Week 5 - papademas.net
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CSC 170 SESSION 5 NOTES
Welcome Message
Welcome to Session 5 .
Course Web Site(s)
Here is the link to the Course Web site ( CSC 170 ) .
http://www.papademas.net/occ/csc170
Here is the link to the Online Course Web site ( CSC 171 / 172 / 173 ) .
http://d2l.oakton.edu
List of Topics for this Session’s Lecture
Some Linux Commands
Using WolframAlpha
Web Page Design Hints and Tips
Using JavaScript
Algorithm Design Hints and Tips
Writing Complete Computer Programs
Review of Binary Numbers
Review of Hexadecimal Numbers
Review of Quiz 2 ( CSC 171 / 172 / 173 )
What about FTP? File Transfer Protocol
A Review of Machine Problem 1 ( MP1 )
( Root Searching by the Secant Method )
A Preview of Machine Problem 2 ( MP2 )
( Root Searching by the Secant Method )
__________________________________________________________________
Some Linux Commands
Linux is an operating system with command - driven capability. We use Linux to complete our
computer laboratory assignments consisting of machine problems.
1
How would you move a file named lab2.f90 that is currently in the labs folder to the
projects folder? The labs directory is in your home directory and the projects folder is
within your labs directory.
mv ~/labs/lab2.f90 ~/projects
Using Mathematica
For Homework 1 ( CAS 1 ) hints are provided within this first homework assignment.
Web Page Design Hints and Tips
Review of HTML Forms
Placing a Fieldset About Form Data
This example demonstrates how to draw a border with a caption around your data.
<html>
<body>
<fieldset>
<legend>
Health information:
</legend>
<form action="">
Height <input type="text" size="3">
Weight <input type="text" size="3">
</form>
</fieldset>
<p>
If there is no border around the input form, your browser is too
old.
</p>
</body>
</html>
2
Example Forms
Sending an email via a form.
<html>
<body>
<form action="MAILTO:bereans@aol.com" method="post"
enctype="text/plain">
<h3>This form sends an e-mail.</h3>
Name:<br>
<input type="text" name="name"
value="yourname" size="20">
<br>
Mail:<br>
<input type="text" name="mail"
value="yourmail" size="20">
<br>
Comment:<br>
<input type="text" name="comment"
value="yourcomment" size="40">
<br><br>
<input type="submit" value="Send">
<input type="reset" value="Reset">
</form>
</body>
</html>
3
Example Forms
Here is an example of using checkboxes in a form.
<html>
<body>
<form name="input" action="html_form_action.asp" method="get">
I have a bike:
<input type="checkbox" name="vehicle" value="Bike"
checked="checked" />
<br />
I have a car:
<input type="checkbox" name="vehicle" value="Car" />
<br />
I have an airplane:
<input type="checkbox" name="vehicle" value="Airplane" />
<br /><br />
<input type="submit" value="Submit" />
</form>
<p>
If you click the "Submit" button, you send your input to a new
page called html_form_action.asp.
</p>
</body>
</html>
4
Using JavaScript
Simple JavaScript ( JavaScript Functions )
Here is an example of calling a function via a button.
<html>
<head>
<script type="text/javascript">
function myfunction()
{
alert("HELLO");
}
</script>
</head>
<body>
5
<form>
<input type="button"
onclick="myfunction()"
value="Call function">
</form>
<p>By pressing the button, a function will be called. The
function will alert a message.</p>
</body>
</html>
Function Returning a Value
Here is an example of a function returning a value.
<html>
<head>
<script type="text/javascript">
function myFunction()
{
return ("Hello, have a nice day!");
}
</script>
</head>
<body>
<script type="text/javascript">
document.write(myFunction())
</script>
6
<p>The script in the body section calls a function.</p>
<p>The function returns a text.</p>
</body>
</html>
Algorithm Design Hints and Tips
The Babylonian Algorithm
The following algorithm, known
root of a number x where x >
step 1
Let y 1 .
step 2
If x > 2 , let y
step 3
Let y = ( y +
as the Babylonian Algorithm, can be used to compute the square
0.
= x / 2.
x / y ) / 2 , which is the average of y and x / y .
2
If y is not close enough to x , then repeat step 3 .
Return y .
step 4
step 5
This algorithm illustrates the importance of an iterative process. An iterative process is one that
continues until a specified condition is met.
Here is an example of how the Babylonian Algorithm can be implemented to approximate
the square root of the number 11 .
Here, x = 11 is the square root of the number that we wish to approximate.
Now let y = 1 be our initial approximation.
Since x = 11 > 2 , we let y = x / 2 = 11 / 2 = 5.5 .
Next compute ( y + x / y ) / 2 = ( 5.5 + 11 / 5.5 ) / 2 = 7.5 / 2 = 3.75 .
This new value y = 3.75 represents our next approximation to the square root of 11 .
Test to verify that 3.75 is close to the actual value of 11 by squaring 3.75 .
2
Since 3.75 = 14.0625 , which is not yet equal to or even close to 11 , the iterative process
continues.
Compute now ( y x / y ) / 2 = ( 3.75 + 11 / 3.75 ) / 2 = 7.5 / 2 = 3.3417 .
2
Since 3.3417 = 11.1670 , which is closer to 11 , we can be satisfied with 3.3417 as
an approximation to the square root of 11 or we can perform another iteration, which
actually yields 3.3167 . Note that 3.3167 2 = 11.0005 , which is very close to 11 .
The Modulus Operation
The modulus of two numbers m and n is the integer remainder when m is divided by n .
In C++ the modulus operation is accomplished via the modulus operator m % n .
In FORTRAN and Basic, the modulus operation is accomplished via a function
Mod ( m , n ) .
Uses of the modulus operation involve bin - packing and reminder analysis.
Examples :
7
C++ / Java
FORTRAN / Basic
MS Excel
12 % 6
Mod (12, 6)
=MOD(12,6)
( the above result is 0 , since 6 evenly divides 12 )
C++ / Java
FORTRAN / Basic
MS Excel
12 % 5
Mod (12, 5)
=MOD(12,5)
( the above result is 2 , since 5 does not evenly divide 12 but leaves a remainder
of 2 )
Writing Complete Computer Programs
Here is a review of selection control structures from MP2 . ( FORTRAN version )
! 4) increase count and test for failure 1, infinite loop
count = count + 1
if (count > MAX_IT) then
write(*,*) "Failed to converge with points"
write(*,*) "(", a, ", ", ya, ") and (", b, ", ", yb, ")"
exit
endif
! 5) test for failure 2, non-existent slope
if (abs(a-b) < 1e-5) then
write(*,*) "Both x values converged to ", a, " with f(x) = ", ya
exit
endif
! 6) test for failure 3, horizontal slope
if (abs(ya-yb) < 1e-5) then
write(*,*) "Both y values converged to f(x) = ", ya, " with a = ", &
a, " and b = ", b
exit
endif
! 7) calculate the next x value
c = b – yb * (b - a)/(yb - ya)
yc = 3.0 * c**2 + sin(c)- exp(c)
!
write(*,*) "At count = ", count, " c = ", c
8
! 8) test for successful search
if (abs(yc) < 1e-5) then
write(*,*) "Successful search"
exit
endif
Review of Computer Logic ( NOT , AND , OR , XOR )
Computer logic involves conditions and operations.
The typical logical operations include:
Excel Function
Return value
NOT function
Type
= NOT ( logic_condition )
AND function
= AND ( condition1 , condition2 )
OR function
= OR ( condition1 , condition2 )
XOR function
not available
the opposite truth value
of the logic condition
false if one or both of the
logic conditions are false,
otherwise true
true if one or both of the
logic conditions are true,
otherwise false
true if exactly one of the
logic conditions is true,
otherwise false
Programming Languages
Type
NOT function
AND function
OR function
XOR function
C++ / Java Logic Symbol
FORTRAN Logic Symbol
!
&&
||
not available
NOT
AND
OR
not available
What is XOR Logic?
Introduction to XOR Encryption
Consider meeting at a traffic intersection: when one traffic light is " green " the
other traffic light must show " red " .
XOR encryption works as a bitwise operation.
XOR = Exclusive OR
1 = True
0 = False
Truth Table
Given Logic Conditions (Operands) P and Q
P
Q
Result
1
1
1
1
0
0
0
1
0
0
0
0
For XOR logic, the output is true only if the inputs disagree.
P XOR Q
0
1
1
0
9
I have an output of “true” or 1 only when my inputs disagree.
Regular OR Operation
Truth Table
Given Logic Conditions (Operands) P and Q
P
Q
Result
1
1
1
1
0
1
0
1
1
0
0
0
P OR Q
1
1
1
0
For Standard OR logic, the output is true except if both inputs disagree.
Review of Hexadecimal Numbers
What is 569 base ten as a hexadecimal number?
Using FTP
What about FTP? File Transfer Protocol
To connect to your personal campus network drive:
http://myfiles.oakton.edu
To connect to the L:\ Samples drive:
http://myfiles.oakton.edu
Review of Quiz 2 ( CSC 171 / 172 / 173 )
Some stages of the software development process in producing the area of a right
triangle.
1.
Problem statement
Define the area of a right triangle as A = ( 1 / 2 ) * base * height
2.
Input / output description
Input: the two legs of the right triangle
Output: the area of the right triangle
10
3.
Hand example
To calculate the area of the right triangle ( 6 , 8 , 10 ) we would compute:
A = (1/2)*(6)*(8)
4.
Algorithm development
(i)
(ii)
(iii)
(iv)
5.
prompt user for and input the base of the right triangle
prompt for and input the height of the right triangle
calculate the area
display the area to the user
Testing
Given any positive values of the base and the height, calculate the area of the
right triangle.
Which of the following are examples of software that would be representative of
application software?
file utilities
operating systems
spreadsheets
graphics user interface environments
word processors
data bases
Root Searching Techniques ( An Example of Conditional Logic )
Root searching techniques involve locating the root or zeros of a function.
Consider a polynomial function
( whose roots or solutions are x = − 3 , x = 2 and x = 5 / 2 ) :
f(x ) =(x − 2)(2x − 5)(x + 3)
or in an un - factored form:
11
f ( x ) = 2 x 3 - 3 x 2 - 17 x 30
which is of the format:
f(x ) =Ax3 + Bx2 + Cx D
a cubic equation has up to three zeros or roots ( refer to Descartes Rule of Signs )
We can use Mathematica to define the function ( note that an _ underscore
character is used in the definition of the function ) .
We can prove that a root exits at, for example, x 3 by typing the following
Mathematica command. A value of zero will be returned.
f[-3]
You can repeat the above command for the other roots x 2 and x 5 / 2 .
To graph this function with a typical graphing calculator, we can use:
. Y= .
. Y1 = 2 x ^ 3 - 3 x ^ 2 - 17 x 30 .
. Zoom .
. ZStandard .
We can also use Mathematica to plot the function.
12
The above graph shows the zeros ( where the graph crosses the x - axis at x 3 ,
x 2 and x 5 / 2 .
13
We can also use MS Excel to plot the function.
14
Goal: How do we locate a root (zero) using a computer algorithm?
Using a graphing calculator, we can view when y is zero using:
. 2nd .
.
. Table .
One algorithm we can use is called the root bi - section method, which uses a divide
and conquer routine.
How does this root bi - section method work?
STEP 1
Choose a reasonable interval where a zero could be possibly located.
For our example, try using the interval a = 0 to b = 3 .
STEP 2
Evaluate the given function f ( x ) at both a = 0 and b = 3 .
f ( 0 ) = 30
f(3)=6
STEP 3
evaluate function at x = 0
evaluate the function at x =3
Multiply f ( a ) and f ( b ) .
If f ( a ) * f ( b ) 0 that means we possibly do not a root between
these points. We say possibly since we may have a root between
even though this product is positive.
15
If f ( a ) * f ( b ) 0 that means we do possibly a root between these
points.
In our example we have no root so far located.
STEP 4
If f ( a ) * f ( b ) 0 , subdivide the interval [ a , b ] .
( 0 + 3 ) / 2 1.5
This gives us two zones to consider [ 0 , 1.5 ] and [ 1.5 , 3 ] .
Let us consider both zones:
[ 0 , 1.5 ]
f ( 0 ) * f ( 1.5 ) 0
root not found
[ 1.5 , 3 ]
f ( 1.5 ) * f ( 3 ) > 0
root not found
For the second case, check if the endpoints f ( 1.5 ) and f ( 3 ) - are
they roots?
( 1.5 + 3 ) / 2 = 2.25
[ 1.5, 2.25 ] and [ 2.25 , 3 ]
[ 1.5 , 2.25 ]
f ( 1.5 ) * f ( 2.25 ) = 4.5 * -0.65625 < 0 root found
[ 2.25 , 3 ]
f ( 2.25 ) * f ( 3 ) = -0.65625 * 6 < 0 another root
found
16
STEP 5
Continue the bi - section process until such time as the difference
between successive approximations is negligible.
__________________________________________________________________
How Do We Learn to Program
The typical approach to learning a computer programming language is with the
standard flow of information given below.
-History of computing
-Types of computers
-Glossary Terms
-Types of Programming Languages
-Syntax Rules
-Declaring Variables and Data Types
-Expressions and Assignments
-Sequential Control Structures
-Selection Control Structures ( MP2 )
-Repetitive Control Structures
-Functions
-Arrays
-file Processing
-Vectors
17
Computer IF Statements ( Conditional Logic )
Computer IF statements are used when a decision has to be made. For example, a
computer program involving the root bi - section method uses IF statements to
determine whether or not a root has been found.
A typical one - way IF statement has this pseudocode form :
Remark: One - Way IF Statement
IF (condition) THEN
Remark: Statements to Execute if the condition is true
END IF
Examples of If Statements ( One Way )
C++ ( CSC 171 )
if ( x > 0 )
cout << " your number is positive ";
FORTRAN ( CSC 172 )
if ( x > 0 ) then
write (*,*) " your number is positive "
end if
Java ( CSC 173 )
if ( x > 0 )
System.out.println(" your number is positive ");
A typical two - way IF statement has this pseudocode form :
Remark: Two - Way IF Statement
IF (condition) THEN
Remark: Statements to Execute if the condition is true
ELSE
Remark: Statements to Execute if the condition is false
END IF
18
Examples of If Statements ( Two Way )
C++ ( CSC 171 )
if ( x > 0 )
cout << " your number is positive ";
else
cout << " your number is negative or zero ";
FORTRAN ( CSC 172 )
if ( x > 0 ) then
write (*,*) " your number is positive "
else
write (*,*) " your number is negative or zero "
end if
Java ( CSC 173 )
if ( x > 0 )
System.out.println(" your number is positive ");
else
System.out.println(" your number is negative or zero ");
A typical multi - path IF statement has this pseudocode form :
Remark: Multi - Path IF Statement Block
IF (condition 1) THEN
Remark: Statements to Execute if the condition 1 is true
ELSE IF (condition 2)
Remark: Statements to Execute if the condition 2 is true
ELSE
Remark: Statements to Execute if both condition 1 and condition 2 are false
END IF
Examples of If Statements ( Multi - Path )
C++ ( CSC 171 )
if ( x > 0 )
cout << " your number is positive ";
else if ( x < 0 )
cout << " your number is negative ";
else
cout << " your number is equal to 0 ";
19
FORTRAN ( CSC 172 )
if ( x > 0 ) then
write (*,*) " your number is positive "
else if ( x < 0 ) then
write (*,*) " your number is negative "
else
write (*,*) " your number is equal to 0 "
end if
Java ( CSC 173 )
if ( x > 0 )
System.out.println(" your number is positive ");
else if ( x < 0 ) then
System.out.println(" your number is negative ");
else
System.out.println(" your number is equal to 0 ");
Note: curly braces { } may be used in some computer languages to execute a block
of code at a time once a condition is true.
Using IF Blocks
If statements are used to make a decision in mathematics especially when we are
solving a system of linear equations and determining the character of the solution.
We know that a system of linear equations can have:
-one unique solution ( lines are intersecting at one point )
-no unique solution ( lines are parallel )
-an infinite number of solutions ( lines are dependent )
Computer If statement blocks are used also when we are solving a system of linear
equations using matrix methods, such as the case of using Cramer’s Rule.
Cramer’s Rule states that a system of two linear equations in two unknowns and
having the form
Ax + By=C
Dx + Ey=F
where A , B , C , D , E and F being real - valued coefficients and x and y being the
variables, has a solution given by:
x = (C×E − B×F)/(A×E − B×D)
y = (A×F − C×D)/(A×E − B×D)
20
If statements help prevent a division by zero error when solving a system of linear
equations using Cramer’s Rule.
If the expression ( A × E − B × D ) is 0 then the system does not have a unique
solution but it instead could have an infinite number of solutions or no solution.
[ Example ] ( Cramer’s Rule for a 2 by 2 System of Linear Equations )
Use Cramer’s Rule to solve this system for x and y .
2x + 3y = 8
5x − y = 3
[ Solution ]
This system has a unique solution x = 1 , y = 2 ) .
Note that the expression ( A × E − B × D ) = ( 2 × ( − 1 ) − 3 × 5 ) = − 17 ≠ 0
and hence the system is determined to have a unique solution.
[ Example ] ( Cramer’s Rule for a 2 by 2 System of Linear Equations )
Use Cramer’s Rule to solve this system for x and y .
5x + 3y = 3
10 x + 6 y = 5
[ Solution ]
This system has no solution - the lines are parallel.
Note that the expression ( A × E − B × D ) = ( 5 × 6 − 3 × 10 ) = 0 and hence
the system is determined not to have a unique solution.
To write a 2 by 2 Cramer’s Rule computer program we would have the following
specifications.
Our input is :
A,B,C,D,E,F
the real - valued coefficients
Our test is :
If ( A × E − B × D ) == 0 )
we test the Cramer’s Rule denominator
21
Our process is :
x = (C×E−B×F)/(A×E − B×D)
the x variable
y = (A×F−C×D)/(A×E − B×D)
the y variable
Our output is :
write " The x solution is " , x
write " The y solution is " , y
Cramer’s Rule C++ Version ( CSC 171 )
#include <iostream>
using namespace std;
int main()
{
//declare the variables
int A = 0, B = 0, C = 0, D = 0, E = 0, F = 0;
double Xnum = 0, Ynum = 0, Den = 0;
//input the values
cout << " please enter the coefficients " << endl;
cin >> A >> B >> C >> D >> E >> F;
//apply the formulas
Xnum = (C * E) - ( B * F);
Ynum = (A * F ) - ( C * D);
Den = (A * E ) - ( B * D);
//test the denominator and display the result
if ( Den == 0)
{
cout << " there is no solution or there is an infinite\n ";
cout << " number of solutions\n ";
}
else
{
cout << " a unique solution exists " << endl;
cout << " x = " << Xnum / Den << endl;
cout << " y = " << Ynum / Den << endl;
}
return 0;
}
22
Cramer’s Rule FORTRAN Version ( CSC 172 )
program Cramer
!declare the variables
integer :: A = 0, B = 0, C = 0, D = 0, E = 0, F = 0
real :: Xnum, Ynum, Den
!input the values
write (*, *) " please enter the coefficients "
read( *, *) A, B, C, D, E, F
!apply the formulas
Xnum = (C * E) - ( B * F)
Ynum = (A * F ) - ( C * D)
Den = (A * E ) - ( B * D)
!test the denominator and display the result
if ( Den == 0 ) then
write (*, *) " there is no solution or an infinite number "
write (*, *) " of solutions "
else
write (*, *) " a unique solution exists "
write (*, *) " x = ", Xnum / Den
write (*, *) " y = ", Ynum / Den
endif
end program Cramer
23
Cramer’s Rule Java Version ( CSC 173 )
import java.util.Scanner;
class Cramer
{
public static void main( String args[] )
{
//declare the variables
Scanner sc = new Scanner(System.in);
int A = 0, B = 0, C = 0, D = 0, E = 0, F = 0;
double Xnum = 0, Ynum = 0, Den = 0;
//input the values
System.out.println( " please enter the coefficients " );
A = sc.nextInt();
B = sc.nextInt();
C = sc.nextInt();
D = sc.nextInt();
E = sc.nextInt();
F = sc.nextInt();
//apply the formulas
Xnum = (C * E) - ( B * F);
Ynum = (A * F ) - ( C * D);
Den = (A * E ) - ( B * D);
//test the denominator
if ( Den == 0)
{
System.out.println(
System.out.println(
}
else
{
System.out.println(
System.out.println(
System.out.println(
}
and display the result
" there is no solution or there is " );
" an infinite number of solutions " );
" a unique solution exists " );
" x = " + Xnum / Den );
" y = " + Ynum / Den );
}//end main
}//end class
24
Testing the Program
Use Cramer’s Rule to solve this system for x and y .
2x + 3y = 8
5x − y = 3
Testing the Program
Use Cramer’s Rule to solve this system for x and y .
5x + 3y = 3
10 x + 6 y = 5
25
__________________________________________________________________
A Preview of MP2 ( Root Searching by the Secant Method )
Sample run of the MP2 problem.
26
