3. control structures and data files structured programming –sequence structures –selection...

3. Control structures and Data Files

• Structured Programming– Sequence structures– Selection structures– Repetition structures

• Pseudocode/Flowchart• Data Files• Linear Regression

3.1 Algorithm Development

Decomposition outline

1. Read the new time value2. Compute the corresponding velocity and

acceleration values3. Print the new velocity and acceleration

Top-Down Design

–Divide and Conquer–Stepwise refinement–Pseudocode/Flowchart–Evaluation of alternative solutions–Error conditions–Test data generation–Program workthrough

Flowchart Pseudocode

Main: read a,f_a;read c,f_c;read b;set f_b to f_a + (f_c-f_a)print f_b

Sequence Structure

b - ac - a

Read a,f_a

f_b=f_a + (f_c - f_a )

print f_b

start main

stop main

Read c,f_c

b - ac - a

Read b

Is|denominator| < 0.0001

?

rdeniminato

numerator fraction

print fraction

print “Denominator close to zero

no yes

Selection StructureIf |denominator| < 0.0001

print “Denominator close to zero”else

set fraction to numerator/denominatorprint fraction

Flowchart

Pseudocode

time = 0

Istime ≤ 10

?

Compute velocity

print velocity

Increment time by 1

yes

no

Repetition Structure

Flowchart Pseudocode

set time to 0while time ≤ 10

compute velocityprint velocityincrement time by 1

Relational operators

3.2 Condition Expressions

Conditions

a < bx + y >= 10.5fabs(denominator) < 0.001

d =b > c;If (d)

cnt ++

Logical operatorsand &&or ||not !

b = 3;c = 5;! ( b == c || b == 5.5 )

b = 3;c = !b;printf(“ b=%i c=%i “, b, c);

• if statement

• if/else statement

3.3 Selection Statement

If (a<50){

++ count;sum += a;

}

If (a<50){

++ count;sum += a;if (b>a)

b = 0;}

If ( fabs(denominator) < 0.0001 )printf(“Denominator close to zero”);

else{

x = numerator / denominator;printf(“x= %f \n”,x);

}

• Indentation

if ( x > y )if ( y < z )

k ++;else

m ++;else

j++;

if ( x > y )if ( y < z )

k ++;else

m ++;

if ( x > y )if ( y < z )

k ++;else

j++;

?

if ( a < b );

elsecount++;

if ( a >= b )count++;

else;

float a;if ( a == 10.5) ( fabs( a-10.5 ) < 10e-5)

printf (“ a = %f”,a);

• Conditional Operator

If ( a < b )count ++;

elsec = a + b;

a < b ? count ++ : (c= a + b) ;

c = a < b ? a : b ;c = a < b ? count++ : (c = a + b);

a = 3.0 ;b = 4.0 ;count = 0 ;

ccount

• Switch statement

If ( code == 10 )printf(“Too hot - turn equipment off \n”);

else{

If ( code == 11 )printf(“Caution - recheck in 5 minutes \n”);else{

If ( code == 13 )printf(“Turn on circulation fan \n”);

elseprintf(“Normal mode of operation \n”);

}}

• Switch statement(cont.)switch( code ){case 10 :

printf(“Too hot - turn equipment off \n”);break;

case 11 :printf(“Caution - recheck in 5 minutes \n”);break;

case 13 :printf(“Turn on circulation fan \n”);break;

default :printf(“Normal mode of operation \n”);break;

}

3.4 Loop Structure

while ( condition ){

statements;}

do{

statements;} while ( condition );

For(exp_1 ; exp_2 ; exp_3){

statements;}

Deg

ree

to R

adia

ns

0

0.00

0000

1

00.

1745

33

20

0.34

9066

:

:

:

:

for ( k =1, j=5 ; k<=10 ; k++, j++){

sum_1 += k ;sum_2 += j ;

}

• break and continue statement

Sum = 0for ( k=1 ; k <=20 ; k ++ ){

scanf ( “ %lf” , &x ) ;if ( x > 10.0 )

break;sum + = x;

}printf ( “sum = &f \n “, sum);

continue ;

3.5 Problem Solving 3.5 Problem Solving Applied:Applied:

Weather BalloonsWeather Balloons

1. PROBLEM STATEMENT• Using the polynomials that represent the altitude and velocity for a

weather balloons, print a table using units of meters and meters/second. Also find the maximum altitude (or height) and its corresponding time.Assume that the following polynomial represents the altitude or height in meters during the first 48 hours following the launch of a weather balloon:

alt(t) = -0.12t4 + 12t3 -380t2 + 4100t + 220– where the units of t are hours.

• The corresponding polynomial model for the velocity in meters per hours of the weather balloon is as follows:

v(t) = -0.48t3 + 36t2 - 760t + 4100

2. INPUT/OUTPUT DESCRIPTION

• INPUT : Starting timeTime incrementEnding time

• OUTPUT : The table of altitude and velocity values

The Maximum altitude and its corresponding time

3. HAND EXAMPLE

• Assume that the starting time is 0 hours, the time increment is 1 hour, and the ending time is 5 hours. To obtain the correct units, we need to divide the velocity value in meters/hour by 3600 in order to get meters/second. Using our calculator, we can then compute the following values:

• We can also determine the maximum altitude for this table, which is 12,645.00 meters, it occurred at 5 hours

Time Altitude(m)Velocity(m/

s)

0 220.00 1.14

1 3,951.88 0.94

2 6,994.08 0.76

3 9,414.28 0.59

4 11,277.28 0.45

5 12,645.00 0.32

4. ALGORITHM DEVELOPMENT

First step : Decomposition outline– breaks the solution into a series of

sequential steps

Decomposition Outline

1.Get user input to specify times for the table.

2.Generate and print conversion table and find maximum height and corresponding time.

3.Print maximum height and corresponding time.

4. ALGORITHM DEVELOPMENT(Cont.)

Second step : Refinement in Pseudocode

main: read initial, increment, final values from keyboard set max_height to zero set max_time to zero print table heading set time to initial while time<=final

compute height and velocity print height and velocity if height>max_height

set max_height to height set max_time to time

add increment to time print max_time and max_height

CodingCoding/*---------------------------------------------------*/ /* Program chapter3_4 *//* *//* This program prints a table of height and */ /* velocity values for a weather balloon. */ #include <stdio.h> #include <stdlib.h> #include <math.h> main(){

/* Declare and initialize variables. */ double initial, increment, final, time, height, velocity, max_time=0, ma

x_height=0;/* Get user input. */ printf("Enter initial value for table (in hours) \n"); scanf("%lf",&initial); printf("Enter increment between lines (in hours) \n"); scanf("%lf",&increment); printf("Enter final value for table (in hours) \n"); scanf("%lf",&final); /* Print report heading. */ printf("\n\nWeather Balloon Information \n"); printf("Time Height Velocity \n"); printf("(hrs) (meters) (meters/s) \n");

Coding(Cont..)Coding(Cont..)/* Compute and print report information. */ for (time=initial; time<=final; time+=increment) {

height = -0.12*pow(time,4) + 12*pow(time,3) - 380*time*time +4100*time + 220;

velocity = -0.48*pow(time,3) + 36*time*time - 760*time + 4100; printf("%6.2f %8.2f %7.2f \n", time,height,velocity/3600);

if (height > max_height) {

max_height = height; max_time = time;

} } /* Print maximum height and corresponding time. */ printf("\nMaximum balloon height was %8.2f meters \n", max_height);

printf("and it occurred at %6.2f hours \n",max_time); /* Exit program. */ return EXIT_SUCCESS;

}

5. TESTINGIf we use the data from the hand example, we have the following interaction with the program:

Enter initial value for table (in hours) 0 Enter increment between lines (in hours) 1 Enter final value for table (in hours) 5

Weather Balloon Information Time Height Velocity (hrs) (meters) (meters/s) 0.00 220.00 1.14 1.00 3951.88 0.94 2.00 6994.08 0.76 3.00 9414.28 0.59 4.00 11277.28 0.45 5.00 12645.00 0.32

Maximum balloon height was 12645.00 meters and it occurred at 5.00 hours.

altitude and velocity of the balloon for a period of 48 hours

Balloon Altitude

Balloon Velocity

3.6 Data FilesFILE *sensor1;sensor1 = fopen ( “ sensor1.dat”, “r”);fscanf ( sensor1, “ %lf %lf”, &t, &motion);

FILE *balloon;balloon = fopen ( “ balloon.dat”, “w”);fprintf (balloon, “%f %f %f \n”,

time, height, velocity);

fclose (sensor1);fclose (balloon);

#define FILENAME “sensor1.dat” …….sensor1 = fopen (FILENAME, “r”);

Reading Data Files

file namedata typesorder of datathe size of data

number of recordstrailer signal(sentinel signal)end of file indicator

Input file : sensor1.dat

100.0 132.50.1 147.20.2 148.30.3 157.30.4 163.20.5 158.20.6 169.30.7 148.20.8 137.60.9 135.9

Output

Number of sensor readings: 10Average reading: 149.77Maximum reading: 169.30Minimum reading: 132.50

Input file : sensor2.dat

0.0 132.50.1 147.20.2 148.30.3 157.30.4 163.20.5 158.20.6 169.30.7 148.20.8 137.60.9 135.9-99 -99

Input file : sensor3.dat

0.0 132.50.1 147.20.2 148.30.3 157.30.4 163.20.5 158.20.6 169.30.7 148.20.8 137.60.9 135.9

Generating Data File

3.7 Numerical Technique: Linear Modeling

Linear Representation

Time(s) Temperature(˚F)

0 0

1 20

2 60

3 68

4 77

5 110

y = m x + b

Equations for the “best” linear fit in terms of least squares

Equations for the “best” linear fit in terms of least squares

3.8 Problem Solving 3.8 Problem Solving Applied:Applied:

Ozone MeasurementsOzone Measurements

Back GroundBack Ground• Satellite sensors can be used to measure many differ

ent pieces of information that help us understand more about the atmosphere, which is composed of a number of layers around the earth. Starting at the earth's surface, the layers are the troposphere, stratosphere, mesosphere, thermosphere and exosphere. Consider a problem in which we have collected set of data measuring the ozone mixing ratio in parts per million volume (ppmv). Over small regions, these data are nearly linear, and thus we can use a linear model to estimate the ozone at altitudes other than ones for which we have specific data.

Atmospheric layers around the earth

(대류권 )(성층권 )( 중간권 )

( 온도권 )(외기권 )

1. PROBLEM STATEMENT

Use the least-squares technique to determine a linear model for estimating the ozone mixing ratio at a specified altitude.

• INOUT : Data file zone1.dat• OUTPUT : Range of altitudes

Linear model for ozone mixing ratio

2. INPUT/OUTPUT DESCRIPTION

3. HAND EXAMPLE• Assume that the data consist of the following four data

points:

• Using the equations, we can now compute the values of m and b:

Altitude(km)

Ozone Mixing Ratio (ppmv)

20 3

24 4

26 5

28 6

m = 0.37b = -4.6

First step : Decomposition outline– breaks the solution into a series of sequential

steps

Decomposition Outline

1. Read data file values and compute corresponding sums and ranges.

2. Compute slope and y intercept. 3. Print range of altitude and linear model.

4. ALGORITHM DEVELOPMENT

Second step : Refinement in Pseudocode

main: set count to zero set sumx, sumy, sumxy, sumx2 to zero while not at end-of-file

read x,y increment count by 1 if count = 1

set first to x add x to sumx add y to sumy add x*x to sumx2 add x*y to sumxy

set last to x compute slope and y intersect print first, last, slope, y intersect

4. ALGORITHM DEVELOPMENT(Cont.)

*-----------------------------------------------------------*/ /* Program chapter3_9 *//* */ /* This program computes a linear model for a set */ /* of altitude and ozone mixing ratio values. */ #include <stdio.h> #include <stdlib.h> #define FILENAME "zone1.dat" main() {

/* Declare and initialize variables. */ int count=0; double x, y, first, last, sumx=0, sumy=0, sumx2=0, sumxy=0, denominator, m, b; FILE *zone1; /* Open input file. */ zone1 = fopen(FILENAME,"r"); /* While not at the end of the file, */ /* read and accumulate information. */ while ((fscanf(zone1,"%lf %lf",&x,&y)) == 2) {

++count; if (count == 1) first = x; sumx += x; sumy += y; sumx2 += x*x; sumxy += x*y;

} last = x;

Coding

Coding(Cont.)/* Compute slope and y-intercept. */ denominator = sumx*sumx - count*sumx2; m = (sumx*sumy - count*sumxy)/denominator; b = (sumx*sumxy - sumx2*sumy)/denominator; /* Print summary information. */ printf("Range of altitudes in km: \n"); printf("%.2f to %.2f \n\n",first,last); printf("Linear model: \n"); printf("ozone-mix-ratio = %.2f altitude + %.2f \n", m,b); /* Close file and exit program. */ fclose(zone1); return EXIT_SUCCESS;

} /*---------------------------------------------------*/

5. TESTING

• Using the data from the hand example as the contents of the data file zone1.dat, we get the following program output:

Range of altitude in km:

20.00 to 28.00

Linear model:

ozone-mix-ratio = 0.37 alitutde + -4.60