T i = indicator random variable of the event that i-th throw results in a tail

Ti = indicator random variable of the event that i-th throw results in a tailE[T] = E[T1] + … + E[T6] = 6*(1/2) = 3

P(T=3) = P(H=3) = binomial(6,3)/26 = 5/16 < 1/2

E[Xi,j] = ½ (consider only i<j)

X=Xi,jE[X]n(n-1) /41 i<j n

T = 1 + (1/2) * 0 + (1/2) * ( T + T )

T = 1 + T

There exists c such that T(n) T(n/2)+T(n/3)+c*n.We need to show that there exists d such that T(n) d*n for all n.

Induction step:T(n) T(n/2) + T(n/3) + c*n d*n/2 + d*n/3 + c*n d*n + (c-d/6)*n d*n, taking d=6c.

l m+1

if B A[m] then

Reverse(a,b) for i from a to a+b do swap(A[i],A[a+b-i]);

Rotate(k) Reverse(1,k) Reverse(k+1,n) Reverse(1,n)

1,….,k,k+1,….,nk,….,1,k+1,….,nk,….,1,n,….,k+1k+1,….,n,1,….,k

1) find the median m of A2)

m mm

sum S3) if S\geq C then recurse on A[n/2..n] else recurse on A[1..n/2] with C’=C-S

3) if S\geq C then recurse on A[n/2..n] else recurse on A[1..n/2] with C’=C-S

T(n) = T(n/2) + O(n)

Coupon collector problem

n coupons to collect

What is the expected number of cereal boxes that you need to buy?

Coupon collector problem

Expected number of darts needed to hit the bull’s eye ?

Assume that a dart throw is uniform inthe circle. Let p beThe fraction occupiedby the bull’s eye.

Coupon collector problem

Expected number of darts needed to hit the bull’s eye ?

Assume that a dart throw is uniform inthe circle. Let p beThe fraction occupiedby the bull’s eye.

1/p

What is the expected number ofboxes that I buy in k-th phase ?

k-th phase = when I have k differentKinds of coupons.

E[X0] = 1…E[Xk] = ?…E[Xn-1] = n

What is the expected number ofboxes that I buy in k-th phase ?

k-th phase = when I have k differentKinds of coupons.

E[X0] = 1…E[Xk] = n/(n-k)…E[Xn-1] = n

What is the expected number ofboxes that I buy in k-th phase ?

k-th phase = when I have k differentKinds of coupons.

X=X0+X1+…+Xn-1 = n nn-kk=0

n-11kk=1

n

= (n ln n)

=

What is the expected number ofboxes that I buy in k-th phase ?

k-th phase = when I have k differentKinds of coupons.

X=X0+X1+…+Xn-1 = n nn-kk=0

n-11kk=1

n

= (n ln n)

=

E[X]=E[X0]+…+E[Xn-1]=

Harmonic numers

1kk=1

n1+ln nln n

Randomized algorithm for “median”

L R

<x =x >x

for random x

2) recurse on the appropriate part

1)

SELECT k-th element

Randomized algorithm for “median”

Las Vegas algorithm

(never makes error, randomnessonly influences running time)

The identity testing algorithm wasMonte Carlo algorithm with 1 sided error.

Markov inequality

P(X > a.E[X]) < 1/a

P(X a.E[X]) 1/a

For non-negative random variable X:

Variance

For a random variable X:

V[ X ] = E[ (X-E[X])2 ]

What is the variance of X=the number on a (6-sided) dice ?

Variance

For a random variable X:

V[ X ] = E[ (X-E[X])2 ]

Y = (X-E[X])2

P( Y > a.E[Y] ) < 1/a

P( (X-E[X])2 > a.V[X] ) < 1/a

P( (X-E[X])2 > b2.E[X]2 ) < V[X]/(b2 E[X]2)

P( |X-E[X]| > b.E[X] ) <V[X]

E[X]2 *1

b2

Chebychev’s inequality

P( |X-E[X]| > b.E[X] ) <V[X]

E[X]2 *1

b2

P( (1-b)*E[X] X (1+b)*E[X] )

>V[X]

E[X]2 *1

b21 -