WHITE BOX TESTING: CYCLOMATIC COMPLEXITY, BASIS PATH, AND CONTROL FLOW

PROBLEM 1: STUDENT GRADE CLASSIFICATION


Algorithm


START

INPUT percentage

IF percentage >= 60 THEN

    grade = "First Class"

    PRINT "First Class with Distinction"

ELSE IF percentage >= 50 AND percentage < 60 THEN

    grade = "First Class"

    PRINT "First Class"

ELSE IF percentage >= 40 AND percentage < 50 THEN

    grade = "Second Class"

    PRINT "Second Class"

ELSE IF percentage >= 35 AND percentage < 40 THEN

    grade = "Pass"

    PRINT "Pass Class"

ELSE

    grade = "Fail"

    PRINT "Fail"

END IF

STOP

 Flow Chart


        [START]

           |

           v

    [Input percentage]

           |

           v

    <percentage >= 60?>

       /        \

     YES         NO

      |           |

      v           v

 [First Class  <percentage >= 50?>

 Distinction]   /           \

      |       YES            NO

      |        |             |

      |        v             v

      |   [First Class]  <percentage >= 40?>

      |        |          /            \

      |        |        YES             NO

      |        |         |              |

      |        |         v              v

      |        |   [Second Class]  <percentage >= 35?>

      |        |         |           /           \

      |        |         |         YES            NO

      |        |         |          |             |

      |        |         |          v             v

      |        |         |      [Pass Class]   [Fail]

      |        |         |          |             |

      \--------+---------+----------+-------------/

                          |

                          v

                      [STOP]

Control Flow Diagram (CFD)


Nodes:

1. Start

2. Input percentage

3. Check percentage >= 60

4. Print "First Class with Distinction"

5. Check percentage >= 50

6. Print "First Class"

7. Check percentage >= 40

8. Print "Second Class"

9. Check percentage >= 35

10. Print "Pass Class"

11. Print "Fail"

12. Stop


Edges:

1 → 2

2 → 3

3 → 4 (True)

3 → 5 (False)

4 → 12

5 → 6 (True)

5 → 7 (False)

6 → 12

7 → 8 (True)

7 → 9 (False)

8 → 12

9 → 10 (True)

9 → 11 (False)

10 → 12

11 → 12

 Cyclomatic Complexity Calculation


Method 1: Using Formula V(G) = E - N + 2

- Edges (E) = 14

- Nodes (N) = 12

- V(G) = 14 - 12 + 2 = 4


Method 2: Using Decision Points V(G) = D + 1

- Decision points (D) = 4 (four IF conditions)

- V(G) = 4 + 1 = 5


Method 3: Using Regions

- Number of regions = 5

- V(G) = 5


 Independent Paths (Basis Path Testing)


1. Path 1: 1 → 2 → 3 → 4 → 12 (percentage >= 60)

   - Test Case: percentage = 65

   - Expected Output: "First Class with Distinction"


2. Path 2: 1 → 2 → 3 → 5 → 6 → 12 (50 <= percentage < 60)

   - Test Case: percentage = 55

   - Expected Output: "First Class"


3. Path 3: 1 → 2 → 3 → 5 → 7 → 8 → 12 (40 <= percentage < 50)

   - Test Case: percentage = 45

   - Expected Output: "Second Class"


4. Path 4: 1 → 2 → 3 → 5 → 7 → 9 → 10 → 12 (35 <= percentage < 40)

   - Test Case: percentage = 37

   - Expected Output: "Pass Class"


5. Path 5: 1 → 2 → 3 → 5 → 7 → 9 → 11 → 12 (percentage < 35)

   - Test Case: percentage = 30

   - Expected Output: "Fail"


 Test Cases for Basis Path Coverage


| Test Case ID | Input (percentage) | Expected Output | Path Covered |

| TC1 | 65 | First Class with Distinction | Path 1 |

| TC2 | 55 | First Class | Path 2 |

| TC3 | 45 | Second Class | Path 3 |

| TC4 | 37 | Pass Class | Path 4 |

| TC5 | 30 | Fail | Path 5 |

| TC6 | 60 | First Class with Distinction | Boundary |

| TC7 | 50 | First Class | Boundary |

| TC8 | 40 | Second Class | Boundary |

| TC9 | 35 | Pass Class | Boundary |



 PROBLEM 2: ODD OR EVEN NUMBER

Algorithm


START

INPUT number

IF number % 2 == 0 THEN

    PRINT "Even"

ELSE

    PRINT "Odd"

END IF

STOP

 Flow Chart

[START]

           |

           v

    [Input number]

           |

           v

    <number % 2 == 0?>

       /        \

     YES         NO

      |           |

      v           v

  [Print "Even"] [Print "Odd"]

      |           |

      \-----------/

           |

           v

        [STOP]

 Control Flow Diagram (CFD)


Nodes:

1. Start

2. Input number

3. Check number % 2 == 0

4. Print "Even"

5. Print "Odd"

6. Stop


Edges:

1 → 2

2 → 3

3 → 4 (True)

3 → 5 (False)

4 → 6

5 → 6

 Cyclomatic Complexity Calculation


Method 1: Using Formula V(G) = E - N + 2

- Edges (E) = 6

- Nodes (N) = 6

- V(G) = 6 - 6 + 2 = 2


Method 2: Using Decision Points V(G) = D + 1

- Decision points (D) = 1

- V(G) = 1 + 1 = 2


 Independent Paths


1. Path 1: 1 → 2 → 3 → 4 → 6 (number is even)

   - Test Case: number = 10

   - Expected Output: "Even"


2. Path 2: 1 → 2 → 3 → 5 → 6 (number is odd)

   - Test Case: number = 7

   - Expected Output: "Odd"


 Test Cases


| Test Case ID | Input (number) | Expected Output | Path Covered |

| TC1 | 10 | Even | Path 1 |

| TC2 | 7 | Odd | Path 2 |

| TC3 | 0 | Even | Boundary |

| TC4 | -4 | Even | Negative Even |

| TC5 | -5 | Odd | Negative Odd |

| TC6 | 2 | Even | Smallest Even |

| TC7 | 1 | Odd | Smallest Odd |


 PROBLEM 3: PRIME NUMBER CHECK


Algorithm


START

INPUT number

IF number <= 1 THEN

    PRINT "Not Prime"

    GOTO STOP

END IF

IF number == 2 THEN

    PRINT "Prime"

    GOTO STOP

END IF

IF number % 2 == 0 THEN

    PRINT "Not Prime"

    GOTO STOP

END IF

SET i = 3

WHILE i * i <= number DO

    IF number % i == 0 THEN

        PRINT "Not Prime"

        GOTO STOP

    END IF

    i = i + 2

END WHILE

PRINT "Prime"

STOP

 Flow Chart


            [START]

               |

               v

        [Input number]

               |

               v

        <number <= 1?>

           /      \

         YES       NO

          |         |

          v         v

    [Not Prime] <number == 2?>

          |       /        \

          |     YES         NO

          |      |           |

          |      v           v

          | [Prime]    <number % 2 == 0?>

          |      |         /        \

          |      |       YES         NO

          |      |        |           |

          |      |        v           v

          |      |   [Not Prime]  [i = 3]

          |      |        |           |

          |      |        |           v

          |      |        |    <i*i <= number?>

          |      |        |       /         \

          |      |        |     YES          NO

          |      |        |      |            |

          |      |        |      v            v

          |      |        | <number % i == 0?> [Prime]

          |      |        |   /        \        |

          |      |        | YES         NO      |

          |      |        |  |          |       |

          |      |        |  v          v       |

          |      |        | [Not Prime] [i=i+2] |

          |      |        |  |          |       |

          |      |        |  |          |       |

          \------+--------+--+----------+-------/

                 |

                 v

              [STOP]

Control Flow Diagram (CFD)


Nodes:

1. Start

2. Input number

3. Check number <= 1

4. Print "Not Prime" (for <= 1)

5. Check number == 2

6. Print "Prime" (for 2)

7. Check number % 2 == 0

8. Print "Not Prime" (for even)

9. i = 3

10. Check i * i <= number

11. Check number % i == 0

12. Print "Not Prime" (divisible)

13. i = i + 2

14. Print "Prime"

15. Stop


Edges:

1 → 2

2 → 3

3 → 4 (True)

3 → 5 (False)

4 → 15

5 → 6 (True)

5 → 7 (False)

6 → 15

7 → 8 (True)

7 → 9 (False)

8 → 15

9 → 10

10 → 11 (True)

10 → 14 (False)

11 → 12 (True)

11 → 13 (False)

12 → 15

13 → 10

14 → 15

Cyclomatic Complexity Calculation


Method 1: Using Formula V(G) = E - N + 2

- Edges (E) = 17

- Nodes (N) = 15

- V(G) = 17 - 15 + 2 = 4


Method 2: Using Decision Points V(G) = D + 1

- Decision points (D) = 5

- V(G) = 5 + 1 = 6

 Independent Paths


1. Path 1: 1 → 2 → 3 → 4 → 15 (number <= 1)

   - Test Case: number = 0

   - Expected Output: "Not Prime"


2. Path 2: 1 → 2 → 3 → 5 → 6 → 15 (number == 2)

   - Test Case: number = 2

   - Expected Output: "Prime"


3. Path 3: 1 → 2 → 3 → 5 → 7 → 8 → 15 (even number)

   - Test Case: number = 4

   - Expected Output: "Not Prime"


4. Path 4: 1 → 2 → 3 → 5 → 7 → 9 → 10 → 11 → 12 → 15 (divisible by odd)

   - Test Case: number = 9

   - Expected Output: "Not Prime"


5. Path 5: 1 → 2 → 3 → 5 → 7 → 9 → 10 → 11 → 13 → 10 → 14 → 15 (prime)

   - Test Case: number = 7

   - Expected Output: "Prime"


6. Path 6: 1 → 2 → 3 → 5 → 7 → 9 → 10 → 14 → 15 (small prime)

   - Test Case: number = 3

   - Expected Output: "Prime"


Test Cases


| Test Case ID | Input (number) | Expected Output | Path Covered | Reason |

| TC1 | 0 | Not Prime | Path 1 | number <= 1 |

| TC2 | 1 | Not Prime | Path 1 | number <= 1 |

| TC3 | -5 | Not Prime | Path 1 | Negative |

| TC4 | 2 | Prime | Path 2 | Only even prime |

| TC5 | 4 | Not Prime | Path 3 | Even composite |

| TC6 | 9 | Not Prime | Path 4 | Odd composite (3²) |

| TC7 | 15 | Not Prime | Path 4 | Odd composite |

| TC8 | 7 | Prime | Path 5 | Odd prime |

| TC9 | 11 | Prime | Path 5 | Odd prime |

| TC10 | 3 | Prime | Path 6 | Smallest odd prime |

| TC11 | 17 | Prime | Path 5 | Larger prime |

| TC12 | 25 | Not Prime | Path 4 | Perfect square |



SUMMARY OF CYCLOMATIC COMPLEXITY


| Problem | Cyclomatic Complexity | Independent Paths | Test Cases Required |

| Student Grade Classification | 5 | 5 | 9 |

| Odd or Even | 2 | 2 | 7 |

| Prime Number | 6 | 6 | 12 |


Note: Cyclomatic Complexity indicates the minimum number of test cases required to achieve basis path coverage and represents the complexity of the program.

Comments

Post a Comment

Popular posts from this blog

BLACK BOX TESTING: BOUNDARY VALUE ANALYSIS AND EQUIVALENCE PARTITIONING

PROBLEM 1: DISCOUNT CALCULATION ON PURCHASE Problem Statement A discount is offered on purchases as follows: | Purchase Range | Discount Percentage | | 1000 Rs to 5000 Rs | 10% | | 5001 Rs to 10000 Rs | 20% | | Above 10000 Rs | 25% |  Equivalence Classes Valid Classes: - VC1: Purchase amount 1000 to 5000 Rs → 10% discount - VC2: Purchase amount 5001 to 10000 Rs → 20% discount - VC3: Purchase amount above 10000 Rs → 25% discount Invalid Classes: - IC1: Purchase amount < 1000 Rs → No discount (0%) - IC2: Purchase amount = 0 Rs → Invalid - IC3: Purchase amount < 0 Rs → Invalid (Negative value) - IC4: Non-numeric input → Invalid  Boundary Value Analysis Boundaries Identified: - Lower boundary of VC1: 1000 Rs - Upper boundary of VC1: 5000 Rs - Lower boundary of VC2: 5001 Rs - Upper boundary of VC2: 10000 Rs - Lower boundary of VC3: 10001 Rs Boundary Values: - Just below 1000: 999 Rs - At boundary 1000: 1000 Rs - Just above 1000: 1001 Rs - Just below 5000: 4999 Rs - At boundar...
Read more