Transcript Analysis of Algorithms

Sorting and Selection
1, c
B 
3, a

3, b
  
7, d
7, g
7, e
 
0 1 2 3 4 5 6 7 8 9
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Lower Bounds
Lower bound: an estimate on a minimum amount
of work needed to solve a given problem
Examples:
number of comparisons needed to find the
largest element in a set of n numbers
number of comparisons needed to sort an array
of size n
number of comparisons necessary for
searching in a sorted array
Lower Bounds (cont.)
Lower bound can be


an exact count
an efficiency class ()
Tight lower bound: there exists an algorithm
with the same efficiency as the lower bound
Problem
sorting (comparison-based)
searching in a sorted array
element uniqueness
n-digit integer multiplication
multiplication of n-by-n matrices
Lower bound Tightness
(nlog n)
(log n)
(nlog n)
(n)
(n2)
yes
yes
yes
unknown
unknown
Decision Trees
Decision tree — a convenient model of algorithms involving
comparisons in which:
 internal nodes represent comparisons
 leaves represent outcomes (or input cases)
Decision tree for 3-element insertion sort
abc
yes
a<b
no
abc
yes
bac
no
b< c
yes
acb
a<b<c
yes
a<c<b
a<c
no
a<c
bca
b<a<c
no
c<a<b
yes
b<c<a
b<c
no
c<b<a
Decision Trees and Sorting Algorithms
Any comparison-based sorting algorithm can be
represented by a decision tree (for each fixed n)
Number of leaves (outcomes)  n!
Height of binary tree with n! leaves  log2n!
Minimum number of comparisons in the worst case 
log2n! for any comparison-based sorting algorithm, since
the longest path represents the worst case and its length
is the height
log2n!  n log2n (by Sterling approximation)
This lower bound is tight (mergesort or heapsort)
Ex. Prove that 5 (or 7) comparisons are necessary and sufficient for
sorting 4 keys (or 5 keys, respectively).
Bucket-Sort
Let be S be a sequence of n
(key, element) items with keys
in the range [0, N - 1]
Bucket-sort uses the keys as
indices into an auxiliary array B
of sequences (buckets)
Phase 1: Empty sequence S by
moving each item (k, o) into its
bucket B[k]
Phase 2: For i = 0, …, N - 1, move
the items of bucket B[i] to the
end of sequence S
Analysis:


Phase 1 takes O(n) time
Phase 2 takes O(n + N) time
Bucket-sort takes O(n + N) time
Algorithm bucketSort(S, N)
Input sequence S of (key, element)
items with keys in the range
[0, N - 1]
Output sequence S sorted by
increasing keys
B  array of N empty sequences
while S.isEmpty()
f  S.first()
(k, o)  S.remove(f)
B[k].insertLast((k, o))
for i  0 to N - 1
while B[i].isEmpty()
f  B[i].first()
(k, o)  B[i].remove(f)
S.insertLast((k, o))
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Example
Key range [0, 9]
7, d
1, c
3, a
7, g
3, b
7, e
Phase 1
1, c
B
3, a

0

1
2
3
3, b



4
5
6
7
7, d


8
9
7, g
7, e
Phase 2
1, c
3, a
3, b
7, d
7, g
7, e
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Properties and Extensions
Key-type Property


The keys are used as
indices into an array
and cannot be arbitrary
objects
No external comparator
Stable Sort Property

The relative order of
any two items with the
same key is preserved
after the execution of
the algorithm
Extensions

Integer keys in the range [a, b]
 Put item (k, o) into bucket
B[k - a]

String keys from a set D of
possible strings, where D has
constant size (e.g., names of
the 50 U.S. states)
 Sort D and compute the rank
r(k) of each string k of D in
the sorted sequence
 Put item (k, o) into bucket
B[r(k)]
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Lexicographic Order
A d-tuple is a sequence of d keys (k1, k2, …, kd), where
key ki is said to be the i-th dimension of the tuple
Example:

The Cartesian coordinates of a point in space are a 3-tuple
The lexicographic order of two d-tuples is recursively
defined as follows
(x1, x2, …, xd) < (y1, y2, …, yd)

x1 < y1  x1 = y1  (x2, …, xd) < (y2, …, yd)
I.e., the tuples are compared by the first dimension,
then by the second dimension, etc.
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Lexicographic-Sort
Let Ci be the comparator
that compares two tuples by
their i-th dimension
Let stableSort(S, C) be a
stable sorting algorithm that
uses comparator C
Lexicographic-sort sorts a
sequence of d-tuples in
lexicographic order by
executing d times algorithm
stableSort, one per
dimension
Lexicographic-sort runs in
O(dT(n)) time, where T(n) is
the running time of
stableSort
Algorithm lexicographicSort(S)
Input sequence S of d-tuples
Output sequence S sorted in
lexicographic order
for i  d downto 1
stableSort(S, Ci)
Example:
(7,4,6) (5,1,5) (2,4,6) (2, 1, 4) (3, 2, 4)
(2, 1, 4) (3, 2, 4) (5,1,5) (7,4,6) (2,4,6)
(2, 1, 4) (5,1,5) (3, 2, 4) (7,4,6) (2,4,6)
(2, 1, 4) (2,4,6) (3, 2, 4) (5,1,5) (7,4,6)
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Radix-Sort
Radix-sort is a
specialization of
lexicographic-sort that
uses bucket-sort as the
stable sorting algorithm
in each dimension
Radix-sort is applicable
to tuples where the
keys in each dimension i
are integers in the
range [0, N - 1]
Radix-sort runs in time
O(d( n + N))
Algorithm radixSort(S, N)
Input sequence S of d-tuples such
that (0, …, 0)  (x1, …, xd) and
(x1, …, xd)  (N - 1, …, N - 1)
for each tuple (x1, …, xd) in S
Output sequence S sorted in
lexicographic order
for i  d downto 1
bucketSort(S, N)
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Radix-Sort for
Binary Numbers
Consider a sequence of n
b-bit integers
x = xb - 1 … x1x0
We represent each element
as a b-tuple of integers in
the range [0, 1] and apply
radix-sort with N = 2
This application of the
radix-sort algorithm runs in
O(bn) time
For example, we can sort a
sequence of 32-bit integers
in linear time
Algorithm binaryRadixSort(S)
Input sequence S of b-bit
integers
Output sequence S sorted
replace each element x
of S with the item (0, x)
for i  0 to b - 1
replace the key k of
each item (k, x) of S
with bit xi of x
bucketSort(S, 2)
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Example
Sorting a sequence of 4-bit integers
1001
0010
1001
1001
0001
0010
1110
1101
0001
0010
1101
1001
0001
0010
1001
0001
1101
0010
1101
1101
1110
0001
1110
1110
1110
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Order Statistics
The ith order statistic in a set of n elements is
the ith smallest element
The minimum is thus the 1st order statistic
The maximum is (duh) the nth order statistic
The median is the n/2 order statistic

If n is even, there are 2 medians
Could calculate order statistics by sorting
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Time: O(n lg n) w/ comparison sort
We can do better
Selection Problem
The selection problem: find the ith
smallest element of a set
Two algorithms:
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A practical randomized algorithm with O(n)
expected running time
A cool algorithm of theoretical interest only
with O(n) worst-case running time
Randomized Selection
Key idea: use partition() from quicksort


But, only need to examine one subarray
This saving shows up in running time: O(n)
 A[q]
p
 A[q]
q
r
Randomized Selection
RandomizedSelect(A, p, r, i)
if (p == r) then return A[p];
q = RandomizedPartition(A, p, r)
k = q - p + 1;
if (i == k) then return A[q];
if (i < k) then
return RandomizedSelect(A, p, q-1, i);
else
return RandomizedSelect(A, q+1, r, i-k);
k
 A[q]
p
 A[q]
q
r
Review: Randomized Selection
Average case

For upper bound, assume ith element always falls
in larger side of partition:
T n  


1 n -1
T max k , n - k - 1 + n 

n k =0
2 n -1
T k  + n 

n k =n / 2
We then showed that T(n) = O(n) by substitution
Linear-Time Median Selection
Given a “black box” O(n) median algorithm,
what can we do?
 ith order statistic:
 Find median x
 Partition input around x
 if (i  (n+1)/2) recursively find ith element of
first half
 else find (i - (n+1)/2)th element in second half
 T(n) = T(n/2) + O(n) = O(n)

Can you think of an application to sorting?
Linear-Time Median Selection
Worst-case O(n lg n) quicksort
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
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Find median x and partition around it
Recursively quicksort two halves
T(n) = 2T(n/2) + O(n) = O(n lg n)