This material is highly important for those who are going for ISTQB training.

This is taken from ..Fenton and Pfleeger, Chapters 7 & 8.

If we talk about white box testing…this is a very important to understand what is the structure of the code..

The more “complex” the structure of code

◦ . . . the harder it is to understand

◦ . . . the more likely it will have a defect

◦ . . . the harder it will be to change

◦ . . . the longer it will take to produce

◦ . . . the more difficult it will be to reuse

Let us understand what is structure..

What is structure?

• control-flow — the sequence of instructions that are

executed

• data-flow — the creation and movement of data between

“components” of the code

• data organisation — the relationship of data items to each

other

let me go back to our engineering …where we have learnt the graph theory…

There are two types of graph mainly available..

1.Directed 2.Undirected.

**Directed Graphs**

• a mathematical structure (with an appealing visual

representation) for representing things that are related

• consists of vertices (or nodes, or points), connected by

edges (or line segments, or arcs).

Vertices: A, B, C, D, E, F

Edges: (A,C), (A,C), (A,E), (B,C), (B,F), (C,D), (D,A), (E,F),

• Also: undirected graphs, rules restricting edges between

vertices, classification of vertices

now the question comes like what is flowgraph?

**Flowgraphs**

• Directed graphs (flowgraphs) can be used to model control

flow of a program

• vertex = statement, edge = (A,B) if control flows from

statement A to B

• properties of flowgraphs may provide information about

properties of the code

in the exam the question comes like find out the complexity of the below written code:

Example:

```
public static boolean isPrime(int n) {
boolean prime = true;
int i = 2;
while (i < n)
{
if (n % i == 0)
{
prime = false;
}
i++;
}
return prime;
}
```

let us put notation over here: as each of the line /statement is under flowgraph…i am putting A,B,C as the notation for the lines.Those will be my nodes where the controll of the pragramme will pass.

the graph looks like:

McCabe’s Cyclomatic Complexity Number (CCN)

• measures the number of linearly independent paths through

the flowgraph

• v(F) = e − n + 2, F the flowgraph of the code, n the number

of vertices, e the number of edges

• Intuition — the larger the CCN the “more complex” the code

• Various sources recommend a CCN of no more than 10-15

• Example: CCN = 8 − 7 + 2 = 3

try to understand this. and I will come back with several other examples.

References:

- https://quizlet.com/41660194/strategic-management-test-2-flash-cards/
- https://pediaa.com/what-is-the-difference-between-directed-and-undirected-graph/
- https://en.wikipedia.org/wiki/Cyclomatic_complexity
- https://epdf.pub/graph-algorithms-and-applications-i.html
- https://homepages.dcc.ufmg.br/~mtov/pmcc/cyclomatic_complexity_tempero.pdf

AnonymousHello Ani !!!!!

Really thanks …. i m very confused in that topic but that article completely solve my problem.

tushar salgaonkar

Anonymousif x<>=z

then stmt 2;

end

According to graph,i got 3 nodes and 2 edges.so,2-3+2=1.

But answer is,3

I couldn't understand.Plz explain

Animesh ChatterjeeHei..we need to consider the else part as well

.

a…if x<>=z then

b…stmt 2

c…end if

Consider a case.. when x=z then it has to go end if i.e the c node

so there are two paths between A to c

1.a->b->c(if part)

2.a->c(else part)

so no of edges=3

No of nodes=3

so the cyclomatic complexity=e-N+2

3-3+2=2

look the second example above…see there is a if statement on D node…look even if there is no else part but that has been taken care.

same case will happen here also.

If you've still doubt, there is an easier equation that will make sense. Cyclomatic complexity can be explained in layman's terms as follows: every decision point in a method (i.e., an if, for, while, or case statement) is counted; additionally, one is added for the method's entry point, resulting in an integer-based measurement denoting a method's complexity.

so in this case there is only one if and only one entry point at node a.

so one if+one entry=Complexity=2

Animesh ChatterjeeSee my second exmaple where there is only one while and one if….

and also one entry point

so CC=1while+i if+1entry point

CC=3

AnonymousThanks Buddy!Now I understood.Thanks for Timely Response.

vinitaint i,j;

if(i>0 && j>0)

{

while(i>j)

{

if(i%2 && j%2)

printf("%dn" ,i);

else

printf("%dn" ,j);

i–;

}

}

The CC will be 4 is it correct?

Animesh ChatterjeeCorrect Vinita…

E=10

N=8

CC=E-N+2

CC=10-8+2

CC=4

let me know if you need any more help!!