Cyclomatic Complexity | Calculation | Examples

Cyclomatic Complexity-

 

Cyclomatic Complexity may be defined as-

• It is a software metric that measures the logical complexity of the program code.
• It counts the number of decisions in the given program code.
• It measures the number of linearly independent paths through the program code.

 

Cyclomatic complexity indicates several information about the program code-

 

Cyclomatic Complexity	Meaning
1 – 10	Structured and Well Written Code High Testability Less Cost and Effort
10 – 20	Complex Code Medium Testability Medium Cost and Effort
20 – 40	Very Complex Code Low Testability High Cost and Effort
> 40	Highly Complex Code Not at all Testable Very High Cost and Effort

 

Importance of Cyclomatic Complexity-

 

• It helps in determining the software quality.
• It is an important indicator of program code’s readability, maintainability and portability.
• It helps the developers and testers to determine independent path executions.
• It helps to focus more on the uncovered paths.
• It evaluates the risk associated with the application or program.
• It provides assurance to the developers that all the paths have been tested at least once.

 

Properties of Cyclomatic Complexity-

 

• It is the maximum number of independent paths through the program code.
• It depends only on the number of decisions in the program code.
• Insertion or deletion of functional statements from the code does not affect its cyclomatic complexity.
• It is always greater than or equal to 1.

 

Calculating Cyclomatic Complexity-

 

Cyclomatic complexity is calculated using the control flow representation of the program code.

 

In control flow representation of the program code,

• Nodes represent parts of the code having no branches.
• Edges represent possible control flow transfers during program execution

 

There are 3 commonly used methods for calculating the cyclomatic complexity-

 

Method-01:

 

Cyclomatic Complexity = Total number of closed regions in the control flow graph + 1

 

Method-02:

 

Cyclomatic Complexity = E – N + 2

 

Here-

• E = Total number of edges in the control flow graph
• N = Total number of nodes in the control flow graph

 

Method-03:

 

Cyclomatic Complexity = P + 1

 

Here,

P = Total number of predicate nodes contained in the control flow graph

 

Note-

 

• Predicate nodes are the conditional nodes.
• They give rise to two branches in the control flow graph.

 

PRACTICE PROBLEMS BASED ON CYCLOMATIC COMPLEXITY-

 

Problem-01:

 

Calculate cyclomatic complexity for the given code-

IF A = 354
   THEN IF B > C
        THEN A = B
        ELSE A = C
   END IF
END IF
PRINT A

 

Solution-

 

We draw the following control flow graph for the given code-

 

 

Using the above control flow graph, the cyclomatic complexity may be calculated as-

 

Method-01:

 

Cyclomatic Complexity

= Total number of closed regions in the control flow graph + 1

= 2 + 1

= 3

 

Method-02:

 

Cyclomatic Complexity

= E – N + 2

= 8 – 7 + 2

= 3

 

Method-03:

 

Cyclomatic Complexity

= P + 1

= 2 + 1

= 3

 

Problem-02:

 

Calculate cyclomatic complexity for the given code-

{ int i, j, k;
  for (i=0 ; i<=N ; i++)
  p[i] = 1;
  for (i=2 ; i<=N ; i++)
  {
     k = p[i]; j=1;
     while (a[p[j-1]] > a[k] {
         p[j] = p[j-1];
         j--;
  }
     p[j]=k;
}

 

Solution-

 

We draw the following control flow graph for the given code-

 

 

Using the above control flow graph, the cyclomatic complexity may be calculated as-

 

Method-01:

 

Cyclomatic Complexity

= Total number of closed regions in the control flow graph + 1

= 3 + 1

= 4

 

Method-02:

 

Cyclomatic Complexity

= E – N + 2

= 16 – 14 + 2

= 4

 

Method-03:

 

Cyclomatic Complexity

= P + 1

= 3 + 1

= 4

 

Problem-03:

 

Calculate cyclomatic complexity for the given code-

begin int x, y, power;
      float z;
      input(x, y);
      if(y<0)
      power = -y;
      else power = y;
      z=1;
      while(power!=0)
      {    z=z*x;
           power=power-1;
      } if(y<0)
      z=1/z;
      output(z);
      end

 

Solution-

 

We draw the following control flow graph for the given code-

 

 

Using the above control flow graph, the cyclomatic complexity may be calculated as-

 

Method-01:

 

Cyclomatic Complexity

= Total number of closed regions in the control flow graph + 1

= 3 + 1

= 4

 

Method-02:

 

Cyclomatic Complexity

= E – N + 2

= 16 – 14 + 2

= 4

 

Method-03:

 

Cyclomatic Complexity

= P + 1

= 3 + 1

= 4

 

To gain better understanding about Cyclomatic Complexity,

Watch this Video Lecture

 

Next Article- Cause Effect Graph Technique

 

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