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CHAPTER 2 Analysis of Algorithms

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CHAPTER 2Analysis of Algorithms

Input

Algorithm

Output

An algorithm is a step-by-step procedure for

solving a problem in a finite amount of time.

Series summations

Logarithms and Exponents

Proof techniques

Basic probability

- properties of logarithms:
logb(xy) = logbx + logby

logb (x/y) = logbx - logby

logbxa = alogbx

logba = logxa/logxb

- properties of exponentials:
a(b+c) = aba c

abc = (ab)c

ab /ac = a(b-c)

b = a logab

bc = a c*logab

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms

Basic computer with sequentially executed instructions

Infinite memory

Has standard operations; addition, multiplication, comparison in 1 time unit unless stated otherwise

Has fixed-size (32bits) integers such that no-fancy operations are allowed!!! eg.matrix inversion,

Analysis of Algorithms

- Most algorithms transform input objects into output objects.
- The running time of an algorithm typically grows with the input size.
- Average case time is often difficult to determine.
- We focus on the worst case running time.
- Easier to analyze
- Crucial to applications such as games, finance, image,robotics, AI, etc.

Analysis of Algorithms

- Write a program implementing the algorithm
- Run the program with inputs of varying size and composition
- Assume a subprogram is written to get an accurate measure of the actual running time
- Plot the results

Analysis of Algorithms

- It is necessary to implement the algorithm, which may be difficult
- Results may not be indicative of the running time on other inputs not included in the experiment.
- In order to compare two algorithms, the same hardware and software environments must be used

Analysis of Algorithms

- Uses a high-level description of the algorithm instead of an implementation
- Characterizes running time as a function of the input size, n.
- Takes into account all possible inputs
- Allows us to evaluate the speed of an algorithm independent of the hardware/software environment

Analysis of Algorithms

Example: find max element of an array

AlgorithmarrayMax(A, n)

Inputarray A of n integers

Outputmaximum element of A

currentMaxA[0]

fori1ton 1do

ifA[i] currentMaxthen

currentMaxA[i]

returncurrentMax

- High-level description of an algorithm
- More structured than English prose
- Less detailed than a program
- Preferred notation for describing algorithms
- Hides program design issues

Analysis of Algorithms

Control flow

if…then… [else…]

while…do…

repeat…until…

for…do…

Indentation replaces braces

Method declaration

Algorithm method (arg [, arg…])

Input…

Output…

Method call

var.method (arg [, arg…])

Return value

returnexpression

Expressions

Assignment(like in Java)

Equality testing(like in Java)

n2Superscripts and other mathematical formatting allowed

Analysis of Algorithms

- Seven functions that often appear in algorithm analysis:
- Constant 1
- Logarithmic log n
- Linear n
- N-Log-N n log n
- Quadratic n2
- Cubic n3
- Exponential 2n

- In a log-log chart, the slope of the line corresponds to the growth rate of the function

Analysis of Algorithms

Basic computations performed by an algorithm

Identifiable in pseudocode

Largely independent from the programming language

Exact definition not important

Assumed to take a constant amount of time in the RAM model

Examples:

Evaluating an expression

Assigning a value to a variable

Indexing into an array

Calling a method

Returning from a method

Analysis of Algorithms

The time require for the following operations are all constants.

Analysis of Algorithms

The time require for the following operations are all constants.

Analysis of Algorithms

Analysis of Algorithms

...

public static int sum(int n) {

int result = 0;

for (int i = 1; i <= n; ++i){

result += i;

}

return result;

}

...

Analysis of Algorithms

y = a [i]

3fetch + [.] + store

- operations
- fetch a (the base address of the array)
- fetch i (the index into the array)
- address calculation
- fetch array element a[i]
- store the result

Analysis of Algorithms

publicclass Horner {

publicstaticvoid main(String[] args) {

Horner h = new Horner();

int[] a = { 1, 3, 5 };

System.out.println("a(1)=" + h.horner(a, a.length - 1, 1));

System.out.println("a(2)=" + h.horner(a, a.length - 1, 2));

}

int horner(int[] a, int n, int x) {

int result = a[n];

for (int i = n - 1; i >= 0; --i) {

result = result * x + a[i];

/**/System.out.println("i=" + i + " result" + result);

}

return result;

}

}

i=1 result8

i=0 result9

a(1)=9

i=1 result13

i=0 result27

a(2)=27

output

Analysis of Algorithms

publicclass FindMaximum {

publicstaticvoid main(String[] args) {

FindMaximum h = new FindMaximum();

int[] a = { 1, 3, 5 };

System.out.println("max=" + h.findMaximum(a));

}

int findMaximum(int[] a) {

int result = a[0];

for (int i = 0; i < a.length; ++i) {

if (result < a[i]) {

result = a[i];

}

System.out.println("i=" + i + " result=" + result);

}

return result;

}

}

i=0 result=1

i=1 result=3

i=2 result=5

max=5

output

3fetch + [.]+ store

fetch + store

(2fetch + <) n

(2fetch + ++ store) (n-1)

(4fetch + [.]+ <) (n-1)

(3fetch + [.]+ store) ?

fetch + store

Analysis of Algorithms

- All timing parameters are expressed in units of clock cycles. In effect, T=1.
- The proportionality constant, k, for all timing parameters is assumed to be the same: k=1.

Analysis of Algorithms

publicclass FindMaximum {

publicstaticvoid main(String[] args) {

FindMaximum h = new FindMaximum();

int[] a = { 1, 3, 5 };

System.out.println("max=" + h.findMaximum(a));

}

int findMaximum(int[] a) {

int result = a[0];

for (int i = 0; i < a.length; ++i) {

if (result < a[i]) {

result = a[i];

}

System.out.println("i=" + i + " result=" + result);

}

return result;

}

}

i=0 result=1

i=1 result=3

i=2 result=5

max=5

out

1

2

3

4

5

6

7

8

9

10

detailed

23fetch + [.]+ store

3afetch + store

3b(2fetch + <) n

3c(2fetch + ++ store) (n-1)

4(4fetch + [.]+ <) (n-1)

6(3fetch + [.]+ store) ?

9fetch + store

simple

25

3a2

3b(3)n

3c(4)(n-1)

4(6)(n-1)

6(5)?

92

Analysis of Algorithms

publicclass GeometrikSeriesSum {

publicstaticvoid main(String[] args) {

System.out.println("1, 4: " + geometricSeriesSum(1, 4));

System.out.println("2, 4: " + geometricSeriesSum(2, 4));

}

publicstaticint geometricSeriesSum(int x, int n) {

int sum = 0;

for (int i = 0; i <= n; ++i) {

int prod = 1;

for (int j = 0; j < i; ++j) {

prod *= x;

}

sum += prod;

}

return sum;

}

}

x=1, n=4: a4=5

x=2, n=4: a4=31

output

1

2

3

4

5

6

7

8

9

10

11

12

simple

1

22

3a2

3b3(n+2)

3c4(n+1)

42(n+1)

5a2(n+1)

5b3 (i+1)

5c4 i

64 i

7

84(n+1)

9

102

11

12

Total11/2 n2 + 47/2 n +27

Analysis of Algorithms

publicclass GeometrikSeriesSumHorner {

publicstaticvoid main(String[] args) {

System.out.println("1, 4: " + geometricSeriesSum(1, 4));

System.out.println("2, 4: " + geometricSeriesSum(2, 4));

}

publicstaticint geometricSeriesSum(int x, int n) {

int sum = 0;

for (int i = 0; i <= n; ++i) {

sum = sum * x + 1;

}

return sum;

}

}

Observe:

Let sum = a_i,

a0=0

a1=1, a2=x+1, a3=x2+x+1,.. an=xn+xn-1+xn-1+...x+1

x=1, n=4, output:1+1+1+1+1=5

x=2, n=4, output:20+21+22+23+24=31

x=1, n=4: a4=5

x=2, n=4: a4=31

output

1

2

3

4

5

6

7

- 22
- 3a2
- 3b3(n+2)
- 3c4(n+1)
- 46(n+1)
- 62
- Total13 n + 22

Analysis of Algorithms

publicclass GeometrikSeriesSumPower {

publicstaticvoid main(String[] args) {

System.out.println("1, 4: " + powerA(1, 4));

System.out.println("1, 4: " + powerB(1, 4));

System.out.println("2, 4: " + powerA(2, 4));

System.out.println("2, 4: " + powerB(2, 4));

}

...

publicstaticint powerA(int x, int n) {

int result = 1;

for (int i = 1; i <= n; ++i) {

result *= x;

}

return result;

}

publicstaticint powerB(int x, int n) {

if (n == 0) {

return 1;

} elseif (n % 2 == 0) { // n is even

return powerB(x * x, n / 2);

} else { // n is odd

return x * powerB(x * x, n / 2);

}

}

}

1, 4: 1

1, 4: 1

2, 4: 16

2, 4: 16

output

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

powerB

1n=0

xn = (x2)n/20<n, n is even

x(x2)n/20<n, n is odd

powerB0<n0<n

n=0 n evenn odd

10333

112--

12-55

13-10+T(n/2)-

15--12+T(n/2)

Total518+T(n/2)20+T(n/2)

powerB

5n=0

xn = 18+T(n/2)0<n, n is even

20+T(n/2)0<n, n is odd

Analysis of Algorithms

Let n = 2k for some k>0.

Since n is even, n/2 = n/2 = 2k-1.

For n = 2k, T(2k) = 18 + T(2k-1).

Using repeated substitution

T(2k) = 18 + T(2k-1)

= 18 + 18 + T(2k-2)

= 18 + 18 + 18 + T(2k-3)

…

= 18j + T(2k-j)

Substitution stops when k = j

T(2k) = 18k + T(1)

= 18k + 20 + T(0)

= 18k + 20 + 5

= 18k + 25.

n = 2k then k = log2 n

T(2k) = 18 log2 n + 25

n is even

powerB

xn = 18+T(n/2)0<n

Analysis of Algorithms

Suppose n = 2k–1 for some k>0.

Since n is odd, n/2 = (2k–1)/2

= (2k–2)/2= 2k-1

For n = 2k-1,

T(2k-1) = 20 + T(2k-1-1), k>1.

Using repeated substitution

T(2k-1) = 20 + T(2k-1-1)

= 20 + 20 + T(2k-2-1)

= 20 + 20 + 20 + T(2k-3-1)

…

= 20j + T(2k-j-1)

Substitution stops when k = j

T(2k-1) = 20k + T(20-1)

= 20k + T(0)

= 20k + 5.

n = 2k-1then k = log2 (n+1)

T(n) = 20 log2 (n+1) + 5

n is odd

Therefore, powerB

5n=0

xn = 18+T(n/2)0<n, n is even

20+T(n/2)0<n, n is odd

Average: 19(log2(n+1) + 1) + 18

Analysis of Algorithms

publicclass GeometrikSeriesSumPower {

publicstaticvoid main(String[] args) {

System.out.println(“s 2, 4: " + geometrikSeriesSumPower (2, 4));

}

...

publicstaticint geometrikSeriesSumPower (int x, int n) {

return powerB(x, n + 1) – 1 / (x - 1);

}

publicstaticint powerB(int x, int n) {

if (n == 0) {

return 1;

} elseif (n % 2 == 0) { // n is even

return powerB(x * x, n / 2);

} else { // n is odd

return x * powerB(x * x, n / 2);

}

}

}

x=2, n=4, a4=31

X must be power of 2

output

Analysis of Algorithms

algorithmT(n)

Sum11/2 n2 + 47/2 n + 27

Horner13n + 22

Power19(log2(n+1) + 1) + 18

sum

Horner

Power

Analysis of Algorithms

By inspecting the pseudocode, we can determine the maximum number of primitive operations executed by an algorithm, as a function of the input size

AlgorithmarrayMax(A, n)

# operations

currentMaxA[0] 2

fori1ton 1do 2n

ifA[i] currentMaxthen2(n 1)

currentMaxA[i]2(n 1)

{ increment counter i }2(n 1)

returncurrentMax 1

Total 8n 2

Analysis of Algorithms

Algorithm arrayMax executes 8n 2 primitive operations in the worst case. Define:

a= Time taken by the fastest primitive operation

b = Time taken by the slowest primitive operation

Let T(n) be worst-case time of arrayMax.Thena (8n 2) T(n)b(8n 2)

Hence, the running time T(n) is bounded by two linear functions

Analysis of Algorithms

- The asymptotic analysis of an algorithm determines the running time in big-Oh notation
- To perform the asymptotic analysis
- We find the worst-case number of primitive operations executed as a function of the input size
- We express this function with big-Oh notation

- Example:
- We determine that algorithm arrayMax executes at most 8n 2 primitive operations
- We say that algorithm arrayMax “runs in O(n) time”

- Since constant factors and lower-order terms are eventually dropped anyhow, we can disregard them when counting primitive operations

Analysis of Algorithms

- Changing the hardware/ software environment
- Affects T(n) by a constant factor, but
- Does not alter the growth rate of T(n)

- The linear growth rate of the running time T(n) is an intrinsic property of algorithm arrayMax

Analysis of Algorithms

- The growth rate is not affected by
- constant factors or
- lower-order terms

- Examples
- 102n+105is a linear function
- 105n2+ 108nis a quadratic function

Analysis of Algorithms

- Given functions f(n) and g(n), we say that f(n) is O(g(n))if there are positive constantsc and n0 such that
f(n)cg(n) for n n0

- Example: 2n+10 is O(n)
- 2n+10cn
- (c 2) n 10
- n 10/(c 2)
- Pick c = 3 and n0 = 10

Analysis of Algorithms

- Example: the function n2is not O(n)
- n2cn
- n c
- The above inequality cannot be satisfied since c must be a constant

Analysis of Algorithms

More Big-Oh Examples

7n-2

7n-2 is O(n)

need c > 0 and n0 1 such that 7n-2 c•n for n n0

this is true for c = 7 and n0 = 1

- 3n3 + 20n2 + 5

3n3 + 20n2 + 5 is O(n3)

need c > 0 and n0 1 such that 3n3 + 20n2 + 5 c•n3 for n n0

this is true for c = 4 and n0 = 21

- 3 log n + 5

3 log n + 5 is O(log n)

need c > 0 and n0 1 such that 3 log n + 5 c•log n for n n0

this is true for c = 8 and n0 = 2

Analysis of Algorithms

- The big-Oh notation gives an upper bound on the growth rate of a function
- The statement “f(n) is O(g(n))” means that the growth rate of f(n) is no more than the growth rate of g(n)
- We can use the big-Oh notation to rank functions according to their growth rate

Analysis of Algorithms

Analysis of Algorithms

Stirlings approximation

Analysis of Algorithms

Theorem

Consider polynomial where am>0.Then f(n) = O(nm). i.e.,

- Drop lower-order terms
- Drop constant factors

- Use the smallest possible class of functions
- Say “2n is O(n)”instead of “2n is O(n2)”

- Use the simplest expression of the class
- Say “3n+5 is O(n)”instead of “3n+5 is O(3n)”

Analysis of Algorithms

- logk n = O(n) for any constant kZ+

f(n) = O (f(n))

c O(f(n)) = O (f(n))

O(f(n)) + O(f(n)) = O(f(n))

O(O(f(n))) = O(f(n))

O(f(n)) O(g(n)) = O(f(n) g(n))

Analysis of Algorithms

Relatives of Big-Oh

- big-Omega
- f(n) is (g(n)) if there is a constant c > 0
and an integer constant n0 1 such that

f(n) c•g(n) for n n0

- f(n) is (g(n)) if there is a constant c > 0
- big-Theta
- f(n) is (g(n)) if there are constants c’ > 0 and c’’ > 0 and an integer constant n0 1 such that c’•g(n) f(n) c’’•g(n) for n n0

Analysis of Algorithms

Big-Oh

- f(n) is O(g(n)) if f(n) is asymptotically less than or equal to g(n)
big-Omega

- f(n) is (g(n)) if f(n) is asymptotically greater than or equal to g(n)
big-Theta

- f(n) is (g(n)) if f(n) is asymptotically equal to g(n)

Analysis of Algorithms

Example Uses of the Relatives of Big-Oh

- 5n2 is (n2)

f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) c•g(n) for n n0

let c = 5 and n0 = 1

- 5n2 is O(n2)

f(n) is O(g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) < c•g(n) for n n0

let c = 5 and n0 = 1

- 5n2 is (n2)

f(n) is (g(n)) if it is (n2) and O(n2). We have already seen the former, for the latter recall that f(n) is O(g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) <c•g(n) for n n0

Let c = 5 and n0 = 1

Analysis of Algorithms

O(1) < O(log n) < O(n) < O(n log n) < O(n2) < O(n3) < O(2n)

Analysis of Algorithms

O(1) < O(log n) < O(n) < O(n log n) < O(n2) < O(n3) < O(2n)

O(1)

O(log n)

O(n)

O(n log n)

O(n2)

O(n3)

O(2n)

Analysis of Algorithms

Worst-case running time if statement 5 executes all the time…

Best-case running time if statement 5 never executes…

On-the-average running time if statement 5 executes half of the time????

int findMaximum(int[] a) {

int result = a[0];

for (int i = 1; i < a.length; ++i) {

if (result < a[i]) {

result = a[i];

}

}

return result;

}

1

2

3

4

5

6

7

8

9

Analysis of Algorithms

- A comparison-based algorithm for searching or sorting a list is
- based on
- making comparisons involving list elements
- then making decisions based on these comparisons.

- SOON U ll know all about these!!!

Analysis of Algorithms

- Solve Exercises in Chapter2 : 1, 7, 9 ,12, 25, 29, 31

Analysis of Algorithms

- We further illustrate asymptotic analysis with two algorithms for prefix averages
- The i-th prefix average of an array X is average of the first (i+ 1) elements of X:
A[i]= (X[0] +X[1] +… +X[i])/(i+1)

- Computing the array A of prefix averages of another array X has applications to financial analysis such as ARMA, ARIMA, FARIMA, models.

Analysis of Algorithms

Prefix Averages (Quadratic) (*)

The following algorithm computes prefix averages in quadratic time by applying the definition

AlgorithmprefixAverages1(X, n)

Inputarray X of n integers

Outputarray A of prefix averages of X#operations

A new array of n integers n

fori0ton 1don

sX[0] n

forj1toido 1 + 2 + …+ (n 1)

ss+X[j] 1 + 2 + …+ (n 1)

A[i]s/(i+ 1)n

returnA 1

Analysis of Algorithms

The running time of prefixAverages1 isO(1 + 2 + …+ n)

The sum of the first n integers is n(n+ 1) / 2

There is a simple visual proof of this fact

Thus, algorithm prefixAverages1 runs in O(n2) time

Analysis of Algorithms

Prefix Averages (Linear) (*)

The following algorithm computes prefix averages in linear time by keeping a running sum

AlgorithmprefixAverages2(X, n)

Inputarray X of n integers

Outputarray A of prefix averages of X #operations

A new array of n integersn

s 0 1

fori0ton 1don

ss+X[i]n

A[i]s/(i+ 1)n

returnA 1

Algorithm prefixAverages2 runs in O(n) time

Analysis of Algorithms

2

1

0

- A CPU
- A potentially unbounded bank of memory cells, each of which can hold an arbitrary number or character

Memory cells are numbered and accessing any cell in memory takes unit time.

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms

Analysis of Algorithms