Presently we can solve problem instances of size 30 in 1 minute using algorithm A, which is a algorithm. On the other hand, we will soon have to solve probleminstances twice this large in 1 minute. Do you think it would help to buy a faster( and more expensive) computer?

Answer:

nooooooo

Explanation:

Answer:

Given Data:

Clock rate of P1 = 4 GHz

Clock rate of P2 = 3 GHz

Average CPI of P1 = 0.9

Number of Instructions = 5.0E9 =  5 × 10^9

Clock rate of P2 = 3 GHz

Average CPI of P2 = 0.75

Number of Instructions = 1.0E9 = 10^9

To find: If the computer with largest clock rate has the largest performance?

Explanation:

Solution:

As given in the question, clock rate of P1 = 4 GHz which is greater than clock rate of P2 = 3 GHz

According to the performance equation:

CPU Time = instruction count * average cycles per instruction/ clock rate

CPU Time = I * CPI / clock rate

Where instruction count refers to the number of instructions.

Performance of P1:

CPU Time (P1) = 5 * 10^9 * 0.9 / (4 * 10^9)

                        = 5000000000 * 0.9 / 4000000000

                        = 4500000000 / 4000000000

                        =  1.125s

Performance of P2:

CPU Time (P2) = 10^9 * 0.75/ (3 * 10^9)

                        = 750000000 / 3000000000

                        = 0.25s

So the Performance of P2 is larger than that of P1,

                          performance (P2) > performance (P1)

                                         0.25 is better than 1.125

But clock rate of P1 was larger than P2

                             clock rate of P1 > clock rate of P2

                                                  4 GHz > 3 GHz

So this is a misconception about P1 and P2.

It is not true that computer with the largest clock rate as having the largest performance.

Answer:

Explanation:

#include <iostream>

#include <string>

#include<vector>

using namespace std;  

vector<int> permute(vector<int>, vector<int>);

string encrypt(vector<int>s1 , vector<int> t1, string p);

string decrypt(vector<int>s1, vector<int> t1, string p);

int main() {

  string plaintext = "cryptology";

  string plaintext2 = "RC4";

  vector<int> S(256);

  vector<int> T(256);

  int key[] = { 1,2,3,6 };

  int key2[] = { 5,7,8,9 };

  int tmp = 0;

  for (int i = 0; i < 256;i++) {

      S[i] = i;

      T[i] = key[( i % (sizeof(key)/sizeof(*key)) )];

  }

  S = permute(S, T);

  for (int i = 0; i < 256 ;i++) {

      cout << S[i] << " ";

      if ((i + 1) % 16 == 0)

          cout << endl;

  }

  cout << endl;

  string p = encrypt(S, T, plaintext);

  cout << "Message: " << plaintext << endl;

  cout << "Encrypted Message: " << " " << p << endl;

  cout << "Decrypted Message: " << decrypt(S, T, p) << endl << endl;

  tmp = 0;

  for (int i = 0; i < 256;i++) {

      S[i] = i;

      T[i] = key2[(i % (sizeof(key) / sizeof(*key)))];

  }

  S = permute(S, T);

  for (int i = 0; i < 256;i++) {

      cout << S[i] << " ";

      if ((i + 1) % 16 == 0)

          cout << endl;

  }  

  cout << endl;

  p = encrypt(S, T, plaintext2);

  cout << "Message: " << plaintext2 << endl;

  cout << "Encrypted Msg: " << p << endl;

  cout << "Decrypted Msg: "<<decrypt(S, T, p) << endl << endl;

  return 0;

}

string decrypt(vector<int>s1, vector<int> t1, string p) {

  int i = 0;

  int j = 0;

  int tmp = 0;

  int k = 0;

  int b;

  int c;

  int * plain = new int[p.length()];

  string plainT;

  for (int r = 0; r < p.length(); r++) {

      i = (i + 1) % 256;

      j = (j + s1[i]) % 256;

      b = s1[i];

      s1[i] = s1[j];

      s1[j] = b;

      tmp = (s1[i] + s1[j]) % 256;

      k = s1[tmp];

      c = ((int)p[r] ^ k);

      plain[r] = c;

      plainT += (char)plain[r];

  }

  return plainT;

}  

string encrypt(vector<int>s1, vector<int> t1, string p) {

  int i = 0;

  int j = 0;

  int tmp = 0;

  int k = 0;

  int b;

  int c;

  int * cipher = new int [p.length()];

  string cipherT;

  cout << "Keys Generated for plaintext: ";

  for (int r = 0; r < p.length(); r++) {

      i = (i + 1) % 256;

      j = (j + s1[i]) % 256;

      b = s1[i];

      s1[i] = s1[j];

      s1[j] = b;

      tmp = (s1[i] + s1[j]) % 256;

      k = s1[tmp];

      cout << k << " ";

      c = ((int)p[r] ^ k);

      cipher[r] = c;  

      cipherT += (char)cipher[r];

  }

  cout << endl;

  return cipherT;

}

vector<int> permute(vector<int> s1, vector<int> t1) {

  int j = 0;

  int tmp;

  for (int i = 0; i< 256; i++) {

      j = (j + s1[i] + t1[i]) % 256;

      tmp = s1[i];

      s1[i] = s1[j];

      s1[j] = tmp;

  }

  return s1;

}

Lists are used in Python to hold multiple values in one variable

(a) Nested list

A nested list is simply a list of list; i.e. a list that contains another list.

It is also called a 2 dimensional list.

An example is:

nested_list = [[ 1, 2, 3, 4]  , [ 5, 6, 7]]

(b) The “*” operator

The "*" operator is used to calculate the product of numerical values.

An example is:

num1 = num2 * num3

List slices

This is used to get some parts of a list; it is done using the ":" sign

Take for instance, you want to get the elements from the 3rd to the 5th index of a list

An example is:

firstList = [1, 2 ,3, 4, 5, 6, 7]

secondList = firstList[2:5]

The “+=” operator

This is used to add and assign values to variables

An example is:

num1 = 5

num2 = 3

num2 += num1

A list filter

This is used to return some elements of a list based on certain condition called filter.

An example that prints the even elements of a list is:

firstList = [1, 2 ,3, 4, 5, 6, 7]

print(list(filter(lambda x: x % 2 == 0, firstList)))

A valid but wrong list operation

The following operation is to return a single list, but instead it returns as many lists as possible

def oneList(x, myList=[]):

   myList.append(x)

   print(myList)

oneList(3)

oneList(4)

Read more about Python listsat:

brainly.com/question/16397886

Answer:

C: the responsible person or organization, date accessed, and URL

Explanation: