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T i = indicator random variable of the event that i-th throw results in a tail E[T] = E[T 1 ] + … + E[T 6 ] = 6*(1/2) = 3 P(T=3) = P(H=3) = binomial(6,3)/2 6 = 5/16 < 1/2
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E[X i,j ] = ½ (consider only i<j) X= X i,j E[X] n(n-1) /4 1 i<j n
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T = 1 + (1/2) * 0 + (1/2) * ( T + T ) T = 1 + T
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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.
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l m+1
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if B A[m] then
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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,….,n k,….,1,k+1,….,n k,….,1,n,….,k+1 k+1,….,n,1,….,k
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1) find the median m of A 2) mm 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
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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)
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Coupon collector problem n coupons to collect What is the expected number of cereal boxes that you need to buy?
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Coupon collector problem Expected number of darts needed to hit the bull’s eye ? Assume that a dart throw is uniform in the circle. Let p be The fraction occupied by the bull’s eye.
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Coupon collector problem Expected number of darts needed to hit the bull’s eye ? Assume that a dart throw is uniform in the circle. Let p be The fraction occupied by the bull’s eye. 1/p
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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[X 0 ] = 1 … E[X k ] = ? … E[X n-1 ] = n
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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[X 0 ] = 1 … E[X k ] = n/(n-k) … E[X n-1 ] = n
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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. X=X 0 +X 1 +…+X n-1 = n n n-k k=0 n-1 1 k k=1 n = (n ln n) =
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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. X=X 0 +X 1 +…+X n-1 = n n n-k k=0 n-1 1 k k=1 n = (n ln n) = E[X]=E[X 0 ]+…+E[X n-1 ] =
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Harmonic numers 1 k k=1 n 1+ln n ln n
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Randomized algorithm for “median” L R <x =x >x for random x 2) recurse on the appropriate part 1) SELECT k-th element
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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.
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Markov inequality P(X > a.E[X]) < 1/a P(X a.E[X]) 1/a For non-negative random variable X:
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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 ?
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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 > b 2.E[X] 2 ) < V[X]/(b 2 E[X] 2 ) P( |X-E[X]| > b.E[X] ) < V[X] E[X] 2 * 1 b2b2
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Chebychev’s inequality P( |X-E[X]| > b.E[X] ) < V[X] E[X] 2 * 1 b2b2 P( (1-b)*E[X] X (1+b)*E[X] ) > V[X] E[X] 2 * 1 b2b2 1 -
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