![Algorithm Analysis
FIND-MAX ( A, n )
1 max ←A[1] ►Store first array element into variable max
2 for j←2 to n do ► Scan remaining elements
3 if ( A[j] > max ) ►Compare an element with max
4 then max ← A[j] ► Replace max with a larger element
5 return max ► Return maximum element
The procedure FIND-MAX returns maximum element in an array. The array A and its
size n, are passed as arguments. The following pseudo code describes the essential steps,
together with comments which are identified by the symbol ►:
Pseudo Code](https://image.slidesharecdn.com/casestudyanalysisofalgorithm-221129152951-230b8250/85/Case-Study-Analysis-of-Algorithm-pdf-5-320.jpg)
![Running Time
Costs of Basic Operations
Cost of returning maximum element
Cr
return max
5
Cost of accessing A[j]
Cost of storing A[j] into max
Ca
Cs
then max ← A[j]
4
Cost of accessing A[j]
Cost of comparing A[j] with max, and branching
Ca
Cc
if ( A[j] > max )
3
Cost of storing 2 into j
Cost of comparing index j with n, and branching
Cost of incrementing j
Cs
Cc
Ci
for j←2 to n do
2
Cost of accessing A[1]
Cost of storing A[1] into max
Ca
Cs
max ←A[1]
1
Remarks
Unit costs
Statement
#
First, we identify basic operations and their associated costs . The table below lists various
operations and costs. Here the term ‘cost’ refers to the time consumed in executing an
operation.
Table of costs of basic operations](https://image.slidesharecdn.com/casestudyanalysisofalgorithm-221129152951-230b8250/85/Case-Study-Analysis-of-Algorithm-pdf-6-320.jpg)
![Running Time
Counts of Basic Operations
1
k
k
n-1
n-1
1
n-1
n-1
1
1
Operations
Count
Cr
Cr
return max
5
(Ca + Cs). k, where k depends on
condition in statement 3 . In general, 0≤ k
≤ n -1
Ca
Cs
then max ← A[j]
4
(n-1).Ca + (n-1).Cc
Ca
Cc
if ( A[j] > max )
3
Cs + (n-1).Cc +(n-1).Ci
Cs
Cc
Ci
for j←2 to n do
2
Ca + Cs
Ca
Cs
max ←A[1]
1
Total Cost
Unit Cost
Statement
#
Next we count the number of the basic operations, and total cost of executing each
statement. The frequency of execution of statement 4 depends on the outcome of statement
3; it will be executed when the condition A[j]>max turns out to be true. This condition ,in
turn, depends on the order of data in the array A. For the purpose of analysis, we assume
that statement 4 is executed k times, where 0 ≤ k ≤ n-1.
Table of frequency and total cost of basic operations](https://image.slidesharecdn.com/casestudyanalysisofalgorithm-221129152951-230b8250/85/Case-Study-Analysis-of-Algorithm-pdf-7-320.jpg)
![Running Time Classification
Best, Worst, Average Cases
The running time T(n) of the algorithm for finding the maximum element of array of size n is given by
T(n) = A + B.k + C.n, where 0 ≤ k ≤ n-1
Here k is the number of times the statement max ← A[j] will be executed. The following possibilities
can arise:
Best Case: Best case occurs when the statement is not executed at all. This happens when
the array maximum element occurs in the first cell. In this case k = 0, and best (minimum ) running time
is given by
Tbest (n)=A + C.n
Average Case: In this case, the statement is executed on an average n / 2 times so that k = n / 2. Thus,
average time running time is given by
Taverage(n) = A + (B / 2 + C).n
Worst Case: In this case, the statement is executed n-1 times; so k = n-1. This happens
when the array is sorted in ascending order .Thus, worst (maximum) running time is given by
Tworst(n) = A-B + (B + C).n](https://image.slidesharecdn.com/casestudyanalysisofalgorithm-221129152951-230b8250/85/Case-Study-Analysis-of-Algorithm-pdf-9-320.jpg)
![Algorithm Analysis
FIND-MAX(A)
1 max ← A[1]
2 for j←2 to n do
3 if( A[j] > max)
4 then max ← A[j]
1 return max
ƒ The correctness of FIND-MAX algorithm, listed below, is established by the
loop invariant method.
ƒ First, we define the loop invariant S as the following statement:
Variable max holds the largest value at all stages of loop execution
• Next, we consider the steps of initialization, maintenance, and termination
.
Correctness](https://image.slidesharecdn.com/casestudyanalysisofalgorithm-221129152951-230b8250/85/Case-Study-Analysis-of-Algorithm-pdf-11-320.jpg)
![Algorithm Analysis
Correctness
• The Initialization condition requires that prior to first iteration the statement S should be
true. This is trivially ( also called vacuously) true, because at this stage max contains the
single element A[1].
• The Maintenance condition requires that if S is true before an iteration of loop, it should
remain true after the iteration It can be easily verified that if max holds the largest of k
elements, after kth iteration, then it holds largest of k+1 elements after the next iteration.
ie.(k+1)st iteration
• The Termination condition requires that post-condition should be true for problem size
i.e, max should return maximum array element. The loop terminates when index j exceeds
n. This implies that just after the last iteration max holds the largest of the first n elements
of the array. Since array has size n, it means that max returns the largest of array elements.](https://image.slidesharecdn.com/casestudyanalysisofalgorithm-221129152951-230b8250/85/Case-Study-Analysis-of-Algorithm-pdf-12-320.jpg)
Case Study(Analysis of Algorithm.pdf
- 2. Case Study This case study is meant to demonstrate the salient features of algorithm design and analysis, as discussed previously. A simple example is used to systematically describe the following steps.: Algorithm Analysis • Problem statement • Algorithm design • Implementation using pseudo code • Analysis of best, worst and average running times • Space complexity • Correctness of algorithm • Visualization of Analysis
- 3. Case Study ¾ Design an algorithm to find maximum element in an array of size n ¾ Analyze the algorithm to determine : • Time efficiency • Space efficiency • Correctness Problem Statement
- 4. Case Study The design features are expressed in plain language. The algorithm for the solution of the problem consists of the following steps: Step #1: Store the first array element in variable max Step #2: Scan array to compare max with other elements Step #3: Replace max with a larger element, when found during the scan Step #4: Return value held by max Algorithm Design
- 5. Algorithm Analysis FIND-MAX ( A, n ) 1 max ←A[1] ►Store first array element into variable max 2 for j←2 to n do ► Scan remaining elements 3 if ( A[j] > max ) ►Compare an element with max 4 then max ← A[j] ► Replace max with a larger element 5 return max ► Return maximum element The procedure FIND-MAX returns maximum element in an array. The array A and its size n, are passed as arguments. The following pseudo code describes the essential steps, together with comments which are identified by the symbol ►: Pseudo Code
- 6. Running Time Costs of Basic Operations Cost of returning maximum element Cr return max 5 Cost of accessing A[j] Cost of storing A[j] into max Ca Cs then max ← A[j] 4 Cost of accessing A[j] Cost of comparing A[j] with max, and branching Ca Cc if ( A[j] > max ) 3 Cost of storing 2 into j Cost of comparing index j with n, and branching Cost of incrementing j Cs Cc Ci for j←2 to n do 2 Cost of accessing A[1] Cost of storing A[1] into max Ca Cs max ←A[1] 1 Remarks Unit costs Statement # First, we identify basic operations and their associated costs . The table below lists various operations and costs. Here the term ‘cost’ refers to the time consumed in executing an operation. Table of costs of basic operations
- 7. Running Time Counts of Basic Operations 1 k k n-1 n-1 1 n-1 n-1 1 1 Operations Count Cr Cr return max 5 (Ca + Cs). k, where k depends on condition in statement 3 . In general, 0≤ k ≤ n -1 Ca Cs then max ← A[j] 4 (n-1).Ca + (n-1).Cc Ca Cc if ( A[j] > max ) 3 Cs + (n-1).Cc +(n-1).Ci Cs Cc Ci for j←2 to n do 2 Ca + Cs Ca Cs max ←A[1] 1 Total Cost Unit Cost Statement # Next we count the number of the basic operations, and total cost of executing each statement. The frequency of execution of statement 4 depends on the outcome of statement 3; it will be executed when the condition A[j]>max turns out to be true. This condition ,in turn, depends on the order of data in the array A. For the purpose of analysis, we assume that statement 4 is executed k times, where 0 ≤ k ≤ n-1. Table of frequency and total cost of basic operations
- 8. Running Time Aggregate Cost • The running time T(n) is obtained by adding the costs in the last column of the table Tn) = Ca + Cs + Cs + (n-1).Cc + (n-1).Ci+ (n-1).Ca + (n-1).Cc + (Ca + Cs). k + Cr • Simplifying and rearranging, we get following expression . T(n) =A + B.k + C.n, where A = 2Cs – 2Cc – Ci + Cr, B = Ca + Cs, C = 2Cc + Ci + Ca • The constants A, B, C depend on the computing environment
- 9. Running Time Classification Best, Worst, Average Cases The running time T(n) of the algorithm for finding the maximum element of array of size n is given by T(n) = A + B.k + C.n, where 0 ≤ k ≤ n-1 Here k is the number of times the statement max ← A[j] will be executed. The following possibilities can arise: Best Case: Best case occurs when the statement is not executed at all. This happens when the array maximum element occurs in the first cell. In this case k = 0, and best (minimum ) running time is given by Tbest (n)=A + C.n Average Case: In this case, the statement is executed on an average n / 2 times so that k = n / 2. Thus, average time running time is given by Taverage(n) = A + (B / 2 + C).n Worst Case: In this case, the statement is executed n-1 times; so k = n-1. This happens when the array is sorted in ascending order .Thus, worst (maximum) running time is given by Tworst(n) = A-B + (B + C).n
- 10. Running Time Classification Plot of Best Worst and Average Cases A plot of running times for different cases is shown below. The running time in all of the three cases increases linearly. However, the slope of line (rate of increase) is different. In the case of worst time, for example, the running time increases at a greater rate.
- 11. Algorithm Analysis FIND-MAX(A) 1 max ← A[1] 2 for j←2 to n do 3 if( A[j] > max) 4 then max ← A[j] 1 return max ƒ The correctness of FIND-MAX algorithm, listed below, is established by the loop invariant method. ƒ First, we define the loop invariant S as the following statement: Variable max holds the largest value at all stages of loop execution • Next, we consider the steps of initialization, maintenance, and termination . Correctness
- 12. Algorithm Analysis Correctness • The Initialization condition requires that prior to first iteration the statement S should be true. This is trivially ( also called vacuously) true, because at this stage max contains the single element A[1]. • The Maintenance condition requires that if S is true before an iteration of loop, it should remain true after the iteration It can be easily verified that if max holds the largest of k elements, after kth iteration, then it holds largest of k+1 elements after the next iteration. ie.(k+1)st iteration • The Termination condition requires that post-condition should be true for problem size i.e, max should return maximum array element. The loop terminates when index j exceeds n. This implies that just after the last iteration max holds the largest of the first n elements of the array. Since array has size n, it means that max returns the largest of array elements.
- 13. Analysis of Algorithm The space analysis of algorithm for finding maximum element is simple and straightforward. It amounts to determining space utilization as function of data structure size. • The total space requirement consists of memory used by the program statements and array element. The former is a fixed and does not depend on array size. • The amount of storage requirement for the array depends on the nature of data type (integer, floating point, strings). It increases in direct proportion to the array size • Thus, space efficiency is given by S(n) = A + B.n Space Efficiency Array space requirement Program space requirement Visualization
- 14. Visualization