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kobusy [5.1K]
3 years ago
7

The most complex part of ssl is the ___________ protocol.

Computers and Technology
2 answers:
Rina8888 [55]3 years ago
7 0
The most complex part of ssl is the handshake protocol. The handshake protocol is part of the Transport Layer Security (TLS) which is responsible for the authentication and ket exchange to have secure sessions. The handshake protocol manages cipher suite negotiation. 
Softa [21]3 years ago
5 0

The most complex part of ssl is the handshake protocol.

When Handshaking Takes Place. In order to establish a connection between a computer and a device like a modem, printer, or server, the handshake process begins the connection by telling both devices how to communicate with each other. A classic example is the noises made when two dial-up modems connect to each other.

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Given n ropes of different lengths, we need to connect these ropes into one rope. We can connect only 2 ropes at a time. The cos
Lelu [443]

Answer:

Given n ropes of different length, the task is to connect all the ropes into one. We need to connect these ropes in minimum cost. The cost of connecting two ropes is the sum of their lengths.

Eg: Given 4 ropes of length { 3, 7, 9, 4 }

We can connect the ropes in the following way:

1) First, connect the ropes of length 3 and 4. Cost of connecting these ropes is 3 + 4 = 7. Now we have ropes of length { 7, 7, 9 }

2) Then, connect the ropes with length 7 and 7. Now the cost of connecting these ropes is 7 + 7 = 14. Now we have ropes of length { 9, 14 }

3) Finally, connect the two last ropes with cost 9 + 14 = 23.

So, Total Cost of connecting these ropes in the above order is 7 + 14 + 23 = 44.

Approach

Given an array rope[] of length n, where rope[i] is the length of ith rope.

First of all, build minHeap from the given array of rope length, because we want to extract ropes with the minimum length of all in every iteration.

In each iteration:

1) Extract two ropes with minimum length from the minHeap.

2) add both the ropes (cost).

3) add the cost to the result and push cost again in the minHeap.

The main reason for selecting ropes with minimum length in each iteration is: The value that is picked first will be included in the final result more than once. Since we want the minimum cost of connecting ropes, we will pick long length ropes at a later stage to minimize its impact on our final cost.

Time Complexity

The complexity of insertion in minHeap is O(logn). Since we are building heap for n ropes, the overall complexity of building minHeap is O(nlogn).

Then in the iteration, we are extracting minimum value from minHeap and pushing the cost of connecting two ropes back to the minHeap. Extract/Insert operation from minHeap costs O(logn).

So, the Overall time complexity of the problem is O(nlogn).

Space Complexity

We are using space for storing minHeap of the ropes which is equal to the number of ropes we have. So Space complexity of the problem is O(n).

Explanation:

Code is:

int main() {

int t;

scanf("%d", &t);

while(t--) {

    int count, first, second;

    int result = 0;

    scanf("%d", &count);

    priority_queue< int, vector<int>, greater<int> > pq;

    for(int i=0; i<count; i++){

        int ropeLen;

        scanf("%d", &ropeLen);

        pq.push(ropeLen);

    }

    while(pq.size() > 1) {

        first= pq.top(); pq.pop();

        second = pq.top(); pq.pop();

        result += (first+second);

        pq.push(first+second);

    }

    cout<<result<<endl;

}

}

4 0
3 years ago
Write a program that converts or calculates values. Use the following guidelines to write your program:
Setler [38]

Answer:

In Section 2.2 “Expressing Units”, we showed some examples of how to ... with other units of the same type to get a numerical value that is easier to comprehend. ... The rules of algebra say that you can change (i.e., multiply or divide or add or ... We see that one-half of a yard equals 3/2, or one and a half, feet—something Explanation:

8 0
3 years ago
Where can i make a 3d animation for free ( pls don't trick me)
Vinvika [58]

Answer:

I don't know if it would work for what you need but you could try scratch. (sorry if this isn't what you needed.)

4 0
3 years ago
A. What is MS-Word ? Write its uses.
solmaris [256]

Answer:

MS word is Microsoft word and you can use MS Word in daily life and business to create professional-looking documents such as resume, letters, applications, forms, brochures, templates, business cards, calendars, reports, eBooks, and newsletters in speed with high quality.

Explanation:

6 0
3 years ago
Read 2 more answers
Build three classes that conform to the following interfaces. Use arrays in creating your classes (e.g., do not use the built-in
daser333 [38]

Explanation:

public class ArrayList {

private Object[] array = new Object[1];

/**

 * Places new element at location

 * @param c, element to be inserted

 * @param i, location it is to be placed

 */

public void insert(char c, int i) {

 if(i>this.size()){

  System.out.println("Index "+i +" outside of list size; max: " +this.size());

  System.exit(0);

 }

 int count = 0;

 try {

  Object[] other = new Object[this.array.length+1];

  switch(i){

     case 0:{

      other[0] = c;

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

       other[j] = this.array[j-1];

       

      this.array = other;

      break;

     }

     default: {

      for(int j=0;j<i;j++)

       other[j] = this.array[j];

      other[i] = c;

      for(int k=i+1;k<this.array.length;k++)

       other[k] = this.array[k-1];

      this.array = other;

      break;

     }

     }

 }

 catch(ArrayIndexOutOfBoundsException exception) {

     largerArray();

     if(++count == 2){

      System.out.println("Something went wrong.");

      System.exit(0);

     }  

 }

}

 

/**

 * Places new element at location

 * @param object, element to be inserted

 * @param index, location it is to be placed

 */

void insert(Object object, int index){

 if(index>this.size()){

  System.out.println("Index "+index +" outside of list size; max: " +this.size());

  System.exit(0);

 }

 int count = 0;

 try {

  Object[] other = new Object[this.array.length+1];

  switch(index){

     case 0:{

      other[0] = object;

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

       other[j] = this.array[j-1];

       

      this.array = other;

      break;

     }

     default: {

      for(int j=0;j<index;j++)

       other[j] = this.array[j];

      other[index] = object;

      for(int k=index+1;k<this.array.length;k++)

       other[k] = this.array[k];

      this.array = other;

      break;

     }

     }

 }

 catch(ArrayIndexOutOfBoundsException exception) {

     largerArray();

     if(++count == 2){

      System.out.println("Something went wrong.");

      System.exit(0);

     }  

 }

}

 

/**

 * Removes element at index

 * @param index, location to remove

 * @return temp, object removed

 */

Object remove(int index){

 if(index>this.size()){

  System.out.println("Index "+index +" outside of list size; max: " +this.size());

  System.exit(0);

 }

 Object temp = this.array[index];

 int count = 0;

 try {

  Object[] other = new Object[this.array.length-1];

  switch(index){

     case 0:{

      other[0] = this.array[1];

      for(int j=1;j<this.size();j++)

       other[j] = this.array[j+1];

       

      this.array = other;

      break;

     }

     default: {

      for(int j=0;j<index;j++)

       other[j] = this.array[j];

      for(int k=index;k<this.size();k++)

       other[k] = this.array[k+1];

      this.array = other;

      break;

     }

     }

 }

 catch(ArrayIndexOutOfBoundsException exception) {

     largerArray();

     if(++count == 2){

      System.out.println("Something went wrong.");

      System.exit(0);

     }  

 }

 return temp;

}

 

/**

 * Copies the array to a new array twice the size.

 */

public void largerArray(){

 Object[] other = new Object[this.array.length*2];

 System.arraycopy(this.array, 0, other, 0, this.array.length);

 this.array = other;

}

 

/**

 * @return true if there are null elements

 */

public boolean isEmpty() {

 boolean bool = true;

 for (Object element : this.array) {

  if(element!=null){

   bool = false;

   break;

  }

  else{

   bool = true;

   break;

 }

 }

 return bool;

}

 

/**

 * Determines the real length of the array

 * @return length of non-null elements

 */

public int size(){

 int count = 0;

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

  if(this.array[i]==null)

   continue;

  else

   count++;

 }

 return count;

}

 

/**

 * Converts array to string.

 * @return the Array as a string

 */

public String toString(){

 String string = "";

 for(int i=0;i<this.size();i++){

  if(i<this.size()-1)

   string += this.array[i]+", ";

  else

   string += this.array[i];

 }

 return string;

}

 

/**

 * Similar to get, imput the character, retrieve index.

 * @param object, what we're looking for

 * @return index

 */

int indexOf(Object object){

 int index = -1;

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

  if(this.array[i]==(object))

   index = i;

  else continue;

 }

 return index;

}

 

/**

 * @param object, Arraylist tyoe

 * @return true, if the elements of the arrays are equivalent.

 */

public boolean equals(ArrayList object){

 boolean bool = false;

 int thisLength = this.array.length;

 int objectLength = object.array.length;

 if(thisLength > objectLength){

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

   if(this.array[i] == object.array[i])

    bool = true;

   else{

    bool = false;

    break;

   }

 }

 else{

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

   if(this.array[i] == object.array[i])

    bool = true;

   else{

    bool = false;

    break;

   }

 }

 return bool;

}

 

/**

 * Getter to check what is at an index

 * @param index, location

 * @return the character at this position

 */

Object get(int index){

 return this.array[index];

}

}

6 0
3 years ago
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