Software Testing Laboratory

SOFTWARE TESTING LABORATORY

1. Design and develop a program in a language of your choice to solve the

triangle problem defined as follows: Accept three integers which are
supposed to be the three sides of a triangle and determine if the three values
represent an equilateral triangle, isosceles triangle, scalene triangle, or they do
not form a triangle at all. Assume that the upper limit for the size of any side is
10. Derive test cases for your program based on boundary- value analysis,
execute the test cases and discuss the results.

2. Design, develop, code and run the program in any suitable language to
solve the commission problem. Analyze it from the perspective of
Boundary value testing, derive different test cases, execute these test cases
and discuss the test results.

3. Design, develop, code and run the program in any suitable language to
implement the NextDate function. Analyze it from the perspective of
boundary value testing, derive different test cases, execute these test cases and
discuss the test results.

4. Design and develop a program in a language of your choice to solve the

triangle problem defined as follows: Accept three integers which are
supposed to be the three sides of a triangle and determine if the three values
represent an equilateral triangle, isosceles triangle, scalene triangle, or they do
not form a triangle at all. Assume that the upper limit for the size of any side is
10. Derive test cases for your program based on equivalence class partitioning,
execute the test cases and discuss the results.

5. Design, develop, code and run the program in any suitable language to solve
the commission problem. Analyze it from the perspective of
equivalence class testing, derive different test cases, execute these test cases
and discuss the test results.

6.Design, develop, code and run the program in any suitable language to
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implement the NextDate function. Analyze it from the perspective of

equivalence class value testing, derive different test cases, and execute these
test cases and discuss the test results.

7. Design and develop a program in a language of your choice to solve the

triangle problem defined as follows: Accept three integers which are
supposed to be the three sides of a triangle and determine if the three values
represent an equilateral triangle, isosceles triangle, scalene triangle, or they do
not form a triangle at all. Derive test cases for your program based on
decision-table approach, execute the test cases and discuss the results.

8. Design, develop, code and run the program in any suitable language to solve
the commission problem. Analyze it from the perspective of decision table-
based testing, derive different test cases, execute these test cases and discuss the
test results.

9. Design, develop, code and run the program in any suitable language to solve
the commission problem. Analyze it from the perspective of dataflow testing,
derive different test cases, execute these test cases and discuss the test results.

10. Design, develop, code and run the program in any suitable language to
implement the binary search algorithm. Determine the basis paths and using
them derive different test cases, execute these test cases and discuss the test
results.

11. Design, develop, code and run the program in any suitable language to
implement the quicksort algorithm. Determine the basis paths and using
them derive different test cases, execute these test cases and discuss the test
results. Discuss the test results.

12. Design, develop, code and run the program in any suitable language to
implement an absolute letter grading procedure, making suitable assumptions.
Determine the basis paths and using them derive different test cases, execute
these test cases and discuss the test results.

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1. Design and develop a program in a language of your choice to solve the

triangle problem defined as follows: Accept three integers which are supposed to be
the three sides of a triangle and determine if the three values represent an
equilateral triangle, isosceles triangle, scalene triangle, or they do not form a
triangle at all. Assume that the upper limit for the size of any side is 10. Derive
test cases for your program based on boundary-value analysis, execute the test
cases and discuss the results.

1.1 REQUIREMENT SPECIFICATIONS

R1. The system should accept 3 positive integer numbers (a, b, c) which
represents 3 sides of the triangle.
R2. Based on the input should determine if a triangle can be formed or
not.
R3. If the requirement R2 is satisfied then the system should determine the
type of the triangle, which can be
• Equilateral (i.e. all the three sides are equal)
• Isosceles (i.e Two sides are equal)
• Scalene (i.e All the three sides are unequal)
R4. Upper Limit for the size of any side is 10

1.2 DESIGN

Algorithm:
Step 1: Input a, b & c i.e three integer values which represent three sides of
the triangle.
Step 2: if (a < (b + c)) and (b < (a + c)) and (c < (a + b) then
do Step 3
else
print not a triangle. do Step 6.
Step 3: if (a=b) and (b=c) then
Print triangle formed is equilateral. do Step 6.
Step 4: if (a ≠ b) and (a ≠ c) and (b ≠ c) then

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Print triangle formed is scalene. do Step 6.

Step 5: Print triangle formed is Isosceles.
Step 6: stop

1.3 PROGRAM CODE:

#include<stdio.h>
#include<ctype.h>
#include<conio.h>
#include<process.h>
int main()
{
int a, b, c;
clrscr();
printf("Enter three sides of the triangle");
scanf("%d%d%d", &a, &b, &c);
if((a > 10) || (b > 10) || (c > 10))
{
printf("Out of range");
getch();
exit(0);
}
if((a<b+c)&&(b<a+c)&&(c<a+b))
{
if((a==b)&&(b==c))
{
printf("Equilateral triangle");
}
else if((a!=b)&&(a!=c)&&(b!=c))
{
printf("Scalene triangle");
}
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else
{
printf("Isosceles triangle");
}
}
else
{
printf("triangle cannot be formed");
}
getch();
return 0;
}

1.4 TESTING

1. Technique used: Boundary value analysis

2. Test Case design

For BVA problem the test cases can be generation depends on the
output and the constraints on the output. Here we least worried on the
constraints on Input domain.

The Triangle problem takes 3 sides as input and checks it for validity,
hence n = 3. Since BVA yields (4n + 1) test cases according to single fault
assumption theory, hence we can say that the total number of test cases will
be (4*3+1) =12+1=13.

The maximum limit of each side a, b, and c of the triangle is 10 units

according to requirement R4. So a, b and c lies between

1≤a≤10
1≤b≤10
1≤c≤10
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Equivalence classes for a:

E1: Values less than 1.
E2: Values in the range.
E3: Values greater than 10.

Equivalence classes for b:

E4: Values less than 1

E5: Values in the range.
E6: Values greater than 10.

Equivalence classes for c:

E7: Values less than 1.
E8: Values in the range.
E9: Values greater than 10.

From the above equivalence classes we can derive the following test cases
using boundary value analysis approach.
TC Test Case Input Data Expected Actual Status
Id Description A b C Output Output
1 For A input is not X 3 6 Not a Triangle
given
2 For B input is not 5 X 4 Not a Triangle
given
3 For C input is not 4 7 X Not a Triangle
given
4 Input of C is in 5 5 -1 Not a Triangle
negative(-)

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5 Two sides are 5 5 1 Isosceles

same one side is
given different
input

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6 All Sides of inputs 5 5 5 Equilateral

are equal
7 Two sides are 5 5 9 Isosceles
same one side is
given different
input
8 The input of C is 5 5 10 Not a Triangle
out of range (i.e.,
range is <10)
9 Two sides are 5 1 5 Isosceles
same one side is
givendifferent input
(i.e., A & C are 5,
B=1)

10 Two sides are 5 2 5 Isosceles

same one side is
givendifferent input
(i.e., A & C are 5,
B=2)

11 Two sides are 5 9 5 Isosceles

same one side is
given different
input (i.e., A &
C are 5, B=9)
12 Two sides are 5 10 5 Not a Triangle
same one side is
given different
input (i.e., A &
C are 5, B=10 so,
it isout of given
range)

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13 Two sides are 1 5 5 Isosceles

same one side is
given different
input (i.e., B &
C are 5, A=1)
14 Two sides are 2 5 5 Isosceles
same one side is
given different
input (i.e., B &
C are 5, A=2)
15 Two sides are 9 5 5 Isosceles
same one side is
given different
input (i.e., B &
16 C
Two sides are 10 5 5 Not a Triangle
same one side is
given different
Table-1: Test case for Triangle Problem

1.5 EXECUTION:

Execute the program and test the test cases in Table-1 against program and
complete the table with for Actual output column and Status column.

Test Report:
1. No of TC’s Executed:
2. No of Defects Raised:
3. No of TC’s Pass:
4. No of TC’s Failed:

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1.6 SNAPSHOTS:
1. Snapshot of Isosceles and Equilateral triangle and triangle can not be
formed.

2. Snapshot for Isosceles and triangle cannot be formed

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3. Snapshot for Isosceles and triangle cannot be formed

4. Output screen for Triangle cannot be formed

1.7 REFERENCES
1. Requirement Specification
2. Assumptions

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2. Design, develop, code and run the program in any suitable language to solve
the commission problem. Analyze it from the perspective of boundary value testing,
derive different test cases, execute these test cases and discuss the test results.

2.1 REQUIREMENT SPECIFICATION

Problem Definition: The Commission Problem includes a salesperson in the former

Arizona Territory sold rifle locks, stocks and barrels made by a gunsmith in
Missouri. Cost includes,
Locks- $45
Stocks- $30
Barrels- $25
The salesperson had to sell at least one complete rifle per month and
production limits were such that the most the salesperson could sell in a month
was 70 locks, 80 stocks and 90 barrels.
After each town visit, the sales person sent a telegram to the Missouri
gunsmith with the number of locks, stocks and barrels sold in the town. At the end of
the month, the salesperson sent a very short telegram showing --1 lock sold. The
gunsmith then knew the sales for the month were complete and computed the
salesperson’s commission as follows:
On sales up to(and including) $1000= 10% On the
sales up to(and includes) $1800= 15% On the sales
in excess of $1800= 20%
The commission program produces a monthly sales report that gave the total number
of locks, stocks and barrels sold, the salesperson’s total dollar sales and finally the
commission

2.2 DESIGN
Algorithm
Step 1: Define lockPrice=45.0, stockPrice=30.0, barrelPrice=25.0
Step 2: Input locks
Step 3: while(locks!=-1) ‘input device uses -1 to indicate end of data goto
Step 12
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Step 4:input (stocks, barrels)

Step 5: compute lockSales, stockSales, barrelSales and sales
Step 6: output(“Total sales:” sales)
Step 7: if (sales > 1800.0) goto Step 8 else goto Step 9
Step 8: commission=0.10*1000.0; commission=commission+0.15 *800.0;
commission = commission + 0.20 * (sales-1800.0)
Step 9: if (sales > 1000.0) goto Step 10 else goto Step 11
Step10: commission=0.10* 1000.0; commission=commission + 0.15 * (sales-
1000.0)
Step 11: Output(“Commission is $”, commission) Step
12: exit

2.3 PROGRAM CODE:

#include<stdio.h>
#include<conio.h>
int main()
{
int locks, stocks, barrels, t_sales, flag = 0;
float commission;
clrscr();
printf("Enter the total number of locks");
scanf("%d",&locks);
if ((locks <= 0) || (locks > 70))
{
flag = 1;

}
printf("Enter the total number of stocks");
scanf("%d",&stocks);
if ((stocks <= 0) || (stocks > 80))
{
flag = 1;
}
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printf("Enter the total number of barrelss");

scanf("%d",&barrels);
if ((barrels <= 0) || (barrels > 90))
{
flag = 1;
}
if (flag == 1)
{
printf("invalid input");
getch();
exit(0);
}
t_sales = (locks * 45) + (stocks * 30) + (barrels * 25);
if (t_sales <= 1000)
{
commission = 0.10 * t_sales;
}
else if (t_sales < 1800)
{
commission = 0.10 * 1000;
commission = commission + (0.15 * (t_sales - 1000));
}
else
{
commission = 0.10 * 1000;
commission = commission + (0.15 * 800);
commission = commission + (0.20 * (t_sales - 1800));
}
printf("The total sales is %d \n The commission is %f",t_sales,
commission);
getch(); return;

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2.4 TESTING

Technique used: Boundary value analysis

‘Boundary value analysis’ testing technique is used to identify errors at
boundaries rather than finding those exist in center of input domain.

Boundary value analysis is a next part of Equivalence partitioning for

designing test cases where test cases are selected at the edges of the
equivalence classes.

BVA: Procedure

1. Partition the input domain using unidimensional partitioning. This leads to as

many partitions as there are input variables. Alternately, a single partition of an
input domain can be created using multidimensional partitioning. We will
generate several sub-domains in this Step.
2. Identify the boundaries for each partition. Boundaries may also be
identified using special relationships amongst the inputs.
3. Select test data such that each boundary value occurs in at least one test input.
4. BVA: Example: Create equivalence classes

Assuming that an item code must be in the range 99...999 and quantity in the
range 1...100,
Equivalence classes for code:
E1: Values less than 99.
E2: Values in the range.
E3: Values greater than 999.

Equivalence classes for qty:

E4:Value less than 1.

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E5:Value in the range.

E6: Value greater than 100.

BVA: Example: Identify boundaries

Equivalence classes and boundaries for find Price. Boundaries are indicated with
an x. Points near the boundary are marked *.

Test Case design

The Commission Problem takes locks, stocks and barrels as input and checks it
for validity. If it is valid, it returns the commission as its output. Here we have
three inputs for the program, hence n = 3.

Since BVA yields (4n + 1) test cases according to single fault assumption
theory, hence we can say that the total number of test cases will be (4*3+1)
=12+1=13.

The boundary value test case can be generated over an output by using fallowing
constraints and these constraints are generated over commission:

C1: Sales up to(and including) $1000= 10% commission

C2: Sales up to(and includes) $1800= 15% commission
C3: Sales in excess of $1800= 20% commission

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Here from these constraints we can extract the test cases using the values of
Locks, Stocks, and Barrels sold in month. The boundary values for commission are
10%, 15% and 20%.
Equivalence classes for 10% Commission:
E1: Sales less than
1000. E2: Sales equals
to 1000.
Equivalence classes for 15% Commission:
E3: Sales greater than 1000 and less than 1800.
E4: Sales equals to 1800
Equivalence classes for 20% Commission:
E5: Sales greater then 1800

From the above equivalence classes we can derive the following test cases using
boundary value analysis approach.
TC Test Case Input Data Expected Actual Status
Id Description Locks Stocks Barrels Sales Output(Commission) Output
1 Input test cases 1 1 1 100 10
for Locks=1,
Stocks=1,
Barrels=1
2 Input test cases 1 1 2 125 12.5
for Locks=1,
Stocks=1,
Barrels=2
3 Input test cases 1 2 1 130 13
for Locks=1,
Stocks=2,
Barrels=1
4 Input test cases 2 1 1 145 14.5
for Locks=2,
Stocks=1,
Barrels=1

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5 Input test cases 5 5 5 500 50

for Locks=5,
Stocks=5,
Barrels=5
6 Input test cases 10 10 9 975 97.5
for Locks=10,
Stocks=10,

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 Barrels=9
7 Input test cases 10 9 10 970 97
for Locks=10,
Stocks=9,
Barrels=10
8 Input test cases 9 10 10 955 95.5
for Locks=9,
Stocks=10,
Barrels=10
9 Input test cases 10 10 10 1000 100
for Locks=10,
Stocks=10,
Barrels=10
10 Input test cases 10 10 11 1025 103.75
for Locks=10,
Stocks=10,
Barrels=11
11 Input test cases 10 11 10 1030 104.5
for Locks=10,
Stocks=11,
Barrels=10
12 Input test cases 11 10 10 1045 106.75
for Locks=11,
Stocks=10,
Barrels=10
13 Input test cases 14 14 13 1400 160
for Locks=14,
Stocks=14,
Barrels=13
14 Input test cases 18 18 17 1775 216.25
for Locks=18,
Stocks=18,
Barrels=17
15 Input test cases 18 17 18 1770 215.5
for Locks=18,
Stocks=17,
Barrels=18
16 Input test cases 17 18 18 1755 213.25
for Locks=17,
Stocks=18,

Barrels=18
17 Input test cases 18 18 18 1800 220
for Locks=18,
Stocks=18,
Barrels=18
18 Input test cases 18 18 19 1825 225
for Locks=18,
Stocks=18,
Barrels=19
19 Input test cases 18 19 18 1830 226
for Locks=18,
Stocks=19,
Barrels=18
20 Input test cases 19 18 18 1845 229
for Locks=19,
Stocks=18,
Barrels=18
21 Input test cases 48 48 48 4800 820
for Locks=48,
Stocks=48,
Barrels=48

Table-1 BVA Test case for commission problem.

This is how we can apply BVA technique to create test cases for our
Commission Problem.

2.5 EXECUTIONS

Execute the program and test the test cases in Table-1 against program and
complete the table with for Actual output column and Status column

TEST REPORT:
1. No of TC’s Executed:
2. No of Defects Raised:
3. No of TC’s Pass:
4. No of TC’s Failed:

2.6 SNAPSHOTS:
1. Snapshot for valid inputs
2. Snapshots when the two inputs are same

3. Snapshots when the two inputs and all the inputs are same
2.7 REFERENCES
1. Requirement Specification
2. Assumptions
3. Design, develop, code and run the program in any suitable language to
implement the NextDate function. Analyze it from the perspective of boundary
value testing, derive different test cases, execute these test cases and discuss the
test results.

3.1 REQUIREMENT SPECIFICATION

Problem Definition: "Next Date" is a function consisting of three variables like:
month, date and year. It returns the date of next day as output. It reads
Current date as input date

The constraints are

C1: 1 ≤ month ≤ 12
C2: 1 ≤ day ≤ 31
C3: 1812 ≤ year ≤ 2012.
If any one condition out of C1, C2 or C3 fails, then this function produces an
output "value of month not in the range 1...12".

Since many combinations of dates can exist, hence we can simply displays one
message for this function: "Invalid Input Date".

A very common and popular problem occurs if the year is a leap year. We have
taken into consideration that there are 31 days in a month. But what happens if
a month has 30 days or even 29 or 28 days?

A year is called as a leap year if it is divisible by 4, unless it is a century year.

Century years are leap years only if they are multiples of 400. So, 1992, 1996 and
2000 are leap years while 1900 is not a leap year.

3.2 DESIGN
Algorithm
Step 1: Input date in format DD.MM.YYYY
Step 2: if MM is 01, 03, 05,07,08,10 do Step 3 else Step 6

Step 3:if DD < 31 then do Step 4 else if DD=31 do Step 5 else

output(Invalid Date);
Step 4: tomorrowday=DD+1 goto Step 18
Step 5: tomorrowday=1; tomorrowmonth=month + 1 goto Step 18
Step 6: if MM is 04, 06, 09, 11 do Step 7
Step 7: if DD<30 then do Step 4 else if DD=30 do Step 5 else
output(Invalid Date);
Step 8: if MM is 12
Step 9: if DD<31 then Step 4 else Step 10
Step 10: tomorrowday=1, tommorowmonth=1, tommorowyear=YYYY+1;
 goto Step 18
Step 11: if MM is 2
Step 12: if DD<28 do Step 4 else do Step 13
Step 13: if DD=28 & YYYY is a leap do Step14 else Step 15
Step 14: tommorowday=29 goto Step 18
Step 15: tommorowday=1, tomorrowmonth=3, goto Step18;
Step 16: if DD=29 then do Step 15 else Step 17
Step 17: output(“Cannot have feb”, DD); Step19
Step 18: output(tomorrowday, tomorrowmonth, tomorrowyear);
Step 19: exit

3.3 PROGRAM CODE:

#include<stdio.h>
#include<conio.h>
main( )
{
int month[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int d,m,y,nd,nm,ny,ndays;
clrscr( );
printf("enter the date,month,year");
scanf("%d%d%d",&d,&m,&y);
ndays=month[m-1];
if(y<=1812 && y>2012)
{
printf("Invalid Input Year");
exit(0);
}
if(d<=0 || d>ndays)
{

printf("Invalid Input Day");

exit(0);
}
if(m<1 && m>12)
{
printf("Invalid Input Month");
exit(0);
}
if(m==2)
{
if(y%100==0)
{
if(y%400==0)
ndays=29;
}
else
if(y%4==0)
ndays=29;
}
nd=d+1;
nm=m;
ny=y;
if(nd>ndays)
{
nd=1;
nm++;
}

if(nm>12)
{
nm=1;
ny++;
}
printf("\n Given date is %d:%d:%d",d,m,y); printf("\n
Next day’s date is %d:%d:%d",nd,nm,ny); getch( );
}

3.4 TESTING

Technique used: Boundary value analysis

‘Boundary value analysis’ testing technique is used to identify errors at

boundaries rather than finding those exist in center of input domain.
Boundary value analysis is a next part of Equivalence partitioning for
designing test cases where test cases are selected at the edges of the
equivalence classes.

BVA: Procedure

1. Partition the input domain using unidimensional partitioning. This leads to as

many partitions as there are input variables. Alternately, a single partition of an
input domain can be created using multidimensional partition. We will generate
several sub-domains in this Step.
2. Identify t h e boundaries for e a c h partition. Boundaries may also be
identified using special relationships amongst the inputs.
3. Select test data such that each boundary value occurs in at least one test input.

BVA: Example: Create equivalence classes

Assuming that an item code must be in the range 99...999 and quantity in the
range 1...100,

Equivalence classes for code:

E1: Values less than 99.
E2: Values in the range.
E3: Values greater than 999.

Equivalence classes for qty:

E4: Values less than 1.
E5: Values in the range.
E6: Values greater than 100.

BVA: Example: Identify boundaries

Equivalence classes and boundaries for find Price. Boundaries are indicated with
an x. Points near the boundary are marked *.

Test Case design

The Next Date program takes date as input and checks it for validity. If it is valid,
it returns the next date as its output. Here we have three inputs for the program,
hence n = 3.

Since BVA yields (4n + 1) test cases according to single fault assumption
theory, hence we can say that the total number of test cases will be (4*3+1)
=12+1=13.

The boundary value test cases can be generated by using following constraints

C1: 1 ≤MM ≤12

C2: 1 ≤ DD ≤ 31
C3: 1812 ≤ YYYY ≤
2012.

Here from these constraints we can extract the test cases using the values of MM,
DD, and YYYY. The following equivalence classes can be generated for each
variable.

Equivalence classes for MM:

E1: Values less than 1.
E2: Values in the range.
E3: Values greater than 12.
Equivalence classes for DD:
E4: Values less than 1.
E5: Values in the range.
E6: Values greater than 31.
Equivalence classes for YYYY:
E7: Values less than 1812.
E8: Values in the range.
E9: Values greater than 2012.

From the above equivalence classes we can derive the following test cases using
boundary value analysis approac h.
TC Test Case Input Data Expected Actual Status
Id Description MM DD YYYY Output Output
1 Testing for Invalid Aa 15 1900 Invalid Input
months with Month
character is typed
2 Testing for Invalid 06 Dd 1901 Invalid Input
Day with character Day
is typed
3 Testing for Invalid 06 15 196y Invalid Input
Year with Year
character is typed
4 Testing for Invalid 03 00 2000 Invalid Input
Day, day with 00 Day
5 Testing for Valid 03 30 2000 03/31/2000
input changing the
day within the
month.
6 Testing for Valid 03 02 2000 03/03/2000
input changing the
day within the
month.

7 Testing for Invalid 03 32 2000 Invalid Input

Day, day with 32 Day
8 Testing for Invalid 00 15 2000 Invalid Input
Day, month with 00 Month
9 Testing for Valid 11 15 2000 11/16/2000
input changing the
day within the
month. MM=11
DD=15
10 Testing for Valid 02 15 2000 02/16/2000
input changing the
day within the
month. MM=02
DD=15
11 Testing for Invalid 13 15 2000 Invalid Input
Month, month with Month
13

12 Testing for Invalid 03 15 1811 Invalid Input

year, year should Year
>=1812
13 Testing for 03 15 2011 03/16/2011
Valid input changing
the day within
the month. MM=03
DD=15 YYYY=2011

14 Testing for Valid 03 15 1813 03/16/1813

input changing the
day within the
month.
15 TestingMM=03
for Invalid 03 15 2013 Invalid Input
year, year should Year
<=2012

Table-1: Test case for Next Date Problem

This is how we can apply BA technique to create test cases for our Next Date
Problem.

3.5 EXECUTIONS

Execute the program and test the test cases in Table-1 against program and
complete the table with for Actual output column and Status column

Test Report:
 1. No of TC’s Executed:
2. No of Defects Raised:
3. No of TC’s Pass:
4. No of TC’s Failed:

3.6 SNAPSHOTS:
1. Snapshot for Invalid Input day and next date

2. Snapshot to show the invalid day when the DD=32

3. Valid Output:

3.7 REFERENCES:
1. Requirement Specification
2. Assumptions
4. Design and develop a program in a language of your choice to solve the
triangle problem defined as follows: Accept three integers which are supposed to be
the three sides of a triangle and determine if the three values represent an
equilateral triangle, isosceles triangle, scalene triangle, or they do not form a
triangle at all. Assume that the upper limit for the size of any side is 10. Derive
test cases for your program based on equivalence class partitioning, execute the
test cases and discuss the results.

4.1 REQUIREMENT SPECIFICATION

R1. The system should accept 3 positive integer numbers (a, b, c) which
represents 3 sides of the triangle.
R2. Based on the input should determine if a triangle can be formed or not. R3.
If the requirement R2 is satisfied then the system should determine the type of
the triangle, which can be
• Equilateral (i.e. all the three sides are equal)
• Isosceles (i.e. two sides are equal)
• Scalene (i.e. All the three sides are unequal)
R4. Upper Limit for the size of any side is 10

4.2 DESIGN

Form the given requirements we can draw the following conditions:

C1: a<b+c?
C2: b<a+c?
C3: c<a+b?
 C4: a=b?
C5: a=c?
C6: b=c?

According to the property of the triangle, if any one of the three conditions C1,
C2 and C3 are not satisfied then triangle cannot be constructed. So only when
C1, C2 and C3 are true the triangle can be formed, then depending on
conditions C4, C5 and C6 we can decide what type of triangle will be
formed(i.e requirements R3)
Algorithm:

Step 1: Input a, b & c i.e three integer values which represent three sides of the
triangle.
Step 2: if (a < (b + c)) and (b < (a + c)) and (c < (a + b) then do
Step 3
else
print not a triangle. do Step 6.
Step 3: if (a=b) and (b=c) then
Print triangle formed is equilateral. do Step 6.
Step 4: if (a ≠ b) and (a ≠ c) and (b ≠ c) then
Print triangle formed is scalene. do Step 6.
Step 5: Print triangle formed is Isosceles.
Step 6: stop

4.3 PROGRAM CODE

#include<stdio.h>
#include<ctype.h>
#include<conio.h>
#include<process.h>
int main()
{
int a, b, c;
clrscr();
printf("Enter three sides of the triangle");
scanf("%d%d%d", &a, &b, &c);
if((a > 10) || (b > 10) || (c > 10))
{
 printf("Out of range");
getch();
exit(0);
}
if((a<b+c)&&(b<a+c)&&(c<a+b))
{
if((a==b)&&(b==c))
{
printf("Equilateral triangle");
}
else if((a!=b)&&(a!=c)&&(b!=c))
{

printf("Scalene triangle");

}
else
printf("Isosceles triangle");
}
else
{
printf("triangle cannot be formed");
} getch(); return 0;
}

4.4 TESTING
1. Technique used: Equivalence class partitioning
2. Test Case design

Equivalence class partitioning technique focus on the Input domain, we can

obtain a richer set of test cases. What are some of the possibilities for the
three integers, a, b, and c? They can all be equal, exactly one pair can be
equal.

The maximum limit of each side a, b, and c of the triangle is 10 units

according to requirement R4. So a, b and c lies between
 1≤a≤10
1≤b≤10
1≤c≤10

First Attempt

Weak normal equivalence class: In the problem statement, we note

that four possible outputs can occur: Not a Triangle, Scalene, Isosceles and
Equilateral. We can use these to identify output (range)
equivalence classes as follows:
R1= {<a,b,c>: the triangle with sides a, b, and c is equilateral}
R2= {<a,b,c>: the triangle with sides a, b, and c is isosceles}
R3= {<a,b,c>: the triangle with sides a, b, and c is scalene}
R4= {<a,b,c>: sides a, b, and c do not form a triangle}

Four weak normal equivalence class test cases, chosen arbitrarily from each class,
and invalid values for weak robust equivalence class test cases are as follows.

TC Test Case Input Data Expected Output Actual Status

Id Description a b c Output
1 WN1 5 5 5 Equilateral

2 WN2 2 2 3 Isosceles

3 WN3 3 4 5 Scalene

4 WN4 4 1 2 Not a Triangle

5 WR1 -1 5 5 Value of a is not in the

range of permitted
values
6 WR2 5 -1 5 Value of b is not in the
range of permitted
values
7 WR3 5 5 -1 Value of c is not in the
range of permitted
values
8 WR4 11 5 5 Value of a is not in the
range of permitted
values
 9 WR5 5 11 5 Value of b is not in the
range of permitted
values
10 WR6 5 5 11 Value of c is not in the
range of permitted
values

Table-1: Weak Normal and Weak Robust Test case for Triangle Problem

Second attempt

The strong normal equivalence class test cases can be generated by

using following possibilities:

D1 = {<a, b, c>: a=b=c}

D2 = {<a, b, c>: a=b, a≠ c}
D3= {<a, b, c>: a=c, a≠ b}
D4 = {<a, b, c>: b=c, a≠ b}
D5 = {<a, b, c>: a≠ b, a≠ c, b≠ c}
D6 = {<a, b, c>: a≥ b+ c}
D7 = {<a, b, c>: b≥ a+ c}
D8 = {<a, b, c>: c≥ a+ b}

TC Test Input Data Expected Output Actual Status

Id a b c Output
Case
1 Descriptio
SR1 -1 5 5 Value of a is not in the
range of permitted
values
2 SR 2 5 -1 5 Value of b is not in the
range of permitted
values
3 SR3 5 5 -1 Value of c is not in the
range of permitted
values
5 SR5 5 -1 -1 Value of b, c is not in the
range of permitted
values
6 SR6 -1 5 -1 Value of a, c is not in the
range of permitted
values
 7 SR7 -1 -1 -1 Value of a, b, c is not in
the range of permitted
values

Table-2: Strong Robust Test case for Triangle Problem

4.5 EXECUTION:

Execute the program and test the test cases in Table-1 and Table-2 against
program and complete the table with for Actual output column and Status
column

Test Report:
1. No of TC’s Executed:
2. No of Defects Raised:
3. No of TC’s Pass:
4. No of TC’s Failed:

4.6 SNAPSHOTS:
1. Snapshot of Equilateral. Isosceles and scalene triangle.
2. Snapshot for Triangle cannot be formed

3. Snapshot for the given range is Out of range and Triangle cannot be formed.
4.7 REFERENCES

1. Requirement Specification
2. Assumptions

5. Design, develop, code and run the program in any suitable language to solve
the commission problem. Analyze it from the perspective of equivalence class
testing, derive different test cases, execute these test cases and discuss the test
results.
5.1 REQUIREMENT SPECIFICATION

Problem Definition: The Commission Problem includes a salesperson in the

former Arizona Territory sold rifle locks, stocks and barrels made by a
gunsmith in Missouri. Cost includes

Locks- $45
Stocks- $30
Barrels- $25
The salesperson had to sell at least one complete rifle per month and
production limits were such that the most the salesperson could sell in a
month was 70 locks, 80 stocks and 90 barrels.
After each town visit, the sales person sent a telegram to the Missouri
gunsmith with the number of locks, stocks and barrels sold in the town. At the end
of the month, the salesperson sent a very short telegram showing --1 lock sold.
The gunsmith then knew the sales for the month were complete and computed
the salesperson’s commission as follows:
On sales up to(and including) $1000= 10% On
the sales up to(and includes) $1800= 15% On
the sales in excess of $1800= 20%
The commission program produces a monthly sales report that gave the total
number of locks, stocks and barrels sold, the salesperson’s total dollar sales and
finally the commission.

5.2 DESIGN
Algorithm:
Step 1: Define lockPrice=45.0, stockPrice=30.0, barrelPrice=25.0
Step2: Input locks

Step3: while(locks!=-1) ‘input device uses -1 to indicate end of data goto

Step 12
Step 4:input (stocks, barrels)
Step 5: compute lockSales, stockSales, barrelSales and sales
Step 6: output(“Total sales:” sales)
Step 7: if (sales > 1800.0) goto Step 8 else goto Step 9
Step 8: commission=0.10*1000.0; commission=commission+0.15 * 800.0;
commission = commission + 0.20 * (sales-1800.0)
 Step 9: if (sales > 1000.0) goto Step 10 else goto Step 11
Step 10: commission=0.10* 1000.0; commission=commission + 0.15 *
(sales-1000.0)
Step 11: Output(“Commission is $”, commission)
Step 12: exit

5.3 PROGRAM CODE:

#include<stdio.h>
#include<conio.h>
int main()
{
int locks, stocks, barrels, t_sales, flag = 0;
float commission;
clrscr();
printf("Enter the total number of locks");
scanf("%d",&locks);
if ((locks <= 0) || (locks > 70))
{
flag = 1;
}
printf("Enter the total number of stocks");
scanf("%d",&stocks);
if ((stocks <= 0) || (stocks > 80))
{
flag = 1;
}
printf("Enter the total number of barrelss");
scanf("%d",&barrels);
if ((barrels <= 0) || (barrels > 90))
{
flag = 1;
}
if (flag == 1)
{
printf("invalid input");
 getch();
exit(0);
}
t_sales = (locks * 45) + (stocks * 30) + (barrels * 25);
if (t_sales <= 1000)
{
commission = 0.10 * t_sales;
}
else if (t_sales < 1800)
{
commission = 0.10 * 1000;
commission = commission + (0.15 * (t_sales - 1000));
}
else
{
commission = 0.10 * 1000;
commission = commission + (0.15 * 800);
commission = commission + (0.20 * (t_sales - 1800));
}
printf("The total sales is %d \n The commission is %f",t_sales,
commission);
getch(); return;
}
5.4 TESTING

Technique used: Equivalence Class testing

Test selection using equivalence partitioning allows a tester to subdivide the input
domain into a relatively small number of sub-domains, say N>1, as shown.
In strict mathematical terms, the sub-domains by definition are disjoint. The
four subsets shown in (a) constitute a partition of the input domain while the
subsets in (b) are not. Each subset is known as an equivalence class.

Example:
Consider an application A that takes an integer denoted by age as input. Let us
suppose that the only legal values of age are in the range [1..120]. The set of
input values is now divided into a set E containing all integers in the range
[1..120] and a set U containing the remaining integers.

Further, assume that the application is required to process all values in the range
[1..61] in accordance with requirement R1 and those in the range [62..120]
according to requirement R2. Thus E is further subdivided into two
regions depending on the expected behavior.
Similarly, it is expected that all invalid inputs less than or equal to 1 are to be
treated in one way while all greater than 120 are to be treated differently.
This leads to a subdivision of U into two categories.
Tests selected using the equivalence partitioning technique aim at targeting faults
in the application under test with respect to inputs in any of the four regions, i.e.
two regions containing expected inputs and two regions containing the
unexpected inputs.
 >120

It is expected that any single test selected from the range [1...61] will reveal any
fault with respect to R1. Similarly, any test selected from the region
[62...120] will reveal any fault with respect to R2. A similar expectation
applies to the two regions containing the unexpected inputs
TEST CASE DESIGN

The input domain of the commission problem is naturally partitioned by the

limits on locks, stocks and barrels. These equivalence classes are exactly
those that would also be identified by traditional equivalence class testing. The
first class is the valid input; the other two are invalid. The input domain
equivalence classes lead to very unsatisfactory sets of test cases. Equivalence
classes defined on the output range of the commission function will be an
improvement.

The valid classes of the input variables are:

L1 = {locks: 1≤locks≤70}
L2 = {locks = -1} (occurs if locks = -1 is used to control input iteration) S1
= {stocks:1≤stocks≤80}
B1 = {barrels: 1≤barrels≤90}
The corresponding invalid classes of the input variables are: L3
= {locks: locks = 0 OR locks < -1}
L4 = {locks: locks > 70}
S2 = {stocks: stocks<1}
S3 = { stocks: stocks>80}
B2 ={barrels: barrels<1}
B3 ={ barrels: barrels>90}

One problem occurs, however. The variables lock are also used as a
sentinel to indicate no more telegrams. When a value of -1 is given for
locks, the while loop terminates, and the values of totallocks, totalstocks
and totalbarrels are used to compute sales, and then commission.
Expect for the names of the variables and the interval endpoint values, this
isidentical to our first version of the NextDate function. therefore we will
have exactly one week normal equivalence class test case – and again, it is
identical to the strong normal equivalence class test case. Note that the case
for locks =-1 just terminates the iteration.

First attempt
We will have eight weak robust test cases.
TC Test Case Input Data Sales Expected Actual Status
Id Description Locks Stocks Barrels Output(Commission) Output
1 WR1 10 10 10 $100 10

2 WR2 -1 40 45 Program Program terminates

terminates
3 WR3 -2 40 45 Values of Values of locks not in
locks not the
inthe range range 1...70
1...70
4 WR4 71 40 45 Values of Values of locks not in
locks not the
inthe range range 1...70
1...70
5 WR5 35 -1 45 Values of Values of stocks not in
stocks not the
inthe range range 1...80
1...80
6 WR6 35 81 45 Values of Values of stocks not in
stocks not the
inthe range range 1...80
1...80
7 WR7 10 9 10 970 97

8 WR8 9 10 10 955 95.5

Second attempt:

Finally, a corner of the cube will be in 3 space of the additional strong

robust equivalence class test cases:

TC Test Case Input Data Sales Expected Actual Status

Id Description Locks Stocks Barrels Output(Commission) Output
1 SR1 -2 40 45 Values of locks Values of locks not in
not in the range the range 1...70
1...70
2 SR2 35 -1 45 Values of stocks Values of stocks not
not in the range in the range 1...80
1...80
3 SR3 35 40 -2 Values of barrels Values of barrels not
not in the range in the range 1...90
1...90
4 SR4 -2 -1 45 Values of locks Values of locks not in
not in the range the range 1...70
1...70 Values of stocks not
Values of stocks in the range 1...80
not in the range
5 SR5 -2 40 -1 1...80
Values of locks Values of locks not in
not in the range the range 1...70
1...70 Values of barrels not
Values of barrels in the range 1...90
not in the range
6 SR6 35 -1 -1 1...90
Values of stocks Values of stocks not
not in the range in the range 1...80
1...80 Values of barrels not
Values of barrels in the range 1...90
not in the range
7 SR7 -2 -1 -1 1...90
Values of locks Values of locks not in
not in the range the range 1...70
1...70 Values of stocks not
Values of stocks in the range 1...80
not in the range Values of barrels not
1...80 in the range 1...90
Values of barrels
not in the range

5.5 EXECUTIONS

Execute the program and test the test cases in Table-1 against program and
complete the table with for Actual output column and Status column

Test Report:

1. No of TC’s Executed:
2. No of Defects Raised:
3. No of TC’s Pass:
4. No of TC’s Failed:

5.6 SNASHOTS
1. Snapshot for invalid inputs
2. Invalid Input and commission for when the all inputs are 10

5.7 REFERENCES
1. Requirement Specification
2. Assumptions
6. Design, develop, code and run the program in any suitable language to
implement the NextDate function. Analyze it from the perspective of
equivalence class value testing, derive different test cases, execute these test
cases and discuss the test results.

6.1 REQUIREMENT SPECIFICATION

Problem Definition: "Next Date" is a function consisting of three variables

like: month, date and year. It returns the date of next day as output. It reads
current date as input date.

The constraints are

C1: 1 ≤ month ≤ 12
C2: 1 ≤ day ≤ 31
C3: 1812 ≤ year ≤ 2012.

If any one condition out of C1, C2 or C3 fails, then this function produces an
output "value of month not in the range 1...12".

Since many combinations of dates can exist, hence we can simply displays one
message for this function: "Invalid Input Date".

A very common and popular problem occurs if the year is a leap year. We have
taken into consideration that there are 31 days in a month. But what happens
if a month has 30 days or even 29 or 28 days ?

A year is called as a leap year if it is divisible by 4, unless it is a century year.

Century years are leap years only if they are multiples of 400. So, 1992, 1996 and
2000 are leap years while 1900 is not a leap year.

Furthermore, in this Next Date problem we find examples of Zipf's law also,
which states that "80% of the activity occurs in 20% of the space". Thus in this
case also, much of the source-code of Next Date function is devoted to the leap
year considerations.

6.2 DESIGN
Algorithm:
Step 1: Input date in format DD.MM.YYYY
Step 2: if MM is 01, 03, 05,07,08,10 do Step 3 else Step 6
Step 3:if DD < 31 then do Step 4 else if DD=31 do Step 5 else output(Invalid
Date);
Step 4: tomorrowday=DD+1 goto Step 18
Step 5: tomorrowday=1; tomorrowmonth=month + 1 goto Step 18
Step 6: if MM is 04, 06, 09, 11 do Step 7
Step 7: if DD<30 then do Step 4 else if DD=30 do Step 5 else output(Invalid
Date);
Step 8: if MM is 12
Step 9: if DD<31 then Step 4 else Step 10
Step 10: tomorrowday=1, tommorowmonth=1, tommorowyear=YYYY+1; goto
Step 18
Step 11: if MM is 2
Step12: if DD<28 do Step 4 else do Step 13
Step 13: if DD=28 & YYYY is a leap do Step 14 else Step 15
Step 14: tommorowday=29 goto Step18
Step 15: tommorowday=1, tomorrowmonth=3, goto Step 18;
Step 16: if DD=29 then do Step15 else Step 17
Step 17: output(“Cannot have feb”, DD); Step 19
Step 18: output(tomorrowday, tomorrowmonth, tomorrowyear);
Step 19: exit

6.3 PROGRAM CODE:

#include<stdio.h>
#include<conio.h>
main( )
{
int month[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int d,m,y,nd,nm,ny,ndays;
clrscr( );
printf("enter the date,month,year");
scanf("%d%d%d",&d,&m,&y);
ndays=month[m-1];
if(y<=1812 && y>2012)
{
 printf("Invalid Input Year");
exit(0);
}

if(d<=0 || d>ndays)
{

printf("Invalid Input Day");

exit(0);
}
if(m<1 && m>12)
{
printf("Invalid Input Month");
exit(0);
}

if(m==2)
{
if(y%100==0)
{

if(y%400==0)
ndays=29;
}
else
if(y%4==0)
ndays=29;
nd=d+1;
nm=m;
ny=y;
}
if(nd>ndays)
{
nd=1;
nm++;
}
if(nm>12)
{
nm=1;
ny++;
}
if(nm>12)
{
nm=1;
ny++;
}
printf("\n Given date is %d:%d:%d",d,m,y); printf("\n
Next day’s date is %d:%d:%d",nd,nm,ny); getch( );
}

6.4 TESTING

Technique used: Equivalence Class testing

Test selection using equivalence partitioning allows a tester to subdivide the input
domain into a relatively small number of sub-domains, say N>1, as shown.

In strict mathematical terms, the sub-domains by definition are disjoint. The four
subsets shown in (a) constitute a partition of the input domain while the
subsets in (b) are not. Each subset is known as an equivalence class.
Example:
Consider an application A that takes an integer denoted by age as input. Let us
suppose that the only legal values of age are in the range [1..120]. The set of
input values is now divided into a set E containing all integers in the range
[1..120] and a set U containing the remaining integers.

Further, assume that the application is required to process all values in the range
[1..61] in accordance with requirement R1 and those in the range [62..120]
according to requirement R2. Thus E is further subdivided into two
regions depending on the expected behavior.
Similarly, it is expected that all invalid inputs less than or equal to 1 are to be
treated in one way while all greater than 120 are to be treated differently.
This leads to a subdivision of U into two categories,

Tests selected using the equivalence partitioning technique aim at targeting faults
in the application under test with respect to inputs in any of the four regions,
i.e. two regions containing expected inputs and two regions containing the
unexpected inputs.

It is expected that any single test selected from the range [1...61] will reveal any
fault with respect to R1. Similarly, any test selected from the region
[62...120] will reveal any fault with respect to R2. A similar expectation
applies to the two regions containing the unexpected inputs.

Test Case design

The NextDate function is a function which will take in a date as input and
produces as output the next date in the Georgian calendar. It uses three
variables (month, day and year) which each have valid and invalid intervals.

First Attempt
A first attempt at creating an equivalence relation might produce intervals
such as these:

Valid Intervals

M1 = {month: 1 ≤ month ≤ 12}

D1 = {day: 1 ≤day ≤31}
Y1 = {year: 1812 ≤ year ≤2012}

Valid Intervals

M1 = {month: 1 ≤ month ≤ 12}

D1 = {day: 1 ≤day ≤31}
Y1 = {year: 1812 ≤ year ≤2012}

Invalid Intervals

M2 = {month: month < 1}

M3 = {month: month > 12}
D2 = {day: day < 1}
D3 = {day: day > 31}
Y2 = {year: year < 1812}
Y3 = {year: year > 2012}

At a first glance it seems that everything has been taken into account and our
day, month and year intervals have been defined well. Using these intervals
we produce test cases using the four different types of Equivalence
Class testing.
Weak and Strong Normal

TC Test Case Input Data Expected Actual Status

Id Description MM DD YYYY Output Output
1 Testing for Valid 6 15 1900 6/16/1900
input changing
the day within the
month.

Table 1: Weak and Strong Normal

Since the number of variables is equal to the number of valid classes, only one
weak normal equivalence class test case occurs, which is the same as the
strong normal equivalence class test case (Table 1).

Weak Robust:

TC Test Case Input Data Expected Actual Status

Id Description MM DD YYYY Output Output
1 Testing for Valid 6 15 1900 6/16/1900
input changing
the day within
the month.
2 Testing for 6 -1 1900 Day not in range
Invalid Day, day
with negative
number it is not
possible
3 Testing for 6 32 1900 Day not in range
Invalid Day, day
with Out of
range i.e.,
DD=32
4 Testing for -1 15 1900 Month not in
Invalid Month, range
month with
negative
number it is not
possible
5 Testing for 13 15 1900 Month not in
Invalid month, range
month with out
of range i.e.,
MM=13 it
should MM<=12
6 Testing for Year, 6 15 1899 Year not in range
year is out of
range
YYYY=1899, it
should <=1812

7 Testing for Year, 6 15 2013 Year not in range

year is out of
range
YYYY=2013, it
should <=2012
 Table 2: Weak Robust

(Table 2) we can see that weak robust equivalence class testing will just test
the ranges of the input domain once on each class. Since we are testing weak
and not normal, there will only be at most one fault per test case (single fault
assumption) unlike Strong Robust Equivalence class testing.

Strong Robust:

This is a table showing one corner of the cube in 3d-space (the three other
corners would include a different combination of variables) since the
complete table would be too large to show.

TC Test Case Input Data Expected Actual Status

Id Description MM DD YYYY Output Output
1 Testing for Month is
not in range MM=-1
i.e., in negative
number there is not
Month not in
possible have to be -1 15 1900 range
month in negative
number
2 Testing for Day is not
in range DD=-1 i.e., in
negative number
there is not possible
6 -1 1900 Day not in range
have to be Day in
negative number
3 Testing for Year is
not in range Year not in
YYYY=1899 i.e., Year 6 15 1899 range
should <=1812
4 Testing for Day and
month is not in range
MM=-1, DD=-1 i.e., in
negative number i) Day not in
there is not possible range
have to be Day and -1 -1 1900 ii) Month not in
Month in negative range
number
5 i) Testing for Day is
not in range and Year
is not in range DD=-1
i.e., in negative
number there is not
possible have to be i) Day not in
Day in negative range
number, and 6 -1 1899 ii) Year not in
ii) YYYY=1899, so the range
range of year is
<=1812

6 i) Testing for Month

is not in range MM=-
1 and i.e., in negative
number there is not
possible have to be i) Month not in
Day in negative range
number, and -1 15 1899 ii) Year not in
ii) Year is not in range
range

YYYY=1899, year
 7 i) Testing for Day is
not in range DD=-1
i.e., in negative
number there is not
possible have to be
Day in negative
number
ii) Testing for Month
i) Day not in
is not in range MM=-
range
1 and i.e., in negative ii) Month not in
number there is not range
possible have to be -1 -1 1899
iii) Year not in
Day in negative
range
number, and
iii) Year is not in
range YYYY=1899,
year should <=1812

Second Attempt

As said before the equivalence relation is vital in producing useful test cases and
more time must be spent on designing it. If we focus more on the
equivalence relation and consider more greatly what must happen to an input date
we might produce the following equivalence classes:

M1 = {month: month has 30 days}

M2 = {month: month has 31 days}
M3 = {month: month is February}

Here month has been split up into 30 days (April, June, September and
November), 31 days (January, March, April, May, July, August, October and
December) and February.

D1 = {day: 1 ≤ day ≤ 28}

D2 = {day: day = 29}
D3 = {day: day = 30}
D4 = {day: day = 31}

Day has been split up into intervals to allow months to have a different
number of days; we also have the special case of a leap year (February 29
days).

Y1 = {year: year = 2000}

Y2 = {year: year is a leap year}
Y3 = {year: year is a common year}

Year has been split up into common years, leap years and the special case the year
2000 so we can determine the date in the month of February.

Here are the test cases for the new equivalence relation using the four types of
Equivalence Class testing.

Weak Normal

TC Test Case Description Input Data Expected Actual Status

Id MM DD YYYY Output Output
1 Testing for all Valid 6 14 2000 6/15/2000
input changing the day
within the month.
2 Testing for Valid input 7 29 1996 7/30/1996
changing the day
within the month.
3 Testing for Leaf year, 2 30 2002 Impossible date
i.e., MM=2 (Feb) the
input DD=30, there is
not possible date 30, in
leaf year only 28 and 29
will occur.
 4 Testing for Impossible 6 31 2000 Impossible input
Date, i.e., MM=6 (June) date
the input DD=31, there
is only 30 days in the
month of June, So,
DD=31 is Impossible
Date.

Table 3: Weak normal

Strong Normal
TC ID Test Case Input Data Expected Actual Status
Description MM DD YYYY Output Output
1 SN1 6 14 2000 6/15/2000

2 SN2 6 14 1996 6/15/1996

3 SN3 6 14 2002 6/15/2002

4 SN4 6 29 2000 6/30/2000

5 SN5 6 29 1996 6/30/1996

6 SN6 6 29 2002 6/30/2002

7 6 30 2000 Invalid Input

SN7
Date
8 6 30 1996 Invalid Input
SN8
Date
9 6 30 2002 Invalid Input
SN9
Date
10 6 31 2000 Invalid Input
SN10
Date
11 6 31 1996 Invalid Input
SN11
Date
12 6 31 2002 Invalid Input
SN12
Date
13 SN13 7 14 2000 7/15/2000

14 SN14 7 14 1996 7/15/1996

15 SN15 7 14 2002 7/15/2002

16 SN16 7 29 2000 7/30/2000

17 SN17 7 29 1996 7/30/1996

18 SN18 7 29 2002 7/30/2002

19 SN19 7 30 2000 7/31/2000

20 SN20 7 30 1996 7/31/1996

21 SN21 7 30 2002 7/31/2002
22 SN22 7 31 2000 8/1/2000
23 SN23 7 31 1996 8/1/1996
24 SN25 7 31 2002 8/1/2002
25 SN24 2 14 2000 2/15/2000
26 SN26 2 14 1996 2/15/1996
27 SN27 2 14 2002 2/15/2002

28 SN28 2 29 2000 Invalid Input

Date
29 SN29 2 29 1996 3/1/1996

30 SN30 2 29 2002 Invalid Input

Date
31 SN31 2 30 2000 Invalid Input
Date
32 SN32 2 30 1996 Invalid Input
Date
33 SN33 2 30 2002 Invalid Input
Date
34 SN34 2 31 2000 Invalid Input
Date
 35 SN35 2 31 1996 Invalid Input
Date
36 SN36 2 31 2002 Invalid Input
Date

Table 4: Strong Normal

6.5 EXECUTIONS

Execute the program and test the test cases in Table-1 against program and
complete the table with for Actual output column and Status column

Test Report:
1. No of TC’s Executed:
2. No of Defects Raised:
3. No of TC’s Pass:
4. No of TC’s Failed:
6.6 SNAPSHOTS:
1. Snapshot to show the nextdate for current date and invalid day is entered

2. Invalid Input
6.7 REFERENCES:
1. Requirement Specification
2. Assumptions

7. Design and develop a program in a language of your choice to solve the

triangle problem defined as follows: Accept three integers which are supposed to be
the three sides of a triangle and determine if the three values represent an
equilateral triangle, isosceles triangle, scalene triangle, or they do not form a
triangle at all. Derive test cases for your program based on decision-table
approach, execute the test cases and discuss the results.

7.1 REQUIREMENT SPECIFICATION:

R1. The system should accept 3 positive integer numbers (a, b, c) which
represents 3 sides of the triangle. Based on the input it should determine if a
triangle can be formed or not.
R2. If the requirement R1 is satisfied then the system should determine the type
of the triangle, which can be
• Equilateral (i.e. all the three sides are equal)
• Isosceles (i.e Two sides are equal)
• Scalene (i.e All the three sides are unequal)
else suitable error message should be displayed. Here we assume that user gives
three positive integer numbers as input.

7.2 DESIGN:
Form the given requirements we can draw the following conditions: C1:
a<b+c?
C2: b<a+c?
C3: c<a+b?
C4: a=b?
C5: a=c?
C6: b=c?
According to the property of the triangle, if any one of the three conditions C1,
C2 and C3 are not satisfied then triangle cannot be constructed. So only when C1,
C2 and C3 are true the triangle can be formed, then depending on
conditions C4, C5 and C6 we can decide what type of triangle will be formed.
(i.e requirement R2).

Algorithm:
Step 1: Input a, b & c i.e three integer values which represent three sides of the
triangle.
Step 2: if (a < (b + c)) and (b < (a + c)) and (c < (a + b) then
do Step 3
else
print not a triangle. do Step 6.
Step 3: if (a=b) and (b=c) then
Print triangle formed is equilateral. do Step 6.
Step 4: if (a ≠ b) and (a ≠ c) and (b ≠ c) then
Print triangle formed is scalene. do Step 6.
Step 5: Print triangle formed is Isosceles.
Step 6: stop

7.3 PROGRAM CODE:

#include<stdio.h>
#include<ctype.h>
#include<conio.h>
#include<process.h>
int main()
{
int a, b, c;
clrscr();
printf("Enter three sides of the triangle");
scanf("%d%d%d", &a, &b, &c);
if((a<b+c)&&(b<a+c)&&(c<a+b))
 {

if((a==b)&&(b==c))
{
printf("Equilateral triangle");
}
else if((a!=b)&&(a!=c)&&(b!=c))
{
printf("Scalene triangle");
}
else
printf("Isosceles triangle");

}
else
{
printf("triangle cannot be formed");
} getch(); return 0;
}

7.4
TESTING:
Technique Used: Decision Table Approach

Decision Table-Based Testing has been around since the early 1960’s; it is
used to depict complex logical relationships between input data. A Decision
Table is the method used to build a complete set of test cases without using the
internal structure of the program in question. In order to create test cases we use
a table to contain the input and output values of a program.The decision table is as
given below:

Condition Entries (Rules)

Conditions R R
R1 R2 R3 R4 R5 R6 R7 R8 R9
10 11
C1: a<b+c? F T T T T T T T T T T
C2: b<a+c? -- F T T T T T T T T T
C3: c<a+b? -- -- F T T T T T T T T
C4: a=b? -- -- -- F T T T F F F T
 C5: a=c? -- -- -- T F T F T F F T
C6: b=c? -- -- -- T T F F F T F T
Actions Action Entries
a1: Not a
X X X
Triangle
a2: Scalene X
a3: Isosceles X X X
a4: Equilateral X
a5: Impossible X X X

The “--“ symbol in the table indicates don’t care values. The table shows the
six conditions and 5 actions. All the conditions in the decision table are binary;
hence, it is called as “Limited Entry decision table”.

Each column of the decision table represents a test case. That is,
The table is read as follows:
Action: Not a Triangle

1. When condition C1 is false we can say that with the given ‘a’ ‘b’ and ‘c’

values, it’s Not a

triangle.

2. Similarly condition C2 and C3, if any one of them are false, we can say that
with the given ‘a’ ‘b’ and ‘c’ values it’s Not a triangle.
Action: Impossible

3. When conditions C1, C2, C3 are true and two conditions among C4, C5, C6
is true, there is no chance of one conditions among C4, C5, C6 failing. So we
can neglect these rules.
Example: if condition C4: a=b is true and C5: a=c is true

Then it is impossible, that condition C6: b=c will fail, so the action
is Impossible.

Action: Isosceles

4. When conditions C1, C2, C3 are true and any one condition among C4, C5
and C6 is true with remaining two conditions false then action is Isosceles
triangle.
Example: If condition C4: a=b is true and C5: a=c and C6: b=c are false, it
 means two sides are equal. So the action will be Isosceles triangle.
Action: Equilateral

5. When conditions C1, C2, C3 are true and also conditions C4, C5 and C6
are true then, the action is Equilateral triangle.
Action: Scalene

6. When conditions C1, C2, C3 are true and conditions C4, C5 and C6 are
false i.e sides a, b and c are different, then action is Scalene triangle.
Number of Test Cases = Number of Rules.
Using the decision table we obtain 11 functional test cases: 3 impossible cases,

3 ways of failing the triangle property, 1 way to get an equilateral triangle, 1

way to get a scalene triangle, and 3 ways to get an isosceles triangle.
Deriving test cases using
Decision Table Approach:
Test Cases:
TC Test Case Expected Actual
a B c Status
ID Description Output Output
Testing for Not a
1 4 1 2
Requirement 1 Triangle
Testing for Not a
2 1 4 2
Requirement 1 Triangle
Testing for Not a
3 1 2 4
Requirement 1 Triangle
Testing for
4 5 5 5 Equilateral
Requirement 2
Testing for
5 2 2 3 Isosceles
Requirement 2
Testing for
6 2 3 2 Isosceles
Requirement 2
Testing for
7 3 2 2 Isosceles
Requirement 2
Testing for
8 3 4 5 Scalene
Requirement 2

7.5 EXECUTION & RESULT DISCUSION

Execute the program against the designed test cases and complete the table for
Actual output column and status column.
Test Report:

1. No of TC’s Executed: 08
 2. No of Defects Raised:

3. No of TC’s Pass:
4. No of TC’s Failed:
The decision table technique is indicated for applications characterised by any of
the following:
Prominent if-then-else logic

Logical relationships among input variables

Calculations involving subsets of the input variables

Cause-and-effect relationship between inputs and outputs

The decision table-based testing works well for triangle problem because a lot of
decision making i.e if-then-else logic takes place.

7.6 SNAPSHOTS:
1. Output screen of Triangle cannot be formed
2. Output screen of Equilateral and Isosceles Triangle.

3. Output screen for Scalene Triangle

7.7. REFERENCES:
1. Requirement Specification
2. Assumption.
8. Design, develop, code and run the program in any suitable language to solve
the commission problem. Analyze it from the perspective of decision table-
based testing, derive different test cases, execute these test cases and discuss the
test results.

8.1 REQUIREMENT SPECIFICATION:

R1: The system should read the number of Locks, Stocks and Barrels sold in a
month.
(i.e 1≤ Locks≤ 70) (i.e
1 ≤ Stocks ≤ 80) (i.e 1
≤ Barrels ≤ 90).
R2: If R1 is satisfied the system should compute the salesperson’s commission
depending on the total number of Locks, Stocks & Barrels sold else it should
display suitable error message. Following is the percentage of commission for the
sales done:
10% on sales up to (and including) $1000

15% on next $800

20% on any sales in excess of $1800

Also the system should compute the total dollar sales. The system should
output salespersons total dollar sales, and his commission.

8.2 DESIGN:

Form the given requirements we can draw the following conditions:

C1: 1≤locks≤70? Locks = -1? (occurs if locks = -1 is used to control input

iteration).
C2: 1≤stocks≤80? Here C1 can be expanded as:
C1a: 1≤locks
C3: 1≤barrels≤90? C1b: locks≤70
C4: sales>1800?
C5: sales>1000?
C6: sales≤1000?

Algorithm:

Step 1: Input 3 integer numbers which represents number of Locks, Stocks and
Barrels sold.
Step 2: compute the total sales =
(Number of Locks sold *45) + (Number of Stocks sold *30) + (Number of
Barrels sold *25)
Step 3: if a totals sale in dollars is less than or equal to $1000
then commission = 0.10* total Sales do Step 6
Step 4: else if total sale is less than $1800
then commission1 = 0.10* 1000
commission = commission1 + (0.15 * (total sales – 1000))
do Step 6
Step 5: else commission1 = 0.10* 1000
commission2 = commission1 + (0.15 * 800))
commission = commission2 + (0.20 * (total sales – 1800)) do
Step 6
Step 6: Print commission.
Step 7: Stop.

8.3 PROGRAM CODE:

#include<stdio.h>
#include<conio.h>
int main()
{
int locks, stocks, barrels, t_sales, flag = 0;
float commission;
clrscr();
printf("Enter the total number of locks");
scanf("%d",&locks);
if ((locks <= 0) || (locks > 70))
{
flag = 1;
}
printf("Enter the total number of stocks");
scanf("%d",&stocks);
if ((stocks <= 0) || (stocks > 80))
{
flag = 1;
}
printf("Enter the total number of stocks");
scanf("%d",&stocks);
if ((stocks <= 0) || (stocks > 80))
 {
flag = 1;
}
printf("Enter the total number of barrelss");
scanf("%d",&barrels);
if ((barrels <= 0) || (barrels > 90))
{
flag = 1;
}
if (flag == 1)
{
printf("invalid input");
getch();
exit(0);
}
t_sales = (locks * 45) + (stocks * 30) + (barrels * 25);
if (t_sales <= 1000)
{
commission = 0.10 * t_sales;
}
else if (t_sales < 1800)
{
commission = 0.10 * 1000;
commission = commission + (0.15 * (t_sales - 1000));
}
else
{
commission = 0.10 * 1000;
commission = commission + (0.15 * 800);
commission = commission + (0.20 * (t_sales - 1800));
}
printf("The total sales is %d \n The commission is %f",t_sales,
commission);
getch();
return;
 }

8.4 TESTING

Technique Used: Decision Table Approach

The decision table is given below

Conditions Condition Entries (Rules)

C1: 1≤locks≤70? F T T T T T
C2: 1≤stocks≤80? -- F T T T T
C3: 1≤barrels≤90? -- -- F T T T
C4: sales>1800? -- -- -- T F F
C5: sales>1000? -- -- -- -- T F
C6: sales≤1000? -- -- -- -- -- T
Actions Action Entries
a1: com1 = 0.10*Sales X
a2: com2 =
X
com1+0.15*(sales-1000)
a3: com3 =
X
com2+0.20*(sales-1800)
a4: Out of Range. X X X

Using the decision table we get 6 functional test cases: 3 cases out of range, 1 case
each for sales greater than $1800, sales greater than $1000, sales less than or
equal to $1000.
DERIVING TEST CASES USING Decision Table Approach:
Test Cases
TC Test Case Expected Actual
Locks Stocks Barrels Status
ID Description Output Output
Testing for
1 Requirement 1 -2 40 45 Out of Range
Condition 1 (C1)
Testing for
2 Requirement 1 90 40 45 Out of Range
Condition 1 (C1)
Testing for
3 Requirement 1 35 -3 45 Out of Range
Condition 2 (C2)
Testing for
4 Requirement 1 35 100 45 Out of Range
Condition 2 (C2)
 Testing for
5 Requirement 1 35 40 -10 Out of Range
Condition 3 (C3)
Testing for
6 Requirement 1 35 40 150 Out of Range
Condition 3 (C3)
Testing for
7 5 5 5 500 a1:50
Requirement 2
Testing for
8 15 15 15 1500 a2: 175
Requirement 2
Testing for
9 25 25 25 2500 a3: 360
Requirement 2

8.5 EXECUTION & RESULT DISCUSION:

Execute the program against the designed test cases and complete the table
for Actual output column and status column.
TEST REPORT:

1. No of TC’s Executed:

2. No of Defects Raised:

3. No of TC’s Pass:

4. No of TC’s Failed:
The commission problem is not well served by a decision table analysis
because it has very little decisional. Because the variables in the equivalence
classes are truly independent, no impossible rules will occur in a decision table
in which condition correspond to the equivalence classes.

8.6 SNAPSHOTS:
1. Snapshot for Total sales and commission when total sales are within 1000 and
1800
2. Snapshot when the inputs all are 25.

8.7 REFERENCES:
1. Requirement Specification
2. Assumptions
9. Design, develop, code and run the program in any suitable language to solve
the commission problem. Analyze it from the perspective of dataflow testing,
derive different test cases, execute these test cases and discuss the test results.

9.1 REQUIREMENT SPECIFICATION

Problem Definition: The Commission Problem includes a salesperson in the

former Arizona Territory sold rifle locks, stocks and barrels made by a
gunsmith in Missouri. Cost includes

Locks- $45
Stocks- $30
Barrels- $25
The salesperson had to sell at least one complete rifle per month and
production limits were such that the most the salesperson could sell in a
month was 70 locks, 80 stocks and 90 barrels.
After each town visit, the sales person sent a telegram to the Missouri
gunsmith with the number of locks, stocks and barrels sold in the town. At the
end of the month, the salesperson sent a very short telegram showing -
-1 lock sold. The gunsmith then knew the sales for the month were
complete and computed the salesperson’s commission as follows:
On sales up to(and including) $1000= 10% On
the sales up to(and includes) $1800= 15% On
the sales in excess of $1800= 20%
The commission program produces a monthly sales report that gave the total
number of locks, stocks and barrels sold, the salesperson’s total dollar sales and
 finally the commission

9.2 DESIGN
Algorithm:
Step 1: Define lockPrice=45.0, stockPrice=30.0, barrelPrice=25.0
Step2: Input locks
Step3: while(locks!=-1) ‘input device uses -1 to indicate end of data goto
Step 12
Step 4:input (stocks, barrels)
Step 5: compute lockSales, stockSales, barrelSales and sales
Step 6: output(“Total sales:” sales)
Step 7: if (sales > 1800.0) goto Step 8 else goto Step 9
Step 8: commission=0.10*1000.0; commission=commission+0.15 * 800.0;
commission = commission + 0.20 * (sales-1800.0)
Step 9: if (sales > 1000.0) goto Step 10 else goto Step 11
Step10: commission=0.10* 1000.0; commission=commission + 0.15 *
(sales-1000.0)
Step 11: Output(“Commission is $”, commission)
Step12: exit

9.3 PROGRAM CODE:

#include<stdio.h>
#include<conio.h>
int main()
{
int locks, stocks, barrels, t_sales, flag = 0;
float commission;
clrscr();
printf("Enter the total number of locks");
scanf("%d",&locks);
if ((locks <= 0) || (locks > 70))
{
flag = 1;
}
printf("Enter the total number of stocks");
scanf("%d",&stocks);
if ((stocks <= 0) || (stocks > 80))
{
flag = 1;
}
printf("Enter the total number of barrelss");
scanf("%d",&barrels);
if ((barrels <= 0) || (barrels > 90))
{
flag = 1;
}
if (flag == 1)
{
printf("invalid input");
getch();
exit(0);
}
t_sales = (locks * 45) + (stocks * 30) + (barrels * 25);
if (t_sales <= 1000)
{
commission = 0.10 * t_sales;
}
else if (t_sales < 1800)
{

commission = 0.10 * 1000;

commission = commission + (0.15 * (t_sales - 1000));
}
else
{
commission = 0.10 * 1000;
commission = commission + (0.15 * 800);
commission = commission + (0.20 * (t_sales - 1800));
}
printf("The total sales is %d \n The commission is %f",t_sales,
commission);
getch(); return; }
9.4 TESTING TECHNIQUE: DATAFLOW TESTING
A structural testing technique
• Aims to execute sub-paths from points where each variable is defined to points
where it is referenced. These sub-paths are called definition-use pairs, or du-pairs
(du-paths, du-chains) Data flow testing is centred on variables (data) Data flow
testing follows the sequences of events related to a given data item with the
objective to detect incorrect sequences It explores the effect of using the value
produced by every and each computation.

Variable definition
Occurrences of a variable where a variable is given a new value (assignment,
input by the user, input from a file, etc.) Variable DECLARATION is NOT its
definition !!!

Variable uses
Occurences of a variable where a variable is not given a new value (variable
DECLARATION is NOT its use)

p-uses (predicate uses)

Occur in the predicate portion of a decision statement such as if-then-else,
while-do etc.

c-uses (computation uses)

All others, including variable occurrences in the right hand side of an
assignment statement, or an output statement

du-path: A sub-path from a variable definition to its use.

Test case definitions based on four groups of coverage
– All definitions.

– All c-uses.

– All p-uses.

– All du-paths.

DATA FLOW TESTING: KEY Steps

Given a code (program or pseudo-code).

1. Number the lines.

2. List the variables.

3. List occurrences & assign a category to each variable.

4. Identify du-pairs and their use (p- or c- ).

5. Define test cases, depending on the required coverage.

line catogary
Definition c-use p-use
1
2
3
4
5
6
7
8
9 locks, stocks, barrels
10 locks, stocks, barrels
11
12 Flag
13
14 flag
15
16
17

18
19
locks, stocks,
20 t_sales barrels
21 t_sales
22
23 commission t_sales
 24
25 t_sales
26
27 commission
28 commission commission, t_sales
29
30
31
32 commission
33 commission commission
34 commission commission, t_sales
35
36 commission
37
38
39
Table: list occurrences & assign a category to each variable

definition - use pair variables()

start line → end line c-use p-use
9→10 locks
9→10 stocks
9→10 barrels
9→20 locks
9→20 stocks
9→20 barrels
12→14 flag
20→21 t_sales
20→23 t_sales
 20→25 t_sales
20→28 t_sales
20→34 t_sales
23→36 commission
27→28 commission
28→36 commission
32→33 commission
33→34 commission
34→36 commission

Table: Define test cases

TEST CASES BASED ON ALL DEFINITION

To achieve 100% All-definitions data flow coverage at least one sub-path from
each variable definition to some use of that definition (either c- or p- use) must be
executed.

Inputs Expected output

Variable(s) du- sub- locks stocks barrels t_sales commiss
pair path ion
locks, 9→20 9,10,20 10 10 10 1000
stocks,
barrels
locks, 9→10 9→10 5 -1 22 Invalid Input
stocks,
barrels
Flag 12→14 12→14 -1 40 45 Invalid Input

t_sales 20→21 20,21 5 5 5 500

t_sales 20→25 20,21,25 15 15 15 1500
commission 23→36 23→36 5 5 5 50
commission 27→36 27,28,36 15 15 15 175
commission 32→36 32,33,34, 25 25 25 360
36

9.5 EXECUTION
Execute the program and test the test cases in above Tables
against program and complete the table with for Actual output column
and
Status column.
 9.6 SNAPSHOTS:
1. Snapshot for Total sales and commission when total sales are within
1000 and Invalid input

2. Invalid Input and Total sales and commission when total sales are within
1000
3. Snapshot for for Total sales and commission when total sales are within
1800 and to find out the total commission 360

4. Snapshot for total sales and commission

9.7 REFERENCES
1. Requirement Specification.
2. Assumptions.

10. Design, develop, code and run the program in any suitable language to
implement the binary search algorithm. Determine the basis paths and using them
derive different test cases, execute these test cases and discuss the test results.
10.1 REQUIREMENTS SPECIFICATION

R1: The system should accept ‘n’ number of elements and key element that is to
be searched among ‘n’ elements..
R2: Check if the key element is present in the array and display the position if
present otherwise print unsuccessful search.

10.2 DESIGN

We use integer array as a data structure to store ‘n’ number of elements.

Iterative programming technique is used.

Algorithm:
Step 1: Input value of ‘n’. Enter ‘n’ integer numbers in array int mid;
Step 2: Initialize low = 0, high = n -1
Step 3: until ( low <= high ) do mid
= (low + high) / 2 if (
a[mid] == key )
then do Step 5
else if ( a[mid] > key )
then do
high = mid - 1
 else low = mid + 1
Step 4: Print unsuccessful search do Step 6.
Step 5: Print Successful search. Element found at position mid+1.
Step 6: Stop.

10.3 PROGRAM CODE:

1 #include<stdio.h>
2 #include<conio.h>
3 int main()
4 {

5 Int a[20],n,low,high,mid,key,I;
6 int flag=0;
7 clrscr();
8 printf("Enter the value of n:\n");
9 scanf("%d",&n);
10 if(n>0)
11 {
12 printf("Enter %d elements in ASCENDING order\n",n);
13 for(i=0;i<n;i++)
14 {
15 scanf("%d",&a[i]);
16 }
17 printf("Enter the key element to be searched\n");
18 scanf("%d",&key);
19 low-=0;
20 high=n-1;
21 while(low<=high)
22 {
23 mid=(low+high)/2;
24 if(a[mid]==key)
25 {
26 flag=1;
27 break;
28 }
29 else if(a[mid]<key)
30 {
31 low=mid+1;

32 }
33 else
34 {
35 high=mid-1;
36 }
37 }
38 if(flag==1)
39 printf("Successful search\n Element found at Location
%d\n",mid+1);
40 else
41 printf(“Key element is not found”);
42 }
43 else
44 printf("Wrong input");
45 getch();
46 return 0;
47 }

10.4 TESTING

Technique Used: Basis Path Testing

Basis path testing is a form of Structural testing (White Box testing).

The method devised by McCabe to carry out basis path testing has four Steps.
These are:
1. Compute the program graph.

2. Calculate the cyclomatic complexity.

3. Select a basis set of paths.

4. Generate test cases for each of these paths.

Below is the program graph of binary search code.

Using the program graph we derive (Decision-to-Decision) DD path graph

Binary search program
Program Graph Nodes DD – Path Name
First 5
A 6,7,8,9,10
B 11
C 12,13,14
D 15,16,17
E 18
F 19,20
G 37
H 21
I 22,23,24,25,26,27
J 28
K 29.30,31
L 32,33,34,35
M 38
N 40
O 41

The cyclomatic complexity of a connected graph is provided by the

formula V(G) = e – n + 2p. The number of edges is represented by e, the
number of nodes by n and the number of connected regions by p. If we
apply this formula to the graph given below, the number of linearly
independent circuits is:
Number of edges = 21

Number of nodes = 15

Number of connected regions = 1

21 – 15 + 2(1) = 4.
Here we are dealing code level dependencies, which are absolutely
incompatible with the latent assumption, that basis path are independent.
McCabe’s procedure successfully identifies basis path that are
topologically independent, but when these contradict semantic
dependencies, topologically possible paths are seen to be logically
infeasible. One solution to this problem is to always require that
flipping a decision result in a semantically feasible path. For this
problem we identify some of the rules:
If node C not traversed, then node M should be traversed.
If node E and node G is traversed, then node M should be traversed.
If node I is traversed, then node N should be traversed.
Taking into consideration the above rules, next step is to find the basis
paths.
According to cyclomatic complexity 4 feasible basis path exists:
P1: A, B, D, E, G, N, O if n value is 0.
P2: A, B, C, B, D, E, F, H, I, G, M, O key element found.
P3: A, B, C B, D, E, F, H, J, K, E, F, H, J, K, E, G, N, O key element not found.
P4: A, B, C, B, D, E, F, H, J, L, E, F, H, J, L, E, G, N, O key element not found.
DERIVING TEST CASES USING BASIS PATH TESTING
The last step is to devise test cases for the basis paths.
TEST CASES
TC Test Case Value array Expected Actual
key St
ID Description of ‘n‘ elements Output Output
Testing for
key not
1 requirement 1 0 -- 5
found
Path P1
Testing for Key found
2 requirement 2 4 2,3,5,6,7 5 at position
Path P2 3
Testing for
key not
3 requirement 2 3 1,2,5 6
found
Path P3
Testing for
key not
4 requirement 2 3 1,2,5 1
found
Path P4
Testing for Key found
5 requirement 2 5 1,2,4,6,7 2 at position
Path P4+P2-P1 2
Testing for key
6 requirement 2 5 4,5,7,8,9 8 found at
Path P3+P2-P1 position

10.5 EXECUTION & RESULT DISCUSION:

Execute the program against the designed test cases and complete the
table for
Actual output column and status column.
Test Report:
1. No of TC’s Executed: 06
2. No of Defects Raised:
3. No of TC’s Pass:
4. No of TC’s Failed:
10.6 SNAPSHOTS:
1. Snapshot to check successful search and not found key element.

2. Snapshot to check successful search and not found key element.

10.7 REFERENCES:
1. Requirement Specification
2. Assumptions
11. Design, develop, code and run the program in any suitable
language to implement the quicksort algorithm. Determine the basis paths
and using them derive different test cases, execute these test cases and
discuss the test results. discuss the test results.

11.1 REQUIREMENTS SPECIFICATION

R1: The system should accept ‘n’ number of elements and key element
that is to be searched among ‘n’ elements.
R2: Check if the key element is present in the array and display the
position if present otherwise print unsuccessful search.
11.2 DESIGN
We use integer array as a data structure to store ‘n’ number of elements.
Iterative programming technique is used.

11.3 PROGRAM CODE:

// An iterative implementation of quick
sort
1 #include <stdio.h>

// A utility function to swap two elements

2 void swap ( int* a, int* b )
{
3 int t = *a;
4 *a = *b;
5 *b = t;
6 }
/* This function is same in both iterative and recursive*/
7 int partition (int arr[], int l, int h)
8 {
9 int x = arr[h]; int i = (l - 1),j;
 11 for (j = l; j <= h- 1; j++)
12 {
13 if (arr[j] <= x)
14 {
15 i++;
16 swap (&arr[i], &arr[j]);
17 }
18 }
19 swap (&arr[i + 1], &arr[h]);
20 return (i + 1);
21 }

/* A[] --> Array to be sorted, l --> Starting index, h --> Ending index */

22 void quickSortIterative (int arr[], int l, int h)

23 {
// Create an auxiliary
24 int stack[10],p;

// initialize top of
25 int top = -1;

// push initial values of l and h to

26 stack stack[ ++top ] = l;
27 stack[ ++top ] = h;

// Keep popping from stack while is not empty

28 while ( top >= 0 )
29 {
// Pop h and l
30 h = stack[ top-- ];
31 l = stack[ top-- ];
 // Set pivot element at its correct position in sorted
array
32 p = partition( arr, l, h );

// If there are elements on left side of pivot, then push left

// side to stack
33 if ( p-1 > l )
34 {
35 stack[ ++top ] = l;
36 stack[ ++top ] = p - 1;
37 }

// If there are elements on right side of pivot, then push right

// side to
38 stack if ( p+1 < h )
39 {
40 stack[ ++top ] = p + 1;
41 stack[ ++top ] = h;
42 }
43 }
44 }
// Driver program to test above
functions int main()
{
int arr[20],n,i;
clrscr();
printf("Enter the size of the array");
scanf("%d",&n);
printf("Enter %d elements",n);
for(i=0;i<n;i++)
scanf("%d",&arr[i]);
quickSortIterative( arr, 0, n - 1 );
printf("Elements of the array are;");
 for(i=0;i<n;i++)
printf("%d",arr[i]);
getch();
return 0;
}

11.4 TESTING
Program Graph for partition:
Using program graph we derive DD path graph for partition()

DD
Path Program Graph
Names
A 50,51,52,53
B 54
C 56
D 57,58,59,60,61,62
E 64,65,66,67,68

Cyclomatic complixity
No. of edges =6
No. of
nodes=5
e-n+2
6-5+2 =3
No. of predicate nodes +1 (i.e., node B and node C)
2+1=3
No. of region + 1
R1 and R2 are two regions
2+1=3
According to cyclomatic complexity 3 basis path exists. They are,

P1: A, B, E
P2: A, B, C,
D, B, E P3:
A, B, C, B, E

Deriving test cases using basis path

testing: Test Cases

Expected output
Test Array Actual
TC Value of i Status
Case element Array output
ID
Descriptio s
1 n for
Testing 5 5 0
path P1
2 Testing 5, 4, 6, 2, 5, 4, 6, 2, 4
for path 7 7
3 P2
Testing 5, 4, 6, 7, 5, 4, 6, 7, 0
for path 5 5
P3
CYCLOMATIC COMPLEXITY

No. of nodes = 8
No. of nodes =10
e-n+2
10-8+2 =4
No. of predicate nodes + 1
3+1=4 (i.e., node B, D & F)
No. of regions+1
3+1=4 (i.e., Region R1, R2 & R3)

According to cyclomatic complexity 4 basis path exists. They are

P1: A, B, C, D, E,
F, G, B, H P2: A,
B, C, D, E, F, B,H
P3: A, B, C, D, F, G, B, H
P4: A, B, C, D, F, B, H
 Deriving test cases using basis path testing
Test cases:

Test Case Array Expected Actual

TC Status
Description elements output output
ID
1 Testing for path 5, 7, 4, 2, 1, 2, 1, 3, 5, 7,
1 3 4
2 Testing for path 5, 4, 8, 2, 7 5, 4, 2, 7, 8
2
3 Testing for path 5, 4, 6, 7, 3 3, 4, 6, 7, 5
3

11.5 EXECUTION
Compile the program and enter inputs Test above table array
elements for test cases.

11.6 SNAPSHOTS:
1. Snapshot of quick sort sorted elements are displayed, when the n=6
2. Snapshot of quick sort sorted elements are displayed, when the n=5

3. Snapshot of quick sort sorted elements are displayed, when the n=5

11.7 REFERENCES:
1. Requirement Specification
2. Assumptions
12. Design, develop, code and run the program in any suitable language
to implement an absolute letter grading procedure, making suitable
assumptions. Determine the basis paths and using them derive different
test cases, execute these test cases and discuss the test results.

12.1 REQUIREMENTS SPECIFICATION:

R1: The system should accept marks of 6 subjects, each marks in the
range 1 to 100.
i.e., for example, 1<=marks<=100
1<=kannada<=100
1<=maths<=100 etc.
R2: If R1 is satisfied compute average of marks scored and percentage of
the same and depending on percentage display the grade.

12.2 DESIGN:
We use the total percentage of marks to grade the student marks.
<35 && >0 of percentage make it as FAIL
avmar<=40 && avmar>35 make it as Grade
C avmar<=50 && avmar>40 make it as
Grade C+ avmar<=60 && avmar>50 make it
as Grade B avmar<=70 && avmar>60 make
it as Grade B+ avmar<=80 && avmar>70
make it as Grade A avmar<=100 &&
avmar>80 make it as Grade A+

12.3 PROGRAM CODE:

#include<stdio.h>
main()
1 {
2 float kan,eng,hindi,maths,science, sst,avmar;
3 printf("Letter Grading\n");
4 printf("SSLC Marks Grading\n");
5 printf("Enter the marks for 6 Kannada:");
6 scanf("%f",&kan);
7 printf("enter the marks for English:");
8 scanf("%f",&eng);
9 printf("enter the marks for Hindi:");
10 scanf("%f",&hindi);
11 printf("enter the marks for Maths");
12 scanf("%f",&maths);
13 printf("enter the marks for Science:");
14 scanf("%f",&science);
15 printf("enter the marks for Social Science:");
16 scanf("%f",&sst);
17 avmar=(kan+eng+hindi+maths+science+sst)/6.25;
18 printf("the average marks are=%f\n",avmar);
19 if((avmar<35)&&(avmar>0))
20 printf("fail");
21 elseif((avmar<=40)&&(avar>35))
22 printf("Grade C");
23 elseif((avmar<=50)&&(avmar>40))
24 printf("Grade C+");
25 elseif((avmar<=60)&&(avmar>50))
26 printf("Grade B");
27 elseif((avmar<=70)&&(avmar>60))
28 printf("Grade B+");
29 elseif((avmar<=80)&&(avmar>70))
30 printf("Grade A");
31 elseif((avmar<=100)&&(avmar>80))
32 printf("Grade A+");
33 else
34 printf(“Invalid”);
35 }
Using the program graph derive DD path graph

DD path
Program Graph Nodes
Names
A 1, 2, 3, 4, 5, 6, 7, 8 . . . 18
B 19
C 20, 21, 22
D 23
E 24
F 25
G 26
H 27
I 28
J 29
K 30
L 31
M 32
N 33
O 34
P 35
Q 37
R 36
CYCLOMATIC COMPLEXITY
No. of nodes = 18
No. of edges = 24
e-n+2
24-18+2=8

No. of predicate nodes + 1

7 + 1 = 8 (i.e., B, D, F, H, J, L, N)

No. of regions + 1
7 + 1 = 8 (i.e., Regions R1, R2, R3, R4, R5, R6, R7)

According to cyclomatic complexity we can derive 8

basis path.
P1: A, B, R q
P2: A, B, C, D, E, q
P3: A, B, C, D, F, G, q
P4: A, B, C, D, F, H, I, q
P5: A, B, C, D, F, H, J, K, q
P6: A, B, C, D, F, H, J, L, M, q
P7: A, B, C, D, F, H, J, L, N, O, q
P8: A, B, C, D, F, H, J, L, N, P, q

Test Cases:
Test Description Expected Actual
TC Input Status
Output Output
ID
K=50
E=50
H=50 Invalid
1 Testing for path P1
M=50 Input
S=50
SST=150
 K=30
E=30
H=30
2 Testing for path P2 M=35 Fa
S=35 il
SST=35
Avg=32.5
K=40
E=38
H=37
3 Testing for path P3 M=40 Grade C
S=40
SST=38
Avg=38.8
3
K=45
E=47
4 Testing for path P4 H=48 Grade
M=46 C+
S=49
SST=50
Avg=47.5

K=55
E=58
H=60
5 Testing for path P5 M=56 Grade B
S=57
SST=60
Avg=57.66

K=65
E=65
H=65
6 Testing for path P6 M=65 Grade B+
S=65
SST=65
Avg=65.0
 K=75
E=72
H=78
7 Testing for path P7 M=75 Grade A
S=80
SST=80
Avg=76.6
K=85
E=90
H=80
8 Testing for path P8 M=95 Grade
S=85 A+
SST=85
Avg=86.6

12.5 EXECUTION
Compile the program and enter inputs for subject marks, then it will
display the Total percentage, depending on the percentage it will shows the
Grade and test the test cases for above table.

12.6 SNAPSHOTS:
1. Snapshot to Show Fail and Grade C
2. Snapshot to show Grade B and Grade C+

3. Snapshot to show the Grade A and Grade B+

4. Snapshot to show the Grade A+

12.7 REFERENCES:
1. Requirement Specification
2. Assumptions
 EXECUTION STEPS IN LINUX

1. Open Terminal
2. Then open VI –Editor using the filename, following command
will shows that
[root@localhost ~]# vi Triangle.c
3. Write the Suitable code for the given program
4. Then compile and execute the program using the command;
[root@localhost ~]# cc triangle.c
5. Then execute the command;
[root@localhost ~]# ./a.out
6. Enter the suitable input for the program.
7. Then will get the suitable output.