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This chapter is from the book

This chapter is from the book

RSA: The Most Used Asymmetric Algorithm

The most common asymmetric cipher currently in use is RSA, which is fully supported by the .NET Security Framework. Ron Rivest, Adi Shamir, and Leonard Adleman invented the RSA cipher in 1978 in response to the ideas proposed by Hellman, Diffie, and Merkel. Later in this chapter, we shall see how to use the high-level implementation of RSA provided by the .NET Security Framework. But first, let's look at how RSA works at a conceptual level.

Underpinnings of RSA

Understanding the underpinnings of RSA will help you to develop a deeper appreciation of how it works. In this discussion we focus on the concepts of RSA, and in Appendix B we look at two examples of implementing RSA from scratch. One of these examples is TinyRSA, which is a toy version that limits its arithmetic to 32-bit integers, and the other is a more realistic, multiprecision implementation named BigRSA. You will probably never implement your own RSA algorithm from scratch, since most cryptographic libraries, including the .NET Security Framework, provide excellent implementations (i.e., probably better than I could do). However, the RSA examples in Appendix B should help you to fully understand what goes on in RSA at a deeper level.

Here is how RSA works. First, we randomly generate a public and private key pair. As is always the case in cryptography, it is very important to generate keys in the most random and therefore, unpredictable manner possible. Then, we encrypt the data with the public key, using the RSA algorithm. Finally, we decrypt the encrypted data with the private key and verify that it worked by comparing the result with the original data. Note that we are encrypting with the public key and decrypting with the private key. This achieves confidentiality. In the next chapter, we look at the flip side of this approach, encrypting with the private key and decrypting with the public key, to achieve authentication and integrity checking.

Here are the steps for generating the public and private key pair.

  1. Randomly select two prime numbers p and q. For the algebra to work properly, these two primes must not be equal. To make the cipher strong, these prime numbers should be large, and they should be in the form of arbitrary precision integers with a size of at least 1024 bits.11

  2. Calculate the product: n = p · q.

  3. Calculate the Euler totient12 for these two primes, which is represented by the Greek letter φ. This is easily computed with the formula φ = (p – 1) · (q – 1).

  4. Now that we have the values n and φ, the values p and q will no longer be useful to us. However, we must ensure that nobody else will ever be able to discover these values. Destroy them, leaving no trace behind so that they cannot be used against us in the future. Otherwise, it will be very easy for an attacker to reconstruct our key pair and decipher our ciphertext.

  5. Randomly select a number e (the letter e is used because we will use this value during encryption) that is greater than 1, less than φ, and relatively prime to φ. Two numbers are said to be relatively prime if they have no prime factors in common. Note that e does not necessarily have to be prime. The value of e is used along with the value n to represent the public key used for encryption.

  6. Calculate the unique value d (to be used during decryption) that satisfies the requirement that, if d · e is divided by φ, then the remainder of the division is 1. The mathematical notation for this is d · e = 1(mod φ). In mathematical jargon, we say that d is the multiplicative inverse of e modulo φ. The value of d is to be kept secret. If you know the value of φ, the value of d can be easily obtained from e using a technique known as the Euclidean algorithm. If you know n (which is public), but not p or q (which have been destroyed), then the value of φ is very hard to determine. The secret value of d together with the value n represents the private key.

Once we have generated a public/private key pair, we can encrypt a message with the public key with the following steps.

  1. Take a positive integer m to represent a piece of plaintext message. In order for the algebra to work properly, the value of m must be less than the modulus n, which was originally computed as p · q. Long messages must therefore be broken into small enough pieces that each piece can be uniquely represented by an integer of this bit size, and each piece is then individually encrypted.

  2. Calculate the ciphertext c using the public key containing e and n. This is calculated using the equation c = me (mod n).

Finally, we can perform the decryption procedure with the private key using the following steps.

  1. Calculate the original plaintext message from the ciphertext using the private key containing d and n. This is calculated using the equation m = cd (mod n).

  2. Compare this value of m with the original m, and you should see that they are equal, since decryption is the inverse operation to encryption.

A Miniature RSA Example

Here is an example of RSA that is almost simple enough to do with pencil and paper. It is similar in scale to the TinyRSA code example discussed in this chapter. The bit size of the numbers used in this example is ridiculously small (32-bit integers) and offers no real security whatsoever, but at a conceptual level, this example provides a complete picture of what actually happens in the RSA algorithm. The advantage of studying this tiny paper and pencil example is that with these very small bit sizes, the underlying concepts are much more tangible and easily visualized. After all, not too many people can do 1024-bit arithmetic in their head! Even working with such tiny 32-bit numbers, the exponentiation step of the algorithm will easily overflow this 32-bit capacity if you are not careful about how you implement it.13

Following the conceptual steps outlined above, we start off by choosing two unequal prime numbers p and q.14 Since we intentionally choose very small values, we prevent subsequent calculations from overflowing the 32-bit integer arithmetic. This also allows us to follow along using the Calculator program provided with Windows to verify the arithmetic.

  1. Assume that the random values for the primes p and q have been chosen as

    • p = 47

    • q = 73

  2. Then the product n of these two primes is calculated:

    • n = p · q = 3431

  3. The Euler totient φ for these two primes is found easily using the following formula:

    • φ = (p – 1) · (q – 1) = 3312

  4. Now that we have n and φ, we should discard p and q, and destroy any trace of their existence.

  5. Next, we randomly select a number e that is greater than 1, less than n, and relatively prime to phi. Of course, there is more than one choice possible here, and any candidate value you choose may be tested using the Euclidian method.15 Assume that we choose the following value for e:

    • e = 425

  6. Then the modular inverse of e is calculated to be the following:

    • d = 1769

  7. We now keep d private and make e and n public.

Now that we have our private key information d and our public key information e and n, we can proceed with encrypting and decrypting data. As you would probably imagine, this data must be represented numerically to allow the necessary calculations to be performed. In a real-life scenario, the plaintext is typically a hash value or a symmetric key, but it could actually be just about any type of data that you could imagine. Whatever form this data takes, it will have to be somehow represented as a sequence of integer numbers, each with a size that will be limited by the key size that you are using. We do not concern ourselves here with the details of encoding and chunking of the data, but instead we focus on the conceptual aspects of RSA. For this reason, this example simply considers a scenario in which the plaintext data is one simple, small integer value.

  1. Assume that we have plaintext data represented by the following simple number:

    • plaintext = 707

  2. The encrypted data is computed by c = me (mod n) as follows:

    • ciphertext = 707^425(mod 3431) = 2142

  3. The ciphertext value cannot be easily reverted back to the original plaintext without knowing d (or, equivalently, knowing the values of p and q). With larger bit sizes, this task grows exponentially in difficulty. If, however, you are privy to the secret information that d = 1769, then the plaintext is easily retrieved using m = c d(mod n) as follows:

    • plaintext = 2142^1769(mod 3431) = 707

If you compile the following code, you will verify that the results shown above are correct. While you look at this code, keep in mind that a realistic RSA implementation uses a much larger modulus than n = 3431, and a realistic message typically contains too many bits to be represented by a tiny number such as m = 707.

int m = 707;  //plaintext
int e = 425;  //encryption exponent
int n = 3431; //modulus
int c = 1;    //ciphertext

//encryption: c = m^e(mod n)
for (int i=0; i<e; i++) //use loop to avoid overflow
   c = c*m;
   c = c%n; //normalize within modulus
//ciphertext c is now 2142

int d = 1769; //decryption exponent
m = 1; //plaintext

//decryption m = c^d(mod n)
for (int i=0; i<d; i++) //use loop to avoid overflow
   m = m*c;
   m = m%n; //normalize within modulus
//plaintext m is now 707 matching original value
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