t i = indicator random variable of the event that i-th throw results in a tail e[t] = e[t 1 ] + …...
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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 -