefficients.

Let a0 = C0, a1 = C1 be the first two terms of

the sequence satisfying the recurrence

relation.


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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,


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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 = (n

2)


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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 + an/2

Result: an = (lg n)


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Merging two sequences

3. Problem: Combine two increasing

sequences into a single increasingsequence (merge two sequences).

Theorem 5.3.7: To merge two sequences

the sum of whose lengths is n, the numberof comparisons required is n-1.


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Merge sort

4. A recursive algorithm is used to sorta sequence into increasing orderusing the algorithm for merging two

increasing sequences into oneincreasing sequence (merge sort).

Theorem 5.3.10: The merge sortalgorithm is (n lg n) in the worstcase.

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