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 tail E[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) /4 1 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]); 1,….,k,k+1,….,n k,….,1,k+1,….,n k,….,1,n,….,k+1 k+1,….,n,1,….,k Rotate(k) Reverse(1,k) Reverse(k+1,n) Reverse(1,n)

find the median m of A m m  m sum 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) 3) if S\geq C then recurse on A[n/2..n] else recurse on A[1..n/2] with C’=C-S

Coupon collector problem n coupons to collect What is the expected number of cereal boxes that you need to buy?

Coupon collector problem Assume that a dart throw is uniform in the circle. Let p be The fraction occupied by the bull’s eye. Expected number of darts needed to hit the bull’s eye ?

Coupon collector problem Assume that a dart throw is uniform in the circle. Let p be The fraction occupied by the bull’s eye. Expected number of darts needed to hit the bull’s eye ? 1/p

What is the expected number of boxes that I buy in k-th phase ? k-th phase = when I have k different Kinds of coupons. E[X0] = 1 … E[Xk] = ? … E[Xn-1] = n

What is the expected number of boxes that I buy in k-th phase ? k-th phase = when I have k different Kinds of coupons. E[X0] = 1 … E[Xk] = n/(n-k) … E[Xn-1] = n

What is the expected number of boxes that I buy in k-th phase ? k-th phase = when I have k different Kinds of coupons. n-1 n n 1 X=X0+X1+…+Xn-1 = n  = n-k k k=0 k=1 =  (n ln n)

What is the expected number of boxes that I buy in k-th phase ? k-th phase = when I have k different Kinds of coupons. n-1 n n 1 X=X0+X1+…+Xn-1 = n  = n-k k k=0 k=1 = =  (n ln n) E[X]=E[X0]+…+E[Xn-1]

Harmonic numers n  1 ln n  1+ln n k k=1

Randomized algorithm for “median” SELECT k-th element for random x 1) =x <x >x R L 2) recurse on the appropriate part

Randomized algorithm for “median” Las Vegas algorithm (never makes error, randomness only influences running time) The identity testing algorithm was Monte Carlo algorithm with 1 sided error.

Markov inequality For non-negative random variable X: P(X > a.E[X]) < 1/a P(X  a.E[X])  1/a

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) V[X] 1 P( |X-E[X]| > b.E[X] ) < * b2 E[X]2

Chebychev’s inequality V[X] 1 P( |X-E[X]| > b.E[X] ) < * b2 E[X]2 P( (1-b)*E[X]  X  (1+b)*E[X] ) > V[X] 1 1 - * b2 E[X]2

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