Cs 1114 - lecture-10

Instructor : Muhammad Haris

All Rights Reserved to Department of Computer Science – GCU Lahore

Programming Fundamentals

Test Yourself

Find sum of all integers between any two numbersHow many times you have to repeat it?

Programming Fundamentals | Lecture-10 2

Consider this Example

Find even and odd numbers from first n natural numbersWhat’s different in this example?

○ Let’s have a look at the solutions of finding even and odd numbers separately


Find first n even numbers


START

count = count + 1

STOP

count < n

count = 0

YesNo

DISPLAY even

even = 0

even = even + 2

All odd numbers are skipped

by this solution

Find first n odd numbers


START

count = count + 1

STOP

count < n

count = 0

YesNo

DISPLAY odd

odd = -1

odd = odd + 2

All even numbers are skipped

by this solution

What’s Required

We have to consider all odd and even numbers up to nIt is quite obvious that our solution needs to

decide whether a number is even or odd○ Hence, we’ll use a decision box while finding a

number in every repetitionWhat should be the decision rule? Or how can we

decide that a number that is even or odd?- If we divide a number by 2 and remainder is 0 (the

number is divisible by 2), it is considered as an even number

- How can we check the remainder of a division?


Remainder Operator

An operator like “+”, “-”, “/” etc. denoted by “%”, that performs the division between two number and gives you the remainderExamples:

○ 4%2 will give 0 as answer because remainder of 4/2 is “0”

○ 9%2 will give you “1”○ 11%3 = ?○ 54%5 = ?



START

count = count + 1

STOP

count < n

count = 0

No

No

DISPLAY count “is odd”

READ n

count % 2 == 0

DISPLAY count “is even”

Yes

Yes

Important Observation

Any kind of decision making can be done while repeatedly doing a taskIt can range from simple decision making to

complex series of decision makings as you learnt in your previous lessons


Try this Yourself

Find all numbers which are divisible by 3 from first n integers

Find all numbers which are divisible by a number from first n integers



START

count = count + 1

STOP

count < n

count = 0

YesNo

READ n

count % 3 == 0

DISPLAY count

YesNo


Find all odd numbers which are divisible by 3 from first n integersHow many decisions are required in every

repetition?



START

count = count + 1

STOP

count < n

count = 0

YesNo

READ n

YesNo count % 2 != 0

count % 3 == 0

DISPLAY count

YesNo


Display “multiplication table” of a number only if the number is not equal to “0”



START

count = count + 1

STOP

count < n

count = 0

YesNo

DISPLAY num * count

READ num, n

num != 0

No

YesNo

Important Observation

“Decision Making” and “Repetitions” can be inter-mixed in either wayRepeatedly doing a task if a certain

condition is true ORWhile repeatedly doing a task, making

certain decisions ORBoth of the above


A Premonition

All of you who are not yet practicing well enough and taking this course “lightly and non-seriously” should:Get ready to face hard-time (as things are

going to get worse for them) ORStart practicing before it gets too late


Tasks (to be done by next lecture) Display “Fibonacci Series” up to n

Display “Fibonacci Series” up to n, if n is not less than 0



BE PREPAREDFOR QUIZ

INNEXT LECTURE

Technology

Cs 1114 - lecture-10