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ICS 353
Design and Analysis of Algorithms
Spring Semester 2006 - 2007 (062)
King Fahd University of Petroleum & Minerals
Information & Computer Science Department
1
Basic Concepts in Algorithmic
Analysis
• Topics
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Introduction
Time Complexity
Space Complexity
Optimal Algorithms
How to estimate the running time of an algorithm
Worst Case Analysis and Average Case Analysis
Amortized Analysis
Input Size and Problem Instance
• Reading Assignment
– All Chapter 1 from the textbook
• In particular, sections 1-3,6,8-14 will be discussed in class.
2
What is an algorithm?
• An algorithm is defined as a finite set of
steps, each of which may require one or
more operations and if carried out on a set
of inputs, will produce one or more
outputs after a finite amount of time.
• Examples of Algorithms
• Examples of computations that are not
algorithms
3
Properties of Algorithms
• Definiteness: It must be clear what should be done.
• Effectiveness: Each step must be such that it can, at
least in principle, be carried out by a person using
pencil and paper in a finite amount of time. E.g.
integer arithmetic.
• An algorithm produces one or more outputs and may
have zero or more externally supplied inputs.
• Finiteness: Algorithms should terminate after a finite
number of operations.
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Our Objective
• Find the most efficient algorithm for solving
a particular problem.
• In order to achieve the objective, we need to
determine:
– How can we find such algorithm?
– What does it mean to be an efficient algorithm?
– How can one tell that it is more efficient than
other algorithms?
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In the First Chapter
• We will answer the following two questions
– What does it mean to be an efficient algorithm?
– How can one tell that it is more efficient than
other algorithms?
based on some easy-to-understand searching
and sorting algorithms that we may have seen
earlier.
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Searching Problem
• Assume A is an array with n elements A[1], A[2],
… A[n]. For a given element x, we must
determine whether there is an index j; 1 ≤ j ≤ n,
such that x = A[j]
• Two algorithms, among others, address this
problem
– Linear Search
– Binary Search
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Linear Search Algorithm
Algorithm: LINEARSEARCH
Input: array A[1..n] of n elements and an element x.
Output: j if x = A[j], 1 ≤ j ≤ n, and 0 otherwise.
1.
2.
3.
4.
5.
j  1
while (j < n) and (x A[j])
j  j + 1
end while
if x = A[j] then return j else return 0
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Analyzing Linear Search
• One way to measure efficiency is to count how many
statements get executed before the algorithm terminates
• One should keep an eye, though, on statements that are
executed “repeatedly”.
• What will be the number of “element” comparisons if x
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–
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First appears in the first element of A
First appears in the middle element of A
First appears in the last element of A
Doesn’t appear in A.
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Binary Search
• We can do “better” than linear search if we knew
that the elements of A are sorted, say in nondecreasing order.
• The idea is that you can compare x to the middle
element of A, say A[middle].
– If x < A[middle] then you know that x cannot be an
element from A[middle+1], A[middle+2], …A[n].
Why?
– If x > A[middle] then you know that x cannot be an
element from A[1], A[2], …A[middle-1]. Why?
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Binary Search Algorithm
Algorithm: BINARYSEARCH
Input: An array A[1..n] of n elements sorted in
nondecreasing order and an element x.
Output: j if x = A[j], 1 ≤ j ≤ n, and 0 otherwise.
1.
2.
3.
4.
5.
6.
7.
8.
low  1; high  n; j  0
while (low ≤ high) and (j = 0)
mid  (low + high)/2
if x = A[mid] then j  mid
else if x < A[mid] then high  mid - 1
else low  mid + 1
end while
return j
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Worst Case Analysis of Binary Search
• What to do: Find the maximum number of element
comparisons
• How to do:
– The number of “element” comparisons is equal to the number
of iterations of the while loop in steps 2-7. HOW?
– How many elements of the input do we have in the
•
•
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•
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First iteration
Second iteration
Third iteration
…
ith iteration
– The last iteration occurs when the size of input we have =
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Theorem
• The number of comparisons performed by
Algorithm BINARYSEARCH on a sorted
array of size n is at most log n  1
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Insertion Sort
Algorithm: INSERTIONSORT
Input: An array A[1..n] of n elements.
Output: A[1..n] sorted in nondecreasing order.
1. for i  2 to n
2.
x  A[i]
3.
j  i - 1
4.
while (j > 0) and (A[j] > x)
5.
A[j + 1]  A[j]
6.
j  j - 1
7.
end while
8.
A[j + 1]  x
9. end for
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Insertion Sort Example
5
2
9
8
4
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Insertion Sort Example
x=2
5
x=9
2 5 9 8 4
x=8
2 5
9
8
4
x=4
2 5
8
9
4
2 4
5
8
9
2
9
8
4
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Analyzing Insertion Sort
• The minimum number of element
comparisons is
which occurs when
• The maximum number of element
comparisons is
which occurs when
• The number of element assignments is
17
Time Complexity
• One way of measuring the performance of
an algorithm is how fast it executes. The
question is how to measure this “time”?
– Is having a digital stop watch suitable?
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Order of Growth
• As measuring time is subjective to many factors, we
look for a more “objective” measure, i.e. the number
of operations
• Since counting the exact number of operations is
cumbersome, sometimes impossible, we can always
focus our attention to asymptotic analysis, where
constants and lower-order terms are ignored.
– E.g. n3, 1000n3, and 10n3+10000n2+5n-1 are all “the same”
– The reason we can do this is that we are always interested in
comparing different algorithms for arbitrary large number of
inputs.
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Example
Growth rate for some function
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Example
Growth rate for same previous functions
showing larger input sizes
21
Running Times for Different Sizes of
Inputs of Different Functions
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Asymptotic Analysis: Big-oh (O())
• Definition: For T(n) a non-negatively valued
function, T(n) is in the set O(f(n)) if there exist
two positive constants c and n0 such that T(n) 
cf(n) for all n > n0.
• Usage: The algorithm is in O(n2) in [best, average,
worst] case.
• Meaning: For all data sets big enough (i.e., n>n0),
the algorithm always executes in less than or equal
to cf(n) steps in [best, average, worst] case.
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Big O()
• O() notation indicates an upper bound.
• Usually, we look for the tightest upper
bound:
– while T(n) = 3n2 is in O(n3), we prefer
O(n2).
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Big O() Examples
• Example 1: Find c and n0 to show that
T(n) = (n+2)/2 is in O(n)
• Example 2: Find c and n0 to show that
T(n)=c1n2+c2n is in O(n2)
• Example 3: T(n) = c. We say this is in O(1).
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Asymptotic Analysis: Big-Omega (())
• Definition: For T(n) a non-negatively valued
function, T(n) is in the set (g(n)) if there exist
two positive constants c and n0 such that T(n) >=
cg(n) for all n > n0.
• Meaning: For all data sets big enough (i.e., n > n0),
the algorithm always executes in more than or
equal to cg(n) steps.
• () notation indicates a lower bound.
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() Example
• Find c and n0 to show that T(n) = c1n2 + c2n
is in (n2) .
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Asymptotic Analysis: Big Theta (())
• When O() and () meet, we indicate this by
using () (big-Theta) notation.
• Definition: An algorithm is said to be (h(n))
if it is in O(h(n)) and it is in (h(n)).
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Example
• Show that log(n!) is in (n log n).
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Complexity Classes and small-oh (o())
• Using () notation, one can divide the functions into
different equivalence classes, where f(n) and g(n)
belong to the same equivalence class if f(n) = (g(n))
• To show that two functions belong to different
equivalence classes, the small-oh notation has been
introduced
• Definition: Let f(n) and g(n) be two functions from
the set of natural numbers to the set of non-negative
real numbers. f(n) is said to be in o(g(n)) if for every
constant c > 0, there is a positive integer n0 such that
f(n) < cg(n) for all n  n0.
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Simplifying Rules
• If f(n) is in O(g(n)) and g(n) is in O(h(n)), then
f(n) is in O(h(n))
• If f(n) is in O(kg(n)) for any constant k > 0, then
f(n) is in ………
• If f1(n) is in O(g1(n)) and f2(n) is in O(g2(n)),
then (f1 + f2)(n) is in ………
• If f1(n) is in O(g1(n)) and f2(n) is in O(g2(n))
then f1(n)f2(n) is in ………
• You can safely “globally” replace O with  or  in the
above, where the above rules will still hold.
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Very Useful Simplifying Rule
• Let f(n) and g(n) be be two functions from the set
of natural numbers to the set of non-negative real
numbers such that:
f ( n)
0  L  lim

n  g ( n)
Then
if L <  then f(n) is in
if L > 0 then f(n) is in
if 0 < L <  then f(n) is in
if L = 0 then f(n) is in
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Space Complexity
• Space complexity refers to the number of memory
cells needed to carry out the computational steps
required in an algorithm excluding memory cells
needed to hold the input.
• Compare additional space needed to carry out
SELECTIONSORT to that of BOTTOMUPSORT
if we have an array with 2 million elements!
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Examples
• What is the space complexity for
–
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Linear search
Binary search
Selection sort
Insertion sort
Merge (that merges two sorted lists)
Bottom up merge sort
34
Optimal Algorithms
• If one can show that there is no algorithm
that solves a certain problem in
asymptotically less than that of a certain
algorithm A, we call A an optimal
algorithm.
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Optimal Algorithms: Example
A decision tree for sorting three elements
Figure 12.1 page 338 from the textbook
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Optimal Algorithms: Example
• Consider the sorting problem of n distinct elements
using element comparison-based sorting
– Using the decision tree model, the number of possible
solutions (leaf nodes) in the binary tree is equal to
................
– You have learnt earlier that a binary tree with n nodes has
height of at least  log n  (Observation 3.3 page 111 of
the textbook)
– Hence, the length of the longest path in a decision tree for
sorting n distinct elements is at least .............
– Therefore,
•
•
•
•
Insertion sort is
Selection sort is
Merge sort is
Quick sort is
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Estimating the Running Time of an Algorithm
• As mentioned earlier, we need to focus on
counting those operations which represent, in
general, the behavior of the algorithm
• This is achieved by
– Counting the frequency of basic operations.
• Basic operation is an operation with highest frequency
to within a constant factor among all other elementary
operations
– Recurrence Relations
38
Counting the Frequency of Basic
Operations
• Sometimes, it is easier to compute the frequency
of an operation that is a good representative of
the overall time complexity of the algorithm
– For example, Algorithm MERGE.
• Counting the number of iterations
– The number of iterations in a while loop and/or a for
loop is a good indication of the total number of
operations
39
Example 1
sum = 0;
for (j=1; j<=n; j++)
for (i=1; i<=j; i++)
sum++;
for (k=0; k<n; k++)
A[k] = k;
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Example 2
for j := 1 to n do
sum[j] := 0;
for i := 1 to j2 do
sum[j] := sum[j] + i;
end for;
end for;
return sum[1..n];
41
Example 3
count := 0;
for i := 1 to
m :=  n/i
for j := 1
count :=
end for;
end for;
return count;
n do

to m do
count + 1 ;
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Example 4
count := 0;
while n >= 1 do
for j := 1 to n do
execute_algorithm_x;
count := count + 1;
end for
n := n / 2;
end while
return count;
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Examples 5 & 6
sum1 = 0;
for (k=1; k<=n; k*=2)
for (j=1; j<=n; j++)
sum1++;
sum2 = 0;
for (k=1; k<=n; k*=2)
for (j=1; j<=k; j++)
sum2++;
44
Example 7
count := 0;
for i := 1 to n do
j := 2;
while j <= n do
j := j2;
count := count + 1;
end while
end for;
return count;
45
Recurrence Relations
• The number of operations can be
represented as a recurrence relation.
• There are very well known techniques,
other than expanding the recurrence
relation, which we will study in order to
solve these recurrences
46
Example
• Recursive Merge Sort
MergeSort(A,p,r)
if p < r then
q := (p+r)/2;
MergeSort(A,p,q);
MergeSort(A,q+1,r);
Merge(A,p,q,r);
end if;
– What is the call to sort an array with n elements?
– Let us assume that the overall cost of sorting n elements is T(n),
assuming that n is a power of two.
•
•
•
•
If n = 1, do we know T(n)?
What is the cost of MergeSort(A,p,q)?
What is the cost of MergeSort(A,q+1,r)?
What is the cost of Merge(A,p,q,r)?
47
Worst Case Analysis
• In worst case analysis of time complexity
we select the maximum cost among all
possible inputs of size n.
– One can do that for the() notation as well as
the O() notation.
– However, it is better use it with the ()
notation.
• Why?
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Average Case Analysis
• Probabilities of all inputs is an important
piece of prior knowledge in order to compute
the number of operations on average
• Usually, average case
 analysis is lengthy and
complicated, even with simplifying
assumptions.
k
i 1
49
Computing the Average Running Time
• The running time in this case is taken to be
the average time over all inputs of size n.
– Assume we have k inputs, where each input
costs Ci operations, and each input can occur
with probability Pi, 1  i  k, the average
running time is given by
k
 PC
i 1
i
i
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Average Case Analysis of Linear Search
• Assume that the probability that key x appears
in any position in the array (1, 2, …, n) or does
not appear in the array is equally likely
– This means that we have a total of ……… different
inputs, each with probability ………
– What is the number of comparisons for each input?
– Therefore, the average running time of linear
search = ………
51
Average Case Analysis of Insertion Sort
• Assume that array A contains the numbers from
1..n ( i.e. elements are distinct)
• Assume that all n! permutations of the input are
equally likely.
• What is the number of comparisons for inserting
A[i] in its proper position in A[1..i]? What about
on average?
• Therefore, the total number of comparisons on
average is
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