Mellissa wants to pursue a career in database administration. Select the requirements needed to achieve this.

What must be the same for each variable in an array in compiled languages?

sukhopar [10]

The data type must be the same for each item in the array.

7 0

3 years ago

Which Internet connection can have speeds around 1 Gbps?<br> Cable<br> DSL<br> Fiber<br> Satellite

Alexus [3.1K]

<h2>Answer:</h2>

<u>Fiber</u> Internet connection has speeds around 1 Gbps.

<h2></h2><h2>Explanation:</h2>

Fiber Internet connection is the modern way of transmitting and receiving data. As obvious from the name Optical Fiber consists of thin fibers that are extremely thin (as a human hair). They transmit data with the beam of light that make this transmission speedy as well as reliable. Therefore they provide the download speed ranging from 25 Mbps to 1 Gbps.

Fiber Internet connection is considered best for the small businesses as it provides high bandwidth for transmission. The best part of this connection is that it is made of glass and plastic that do not conduct heat or electricity that means the fiber is safe from the damages as well.

5 0

4 years ago

) Suppose algorithm A takes 10 seconds to handle a data set of 1000 records. Suppose the algorithm A is of complexity O(n2). Ans

Mice21 [21]

Answer:

i) The time taken for 1500 records = 15 seconds.

ii) The time taken for 1500 records = 50 seconds.

Explanation:

A is an O(n) algorithm.

An algorithm with O(n) efficiency is described as a linearly increasing algorithm, this mean that the rate at which the input increases is linear to the time needed to compute that algorithm with n inputs.

From the question, it is given that for 1000 records, the time required is: 10 seconds.

Algorithm time taken is O(n)

Hence,

1) For 1,500 records

=> 10/1000 = x/1500

=> 10x1500/1000 = x

x = 15 seconds

Thus, the time taken for 1500 records = 15 seconds.

2) For 5,000 records

=> 10/1000 = x/5000

=> 10x5000/1000 = x

x = 50 seconds

Thus, the time taken for 1500 records = 50 seconds.

8 0

3 years ago

Implement the RC4 stream cipher in C++. User should be able to enter any key that is 5 bytes to 32 bytes long. Be sure to discar

gladu [14]

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;

}

6 0

3 years ago

Moving through a neighborhood trying to locate open wireless access points is called ________.

disa [49]

<span>Moving through a neighborhood trying to locate open wireless access points is called wardriving.
</span> The action wardriving is performed by a person in a moving vehicle (car, bike..) who is using a laptop or smartphone and wants to catch a Wireless LAN (WiFi) network. For this purpose there is software <span>freely available on the Internet.</span>

7 0

3 years ago