Various examples in Basis Path Testing

Prerequisite – Basis Path Testing
We have seen the steps involved in designing the test cases for a program using the basis path testing in the previous article. Now, let’s solve an example following the same steps.

Question : Consider the given program that checks if a number is prime or not. For the following program :

  1. Draw the Control Flow Graph
  2. Calculate the Cyclomatic complexity using all the methods
  3. List all the Independent Paths
  4. Design test cases from independent paths
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int main()
{
    int n, index;
    cout << "Enter a number: " << endl;
    cin >> n;
    index = 2;
    while (index <= n - 1) {
        if (n % index == 0) {
            cout << "It is not a prime number" << endl;
            break;
        }
        index++;
    }
    if (index == n)
        cout << "It is a prime number" << endl;
} // end main

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Solution :
1. Draw the Control Flow Graph –

  • Step-1:
    Start numbering the statements after declaration of the variables (if no variables have been initialized in that statement). However, if a variable has been initialized and declared in the same line, then numbering should start from that line itself.

    For the given program, this is how numbering will be done:



    int main()
    {
        int n, index;
    1    cout << "Enter a number: " <> n;
    3    index = 2;
    4    while (index <= n - 1) 
    5    {
    6        if (n % index == 0) 
    7        {
    8            cout << "It is not a prime number" << endl;
    9            break;
    10        }
    11        index++;
    12    }
    13    if (index == n)
    14        cout << "It is a prime number" << endl;
    15  } // end main
    
  • Step-2:
    Put the sequential statements into one single node. For example, statements 1, 2 and 3 are all sequential statements and hence should be combined into a single node. And for other statements, we will follow the notations as discussed here.

    Note –
    Use alphabetical numbering on nodes for simplicity.

    The graph obtained will be as follows :

2. Calculate the Cyclomatic complexity :

  • Method-1:
    V(G) = e - n + 2*p 

    In the above control flow graph,

    where, e = 10, n = 8 and p = 1
    
    Therefore, 
    Cyclomatic Complexity V(G) 
    = 10 - 8 + 2 * 1 
    = 4
    
  • Method-2:
    V(G) = d + p 

    In the above control flow graph,

    where, d = 3 (Node B, C and F) and p = 1
    
    Therefore, 
    Cyclomatic Complexity V(G) 
    = 3 + 1 
    = 4
    
  • Method-3:
    V(G) = Number of Regions 

    In the above control flow graph, there are 4 regions as shown below :



    Therefore, there are 4 regions: R1, R2, R3 and R4
    
    Cyclomatic Complexity V(G) 
    = 1 + 1 + 1 + 1 
    = 4
    

It is important to note that all three methods give same value for cyclomatic complexity V(G).

3. Independent Paths :
As the cyclomatic complexity V(G) for the graph has come out to be 4, therefore there are 4 independent paths.
Edges covered (marked with red) by Path 1 are:

Path 1 : A - B - F - G - H 

Edges covered by Path 1 and Path 2 are shown below :

Path 2 : A - B - F - H 

Edges covered by Path 1, Path 2 and Path 3 are :

Path 3 : A - B - C - E - B - F - G - H 

Now only 2 edges are left uncovered i.e. edge C-D and edge D-F. Hence, Path 4 must include these two edges.

Path 4 : A - B - C - D - F - H 

Each of these paths have introduced at least one new edge which has not been traversed before.

Note –
Independent paths are not necessarily unique.

4. Test cases :
To derive test cases, we have to use the independent paths obtained previously. To design a test case, provide input to the program such that each independent path is executed.
For the given program, the following test cases will be obtained:

Test case ID Input Number Output Independent Path covered
1 1 No output A-B-F-H
2 2 It is a prime number A-B-F-G-H
3 3 It is a prime number A-B-C-E-B-F-G-H
4 4 It is not a prime number A-B-C-D-F-H

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