Software-Engineering
October 5, 2023Software-Engineering
October 5, 2023Software-Engineering
| Question 27 |
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
| 2 | |
| 3 | |
| 4 | |
| 5 |
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.
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.
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