Lec2b [Compatibility Mode]

8/2/2019 Lec2b [Compatibility Mode]

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SET THEORY Set: It is an unordered collection of distinct

.

Finite Set: A set is finite if it contains a finite

A = {1, 3, 5, 7, 9, 11, 15, 17, 19}

infinite number of elements

B = 0 1 2 3 4 .

Empty Set: A set is empty if it contain noelements

= { }


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If X is a set:

x X x belon s to Xx X x does not belong to X{ x x N and x is odd } The set of all x such that x belongs to N

Subset: If every element of A is also an element of BA B

Proper Subset: A is proper subset of B if:

(i) A is subset of B.

A B

Improper Subset: A is an improper subset of B if:-

(i) A is a subset of B

(ii) There does not exist any element in B which does not belongto A.

Every set is an improper subset of itself


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Equal Sets: A and B are equal sets if:-

or

Union of X & Y:=

Intersection of X & Y:

Difference of sets:

=

Cartesian Product:

= ,


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INTEGERS, REALS AND INTERVALS Integers:

Set of integers : Z = { 0, 1, 2, 3, 4, .}Set of natural integers : N = { 0,1, 2, 3, 4, }

Set of positive integers : N+ = { 1, 2, 3, 4, }

Set of ne ative inte ers : N- = { -1, -2, -3, -4, }

Real Numbers:

Set of positive real numbers: R+ = { x R x > 0 }-v u : = x x n ]

For every natural number, there exists anothernatural number (m) greater than (n).

( m N) ( n N ) [ m > n ]

There is an integer m that is larger that everynatura num er nc u ng m tse .

It is obviously false

the order in which the quantifiers are presented isimportant.


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Sums and ProductsSums )2()1()( +++= f(n).....ffif

n

Sum of the values taken b on the first n ositive inte ers

1=i

i = 1

)(ifn

P(i)

Sum of the values taken by f on those integers between 1 and nfor which ro ert P holds.

If n = 10 and P(i) is odd

10

i = 1 P(i) is odd

2597531 =++++= i


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Productsn

(i) = (1) x (2) x (3) x .. x (n)

Product of (i) as i goes from 1 to n.If n = 0, the product is defined to be one.

Miscellaneous Notations

Logbx=y Where b1 and x are +ve realy = x

log10 1000 = 3 or 103 = 1000

numbers (b&x must be +ve)

natural logarithm (ln)

Loga(xy) = logax + logay

oga x y = ogax ogay

Logaxy = ylog

ax

a

xx

b

ba

log

loglog =


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729log729log)( 39 =i

4log

16log16log)(

2

24 =ii

xy bb

yx

loglog

=

Floor of x: If x is a real number, [x] denotes the

largest integer that is not larger than x.

3

2

13

=

=

x

x

2

13=x

77,95.8,25,3]14.3[

4

====

=

Floor

x


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Ceiling of x: If x is a real number, [x] denotesthe smallest integer that is not smaller than x.

213=x

13

4

=

=

x

x

77,85.8,35,4]14.3[

3

====

=

Ceilin

x


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Algorithm to Find a New Prime Number

uc e o

Function NewPrime (P : set of integers)-

primes}

x Product of the elements in Py x + 1d 1Repeat d d+1 until d divides yreturn d


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Another Algorithm to Find a New

r me um er

Function DumpEuclid (P : set of integers)-

primes}

x The lar est element in Prepeat x x + 1 until x is primereturn x


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PROOF

(1) Proof by contradiction

y

It is also known as indirect proof.eorem : There exist two irrational numbers x

and y such that xy is rational.

, =we know 2 is a irrational number

2

2=zLet


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By our assumption z is irrational

2zwLet =

s aga n rrat ona y our assumpt on

but

( )22 2222 =

==

zw

( ) 22 2 ==Here conclusion says that 2 is irrational which is false.

Thus, our assumption was false. So it is possible to

an irrational power. It is a proof by contradiction and isavoided in the context of algorithmics.


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Proof By Mathematical Induction(Very useful tool in algorithmics)

Induction Approach: inferring of general law from.

Deduction Approach: an inference from general toparticular (always valid if it is applied properly)

Induction Approach

(i) Eulers Conjecture (1769)

=

Enuler conjectured that this equation can never besatisfied.

Frye (after two centuries) using computing machines forhundreds of hours found that:

4+ 4+ 4= 4 ( -


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Proof By Mathematical Induction

(ii) Polynomial p(n) = n2 + n +41

+ 1 + 2 + + 4 + +

+p(7) + p(8) + p(9) + p(10) = 41, 43, 47, 53, 61,71, 83, 97, 113, 131 and 151.

It is natural to infer by induction that p(n) is

prime for all integer values of n. But p(40) = 1681 = 412 so induction has gone

wrong.


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(i) Sum of the cubes of the first n positive

integers is always a perfect square.13 = 1 = 12

13+23 = 9 = 32

13+23+33+43 = 100 = 102

13+23+33+43+53 = 225 = 152


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(ii) Algorithm for Mathematical Induction

function sq (n)

if n=0 then return 0

else return 2n + sq(n-1)-1

Check:

sq = , sq = + - =

sq(2)=4+1-1=4, sq(3)=6+4-1=9

= + - =

Proof:

= - -sq(n)=2n+n2-2n+1-12-

sq(n)=n2


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Tilin Pr l m m squares in each row and column where m

1 square is distinguished as special square.t s not covere y any t e.

1 tile takes 3 squares 2 x 2 board is formed


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The tiling problem

(a) Board with special square (c) One tile

(c) Placing the first tile (d) solution


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Proof by Mathematical Indirection

m = 2n where n is an integer

if n=0m=20=11x1 board1 square is special

Board is tiled by doing nothingif n=1

m=21=22x2 board

square s spec aBoard is tiled by putting 1 tile


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(ii) Induction Step

ons er nm = 2n

According to induction hypothesis theorem is true for2n-1 x 2n-1 boards

Suppose a m x m board containing one special square

Place the tile in the middle of the original board so to cover 1

square of three sub-boards. These three squares are also.

By induction hypothesis, each of these sub-boards can betiled

Thus theorem is for m=2 , and since its truth for m=2n

followsfrom its assumed truth for m=2n-1 for all n 1, it follows fromthe principle of mathematical induction that the theorem istrue or a m prov e m s a power o .

A suitable algorithm can be obtained to solve tiling problem.


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Construction Induction It can also be used to prove the truth of a partially

specified assertion and discover the missing

This technique can be illustrated featuring the Fibonaccise uence 12th centur , Italian mathematician .

Every month a breeding pair of rabbits produce a pair of

offspring. The offspring will in their turn start breeding after two

months and so on.Month-4

1

1

Month-2Month-1

Month-31

1

1

1

11 1 1

(2 pairs)(1 pair) (3 pairs)(5 pairs) and so on


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Format definition

=

2;0 2110 >=+=== noran nnn

, , , , , , , , , , , , ..

Constructive induction can be useful in the analysis of algorithms.

gor m or compu ng e onaccfunction Fibonacci (n)

if n


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Elementary Probability It is concerned with the study of random phenomena

whose future is not predictable with certainty.

xamp e

(i) Throwing of a dice

point in a given period of time

(iii) Measuring the response time of a computersystem

The set of all possible outcomes of a random experiment is

The individual outcomes are called sample points orelementary events.


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Possible outcomes of throwing an ordinary dice=6, namely1 to 6.

Sample space = S = {1, 2, 3, 4, 5, 6,} for throwing a dice

S = {0, 1, 2, 3, } for counting the carspassing a given point

(S is infinite but discrete)

for measuring the= >

(S is continuous)

An event is subset of S as it is a collection of sam le oints

response me o a

computer system

Universal Event: The entire sample space (S) is anevent called the universal event.

Impossible Event: The empty set () is an event called

the impossible event.


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Outline of the Basic Procedure

for Solving Problems

1. Identify the sample space, S..

elements in S.

. .

4. Compute the desired probabilities


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Example: To know probability that a random

num er genera or w pro uce a va ue a sprime (range 09999)

. = , , , , ,

2. Pr [1] = Pr [2] = Pr [3] = .. = Pr [9999]=1/10000. , , , .., ,

(Prime numbers,Total = 1229)

4. Probabilit of elementar events in A

1229

.10000 =


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Horse-Race with Five runners

Outcome : name of the winner

.

S : { A, B, C, D, E }-

Outcome Assigned Probability Winnings(amount)

(W)A 0.10 50

B 0.05 100

C 0.25 -30

D 0.50 -30

E 0.10 15

W is a random variable, amount to win or lose is shown


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3. Assigning Probabilities

W(A) = 50, W(B) =100, W(C) = -30 etc.

P(-30) = Pr(C)+Pr(D)= 0.25+0.50 = 0.75

P(15) = Pr(E) = 0.10

P(50) = Pr(A) = 0.10

P(100) = Pr(B) = 0.054. Expected winnings E(W)

E(W) = -30p(-30)+15p(15)+50p(50)+100p(100)

= -30(0.75)+15(0.10)+50(0.10)+100(0.05)

= -22.5+1.50+5.0+5.0= -22.50+11.50

= -11

Variance of X = Var[X] = E[(x-E[X])2] = p(x)(x-E[X])2

Var[W] = p(-30)x192 +p(15)x262+p(50)x612+p(100)x1112= 1326.5

Standard deviation of W = s rt 1326.5 =36.42

E[W] allows to predict the average observed value of W and the

variance serves to quantify how good this prediction is likely to be.

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