Caeser Cipher

Recently i have been playing around with caeser cipher and i came up with this:

Code:

---

    string a = "thisisit";
    string ciphertext;
    int b = a.length();
    int offset = 3;
    int offset2 = 5;
    int offset3 = 2;
    for (int i = 0; i < b; ++i)
    {
        ciphertext += ( ( (a[i] - 'a') + (offset - offset2) / (offset3 * offset) - offset2) %26) + 'a';
    }
    cout<< ciphertext <<endl;
    cin.get();

---

And it works to encrypt the data but how do i decrypt it after that? I was thinking maybe something along the lines of reversing this part of the code:

[code]
ciphertext += ( ( (a[i] - 'a') + (offset - offset2) / (offset3 * offset) - offset2) %26) + 'a';
[code]

P.S i am quite a newb and have never done anything like this before so be gentile(lol).

Break your expression down into its component parts. It's way too complicated, especially if you're a beginner. Once it's simplified, you can find ways to optimize it and easily reverse it.

My other reply got deleted. I don't know what those offset things are, but I think encryption is just

Code:

---

ciphertext += (((a[i] - 'a') + shift_value) % 26) + 'a';

---

You get the offset in the alphabet, shift it, modulus to make it wrap, and then make it ASCII again.

And then decryption, you can subtract the shift value from (a[i] - 'a'), check if it's less than 0, and if it is add it to 26. You can make 'a' in your code into a var that is 'A' or 'a' based on whether it's upper or lower case.

I just cant do it i ave tryed reversing my code e.g:

Code:

---

decrypted += ( ( (a[i] + 'a') - (offset + offset2) * (offset3 / offset) + offset2) %26) - 'a';

---

But that doesn't do it so i drastically need help with this i am not a complete newb but at crpyt. i am.

the modulus operator will not work in this case since a % b is a negative number if a is negative (if I recall correctly, but if a % b was always in the range [0, b - 1] simple subtraction should work)

Anyway since your addition is done in modulus arithmetic
(a - b) % c should be the same as (a + (c - b)) % c.
If you use the second expression the result will always be a positive (if b < c) number and thus will work correctly.

The other solution is to check for negative numbers before the modulus operator is applied and add 26 repeatedly until the number is again positive.

Remember that an aribtrary shift s is an effective shift of s % c (c = 26 in your case) so the limitation of b < c is in fact irrelevent

Another note is DON'T USE THIS TO ENCRYPT IMPORTANT STUFF. A Caesar code can be broken by stunned chimpmunk high on morphine. Really it is that easy to break so don't use it for anything but toying around. (If you knew this already please ignore me, but since you said you were just starting with cryptography I thout it best to warn you)

Break it down like prelude said (no one listens anymore)

Code:

---

for (int i = 0; i < b; ++i)
{
    int ch = a[i];
    ch = ch + 1;
    ch = ch % 26;
    ch = ch + 'a';
    ciphertext += ch;
}

---

Now write the decrypt function for this as well.

Then refine BOTH at the same time, by adding your next step in the calculation to both sides and checking that it still works.

You'll also find it a hell of a lot easier to step through using the debugger rather than having "BOOM" the answer in one single statement, but you still being as clueless as ever as to what actually happened.

I know that the caeser cipher is incredibaly easy to crack for anyone that knows how to however i am just using it as something to start on as i have aways been really interested in cryptography and have always wanted to be able to do it.So this is where i begin (I hope).

I still cant get the hand of this though i have tried loads of different things and thinking of how i could do it but i just cant i was wondering if i post a section of code could someone tell me how to reverse it as remember i am fairly new to this so i dont really understand anything you lot are saying(lol).

Code:

---

for (int i = 0; i < b; ++i)
{
  ciphertext += ( ( (a[i] - 'a') + (offset - offset2) / (offset3 * offset) - offset2) %26) + 'a';
}
cout<< ciphertext <<endl;

---

Thanx for all of your help on this and i am sorry that i dont understand.

Ok, let's try this again, step by step.

Any caesar cypher performs a transformation

Code:

---

cryptchar = (plainchar + shift) % alphabet_size

---

the inverse of this transformation is

Code:

---

plainchar = (cryptchar - shift) % alphabet_size

---

since the we are working with ASCII encoded characters we need to shift the origin from 0 to a before the transformation and back to 0 afterwards

Code:

---

plainchar = plainchar1 - 'a' // shift origin to a
cryptchar = (plainchar + shift) % alphabet_size
cryptchar1 = cryptchar + 'a' // shift origin back to 0

---

So far it's all really simple. The tricky bit is due to c++ implementation of the % operator. Usually modulus operations have range [0, n-1]. The c++ % operator has range [0, n-1] for positive input (the a in a % b) and [-n+1, 0] for negative input.
This is not an issue in the encryption since plainchar + shift is always positive (if the shift is positive, I'll show later that restricting the shift to the range [0, alphabet_size-1] still gives all possible shifts).
On the other hand on the decryption step there might be problems. If shift > cryptchar1 then cryptchar - shift will be a negative number which will give you a final result that is out of the intended range.
To solve this problem you need to ensure that cryptchar1 - shift is never negative.
One solution is to repeatedly add alphabet_size to the result untile the result is positive (since you are working in modulus arithmetic adding an integer multiple of alphabet_size doesn't change the final result).
The other soultion is to recognize the fact that (a - b) % c = (a + c - b) % c (remember adding any integer multiple of c doesn't change the result)
Thus the decryption routine can be written as

Code:

---

plainchar = (cryptchar + alphabet_size - shift) % alphabet_size

---

alphabet_size - shift is never negative as long as shift < alphabet_size

Why do you only need shifts in the range [0, alphabet_size - 1]?
Assume we have a bad_shift that is outside that range. We can break down bad shift into

Code:

---

bad_shift = a * alphabet_size + shift

---

where a is an arbitrary integer and 0 <= shift < alphabet_size.
Since we are working in modulus arithmetic the a * alphabet_size component has no effect to the final reusult and we can safely ignore it.

Hope this helps

PS

Quote:

---

Originally Posted by L_U_K_E

Code:

---

  ciphertext += ( ( (a[i] - 'a') + (offset - offset2) / (offset3 * offset) - offset2) %26) + 'a';

---

---

I hope you realise that all those calculations with the 3 offsets could be done once and stored in a single variable

Quote:

---

Originally Posted by sigfriedmcwild

Hope this helps

---

It won't, I'm afraid. Not untill Luke understands he needs to actually read the answers.

Is there a FAQ for these things? Something like, "Should I read the answers to my Post?"...

Oh well...
On a slightly different topic, can anyone explain why c++ (and I guess c originally) use such an obtuse implementation of the % operator?

The output of the modulos operator is machine-dependent if one of the operands is negative and another is positive. In fact both the result and the sign of the result are machine-dependent. This is not a C++ thing. But instead the way different machines deal with arithmetics.

I guess on modern platforms all the basic operators +-*/%=&|^~ just call the processor to do it for you. That's a Good Thing, you know.

And LUKE, for more cryptography go brush up on math, especially on number theory. If you have some dough, get Elementary Number Theory and its Applications, Kenneth H. Rosen.

could someone tell me whats the matter with this please:

Code:

---

#include <iostream>
#include <fstream>

using namespace std;

int main()
{
    string cryptchar;
    string plainchar = "luke";
    int shift = 2;
    int alphabet_size = plainchar.length();
    for (int i = 0; i < alphabet_size; ++i)
    {
        cryptchar = (plainchar + shift) %alphabet_size;
        plainchar = (cryptchar - shift) %alphabet_size;
    }
    cout<< "Encrypted: " << cryptchar <<endl;
    cin.get();
    cout<< "Decrypted: " << plainchar <<endl;
    cin.get();
    return 0;
}

---

When i get to the

Code:

---

cryptchar = (plainchar + shift) %alphabet_size;

---

bit i just get the message 'no match for operator+ ' in 'plainchar + shift'.Thanks for all your help so far i just hope i can get the hang of it.

P.S Thanx loads Sigfried for all of your help and patience with me(lol).

plainchar is a string, shift is an int. Perhaps you want to use plainchar[i] instead.

I changed but alli get is 1 little heart on the encrypted part and decrypted = 1 little smily face!

You forgot the +/- 'a' bit of the code to make 0..25 into 'a' to 'z'

KK i finally understand all of this after playing around with it a bit however when i still try to decrypt this just prints the right amount of characters this time except it prints not "luke" like it should do.Anyway here's the code i was hoping someone could have a look at it and tell me whats the matter with the decrypt part as i thought it would work as it is similar to what Sifried showed me:

Code:

---

#include <iostream>
#include <fstream>

using namespace std;

int main()
{
    string cyphertext;
    string plainchar = "luke";
    int shift = 2;
    int alphabet_size = plainchar.length();
    for (int i = 0; i < alphabet_size; ++i)
    {
        cyphertext += ( ( (plainchar[i] - 'a') + shift) %26) + 'a';
        plainchar += ( ( (plainchar[i] + 'a') - shift) %26) - 'a';
    }
    cout<< "Encrypted: " << cyphertext <<endl;
    cin.get();
    cout<< "Decrypted: " << plainchar <<endl;
    cin.get();
    return 0;
}

---

*sigh* One last time...

Decompose the long expressions and you will be able to test it yourself!

decompose definiton = to decay, putrefy.(lol) look i have tried this before what makes you think that it will work this time?

Done it sorry 4 the arigance but it works except i cant remember how to flush a string?

Code:

---

std::cout << "flush me" << std::flush; //just flushes

---

or

Code:

---

std::cout << "flush me" << std::endl; //outputs a newline and flushes

---

or

Code:

---

std::cout << "flush me" << std::ends; //outputs a null and flushes

---

or

Code:

---

std::cout << "flush me";
std::cin >> val;  //flushes cout by calling cin

---

or

Code:

---

int main() {
    std:: cout << "flush me";
    return 0;
}  //flushes by function termination

---

or

Code:

---

pull the lever or push the button

---

Quote:

---

Originally Posted by L_U_K_E

decompose definiton = to decay, putrefy.(lol) look i have tried this before what makes you think that it will work this time?

---

If you don't understand mathematical terms, I suggest you move out from cyphers and do something else in your code.

I sorted it anyway by just creating a new string and using that thanx for all of your help.Here is my code:

Code:

---

#include <iostream>
#include <fstream>

using namespace std;

int main()
{
    string cyphertext;
    string plainchar = "luke";
    string plainchar2;
    int shift = 2;
    int alphabet_size = plainchar.length();
    for (int i = 0; i < alphabet_size; ++i)
    {
        cyphertext += ( ( (plainchar[i] - 'a') + shift) %26) + 'a';
        plainchar2 += cyphertext[i] - shift;
    }
    cout<< "Encrypted: " << cyphertext <<endl;
    cin.get();
    cout<< "Decrypted: " << plainchar2 <<endl;
    cin.get();
    return 0;
}

---

However I am still having trouble when it comes to capitals, spaces and the letters 'a' and 'z' is that because i havn't linked it round?

You don't need the fstream header.

Place a cin.ignore() before your last cin.get(). If you input test on your first cin.get(), the second will not execute as you expected.

It's also probably better you get used to replace the less-than operator in your for loops with the not-equal one. It's more intuitive. Although not necessarily for all situations.

As a curiosity, if you want to have the actual caeser's cypher, shift by 3, not 2. That's how he did it. Or shift by 13 to produce the ROT13 cypher made common in signatures and such.

Thanx for all your help and i ahve made those changes hopwever i get problems if i change "luke" to numbers or the letters 'a' or 'z' is that because i havn't linked it round?

Time you try to do it yourself... give it a shot

No, the problem is that your entire implementation relies on the assumption that there won't be anything but lower-case alphabetic characters in the string. It's the -'a' and +'a' parts.
To do this properly, you have to treat each character according to its class: differ between upper-case, lower-case, digits and special characters, changing the class shift for the classes and leaving specials alone.

So do you mean that instead of typing:

Code:

---

cyphertext += ( ( (plainchar[i] - 'a') + shift) %26) + 'a';

---

I would type:

Code:

---

        cyphertext += ( ( (plainchar[i] - '1') + shift) %9) + '1';

---

For numbers or is that like completely wrong?

'0', but the principle is right. Now you just need to find a way to abstract the common code out of it.

what do you mean by 'abstract the common code out of it'?(sorry about this but by asking questions it makes it easier to learn and not make the same mistakes again).

Well, you could repeat the cipher code for every class of characters.
Or you can find a way to write the common code once in a way that allows you to reuse it.

Ahh i get you thanx for all of your help by the way.

Do you think i could do something like:

Code:

---

for (int i = 0; i != alphabet_size; ++i)
{
      cyphertext += ( ( (plainchar[i] - '0' | 'A' | 'a') + shift) %) + '0' | 'A' | 'a';
      plainchar2 += cyphertext[i] - shift;
}

---

But then i would be puzzeled about what goes after the modulus.

No, it could not be. How on earth is the compiler supposed to know when to use what?
You need a function. Learn about functions. And you need the contents of the <cctype> standard header to classify characters.
After the modulus you need the size of the class, i.e. 26 for alphabetic characters, 10 for digits.

It was just a guess really.I guess i could try read in each char 1 at a time and use its corrisponding type('a' | 'A' | '0').What do you think of that?

Quote:

---

Originally Posted by L_U_K_E

I sorted it anyway by just creating a new string and

Code:

---

    int alphabet_size = plainchar.length();

---

---

???

And yes, for each character type you need to determine what kind of character it is and do the shift for that. Or you could Aplly to ceasar cypher to entiere ascii set (or better the entire ascii set apart from null), it wouldn't necessarily be printable but as far as the information stored goes it would work.

Well, you'll need to test if a character is

# 'a' to 'z'
# 'A' to 'Z'
# '0' to '9'

And do a seperate dingo for each one. Again, when you decrypt (or more accurately decode), check if the character is

# 'a'+3 to 'z'+3

and so on.