Kennesaw State University CS 4491 Assigment 1
Kennesaw
State University
Department
of Computer Science
Advanced
Topic in Computer Science CS 4491/2
Assignment
#1 / Computing with C / Scientific Model
Dalibor
Labudovic
01/16/2013
Initial
Problem Statement:
1.)
- Elaborate on and expand the concept of multi-dimensional abstraction.
- List several arguments that support the concept of multi-dimensional abstraction.
- How can multi-dimensional abstraction help in multi- disciplinary computation modeling?
2.)
Estimate the height of a building, given the height of a person, h,
the distance from the building, D,
and the elevation angle, θ,
in degrees. Develop a C program to compute the height of a building
using the mathematical formula: bh
= h + D x tan(θπ/180)
3.)
Develop a C program to compute the distance between two points on a
circle in the X-Y plane. Use the mathematical expression for
computing the distance between two points in the X-Y plane: d
=![]()
Summary
and purpose of the assignment activity:
The
purpose of the assignment is explore the different resources a
computer programmer/scientist has at his disposal in order to achieve
an objective. Defining the term multi-disciplinary abstraction is a
summary of individual skills combined to provide a fundamental and
essential principle to solve a problem and software development.
Detail
description of the solution and used in the project:
On
this assignment, I used shell to write a file dot C program to
computerize a mathematical solution to the initial problem presented.
Once the program has been written, saved, compiled and tested, the
program has presented the desired solution. Variables have been
declared and initiated as a double integer. To test the estimate of a
building height, the math.h library had been initialed, but the
library had to be linked. To link a library to the file.c, script
command required to link is as followed: gcc filename –o outputfile
–lm.
Table
of results:
1.a)
To elaborate and expand on the multi-dimensional abstraction, is an
individual skill set combined with other individuals to simplify
and organize a complex system into a sensible model. A sensible model
is a working prototype but it only contains the essentials of the
planed/proposed solution.
1.b)
The concept of multi-dimension abstraction can be supported with
several arguments . Each technical part of a system, such as
mathematical computing may require higher level of math, where a
mathematician determines and develops a math model and computer
scientist translates it into a computer model. A complex system
requires input from couple of different fields of science and
expertise in order to provide a computing solution.
1.c)
A multi-dimensional abstraction process can determine a solution to a
computing issue with the help of multidisciplinary computing. With
multidisciplinary computing, each individual retains their own
methods to the solution without intermixture with other individuals.
The advantage of the multidisciplinary computing is the fresh look
from each perspective on the same issue. With each discipline, there
are advantages and disadvantages but the best part of
multidisciplinary computing is the ability to pick and choose best
methodologies from each individual and then compile the best system
to model a solution.
2)
The mathematical formula for estimated height, the solution is
straight forward. The solution is easily found with the three known
variables.
3)
The mathematical formula for distance between two points on a X-Y
plane is appropriate for computing, as it is simple mathematical
solution. The computing program is useful in situations with two
known points.
Discussion
of how the system modeled can be improved and extended to include
other service:
The
system model can be improved with addition of graphics and easy to
use with UGI. The simplification and performance could be improved
with form and event driven programming.
Comments
and Conclusion:
During
the assignment, I have learned the importance of other fields input
in computing disciplines and finished product. The complexity of a
system has to be simplified with a model.
List
of References:
Garrido
C., Jose. Using Numerical Libraries and C programming for Computing
Modeling.
CS 4491/02.Web.Science.kennesaw.edu/~jgarrido/CS4491_notes/CS4491_sp13_1_slides.pdf
01/16/2013
Script:
#include
<stdio.h>
#include
<math.h>
#define
PI 3.14159
int
main(){
double
h;//height
double
D;//distance
double
a;//angle
double
bh;// building height
//user
input
printf("\nEnter
person height: \n");
scanf("\n%lf",
&h);
printf("\nEnter
distance form building: \n");
scanf("\n%lf",
&D);
printf("\nEnter
angle: \n");
scanf("\n%lf",
&a);
//calculation
bh
= h + D * tan(a*(PI/180));
//print
solution
printf("\nThe
estimated distance of the building is : %lf", bh);
}
#include
<stdio.h>
#include
<math.h>
int
main(){
double
x1, y0, x2, y2, distance;
printf("\nEnter
first point: (x,y)\n");
scanf("%lf%lf",
&x1, &y0);
printf("\nEnter
second point: (x,y)\n");
scanf("%lf%lf",
&x2, &y2);
//calculate
distance
= ((x2-x1) + (y2-y0));
//print
solution
printf("\nDistance
between (%lf,%lf) and (%lf,%lf) is : %lf",x1,
y0,x2,y2,distance);
}
Script
started on Tue 29 Jan 2013 05:44:38 PM EST
[dlabudov@cs3
~]$ nano distance.c
#(B#)0#[1;28r#[m##[?7h#[?1h#=#[?1h#=#[?1h#=#[H#[J#[0;7m#
GNU nano 1.3.12 File: distance.c
#[3;1H#[m##include
<stdio.h>
#[1B#include
<math.h>
#[2Bint
main(){
#[1B double
x1, y0, x2, y2, distance;#[9;9Hprintf("\nEnter first point:
(x,y)\n");#[10;9Hscanf("%lf%lf", &x1,
&y0);#[11;9Hprintf("\nEnter second point:
(x,y)\n");#[12;9Hscanf("%lf%lf", &x2,
&y2);#[14;9H//calculate#[15;9Hdistance = ((x2-x1) +
(y2-y0));#[17;9H//print solution#[18;9Hprintf("\nDistance
between (%lf,%lf) and (%lf,%lf) is : %lf",x1,
y0,x2,y2,distance);
#[1B}#[26;53H#[0;7m#[
Read 17 lines ]
#[1B^G#[m#
Get Help#[27;21H#[0;7m#^O#[m# WriteOut#[27;41H#[0;7m#^R#[m# Read
File#[27;61H#[0;7m#^Y#[m# Prev Page#[27;81H#[0;7m#^K#[m# Cut
Text#[27;101H#[0;7m#^C#[m# Cur Pos
#[1B#[0;7m#^X#[m#
Exit#[28;21H#[0;7m#^J#[m# Justify#[28;41H#[0;7m#^W#[m# Where
Is#[28;61H#[0;7m#^V#[m# Next Page#[28;81H#[0;7m#^U#[m# UnCut
Text#[7C#[0;7m#^T#[m# To Spell
#[25A##[1;112H#[0;7m#Modified
#[2B#[m#include
<stdio.h>
#include
<stdio.h>
##[23B##[0;7m#Save
modified buffer (ANSWERING "No" WILL DESTROY CHANGES) ?
#[27;1H Y#[m#
Yes#[K
#[1B#[0;7m#
N#[m# No #[28;17H#[0;7m#^C#[m# Cancel#[K#[26;62H
#[0;7m#File
Name to Write: distance.c
#[1B^G#[m#
Get Help#[27;31H#[0;7m#^T#[m# To Files#[27;61H#[0;7m#M-M#[m# Mac
Format#[27;91H#[0;7m#M-P#[m# Prepend
#[1B#[0;7m#^C#[m#
Cancel #[0;7m#M-D#[m# DOS Format#[28;61H#[0;7m#M-A#[m#
Append#[28;91H#[0;7m#M-B#[m# Backup File#[26;31H
#[1B#[J#[1;112H#[0;7m#
#[26;51H#[m##[1K #[0;7m#[ Wrote 17 lines
]#[m##[K#[2B#[51C#[28;1H
#[?1l#>[dlabudov@cs3
~]$ gcc -Wall distance.c -lm
distance.c:
In function ‘main’:
distance.c:17:
warning: control reaches end of non-void function
[dlabudov@cs3
~]$ ./a.out
Enter
first point: (x,y)
3
4
Enter
second point: (x,y)
7
8
Distance
between (3.000000,4.000000) and (7.000000,8.000000) is :
8.000000[dlabudov@cs3 ~]$ nano e##[Kheight.c
#(B#)0#[1;28r#[m##[?7h#[?1h#=#[?1h#=#[?1h#=#[H#[J#[0;7m#
GNU nano 1.3.12 File: height.c
#[3;1H#[m##include
<stdio.h>
#[1B#include
<math.h>
#[1B#define
PI 3.14159
#[2Bint
main(){
#[2B double
h;//height#[10;9Hdouble D;//distance#[11;9Hdouble
a;//angle#[12;9Hdouble bh;// building height#[14;9H//user
input#[15;9Hprintf("\nEnter person height:
\n");#[16;9Hscanf("\n%lf", &h);#[17;9Hprintf("\nEnter
distance form building: \n");#[18;9Hscanf("\n%lf",
&D);#[19;9Hprintf("\nEnter angle: \n");#[20;9Hscanf("\n%lf",
&a);#[22;9H//calculation#[23;9Hbh = h + D *
tan(a*(PI/180));#[25;9H//print solution#[26;53H#[0;7m#[ Read 25 lines
]
#[1B^G#[m#
Get Help#[27;21H#[0;7m#^O#[m# WriteOut#[27;41H#[0;7m#^R#[m# Read
File#[27;61H#[0;7m#^Y#[m# Prev Page#[27;81H#[0;7m#^K#[m# Cut
Text#[27;101H#[0;7m#^C#[m# Cur Pos
#[1B#[0;7m#^X#[m#
Exit#[28;21H#[0;7m#^J#[m# Justify#[28;41H#[0;7m#^W#[m# Where
Is#[28;61H#[0;7m#^V#[m# Next Page#[28;81H#[0;7m#^U#[m# UnCut
Text#[7C#[0;7m#^T#[m# To Spell
#[25A##[1;112H#[0;7m#Modified
#[2B#[m#include
<stdio.h>
#include
<stdio.h>
##[23B##[0;7m#Save
modified buffer (ANSWERING "No" WILL DESTROY CHANGES) ?
#[27;1H Y#[m#
Yes#[K
#[1B#[0;7m#
N#[m# No #[28;17H#[0;7m#^C#[m# Cancel#[K#[26;62H
#[0;7m#File
Name to Write: height.c
#[1B^G#[m#
Get Help#[27;31H#[0;7m#^T#[m# To Files#[27;61H#[0;7m#M-M#[m# Mac
Format#[27;91H#[0;7m#M-P#[m# Prepend
#[1B#[0;7m#^C#[m#
Cancel #[0;7m#M-D#[m# DOS Format#[28;61H#[0;7m#M-A#[m#
Append#[28;91H#[0;7m#M-B#[m# Backup File#[26;29H
#[1B#[J#[1;112H#[0;7m#
#[26;51H#[m##[1K #[0;7m#[ Wrote 25 lines
]#[m##[K#[2B#[51C#[28;1H
#[?1l#>[dlabudov@cs3
~]$ gcc -Wall height.c -lm
height.c:
In function ‘main’:
height.c:25:
warning: control reaches end of non-void function
[dlabudov@cs3
~]$ ./a.out
Enter
person height:
2
Enter
distance form building:
545
Enter
angle:
45
The
estimated distance of the building is : 546.999277[dlabudov@cs3 ~]$
exit
exit
Script
done on Tue 29 Jan 2013 05:45:38 PM EST
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