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@AArnott
Created February 7, 2016 06:17
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Calculate full RSA private key parameters from the P and Q parameters
using System;
using System.Numerics;
using System.Security.Cryptography;
using System.Text;
using Xunit;
public class RSA
{
[Fact]
public void WikipediaExample()
{
var p = new BigInteger(61);
var q = new BigInteger(53);
var e = 17;
var n = p * q;
Assert.Equal(n, 3233);
var phiOfN = (p - 1) * (q - 1);
Assert.Equal(phiOfN, 3120);
var d = ModInverse(e, phiOfN);
Assert.Equal(d, 2753);
var m = 65;
var c = Encrypt(m, n, e);
Assert.Equal(2790, c);
var m_ = Decrypt(c, d, n);
Assert.Equal(m, m_);
}
[Fact]
public void RSACompatibility()
{
var rsa = new RSACryptoServiceProvider(512);
var parameters = rsa.ExportParameters(true);
var recreatedParameters = Create(parameters.P, parameters.Q, parameters.Exponent, parameters.Modulus);
var recreatedRSA = new RSACryptoServiceProvider();
recreatedRSA.ImportParameters(recreatedParameters);
byte[] message = Encoding.UTF8.GetBytes("hello");
var ciphertext = rsa.Encrypt(message, false);
var plaintext = recreatedRSA.Decrypt(ciphertext, false);
string recreatedMessage = Encoding.UTF8.GetString(plaintext);
}
private static RSAParameters Create(byte[] p, byte[] q, byte[] exponent, byte[] modulus)
{
var addlParameters = GetFullPrivateParameters(
p: new BigInteger(CopyAndReverse(p)),
q: new BigInteger(CopyAndReverse(q)),
e: new BigInteger(CopyAndReverse(exponent)),
modulus: new BigInteger(CopyAndReverse(modulus)));
return new RSAParameters
{
P = p,
Q = q,
Exponent = exponent,
Modulus = modulus,
D = addlParameters.D,
DP = addlParameters.DP,
DQ = addlParameters.DQ,
InverseQ = addlParameters.InverseQ,
};
}
private static RSAParameters GetFullPrivateParameters(BigInteger p, BigInteger q, BigInteger e, BigInteger modulus)
{
var n = p * q;
var phiOfN = n - p - q + 1; // OR: (p - 1) * (q - 1);
var d = ModInverse(e, phiOfN);
Assert.Equal(1, (d * e) % phiOfN);
var dp = d % (p - 1);
var dq = d % (q - 1);
var qInv = ModInverse(q, p);
//Assert.Equal(1, (qInv * q) % p);
return new RSAParameters
{
D = CopyAndReverse(d.ToByteArray()),
DP = CopyAndReverse(dp.ToByteArray()),
DQ = CopyAndReverse(dq.ToByteArray()),
InverseQ = CopyAndReverse(qInv.ToByteArray()),
};
}
private static BigInteger Encrypt(BigInteger m, BigInteger n, BigInteger e)
{
return BigInteger.ModPow(m, e, n);
}
private static BigInteger Decrypt(BigInteger mEnc, BigInteger d, BigInteger n)
{
return BigInteger.ModPow(mEnc, d, n);
}
/// <summary>
/// Calculates the modular multiplicative inverse of <paramref name="a"/> modulo <paramref name="m"/>
/// using the extended Euclidean algorithm.
/// </summary>
/// <remarks>
/// This implementation comes from the pseudocode defining the inverse(a, n) function at
/// https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm
/// </remarks>
public static BigInteger ModInverse(BigInteger a, BigInteger n)
{
BigInteger t = 0, nt = 1, r = n, nr = a;
if (n < 0)
{
n = -n;
}
if (a < 0)
{
a = n - (-a % n);
}
while (nr != 0)
{
var quot = r / nr;
var tmp = nt; nt = t - quot * nt; t = tmp;
tmp = nr; nr = r - quot * nr; r = tmp;
}
if (r > 1) throw new ArgumentException(nameof(a) + " is not convertible.");
if (t < 0) t = t + n;
return t;
}
private static byte[] CopyAndReverse(byte[] data)
{
byte[] reversed = new byte[data.Length];
Array.Copy(data, 0, reversed, 0, data.Length);
Array.Reverse(reversed);
return reversed;
}
}
@ingfraga
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ingfraga commented May 6, 2019

Can you explain to me, why is the reverse of the bytes needed when creating a BigInteger?

@roa-nyx
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roa-nyx commented Apr 4, 2022

Note: BigInteger.ToByteArray() will sometimes give an array that is not appropriately sized to the expected byte array size of the RSA parameters. I ended up creating a function that takes the output of ToByteArray, and sizes it up prepending zeros to the array so that it's the expected size. This then worked in all cases for me.

Also worth noting that, at least in .Net 6.0, ToByteArray and the BigInteger constructors now have overloads that work in BigEndian mode, entirely eliminating the need for the CopyAndReverse operation.

Thanks for your work. This was an excellent starting point for me.

@AArnott
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AArnott commented Apr 5, 2022

Great. Thanks for sharing your tips, @roa-nyx.

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