How to encrypt in RSA?

Question

I want to write my own RSA encrypter without libaries!

Code:

import java.util.Random;

public class Main {

public static void main(String[] args) {
    System.out.println(createPrime());
}

private static byte encrypt(byte message) {
    double p = createPrime();
    double q = createPrime();
    double e = 2 ^ new Random().nextInt(100) % (p * q);
    byte ciphered = Math.pow(message, e);
    return ciphered;
}

private static double createPrime() {
    double testPow;
    do {
    int test = new Random().nextInt(20);
    int power = new Random().nextInt(20);
    test += 1;
    power +=1;
    testPow = Math.pow(test, power);
    System.out.println("Double Math.pow: " + testPow);
    } while (!testPrime(testPow));
    return testPow;
}

private static Boolean testPrime(Double test) {
    int factor = 2;
    int lastFactor = 1;
    while (test > 1) {
        if (test % factor == 0) {
            lastFactor = factor;
            test /= factor;
            while (test % factor == 0) {
                test /= factor;
            }
        }
        factor++;
    }
    Boolean isPrime = false;
    if (test == lastFactor) {
        isPrime = true;
    }
    return isPrime;
 }
}

This is what I have for encrypting. I don't know what I should do to correct this, but I have pretty much done this by hand before trying this.

So I know that the equation to encrypt is c = m^e (mod N) and decryption m = c^d (mod N)

where p, q are prime numbers - m is message - c is ciphered text - e is totient of N - N is p times q - totient of N is (p-1)(q-1)

Any help is appreciated

norbert thocker · Accepted Answer · 2011-02-26 22:01:29Z

4

The first thing to do is to have a look at the class java.math.BigInteger. This class will help you a lot to implement "School-book" RSA.

answered Feb 26, 2011 at 22:01

norbert thocker

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Comments

BlueRaja - Danny Pflughoeft · Accepted Answer · 2011-02-26 22:01:13Z

2

You didn't ask a real question, but I see a couple of problems anyways

double e = 2 ^ new Random().nextInt(100) % (p * q);

I don't know what this is supposed to do, but it's wrong. Did you mean Math.Pow() rather than ^? In any case, usually you just use some small constant with very few set bits for e to make encryption fast. e=3 or e=65 would work fine.

You don't seem to be calculating the private key (d), or even storing the public key (e, p*q) at all.

When you start using large numbers, int and double (??) will not be able to hold them. Use BigInteger instead.

do {
    testPow = Math.pow(test, power);
} while (!testPrime(testPow));

If power > 1, testPow will never be prime...

I have not looked at testPrime(), but you should be able to write some quick unit tests to convince yourself whether or not it probably works.

edited Feb 26, 2011 at 22:01

answered Feb 26, 2011 at 21:55

BlueRaja - Danny Pflughoeft

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1 Comment

Paŭlo Ebermann Over a year ago

RSA-like encryption works with integer modular exponentation, Math.pow is not really useful here. But yeah, BigInteger has some methods which do (quite efficient) modular exponentation.

ykaganovich · Accepted Answer · 2011-02-26 22:10:51Z

1

I'd suggest reading/copying an existing implementation for reference, such as BouncyCastle: http://www.docjar.com/html/api/org/bouncycastle/crypto/engines/RSACoreEngine.java.html

By the way, if you want this to be at all secure, you should be using java.security.SecureRandom, not java.util.Random

answered Feb 26, 2011 at 22:10

ykaganovich

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Comments

Bozho · Accepted Answer · 2011-02-26 20:08:42Z

1

Java has built-in encryption algorithms under the java.security package. Check this. So no need for external libraries. I see no production need to implement it yourself. Only if it is homework (which you didn't tag)

answered Feb 26, 2011 at 20:08

Bozho

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2 Comments

Zeveso Over a year ago

its homework, well not really... but lets pretend it is because I WANT to learn this...

Bozho Over a year ago

@user516664 you can look at the java code perhaps. It's open-source.

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