In C: Find a string within a string Given two strings S1 & S2, search for an occurrence of the second string within a first string. Note: Do not use system library for the implementation. Input: Code Zinger University Zinger where, First line represents string S1. Second line represents string S2. Output: 5 Here 'Zinger' word starts at 5th index within 'Code Zinger University’. Assume that, The length of strings S1 & S2 are within the range [1 to 10000]. Character comparisons will be case-sensitive.

Develop a test strategy for testing the entire application chosen in unit 3 for analysis and design. Keep in mind that testing that involves users should minimize their time commitment while obtaining essential information from their involvement. Specifically define roles, responsibilities, timing, and test strategy for each level of testing.

• Consider a simple application-level protocol built on top of UDP that allows a client to retrieve a file from a remote server residing at a well-known address. The client first sends a request with a file name, and the server responds with a sequence of data packets containing different parts of the requested file. To ensure reliability and sequenced delivery, client and server use a stop-and-wait protocol. Ignoring the obvious performance issue, do you see a problem with this protocol? Think carefully about the possibility of processes crashing.

Write a linear-search Java method to test if a String array is already sorted alphabetically. The prototype should be static boolean ifsorted(String [] s)

{

}

Java programming extra credit.

The following program will be used to test your knowledge on Implementing with tester classes. Included is the Details class. Your assignment is to create a class named DetailsTester using info from the two classes(Procedure class is only included to know the details of the procedure, which includes desc, date, and cost) Patient class is included as well.That will properly implement the class.

package doctorOffice;

import java.util.ArrayList;
import java.time.LocalDate;

public class Details
{
//Instance variables
private Patient patient;
private double height;
private double weight;
private ArrayList procedures = new ArrayList();
private LocalDate lastVisit;

Details(Patient patient,
double height,
double weight)
{
this.patient = patient;
this.height = height;
this.weight = weight;
}

//Getters--------------------------------------
public Patient getPatient()
{
return patient;
}

public double getHeight()
{
return height;
}

public double getWeight()
{
return weight;
}

public ArrayList getProcedures()
{
return procedures;
}

public LocalDate getLastVisit()
{
return lastVisit;
}

//Setters--------------------------------------
public void setHeight(double height)
{
this.height = height;
}

public void setWeight(double weight)
{
this.weight = weight;
}

public void setLastVisit(LocalDate lastVisit)
{
this.lastVisit = lastVisit;
}

//Methods--------------------------------------
public boolean addProcedure(Procedure toAdd)
{
//LOGIC HERE
return true;
}
}

*******

package doctorOffice;

import java.time.LocalDate;
import java.time.format.DateTimeFormatter;

//Procedure class holds information about any procedures performed on the patients
public class Procedure
{
   private String desc;
   private double cost;
   private LocalDate date;
  
   public Procedure(String desc, double cost, String date) throws Exception
   {
       this.desc = desc;
       this.cost = cost;
       this.date = LocalDate.parse(date, DateTimeFormatter.ofPattern("MM/dd/yyyy"));
   }
  
   //getters
   public String getDesc()
   {
       return this.desc;
   }
  
   public double getCost()
   {
       return this.cost;
   }
  
   public LocalDate getDate()
   {
       return this.date;
   }

   //toString override
   public String toString()
   {
       String output = "Description: " + desc;
       output += "\n\nCost: " + cost;
       output += "\n\nDate: " + date.toString();
       return output;
   }
}

1. Assuming standard 1500 byte Ethernet max payloads: how many IPv4 fragments will be needed to transfer 2000 bytes of user data with a single UDP send? And, how do the 2000 bytes get split over the frags?
2. Despite its conceptual elegance, RPC (Remote Procedure Call) has a few problems. Discuss any 3 of those in brief.
3. Why is timestamping needed in real-time applications? This is in the context of Real-time Transport Protocol (RTP).
4. Why does UDP exist? Would it not have been enough to just let user processes send raw IP packets?
5. Explain how QUIC eliminates a couple of RTTs usually needed at the start of a secure web connection.

I'm looking for a program, written in vPython, that simulates the movement of a spherical mass, suspended from a ceiling, tied to a spring. We are supposed to take the gravitaional force into account but are allowed to ignore the effects of air-resistance (for now).

I am a newbie at vPython so comments within the program to explain what certain parts of the code do would be appreciated!

An increasing-order integer array may contain duplicate elements. Write a Java method that does a binary search through the array for finding a specific target and return an array of two integer indices that specifies the close interval of indices at which the target is found or null if the target is not present. Use this test driver to test

public class MultBS

{

int [] multBinarySearch(int [] x, int target)

{

// your codes go here

}

public static void main(String [] args)

{ int [] x = new int[]{2,3,3,4,4,4,5,6,7,7,7};

int [] range = multBinarySearch(x, 3);

System.out.println("Target was found from position " + range[0] + " to " + range[1]);

}

}

test your program to make sure you see

the printout is

"Target was found from position 1 to 2"

I have the following code:

//Set.java

import java.util.ArrayList;

public class Set<T> {
    //data fields
    private ArrayList<T> myList;
    // constructors
    Set(){
        myList = new ArrayList<T>();
    }
    // other methods
    public void add(T item){
        if(!membership(item))
            myList.add(item);
    }

    public void remove(T item){
        if(membership(item))
            myList.remove(item);
    }

    public Boolean membership(T item){
        for (T t : myList)
            if (t.equals(item))
                return true;
        return false;
    }

    public String toString(){
        StringBuilder str = new StringBuilder("[ ");

        for(int i = 0; i < myList.size() - 1; i++)
            str.append(myList.get(i).toString()).append(", ");

        if(myList.size() > 0)
            str.append(myList.get(myList.size() - 1).toString());

        str.append(" ]");

        return str.toString();
    }

}

(20 points) Fill the following table with the big-O running times of each of the methods implemented, as well as the running times for the same methods if they were implemented with LinkedList instead. (No need to explain here, only the big-O expression.)

(20 points) Explain each of the eight running times in the table applying the rules of counting seen in class, and for the method calls, if 4 we have covered them already, directly refer to the running times of those methods. (Partial credit if you do not explain all eight. No points for ambiguous explanations. For instance, if you implement add in the class Set using addFirst in the class ArrayList, then you should say something like the following. “The add method of Set is implemented with just one method call to addFirst of ArrayList. addFirst of ArrayList is in O(n) in the worst case. Hence, the same bound on the running time applies to add”. )

Suppose a function receives a pointer as an argument. Explain how this function is declared within its calling function. In particular, explain how the data type of the pointer argument is represented. C++

What are the application and uses of Telemedicine?

please give complete code in python using def function thank you!

Validate Credit Card Numbers

Credit card numbers can be quickly validated by the Luhn checksum algorithm.

1. Start at the rightmost digit save one (the last being the checksum digit). Moving left, double the value of every second digit. If the doubled value is greater than nine, then subtract 9 from the doubled value to renormalize within the range 0–9.
2. Sum all the digits including the checksum digit.
3. If this total modulo 10 is zero, the number is valid.

For instance, Ned Flander's credit card number is 8525.4941.2525.4158.

1. Here are the digits with checksum underlined:

   8525494125254158_

2. Here are the digits doubled back:

   16 5 4 5 8 9 8 1 4 5 4 5 8 1 10 8_

   Here are the two-digit results subtracted nine:

   7 5 4 5 8 9 8 1 4 5 4 5 8 1 1 8_

3. Here is the sum:

   7+5+4+5+8+9+8+1+4+5+4+5+8+1+1+8=83

4. Here is the modulo 10 result: 83%10=3 and this is not a valid credit card number.

Compose a function luhn which correctly validates a given candidate 16-digit credit-card number (as a numeric input in int).

hi! I have this code. when I run it, it works but in the print(the number we are searching for is ) it prints the number twice. ex. for 5 it prints 55 etc. apart from this, it works

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

int main(){
int n,i;
printf("Give me the size of the table : \n");
scanf("%d", &n);
int A[n],x,y;

for (i=0;i<n;i++)
A[i]=rand() % n;

for (i=0;i<n;i++)
printf("%d\n", A[i]);

srand(time(NULL));
y=rand()% n ;

printf("The number we are searching for is %d", y);

for (i=0;i<n;i++)
{
    if (A[i]==y)
    {
      printf("%d is present at location %d.\n", y,i+1);
      break;
    }
}
if (i == n)
    printf("%d isn't present in the array.\n", y);

return 0;
}

2. Consider the following variable declarations

hours = 40;

status= True;

overtime = False

What is the value of each of the following Boolean expressions?       [4 marks]

1. status and !status
2. status or  !status
3. overtime == (hours >= 40)
4. False == hours > 40

Rewrite the following Matlab code so it does the same thing but with complete new variable names and structure.

function stump = stumpGenerator(dataX, dataY, Dt)
intervals = 100;

rangex1 = max(dataX(:,1)) - min(dataX(:,1));
rangex2 = max(dataX(:,2)) - min(dataX(:,2));

width = (rangex1/intervals);
height = (rangex2/intervals);

starterx1 = min(dataX(:,1)) - (width/2);
starterx2 = min(dataX(:,2)) - (height/2);

currepsilon = inf;

stump = [0,0,0,1,0,0];

for i = 1:(intervals + 1)
  
horzRightError = sum(Dt(find(((dataX(:,1) - starterx1) .* dataY) < 0)));
horzLeftError = sum(Dt(find(((dataX(:,1) - starterx1) .* dataY) > 0)));
vertUpError = sum(Dt(find(((dataX(:,2) - starterx2) .* dataY) < 0)));
vertDownError = sum(Dt(find(((dataX(:,2) - starterx2) .* dataY) > 0)));
  
if (horzRightError <= horzLeftError)
horzError = horzRightError;
else
horzError = -horzLeftError;
end
  
if (vertUpError <= vertDownError)
vertError = vertUpError;
else
vertError = -vertDownError;
end
  
  
if (abs(horzError) <= abs(vertError))
if (currepsilon > abs(horzError))
currepsilon = abs(horzError);
stump(1) = 1;
stump(2) = 0;
stump(3) = starterx1;
stump(4) = horzError/(abs(horzError));
stump(5) = (log((1 - currepsilon)/currepsilon))/2;
stump(6) = currepsilon;
else
% do nothing, we already have the best stump
end
  
else
if (currepsilon > abs(vertError))
currepsilon = abs(vertError);
stump(1) = 0;
stump(2) = 1;
stump(3) = starterx2;
stump(4) = vertError/(abs(vertError));
stump(5) = (log((1/currepsilon) - 1))/2;
stump(6) = currepsilon;
else
% do nothing, we already have the best stump
end
end

starterx1 = starterx1 + width;
starterx2 = starterx2 + height;
end

end