Thread: Encrypt a text message

I need to write a code to encrypt any sentence in a square caesar box. So if the sentence has 28 words, the 'box'(array) should have 36 spaces. I should change s.length in my code to something that works this out, but I don't know how.

 
	public static String encrypt(String plainText) {
 
 
	String s = plainText.toLowerCase();
	s = s.replaceAll("\\s","");
	s.toCharArray();
	s.charAt(1); // This is the sencence as an array
	int n = s.length();// This is the size of the box
	int counter = 0;
	string [] [] box = new string [n] [];//This is the box
	for (int i = 0; i<n; i++){
		for(int j=0; j<n; j++){
			box[i][j] = s.charAt(counter);
			counter++;
		}
	}
 
 
 
 
 
	return box;
 
 
 
	}

Any ideas?

hi
you dnt need to replace string s .
instead take n as full string with space. And check whether the string length gives perfect square.If its yes then
take char[][] box = new char[size][size] where size is the squareroot obtained.Then u can run the loop
for (int i = 0; i<size; i++){
for(int j=0; j<size; j++){
box[j][i] = s.charAt(0);
}
}

then run a loop say taking a string stobtained
for (int i = 0; i < size; i++)
{
for (int j = 0; j < size; j++)
{
stobtained += box[i][j];
}
}
System.out.println("string obtained==" + stobtained);
or return stobtained

Check this .. hope this solves your problem.

Thanks

The asssignment requires me to take out blank spaces first

My main problem is how to proceed when the size is not squareroot, because the code should run for any sentence the user sets up. Any ideas?

--- Update ---

ps: perfect square root

hi

so you need to check after removing spaces that if its not a square root and proceed by adding that much space to make it a square root?
if no then you can jst keep a check like if its not a square root then the print the message that "cannot be encrypted.Not a square root"

It needs to be encypted even if the number of letters is not a square root and in this case there will be extra spaces in the box. I just don't know how to set the size of the box so it will always be a square box even when the number of letters is not a perfect square. Any ideas?

--- Update ---

ps:perfect square root

hi
after removing the spaces if its not a squareroot then keep a check like
find squareroot of n and round the result up for eg if your string length is 7 then when you find squareroot you will get 2.64.Round this and you will get 3. and square the result.
Now take the boxsize to the result you obtained when you square the no

Yes, I thought about this pseudocode. I just don't know the correct syntax in Java for it. How to calculate square root?

I wanted to write something like if squareroot of n is not an integer then increment n. But the problem is that squareroot must return a double so the way I wrote below won't work.



[QUOTE]double n = s.length();// This is the size of the sentence
if(sqrt(n)!=int){
n++
}
/QUOTE]

How could I fix it?

--- Update ---

sorry

int n = s.length();// This is the size of the sentence
	if(sqrt(n)!=int){
	n++	
	}

I did.


public static double sqrt(double a)
Returns the correctly rounded positive square root of a double value.

So it always returns a rounded double. Does it mean it might round down so I will get less spaces than s.length()?

No, the code below returns error: Syntax error on token "int", invalid Expression

int n = s.length();// This is the size of the sentence
	if(sqrt(n)!=int){
	n++;	
	}

Yes, sqrt(n) is a double, but how could I test if this double will be perfect square root if it "Returns the correctly rounded positive square root of a double value" ?

I have tried for only one input of 27 letters and it is not working. I want it to return 36 and it is returning 33, because when n gets to 33 the square root is 6 and the if test is not satisfied anymore. How can I make it not round up the square root?

double n = s.length();// This is the size of the sentence
 
	double sqrt = (double) Math.sqrt(n);// This is the square root of number of letters
 
	if(sqrt*sqrt!=n){
	n++;
	}

No, I did not get this error. And I tried with int as well

int n = s.length();// This is the size of the sentence
 
	int sqrt = (int) Math.sqrt(n);// This is the square root of number of letters
 
	if(sqrt*sqrt!=n){
	n++;
	}
 
    System.out.println(n);

I need to find a way sqr won't round up because this is the reason why n is not a number with perfect square root.