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ALGORITHMS
Dr. Yasir Ali
College of E&ME,
National University of Sciences and Technology,
Islamabad, Pakistan.
December 16, 2015
ALGORITHM
An algorithm is a finite sequence of precise instructions for
performing a computation or for solving a problem.
ALGORITHM
An algorithm is a finite sequence of precise instructions for
performing a computation or for solving a problem.
ALGORISM ABU JA’FAR MOHAMMED IBN MUSA
AL-KHOWARIZMI (C. 780 - C. 850)
ALGORITHM
An algorithm is a finite sequence of precise instructions for
performing a computation or for solving a problem.
ALGORISM ABU JA’FAR MOHAMMED IBN MUSA
AL-KHOWARIZMI (C. 780 - C. 850)
Algorism was used for the rules for performing arithmetic using
decimal notation.
ALGORITHM
An algorithm is a finite sequence of precise instructions for
performing a computation or for solving a problem.
ALGORISM ABU JA’FAR MOHAMMED IBN MUSA
AL-KHOWARIZMI (C. 780 - C. 850)
Algorism was used for the rules for performing arithmetic using
decimal notation.
Algorism evolved into the word algorithm by the eighteenth
century. With the growing interest in computing machines, the
concept of an algorithm was given a more general meaning, to
include all definite procedures for solving problems, not just the
procedures for performing arithmetic.
Describe an algorithm for finding the maximum (largest)
value in a finite sequence of integers.
1. Set the temporary maximum equal to the first integer in the
sequence.
Describe an algorithm for finding the maximum (largest)
value in a finite sequence of integers.
1. Set the temporary maximum equal to the first integer in the
sequence. (The temporary maximum will be the largest
integer examined at any stage of the procedure.)
Describe an algorithm for finding the maximum (largest)
value in a finite sequence of integers.
1. Set the temporary maximum equal to the first integer in the
sequence. (The temporary maximum will be the largest
integer examined at any stage of the procedure.)
2. Compare the next integer in the sequence to the temporary
maximum, and if it is larger than the temporary maximum,
set the temporary maximum equal to this integer.
Describe an algorithm for finding the maximum (largest)
value in a finite sequence of integers.
1. Set the temporary maximum equal to the first integer in the
sequence. (The temporary maximum will be the largest
integer examined at any stage of the procedure.)
2. Compare the next integer in the sequence to the temporary
maximum, and if it is larger than the temporary maximum,
set the temporary maximum equal to this integer.
3. Repeat the previous step if there are more integers in the
sequence.
Describe an algorithm for finding the maximum (largest)
value in a finite sequence of integers.
1. Set the temporary maximum equal to the first integer in the
sequence. (The temporary maximum will be the largest
integer examined at any stage of the procedure.)
2. Compare the next integer in the sequence to the temporary
maximum, and if it is larger than the temporary maximum,
set the temporary maximum equal to this integer.
3. Repeat the previous step if there are more integers in the
sequence.
4. Stop when there are no integers left in the sequence. The
temporary maximum at this point is the largest integer in the
sequence.
Finding the Maximum Element in a Finite Sequence.
procedure max(a1 , a2 , ..., an : integers)
max := a1
for i := 2 to n
if max < ai then max := ai
return max{max is the largest element}
PROPERTIES OF ALGORITHMS
Input. An algorithm has input values from a specified set.
Output. From each set of input values an algorithm produces
output values from a specified set. The output values
are the solution to the problem.
Definiteness. The steps of an algorithm must be defined precisely.
Correctness. An algorithm should produce the correct output
values for each set of input values.
Finiteness. An algorithm should produce the desired output after
a finite (but perhaps large) number of steps for any
input in the set.
Effectiveness. It must be possible to perform each step of an
algorithm exactly and in a finite amount of time.
Generality. The procedure should be applicable for all problems
of the desired form, not just for a particular set of
input values
Searching Algorithms
The general searching problem can be described as follows:
Searching Algorithms
The general searching problem can be described as follows:
Locate an element x in a list of distinct elements a1 , a2 , · · · , an , or
determine that it is not in the list.
Searching Algorithms
The general searching problem can be described as follows:
Locate an element x in a list of distinct elements a1 , a2 , · · · , an , or
determine that it is not in the list.
The solution to this search problem is
Searching Algorithms
The general searching problem can be described as follows:
Locate an element x in a list of distinct elements a1 , a2 , · · · , an , or
determine that it is not in the list.
The solution to this search problem is
the location of the term in the list that equals x (that is, i is the
solution if x = ai ) and is 0 if x is not in the list.
THE LINEAR SEARCH
The linear search, or sequential search algorithm.
The linear search algorithm begins by comparing x and a1 . When
x = a1 , the solution is the location of a1 , namely, 1.
THE LINEAR SEARCH
The linear search, or sequential search algorithm.
The linear search algorithm begins by comparing x and a1 . When
x = a1 , the solution is the location of a1 , namely, 1.
When x 6= a1 , compare x with a2 .
THE LINEAR SEARCH
The linear search, or sequential search algorithm.
The linear search algorithm begins by comparing x and a1 . When
x = a1 , the solution is the location of a1 , namely, 1.
When x 6= a1 , compare x with a2 . If x = a2 , the solution is the
location of a2 , namely, 2.
THE LINEAR SEARCH
The linear search, or sequential search algorithm.
The linear search algorithm begins by comparing x and a1 . When
x = a1 , the solution is the location of a1 , namely, 1.
When x 6= a1 , compare x with a2 . If x = a2 , the solution is the
location of a2 , namely, 2.
When x 6= a2 , compare x with a3 ....
THE LINEAR SEARCH
The linear search, or sequential search algorithm.
The linear search algorithm begins by comparing x and a1 . When
x = a1 , the solution is the location of a1 , namely, 1.
When x 6= a1 , compare x with a2 . If x = a2 , the solution is the
location of a2 , namely, 2.
When x 6= a2 , compare x with a3 ....
Continue this process, comparing x successively with each term of
the list until a match is found, where the solution is the location of
that term, unless no match occurs. If the entire list has been
searched without locating x, the solution is 0.
THE LINEAR SEARCH
The linear search, or sequential search algorithm.
The linear search algorithm begins by comparing x and a1 . When
x = a1 , the solution is the location of a1 , namely, 1.
When x 6= a1 , compare x with a2 . If x = a2 , the solution is the
location of a2 , namely, 2.
When x 6= a2 , compare x with a3 ....
Continue this process, comparing x successively with each term of
the list until a match is found, where the solution is the location of
that term, unless no match occurs. If the entire list has been
searched without locating x, the solution is 0.
The pseudo code for the linear search algorithm ?
ALGORITHM - The Linear Search Algorithm.
procedure linear search(x : integer , a1 , a2 , ..., an : distinctintegers)
i := 1
while (i ≤ n and x 6= ai )
i := i + 1
if i ≤ n then location := i
else location := 0
return
location {location is the subscript of the term that equals x ,
or is 0 if x is not found}
THE BINARY SEARCH
This algorithm can be used when the list has terms occurring in
order of increasing size
It proceeds by comparing the element to be located to the middle
term of the list.
THE BINARY SEARCH
This algorithm can be used when the list has terms occurring in
order of increasing size
It proceeds by comparing the element to be located to the middle
term of the list.
The list is then split into two smaller sublists of the same size, or
where one of these smaller lists has one fewer term than the other.
THE BINARY SEARCH
This algorithm can be used when the list has terms occurring in
order of increasing size
It proceeds by comparing the element to be located to the middle
term of the list.
The list is then split into two smaller sublists of the same size, or
where one of these smaller lists has one fewer term than the other.
The search continues by restricting the search to the appropriate
sublist based on the comparison of the element to be located and
the middle term.
THE BINARY SEARCH
This algorithm can be used when the list has terms occurring in
order of increasing size
It proceeds by comparing the element to be located to the middle
term of the list.
The list is then split into two smaller sublists of the same size, or
where one of these smaller lists has one fewer term than the other.
The search continues by restricting the search to the appropriate
sublist based on the comparison of the element to be located and
the middle term.
To search for 19 in the list
1 2 3 5 6 7 8 10 12 13 15 16 18 19 20 22,
First split this list, which has 16 terms, into two smaller lists with
eight terms each, namely,
1 2 3 5 6 7 8 10
12 13 15 16 18 19 20 22.
Then, compare 19 and the largest term in the first list.
First split this list, which has 16 terms, into two smaller lists with
eight terms each, namely,
1 2 3 5 6 7 8 10
12 13 15 16 18 19 20 22.
Then, compare 19 and the largest term in the first list.
10 < 19,
Because
First split this list, which has 16 terms, into two smaller lists with
eight terms each, namely,
1 2 3 5 6 7 8 10
12 13 15 16 18 19 20 22.
Then, compare 19 and the largest term in the first list. Because
10 < 19, the search for 19 can be restricted to the list containing
the 9th through the 16th terms of the original list.
First split this list, which has 16 terms, into two smaller lists with
eight terms each, namely,
1 2 3 5 6 7 8 10
12 13 15 16 18 19 20 22.
Then, compare 19 and the largest term in the first list. Because
10 < 19, the search for 19 can be restricted to the list containing
the 9th through the 16th terms of the original list.
Next, split this list, which has eight terms, into the two smaller
lists of four terms each, namely,
12 13 15 16
18 19 20 22.
First split this list, which has 16 terms, into two smaller lists with
eight terms each, namely,
1 2 3 5 6 7 8 10
12 13 15 16 18 19 20 22.
Then, compare 19 and the largest term in the first list. Because
10 < 19, the search for 19 can be restricted to the list containing
the 9th through the 16th terms of the original list.
Next, split this list, which has eight terms, into the two smaller
lists of four terms each, namely,
12 13 15 16
Because 16 < 19.
18 19 20 22.
First split this list, which has 16 terms, into two smaller lists with
eight terms each, namely,
1 2 3 5 6 7 8 10
12 13 15 16 18 19 20 22.
Then, compare 19 and the largest term in the first list. Because
10 < 19, the search for 19 can be restricted to the list containing
the 9th through the 16th terms of the original list.
Next, split this list, which has eight terms, into the two smaller
lists of four terms each, namely,
12 13 15 16
18 19 20 22.
Because 16 < 19. The list 18 19 20 22 is split into two lists,
namely,
18 19
20 22.
First split this list, which has 16 terms, into two smaller lists with
eight terms each, namely,
1 2 3 5 6 7 8 10
12 13 15 16 18 19 20 22.
Then, compare 19 and the largest term in the first list. Because
10 < 19, the search for 19 can be restricted to the list containing
the 9th through the 16th terms of the original list.
Next, split this list, which has eight terms, into the two smaller
lists of four terms each, namely,
12 13 15 16
18 19 20 22.
Because 16 < 19. The list 18 19 20 22 is split into two lists,
namely,
18 19
20 22.
Because 19 is not greater than the largest term of the first of these
two lists,
First split this list, which has 16 terms, into two smaller lists with
eight terms each, namely,
1 2 3 5 6 7 8 10
12 13 15 16 18 19 20 22.
Then, compare 19 and the largest term in the first list. Because
10 < 19, the search for 19 can be restricted to the list containing
the 9th through the 16th terms of the original list.
Next, split this list, which has eight terms, into the two smaller
lists of four terms each, namely,
12 13 15 16
18 19 20 22.
Because 16 < 19. The list 18 19 20 22 is split into two lists,
namely,
18 19
20 22.
Because 19 is not greater than the largest term of the first of these
two lists, the search is restricted to the first list: 18 19, this list of
two terms is split into two lists of one term each: 18 and 19.
Because 18 < 19,
First split this list, which has 16 terms, into two smaller lists with
eight terms each, namely,
1 2 3 5 6 7 8 10
12 13 15 16 18 19 20 22.
Then, compare 19 and the largest term in the first list. Because
10 < 19, the search for 19 can be restricted to the list containing
the 9th through the 16th terms of the original list.
Next, split this list, which has eight terms, into the two smaller
lists of four terms each, namely,
12 13 15 16
18 19 20 22.
Because 16 < 19. The list 18 19 20 22 is split into two lists,
namely,
18 19
20 22.
Because 19 is not greater than the largest term of the first of these
two lists, the search is restricted to the first list: 18 19, this list of
two terms is split into two lists of one term each: 18 and 19.
Because 18 < 19, the search is restricted to the second list: which
is 19.
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
SECOND It should be efficient.
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
SECOND It should be efficient.
How can the efficiency of an algorithm be analyzed?
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
SECOND It should be efficient.
How can the efficiency of an algorithm be analyzed?
We have following two measures for an algorithm when input
values are of a specified size:
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
SECOND It should be efficient.
How can the efficiency of an algorithm be analyzed?
We have following two measures for an algorithm when input
values are of a specified size:
1. the time used by a computer to solve a problem using the
algorithm
2. the amount of computer memory required to implement the
algorithm
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
SECOND It should be efficient.
How can the efficiency of an algorithm be analyzed?
We have following two measures for an algorithm when input
values are of a specified size:
1. the time used by a computer to solve a problem using the
algorithm
2. the amount of computer memory required to implement the
algorithm
I
Computational Complexity
Time Complexity An analysis of the time required to solve a
problem of a particular size involves
Space Complexity An analysis of the computer memory
required involves the space complexity of the
algorithm.
The Growth of Functions
How many number of operations used by algorithm?
The Growth of Functions
How many number of operations used by algorithm?
The time required to solve a problem depends on the number of
operations it uses.
The Growth of Functions
How many number of operations used by algorithm?
The time required to solve a problem depends on the number of
operations it uses.
Using big-O notation, we can estimate the growth of a function.
The Growth of Functions
How many number of operations used by algorithm?
The time required to solve a problem depends on the number of
operations it uses.
Using big-O notation, we can estimate the growth of a function.
Big-O notation is used extensively to estimate the number of
operations an algorithm uses as its input grows. With the help of
this notation, we can determine whether it is practical to use a
particular algorithm to solve a problem as the size of the input
increases.
The Growth of Functions
How many number of operations used by algorithm?
The time required to solve a problem depends on the number of
operations it uses.
Using big-O notation, we can estimate the growth of a function.
Big-O notation is used extensively to estimate the number of
operations an algorithm uses as its input grows. With the help of
this notation, we can determine whether it is practical to use a
particular algorithm to solve a problem as the size of the input
increases.
Furthermore, using big-O notation, we can compare two algorithms
to determine which is more efficient as the size of the input grows.
Graph of a Function
Graph of Some Power Functions
Graph of a Multiple of a Function
Graph of a Multiple of a Function
Increasing and Decreasing Functions
Comparison of Functions
Comparison of Functions
Big-O Notation
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers. We say that
f (x) is O(g (x)) if there are constants C and k such that
|f (x)| ≤ C |g (x)|
whenever x > k.
[This is read as f (x) is big-oh of g (x).]
Big-O Notation
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers. We say that
f (x) is O(g (x)) if there are constants C and k such that
|f (x)| ≤ C |g (x)|
whenever x > k.
[This is read as f (x) is big-oh of g (x).]
Intuitively, the definition that f (x) is O(g (x)) says that f (x)
grows slower that some fixed multiple of g (x) as x grows
without bound.
Big-O Notation
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers. We say that
f (x) is O(g (x)) if there are constants C and k such that
|f (x)| ≤ C |g (x)|
whenever x > k.
[This is read as f (x) is big-oh of g (x).]
Intuitively, the definition that f (x) is O(g (x)) says that f (x)
grows slower that some fixed multiple of g (x) as x grows
without bound.
The constants C and k in the definition of big-O notation are
called witnesses to the relationship f (x) is O(g (x)).
Big-O Notation
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers. We say that
f (x) is O(g (x)) if there are constants C and k such that
|f (x)| ≤ C |g (x)|
whenever x > k.
[This is read as f (x) is big-oh of g (x).]
Intuitively, the definition that f (x) is O(g (x)) says that f (x)
grows slower that some fixed multiple of g (x) as x grows
without bound.
The constants C and k in the definition of big-O notation are
called witnesses to the relationship f (x) is O(g (x)).
To establish that f (x) is O(g (x)) we need only one pair of
witnesses to this relationship. That is, to show that f (x) is
O(g (x)), we need find only one pair of constants C and k, the
witnesses, such that
|f (x)| ≤ C |g (x)|
whenever x > k
Big-O f (x) = x 2 + 2x + 1 is O(x 2 )
Big-O f (x) = x 2 + 2x + 1 is O(x 2 )
graph of x 2 + 2x + 1, x 2 and 2x 2
Big-O f (x) = x 2 + 2x + 1 is O(x 2 )
graph of x 2 + 2x + 1, x 2 and 2x 2
Big-O
Show that f (x) = x 2 + 2x + 1 is O(x 2 )
Big-O
Show that f (x) = x 2 + 2x + 1 is O(x 2 )
when x > 1 ? Find witness?
Big-O
Show that f (x) = x 2 + 2x + 1 is O(x 2 )
when x > 1 ? Find witness?
when x > 2 ? Find witness?
Big-O
Show that f (x) = x 2 + 2x + 1 is O(x 2 )
when x > 1 ? Find witness?
when x > 2 ? Find witness?
Show that 7x 2 is O(x 3 ).
Big-O
Show that f (x) = x 2 + 2x + 1 is O(x 2 )
when x > 1 ? Find witness?
when x > 2 ? Find witness?
Show that 7x 2 is O(x 3 ). Note that if C and k are one pair of
witnesses, then any pair C 0 and k 0 , where C < C 0 and k < k 0 , is
also a pair of witnesses, because |f (x)| ≤ C |g (x)| ≤ C 0 |g (x)|
whenever x > k > k 0 .
Big-Omega, (Ω)
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers.
We say that f (x) is Ω(g (x)) if there are positive constants C and
k such that
Big-Omega, (Ω)
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers.
We say that f (x) is Ω(g (x)) if there are positive constants C and
k such that
|f (x)|≥C |g (x)|
whenever x > k.
This is read as “f (x) is big-Omega of g (x)”
Big-Omega, (Ω)
f (x) = 8x 3 + 5x 2 + 7 is Ω(g (x)), where g (x) is the function
g (x) = x 3
Big-Omega, (Ω)
f (x) = 8x 3 + 5x 2 + 7 is Ω(g (x)), where g (x) is the function
g (x) = x 3
This is easy to see because
Big-Omega, (Ω)
f (x) = 8x 3 + 5x 2 + 7 is Ω(g (x)), where g (x) is the function
g (x) = x 3
This is easy to see because f (x) = 8x 3 + 5x 2 + 7 ≥ 8x 3 for all
positive real number x.
O vs Ω
O vs Ω
f (x) is Ω(g (x)) if and only if g (x) is O(f (x)).
Big-Theta,(Θ)
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers.
Big-Theta,(Θ)
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers.
We say that f (x) is Θ(g (x)) if
Big-Theta,(Θ)
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers.
We say that f (x) is Θ(g (x)) if f (x) is O(g (x)) and f (x) is
Ω(g (x)).
Big-Theta,(Θ)
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers.
We say that f (x) is Θ(g (x)) if f (x) is O(g (x)) and f (x) is
Ω(g (x)). Then f (x) is Θ(g (x)).
Big-Theta,(Θ)
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers.
We say that f (x) is Θ(g (x)) if f (x) is O(g (x)) and f (x) is
Ω(g (x)). Then f (x) is Θ(g (x)).
We say that f is big-Theta of g (x),
Big-Theta,(Θ)
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers.
We say that f (x) is Θ(g (x)) if f (x) is O(g (x)) and f (x) is
Ω(g (x)). Then f (x) is Θ(g (x)).
We say that f is big-Theta of g (x),that f (x) is of order g (x), and
that f (x) and g (x) are of the same order.
Big-Theta,(Θ)
Let f and g be functions from the set of integers or the set of real
numbers to the set of real numbers.
We say that f (x) is Θ(g (x)) if f (x) is O(g (x)) and f (x) is
Ω(g (x)). Then f (x) is Θ(g (x)).
We say that f is big-Theta of g (x),that f (x) is of order g (x), and
that f (x) and g (x) are of the same order.
f (x) is Θ(g (x)) if and only if there are real numbers C1 and C2
and a positive real number k such that
C1 |g (x)| ≤ |f (x)| ≤ C2 |g (x)|
whenever x > k.
The existence of the constants C1 , C2 , and k tells us that f (x) is
Ω(g (x)) and that f (x) is O(g (x)), respectively.
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
SECOND It should be efficient.
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
SECOND It should be efficient.
How can the efficiency of an algorithm be analyzed?
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
SECOND It should be efficient.
How can the efficiency of an algorithm be analyzed?
We have following two measures for an algorithm when input
values are of a specified size:
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
SECOND It should be efficient.
How can the efficiency of an algorithm be analyzed?
We have following two measures for an algorithm when input
values are of a specified size:
1. the time used by a computer to solve a problem using the
algorithm
2. the amount of computer memory required to implement the
algorithm
Complexity Of Algorithms
When does an algorithm provide a satisfactory solution to a
problem?
FIRST It must always produce the correct answer.
SECOND It should be efficient.
How can the efficiency of an algorithm be analyzed?
We have following two measures for an algorithm when input
values are of a specified size:
1. the time used by a computer to solve a problem using the
algorithm
2. the amount of computer memory required to implement the
algorithm
I
Computational Complexity
Time Complexity An analysis of the time required to solve a
problem of a particular size involves
Space Complexity An analysis of the computer memory
required involves the space complexity of the
algorithm.
Time Complexity
The time complexity of an algorithm can be expressed in terms of
the number of operations used by the algorithm when the input
has a particular size.
Time Complexity
The time complexity of an algorithm can be expressed in terms of
the number of operations used by the algorithm when the input
has a particular size.
The operations used to measure time complexity can be the
comparison of integers, the addition of integers, the multiplication
of integers, the division of integers, or any other basic operation
Let A be an algorithm. Time complexity of A?
1. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on n alone
and not on the nature of the input data;
say it equals f (n).
Let A be an algorithm. Time complexity of A?
1. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on n alone
and not on the nature of the input data;
say it equals f (n). If f (n) is Θ(g (n)), we say that A is
Θ(g (n))
Let A be an algorithm. Time complexity of A?
1. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on n alone
and not on the nature of the input data;
say it equals f (n). If f (n) is Θ(g (n)), we say that A is
Θ(g (n)) or A is of order g (n).
Let A be an algorithm. Time complexity of A?
1. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on n alone
and not on the nature of the input data;
say it equals f (n). If f (n) is Θ(g (n)), we say that A is
Θ(g (n)) or A is of order g (n).
2. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on the
nature of the input data as well as on n.
Let A be an algorithm. Time complexity of A?
1. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on n alone
and not on the nature of the input data;
say it equals f (n). If f (n) is Θ(g (n)), we say that A is
Θ(g (n)) or A is of order g (n).
2. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on the
nature of the input data as well as on n.
a- Let b(n) be the minimum number of elementary
operations required to execute A for all possible input sets
of size n.
Let A be an algorithm. Time complexity of A?
1. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on n alone
and not on the nature of the input data;
say it equals f (n). If f (n) is Θ(g (n)), we say that A is
Θ(g (n)) or A is of order g (n).
2. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on the
nature of the input data as well as on n.
a- Let b(n) be the minimum number of elementary
operations required to execute A for all possible input sets
of size n.If b(n) is Θ(g (n)), we say that in the best case,
A is Θ(g (n))
Let A be an algorithm. Time complexity of A?
1. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on n alone
and not on the nature of the input data;
say it equals f (n). If f (n) is Θ(g (n)), we say that A is
Θ(g (n)) or A is of order g (n).
2. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on the
nature of the input data as well as on n.
a- Let b(n) be the minimum number of elementary
operations required to execute A for all possible input sets
of size n.If b(n) is Θ(g (n)), we say that in the best case,
A is Θ(g (n)) or A has a best-case order of g (n).
Let A be an algorithm. Time complexity of A?
1. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on n alone
and not on the nature of the input data;
say it equals f (n). If f (n) is Θ(g (n)), we say that A is
Θ(g (n)) or A is of order g (n).
2. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on the
nature of the input data as well as on n.
a- Let b(n) be the minimum number of elementary
operations required to execute A for all possible input sets
of size n.If b(n) is Θ(g (n)), we say that in the best case,
A is Θ(g (n)) or A has a best-case order of g (n).
b- Let w (n) be the maximum number of elementary
operations required to execute A for all possible input sets
of size n.
Let A be an algorithm. Time complexity of A?
1. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on n alone
and not on the nature of the input data;
say it equals f (n). If f (n) is Θ(g (n)), we say that A is
Θ(g (n)) or A is of order g (n).
2. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on the
nature of the input data as well as on n.
a- Let b(n) be the minimum number of elementary
operations required to execute A for all possible input sets
of size n.If b(n) is Θ(g (n)), we say that in the best case,
A is Θ(g (n)) or A has a best-case order of g (n).
b- Let w (n) be the maximum number of elementary
operations required to execute A for all possible input sets
of size n.If w (n) is Θ(g (n)), we say that in the worst case,
A is Θ(g (n))
Let A be an algorithm. Time complexity of A?
1. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on n alone
and not on the nature of the input data;
say it equals f (n). If f (n) is Θ(g (n)), we say that A is
Θ(g (n)) or A is of order g (n).
2. Suppose the number of elementary operations performed
when A is executed for an input of size n depends on the
nature of the input data as well as on n.
a- Let b(n) be the minimum number of elementary
operations required to execute A for all possible input sets
of size n.If b(n) is Θ(g (n)), we say that in the best case,
A is Θ(g (n)) or A has a best-case order of g (n).
b- Let w (n) be the maximum number of elementary
operations required to execute A for all possible input sets
of size n.If w (n) is Θ(g (n)), we say that in the worst case,
A is Θ(g (n)) or A has a worst-case order of g (n).
Describe the time complexity of Algorithm for finding the
maximum element in a finite set of integers.1
1
The number of comparisons will be used as the measure of the time
complexity of the algorithm, because comparisons are the basic operations used.
Describe the time complexity of Algorithm for finding the
maximum element in a finite set of integers.2
2
The number of comparisons will be used as the measure of the time
complexity of the algorithm, because comparisons are the basic operations used.
Describe the time complexity of Algorithm for finding the
maximum element in a finite set of integers.3
3
The number of comparisons will be used as the measure of the time
complexity of the algorithm, because comparisons are the basic operations used.
Describe the time complexity of Algorithm for finding the
maximum element in a finite set of integers.4
4
The number of comparisons will be used as the measure of the time
complexity of the algorithm, because comparisons are the basic operations used.
Describe the time complexity of Algorithm for finding the
maximum element in a finite set of integers.4
Exactly 2n − 2 + 1 = 2n − 1 comparisons are used whenever this
algorithm is applied.
4
The number of comparisons will be used as the measure of the time
complexity of the algorithm, because comparisons are the basic operations used.
Describe the time complexity of Algorithm for finding the
maximum element in a finite set of integers.4
Exactly 2n − 2 + 1 = 2n − 1 comparisons are used whenever this
algorithm is applied. Hence, the algorithm for finding the
maximum of a set of n elements has time complexity Θ(n),
measured in terms of the number of comparisons used.
4
The number of comparisons will be used as the measure of the time
complexity of the algorithm, because comparisons are the basic operations used.
Describe the time complexity of the binary search
algorithm5
5
ignoring the time required to compute m = b(i + j)/2c in each iteration of
the loop in the algorithm
Describe the time complexity of the binary search
algorithm6
6
ignoring the time required to compute m = b(i + j)/2c in each iteration of
the loop in the algorithm
Describe the time complexity of the binary search
algorithm7
7
ignoring the time required to compute m = b(i + j)/2c in each iteration of
the loop in the algorithm
Describe the time complexity of the binary search
algorithm8
8
ignoring the time required to compute m = b(i + j)/2c in each iteration of
the loop in the algorithm
Describe the time complexity of the binary search
algorithm9
9
ignoring the time required to compute m = b(i + j)/2c in each iteration of
the loop in the algorithm
Describe the time complexity of the binary search
algorithm10
10
ignoring the time required to compute m = b(i + j)/2c in each iteration of
the loop in the algorithm
Now n = 2k or k = log2 n
Hence, at most 2k + 2 = 2 log n + 2 comparisons are required to
perform a binary search when the list being searched has 2k
elements.
Now n = 2k or k = log2 n
Hence, at most 2k + 2 = 2 log n + 2 comparisons are required to
perform a binary search when the list being searched has 2k
elements.
If n is not a power of 2, the original list is expanded to a list with
2k+1 terms, where k = blog nc,
Now n = 2k or k = log2 n
Hence, at most 2k + 2 = 2 log n + 2 comparisons are required to
perform a binary search when the list being searched has 2k
elements.
If n is not a power of 2, the original list is expanded to a list with
2k+1 terms, where k = blog nc,
The search requires at most 2dlog ne + 2 comparisons.
Now n = 2k or k = log2 n
Hence, at most 2k + 2 = 2 log n + 2 comparisons are required to
perform a binary search when the list being searched has 2k
elements.
If n is not a power of 2, the original list is expanded to a list with
2k+1 terms, where k = blog nc,
The search requires at most 2dlog ne + 2 comparisons.
It follows that in the worst case, binary search requires O(log n)
comparisons.
Now n = 2k or k = log2 n
Hence, at most 2k + 2 = 2 log n + 2 comparisons are required to
perform a binary search when the list being searched has 2k
elements.
If n is not a power of 2, the original list is expanded to a list with
2k+1 terms, where k = blog nc,
The search requires at most 2dlog ne + 2 comparisons.
It follows that in the worst case, binary search requires O(log n)
comparisons.
Note that in the worst case, 2 log n + 2 comparisons are used by
the binary search. Hence, the binary search uses Θ(log n)
comparisons in the worst case,
Now n = 2k or k = log2 n
Hence, at most 2k + 2 = 2 log n + 2 comparisons are required to
perform a binary search when the list being searched has 2k
elements.
If n is not a power of 2, the original list is expanded to a list with
2k+1 terms, where k = blog nc,
The search requires at most 2dlog ne + 2 comparisons.
It follows that in the worst case, binary search requires O(log n)
comparisons.
Note that in the worst case, 2 log n + 2 comparisons are used by
the binary search. Hence, the binary search uses Θ(log n)
comparisons in the worst case, because 2 log n + 2 = Θ(logn).
Let f and g be real-valued functions defined on the same set of
non-negative real numbers.
Let f and g be real-valued functions defined on the same set of
non-negative real numbers.
1. f (x) is Ω(g (x)) and f (x) is O(g (x))
Let f and g be real-valued functions defined on the same set of
non-negative real numbers.
1. f (x) is Ω(g (x)) and f (x) is O(g (x)) if, and only if f (x) is
Θ(g (x)).
Let f and g be real-valued functions defined on the same set of
non-negative real numbers.
1. f (x) is Ω(g (x)) and f (x) is O(g (x)) if, and only if f (x) is
Θ(g (x)).
2. f (x) is Ω(g (x)) if, and only if,
Let f and g be real-valued functions defined on the same set of
non-negative real numbers.
1. f (x) is Ω(g (x)) and f (x) is O(g (x)) if, and only if f (x) is
Θ(g (x)).
2. f (x) is Ω(g (x)) if, and only if, g (x) is O(f (x)).
Let f and g be real-valued functions defined on the same set of
non-negative real numbers.
1. f (x) is Ω(g (x)) and f (x) is O(g (x)) if, and only if f (x) is
Θ(g (x)).
2. f (x) is Ω(g (x)) if, and only if, g (x) is O(f (x)).
3. f (x) is O(g (x)) and g (x) is O(h(x)), then f (x) is O(h(x))
Let f and g be real-valued functions defined on the same set of
non-negative real numbers.
1. f (x) is Ω(g (x)) and f (x) is O(g (x)) if, and only if f (x) is
Θ(g (x)).
2. f (x) is Ω(g (x)) if, and only if, g (x) is O(f (x)).
3. f (x) is O(g (x)) and g (x) is O(h(x)), then f (x) is O(h(x))
For any rational numbers r and s,
if x > 1 and r < s, then x r < x s
if r < s, then x r is O(x s )
Let f and g be real-valued functions defined on the same set of
non-negative real numbers.
1. f (x) is Ω(g (x)) and f (x) is O(g (x)) if, and only if f (x) is
Θ(g (x)).
2. f (x) is Ω(g (x)) if, and only if, g (x) is O(f (x)).
3. f (x) is O(g (x)) and g (x) is O(h(x)), then f (x) is O(h(x))
For any rational numbers r and s,
if x > 1 and r < s, then x r < x s
if r < s, then x r is O(x s )
Show that for any real number x, if x > 1, then
3x 3 + 2x + 7 ≤ 12x 3 .
To show that 2x 4 + 3x 3 + 5 is Θ(x 4 ).
Define functions f and g as follows. For all non-negative real
numbers x,
To show that 2x 4 + 3x 3 + 5 is Θ(x 4 ).
Define functions f and g as follows. For all non-negative real
numbers x,
f (x) = 2x 4 + 3x 3 + 5
Observe that for all real numbers x > 0,
g (x) = x 4
To show that 2x 4 + 3x 3 + 5 is Θ(x 4 ).
Define functions f and g as follows. For all non-negative real
numbers x,
f (x) = 2x 4 + 3x 3 + 5
g (x) = x 4
Observe that for all real numbers x > 0, 2x 4 ≤ 2x 4 + 3x 3 + 5
To show that 2x 4 + 3x 3 + 5 is Θ(x 4 ).
Define functions f and g as follows. For all non-negative real
numbers x,
f (x) = 2x 4 + 3x 3 + 5
g (x) = x 4
Observe that for all real numbers x > 0, 2x 4 ≤ 2x 4 + 3x 3 + 5
Let C = 2 and k = 0, gives 2|x 4 | ≤ |2x 4 + 3x 3 + 5| implies
2x 4 + 3x 3 + 5 is of Ω(x 4 )
To show that 2x 4 + 3x 3 + 5 is Θ(x 4 ).
Define functions f and g as follows. For all non-negative real
numbers x,
f (x) = 2x 4 + 3x 3 + 5
g (x) = x 4
Observe that for all real numbers x > 0, 2x 4 ≤ 2x 4 + 3x 3 + 5
Let C = 2 and k = 0, gives 2|x 4 | ≤ |2x 4 + 3x 3 + 5| implies
2x 4 + 3x 3 + 5 is of Ω(x 4 )
Also for x > 1,
2x 4 + 3x 3 + 5 ≤ 2x 4 + 3x 4 + 5x 4 = 10x 4 .
To show that 2x 4 + 3x 3 + 5 is Θ(x 4 ).
Define functions f and g as follows. For all non-negative real
numbers x,
f (x) = 2x 4 + 3x 3 + 5
g (x) = x 4
Observe that for all real numbers x > 0, 2x 4 ≤ 2x 4 + 3x 3 + 5
Let C = 2 and k = 0, gives 2|x 4 | ≤ |2x 4 + 3x 3 + 5| implies
2x 4 + 3x 3 + 5 is of Ω(x 4 )
Also for x > 1,
2x 4 + 3x 3 + 5 ≤ 2x 4 + 3x 4 + 5x 4 = 10x 4 .
|2x 4 + 3x 3 + 5| ≤ 10|x 4 | whenever x > 1
implies 2x 4 + 3x 3 + 5 is of O(x 4 ).
To show that 2x 4 + 3x 3 + 5 is Θ(x 4 ).
Define functions f and g as follows. For all non-negative real
numbers x,
f (x) = 2x 4 + 3x 3 + 5
g (x) = x 4
Observe that for all real numbers x > 0, 2x 4 ≤ 2x 4 + 3x 3 + 5
Let C = 2 and k = 0, gives 2|x 4 | ≤ |2x 4 + 3x 3 + 5| implies
2x 4 + 3x 3 + 5 is of Ω(x 4 )
Also for x > 1,
2x 4 + 3x 3 + 5 ≤ 2x 4 + 3x 4 + 5x 4 = 10x 4 .
|2x 4 + 3x 3 + 5| ≤ 10|x 4 | whenever x > 1
implies 2x 4 + 3x 3 + 5 is of O(x 4 ).
Since 2x 4 + 3x 3 + 5 is of Ω(x 4 ) and of O(x 4 ) therefore,
2x 4 + 3x 3 + 5 is of Θ(x 4 )
Show that 3x 3 − 1000x − 200 is O(x 3 )11 .
|3x 3 − 1000x − 200| ≤ C |x 3 | whenever x > k.
11
Triangular Inequality |a + b| ≤ |a| + |b|
Show that 3x 3 − 1000x − 200 is O(x 3 )11 .
|3x 3 − 1000x − 200| ≤ C |x 3 | whenever x > k.
Show 3x 3 − 1000x − 200 is O(x s ) for s ≥ 3.
11
Triangular Inequality |a + b| ≤ |a| + |b|
Show that 3x 3 − 1000x − 200 is O(x 3 )11 .
|3x 3 − 1000x − 200| ≤ C |x 3 | whenever x > k.
Show 3x 3 − 1000x − 200 is O(x s ) for s ≥ 3.
Show 3x 3 − 1000x − 200 is Ω(x 3 ).
11
Triangular Inequality |a + b| ≤ |a| + |b|
Show that 3x 3 − 1000x − 200 is O(x 3 )11 .
|3x 3 − 1000x − 200| ≤ C |x 3 | whenever x > k.
Show 3x 3 − 1000x − 200 is O(x s ) for s ≥ 3.
Show 3x 3 − 1000x − 200 is Ω(x 3 ).
Show 3x 3 − 1000x − 200 is Ω(x r ) for r < 3.
11
Triangular Inequality |a + b| ≤ |a| + |b|
Show that 3x 3 − 1000x − 200 is O(x 3 )11 .
|3x 3 − 1000x − 200| ≤ C |x 3 | whenever x > k.
Show 3x 3 − 1000x − 200 is O(x s ) for s ≥ 3.
Show 3x 3 − 1000x − 200 is Ω(x 3 ).
Show 3x 3 − 1000x − 200 is Ω(x r ) for r < 3.
Theorem
Suppose a0 , a1 , a2 , ..., an are real numbers and an 6= 0.
1. an x n + an−1 x n−1 + · · · + a1 x + a0 is O(x s ) for all integers
s≥n
2. an x n + an−1 x n−1 + · · · + a1 x + a0 is Ω(x r ) for all integers
r ≤n
3. an x n + an−1 x n−1 + · · · + a1 x + a0 is Θ(x n )
11
Triangular Inequality |a + b| ≤ |a| + |b|
The Growth of Combinations of Functions
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 + f2 )(x) is O(max(|g1 (x)|, |g2 (x)|)).
The Growth of Combinations of Functions
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 + f2 )(x) is O(max(|g1 (x)|, |g2 (x)|)).
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 .f2 )(x) is O((|g1 (x)|.|g2 (x)|)).
Example
Give a big-O estimate for f (x) = (x + 1) log(x 2 + 1) + 3x 2 . Here
f1 (x) = (x + 1) log(x 2 + 1) and f2 (x) = 3x 2
f1 (x) : f1 (x) is product of two functions,
The Growth of Combinations of Functions
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 + f2 )(x) is O(max(|g1 (x)|, |g2 (x)|)).
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 .f2 )(x) is O((|g1 (x)|.|g2 (x)|)).
Example
Give a big-O estimate for f (x) = (x + 1) log(x 2 + 1) + 3x 2 . Here
f1 (x) = (x + 1) log(x 2 + 1) and f2 (x) = 3x 2
f1 (x) : f1 (x) is product of two functions,
I f ∗ = x + 1 is O(x)
1
The Growth of Combinations of Functions
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 + f2 )(x) is O(max(|g1 (x)|, |g2 (x)|)).
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 .f2 )(x) is O((|g1 (x)|.|g2 (x)|)).
Example
Give a big-O estimate for f (x) = (x + 1) log(x 2 + 1) + 3x 2 . Here
f1 (x) = (x + 1) log(x 2 + 1) and f2 (x) = 3x 2
f1 (x) : f1 (x) is product of two functions,
I f ∗ = x + 1 is O(x)
1
I f ∗∗ = log(x 2 + 1). For x > 2,
1
log(x 2 + 1) ≤ log(2x 2 ) = log 2 + 2 log x ≤ 3 log x
Hence f1∗∗ is of O(log x).
The Growth of Combinations of Functions
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 + f2 )(x) is O(max(|g1 (x)|, |g2 (x)|)).
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 .f2 )(x) is O((|g1 (x)|.|g2 (x)|)).
Example
Give a big-O estimate for f (x) = (x + 1) log(x 2 + 1) + 3x 2 . Here
f1 (x) = (x + 1) log(x 2 + 1) and f2 (x) = 3x 2
f1 (x) : f1 (x) is product of two functions,
I f ∗ = x + 1 is O(x)
1
I f ∗∗ = log(x 2 + 1). For x > 2,
1
log(x 2 + 1) ≤ log(2x 2 ) = log 2 + 2 log x ≤ 3 log x
Hence f1∗∗ is of O(log x). f1 (x) is O(x log x)
The Growth of Combinations of Functions
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 + f2 )(x) is O(max(|g1 (x)|, |g2 (x)|)).
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 .f2 )(x) is O((|g1 (x)|.|g2 (x)|)).
Example
Give a big-O estimate for f (x) = (x + 1) log(x 2 + 1) + 3x 2 . Here
f1 (x) = (x + 1) log(x 2 + 1) and f2 (x) = 3x 2
f1 (x) : f1 (x) is product of two functions,
I f ∗ = x + 1 is O(x)
1
I f ∗∗ = log(x 2 + 1). For x > 2,
1
log(x 2 + 1) ≤ log(2x 2 ) = log 2 + 2 log x ≤ 3 log x
Hence f1∗∗ is of O(log x). f1 (x) is O(x log x)
f2 (x) : f2 (x) = 3x 2 is O(x 2 )
The Growth of Combinations of Functions
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 + f2 )(x) is O(max(|g1 (x)|, |g2 (x)|)).
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 .f2 )(x) is O((|g1 (x)|.|g2 (x)|)).
Example
Give a big-O estimate for f (x) = (x + 1) log(x 2 + 1) + 3x 2 . Here
f1 (x) = (x + 1) log(x 2 + 1) and f2 (x) = 3x 2
f1 (x) : f1 (x) is product of two functions,
I f ∗ = x + 1 is O(x)
1
I f ∗∗ = log(x 2 + 1). For x > 2,
1
log(x 2 + 1) ≤ log(2x 2 ) = log 2 + 2 log x ≤ 3 log x
Hence f1∗∗ is of O(log x). f1 (x) is O(x log x)
f2 (x) : f2 (x) = 3x 2 is O(x 2 )
f (x) is O(max(x log x), (x 2 )).
The Growth of Combinations of Functions
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 + f2 )(x) is O(max(|g1 (x)|, |g2 (x)|)).
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 .f2 )(x) is O((|g1 (x)|.|g2 (x)|)).
Example
Give a big-O estimate for f (x) = (x + 1) log(x 2 + 1) + 3x 2 . Here
f1 (x) = (x + 1) log(x 2 + 1) and f2 (x) = 3x 2
f1 (x) : f1 (x) is product of two functions,
I f ∗ = x + 1 is O(x)
1
I f ∗∗ = log(x 2 + 1). For x > 2,
1
log(x 2 + 1) ≤ log(2x 2 ) = log 2 + 2 log x ≤ 3 log x
Hence f1∗∗ is of O(log x). f1 (x) is O(x log x)
f2 (x) : f2 (x) = 3x 2 is O(x 2 )
f (x) is O(max(x log x), (x 2 )). Because x log x ≤ x 2 for x > 1.
The Growth of Combinations of Functions
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 + f2 )(x) is O(max(|g1 (x)|, |g2 (x)|)).
Suppose that f1 (x) is O(g1 (x)) and that f2 (x) is O(g2 (x)). Then
(f1 .f2 )(x) is O((|g1 (x)|.|g2 (x)|)).
Example
Give a big-O estimate for f (x) = (x + 1) log(x 2 + 1) + 3x 2 . Here
f1 (x) = (x + 1) log(x 2 + 1) and f2 (x) = 3x 2
f1 (x) : f1 (x) is product of two functions,
I f ∗ = x + 1 is O(x)
1
I f ∗∗ = log(x 2 + 1). For x > 2,
1
log(x 2 + 1) ≤ log(2x 2 ) = log 2 + 2 log x ≤ 3 log x
Hence f1∗∗ is of O(log x). f1 (x) is O(x log x)
f2 (x) : f2 (x) = 3x 2 is O(x 2 )
f (x) is O(max(x log x), (x 2 )). Because x log x ≤ x 2 for x > 1.
Therefore f (x) is O(x 2 ).
Quiz # 2
1. Show that x 5 is not O(x 2 ).
Quiz # 2
1. Show that x 5 is not O(x 2 ).
2. Show 7x 4 − 95x 3 + 3 is Ω(x 4 )
Quiz # 2
1. Show that x 5 is not O(x 2 ).
2. Show 7x 4 − 95x 3 + 3 is Ω(x 4 )
3. How can big-O notation be used to estimate the sum of the
first n positive integers?
4. Give big-O estimates for the factorial function and the
logarithm of the factorial function.
Quiz # 2
1. Show that x 5 is not O(x 2 ).
2. Show 7x 4 − 95x 3 + 3 is Ω(x 4 )
3. How can big-O notation be used to estimate the sum of the
first n positive integers?
4. Give big-O estimates for the factorial function and the
logarithm of the factorial function.
5. Show that 3x 2 + 8x log x is Θ(x 2 ).