Outline
• Applications of Median
Bucket Sort,
Applications of Median
Recitation 4: Bucket Sort,
Applications of Median
– Median Partitioned Quicksort
• Bucket Sort
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Recitation 4: Bucket Sort,
Applications of Median
Median Partitioned Quicksort
Bucket Sort
• Quicksort has average running time Θ(n lg n) but
a worst case running time Θ(n2).
• Can we improve the worst case?
• Linear time sorts:
– Counting sort : Bound on the size of the
numbers known
– Radix sort : Bound on the number of bits
known
– partition around the median!!
• Finding the median runs in Θ(n) worst case!
– Recursively sort both halves.
• T(n) = 2T(n/2) + Θ(n)
= Θ(n lg n) by Master Theorem case 2.
• Theoretically interesting but not used much in
practice - constants in Θ(n) for finding median is
large.
Recitation 4: Bucket Sort,
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• Linear time sorting of real number
– Bucket sort: Expected run time is Θ(n) when
the input is generated independently by a
distribution over [0,1).
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Recitation 4: Bucket Sort,
Applications of Median
Applications of Median
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Bucket Sort
BUCKET − SORT(A)
1
n ← length[A]
2
for i ← 1 to n
• Partition [0,1) into n equal-sized intervals
(buckets).
• Distribute the inputs into the buckets.
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4
5
6
– A linked list normally used
• Sort each bucket using insertion sort
• Go through the buckets in order, listing the
elements.
Recitation 4: Bucket Sort,
Applications of Median
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do insert A[i] into B[ ⎣nA[i]⎦
for i ← 0 to n - 1
do sort list B[i] with insertion sort
concatenate the lists B[0], B[1], ..., B[n - 1] together in order
Recitation 4: Bucket Sort,
Applications of Median
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1
Running time
Analysis
n −1
T(n) = Θ(n) + ∑ O(n 2 )
• Worst case running time
i =0
– All elements fall into the same bucket
– O(n2)
n −1
E[T(n)] = E[Θ(n) + ∑ O(n i2 )] (linearity of expectation)
i =0
• Average case running time
n −1
= Θ(n) + ∑ O(E[n i2 ])
– Θ(n)
i =0
• We claim that E[ni2] = 2 - 1/n
• So E[T(n)] = Θ(n)
Recitation 4: Bucket Sort,
Applications of Median
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Recitation 4: Bucket Sort,
Applications of Median
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• Let
⎧1
X ij = ⎨
⎩0
n
if A[j] falls in bucket i
E[ni2 ] = E[(∑ X ij ) 2 ]
otherwise
j =1
⎡ n n
⎤
= E ⎢∑∑ X ij X ik ⎥
⎣ j =1 k =1
⎦
⎡ n
⎤
= E ⎢∑ X ij2 + ∑ ∑ X ij X ik ⎥
⎢ j =1
⎥
1≤ j ≤ n 1≤ k ≤ n
k≠ j
⎣⎢
⎦⎥
for i = 0,1,...,n-1 and j = 1,2,...,n (indicator random
variable).
• We have
n
ni = ∑ X ij
j =1
n
= ∑ E[ X ij2 ] +
j =1
Recitation 4: Bucket Sort,
Applications of Median
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X ik ]
Recitation 4: Bucket Sort,
Applications of Median
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k≠ j
1
1
= n. + n(n − 1). 2
n
n
n −1
= 1+
n
1
= 2−
n
• When k≠j, Xij and Xik are independent
E[ X ij X ik ] = E[ X ij ]E[ X ik ]
11 1
=
n n n2
Recitation 4: Bucket Sort,
Applications of Median
ij
n
1
1
E[ ni2 ] = ∑ + ∑ ∑ 2
j =1 n
1≤ j ≤ n 1≤ k ≤ n n
1
1
1
E[ X ij2 ] = 1. + 0.(1 − ) =
n
n
n
=
∑ ∑ E[ X
1≤ j ≤ n 1≤ k ≤ n
k≠ j
• Even if input distribution is not uniform,
expected running time is still linear if
E[ni2]=O(1).
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Recitation 4: Bucket Sort,
Applications of Median
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2
Exercise
Summary
A ship arrives at a port, and the 40 sailors on
board go ashore for revelry. When they
return, each chooses a random cabin in their
state of drunkenness. What is the expected
number of sailors in their own cabins?
Recitation 4: Bucket Sort,
Applications of Median
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• Application of median:
– Good worst case theoretical bound for median
partitioned quicksort
• Bucket sort:
– Linear time sorting of real number
– Only for expected run time
– Only for certain distributions of inputs
Recitation 4: Bucket Sort,
Applications of Median
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