This presentation discusses path testing, which is a white box testing technique that ensures every path in a program has been executed at least once. It requires complete knowledge of the program structure. The steps of path testing include converting the program language to a directed graph, determining the cyclomatic complexity, and determining the basis set of independent paths. For the example HR application program shown, it is determined to have 6 paths based on its cyclomatic complexity of 6. The 6 paths that cover the program are then listed.

1. Presentation on
Path Testing
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AGENDA

2. Path Testing
Path testing is a White box testing technique which ensures that every path in a program has
been executed at least once.
• It requires complete knowledge of program structure.
• It is most often used by programmers to test their own code.
Why do we need Path Testing?
Goal of the path testing is to pick enough paths to ensure that every source statement is
executed at least once and to eliminate the redundant code.
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3. • Converting the Programming Language to directed graph
• Converting Directed graph to DD-Path (Decision to Decision)
• Determine the Cyclomatic Complexity
• Determine the Basis set of Independent paths
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Path Testing Steps

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• 1 setDays = 0
• 2 availDays = 4
• 3 input = InputBox ("enter the selection 1: Paid Leave 2: LOP")
• 4 if (input=1) then
• 5 setDays = InputBox ("Enter the required number of paid leave")
• 6 if (setDays<=4 and setDays>0) Then
• 7 availDays = availDays - setDays
• 8 MsgBox("Paid Leave approved." &chr(13) &"Remaining Paid Leaves : "&availDays)
• 9 elseIf ((setDays > availDays) AND (availDays<>0)) Then
• 10 MsgBox ("Enter Valid days for leave")
• 11 elseif (setDays<=0) then
• 12 MsgBox("Enter Valid days for leave")
• 13 End If
• 14 elseif (input=2) then
• 15 setDays = InputBox ("Enter the required number of LOP ")
• 16 if (setDays>0) then
• 17 MsgBox ("LOP approved.")
• 18 else MsgBox ("Enter valid days for leave")
• 19 end if
• 20 else msgbox ("no selection")
• 21 end if
HRM Application with Directed Graph

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DD Path
Directed Graph DD Path

6. Case1: It consist of a single node with indeg=0
Case2: It consist of a single node with outdeg=0
Case3: It consist of a single node with indeg>=2 or Outdeg>=2
Case4: It consist of a single node with indeg=1 and Outdeg=1
Case5: It is maximal chain of length>=1
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DD path case:

7. Cyclomatic complexity is a software metric (measurement), used to
indicate the complexity of a program.
V(G)=E-N+2P
E-Number of Edges
N-Number of Nodes
P-Number of connected Region
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Determine Cyclomatic Complexity:

8. No of Edges=20
No of Nodes=16
Connected Region=1
V(G)=E-N+2P
V(G)=20-16+2(1)
= 4+2
V(G) = 6
Hence the number of paths for the given program will be
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Determine Cyclomatic Complexity:

9. i. P1: First, A,B,C,D,E,H,Last
ii. P2: First, A,B,C,D,F,H,Last
iii. P3: First, A,B,C,D,G,H,Last
iv. P4: First,A,B,I,J,K,M,Last
v. P5: First,A,B,I,J,L,M,Last
vi. P6: First,A,B,N,Last
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Paths covered in the program

10. THANK YOU
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