# 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 :

- Draw the Control Flow Graph
- Calculate the Cyclomatic complexity using all the methods
- List all the Independent Paths
- 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 |