Cyclomatic-metric

Question 1

The cyclomatic complexity of the flow graph of a program provides

A
an upper bound for the number of tests that must be conducted to ensure that all statements have been executed at most once
B
a lower bound for the number of tests that must be conducted to ensure that all statements have been executed at most once
C
an upper bound for the number of tests that must be conducted to ensure that all statements have been executed at least once
D
a lower bound for the number of tests that must be conducted to ensure that all statements have been executed at least once
Question 1 Explanation: 
Note: Out of syllabus.
Question 2

With respect to software testing, consider a flow graph G with one connected component. Let E be the number of edges, N be the number of nodes, and P be the number of predicate nodes of G. Consider the following four expressions:
1. E - N + P
2. E - N + 2
3. P + 2
4. P + 1
The cyclomatic complexity of G is given by

A
1 or 3
B
2 or 3
C
2 or 4
D
1 or 4
Question 2 Explanation: 
Note: Out of syllabus.
Question 3

Consider the following pseudo-code:

 IF ((A > B) AND (C > D)) THEN
       A = A + 1
       B = B + 1
       ENDIF

The cyclomatic complexity of the pseudo-code is

A
2
B
3
C
4
D
5
Question 3 Explanation: 
The cyclomatic complexity of a program can be calculated using the formula: V(G) = E - N + 2P Where: V(G) is the cyclomatic complexity. E is the number of edges in the control flow graph. N is the number of nodes in the control flow graph. P is the number of connected components (usually 1). In your pseudo code, there are two conditional statements (A > B and C > D), and an "ENDIF" which represents the end of the conditional block. This forms a basic control flow structure. Let's break it down into a control flow graph: Start (Node 1) Condition A > B (Node 2) Condition C > D (Node 3) A = A + 1 (Node 4) B = B + 1 (Node 5) EndIF (Node 6) Now, count the number of edges (E) and nodes (N): E = 7 (7 transitions between nodes) N = 6 (6 nodes) Since there is only one connected component (P = 1), we have: V(G) = E - N + 2P = 7 - 6 + 2 * 1 = 3 So, the cyclomatic complexity of your pseudo code is 3.
Question 4

Consider the following C program segment.

while (first <= last)
{
   if (array [middle] < search)
      first = middle +1;
   else if (array [middle] == search)
      found = True;
   else last = middle – 1;
   middle = (first + last)/2;
}
if (first < last) not Present = True;

The cyclomatic complexity of the program segment is __________.

A
5
B
6
C
7
D
8
Question 4 Explanation: 
Note: Out of syllabus.
There are 4 questions to complete.

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