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

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

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