One line of code is missing (marked in

The following method public String removeFromString(String old, String frag) removes all occurences of the string frag from the

Mashutka [201]

private static String removeFromString(String old, String frag)

{

int i = old.indexOf(frag);

while (i> -1) {

String rest = old.substring(i + frag.length());

System.out.println("rest = " + rest);

old = old.substring(0, i);

System.out.println("rest = " + old);

old = old + rest;

System.out.println("rest = " + old);

i = old.indexOf(frag);

System.out.println("i = "+ i);

}

return old;

}

Here the index of first occurrence is obtained outside the “while loop” and if this loop runs until index value is >-1. It extracts the rest of the characters other than “frag” from the index and all the characters for which the given set of characters are removed.

4 0

3 years ago

100 POINTS!!!!!!

Veronika [31]

Answer:

The fundamental limitation of symmetric (secret key) encryption is ... how do two parties (we may as well assume they are Alice and Bob) agree on a key? In order for Alice and Bob to communicate securely they need to agree on a secret key. In order to agree on a secret key, they need to be able to communicate securely. In terms of the pillars of IA, To provide CONFIDENTIALITY, a secret key must first be shared. But to initially share the key, you must already have CONFIDENTIALITY. It's a whole chicken-and-egg problem.

This problem is especially common in the digital age. We constantly end up at websites with whom we decide we want to communicate securely (like online stores) but with whom we there is not really an option to communicate "offline" to agree on some kind of secret key. In fact, it's usually all done automatically browser-to-server, and for the browser and server there's not even a concept of "offline" — they only exist online. We need to be able to establish secure communications over an insecure channel. Symmetric (secret key) encryption can't do this for us.

Asymmetric (Public-key) Encryption

Yet one more reason I'm barred from speaking at crypto conferences.

xkcd.com/177/In asymmetric (public key) cryptography, both communicating parties (i.e. both Alice and Bob) have two keys of their own — just to be clear, that's four keys total. Each party has their own public key, which they share with the world, and their own private key which they ... well, which they keep private, of course but, more than that, which they keep as a closely guarded secret. The magic of public key cryptography is that a message encrypted with the public key can only be decrypted with the private key. Alice will encrypt her message with Bob's public key, and even though Eve knows she used Bob's public key, and even though Eve knows Bob's public key herself, she is unable to decrypt the message. Only Bob, using his secret key, can decrypt the message ... assuming he's kept it secret, of course.

Explanation:

3 0

2 years ago

Ashley Baker has been the webmaster for Berryhill Finance only ten days when she received an e-mail that threatened to shut down

Natasha_Volkova [10]

Answer:

cyber-extortion

Explanation:

Ashley Baker has been the webmaster for Berryhill Finance only ten days when she received an e-mail that threatened to shut down Berryhill's website unless Ashley wired payment to an overseas account. Ashley was concerned that Berryhill Finance would suffer huge losses if its website went down, so she wired money to the appropriate account. The author of the e-mail successfully committed cyber-extortion.

7 0

3 years ago

a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv

Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

// (if you want better performance: use array instead of unordered_map)

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]

bool helper(vector<string>& words, string& result, int digit, int widx, int sum) {

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

// if summation at digit position complete, validate it with result[digit].

if (widx == words.size()) {

if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {

if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

bool tmp = helper(words, result, digit+1, 0, sum/10);

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

} else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

return helper(words, result, digit+1, 0, sum/10);

} else {

return false;

}

// if word[widx] length less than digit, ignore and go to next word

if (digit >= words[widx].length()) {

return helper(words, result, digit, widx+1, sum);

}

// if word[widx][digit] already mapped to a value

if (c2i.count(words[widx][digit])) {

if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0)

return false;

return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);

}

// if word[widx][digit] not mapped to a value yet

for (int i = 0; i < 10; i++) {

if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

bool tmp = helper(words, result, digit, widx+1, sum+i);

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

void isSolvable(vector<string>& words, string result) {

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

vector<string> words={"GREEN" , "BLUE"} ;

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

See more about C++ code at brainly.com/question/19705654

#SPJ1

3 0

2 years ago

How many panes can Pablo view in a spreadsheet if he chooses the split worksheet option?

ddd [48]

Pablo can be a minimum of two pains and a maximum of four panes in a split screen worksheet.

6 0

3 years ago

Read 2 more answers