administrative sep. 25 (today) – hw3 (=quiz #1) due sep. 27 – hw4 due sep. 28 8am – problem...
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![Page 1: Administrative Sep. 25 (today) – HW3 (=QUIZ #1) due Sep. 27 – HW4 due Sep. 28 8am – problem session Oct. 2 Oct. 4 – QUIZ #2 (pages 45-79 of DPV)](https://reader036.vdocument.in/reader036/viewer/2022062715/56649d805503460f94a6427d/html5/thumbnails/1.jpg)
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)
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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:
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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)
2
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Finding the k-th smallest element
k = n/2 = MEDIAN
Split(A[1..n],x)
x x
runs in time O(n)
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Finding the k-th smallest element
k = n/2 = MEDIAN
Split(A[1..n],x)
x x
j
j k k-th smallest on leftj<k (k-j)-th smallest on right
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Finding the k-th smallest element
631
87
261
85
891
32
6 3 1 87 2 6 1 85 8 9 1 32
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Finding the k-th smallest element
631
87
261
85
891
32
1) sort each 5-tuple
6 3 1 87 2 6 1 85 8 9 1 32
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Finding the k-th smallest element
631
87
261
85
891
32
1) sort each 5-tuple
![Page 9: Administrative Sep. 25 (today) – HW3 (=QUIZ #1) due Sep. 27 – HW4 due Sep. 28 8am – problem session Oct. 2 Oct. 4 – QUIZ #2 (pages 45-79 of DPV)](https://reader036.vdocument.in/reader036/viewer/2022062715/56649d805503460f94a6427d/html5/thumbnails/9.jpg)
Finding the k-th smallest element
136
87
261
85
891
32
1) sort each 5-tuple
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Finding the k-th smallest element
136
87
261
85
891
32
1) sort each 5-tuple
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Finding the k-th smallest element
136
87
125
86
891
32
1) sort each 5-tuple
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Finding the k-th smallest element
136
87
125
86
123
97
1) sort each 5-tuple
TIME = ?
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Finding the k-th smallest element
136
87
125
86
123
97
1) sort each 5-tuple
TIME = (n)
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Finding the k-th smallest element
136
87
125
86
123
97
2) find median of the middle n/5 elements
TIME = ?
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Finding the k-th smallest element
136
87
125
86
123
97
2) find median of the middle n/5 elements
TIME = T(n/5)
We will use this element as the pivot
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Finding the k-th smallest element
136
87
125
86
123
97
At least ? Many elements in the array are X
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Finding the k-th smallest element
123
97
At least ? Many elements in the array are X
125
86
136
87
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Finding the k-th smallest element
123
97
At least 3n/10 elements in the array are X
125
86
136
87
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Finding the k-th smallest element
123
97
At least 3n/10 elements in the array are X
125
86
136
87
6 3 1 87 2 6 1 85 8 9 1 32
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Finding the k-th smallest element
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
X X
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Finding the k-th smallest element
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
X X
Recurse, time ?
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Finding the k-th smallest element
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
X X
Recurse, time T(7n/10)
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Finding the k-th smallest element
631
87
261
85
891
32
6 3 1 87 2 6 1 85 8 9 1 32
136
87
125
86
123
97
136
87
125
86
123
97
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
Split
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Finding the k-th smallest element
631
87
261
85
891
32
6 3 1 87 2 6 1 85 8 9 1 32
136
87
125
86
123
97
136
87
125
86
123
97
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
n)
T(n/5)
n)
T(7n/10)
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Finding the k-th smallest element
T(n) T(n/5) + T(7n/10) + O(n)
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Finding the k-th smallest element
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
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Why 5-tuples?
317
615
912
3 1 7 16 5 9 1 2
137
156
129
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
n)n)
137
156
129
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Why 5-tuples?
317
615
912
3 1 7 16 5 9 1 2
137
156
129
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
n)n)
T(2n/3)137
156
129
T(n/3)
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Why 5-tuples?
T(n) T(n/3) + T(2n/3) + (n)
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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
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Quicksort(A[b..c])
Quicksort(A[b..i]);Quicksort(A[j..c]);
Split(A[b..c],x)
x xxi jb c
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Quicksort(A[b..c])
Worst-case running time?
How to make the worst-case running time O(n.log n) ?
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Quicksort(A[b..c])
if pivot = medianthen the worst-case running time satisfies
T(n) = 2T(n/2) + O(n)
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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]
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Finding the k-th smallest element
k = n/2 = MEDIAN
Split(A[1..n],x)
x x
j
j k k-th smallest on leftj<k (k-j)-th smallest on right
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Finding the k-th smallest element
Select(k,A[c..d])
Split(A[c..d],x)
x x
j
j k k-th smallest on leftj<k (k-j)-th smallest on right
x=random element from A[c..d]
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Finite probability space
set (sample space)function P: R+ (probability distribution)
P(x) = 1x
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Finite probability space
set (sample space)function P: R+ (probability distribution)
elements of are called atomic eventssubsets of are called events
probability of an event A is
P(x)xA
P(A)=
P(x) = 1x
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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?
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Union bound
P(A B) P(A) + P(B)
P(A1 A2 … An) P(A1) + P(A2)+…+P(An)
LEMMA:
More generally:
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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 = ?
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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 = A1A2…A100
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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?
P(A1A2…A100) 100*10-6 = 10-4
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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?
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
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Independence Events A,B are independent if
P(A B) = P(A) * P(B)
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Independence Events A,B are independent if
P(A B) = P(A) * P(B)
“observing whether B happened gives no information on A”
B
A
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Independence Events A,B are independent if
P(A B) = P(A) * P(B)
“observing whether B happened gives no information on A”
B
AP(A|B) = P(AB)/P(B)conditional probability of A, given B
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Independence Events A,B are independent if
P(A B) = P(A) * P(B)
P(A|B) = P(A)
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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?
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Examples
A
BC
Are A,B independent ?Are A,C independent ?Are B,C independent ?Is it true that P(ABC)=P(A)P(B)P(C)?
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Examples
A
BC
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
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Full independence
Events A1,…,An are (fully) independentIf for every subset S[n]:={1,2,…,n}
P ( Ai ) = P(Ai)iS iS
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Random variable
set (sample space)function P: R+ (probability distribution)
P(x) = 1x
A random variable is a function Y : RThe expected value of Y is
E[X] := P(x)* Y(x) x
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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
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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
Y:
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Linearity of expectation
E[X Y] E[X] + E[Y]
E[X1 X2 … Xn] E[X1] + E[X2]+…+E[Xn]
LEMMA:
More generally:
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Linearity of expectation
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?
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Linearity of expectation
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.
X1 = -9 if player 1 gets his card back 1 otherwise
E[X1] = ?
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Linearity of expectation
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.
X1 = -9 if player 1 gets his card back 1 otherwise
E[X1] = -9/n + 1*(n-1)/n
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Linearity of expectation
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.
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.
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Expected number of coin-tosses until HEADS?
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Expected number of coin-tosses until HEADS?
1/2 1 1/4 21/8 31/16 4….
n.2-n = 2
n=1
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Expected number of dice-throws until you get “6” ?