- 74 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Administrative' - fulton-howard

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Sep. 27 (today) – HW4 due

Sep. 28 8am – problem session

Oct. 2

Oct. 4 – QUIZ #2

(pages 45-79 of DPV)

algorithm for k-select with

O(n) worst-case running time

modification of quick-sort which

has O(n.log n) worst-case

running time

randomized k-select

GOAL: O(n) expected running-time

Finding the k-th smallest element

Select(k,A[c..d])

x=random element from A[c..d]

Split(A[c..d],x)

>x

x

j

j k k-th smallest on left

j<k (k-j)-th smallest on right

set (sample space)

function P: R+ (probability distribution)

P(x) = 1

x

elements of are called atomic events

subsets of are called events

probability of an event A is

P(x)

P(A)=

xA

A

B

C

Are A,B independent ?Are A,C independent ?

Are B,C independent ?

Is it true that P(ABC)=P(A)P(B)P(C)?

Events A,B,C are

pairwise independent

but not

(fully) independent

A

B

C

Are A,B independent ?Are A,C independent ?

Are B,C independent ?

Is it true that P(ABC)=P(A)P(B)P(C)?

Events A1,…,An are (fully) independent

If for every subset S[n]:={1,2,…,n}

P ( Ai ) = P(Ai)

iS

iS

set (sample space)

function P: R+ (probability distribution)

P(x) = 1

x

A random variable is a function

Y : R

The expected value of Y is

E[X] := P(x)* Y(x)

x

Roll two dice. Let S be their sum.

If S=7 then player A gives player B $6

otherwise player B gives player A $1

2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12

Roll two dice. Let S be their sum.

If S=7 then player A gives player B $6

otherwise player B gives player A $1

2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12

-1 , -1,-1 ,-1, -1, 6 ,-1 ,-1 , -1 , -1 , -1

Y:

Expected income for B

E[Y] = 6*(1/6)-1*(5/6)= 1/6

LEMMA:

E[X + Y] = E[X] + E[Y]

More generally:

E[X1+ X2+ … + Xn] = E[X1] + E[X2]+…+E[Xn]

Everybody pays me $1 and writes their

name on a card. I mix the cards and

give everybody one card. If you get back

the card with your name – I pay you $10.

Let n be the number of people in the class.

For what n is the game advantageous for me?

Everybody pays me $1 and writes their

name on a card. I mix the cards and

give everybody one card. If you get back

the card with your name – I pay you $10.

X1 = -9 if player 1 gets his card back

1 otherwise

E[X1] = ?

Everybody pays me $1 and writes their

name on a card. I mix the cards and

give everybody one card. If you get back

the card with your name – I pay you $10.

X1 = -9 if player 1 gets his card back

1 otherwise

E[X1] = -9/n + 1*(n-1)/n

Everybody pays me $1 and writes their

name on a card. I mix the cards and

give everybody one card. If you get back

the card with your name – I pay you $10.

X1 = -9 if player 1 gets his card back

1 otherwise

X2 = -9 if player 2 gets his card back

1 otherwise

E[X1+…+Xn] = E[X1]+…+E[Xn] =

n ( -9/n + 1*(n-1)/n ) = n – 10.

Expected number of coin-tosses

until HEADS?

n.2-n = 2

n=1

Expected number of dice-throws

until you get “6” ?

Finding the k-th smallest element

Select(k,A[c..d])

x=random element from A[c..d]

Split(A[c..d],x)

>x

x

j

j k k-th smallest on left

j<k (k-j)-th smallest on right

Polynomial of degree d

p(x) = a0 + a1 x + ... + ad xd

Polynomial of degree d’

q(x) = b0 + b1 x + ... + bd’ xd’

p(x)q(x) = (a0b0) + (a0b1 + a1b0) x +

.... + (adbd’) xd+d’

Polynomial of degree d

p(x) = a0 + a1 x + ... + ad xd

THEOREM:

A non-zero polynomial of

degree d has at most d roots.

COROLLARY:

A polynomial of degree d is determined

by its value on d+1 points.

A polynomial of degree d is determined

by its value on d+1 points.

Find a polynomial p of degree d

such that

p(a0) = 1

p(a1) = 0

....

p(ad) = 0

A polynomial of degree d is determined

by its value on d+1 points.

Find a polynomial p of degree d

such that

p(a0) = 1

p(a1) = 0

....

p(ad) = 0

(x-a1)(x-a2)...(x-ad)

(a0-a1)(a0-a2)...(a0-ad)

Representing polynomial of degree d

the coefficient representation

d+1 coefficients

evaluation

interpolation

the value representation

evaluation on d+1 points

p(x) = 7 + x + 5x2 + 3x3 + 6x4 + 2x5

p(z) = 7 + z + 5z2 + 3z3 + 6z4 + 2z5

p(-z) = 7 – z + 5z2 – 3z3 + 6z4 – 2z5

p(x) = (7+5x2 + 6x4) + x(1+3x2 + 2x4)

p(x) = pe(x2) + x po(x2)

p(-x) = pe(x2) – x po(x2)

p(x) = a0 + a1 x + a2 x2 + ... + ad xd

p(x) = pe(x2) + x po(x2)

p(-x) = pe(x2) – x po(x2)

To evaluate p(x) on

-x1,x1,-x2,x2,...,-xn,xn

we only evaluate pe(x) and po(x) on

x12,...,xn2

To evaluate p(x) on

-x1,x1,-x2,x2,...,-xn,xn

we only evaluate pe(x) and po(x) on

x12,...,xn2

To evaluate pe(x) on

x12,...,xn2

we only evaluate pe(x) on ?

2ik/n

= k

e

FACT 1:

n = 1

k . l = k+l

0 + 1 + ... + n-1 = 0

FACT 2:

FACT 3:

FACT 4:

k = -k+n/2

FFT (a0,a1,...,an-1,)

(s0,...,sn/2-1)= FFT(a0,a2,...,an-2,2)

(z0,...,zn/2-1) = FFT(a1,a3,...,an-1,2)

s0 + z0

s1 + z1

s2 + 2 z2

....

s0 – z0

s1 - z1

s2 - 2 z2

....

on multiple points

1

xn

xn2

.

.

xnd

1

x1

x12

.

.

x1d

1

x2

x22

.

.

x2d

(a0,a1,a2,...,ad)

. . .

Vandermonde matrix

Download Presentation

Connecting to Server..