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Algorithm Design and Analysis
Summer 2018 Amo G. Tong 1
Lecture 4Randomized Algorithmsβ’ Randomized Quick Sort
β’ Randomized Selection
Summer 2018 Amo G. Tong 2
β’ Quick Sort:β’ Take an element π₯ = π΄[π] in π΄[1,β¦ , π].
β’ Partition into two subarrays π΄[1,β¦ , π β 1] and π΄[π + 1,β¦ , π] such that each element of π΄[1,β¦ , π β 1] is less than or equal to π₯ which is less than or equal to each element of π΄[π + 1,β¦ , π].
β’ Sort the subarrays recursively.
β’ No need to combine.
Randomized Algorithms
Summer 2018 Amo G. Tong 3
Input: an array π΄[1,β¦ , π] of numbers
β’ Quick Sort:
Randomized Algorithms
Summer 2018 Amo G. Tong 4
Input: an array π΄[1,β¦ , π] of numbers
PARTITION (π΄, π, π)1 select an element π₯ = π΄[π] in π΄[π. . π]2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
β’ Quick Sort:
Randomized Algorithms
Summer 2018 Amo G. Tong 5
Input: an array π΄[1,β¦ , π] of numbers
0 9 5 2 1 π = 0, exchange 5 with 1
0 1 2 5 9
PARTITION (π΄, π, π)1 select an element π₯ = π΄[π] in π΄[π. . π]2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
β’ Quick Sort:
Randomized Algorithms
Summer 2018 Amo G. Tong 6
Input: an array π΄[1,β¦ , π] of numbers
0 9 5 2 10 9 1 2 5
π = 0, exchange 5 with 1
π = 1, exchange 0 with 0
0 1 2 5 9
PARTITION (π΄, π, π)1 select an element π₯ = π΄[π] in π΄[π. . π]2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
β’ Quick Sort:
Randomized Algorithms
Summer 2018 Amo G. Tong 7
Input: an array π΄[1,β¦ , π] of numbers
0 9 5 2 10 9 1 2 50 9 1 2 5
π = 0, exchange 5 with 1
π = 1, exchange 0 with 0
π = 1
0 1 2 5 9
PARTITION (π΄, π, π)1 select an element π₯ = π΄[π] in π΄[π. . π]2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
β’ Quick Sort:
Randomized Algorithms
Summer 2018 Amo G. Tong 8
Input: an array π΄[1,β¦ , π] of numbers
0 9 5 2 10 9 1 2 50 9 1 2 50 9 1 2 5
π = 0, exchange 5 with 1
π = 1, exchange 0 with 0
π = 1
π = 2, exchange 9 with 1
0 1 2 5 9
PARTITION (π΄, π, π)1 select an element π₯ = π΄[π] in π΄[π. . π]2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
β’ Quick Sort:
Randomized Algorithms
Summer 2018 Amo G. Tong 9
Input: an array π΄[1,β¦ , π] of numbers
0 9 5 2 10 9 1 2 50 9 1 2 50 9 1 2 50 1 9 2 5
π = 0, exchange 5 with 1
π = 1, exchange 0 with 0
π = 1
π = 2, exchange 9 with 1
π = 3, exchange 9 with 2
0 1 2 5 9
PARTITION (π΄, π, π)1 select an element π₯ = π΄[π] in π΄[π. . π]2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
β’ Quick Sort:
Randomized Algorithms
Summer 2018 Amo G. Tong 10
Input: an array π΄[1,β¦ , π] of numbers
0 9 5 2 10 9 1 2 50 9 1 2 50 9 1 2 50 1 9 2 50 1 2 9 5
π = 0, exchange 5 with 1
π = 1, exchange 0 with 0
π = 1
π = 2, exchange 9 with 1
π = 3, exchange 9 with 2
exchange 9 with 5
0 1 2 5 9
PARTITION (π΄, π, π)1 select an element π₯ = π΄[π] in π΄[π. . π]2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
β’ Quick Sort:
Randomized Algorithms
Summer 2018 Amo G. Tong 11
Input: an array π΄[1,β¦ , π] of numbers
0 9 5 2 10 9 1 2 50 9 1 2 50 9 1 2 50 1 9 2 50 1 2 9 50 1 2 5 9
π = 0, exchange 5 with 1
π = 1, exchange 0 with 0
π = 1
π = 2, exchange 9 with 1
π = 3, exchange 9 with 2
exchange 9 with 5
Return 4
0 1 2 5 9
PARTITION (π΄, π, π)1 select an element π₯ = π΄[π] in π΄[π. . π]2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
β’ Quick Sort:
Randomized Algorithms
Summer 2018 Amo G. Tong 12
Input: an array π΄[1,β¦ , π] of numbers
QUICKSORT (π΄, π, π)1 if π < π2 π =PARTITION(A, p, r)3 QUICKSORT (π΄, π, π β 1);4 QUICKSORT (π΄, π + 1, π);5 end
PARTITION (π΄, π, π)1 select an element π₯ = π΄[π] in π΄[π. . π]2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
0 9 5 2 1
0 1 2 5 9
β’ Quick Sort:
Randomized Algorithms
Summer 2018 Amo G. Tong 13
Input: an array π΄[1,β¦ , π] of numbers
QUICKSORT (π΄, π, π)1 if π < π2 π =PARTITION(A, p, r)3 QUICKSORT (π΄, π, π β 1);4 QUICKSORT (π΄, π + 1, π);5 end
PARTITION (π΄, π, π)1 select an element π₯ = π΄[π] in π΄[π. . π]2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
Which element should we take?
β’ Quick Sort:β’ Take an element π₯ in π΄[1,β¦ , π].
β’ Partition into two subarrays π΄1[1, β¦ , π] and π΄2[1, β¦ , π] such that each element of π΄1 is less than or equal to x which is less than or equal to each element of π΄2. (π + π = π β 1)
β’ Sort the subarrays recursively.
β’ No need to combine.
β’ π₯ decides the size of the two subarrays.β’ π(π) = π(π) + π(π) + Ξ(π);
Randomized Algorithms
Summer 2018 Amo G. Tong 14
β’ π₯ decides the size of the two subarrays.β’ π π = π π + π π + Ξ π .
β’ If the array is always evenly partitioned:β’ π₯=median
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Summer 2018 Amo G. Tong 15
β’ π₯ decides the size of the two subarrays.β’ π π = π π + π π + Ξ π .
β’ If the array is always evenly partitioned:β’ π₯=median
β’ π π = π π/2 + π π/2 + Ξ π .
β’ π(π) = Ξ(π log π)
Randomized Algorithms
Summer 2018 Amo G. Tong 16
β’ π₯ decides the size of the two subarrays.β’ π π = π π + π π + Ξ π .
β’ If the array is always evenly partitioned:β’ π₯=median
β’ π π = π π/2 + π π/2 + Ξ π .
β’ π(π) = Ξ(π log π)
β’ If the array is always partitioned with a fixed proportion π:β’ x= the ππ -th smallest element
β’ π(π) = π(ππ) + π((1 β π)π) + Ξ(π) where 0 < π < 1;
β’ π(π) = Ξ(π log π)
Randomized Algorithms
Summer 2018 Amo G. Tong 17
β’ π₯ decides the size of the two subarrays.β’ π π = π π + π π + Ξ π .
β’ If one of the subarray is always empty:β’ π₯= the maximum or the minimum
β’ π(π) = π(π β 1) + π(0) + Ξ(π)
β’ π(π) = Ξ(π2)
Randomized Algorithms
Summer 2018 Amo G. Tong 18
β’ π₯ decides the size of the two subarrays.β’ π π = π π + π π + Ξ π .
β’ If one of the subarray is always empty:β’ π₯= the maximum or the minimum
β’ π(π) = π(π β 1) + π(0) + Ξ(π)
β’ π(π) = Ξ(π2)
β’ If one of the subarray has a constant length π:β’ x= the a-th smallest element
β’ π(π) = π(π β 1 β π) + π(π) + Ξ(π) where 0 < π < π;
β’ π(π) =? ? (try it yourself)
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Summer 2018 Amo G. Tong 19
β’ If we always select the last element as the pivot:
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Summer 2018 Amo G. Tong 20
β’ If we always select the last element as the pivot:β’ Worst-case running time:
β’ π(π) = π(π β 1) + π(0) + Ξ(π)
β’ π(π) = π(π2)
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Summer 2018 Amo G. Tong 21
β’ If we always select the last element as the pivot:β’ Worst-case running time:
β’ π(π) = π(π β 1) + π(0) + Ξ(π)
β’ π(π) = π(π2)
β’ Expected running time:β’ Assume each of the possible permutations appears with the same
probability.
Randomized Algorithms
Summer 2018 Amo G. Tong 22
β’ If we always select the last element as the pivot:β’ Worst-case running time:
β’ π(π) = π(π β 1) + π(0) + Ξ(π)
β’ π(π) = π(π2)
β’ Expected running time:β’ Assume each of the possible permutations appears with the same
probability.
β’ With probability 1/n, the last element is the π-th smallest.
Randomized Algorithms
Summer 2018 Amo G. Tong 23
β’ If we always select the last element as the pivot:β’ Worst-case running time:
β’ π(π) = π(π β 1) + π(0) + Ξ(π)
β’ π(π) = π(π2)
β’ Expected running time:β’ Assume each of the possible permutations appears with the same
probability.
β’ With probability 1/n, the last element is the π-th smallest.
β’ If the last element is the π-th smallest, π π = T(π β 1) + π(π β π) + ππ
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Summer 2018 Amo G. Tong 24
β’ If we always select the last element as the pivot:β’ Worst-case running time:
β’ π(π) = π(π β 1) + π(0) + Ξ(π)
β’ π(π) = π(π2)
β’ Expected running time:β’ Assume each of the possible permutations appears with the same
probability.
β’ With probability 1/n, the last element is the π-th smallest.
β’ If the last element is the π-th smallest, π π = T(π β 1) + π(π β π) + ππ
β’ The expected value is
πΈ π π = Οπ=1π 1
π(πΈ[π π β 1 ] + πΈ[π π β π ] + ππ)
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Summer 2018 Amo G. Tong 25
Randomized Algorithms
Summer 2018 Amo G. Tong 26
πΈ π π = Οπ=1π 1
π(π(π β 1) + π(π β π) + ππ)β πΈ π π = O(n log n)
Proof: Let π π β€ π for π < 2.
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Summer 2018 Amo G. Tong 27
πΈ π π = Οπ=1π 1
π(π(π β 1) + π(π β π) + ππ)β πΈ π π = O(n log n)
Proof: Let π π β€ π for π < 2.Inductive hypothesis: πΈ π π β€ π π logπ
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Summer 2018 Amo G. Tong 28
πΈ π π = Οπ=1π 1
π(π(π β 1) + π(π β π) + ππ)β πΈ π π = O(n log n)
Proof: Let π π β€ π for π < 2.Inductive hypothesis: πΈ π π β€ π π logπ
Basic: πΈ[π(2)] =1
2(π(0) + π(1) + π β 2) +
1
2(π(1) + π(0) + π β 2) = 2π + 2π.
We need 2π + 2π < 2π.
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Summer 2018 Amo G. Tong 29
πΈ π π = Οπ=1π 1
π(π(π β 1) + π(π β π) + ππ)β πΈ π π = O(n log n)
Proof: Let π π β€ π for π < 2.Inductive hypothesis: πΈ π π β€ π π logπ
Basic: πΈ[π(2)] =1
2(π(0) + π(1) + π β 2) +
1
2(π(1) + π(0) + π β 2) = 2π + 2π.
We need 2π + 2π < 2π.Induction: Suppose πΈ π π β€ π π log π for all 2 < π < π.
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Summer 2018 Amo G. Tong 30
πΈ π π = Οπ=1π 1
π(π(π β 1) + π(π β π) + ππ)β πΈ π π = O(n log n)
Proof: Let π π β€ π for π < 2.Inductive hypothesis: πΈ π π β€ π π logπ
Basic: πΈ[π(2)] =1
2(π(0) + π(1) + π β 2) +
1
2(π(1) + π(0) + π β 2) = 2π + 2π.
We need 2π + 2π < 2π.Induction: Suppose πΈ π π β€ π π log π for all 2 < π < π.
π¬[π»(π)] = Οπ=ππ π
π(π¬[π» π β π ] + π¬[π» π β π ] + ππ)
Randomized Algorithms
Summer 2018 Amo G. Tong 31
πΈ π π = Οπ=1π 1
π(π(π β 1) + π(π β π) + ππ)β πΈ π π = O(n log n)
Proof: Let π π β€ π for π < 2.Inductive hypothesis: πΈ π π β€ π π logπ
Basic: πΈ[π(2)] =1
2(π(0) + π(1) + π β 2) +
1
2(π(1) + π(0) + π β 2) = 2π + 2π.
We need 2π + 2π < 2π.Induction: Suppose πΈ π π β€ π π log π for all 2 < π < π.
πΈ[π(π)] = Οπ=1π 1
π(πΈ[π π β 1 ] + πΈ[π π β π ] + ππ)
=π
π(πΟπ=π
πβππ¬[π» π ] + πππ)
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Summer 2018 Amo G. Tong 32
πΈ π π = Οπ=1π 1
π(π(π β 1) + π(π β π) + ππ)β πΈ π π = O(n log n)
Proof: Let π π β€ π for π < 2.Inductive hypothesis: πΈ π π β€ π π logπ
Basic: πΈ[π(2)] =1
2(π(0) + π(1) + π β 2) +
1
2(π(1) + π(0) + π β 2) = 2π + 2π.
We need 2π + 2π < 2π.Induction: Suppose πΈ π π β€ π π log π for all 2 < π < π.
πΈ[π(π)] = Οπ=1π 1
π(πΈ[π π β 1 ] + πΈ[π π β π ] + ππ)
=1
π(2Οπ=0
πβ1πΈ[π π ] + ππ2)
=π
π(πΟπ=π
πβππ¬[π» π ] + ππ»(π) + ππ»(π) + πππ)
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Summer 2018 Amo G. Tong 33
πΈ π π = Οπ=1π 1
π(π(π β 1) + π(π β π) + ππ)β πΈ π π = O(n log n)
Proof: Let π π β€ π for π < 2.Inductive hypothesis: πΈ π π β€ π π logπ
Basic: πΈ[π(2)] =1
2(π(0) + π(1) + π β 2) +
1
2(π(1) + π(0) + π β 2) = 2π + 2π.
We need 2π + 2π < 2π.Induction: Suppose πΈ π π β€ π π log π for all 2 < π < π.
πΈ[π(π)] = Οπ=1π 1
π(πΈ[π π β 1 ] + πΈ[π π β π ] + ππ)
=1
π(2Οπ=0
πβ1πΈ[π π ] + ππ2)
=1
π(2Οπ=2
πβ1πΈ[π π ] + 2π(0) + 2π(1) + ππ2)
β€π
π(πΟπ=π
πβππ π π₯π¨π π) + (ππ + πππ)/π
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Summer 2018 Amo G. Tong 34
πΈ π π = Οπ=1π 1
π(π(π β 1) + π(π β π) + ππ)β πΈ π π = O(n log n)
Proof: Let π π β€ π for π < 2.Inductive hypothesis: πΈ π π β€ π π logπ
Basic: πΈ[π(2)] =1
2(π(0) + π(1) + π β 2) +
1
2(π(1) + π(0) + π β 2) = 2π + 2π.
We need 2π + 2π < 2π.Induction: Suppose πΈ π π β€ π π log π for all 2 < π < π.
πΈ[π(π)] = Οπ=1π 1
π(πΈ[π π β 1 ] + πΈ[π π β π ] + ππ)
=1
π(2Οπ=0
πβ1πΈ[π π ] + ππ2)
=1
π(2Οπ=2
πβ1πΈ[π π ] + 2π(0) + 2π(1) + ππ2)
β€2π
π(Οπ=2
πβ1 π log π) + (4π + ππ2)/π
(see additional reading material)
β€2b
π(
π2
2log π β
π2
4+
1
4) + (4π + ππ2)/π
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Summer 2018 Amo G. Tong 35
We need this to be β€ π π log π.
πΈ π π = Οπ=1π 1
π(π(π β 1) + π(π β π) + ππ)β πΈ π π = O(n log n)
Proof: Let π π β€ π for π < 2.Inductive hypothesis: πΈ π π β€ π π logπ
Basic: πΈ[π(2)] =1
2(π(0) + π(1) + π β 2) +
1
2(π(1) + π(0) + π β 2) = 2π + 2π.
We need 2π + 2π < 2π.Induction: Suppose πΈ π π β€ π π log π for all 2 < π < π.
πΈ[π(π)] = Οπ=1π 1
π(πΈ[π π β 1 ] + πΈ[π π β π ] + ππ)
=1
π(2Οπ=0
πβ1πΈ[π π ] + ππ2)
=1
π(2Οπ=2
πβ1πΈ[π π ] + 2π(0) + 2π(1) + ππ2)
β€2π
π(Οπ=2
πβ1 π log π) + (4π + ππ2)/π
(see additional reading material)
β€2b
π(
π2
2log π β
π2
4+
1
4) + (4π + ππ2)/π
β’ If we always select the last element as the pivot:β’ Worst-case running time:
β’ π(π) = π(π β 1) + π(0) + Ξ(π)
β’ π(π) = π(π2)
β’ Expected running time:β’ Assume each of the possible permutations appears with the same
probability.
β’ The expected running time is
πΈ π π = Ξ(π log π)
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Summer 2018 Amo G. Tong 36
β’ If we always select the last element as the pivot:β’ Worst-case running time:
β’ π(π) = π(π β 1) + π(0) + Ξ(π)
β’ π(π) = π(π2)
β’ Expected running time:β’ Assume each of the possible permutations appears with the same
probability.
β’ The expected running time is
πΈ π π = Ξ(π log π)
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Summer 2018 Amo G. Tong 37
This is not always true.We want a Ξ(π log π)algorithm which does not rely on this assumption
β’ Randomized Quick Sort;β’ Select the pivot uniformly at random from the given array.
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Summer 2018 Amo G. Tong 38
β’ Randomized Quick Sort;β’ Select the pivot uniformly at random from the given array.
β’ With probability 1/π, π΄[π] is selected as the pivot.
β’ With probability 1/π, the π-th smallest element is selected.
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Summer 2018 Amo G. Tong 39
β’ Randomized Quick Sort;β’ Select the pivot uniformly at random from the given array.
β’ With probability 1/π, π΄[π] is selected as the pivot.
β’ With probability 1/π, the π-th smallest element is selected.
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Summer 2018 Amo G. Tong 40
RANDOM-PARTITION (π΄, π, π)1 randomly select an element x = π΄[π] in π΄[π. . π],
where each element has the same probability to be selected.2 exchange A[k] with A[r];3 π = π β 14 for π = π to π β 15 if π΄ π β€ π₯6 π = π + 17 exchange π΄[π] with π΄[π]8 end9 end10 exchange π΄[π + 1] with π΄[π]11 return π + 1;
β’ Randomized Quick Sort;β’ Select the pivot uniformly at random from the given array.
β’ With probability 1/π, π΄[π] is selected as the pivot.
β’ With probability 1/π, the π-th smallest element is selected.
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Summer 2018 Amo G. Tong 41
QUICKSORT (π΄, π, π)1 if π < π2 π =PARTITION(A, p, r)3 QUICKSORT (π΄, π, π β 1);4 QUICKSORT (π΄, π + 1, π);5 end
RANDOM-QUICKSORT (π΄, π, π)1 if π < π2 π =RANDOM-PARTITION(A, p, r)3 RANDOM-QUICKSORT (π΄, π, π β 1);4 RANDOM-QUICKSORT (π΄, π + 1, π);5 end
β’ Randomized Quick Sort;β’ Select the pivot uniformly at random from the given array.
β’ With probability 1/π, π΄[π] is selected as the pivot.
β’ With probability 1/π, the π-th smallest element is selected.
β’ Expected running time:β’ No assumption on the input.
β’ The expected running time is
πΈ π π = Οπ=1π 1
π(πΈ[π π β 1 ] + πΈ[π π β π ] + ππ)
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Summer 2018 Amo G. Tong 42
The same as that in page 24.
β’ Selection:
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Summer 2018 Amo G. Tong 43
Input: an array π΄[1,β¦ , π] of distinct numbers and an integer π, 1 β€ π β€ π.
Output: the π-th smallest element in π΄[1,β¦ , π] .
β’ Selection:
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Pseudocode:RANDOMIZED-SELECT(π΄, π, π, π)1 if π == π2 return A[p]3 end4 q=RANDOM-PARTITION(π΄, π, π)5 π = π β π + 1;6 if π == π7 return π΄[π]8 end9 if π < π10 return RANDOM-SELECT(π΄, π, π β 1, π)11 else12 return RANDOM-SELECT(π΄, π + 1, π, π β π)13 end
β’ Selection:
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Summer 2018 Amo G. Tong 45
Pseudocode:RANDOMIZED-SELECT(π΄, π, π, π)1 if π == π2 return A[p]3 end4 π =RANDOM-PARTITION(π΄, π, π)5 π = π β π + 1;6 if π == π7 return π΄[π]8 end9 if π < π10 return RANDOMIZED-SELECT(π΄, π, π β 1, π)11 else12 return RANDOMIZED-SELECT(π΄, π + 1, π, π β π)13 end
Ξ(π)
π(max(π β π, π β π))
T(π)
β’ Randomized selection;β’ Select the pivot uniformly at random from the given array.
β’ With probability1/π, π΄[π] is selected as the pivot.
β’ With probability1/π, the π-th smallest element is selected.
β’ Running time:β’ If the π-th smallest element is selected as the pivot, the running time is
π π = π max π β 1, π β π + Ξ(π)
β’ So the expected running time is:
πΈ π π =
π=0
πβ11
π(πΈ[π max π β 1, π β π ] + Ξ(π) )
Randomized Algorithms
Summer 2018 Amo G. Tong 46
Randomized Algorithms
Summer 2018 Amo G. Tong 47
πΈ π π
=
π=1
π1
π(πΈ[π max π β 1, π β π ] + Ξ(π) )
=
π=1
π/2 β11
π(πΈ[π max π β 1, π β π ] + Ξ(π) ) +
π= π/2 β1
π1
π(πΈ[π max π β 1, π β π ] + Ξ(π) )
=
π=1
π/2 β11
π(πΈ[π π β π ] + Ξ(π) ) +
π= π/2
π1
π(πΈ[π π β 1 ] + Ξ(π) )
=
π= π/2
π2
π(πΈ[π π β 1 ] + Ξ(π) )
πΈ π π =
π= π/2
π2
π(πΈ[π π β 1 ] + Ξ(π) )
Prove πΈ[π(π)] = π(π)
β’ Expected running time:β’ The algorithm is not randomized. The expectation is taken over the
distribution of the inputs.
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Summer 2018 Amo G. Tong 48
β’ Expected running time:β’ The algorithm is not randomized. The expectation is taken over the
distribution of the inputs.
β’ The algorithm is randomized. The expected running time is πΆ(π(π)) for any given input.
Randomized Algorithms
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β’ Expected running time:β’ The algorithm is not randomized. The expectation is taken over the
distribution of the inputs.
β’ The algorithm is randomized. The expected running time is π(π(π)) for any given input.
β’ The algorithm is randomized and the input is also sampled.
Randomized Algorithms
Summer 2018 Amo G. Tong 50
β’ Expected running time:β’ The algorithm is not randomized. The expectation is taken over the
distribution of the inputs.
β’ The algorithm is randomized. The expected running time is π(π(π)) for any given input.
β’ The algorithm is randomized and the input is also sampled.β’ Consider the Randomized Quick Sort with an input uniformly sampled
from {all the permutations over {1, β¦ , π}}.
β’ What is the running time? (exercise)
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