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katovenus [111]
2 years ago
15

What do conditions do for programs?

Computers and Technology
1 answer:
Luba_88 [7]2 years ago
7 0

Answer:

add versatility thank you

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Write a Java program to encrypt and decrypt a phrase using two similar approaches, each insecure by modern standards. The first
xeze [42]

Answer:

See explaination

Explanation:

//CryptoManager.java

public class CryptoManager {

static int LOWER_BOUND=32;

static int UPPER_BOUND=95;

/*This method determines if a string is within the allowable bounds of ASCII

codes according to the LOWER_BOUND and UPPER_BOUND characters. The parameter

plainText is the string to be encrypted. The method returns true if all

characters are within the allowable bounds, false if any character is outside.*/

public static boolean stringInBounds (String plainText)

{

boolean flag=true;

//determines if a string is within the allowable bounds of ASCII

//codes according to the LOWER_BOUND and UPPER_BOUND characters.

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

{

if(!((int)plainText.charAt(i)>=LOWER_BOUND && (int)plainText.charAt(i)<=UPPER_BOUND))

{ //false if any character is outside the bounds

flag=false;

break;

}

}

//returns true if all characters are within the allowable bounds

return flag;

}

/*This method encrypts a string according to the Caesar Cipher. The integer key

specifies an offset and each character in plainText is replaced by the character

the specified distance away from it. The parameter plainText is an uppercase

string to be encrypted. The parameter key is an integer that specifies the

offset of each character. The method returns the encrypted string.*/

public static String encryptCaesar(String plainText, int key)

{

//Wrap around the key, if it is greater than the UPPER_BOUND

key=Wrap_around(key);

//encrypted text

String res="";

//encryption

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

{

res+=Character.toString((char) ((int)plainText.charAt(i)+key));

}

//return result

return res;

}

/* This method decrypts a string according to the Caesar Cipher. The integer

key specifies an offset and each character in encryptedText is replaced by

the character "offset" characters before it. This is the inverse of the

encryptCaesar method. The parameter encryptedText is the encrypted string

to be decrypted, and key is the integer used to encrypt the original text.

The method returns the original plain text string.*/

public static String decryptCaesar(String encryptedText, int key){

//Wrap around the key, if it is greater than the UPPER_BOUND

key=Wrap_around(key);

//decrypted text

String org="";

//encryption

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

{

org+=Character.toString((char) ((int)encryptedText.charAt(i)-key));

}

//return result

return org;

}

public static int Wrap_around(int key)

{

while(key>UPPER_BOUND)

{

key-=(UPPER_BOUND-LOWER_BOUND);

}

return key;

}

/* This method encrypts a string according to the Bellaso Cipher. Each character

in plainText is offset according to the ASCII value of the corresponding

character in bellasoStr, which is repeated to correspond to the length of

plaintext. The method returns the encrypted string.*/

public static String encryptBellaso(String plainText, String bellasoStr)

{

//encrypted text

String res="";

//Adjust length of bellasoStr to plainText

while(bellasoStr.length()<plainText.length())

{

bellasoStr+=bellasoStr.substring(0,(plainText.length()-bellasoStr.length()));

}

//encryption

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

{

char c=(char)Wrap_around((int)plainText.charAt(i)+(int)bellasoStr.charAt(i) );

res+=Character.toString(c);

}

//return result

return res;

}

/*

This method decrypts a string according to the Bellaso Cipher. Each character

in encryptedText is replaced by the character corresponding to the character in

bellasoStr, which is repeated to correspond to the length of plainText. This is

the inverse of the encryptBellaso method. The parameter encryptedText is the

encrypted string to be decrypted, and bellasoStr is the string used to encrypt

the original text. The method returns the original plain text string.*/

public static String decryptBellaso(String encryptedText, String bellasoStr)

{

//decrypted text

String res="";

//Adjust length of bellasoStr to plainText

while(bellasoStr.length()<encryptedText.length())

{

bellasoStr+=bellasoStr.substring(0,(encryptedText.length()-bellasoStr.length()));

}

//decryption

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

{

char c=(char)Wrap_around((int)encryptedText.charAt(i)-(int)bellasoStr.charAt(i) );

res+=Character.toString(c);

}

//return result

return res;

}

}

6 0
3 years ago
There are many modes of remote visual communication. Which is the most common mode?
Neko [114]
Videoconferencing, as most people use that. <span />
3 0
3 years ago
Read 2 more answers
Suppose you have a program P and 90 percent of P can be parallelized, but 10 percent of P is inherently sequential and cannot be
EleoNora [17]

Answer:

Means no matter how many processors you use, speed up never increase from 10 times.

Explanation:

If a problem of size W has a serial component Ws,then performance using parallelism:

Using Amdahl's Law:

Tp = (W - Ws )/ N + Ws

Here, Ws = .1,

W - Ws = .9

Performance Tp = (.9 / N) + .1

---------------------------------------------------------

Speed Up = 1 / ( (.9 / N) + .1)

If N -> infinity, Speed Up <= 10

Means no matter how many processors you use, speed up never increase from 10 times.

5 0
3 years ago
Assume there is a class AirConditioner that supports the following behaviors: turning the air conditioner on and off, and checki
vaieri [72.5K]

Answer:

yes?

Explanation:

8 0
3 years ago
Draw the 2-3 tree that results when you insert the keys E A S Y Q U T I O N in that order into an initially empty tree
mr_godi [17]

Answer:

If its any easy question, answer it yourself.

Explanation:

:)

6 0
2 years ago
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