Administrative Sep. 25 (today) – HW3 (=QUIZ #1) due Sep. 27 – HW4 due Sep. 28 8am – problem...

AdministrativeSep. 25 (today) – HW3 (=QUIZ #1) due Sep. 27 – HW4 dueSep. 28 8am – problem session

Oct. 2Oct. 4 – QUIZ #2

(pages 45-79 of DPV)

Administrative

You may work with one other person on homeworks, but you must each write up your

solutions separately (without any written aid). If you work with another person, indicate who you worked with on your solution (thus you should each indicate each other).

Homework rules:

All cases of suspected dishonesty must be reported to the Board, either through a shortform resolution or by forwarding a case to the Board for a hearing. Faculty may not come to an understanding with a student on their own in a case of suspected dishonesty, but must use the short form resolution or submit a case.

University Academic Honesty Policy:

Recurrences T(n) = a T(n/b) + f(n)

• If f(n) = O(nc-) then T(n) =(nc)• If f(n) = (nc) then T(n) =(nc.log n)• If f(n) = (nc+) then T(n)=(f(n)) if a.f(n/b) d.f(n) for some d<1 and n>n0

c=logb a

T(n) = 3 T(n/2) + (n) T(n) = (nlog 3)

T(n) = 2T(n/2) + (n) T(n) = (n.log n)

Finding the k-th smallest element

k = n/2 = MEDIAN

Split(A[1..n],x)

runs in time O(n)

k = n/2 = MEDIAN

Split(A[1..n],x)

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

6 3 1 87 2 6 1 85 8 9 1 32

1) sort each 5-tuple

6 3 1 87 2 6 1 85 8 9 1 32

TIME = ?

TIME = (n)

2) find median of the middle n/5 elements

TIME = ?

2) find median of the middle n/5 elements

TIME = T(n/5)

We will use this element as the pivot

At least ? Many elements in the array are X

At least 3n/10 elements in the array are X

6 3 1 87 2 6 1 85 8 9 1 32

63 1 32 2 1 1 85 8 9 6 87

6 3 1 87 2 6 1 85 8 9 1 32

63 1 32 2 1 1 85 8 9 6 87

Recurse, time ?

6 3 1 87 2 6 1 85 8 9 1 32

63 1 32 2 1 1 85 8 9 6 87

Recurse, time T(7n/10)

6 3 1 87 2 6 1 85 8 9 1 32

63 1 32 2 1 1 85 8 9 6 87X X

recurse

6 3 1 87 2 6 1 85 8 9 1 32

63 1 32 2 1 1 85 8 9 6 87X X

recurse

T(n/5)

T(7n/10)

T(n) T(n/5) + T(7n/10) + O(n)

T(n) d.n

Induction step:

T(n) T(n/5) + T(7n/10) + O(n) d.(n/5) + d.(7n/10) + O(n) d.n + (O(n) – dn/10) d.n

Why 5-tuples?

3 1 7 16 5 9 1 2

6 3 1 87 2 6 1 85 8 9 1 32

63 1 32 2 1 1 85 8 9 6 87X X

recurse

Why 5-tuples?

3 1 7 16 5 9 1 2

6 3 1 87 2 6 1 85 8 9 1 32

63 1 32 2 1 1 85 8 9 6 87X X

recurse

T(2n/3)137

T(n/3)

Why 5-tuples?

T(n) T(n/3) + T(2n/3) + (n)

Why 5-tuples?

T(n) T(n/3) + T(2n/3) + (n)

T(n) c.n.ln n

Induction step:

T(n) = T(n/3) + T(2n/3) + (n) c.(n/3).ln (n/3) + c.(2n/3).ln (2n/3) + (n) c.n.ln n - c.n.((1/3)ln 3+(2/3)ln 3/2)+(n)c.n.ln n

Quicksort(A[b..c])

Quicksort(A[b..i]);Quicksort(A[j..c]);

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

x xxi jb c

Quicksort(A[b..c])

Worst-case running time?

How to make the worst-case running time O(n.log n) ?

Quicksort(A[b..c])

if pivot = medianthen the worst-case running time satisfies

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

Quicksort(A[b..c])

Quicksort(A[b..i]);Quicksort(A[j..c]);

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

x xxi jb c

x = random element of A[b..c]

k = n/2 = MEDIAN

Split(A[1..n],x)

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

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

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

Finite probability space

set (sample space)function P: R+ (probability distribution)

P(x) = 1x

Finite probability space

elements of are called atomic eventssubsets of are called events

probability of an event A is

P(x)xA

P(x) = 1x

Examples

1. Roll a (6 sided) dice. What is the probability that the number on the dice is even?

2. Flip two coins, what is the probability thatthey show the same symbol?

3. Flip five coins, what is the probability thatthey show the same symbol?

4. Mix a pack of 52 cards. What is the probability that all red cards come before all black cards?

Union bound

P(A B) P(A) + P(B)

P(A1 A2 … An) P(A1) + P(A2)+…+P(An)

LEMMA:

More generally:

Union bound P(A1 A2 … An) P(A1) + P(A2)+…+P(An)

Suppose that the probability of winning ina lottery is 10-6. What is the probability thatsomebody out of 100 people wins?

Ai = i-th person wins

somebody wins = ?

Ai = i-th person wins

somebody wins = A1A2…A100

P(A1A2…A100) 100*10-6 = 10-4

P(A1A2…A100) = 1–P(AC

1 AC2… AC

100) =1-P(AC

1)P(AC2)…P(AC

100) =1-(1-10-6)100 0.99*10-4

Independence Events A,B are independent if

P(A B) = P(A) * P(B)

“observing whether B happened gives no information on A”

P(A B) = P(A) * P(B)

“observing whether B happened gives no information on A”

AP(A|B) = P(AB)/P(B)conditional probability of A, given B

P(A B) = P(A) * P(B)

P(A|B) = P(A)

Examples

Roll two (6 sided) dice. Let S be their sum. 1) What is that probability that S=7 ? 2) What is the probability that S=7, conditioned on S being odd ? 3) Let A be the event that S is even and B the event that S is odd. Are A,B independent? 4) Let C be the event that S is divisible by 4. Are A,C independent? 5) Let D be the event that S is divisible by 3. Are A,D independent?

Examples

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

Examples

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

Full independence

Events A1,…,An are (fully) independentIf for every subset S[n]:={1,2,…,n}

P ( Ai ) = P(Ai)iS iS

Random variable

P(x) = 1x

A random variable is a function Y : RThe expected value of Y is

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

Examples

Roll two dice. Let S be their sum.

If S=7 then player A gives player B $6otherwise player B gives player A $1

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

Examples

Roll two dice. Let S be their sum.

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

Expected income for B E[Y] = 6*(1/6)-1*(5/6)= 1/6

Linearity of expectation

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

E[X1 X2 … Xn] E[X1] + E[X2]+…+E[Xn]

LEMMA:

More generally:

Everybody pays me $1 and writes their name on a card. I mix the cards and give everybody one card. If you get backthe 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?

X1 = -9 if player 1 gets his card back 1 otherwise

E[X1] = ?

X1 = -9 if player 1 gets his card back 1 otherwise

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

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?

1/2 1 1/4 21/8 31/16 4….

n.2-n = 2

Expected number of dice-throws until you get “6” ?

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