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

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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

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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

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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

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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:

Question:

Solution: