lec2byuuryu compatibility mode

8/18/2019 Lec2byuuryu 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

Ø = { }


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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 B A ⊆ 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 belong

to 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 negative integers : N- = { -1, -2, -3, -4, …}

• Real Numbers:

Set of positive real numbers: R+ = { x ∈ R⎥ x > 0 }- = ∈


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• n n erva s a se o num ers y ngbetween two bounds.

– Open interval = { x ∈ R ⎥ a < x < b} – Semi-o en interval = x ∈ R ⎥ a < x ≤ b

or { x ∈ R ⎥ a ≤ x < b }

– Empty interval = { x ∈ R ⎥ a < x < b } where a=b


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FUNCTIONS AND RELATIONS• X & Y be two sets

• e re a on e ween an s ca e unc on :

for each x ∈ X, there exists one and only oney ∈ suc a x, y ∈ .

ƒ : X→Y⇒ ƒ is a function from X to Y.• n re at on x, y ∈ ƒ ⇒ xThe set X is called the domain of the function

The set Y is its imageThe set ƒ[X] = { ƒ(x)⎟ x ∈ X} is its range


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• Example(i) C={(x,1), (y,1)}Domain C={x, y}

(ii) D={(1, 1), (1, 2), (2, 1)}

Domain D= 1 2

Range D={1, 2}

• Injective Function (one-to-one): A function is injectiveere o no ex s wo s nc x1, x2 ∈

such that ƒ(x1) = ƒ(x2).

A={2, 3, 5} B={a, b, x, y}

ƒ={(2, a), (3, x), (5,y)}Domain ƒ={2, 3, 5}Range ƒ={a, x, y}


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A B

“ ”23

5

b

x

Domain ƒ=ARange ƒ≠B

(i)

ange ≠ mage

2 1

A B “A into B”Domain ƒ={2, 4, 6}

=

6 5

,

(Range ≠ image)

ƒ={(2,1), (4,3), {6, 3)}Domain ƒ = A


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Surjective function (onto)

2 1 A B “A onto B”

4

6

3

5

, ,

Range ƒ={1, 3, 5}

ƒ={(2,1), (4,3), (6, 5)}Domain ƒ = A ( ∴Range = image)


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(ii)23

23

“A onto A”Domain ƒ=A=

= 2 2 3 5 5 3

5 5

Domain ƒ = {2, 3, 5}Range ƒ= {2, 3, 5}

( ∴Range = image)


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•both injective and surjective

-1⇒ inverse

ƒ(ƒ-1 (y)) = y for all y ∈ Y

2

3

2

3

“A onto A”

and“ ”

= 2 2 3 5 5 3

5 5

Domain ƒ=ARange ƒ=A

Domain ƒ = {2, 3, 5}Range ƒ= {2, 3, 5}


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• Predicate: A function P: X→ {true, false} is called a predicate on X.

so

P(true, false, true) = true

,, ¬

when P is a predicate on X, we sometimes say P is a property of X.

• Quantifiers“ ”⇒

∃ ⇒ “there exists”

(∀ x ∈ X) [P(x)] ⇒ every x in X has property P.(∃ x ∈ X) [P(x)] ⇒ there exists at least one x in X that has property P.

(∃ ! x ∈ X) [P(x)] ⇒ there exists exactly one x in X that has property P.∀ x ∈ X P x ⇒ is alwa s true if X is an em t set.

(∃ x ∈ X) [P(x)] ⇒ is always false if X is an empty set.


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Alternation of Quantifiers(∀ n ∈ N) (∃ m ∈ N ) [ m > n ]

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

(∃ m ∈ N) (∀ n ∈ N ) [ m > n ]

There is an integer m that is larger that everyna ura num er nc u ng m se .

It is obviously false

∴the order in which the quantifiers are presented isimportant.


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

n

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

1=i

i = 1

)(i f n

Σ

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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729log729log)( 39 =i

4log

16log16log)(

2

24 =ii

x y bb

y x

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] denotest e sma est nteger t at s not sma er t an x.

213= x

13

4

−=

=

x

x

77,85.8,35,414.3

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 P

y x + 1

d 1

Repeat d d+1 until d divides y

return 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 P

repeat x x + 1 until x is prime

return 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= z Let


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By our assumption z is irrational

2 z w Let =

s aga n rra ona y our assump on

but

( )22 2222 =⎟ ⎠

⎜⎝

== ×

z w

( ) 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 is

avoided 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) Euler’s Conjecture (1769)

Enuler conjectured that this equation can never besatisfied.

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

4+ 4+ 4= 4 ( -


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Proof By Mathematical Induction

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

0 + 1 + 2 + 3 + 4 + 5 + 6

+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 Problem• 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=1⇒1x1 board1 square is special

Board is tiled by doing nothingif n=1

m=21=2⇒2x2 board

square s spec aBoard is tiled by putting 1 tile


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(ii) Induction Step

•m = 2n

• According to induction hypothesis theorem is true for

2n-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 be

tiled

• us t eorem s or m= , an s nce ts trut or m=n

o owsfrom its assumed truth for m=2n-1 for all n ≥ 1, it follows fromthe principle of mathematical induction that the theorem is

rue 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

• =

21;0 2110 >=+=== −− nor an nnn

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

• Constructive induction can be useful in the analysis of algorithms.

function 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 or

elementary 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 0……9999)

. , , , , ……,

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 - = Pr +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.

lec2byuuryu compatibility mode

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