Knapsack Problem - Theory, Algorithm, Example, Math for Computer Science Students

Algorithm Greedy Knapsack(m, n)
// P[1:n] and W[1:n] contain the profit and the weight respectively
// of the n objectives such that P[i]/W[i] >= P[i+1]/W[i+1]
// m is the knapsack size and x[1:n] is the solution vector
{
    for i = 1 to n do x[i] = 0.0
    U = m;
    for i = 1 to n do
    {
        if (W[i] > U) then break
        x[i] = 1.0; U = U - W[i]
    }
  if (i <= n) then x[i] = U/W[i]
}

Data Mining. Concepts and Techniques, 3rd Edition Free Download.pdf

Book Name : Data Mining. Concepts and Techniques
Book Edition: 3rd Edition
Book Author : Jiawei Han, Micheline Kamber, Jian Pei
Book Download Link :
Download Link 1 - Download Data Mining Book From BDUpload
Download Link 2 - Download From Media Fire via Adf.ly
Download Link 3 -
Book Download Size - 12.5MB


Book Front Page -

So, Download Differential Equation Book By Clicking
Download Link 1 - Download Data Mining Book From BDUpload
Download Link 2 - Download From Media Fire via Adf.ly
 Button.

Some Demo Pages 

Of Data Mining. Concepts and Techniques, 3rd Edition Book Screenshot. Click download link1 or download link2 to download

Note:

If you have face any problem to download Data Mining. Concepts and Techniques, 3rd Edition Free Download By Jiawei Han, Micheline Kamber, Jian Pei.pdf then comment below. A direct link of this book will send to you immediately.



Tags:
Data Mining. Concepts and Techniques, 3rd Edition Free Download By Jiawei Han, Micheline Kamber, Jian Pei.pdf, Data Mining. Concepts and Techniques, 3rd Edition, Data Mining free pdf, data mining book download link, data mining concept and techniques book free download

বাংলায় সি প্রোগ্রামিং শিখুন - ৫৫ মিনিটের ভিডিও টিউটরিয়াল

#include <stdio.h>

main()
{
    printf("Welcome To Our C Programming");
    return 0;
}

#include <stdio.h>

main()
{
    int number;
    float price;
    double total_price;

    //Give a number
    printf("Enter a number = ");
    scanf("%d", &number);

    printf("Enter Product Price = ");
    scanf("%f", &price);

    printf("Enter total price = ");
    scanf("%lf", &total_price);

    printf("Number = %d\t Price = %.2f\tTotal Price = %.3lf\n", number, price, total_price);


}

#include <stdio.h>

main()
{
    int number;
    printf("Please enter a number = ");
    scanf("%d", &number);

    //Positive or negative check
    if (number > 0){
        printf("Number is Positive\n");
    }else if(number == 0){
        printf("ooops... This is zero\n");
    }
    else{
        printf("Number is Negative\n");
    }


    // Even Odd Check
    if (number % 2 == 0){
        printf("Even number\n");
    }else{
        printf("Odd number\n");
    }

}

#include <stdio.h>

main()
{
    int number, i, sum = 0;

    printf("Please enter upto number = ");
    scanf("%d", &number);

    //Even
    printf("\nEven Numbers = ");
    for(i = 0; i <= number; i = i+2){
        printf("%d\t", i);
    }

    //Odd
    printf("\nOdd Numbers = ");
    for(i = 1; i <= number; i = i+2){
        printf("%d\t", i);
    }

    //3 = 1+2+3 = 6
    //Summing Using For Loop
    for(i = 1; i <= number; i++){
        sum = sum + i;
    }
    printf("\nTotal Summation = %d\n", sum);
}

#include <stdio.h>

main()
{
    int number, i = 0, sum = 0;

    printf("Please enter upto number = ");
    scanf("%d", &number);

    //Summing Using While Loop

    while(i <= number)
    {
        sum = sum + i;
        i++;
    }

    printf("\nTotal Summation = %d\n", sum);
}

#include <stdio.h>

main()
{
    int number, i;
    printf("Enter upto = ");
    scanf("%d", &number);

    int numbers[10000];


    for(i = 0; i < number; i++){
        printf("Element %d = ", i);
        scanf("%d", &numbers[i]);
    }

    printf("Full numbers are = ");
    for(i = 0; i < number; i++){
        printf("%d\n", numbers[i]);
    }

}

Python Code Example - Fibonacci Series in Python

Code Part

n_string = input('How many fibonacci number do want to get = ')
n = int(n_string)

fib0 = 0
fib1 = 1
fib = 0
i = 0

while i < n:
    fib = fib0+fib1
    print(fib)
    fib0 = fib1
    fib1 = fib
    i = i+1

Logic behind the Fibonacci Series

Code Explanation Line By Line

n_string = input('How many gibonacci number do want to get = ')
n = int(n_string)

fib0 = 0
fib1 = 1
fib = 0
i = 0

while i < n:
    fib = fib0+fib1
    print(fib)
    fib0 = fib1
    fib1 = fib
    i = i+1

Factorial Number generate C program Bangla Tutorials - সি প্রোগ্রাম দিয়ে কিভাবে ফ্যাক্টরিয়াল সংখ্যা বের করা যায় তার প্রোগ্রাম

কোনো সংখ্যার ফ্যাকটরিয়াল বের করার লজিক

    double factorial = 1;

    for(i = 1; i <= number; i++)
    {
        factorial = factorial * i;
    }

কোড পার্ট

#include<stdio.h>

main()
{
    int number;
    double factorial = 1;
    int i;

    printf("Enter Number = ");
    scanf("%d", &number);

    for(i = 1; i <= number; i++)
    {
        factorial = factorial * i;
    }

    printf("%d! = %.2lf\n", number, factorial);
}

আউটপুটঃ

ট্যাগসমূহঃ

Even, Odd Number generate C program Bangla Tutorials - সি প্রোগ্রাম দিয়ে কিভাবে জোড় বিজোড় সংখ্যা বের করা যায় তার প্রোগ্রাম

কোড পার্ট

#include <stdio.h>

main ()
{
    int number, i;

    printf("How many odd number you want to generate = ");
    scanf("%d", &number);

    for(i = 1; i <= number; i = i+2)
    {
        printf("%d\n", i);
    }

}

আউটপুটঃ

ট্যাগসমূহঃ

Summation, Subtraction, Multiplication, Division - সহজ সি প্রোগ্রাম ভিডিও টিউটোরিয়াল + কোড 

সি প্রোগ্রামিং এই এক্সাম্পলটির ভিডিও ইউটিউবে দেখুনঃ

কোড পার্ট

#include <stdio.h>

main()
{
    float number1, number2, result;

    printf("Enter number 1 = ");
    scanf("%f", &number1);

    printf("Enter number 2 = ");
    scanf("%f", &number2);

    result = number1+number2;
    printf("\n%f + %f = %.3f", number1, number2, result);

    result = number1-number2;
    printf("\n%f - %f = %.3f", number1, number2, result);

    result = number1*number2;
    printf("\n%f * %f = %.3f", number1, number2, result);

    result = number1/number2;
    printf("\n%f / %f = %.3f\n", number1, number2, result);
}

আউটপুটঃ

ট্যাগসমূহঃ

Summation, Subtraction, Multiplication, Division - সি প্রোগ্রাম ভিডিও টিউটোরিয়াল + কোড 

সরাসরি ভিডিও দেখুন ইউটিউবেঃ

কোড পার্ট

#include<stdio.h>
void main(){
    char option;
    float number1, number2, result;

    printf("Please enter number 1 = ");
    scanf("%f", &number1);

    printf("Please enter number 2 = ");
    scanf("%f", &number2);

    printf("\nPress + to make addition\nPress - for subtraction\nPress * for multiplication\nPress / For division = ");

    printf("\nEnter Option = ");
    scanf(" %c", &option);

    if(option == '+'){
        result = number1+number2;
        printf("\nSummation is = %f\n", result);
    }else if(option == '-'){
        result = number1-number2;
        printf("\nSubtraction is = %f\n", result);
    }else if(option == '*'){
        result = number1*number2;
        printf("\nMultiplication is = %f\n", result);
    }else if(option == '/'){
        result = number1/number2;
        printf("\nDivision is = %f\n", result);
    }else{
        printf("\nSorry !! You've pressed a wrong button, Try again\n");
    }

    return 0;
}

আউটপুটঃ

ট্যাগসমূহঃ

Matlab 13 Codes for Simulation and Modeling

1. Mid Square Random Number Generator Matlab code
2. Residue Method / Mixed Multiplicative Random Number Generator Matlab code
3. Additive Congruential Random Number Generator Matlab code
4. Multiplicative Congruential Random Number Generator Matlab code
5. Arithmetic congruential Method Random Number Generator Matlab code
6. Binomial Distribution Matlab code
7. Poison Distribution Matlab code
8. Chi Square Test for a given set of random numbers
9. Monte Carlo - Pure Pursuit Code For Matlab
10. Monte Carlo - Gambling Game Code For Matlab
11. Monte Carlo - Monte Carlo Integration Code For Matlab
12. Monte Carlo - Monte Carlo Pi Code For Matlab
13. Single Server Queuing System Matlab Code

Mid Square Method Random Number Generator Code

totalNumber = input('Number of Random Numbers want to generate : ');
choose = input('Enter the seed : ');
disp('Total Random Numbers are : ');
for i = 1:totalNumber
    random = choose ^ 2;
    random = random / 100; % Find the dividend
    random = rem(random, 10000); % Get the reminder;
    choose = random;
    fprintf('%.2f ', random);
end
fprintf('\n');

Output Mid square random Number generator:

Residue Method / Mixed Multiplicative Random Number Generator Code

clc;
a=input('Please Enter the value of a : ');
b=input('Please Enter the value of b : ');
m=input('Please Enter the value of m : ');
r=input('Please Enter the value of seed or r : ');

totalNumber = input('Number of Random Numbers want to generate : ');

for i = 1:totalNumber
    r = mod(a*r +b, m);
    fprintf('%4.0f ', r);
end
fprintf('\n');

Output Residue Method random Number generator:

Additive Congruential Method Random Number Generator Code

%% r(i+1) = (r(i) + b) mod m
clc;
b=input('Please Enter the value of b : ');
m=input('Please Enter the value of m : ');
r=input('Please Enter the value of seed or r : ');

totalNumber = input('Number of Random Numbers want to generate : ');

for i = 1:totalNumber
    r = mod(r +b, m);
    fprintf('%4.0f ', r);
end
fprintf('\n');

Output Additive Congruential Method  random Number generator:

Multiplicative Congruential Method Random Number Generator Code

%% r(i+1) = (ar(i)) mod m
clc;
a=input('Please Enter the value of a : ');
m=input('Please Enter the value of m : ');
r=input('Please Enter the value of seed or r : ');

totalNumber = input('Number of Random Numbers want to generate : ');

for i = 1:totalNumber
    r = mod(a*r, m);
    fprintf('%4.0f ', r);
end
fprintf('\n');

Output Multiplicative Congruential Method  random Number generator:

Arithmetic congruential Random Number Generator Code

%% r(i + 1) = (r(i) + r(i - 1)) percent m
clc;
N = input('Enter How Many Number You want to generate = ');
r0 = input('Enter First Random Number = ');
r1 = input('Enter Second Random Number = ');
m = input('Enter Maximum Range Of Random Number = ');

for i = 1:N
    r2 = r0 + r1;
    random = mod(r2, m);
    fprintf('%d ', random);
    r0 = r1;
    r1 = r2;
end

Output Arithmetic congruential random Number generator:

Binomial Distribution Code

clc;
trial = input ('Type the number of trial = ');
p = input('Type the value of p(<1.0) the probability of success = ');
variate = input('Type the number of variates to be generated = ');
fprintf('Here,You gave \t Number of trial = %d, Probability of success =  %d\nNumber of variates = %d\n',trial,p,variate);
x = 0;  n = trial;
for i = 1:n
    y = rand();
    if(y<p)
        x = x+1;
    end
    fprintf('x = %d\n', x);
end

Output Binomial Distribution :

Poison Distribution Code

clc;
lamda_value = input('Value of Lamda = ');
number_values = input('Number Of Values = ');


for i=1:number_values
   count = 0;
   p = 0;
   fact = 1;
   y = rand;
   while(y > p) 
        prob = (power(lamda_value,count)*power(2.718282,-lamda_value))/fact;  %Use the poison rules
        p = p + prob;
        count = count + 1;
        fact = fact * count;
   end
   fprintf('%.f\t',count);
   fprintf('\n');
end
fprintf('\n');

Output Poison Distribution :

Chi Square Test for a given set of random numbers Matlab Code

clc;
r = [36 91 51 02 54 06 58 06 58 02 54 01 48 97 43 22 83 25 79 95 42 87 73 17 02 42 95 38 79 29 65 09 55 97 39 83 31 77 17 67 62 03 49 90 37 13 17 58 11 51 92 33 78 21 66 09 54 49 90 35 84 26 74 22 62 12 90 36 83 32 75 31 94 34 87 40 07 58 05 56 22 58 77 71 10 73 23 57 13 36 89 22 68 02 44 99 27 81 26 85 ];
r_length = length(r);
fprintf('Total Students = %d\n', r_length);
for i = 1:r_length
    fprintf(' %d ', r(i));
end
fprintf('\n');

Acceptance_value = input('Enter acceptance value at that confidence level = ');
Expected = input('Enter Expected value = ');
E = Expected;
Range = Expected * Expected;
frequency = 0;
diff_square = 0;
i = 0;


while i < Range
    for j = 1:r_length
        if((r(j) > i) && (r(j) <= Expected))
            frequency = frequency + 1;
        end
    end
     O = frequency;
     diff_square = diff_square + (abs(O - E)) ^ 2;

    fprintf('Difference Square = %d\n', frequency);
    i = i + E;
    Expected = Expected + E;
    frequency = 0;
end
 chi_square = double(diff_square / E);
 fprintf('\n\nChi Square value is = %f\n', chi_square);
 
 if (Acceptance_value > chi_square)
     fprintf('\nThe random numbers follows the chi-square test. Because Chi Square value %.2f is less than Acceptance value %.2f\n', chi_square, Acceptance_value);
 else
     fprintf('\nThe random numbers follows the chi-square test. Because Chi Square value %.2f is greater than Acceptance value %.2f\n', chi_square, Acceptance_value);
 end
 
 fprintf('\n\n');

Output Chi Square Test Code :

Monte Carlo - Pure pursuit Bombers and Fighters Matlab Code

clc;
hold all;
xb=[100 110 120 129 140 149 158 168 179 188 198 209 219 226 234 240];
yb=[0 3 6 10 15 20 26 32 37 34 30 27 23 19 16 14];

xf = [];
yf = [];
xf(1)=0;
yf(1)=50;
s=20;
dist=0;

for i=1:15
   pause on;
    plot(xb(i),yb(i),'r*');
    title('Pure Pursuit Problem');
    pause(1);
    plot(xf(i),yf(i),'g*');
    y=yb(i)-yf(i);
    x=xb(i)-xf(i);
    
  dist=sqrt(y^2+x^2);

  if(dist<=12)
        fprintf('Bomber destroyed at  %d s',i);
        break;
  end
  
  xf(i+1)=xf(i)+s*((xb(i)-xf(i))/dist);
  yf(i+1)=yf(i)+s*((yb(i)-yf(i))/dist);
end

Output Monte Carlo - Pure Pursuit Bombers and fighters Code :

Gambling Game - Monte Carlo Matlab Code

clc;
n = input('How many time you want to play the game : ');

difference = 0;
tail = 0;
head = 0;

fprintf('\n---------------------------------------------------------------\n');
fprintf('Game No.\tSl No.\tRandom Number\tHeads\tTails\tDifference');
fprintf('\n---------------------------------------------------------------\n');

for i = 1:n
    j = 1;
    
    while 1
        toss = int8((10-0)*rand(1) + 0); %Generate a random number between 0 to 10%
        
        if (toss > 4)
            tail = tail + 1;
        else
            head = head + 1;
        end
        
        difference = abs((head - tail));
        fprintf('\n\t%d\t\t%d\t\t\t%d\t\t\t%d\t\t%d\t\t%d\t\t%d', i, j, toss, head, tail, difference);
        
        if(difference == 3) %The rules of the Gambling - If difference = 3 then Count the lose or won coins
            result = 8 - j;
            if(result < 0)
                fprintf('\n\t\tYou lose %d coins\n', abs(result));
            else
                fprintf('\n\t\tYou won %d coins\n', abs(result));
            end
            break;
        end
        j = j + 1;
    end % End while
end  % End For

Output Gambling Game - Monte Carlo Matlab Code :

Monte Carlo Integration - Monte Carlo Matlab Code

clc;
max_dot = input('How many times do you want to put dot : ');

inside = 0;
a = input('Enter a : ');
b = input('Enter b : ');
h = input('Enter h : ');

%Draw Plot
for x = 1: .01:10
    y = x^3;
    plot(x,y, 'y.');
    hold on;
end

for i = 1:max_dot
    x = a + rand * (b-a);
    y = rand * h;
    
    if y <= x^3
        inside = inside+1;
        plot(x,y, 'r.');
    else
        plot(x,y, 'g.');
    end
    
    hold on;
end
    
integration = (inside/max_dot) * ((b-a) * h);
fprintf('\nIntegration is : %f\n', integration);

Output Monte Carlo Integration - Monte Carlo Matlab Code :

Monte Carlo Pi  - Monte Carlo Matlab Code

clc;
maximum_dot = input('How many times want to put Dot : ');
inside = 0;
a = input('Enter a : ');
b = input('Enter b : ');
h = input('Enter h : ');

for i = 1:maximum_dot
    x = rand;
    y = rand;
    if sqrt(x^2 + y^2) <= 1
        plot(x,y,'r.');
        inside = inside + 1;
    else 
        plot(x,y,'b.');
        hold on;
    end
end

fprintf('Pi = %f\n', inside / maximum_dot * 4);

Output Monte Carlo Pi - Monte Carlo Matlab Code :

Code - Single Server Queuing System Matlab Code

total=0; busy=0;
%a=randi([0 8],1,8);
 a=[.4 1.6 2.1 3.8 4.0 5.6 5.8 7.2];
a_length=length(a);
arr=zeros(a_length);

%d=randi([0 9],1,5);
d=[2.4 3.1 3.3 4.9 8.6];
b=union(a,d);
l=length(b);
a_t=0;d_t=0;q=0;
axis([0 b(length(b))+1 0 length(a)]);


%Queue delay Time
figure(1);
title('Queue Delay Time');
for i=1:l-1
   a_m=ismember(a,b(i));
   d_m=ismember(d,b(i));
   if sum(a_m)>=1
       a_t=a_t+1;
   end
   if sum(d_m)>=1
       d_t=d_t+1;
   end
   dif=a_t-d_t;
   if dif>1
       rectangle('Position',[b(i) 0 b(i+1)-b(i) dif-1],'FaceColor',[1 0 0]);
       arr(dif)=arr(dif)+(b(i+1)-b(i));
   end
   if dif>0
       busy=busy+(b(i+1)-b(i));
   end
end


%Server Busy Time
b_t=0;
figure(2);
title('Server Busy Time');
axis([0 b(length(b))+1 0 length(a)]);
for i=1:l-1
   a_m=ismember(a,b(i));
   d_m=ismember(d,b(i));
   if sum(a_m)>=1
       b_t=b_t+1;
   end
   if sum(d_m)>=1
       b_t=b_t-1;
   end
   if b_t>0
       rectangle('Position',[b(i) 0 b(i+1)-b(i) 1],'FaceColor',[0 1 0]);
   end
end
for i=1:a_length
    total=total+arr(i)*(i-1);
end


disp(total);
disp(total/d(length(d)));


disp(busy);
disp(busy/d(length(d)));

Output Single Server Queuing System Matlab Code

Gambling Game Code Implementation in Matlab and C - Simulation

Problem:

Logic / Algorithm for Gambling game code:

Initialize head = 0, tail = 0, difference = 0
while (1)
toss the coin and make value from 0 to 10 using reminder
  if (toss > 4)
     tail = tail + 1
  else
     head = head + 1
  take difference = head - tail
  if (difference == 3)
    result = 8 - upto_time
    if (result < 0)
       lose
    else
       won
  else

Gambling game Code in Matlab

clc;
n = input('How many time you want to play the game : ');

difference = 0;
tail = 0;
head = 0;

fprintf('\n---------------------------------------------------------------\n');
fprintf('Game No.\tSl No.\tRandom Number\tHeads\tTails\tDifference');
fprintf('\n---------------------------------------------------------------\n');

for i = 1:n
    j = 1;
    
    while 1
        toss = int8((10-0)*rand(1) + 0); %Generate a random number between 0 to 10%
        
        if (toss > 4)
            tail = tail + 1;
        else
            head = head + 1;
        end
        
        difference = abs((head - tail));
        fprintf('\n\t%d\t\t%d\t\t\t%d\t\t\t%d\t\t%d\t\t%d\t\t%d', i, j, toss, head, tail, difference);
        
        if(difference == 3) %The rules of the Gambling - If difference = 3 then Count the lose or won coins
            result = 8 - j;
            if(result < 0)
                fprintf('\n\t\tYou lose %d coins\n', abs(result));
            else
                fprintf('\n\t\tYou won %d coins\n', abs(result));
            end
            break;
        end
        j = j + 1;
    end % End while
end  % End For

Output Gambling game Code in Matlab

Gambling game Code in C Language

#include<stdio.h>
#include<stdlib.h>

int main()
{
    int tail,head,i,n,j,difference=0,result,toss;
    printf("How many time you want to play ? ");
    scanf("%d",&n);
    printf("\n");

    tail = 0;
    head = 0;

    printf("Serial \t Head: \t Tail: \t Difference: \t\n");

    for(i=1; i<=n; i++)
    {
        j = 1;


        while(1)
        {
            // taking random number
            toss = rand() % 10;
            if(toss > 4)
            {
                // if(5,6,7,8,9) then tail will increase
                tail = tail + 1;
            }
            else
            {
                // if(,1,2,3,4) then head will increase
                head = head + 1;
            }
            // difference between head and tail
            difference = abs((head - tail));
            printf("%d\t%d\t%d\t%d\n",j,head,tail,difference);
            if(difference == 3)
            {
                // counting result suppose the difference match after 10 step then difference = 8-10=2
                result = 8 - j;
                if(result < 0)
                {
                    // if answer negative then you are in lose this is the amount of losed coin
                    printf("You lose %d Coin\n",abs(result));
                }
                else
                {
                    // if answer positive then you are in benefit this is the amount of benefited/win coin
                    printf("You win %d Coin\n",abs(result));
                }
                break;
            }
            // increament the count like serial number
            j++;
        }
    }
    return 0;
}

Gambling game Code in C Language Output

Tags:

Pure Pursuit Problem Code - Matlab and C code of Pure pursuit problem

What is Pure pursuit:

What is mainly Pure pursuit is:

1. Bomber Aircraft and a Fighter Aircraft are flying in the a horizontal plane. 
2. Fighter aircraft and bomber aircraft both are moving inside the rectangular range.  
3. The fighters and bombers have a velocity given, suppose s = 20 in our code.
4. When the distance of the Bomber and the Fighter is less than 12 units, it is assumed that the Bomber is shot down or destroyed.
5. The distance between this the bombers and fighter follows the distance rule - dist[t] = √( (yb[t]-yf[t])² + (xb[t]-yf[t])² ). 
6. In matlab the distance will be dist = sqrt(y^2+x^2). 
7. Look the pure pursuit problem code now and hope it'll clear to you.

Pure Pursuit Problem MatLab Code

clc;
hold all;
xb=[100 110 120 129 140 149 158 168 179 188 198 209 219 226 234 240];
yb=[0 3 6 10 15 20 26 32 37 34 30 27 23 19 16 14];

xf = [];
yf = [];
xf(1)=0;
yf(1)=50;
s=20;
dist=0;

for i=1:15
   pause on;
    plot(xb(i),yb(i),'r*');
    title('Pure Pursuit Problem');
    pause(1);
    plot(xf(i),yf(i),'g*');
    y=yb(i)-yf(i);
    x=xb(i)-xf(i);
    
  dist=sqrt(y^2+x^2);

  if(dist<=12)
        fprintf('Bomber destroyed at  %d s',i);
        break;
  end
  
  xf(i+1)=xf(i)+s*((xb(i)-xf(i))/dist);
  yf(i+1)=yf(i)+s*((yb(i)-yf(i))/dist);
end

Pure Pursuit Problem Output - Matlab:

Pure Pursuit problem in C language:

Code:

#include < stdio.h >
#include < math.h >
#include < stdlib.h >

void main()
  {
   float xf,yf, xb,yb,d,distance;
   int flag=0,
   vf=20,
   time=0;
   randomize();
   xf=rand()%1001;
   yf=rand()%1001;
   xb=rand()%1001;
   yb=rand()%1001;
   while(flag==0)
    {
 d= (yb-yf)*(yb-yf)+(xb-xf)*(xb-xf);
 distance=sqrt(d);
 printf("time=%d   xf=%5.2f  yf=%5.2f  xb=%5.2f  yb=%5.2f  distance=%5.2f\n\n",time,xf,yf,xb,yb,distance);
 if(distance >100)
   {
    printf("The bomber plain was shot down at %d second\n",time);
    flag=1;
   }
 else if(distance>900)
   {
     printf("The bomber plane escaped from sight at %d second\n", time);
     flag=1;
   }
 else
   {
     xf=xf+vf*(xb-xf)/distance;
     yf=yf+vf*(yb-yf)/distance;
     xb=rand()%1001;
     yb=rand()%1001;
     time=time+1;
   }
    }
    getch();
}

Output C code Pure Pursuit:

CASE 1:  BOMBER  IS SHOT DOWN BY FIGHTER

time=0   xf=688.00  yf=796.00  xb=366.00  yb=119.00  distance=749.68

time=1   xf=679.41  yf=777.94  xb=563.00  yb=771.00  distance=116.62

time=2   xf=659.45  yf=776.75  xb=419.00  yb=939.00  distance=290.07

time=3   xf=642.87  yf=787.94  xb=87.00  yb=931.00  distance=573.98

time=4   xf=623.50  yf=792.92  xb=960.00  yb=247.00  distance=641.30

time=5   xf=633.99  yf=775.90  xb=197.00  yb=203.00  distance=720.54

time=6   xf=621.86  yf=759.99  xb=799.00  yb=891.00  distance=220.32

time=7   xf=637.94  yf=771.89  xb=207.00  yb=310.00  distance=631.70

time=8   xf=624.30  yf=757.26  xb=193.00  yb=915.00  distance=459.24

time=9   xf=605.52  yf=764.13  xb=311.00  yb=991.00  distance=371.76

time=10   xf=589.67  yf=776.34  xb=816.00  yb=304.00  distance=523.76

time=11   xf=598.31  yf=758.30  xb=804.00  yb=587.00  distance=267.68

time=12   xf=613.68  yf=745.50  xb=122.00  yb=530.00  distance=536.84

time=13   xf=595.36  yf=737.47  xb=924.00  yb=705.00  distance=330.24

time=14   xf=615.27  yf=735.51  xb=100.00  yb=508.00  distance=563.26

time=15   xf=596.97  yf=727.43  xb=48.00  yb=39.00  distance=880.51

time=16   xf=584.50  yf=711.79  xb=990.00  yb=710.00  distance=405.50

time=17   xf=604.50  yf=711.70  xb=523.00  yb=629.00  distance=116.11

time=18   xf=590.46  yf=697.46  xb=894.00  yb=148.00  distance=627.72

time=19   xf=600.13  yf=679.95  xb=463.00  yb=537.00  distance=198.09

time=20   xf=586.29  yf=665.52  xb=308.00  yb=709.00  distance=281.67

time=21   xf=566.53  yf=668.61  xb=488.00  yb=605.00  distance=101.06

time=22   xf=550.99  yf=656.02  xb=107.00  yb=522.00  distance=463.77

time=23   xf=531.84  yf=650.24  xb=305.00  yb=777.00  distance=259.86

time=24   xf=514.38  yf=659.99  xb=250.00  yb=794.00  distance=296.40

time=25   xf=496.54  yf=669.04  xb=448.00  yb=950.00  distance=285.13

time=26   xf=493.14  yf=688.74  xb=252.00  yb=285.00  distance=470.27

time=27   xf=482.88  yf=671.57  xb=623.00  yb=646.00  distance=142.43

time=28   xf=502.56  yf=667.98  xb=640.00  yb=327.00  distance=367.64

time=29   xf=510.03  yf=649.43  xb=357.00  yb=811.00  distance=222.54

time=30   xf=496.28  yf=663.95  xb=623.00  yb=353.00  distance=335.78

time=31   xf=503.83  yf=645.43  xb=156.00  yb=371.00  distance=443.06

time=32   xf=488.13  yf=633.04  xb=533.00  yb=117.00  distance=517.99

time=33   xf=489.86  yf=613.12  xb=301.00  yb=627.00  distance=189.37

time=34   xf=469.91  yf=614.59  xb=143.00  yb=263.00  distance=480.09

time=35   xf=456.29  yf=599.94  xb=780.00  yb=654.00  distance=328.19

time=36   xf=476.02  yf=603.23  xb=928.00  yb=422.00  distance=486.96

time=37   xf=494.58  yf=595.79  xb=757.00  yb=44.00  distance=611.01

time=38   xf=503.17  yf=577.73  xb=734.00  yb=192.00  distance=449.52

time=39   xf=513.44  yf=560.57  xb=49.00  yb=949.00  distance=605.47

time=40   xf=498.10  yf=573.40  xb=515.00  yb=893.00  distance=320.05

time=41   xf=499.16  yf=593.37  xb=111.00  yb=765.00  distance=424.41

time=42   xf=480.87  yf=601.46  xb=463.00  yb=633.00  distance=36.25

The bomber plain was shot down at 42 second

CASE 2: BOMBER  ESCAPES FROM  THE SIGHT OF FIGHTER

time=0   xf=642.00  yf=902.00  xb=788.00  yb=709.00  distance=242.00

time=1   xf=654.07  yf=886.05  xb=585.00  yb=997.00  distance=130.69

time=2   xf=643.50  yf=903.03  xb=587.00  yb= 6.00  distance=898.81

time=3   xf=642.24  yf=883.07  xb=11.00  yb=162.00  distance=958.33

The bomber plane escaped from sight at 3 second

Links Where you can learn more on Pure pursuit problem:

1. Wikipedia - Pure Pursuit
2. Mathworks - Pure Pursuit

Tags:

Single Server Queuing System - MatLab and C code Implementation

What:

1. How many times a user need to wait in waiting  & Total waiting time
2. How many times user take in service time & Total service time
3. How many users are in the Queue & Total queue time
4. How many users has completed their work in that system yet.

Where Single server queuing system is used

1. In banking system, taking money from the bank. User needs to stand in a line and take money one by one
2. A Bus, plane ticketing system
3. Any system with line and who needs to put user in the queue

Real time example of a server queuing system - Patient Data:

Single Server Queuing Example of  patient and doctor

Single Server Queuing MatLab Code implementation without input (with figure):

total=0; busy=0;
%a=randi([0 8],1,8);
 a=[.4 1.6 2.1 3.8 4.0 5.6 5.8 7.2];
a_length=length(a);
arr=zeros(a_length);

%d=randi([0 9],1,5);
d=[2.4 3.1 3.3 4.9 8.6];
b=union(a,d);
l=length(b);
a_t=0;d_t=0;q=0;
axis([0 b(length(b))+1 0 length(a)]);


%Queue delay Time
figure(1);
title('Queue Delay Time');
for i=1:l-1
   a_m=ismember(a,b(i));
   d_m=ismember(d,b(i));
   if sum(a_m)>=1
       a_t=a_t+1;
   end
   if sum(d_m)>=1
       d_t=d_t+1;
   end
   dif=a_t-d_t;
   if dif>1
       rectangle('Position',[b(i) 0 b(i+1)-b(i) dif-1],'FaceColor',[0 .5 .5]);
       arr(dif)=arr(dif)+(b(i+1)-b(i));
   end
   if dif>0
       busy=busy+(b(i+1)-b(i));
   end
end


%Server Busy Time
b_t=0;
figure(2);
title('Server Busy Time');
axis([0 b(length(b))+1 0 length(a)]);
for i=1:l-1
   a_m=ismember(a,b(i));
   d_m=ismember(d,b(i));
   if sum(a_m)>=1
       b_t=b_t+1;
   end
   if sum(d_m)>=1
       b_t=b_t-1;
   end
   if b_t>0
       rectangle('Position',[b(i) 0 b(i+1)-b(i) 1],'FaceColor',[0 .5 .5]);
   end
end
for i=1:a_length
    total=total+arr(i)*(i-1);
end
disp(total);
disp(total/d(length(d)));
disp(busy);
disp(busy/d(length(d)));

Matlab Complaete Example - Single Server Queuing system

Single Server Queuing MatLab Code implementation with input:

N = input('Enter the Value of N: ');

% Variable Declaration
AT = [];
ST = [];
WT = [];
QL = [];
IDT = [];
CAT = [];
CDT = [];
CLK = 0;

% Initialization
AT(1) = 0;
for k = 2:N
AT(k) = input('Enter interarrival time : ');
end
for k = 1:N
ST(k) = input('Enter Service time : ');
end
CAT(1) = AT(1);
CDT(1) = ST(1);
for k = 1:N
WT(k) = 0;
IDT(k) = AT(1);
QL(k) = 0;
end

% Calculation
for i = 2:N
CAT(i) = CAT(i-1) + AT(i);
WT(i) = CDT(i-1) - CAT(i);
if WT(i) < 0
WT(i) = 0;
end
   
DIF = CAT(i) - CDT(i-1);
if DIF < 0
CDT(i) = CDT(i-1) + ST(i) ;
if i>2
if (CAT(i) < CDT(i-2) || QL(i-1) == 0)
QL(i) = QL(i-1) + 1;
else
QL(i) = QL(i-1);
end
else
QL(i) = QL(i-1) + 1;
end
elseif DIF > 0
CDT(i) = CAT(i) + ST(i) ;
if QL(i-1) > 0
QL(i) = QL(i-1) - 1;
end
else
QL(i) = QL(i-1);
end
if QL(i) == 0
IDT(i) = CAT(i) - CDT(i-1);
end
end

% Display Results
CAT
CDT
WT
IDT

QL

Enter the Value of N: 8
Enter interarrival time : 10
Enter interarrival time : 15
Enter interarrival time : 35
Enter interarrival time : 30
Enter interarrival time : 10
Enter interarrival time : 5
Enter interarrival time : 5
Enter Service time : 20
Enter Service time : 15
Enter Service time : 10
Enter Service time : 5
Enter Service time : 15
Enter Service time : 15
Enter Service time : 10
Enter Service time : 10

CAT =
     0    10    25    60    90   100   105   110

CDT =
    20    35    45    65   105   120   130   140

WT =
     0    10    10     0     0     5    15    20

IDT =
     0     0     0    15    25     0     0     0

QL =

     0     1     1     0     0     1     1     2

>> 

Single Server Queuing C Code implementation:

Single Server Queuing system in Java language:

import java.util.Random;
 
/**
*
* @author Hizbul Bahar
*/
public class SingleServer {
static int Queue_size = 32000;
static int next_event_type;
static int num_custs_delayed;
static int num_events=2;
static int num_in_q;
static int server_status;
static int num_delays_required;
static double area_num_in_q;
static double area_server_status;
static double sim_time;
static double time_last_event;
static double total_of_delays;
static double mean_interarrival;
static double mean_service;
 
static double[] time_arrival = new double[Queue_size];
static double[] time_next_event=new double[3];
 
static Random random = new Random(10000);
 
static void initialize()
{
sim_time = 0;
 
server_status = 0;
num_in_q = 0;
time_last_event = 0;
 
num_custs_delayed = 0;
total_of_delays = 0;
area_num_in_q = 0;
area_server_status = 0;
 
time_next_event[1] = sim_time + expon(mean_interarrival);
time_next_event[2] = 1.0e+30;
}
 
static void timing()
{
 
if (time_next_event[1] < time_next_event[2])
next_event_type = 1;
else
next_event_type = 2;
 
sim_time = time_next_event[next_event_type];
}
 
static void arrive()
{
double delay;
 
time_next_event[1] = sim_time + expon(mean_interarrival);
 
if (server_status == 1)
{
num_in_q++;
time_arrival[num_in_q] = sim_time;
}
 
else
{
delay = 0;
total_of_delays += delay;
 
num_custs_delayed++;
server_status = 1;
 
time_next_event[2] = sim_time + expon(mean_service);
}
}
 
static void depart()
{
if (num_in_q == 0)
{
server_status = 0;
time_next_event[2] = 1.0e+30;
}
else
{
num_in_q--;
 
num_custs_delayed++;
time_next_event[2] = sim_time + expon(mean_service);
 
for (int i = 1; i <= num_in_q; i++)
time_arrival[i] = time_arrival[i+1];
}
}
 
static void report()
{
System.out.println( "TOtal customer uses this server " + num_custs_delayed + "\n");
System.out.println( "Average delay in queue minutes  " + total_of_delays / num_custs_delayed + "\n");
System.out.println( "Average number in queue  " + area_num_in_q / sim_time + "\n");
System.out.println( "Server utilization  " + area_server_status / sim_time + "\n");
}
 
static void update_time_avg_stats()
{
double time_since_last_event;
 
time_since_last_event = sim_time - time_last_event;
time_last_event = sim_time;
 
area_num_in_q += num_in_q * time_since_last_event;
 
area_server_status += server_status * time_since_last_event;
}
 
static double expon(double  mean)
{
return -mean * Math.log(random.nextDouble());
}
 
public static void main(String[] args) {
 
timing();
update_time_avg_stats();
 
switch (next_event_type)
{
case 1: arrive();
break;
case 2: depart();
}
}
}

1. Wikipedia - Queuing Theory
2. A single Server Queuing system.pdf

Tags:

Question is:

Solution:

Laplace Transform Derivatives Theorem Proofs - First, Second, Third Order Proofs

Laplace Transform Derivatives First Order Proof:

We'll now find the first order derivatives theorem proof of Laplace transform here:

First Order Derivative theorem is = 

Proof Of First Order Derivative theorem Laplace Transform:

Laplace Transform Derivatives Second Order Proof:

We'll now find the second order derivatives theorem proof of Laplace transform here:

If F(t) is a twice differential function of t then

Proof :

I've written here the complete works , we can also define that we can get one equation from previous Laplace transform theorem. So, we've proved Laplace transform second order derivatives theorem.

Laplace Transform Derivatives Third Order Proof:

We'll now find the third order derivatives theorem proof of Laplace transform here:

If F(t) is a thrice differential function of t then

Proof  Using Second Order Derivative:

Proof  Third Order Full demonstration:

Now get third order derivative a more explained solution or proof

So, now we've proved all of the three basic Laplace Transform Derivatives theorem. Having any problem comment below.

Download Full Tutorial as PDF

Tags:

Laplace Transform Derivatives Theorem Proofs - First, Second, Third Order Proofs, Laplace Transform Derivatives Theorem Proofs First Order, Laplace Transform Derivatives Theorem Second Order derivatives Proof, Laplace Transform Derivatives Third Order Derivative Proof

Mathematics Book Differential Equation by Kedar Nath Ram Nath Download Link

Note:

Bash Scripting at a glance - Learn Bash scripting

name="Akash"
echo "Welcome $name"

Welcome Akash

#Simple If else in Bash

echo "Please Enter your age : "
read age

if [ $age < 10 ]; 
    then echo "You are really small.."
else
    echo "You are growing.."
fi

# Printing 1 to 100 using for loop

for((i=1;i<=100;i++))
do
    echo $i
done

for i in `seq 1 100`
do
    echo $i
done

# Printing 1 to 100 even numbers using for loop
for((i=0;i<=100;i=i+2))
do
    echo $i
done

# Printing 1 to 100 odd numbers using for loop
for((i=1;i<=100;i=i+2))
do
    echo $i
done

Multiplication program upto 10 using for double for loop Bash scripting [Bash Programming Language]

0 to 100 even numbers using Bash Scripting

# Multiplication Program 1 to 10 in Bash

for((i=1;i<=10;i++))
do
    for((j=1;j<=10;j++))
    do
        echo $i "X" $j "=" $(($i*$j))
    done
    echo
done

Simple Output with Code

Live Link - Live Demo

echo "Please Enter your File Name : "
read file_name

if [ -f $file_name ];
    then  clear
    rm $file_name
    echo "File $file_name is deleted successfully"
    
else
    clear
    touch $file_name
    echo "No file found for your search $file_name"
fi

echo "All files are Now : "
ls

echo "How many numbers do you want to add in the array : "
read num

#Take value in the array and put it in a sum variable
arr=()  #Empty Array
sum=0

for ((i=0;i<num;i++));
do
    echo "Array[$i] = "
    read arr[$i]
    sum = $((sum+arr[$i]))
done

echo "Summation Is : $sum"

##
#   Delete a value from array
##

echo "Please enter a value to delete from the array = "
read delete_value;

for ((i=0;i<num;i++));
do
    if [ $((arr[$i])) = $delete_value ]; then 
        unset arr[$i]   #Delete using unset
    fi
done


#Show final array
echo  "Array is = ${arr[@]}"

n=123456
rev=0

while [ $n -gt 0 ]
do
    reminder=$(( $n % 10 ))
    rev=$(( rev*10 + reminder  ))
    n=$(( $n / 10 ))
done

echo $rev

n=123456
echo $n | rev

num=1289
i=2

while [ $i -lt $num ]
do
  if [ $(( num%i )) = 0 ];
  then
      echo "$num is not a prime number"
      echo "It is divisible by $i"
      exit
  fi
  $i=$(( i+1 ))
done

echo "$num is a prime number "

Windows Every Short Command by Windows cmd

How to see the every command list in windows - windows command:

help

cd

dir

mkdir dir_name

rmdir dir_name

type NUL > test.txt
type NUL > another.c

rename test.txt demo.txt

cls

rename *.rtf *.txt

Tutorial on:

FCFS - First Come First Serve

Algorithm of FCFS:

• Which process comes first to CPU, the process will serve then.
• It is mainly a non preemptive scheduling algorithm. That means, no stop of a process when that's running.
• FCFS implements the First In First Out (FIFO) algorithm. That means which come first to the queue,  that will out first from the queue.

C Code of FCFS / First Come First Serve C implementation:

#include<stdio.h>

int main()
{
    int n, burst_time[100], waiting_time[100], turn_around_time[100];
    int average_wating_time=0,average_turn_around_time=0;

    printf("Total Processes : ");
    scanf("%d",&n);

    int i, j;
    printf("\nEnter Burst Time : \n");
    for(i=0; i<n; i++)
    {
        printf("P%d = ",i+1);
        scanf("%d",&burst_time[i]);
    }

    waiting_time[0]=0;
    for(i=1; i<n; i++)
    {
        waiting_time[i]=0;
        for(j=0; j<i; j++){
            waiting_time[i] = waiting_time[i] + burst_time[j];
        }

    }


    printf("\n--------------------------------------------------------------");
    printf("\nProcess\t\tBurst Time\tWaiting Time\tTurn around Time");
    printf("\n--------------------------------------------------------------\n");

    for(i=0; i<n; i++)
    {
        turn_around_time[i] = burst_time[i] + waiting_time[i];
        average_wating_time += waiting_time[i];
        average_turn_around_time += turn_around_time[i];
        printf("\nProcess = %d\t\t%d\t\t%d\t\t%d", i+1, burst_time[i], waiting_time[i], turn_around_time[i]);
    }

    average_wating_time = average_wating_time/n;
    average_turn_around_time = average_turn_around_time / n;


    printf("\n\nAverage Waiting Time:%d", average_wating_time);
    printf("\nAverage Turnaround Time:%d\n\n", average_turn_around_time);

    return 0;
}

Output of FCFS:

Explanation of First Come First Serve Algorithm (FCFS)

    for(i=0; i<n; i++)
    {
        printf("P%d = ",i+1);
        scanf("%d",&burst_time[i]);
    }

    waiting_time[0]=0;

    for(i=1; i<n; i++)
    {
        waiting_time[i]=0;
        for(j=0; j<i; j++){
            waiting_time[i] = waiting_time[i] + burst_time[j];
        }

    }

    for(i=0; i<n; i++)
    {
        turn_around_time[i] = burst_time[i] + waiting_time[i];
        average_wating_time += waiting_time[i];
        average_turn_around_time += turn_around_time[i];
        printf("\nProcess = %d\t\t%d\t\t%d\t\t%d", i+1, burst_time[i], waiting_time[i], turn_around_time[i]);
    }

    average_wating_time = average_wating_time/n;
    average_turn_around_time = average_turn_around_time / n;

Advantages of FCFS or First Come First Serve Algorithm:

• FCFS is suitable for batch system, basically.
• FCFS is easier to implement.

Disadvantages of FCFS or First Come First Serve Algorithm:

Note:Having any problems in this First Come First Serve algorithm or code, please comment here and don't hesitate.