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X i = indicator random variable of the event that i-th person gets his hat back.

Date post: 16-Mar-2016
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X i = indicator random variable of the event that i-th person gets his hat back. E[X i ]=1/20 X=X 1 +…+X 20 E[X] = E[X 1 ] +…+ E[X 20 ] = 1. derangement = permutation with no fixed points  n! / e . - PowerPoint PPT Presentation
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indicator random variable of the event that i-th pe gets his hat back. E[X i ]=1/20 …+X 20 E[X] = E[X 1 ] +…+ E[X 20 ] =
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Page 1: X i  = indicator random variable of the event that i-th person            gets his hat back.

Xi = indicator random variable of the event that i-th person gets his hat back.

E[Xi]=1/20X=X1+…+X20

E[X] = E[X1] +…+ E[X

20] = 1

Page 2: X i  = indicator random variable of the event that i-th person            gets his hat back.

derangement = permutation with no fixed points

n! / e permutations with 1 fixed point

number =

Page 3: X i  = indicator random variable of the event that i-th person            gets his hat back.

derangement = permutation with no fixed points

n! / e permutations with 1 fixed point

number =

n (n-1)! / e

n (n-1)! / e n!

0.3678 0.368

Page 4: X i  = indicator random variable of the event that i-th person            gets his hat back.
Page 5: X i  = indicator random variable of the event that i-th person            gets his hat back.

Claim: Alice wins only on HHH game.

1/8 - Alice wins, gets $67/8 - Alice loses, pays $1= gets -$1

Alice’s expected payoff: (1/8)* 6 + (7/8) * (-1) = - 1/8

Bob has the advantage.

Page 6: X i  = indicator random variable of the event that i-th person            gets his hat back.
Page 7: X i  = indicator random variable of the event that i-th person            gets his hat back.

Heap MIN-HEAP-INSERT O(log k)HEAP-EXTRACT-MIN O(log k)

P1

Pk

Page 8: X i  = indicator random variable of the event that i-th person            gets his hat back.

We will find the array Ai whose first element e is the smallest, output e to B, remove e from Ai, and repeat. We will use a heap H as follows: we find e using Heap-Extract-Min procedure and then add the next element from Ai to Husing Min-Heap-Insert procedure. We make n calls to Min-Heap-Insert and n calls to Heap-Extract-Min. Hencethe running time is O(n.log k). To simplify the exposition we add at the end of each array. for i from 1 to k do

Pi 1; Max-Heap-Insert( H,[Ai[1],i] );for j from 1 to n do [e,i] Heap-Extract-Max (H); Pi Pi + 1; Max-Heap-Insert( H, [Ai [Pi],i] ); add e to B

Page 9: X i  = indicator random variable of the event that i-th person            gets his hat back.
Page 10: X i  = indicator random variable of the event that i-th person            gets his hat back.

<

Page 11: X i  = indicator random variable of the event that i-th person            gets his hat back.

<

LA1,RAn,LB1,RBnwhile LA<R

A do

MA (LA + RA)/2 MB (L + R)/2 if A[MA]<B[M] then LA M+ 1, R M- 1else RA M, L M

output smaller of A[LA],B[LB]

Page 12: X i  = indicator random variable of the event that i-th person            gets his hat back.

Randomized algorithm for “median”

L R<x =x >x

for random x

2) recurse on the appropriate part

1)

SELECT k-th element

Page 13: X i  = indicator random variable of the event that i-th person            gets his hat back.

Quick-sort

L R<x =x >x

for random x

2) recurse on both parts

1)

PARTITION

Page 14: X i  = indicator random variable of the event that i-th person            gets his hat back.

Quick-sortR-QUICK-SORT (A, l, r) x random element of A[l,r]; q PARTITION(A,x,l,r); R-QUICK-SORT(A,l,q-1); R-QUICK-SORT(A,q+1,r);

Page 15: X i  = indicator random variable of the event that i-th person            gets his hat back.

Quick-sortR-QUICK-SORT (A, l, r) x random element of A[l,r]; q PARTITION(A,x,l,r); R-QUICK-SORT(A,l,q-1); R-QUICK-SORT(A,q+1,r);

How many times is R-QUICK-SORT called?

Page 16: X i  = indicator random variable of the event that i-th person            gets his hat back.

Quick-sort R-QUICK-SORT (A, l, r) x random element of A[l,r]; q PARTITION(A,x,l,r); R-QUICK-SORT(A,l,q-1); R-QUICK-SORT(A,q+1,r);

Time spent in PARTITION?

Page 17: X i  = indicator random variable of the event that i-th person            gets his hat back.

Quick-sort R-QUICK-SORT (A, l, r) x random element of A[l,r]; q PARTITION(A,x,l,r); R-QUICK-SORT(A,l,q-1); R-QUICK-SORT(A,q+1,r);

Time spent in PARTITION?

compare x with all elements in A[l,r]

we will count the number of comparisons

Page 18: X i  = indicator random variable of the event that i-th person            gets his hat back.

Quick-sort R-QUICK-SORT (A, l, r) x random element of A[l,r]; q PARTITION(A,x,l,r); R-QUICK-SORT(A,l,q-1); R-QUICK-SORT(A,q+1,r);

Time spent in PARTITION?

Let the elements of A after sorting be b1 < b2 < … < bn

Let Xi,j be the indicator random variable for the event bi is compared to bj.

Page 19: X i  = indicator random variable of the event that i-th person            gets his hat back.

Quick-sortTime spent in PARTITION? Let the elements of A after sorting be b1 < b2 < … < bn

Let Xi,j be the indicator random variable for the event bi is compared to bj.

What is the probability that bi and bj arecompared in the first round ?

Page 20: X i  = indicator random variable of the event that i-th person            gets his hat back.

Quick-sortTime spent in PARTITION? Let the elements of A after sorting be b1 < b2 < … < bn

Let Xi,j be the indicator random variable for the event bi is compared to bj.

What is the probability that bi and bj arecompared in the first round ?

2/n (the pivot has to be bi or bj)

Page 21: X i  = indicator random variable of the event that i-th person            gets his hat back.

Quick-sortTime spent in PARTITION? What is the probability that bi and bj arecompared ?

2/(j-i+1)

Let bk be the first pivot such that ik j.

bi, bj get compared k=i or k=j

k is uniformly random in {i,…,j}

Page 22: X i  = indicator random variable of the event that i-th person            gets his hat back.

Quick-sortTime spent in PARTITION? What is the probability that bi and bj arecompared ?

2/(j-i+1)E[Xi,j] = 2/(j-i+1)

X= Xi,j1i<j n

E[X]=n = O(n ln n)2j-i+1

1i<j n k=2

n 2 k


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