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miskamm [114]
2 years ago
6

Given two objects represented by the tuples (22, 1, 42, 10) and (20, 0, 36, 8):

Computers and Technology
1 answer:
Rudik [331]2 years ago
4 0

Answer:

Given,

P = (22, 1, 42, 10)

Q = (20, 0, 36, 8)

a. Formula for Euclidean Distance :

distance = ((p1-q1)^2 + (p2-q2)^2 + ... + (pn-qn)^2)^(1/2)

Now,

distance = ( (22-20)^2 + (1-0)^2 + (42 - 36)^2 + (10-8)^2) ) ^(1/2)

=( (2)^2 + (1)^2 + (6)^2 + (2)^2 ) ) ^(1/2)

=(4+1+36+4)^(1/2)

=45^(1/2)

Distance = 6.7082

b.Manhattan distance :

d = |x1 - x2| + |y1 - y2|

d = |22- 20| + |1 - 0|

d = |2| + |1|

Explanation:

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Answer:

Check the explanation

Explanation:

Linear search in JAVA:-

import java.util.Scanner;

class linearsearch

{

  public static void main(String args[])

  {

     int count, number, item, arr[];

     

     Scanner console = new Scanner(System.in);

     System.out.println("Enter numbers:");

     number = console.nextInt();

   

     arr = new int[number];

     System.out.println("Enter " + number + " ");

     

     for (count = 0; count < number; count++)

       arr[count] = console.nextInt();

     System.out.println("Enter search value:");

     item = console.nextInt();

     for (count = 0; count < number; count++)

     {

        if (arr[count] == item)

        {

          System.out.println(item+" present at "+(count+1));

         

          break;

        }

     }

     if (count == number)

       System.out.println(item + " doesn't found in array.");

  }

}

Kindly check the first attached image below for the code output.

Selection Sort in JAVA:-

public class selectionsort {

   public static void selectionsort(int[] array){

       for (int i = 0; i < array.length - 1; i++)

       {

           int ind = i;

           for (int j = i + 1; j < array.length; j++){

               if (array[j] < array[ind]){

                   ind = j;

               }

           }

           int smaller_number = array[ind];  

           array[ind] = array[i];

           array[i] = smaller_number;

       }

   }

     

   public static void main(String a[]){

       int[] arr = {9,94,4,2,43,18,32,12};

       System.out.println("Before Selection Sort");

       for(int i:arr){

           System.out.print(i+" ");

       }

       System.out.println();

         

       selectionsort(arr);

       

       System.out.println("After Selection Sort");

       for(int i:arr){

           System.out.print(i+" ");

       }

   }

}  

Kindly check the second attached image below for the code output.

Bubble Sort in JAVA:-

public class bubblesort {

   static void bubblesort(int[] array) {

       int num = array.length;

       int temp = 0;

        for(int i=0; i < num; i++){

                for(int j=1; j < (num-i); j++){

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                }

        }

   }

   public static void main(String[] args) {

               int arr1[] ={3333,60,25,32,55,620,85};

               

               System.out.println("Before Bubble Sort");

               for(int i=0; i < arr1.length; i++){

                       System.out.print(arr1[i] + " ");

               }

               System.out.println();

                 

               bubblesort(arr1);

               

               System.out.println("After Bubble Sort");

               for(int i=0; i < arr1.length; i++){

                       System.out.print(arr1[i] + " ");

               }

 

       }

}  

Kindly check the third attached image below for the code output.

Binary search in JAVA:-

public class binarysearch {

  public int binarySearch(int[] array, int x) {

     return binarySearch(array, x, 0, array.length - 1);

  }

  private int binarySearch(int[ ] arr, int x,

        int lw, int hg) {

     if (lw > hg) return -1;

     int middle = (lw + hg)/2;

     if (arr[middle] == x) return middle;

     else if (arr[middle] < x)

        return binarySearch(arr, x, middle+1, hg);

     else

        return binarySearch(arr, x, lw, middle-1);

  }

  public static void main(String[] args) {

     binarysearch obj = new binarysearch();

     int[] ar =

       { 22, 18,12,14,36,59,74,98,41,23,

        34,50,45,49,31,53,74,56,57,80,

        61,68,37,12,58,79,904,56,99};

     for (int i = 0; i < ar.length; i++)

        System.out.print(obj.binarySearch(ar,

           ar[i]) + " ");

     System.out.println();

     System.out.print(obj.binarySearch(ar,19) +" ");

     System.out.print(obj.binarySearch(ar,25)+" ");

     System.out.print(obj.binarySearch(ar,82)+" ");

     System.out.print(obj.binarySearch(ar,19)+" ");

     System.out.println();

  }

}

Kindly check the fourth attached image below for the code output

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In the Stop-and-Wait flow-control protocol, what best describes the sender’s (S) and receiver’s (R) respective window sizes?
kolbaska11 [484]

Answer:

The answer is "For the stop and wait the value of S and R is equal to 1".

Explanation:

  • As we know that, the SR protocol is also known as the automatic repeat request (ARQ), this process allows the sender to sends a series of frames with window size, without waiting for the particular ACK of the recipient including with Go-Back-N ARQ.  
  • This process is  mainly used in the data link layer, which uses the sliding window method for the reliable provisioning of data frames, that's why for the SR protocol the value of S =R and S> 1.
3 0
3 years ago
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