Consider three software items: Program-$X$, Control Flow Diagram of Program-$Y$ and Control Flow Diagram of Program-$Z$ as shown below The values of McCabe's Cyclomatic complexity of Program-$X$, Program-$Y$, and Program-$Z$ respectively are

McCabe's Cyclomatic Complexity $V(G)$ of a control flow graph can be calculated using several standard methods [6]:

1. $V(G) = P + 1$, where $P$ is the number of predicate (decision) nodes.

2. $V(G) = E - N + 2$, where $E$ is the number of edges and $N$ is the number of nodes.

3. $V(G) = R$, where $R$ is the number of bounded and unbounded regions in the planar graph representation.

### 1. Cyclomatic Complexity of Program-$X$ Program-$X$ contains three decision statements:

• if (value < 0)

• while ((i < value) AND (result <= maxint))

• if (result <= maxint)

Number of predicate nodes $P = 3$.

### 2. Cyclomatic Complexity of Program-$Y$ From the control flow diagram for Program-$Y$:

• Total number of nodes $N = 8$

• Total number of directed edges $E = 10$

*(Alternatively, counting the regions in the graph: 3 enclosed regions $+ 1$ external region $= 4$.)*

### 3. Cyclomatic Complexity of Program-$Z$ Program-$Z$ is formed by connecting the exit of Program-$X$ to the entry of Program-$Y$ via a single directed edge:

• $N_Z = N_X + N_Y$

• $E_Z = E_X + E_Y + 1$

Thus, the cyclomatic complexities are 4, 4, 7, which corresponds to option (A).