1.1
Fundamental Control Structures
All algorithms, no matter how complex, can be reduced to combinations of three
fundamental control structures which are independent of the programming language.
1. SEQUENCE structure
One operation, or a group of operations, is performed one after another
2. SELECTION structure
A condition is tested to determine which steps are to be performed next.
Different operations are performed depending on whether the condition tested
was found to be true or false.
3. REPETITION
A series of steps is repeats. In a conditional loop, the steps are repeated as
longs as a certain condition is true. In a counting loop the steps are repeated
for a predetermined number of times.
1.
Sequential Blocks
2.
Branch (or selection) block
3.
Repetition( or Loop)
block
Why are the three basic control structures special and what other control structures
are useful? If we could select our own control structures, the following criteria would
apply:
1.1.1 Criteria for choosing a control structure are:
1. Minimality
The idea of minimality is that we choose the minimal amount of the simplest
structures that we need.
2. Standardisation
It is desirable to have program structures that have the same external interface.
Programs can then be constructed by assembling these components. The three
constructs have only one entry point at the top and one exit point at the bottom so
they can be “plugged” easily together
3. Clarity and expressive power
When choosing control constructs we need a large enough range to allow us to
express clearly what we want the program to do. For this reason C and Java
provide us with do-while and for statements as well as while statements for
repetitions and the switch statement which can be used as an alternative to an
if-else ladder selection. A balance is needed between minimality and the need
for expressive power
4. Abstraction
This is one of the most important ideas in structured programming. To allow us
to understand a complex system, it is necessary to be able to express components
of the system as a hierarchy of interconnected black boxes whose implementation
details is hidden at each level.
1.1.2 Sequence structure
The sequential structure consists of steps that are performed sequentially. The figure
below represents the flowchart and pseudocode of a sequence structure with three
tasks.
Flowchart
Task 1
Task 2
Task 3
Pseudocode
BEGIN
Task 1
Task 2
Task 3
END
In the pseudocode the keywords BEGIN and END mark the beginning and end points
of the structure. The operations performed in the sequence structure are usually
input, output and computations.
Assignment Statements
In pseudocode assignment is indicated by a left arrow (), for example:
Algorthim:
 Assign the product of PI, radius and height to the variable area”
In pseudocode this would be written as follows:
area  PI * radius * height
The flowchart would be
area = PI * radius * height
Input Statements
In pseudocode Input Statements are specified by the word READ, for example:
READ radius, height
The flowchart would be
READ radius, height
Ouput Statements
In pseudocode Output Statements are specified by the word PRINT, for example:
PRINT “Area = “, area
This pseudocode prints a message followed by the value of the variable area.
The flowchart would be
PRINT “Area = “ area
1.1.3 Selection structure
The selection structure tests a condition to determine which steps are to be performed
next. The two forms of selection structure are .
Open Branch - Basic IF Structure
IF is followed by directly by statement or set of statements which is only executed if
the condition tested is true
Flowchart
Pseudocode
true
false
condition
IF condition
True alternative task
ENDIF
true alternative
task
Closed Branch - IF-ELSE Structure
In a closed branch, the flow of the program always goes forward to new information
and the program will do one of two options. If the condition tested is true one set of
operations will be performed. If the condition tested is false, the other set of
operations is executed.
Flowchart
false
false alternative
task
Pseudocode
true
condition
true alternative
task
IF condition
True alternative task
ELSE
False alternative task
ENDIF
Example of Open Branch
Step 3of the following Algorithm to calculate the absolute value of a number.
Read in the number
Check the sign of the number
If the number is negative, then multiply it by -1
Print the absolute value of the number
The flowchart and Pseudocode are as follows:
Flowchart
false
number < 0?
Pseudocode
IF number < 0
true
number
number = (-1) * number
(-1) * number
ENDIF
Example of Closed Branch
Step 3of the following Algorithm to calculate the absolute value of a number.
1. Read in the number
2. Check the sign of the number
3. If the number is greater than or equal to zero proceed to step 4,
otherwise, print a message and stop;
4. Compute the square root of the number.
5. Print the number and the square root on the monitor
The flowchart and Pseudocode are as follows:
Flowchart
false
Pseudocode
true
x >= 0 ?
PRINT
“Number is negative”
y = sqrt(x)
IF number >= 0
y
sqrt (x)
ELSE
PRINT “Number is negative”
ENDIF
IF-ELSEIF Structure and the CASE SELECT Structure
Structures used for multiple choice decisions:
Flowchart for IF ELSEIF Structure
condition 1
true
Pseudocode for IF-ELSEIF
fasle
condition 2
false
task1
true
false
condition 3
task 2
true
IF condition 1
Task1
ELSEIF condition 2
Task 2
ELSEIF condition 3
Task 3
ELSE
Default task
ENDIF
Invalid Selection
task 3
Flowchart for CASE SELECT Structure
Pseudocode for CASE SELECT
SELECTION
CHOICE1’
YES
NO
CHOICE2
NO
ACTION 1
YES
NO
CHOICE3
ACTION 2
YES
ACTION 3
Invalid Selection
SWITCH selection
CASE choice 1
Task1
CASE choice 2
Task2
CASE choice 3
Task3
DEFAULT
Default task
ENDCASE
1.1.4 Repetition structure
3 types of loops:
1. The while Loop
2. The do while Loop
3. The for Loop
WHILE Loop
Flowchart
condition
Pseudocode
false
WHILE condition
Tasks
ENDWHILE
true
Tasks
DOWHILE Loop
Flowchart
Pseudocode
Tasks
DO
Tasks
WHILE condition
condition
true
false
FOR Loop
Flowchart
Pseudocode
Initial Task
false
condition
End task
initial task
FOR condition DO
Tasks
ENDFOR end task
true
Tasks
FOR
control variable {FROM start} TO finish {INSTEPS OF value} D
Tasks
ENDFOR
Flowchart
initialise
index
condition
increment
index
false
true
Tasks
ADVANTAGES OF USING FLOWCHARTS
The benefits of flowcharts are as follows:
1. Communication: Flowcharts are better way of communicating the logic of a
system to all concerned.
2. Effective analysis: With the help of flowchart, problem can be analysed in
more effective way.
3. Proper documentation: Program flowcharts serve as a good program
documentation, which is needed for various purposes.
4. Efficient Coding: The flowcharts act as a guide or blueprint during the systems
analysis and program development phase.
5. Proper Debugging: The flowchart helps in debugging process.
6. Efficient Program Maintenance: The maintenance of operating program
becomes easy with the help of flowchart. It helps the programmer to put
efforts more efficiently on that part
LIMITATIONS OF USING FLOWCHARTS
1. Complex logic: Sometimes, the program logic is quite complicated. In that
case, flowchart becomes complex and clumsy.
2. Alterations and Modifications: If alterations are required the flowchart may
require re-drawing completely.
3. Reproduction: As the flowchart symbols cannot be typed, reproduction of
flowchart becomes a problem.
4. The essentials of what is done can easily be lost in the technical details of how
it is done.
2.
Exercises
Exercise 1:
Draw the flowchart and give pseudocode for the following algorithm:
Algorithm:
1. Read in the number
2. Check the sign of the number
3. If the number is negative, then print a message and stop; otherwise, proceed to step 4;
4. Compute the square root of the number.
5. Print the number and the square root on the monitor
Exercise 2:
Draw the flowchart for the following algorithm:
Algorithm:
1. Input the radius into variable Radius.
2. Input the angle into variable Theta.
3. Compute the arc length from and save the result in the variable ArcLength.
3.1 Assign the product of PI, Radius and Theta
divided by the value 180 to the variable ArcLength
4. Compute the sector area and save the result in the variable SectorArea
4.1 Assign the product of ArcLength and Radius divided by the value 2
to the variable SectorArea
5. Print the values ArcLength and SectorArea on the monitor
5.1 Print the value of the variable ArcLength on the monitor
5.2 Print the value of the variable SectorArea on the monitor
3.
Examples
Example 1:
Draw the flowchart for the following algorithm for a program to compute the gross
salary for an employee given the number of hours worked and the hourly rate. If the
number of hours worked is greater than 40, the hourly rate shall be 1.5 times the
normal hourly rate for all overtime hours. The program should print the number of
overtime hours, the regular salary, the overtime salary, and the gross salary for the
employee :
PROGRAM ANALYSIS:
The main functional requirements are as follows:
 Read in the number of hours worked and the hourly rate
 Determine if the number of hours worked is greater than 40
 Calculate the gross salary if hours worked are less than or equal to 40
 Calculate the gross salary if hours worked are greater than 40
 Print out the number of overtime hours, the regular salary, the overtime
salary, and the gross salary for the employee
Input Data:
 number of hours worked (float)
 hourly rate (float)
Output Data:
 number of overtime hours (float)
 regular salary (float)
 overtime salary (float)
 gross salary (float)
Relationship Input-Output Data:
If hours Number of hours worked <=40 ->
gross salary = hourly rate * number of hours worked
regular salary = hourly rate * number of hours worked
number of overtime hours=0
overtime salary = 0
If hours Number of hours worked >40 ->
regular salary= 40* hourly rate
number of overtime hours= number of hours worked - 40
overtime salary = 1.5*hourly rate * number of overtime hours
gross salary = 40* hourly rate + 1.5*hourly rate * number of
overtime hours
ALGORITHM:
1. Read in the number of hours worked and hourly rate
2. Calculate the number of overtime hours, regular salary and overtime salary as
follows:
If number of hours worked is less than or equal to 40
2.1
Set overtime hours equal to 0
2.2
Calculate regular salary equal to hourly rate * number of hours
worked
2.3
Set overtime salary equal to 0
else do the following
2.4
Calculate overtime hours equal to number of hours worked minus 40
2.5
Calculate regular salary equal to 40* hourly rate
2.6
Set overtime salary equal to 1.5*hourly rate * number of overtime
hours
3. Calculate gross salary equal to regular salary + overtime salary
4. Print number of hours worked
5. Print the regular salary, the overtime salary, and the gross salary for the employee
c
PSEUDOCODE
BEGIN
READ hours, rate
IF hours <= 40
regular_pay  hours * rate
Overtime  0
Overtime_pay  0
ELSE
regular_pay  40 rate
Overtime  hours - 40
Overtime_pay  overtime * 1.5 * rate
ENDIF
Gross_pay  regular_pay + overtime_pay
PRINT hours
PRINT overtime, reguler_pay, overtime_pay, gross_pay
END
FLOWCHART
START
READ hours,rate
Hours
<=40
regular_pay = hours*rate
overtime = 0
overtime_pay=0
regular_pay = 40*rate
overtime = hours - 40
overtime_pay=1.5*overtime*rate
gross_pay = regular_pay +overtime_pay
PRINT hours
PRINT overtime, reguler_pay,
PRINT overtime_pay, gross_pay
STOP
PSEUDOCODE
BEGIN
sum_max  0
sum_min  0
count  1
READ n
WHILE count <= n
READ max_temp, min_temp
sum_min  sum_min + min_temp
sum_max  sum_max + max_temp
count  count + 1
ENDWHILE
avg_min  sum_min/n
avg_max  sum_max/n
PRINT avg_max, avg_min
END
FLOWCHART
START
sum_min = 0
sum_max = 0
count = 1
READ n
count <=n
TRUE
READ min_temp, max_temp
sum_min = sum_min + min_temp
sum_max = sum_max + max_temp
count = count + 1
avg_min = sum_min/n
avg_max = sum_max/n
PRINT avg_min, avg_max
STOP
FALSE
Example 3:
Requirement Specification:
A college is running a number of courses and wishes to find out the age of the oldest
student on each course.
1. The main functional requirements are as follows:
 The program should take as input the age of each student on a particular
course and output the age of the oldest student.
2. Constraints
 The program must be written in C
 The program must be run on a IBM compatible PC
3. Clarifications/Assumption/Guidelines
 Maximum number of students is a class is 50
 Assume the inputs entered are correct and no input checking is required
4. Data Specification
Input Data:
 number of student - (integer)
 age of students (integer array[0..49])
Output Data:
 age of oldest student (integer)
Relationship Input-Output Data:
 age of oldest student > age of each other student
Design: Algorithm:
1. Read in the number of students
2. Read in the ages of the students
3. Calculate the ages of the oldest student
4. Output the age of the oldest student
Design: Refined Algorithm:
1. Read in the number of students
1.1 Print “enter number of students”
1.2 Read number of students, num
2. Read in the ages of the students
For each student
2.1 Print “Enter age of student”
2.2 Read age of student
3. Calculate the ages of the oldest student
3.1 assign oldest to first student
For each student
3.2 if age > oldest then assign oldest to this student
3.2 else do nothing
4. Output the age of the oldest student
PSEUDOCODE
BEGIN
PRINT “enter number of students”
READ num
FOR temp = 1 TO num DO
PRINT “enter age of student number”, temp
READ age[temp-1]
ENDFOR
oldest  age[0]
FOR temp = 1 TO num-1 DO
IF age[temp] > oldest
Oldest  age[temp]
ENDIF
ENDFOR
START
PRINT “oldest student is”, oldest
END
PRINT “ enter
“
READ num
temp = 1
temp num?
FALSE
temp = temp+1
TRUE
PRINT “enter age of
“
READ age[temp-1]
oldest = age[0]
temp = 1
temp num-1?
FALSE
temp = temp+1
TRUE
age[temp]>oldest?
TRUE
oldest = age[temp]
PRINT “
“
STOP
FALSE