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Quiz Two

• When – Mon (23 Sep starting at 12noon) till Sun (29 Sep ending at 11.55pm).

• You have up to 5 attempts• Remember to press “submit” after you have

completed each quiz attempt. Pressing “save” WILL NOT register your quiz mark.

Mid Semester Test

• When – Next Mon (23 Sep) at 11am• Three main questions (40 mins)

Q1 : Fundamentals (14 marks)Q2 : Control (13 marks)Q3 : Repetition (13 marks)

• Past year test papers are available in Blackboard

Engineering Programming 100Lecture 6: Repetition

Dr Tele Tan

This week

• Explore the various methods to repeat sections of code– For loops– Do while loops– While do loops

• Explore the methods using an engineering/mathematical example

Last week example: what if enter a negative time?

• Need an if statement to control whether to compute distanceif (time <= 0.0)

{distance = 0.0;printf("warning, time <= 0.0\n");

}else{

conversion = 3.141592 / 180.0;theta_2 = conversion * theta;distance = 0.5 * time * time * 9.81 * sin (theta_2);

}

• Problem? Need to re-run program to enter correct value

Simple program

• List of statements executed in order:int main(void) 1

{

int x; 2

int y; 3

int z; 4

x = 5; 5

y = 7; 6

z = x + y; 7

printf("z: %d\n",z); 8

} exit

Repeating statements

• Choose how many times to execute a block of code - called iteration

• Examples (from last lecture):– Compute distance a block has slid for a number of values of

friction coefficients and plot graph of results– Convert student marks to grades for a number of students

• Types of looping:– for loop– do while (also known as repeat until) loop– while do loop

Last week example – solution (1)

• Use a while loop to make sure time > 0.0time = -1.0; // force into loop by making -vewhile (time <= 0.0){

printf("enter a time: ");scanf("%f", &time);if(time <= 0.0)

printf("error, time must be +ve and greater than zero\n");}conversion = 3.141592 / 180.0;theta_2 = conversion * theta;distance = 0.5 * time * time * 9.81 * sin (theta_2);

• Use of relation (i.e. time <= 0.0) and logical (i.e. &&, || or !) operators as in last week

• The loop exits only if time is > 0.0

while_example.c

Last week example – solution (2)

• Use a do loop to make sure time > 0.0

do{

printf("enter a time: ");scanf("%f", &time);if(time <= 0.0)

printf("error, time must be +ve and greater than zero\n");} while (time <= 0.0);conversion = 3.141592 / 180.0;theta_2 = conversion * theta;distance = 0.5 * time * time * 9.81 * sin (theta_2);

• The loop exits only if time is > 0.0

do_example.c

Example - flow of liquid in a pipe

• An engineer is responsible for recommending a certain size of pipe for a new plant. The internal radius of the pipe can vary between 10 and 50 cm in steps of 5 cm. The size of the pipe required is dependent on the volume of liquid flowing out of the pipe in m3 per second.

Flow of liquid

• We need to design a program that prints out a table of values showing the radius of the pipe and the flow volume:-----------------------------------

Volume of liquid flowing from pipes

Flow is constant at 2.0m/s

-----------------------------------

Radius in cm Volume in cubic m/s

10.0 0.063

15.0 0.141

20.0 0.251

25.0 0.393

30.0 0.565

35.0 0.770

40.0 1.005

45.0 1.272

50.0 1.571

-----------------------------------

• If need a volume rate of 0.5m3/sec then choose radius = 30.0cm.

Using a “while do” loopint i = 0;

float radius, volume_flow;

float initial_radius;

float increase_radius;

float flow;

float conversion = 0.01;

initial_radius = 10.0;

increase_radius = 5.0;

flow = 2.0;

while ( i < 9 )

{

radius = initial_radius + increase_radius * i;

volume_flow = PI * radius * conversion * radius * conversion * flow;

printf("%.1f %.3f\n", radius, volume_flow);

i = i + 1;

}

• Executes for i = 0, 1, 2, 3, 4, 5, 6, 7, 8 (9 iterations)

Pseudo code:

while condition truebeginexecute commandsend

Code6_1.c

“while do” flowchart

testcondition

statement

true

false

Using “do while” loopint i = 0;

float radius, volume_flow;

float initial_radius;

float increase_radius;

float flow;

float conversion = 0.01;

initial_radius = 10.0;

increase_radius = 5.0;

flow = 2.0;

do

{

radius = initial_radius + increase_radius * i;

volume_flow = PI * radius * conversion * radius * conversion * flow;

printf("%.1f %.3f\n", radius, volume_flow);

i = i + 1;

} while ( i < 9 );

• Executes for i = 0, 1, 2, 3, 4, 5, 6, 7, 8 (9 iterations)

Pseudo code:

dobeginexecute commandsend

while condition true

Code6_2.c

“do while” flowchart

testcondition

statement

true

false

Using a “for” loopint i = 0;float radius, volume_flow;float initial_radius;float increase_radius;float flow;float conversion = 0.01;

initial_radius = 10.0;increase_radius = 5.0;flow = 2.0;

for ( i=0; i<9; i++){

radius = initial_radius + increase_radius * i;volume_flow = PI * radius * conversion * radius * conversion *

flow;printf("%.1f %.3f\n", radius, volume_flow);

}

• Executes for i = 0, 1, 2, 3, 4, 5, 6, 7, 8 (9 iterations)

Code6_3.c

The “for” loop

• Pseudo code:Do for i from start to finish, incrementing by inc

beginstatements

end

• Variables: start = 3, finish = 8, inc = 1Values of i:• 3, 4, 5, 6, 7, 8

• Variables: start = 3, finish = 12, inc = 3Values of i:• 3, 6, 9, 12

• Variables: start = 12, finish =-6, inc = -3Values of i:• 12, 9, 6, 3, 0, -3, -6

for ( i = 3; i <= 8; i++)

for ( i = 3; i <= 12; i+=3)

for ( i = 12; i >= -6; i-=3)

The “for” loop

• Variables: start = 3, finish = 7, inc = 1Values of i:• 3, 4, 5, 6, 7

• Variables: start = 3, finish = 9, inc = 3Values of i:• 3, 6, 9

• Variables: start = 12, finish =-3, inc = -3Values of i:• 12, 9, 6, 3, 0, -3

The “for” loop

• Variables: start = 3, finish = 7, inc = 1• Values of i:

• 3, 4, 5, 6, 7

• Variables: start = 3, finish = 9, inc = 3• Values of i:

• 3, 6, 9

• Variables: start = 12, finish =-3, inc = -3• Values of i:

• 12, 9, 6, 3, 0, -3

for ( i = 3; i < 8; i++) orfor ( i = 3; i <= 7; i++)

for ( i = 3; i < 12; i+=3) orfor ( i = 3; i <= 9; i=i+3)

for ( i = 12; i > -6; i-=3) orfor ( i = 12; i >= -3; i=i-3)

Using a “for” loop - alternativefloat radius, volume_flow;

float initial_radius;

float increase_radius;

float conversion = 0.01;

initial_radius = 10.0;

increase_radius = 5.0;

for (radius=initial_radius; radius<=50.0; radius+=increase_radius)

{

volume_flow = PI * radius * conversion * radius * conversion * flow;

printf("%.1f %.3f\n", radius, volume_flow);

}

• Executes for radius = 10, 15, 20, 25, 30, 35, 40, 45, 50 (9 iterations)

Which looping method to use?

• Common features:– Loop variable– Initialisation– Termination test– Updating expression

• Counting (for) or general (while or do while)• If a fixed number of iterations,

– Use a for loop: for(...; ...; ...)– Don’t change the loop variable in the loop!

• Execute the loop at least once:– Use a do while loop: do{...}(while...)– Post tests the loop variable

• Allow no execution of the loop:– Use a while do loop: – Pre tests the loop variable

The Bisection method• Find the root of an equation given the

function

• E.g.: Find x given F(x) = x2-5 = 0

• Iterate through values of x to find solution

• Bisection method:1 Choose a=a1 and b=b12 Find midpoint: (a + b)/23 If F(a) * F(midpoint) +ve4 make a = midpoint5 else6 make b = midpoint7 Repeat from 1 until a=b

From en.wikipedia.org/wiki/Bisection_method

Pseudo code versionwhile fabs(x_right - x_left) > epsilon

x_midpoint = (x_right + x_left) / 2if ( ( f(x_left) * f(x_midpoint) ) > 0) then

x_left = x_midpointelse

x_right = x_midpointend if

end while

• epsilon stops the iteration• epsilon - Why not just: (x_left - x_right) == 0 ?

– Rounding errors - may never equal 0

• Computer representation of real numbers is an approximation

fabs() – absolute value function

C version (not complete)double x_midpoint, x_left, x_right;double f_midpoint, f_left, f_right;double epsilon;

x_left = 0.0;x_right = 100.0;epsilon = 0.0000000001;

while (fabs(x_right - x_left) > epsilon) /* while (x_right - x_left) > epsilon */

{x_midpoint = (x_right + x_left) / 2.0; /* x_midpoint = (x_right+x_left) / 2 */f_left = x_left * x_left - 5.0; /* Calculate f(x_left) */

f_midpoint = x_midpoint * x_midpoint - 5.0; /* Calculate f(x_midpoint) */if ((f_left * f_midpoint) > 0) /* if ((f(x_left)*f(x_midpoint)) > 0) then */

{ x_left = x_midpoint;

} else /* else ((f(x_left)*f(x_midpoint)) <= 0)*/

{x_right = x_midpoint;

} }

Code6_4.c

SolutionSqrt(5) = 2.2360679774

Example of approximations/* simple program demonstrating rounding errors */

#include <stdio.h>

int main(void)

{

float f = 0.37;

float g = 2.01 - 2.0;

printf("%.10f\n", f);

printf("%.10f\n", g);

printf("%.16f\n", (2.01 - 2.0) );

return 0;

}

• Outputs (note can change depending on the computer architecture and compiler:

0.3700000048

0.0099999998

0.0099999999999998

Solutions to approximation errors

• Comparing whole numbers:– convert type

if ( (int) amount == 0 ) /* e.g. amount is float */

• removes decimal part • e.g. 2.4 => 2• e.g. 13.6 => 13

– Use the round function in the maths libraryif ( (int) round(amount) == 0)

• rounds to nearest integer• 2.4 => 2• 2.6 => 3• 2.5 => 3

• Use a tolerance (precision) - must be valid for the application

• Two or three decimal places may be okay for money but not for submarine propeller blades!

Today we discussed…

• Looping in a program• Explored different ways to do it• Constructed code using some of the methods

Mid Semester Test

• When – Next Mon (23 Sep) at 11am• Three main questions (40 mins)

Q1 : Fundamentals (14 marks)Q2 : Control (13 marks)Q3 : Repetition (13 marks)

• Past year test papers are available in Blackboard