当前位置: 首页>>代码示例>>C#>>正文


C# BigInteger.ModInverse方法代码示例

本文整理汇总了C#中BigInteger.ModInverse方法的典型用法代码示例。如果您正苦于以下问题:C# BigInteger.ModInverse方法的具体用法?C# BigInteger.ModInverse怎么用?C# BigInteger.ModInverse使用的例子?那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在BigInteger的用法示例。


在下文中一共展示了BigInteger.ModInverse方法的10个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的C#代码示例。

示例1: RsaSecretBcpgKey

		public RsaSecretBcpgKey(
			BigInteger d,
			BigInteger p,
			BigInteger q)
		{
			// PGP requires (p < q)
			int cmp = p.CompareTo(q);
			if (cmp >= 0)
			{
				if (cmp == 0)
					throw new ArgumentException("p and q cannot be equal");

				BigInteger tmp = p;
				p = q;
				q = tmp;
			}

			this.d = new MPInteger(d);
			this.p = new MPInteger(p);
			this.q = new MPInteger(q);
			this.u = new MPInteger(p.ModInverse(q));

			this.expP = d.Remainder(p.Subtract(BigInteger.One));
			this.expQ = d.Remainder(q.Subtract(BigInteger.One));
			this.crt = q.ModInverse(p);
		}
开发者ID:Xanagandr,项目名称:DisaOpenSource,代码行数:26,代码来源:RsaSecretBcpgKey.cs

示例2: TestModInverse

        public void TestModInverse()
        {
            var value = new BigInteger(4);
            var modulus = new BigInteger(5);
            var result = value.ModInverse(modulus);

            Assert.AreEqual(new BigInteger(4), result);
        }
开发者ID:hanson-huang,项目名称:Encore,代码行数:8,代码来源:BigIntegerTest.cs

示例3: GenerateSignature

		/**
		 * Generate a signature for the given message using the key we were
		 * initialised with. For conventional DSA the message should be a SHA-1
		 * hash of the message of interest.
		 *
		 * @param message the message that will be verified later.
		 */
		public BigInteger[] GenerateSignature(
			byte[] message)
		{
			DsaParameters parameters = key.Parameters;
			BigInteger q = parameters.Q;
			BigInteger m = calculateE(q, message);
			BigInteger k;

			do
			{
				k = new BigInteger(q.BitLength, random);
			}
			while (k.CompareTo(q) >= 0);

			BigInteger r = parameters.G.ModPow(k, parameters.P).Mod(q);

			k = k.ModInverse(q).Multiply(
				m.Add(((DsaPrivateKeyParameters)key).X.Multiply(r)));

			BigInteger s = k.Mod(q);

			return new BigInteger[]{ r, s };
		}
开发者ID:Xanagandr,项目名称:DisaOpenSource,代码行数:30,代码来源:DsaSigner.cs

示例4: TestModInverse

        public void TestModInverse()
        {
            for (int i = 0; i < 10; ++i)
            {
                BigInteger p = BigInteger.ProbablePrime(64, Rnd);
                BigInteger q = new BigInteger(63, Rnd).Add(One);
                BigInteger inv = q.ModInverse(p);
                BigInteger inv2 = inv.ModInverse(p);

                Assert.AreEqual(q, inv2);
                Assert.AreEqual(One, q.Multiply(inv).Mod(p));
            }
        }
开发者ID:ChemicalRocketeer,项目名称:BigMath,代码行数:13,代码来源:BigIntegerTest.cs

示例5: VerifySignature

		/**
		 * return true if the value r and s represent a DSA signature for
		 * the passed in message for standard DSA the message should be a
		 * SHA-1 hash of the real message to be verified.
		 */
		public bool VerifySignature(
			byte[]		message,
			BigInteger	r,
			BigInteger	s)
		{
			DsaParameters parameters = key.Parameters;
			BigInteger q = parameters.Q;
			BigInteger m = calculateE(q, message);

			if (r.Sign <= 0 || q.CompareTo(r) <= 0)
			{
				return false;
			}

			if (s.Sign <= 0 || q.CompareTo(s) <= 0)
			{
				return false;
			}

			BigInteger w = s.ModInverse(q);

			BigInteger u1 = m.Multiply(w).Mod(q);
			BigInteger u2 = r.Multiply(w).Mod(q);

			BigInteger p = parameters.P;
			u1 = parameters.G.ModPow(u1, p);
			u2 = ((DsaPublicKeyParameters)key).Y.ModPow(u2, p);

			BigInteger v = u1.Multiply(u2).Mod(p).Mod(q);

			return v.Equals(r);
		}
开发者ID:Xanagandr,项目名称:DisaOpenSource,代码行数:37,代码来源:DsaSigner.cs

示例6: GenerateKeyPair

		public AsymmetricCipherKeyPair GenerateKeyPair()
        {
            BigInteger p, q, n, d, e, pSub1, qSub1, phi;

            //
            // p and q values should have a length of half the strength in bits
            //
			int strength = param.Strength;
            int pbitlength = (strength + 1) / 2;
            int qbitlength = (strength - pbitlength);
			int mindiffbits = strength / 3;

			e = param.PublicExponent;

			// TODO Consider generating safe primes for p, q (see DHParametersHelper.generateSafePrimes)
			// (then p-1 and q-1 will not consist of only small factors - see "Pollard's algorithm")

			//
            // Generate p, prime and (p-1) relatively prime to e
            //
            for (;;)
            {
				p = new BigInteger(pbitlength, 1, param.Random);

				if (p.Mod(e).Equals(BigInteger.One))
					continue;

				if (!p.IsProbablePrime(param.Certainty))
					continue;

				if (e.Gcd(p.Subtract(BigInteger.One)).Equals(BigInteger.One)) 
					break;
			}

            //
            // Generate a modulus of the required length
            //
            for (;;)
            {
                // Generate q, prime and (q-1) relatively prime to e,
                // and not equal to p
                //
                for (;;)
                {
					q = new BigInteger(qbitlength, 1, param.Random);

					if (q.Subtract(p).Abs().BitLength < mindiffbits)
						continue;

					if (q.Mod(e).Equals(BigInteger.One))
						continue;

					if (!q.IsProbablePrime(param.Certainty))
						continue;

					if (e.Gcd(q.Subtract(BigInteger.One)).Equals(BigInteger.One)) 
						break;
				}

                //
                // calculate the modulus
                //
                n = p.Multiply(q);

                if (n.BitLength == param.Strength)
					break;

                //
                // if we Get here our primes aren't big enough, make the largest
                // of the two p and try again
                //
                p = p.Max(q);
            }

			if (p.CompareTo(q) < 0)
			{
				phi = p;
				p = q;
				q = phi;
			}

            pSub1 = p.Subtract(BigInteger.One);
            qSub1 = q.Subtract(BigInteger.One);
            phi = pSub1.Multiply(qSub1);

            //
            // calculate the private exponent
            //
            d = e.ModInverse(phi);

            //
            // calculate the CRT factors
            //
            BigInteger dP, dQ, qInv;

            dP = d.Remainder(pSub1);
            dQ = d.Remainder(qSub1);
            qInv = q.ModInverse(p);

            return new AsymmetricCipherKeyPair(
//.........这里部分代码省略.........
开发者ID:Xanagandr,项目名称:DisaOpenSource,代码行数:101,代码来源:RsaKeyPairGenerator.cs

示例7: VerifySignature

		/**
		 * return true if the value r and s represent a GOST3410 signature for
		 * the passed in message (for standard GOST3410 the message should be
		 * a GOST3411 hash of the real message to be verified).
		 */
		public bool VerifySignature(
			byte[]		message,
			BigInteger	r,
			BigInteger	s)
		{
			byte[] mRev = new byte[message.Length]; // conversion is little-endian
			for (int i = 0; i != mRev.Length; i++)
			{
				mRev[i] = message[mRev.Length - 1 - i];
			}

			BigInteger e = new BigInteger(1, mRev);
			BigInteger n = key.Parameters.N;

			// r in the range [1,n-1]
			if (r.CompareTo(BigInteger.One) < 0 || r.CompareTo(n) >= 0)
			{
				return false;
			}

			// s in the range [1,n-1]
			if (s.CompareTo(BigInteger.One) < 0 || s.CompareTo(n) >= 0)
			{
				return false;
			}

			BigInteger v = e.ModInverse(n);

			BigInteger z1 = s.Multiply(v).Mod(n);
			BigInteger z2 = (n.Subtract(r)).Multiply(v).Mod(n);

			ECPoint G = key.Parameters.G; // P
			ECPoint Q = ((ECPublicKeyParameters)key).Q;

			ECPoint point = ECAlgorithms.SumOfTwoMultiplies(G, z1, Q, z2);

			BigInteger R = point.X.ToBigInteger().Mod(n);

			return R.Equals(r);
		}
开发者ID:Xanagandr,项目名称:DisaOpenSource,代码行数:45,代码来源:ECGOST3410Signer.cs

示例8: GenerateSignature

		// 5.3 pg 28
		/**
		 * Generate a signature for the given message using the key we were
		 * initialised with. For conventional DSA the message should be a SHA-1
		 * hash of the message of interest.
		 *
		 * @param message the message that will be verified later.
		 */
		public BigInteger[] GenerateSignature(
			byte[] message)
		{
			BigInteger n = key.Parameters.N;
			BigInteger e = calculateE(n, message);

			BigInteger r = null;
			BigInteger s = null;

			// 5.3.2
			do // Generate s
			{
				BigInteger k = null;

				do // Generate r
				{
					do
					{
						k = new BigInteger(n.BitLength, random);
					}
					while (k.Sign == 0 || k.CompareTo(n) >= 0);

					ECPoint p = key.Parameters.G.Multiply(k);

					// 5.3.3
					BigInteger x = p.X.ToBigInteger();

					r = x.Mod(n);
				}
				while (r.Sign == 0);

				BigInteger d = ((ECPrivateKeyParameters)key).D;

				s = k.ModInverse(n).Multiply(e.Add(d.Multiply(r))).Mod(n);
			}
			while (s.Sign == 0);

			return new BigInteger[]{ r, s };
		}
开发者ID:Xanagandr,项目名称:DisaOpenSource,代码行数:47,代码来源:ECDsaSigner.cs

示例9: VerifySignature

		// 5.4 pg 29
		/**
		 * return true if the value r and s represent a DSA signature for
		 * the passed in message (for standard DSA the message should be
		 * a SHA-1 hash of the real message to be verified).
		 */
		public bool VerifySignature(
			byte[]		message,
			BigInteger	r,
			BigInteger	s)
		{
			BigInteger n = key.Parameters.N;

			// r and s should both in the range [1,n-1]
			if (r.Sign < 1 || s.Sign < 1
				|| r.CompareTo(n) >= 0 || s.CompareTo(n) >= 0)
			{
				return false;
			}

			BigInteger e = calculateE(n, message);
			BigInteger c = s.ModInverse(n);

			BigInteger u1 = e.Multiply(c).Mod(n);
			BigInteger u2 = r.Multiply(c).Mod(n);

			ECPoint G = key.Parameters.G;
			ECPoint Q = ((ECPublicKeyParameters) key).Q;

			ECPoint point = ECAlgorithms.SumOfTwoMultiplies(G, u1, Q, u2);

			BigInteger v = point.X.ToBigInteger().Mod(n);

			return v.Equals(r);
		}
开发者ID:Xanagandr,项目名称:DisaOpenSource,代码行数:35,代码来源:ECDsaSigner.cs

示例10: modinv

 private BigInteger modinv(BigInteger a, BigInteger b)
 {
     return a.ModInverse(b);
 }
开发者ID:mzkabbani,项目名称:cSharpProjects,代码行数:4,代码来源:SignatureDSA.cs


注:本文中的BigInteger.ModInverse方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。