Lecture 4: Calculating by Iterating

The while Repetition StatementRepetition structure

Programmer specifies an action to be repeated while some condition remains true.Pseudocode (go shopping)

While there are more items on my shopping list Purchase next item and cross it off my list

while loop repeated until condition becomes false

Can loop a known or unknown number of times for repeating calculations or function calls (printf).Flowchart

condition

The body of the while

condition while body

Counter-Controlled RepetitionProblemWrite a power algorithm (ex. 2x, where x is a user defined number)Formulating algorithm

PseudocodeSet product to 1Type in exponentSet counter to 1While counter is less than or equal to exponent product = 2 * product

add one to the counterPrint out product

1 Product 20

2*product product 21

x times

2*product product 22

2*product product 23

2*product product 24

2*product product 25

Counter-Controlled RepetitionFormulating algorithm

Flowchart

begin

product = 1;counter = 1;

Input exponent

counter< exponentproduct = 2*product;counter = counter+1;

true

false

Print product

end

C code

Counter-Controlled RepetitionPseudocode

Set product to 1Type in exponentSet counter to 1While counter is less than or

equal to exponent product = 2 * product

add one to the counterPrint out product

Counter to control while loop

/* Computing 2^x, where x is a user defined number */

#include <stdio.h>

/* function main begins program execution */int main( void ){ int product; /* product of 2^x */ int counter; /* number of 2's multipled */ int exponent; /* user defined the power of 2 */

product = 1; counter = 1;

printf( "Enter the exponent:\n" ); /* prompt */ scanf( "%d", &exponent ); /* read an integer */

while ( counter <= exponent ) { product = 2 * product; counter = counter + 1; }

printf("2^%d equals to %d\n", exponent, product);

return 0; /* indicate that program ended successfully */

} /* end function main */

Initialize counter to 1

while loop iterates as long as counter <= exponent

Increment the counter

Do we really need the variable power? Can we save one variable?

/* Computing 2^x, where x is a user defined number */

#include <stdio.h>

/* function main begins program execution */int main( void ){ int product; /* product of 2^x */ int exponent; /* user defined the power of 2 */

product = 1;

printf( "Enter the exponent:\n" ); /* prompt */ scanf( "%d", &exponent ); /* read an integer */

printf("2^%d equals to ", exponent ); while ( exponent > 0 ) { product = 2 * product; exponent = exponent - 1; }

printf("%d\n", product);

return 0; /* indicate that program ended successfully */

} /* end function main */

C code - using one less variable

Counter-Controlled Repetition

Loop repeated until counter reaches a certain valueDefinite repetition: number of repetitions is known

Counter-Controlled Repetition

Write a program that completes the following:Read in an integer from the user.Print out that number of numbers from the Fibonacci sequence.  This would require you to calculate the values of the sequence, not hard code them in your program. (Hint: You should calculate the Fibonacci sequence using two variables.)Use a while loop counter to print the values (you should need only 1 print statement in your loop).(Hint: The Fibonacci number looks like this: 1 1 2 3 5 8 13 21 34... )

Submit your program in the dropbox called Fibonacci Sequence and complete Program Quiz 3 using your program. 

In-Class Programming Exercise

1 1 2 3 5 8 13 21 34 ...

F1 = 1

F2 = 1

Fn = Fn-1 + Fn-2

Fibonacci Sequence

Fibonacci Sequencebegin

Initializing first two Fibonacci numbers;

Enter # of Fibonacci numbers, num

counter<=num

Compute next Fibonacci number;Print the Fibonacci number;counter = counter +1;

true

false

end

Print first two Fibonacci numbers

counter = 3;

Program assumes that input of 2 or higher is used.

Fibonacci Sequencebegin

last = 1;current = 1;

Enter # of Fibonacci numbers, num

counter<=num

next = last + current;Print the Fibonacci number next;last = current;current = next;counter = counter +1;

true

false

end

Print first two Fibonacci numbers

counter = 3;

Program assumes that input of 2 or higher is used.