后量子密码 区块链
Public key encryption is all about setting up a trapdoor function, where someone who knows a given secret will fall through a function. As much as possible we need a difficult problem to solve, and one which is know to be hard to solve. Knowing a secret allows the hard problem to become simple. For example, I might have a secret value of 6, and then use a secret multiplier of 5. My cipher method can then be to multiply my message by the secret to give:
公钥加密是关于设置活板门功能的，任何知道给定机密的人都将通过该功能。 我们需要尽可能解决一个棘手的问题，而这是一个很难解决的问题。 知道秘密可以使难题变得简单。 例如，我可能有一个秘密值6，然后使用秘密乘数5。然后，我的密码方法可以是将我的消息乘以秘密以得到：
Cipher = 6 x 5 = 30
密码= 6 x 5 = 30
Now, my trap door is now the inverse of my secret: 1/5. If you know that, you can now multiply the Cipher by the inverse value. But in this case, the inverse of 5 is easy to find, so we need tougher ways.
现在，我的陷阱门现在是我秘密的反面：1/5。 如果知道这一点，您现在可以将密码乘以倒数。 但是在这种情况下，很容易找到5的倒数，因此我们需要更艰难的方法。
In public key encryption, such as RSA, we encrypt with C=M^e (mod N) and decrypt with M=C^d (mod N). We pick a value of e, and then must find its inverse, by finding a value of d when:
在公共密钥加密(例如RSA)中，我们使用C = M ^ e(mod N)进行加密，并使用M = C ^ d(mod N)进行解密。 我们选择e的值，然后必须通过在以下情况下找到d的值来找到它的逆值：
d e(mod N)=1
de (mod N )= 1
For this we need to perform an inverse mod operation:
为此，我们需要执行逆模运算：
d=e^{−1} (mod N) =1
d = e ^ {-1}(mod N )= 1
Unforunately RSA, along with Elliptic Curve Cryptography, are at risk from quantum computers, so we must find another way. One of the methods proposed is lattice cryptography, and which uses polynomials. With polynominals — such as 1+6x — 5x²+2x² - we will operate with our mathematical operations, such as add :
不幸的是，RSA与椭圆曲线密码术一起受到量子计算机的威胁，因此我们必须找到另一种方法。 提出的方法之一是晶格密码术，它使用多项式。 对于多项式-例如1 + 6x-5x²+2x²-我们将使用数学运算进行运算，例如add：
(a+bx+c x²) + (d + ex+fx²) = (a+d) + (b+e) x + (c+f)x²
(a + bx + cx²)+(d + ex +fx²)=(a + d)+(b + e)x +(c + f)x²
and subtract:
并减去：
(a+bx+c x²) -(d + ex+fx²) = (a-d) + (b-e) x + (c-f)x²
(a + bx + cx²)-(d + ex +fx²)=(ad)+(be)x +(cf)x²
and multiply:
并乘以：
(a+bx+c x²) x (d + ex+fx²) = (ad) + ae x + af x² + bd x + be x² + bf x³ + cd x² + ce x³ + cf x⁴
(a + bx + cx²)x(d + ex +fx²)=(ad)+ ae x + afx²+ bd x +是x²+ bfx³+ cdx²+ cex³+ cfx⁴
We can also divide using a long-division method. By one important operation is the inverse mod p operation, for where we find f^{-1} for f (mod p). The method most often used to find the inverse mod is the Extended Euclidean method applied to polynomial values. In lattice methods we use polynomials, such as:
我们还可以使用长除法进行除法。 一个重要的运算是逆模p运算，在这里我们找到f的f ^ {-1}(模p)。 查找反模的最常用方法是应用于多项式值的扩展欧几里得方法。 在晶格方法中，我们使用多项式，例如：
f=−1+x²+x³ (mod p)
F = -1 + X 2 + X³(mod p)的
To define a key, we often use trinary values of {−1,0,+1}as our factors. Then we need to find f^{-1}(mod p) in order to provide the reverse operation. Luckily we can do with the Extended Euclidean method applied to polynomials. So let’s take an example, and prove it. With {-1,0,1,1}, we can represent with a polynomial of:
为了定义键，我们经常使用{−1,0，+ 1}的三进制值作为我们的因子。 然后我们需要找到f ^ {-1}(mod p )以提供相反的操作。 幸运的是，我们可以将扩展的欧几里得方法应用于多项式。 因此，让我们举一个例子并加以证明。 使用{-1,0,1,1}，我们可以用以下多项式表示：
f=−1+x²+x³ (mod p)
F = -1 + X 2 + X³(mod p)的
and want to find the inverse mod of this for (mod 32), we can determine it as [solution]:
并想找到它的逆模(模32)，我们可以将其确定为[ 解 ]：
f′=19+17x+13x²+26x³
F'= 19 + 17×13 X 2 26 X³
Now let’s provide that this is true. For this we can do a multiplication of the original with the inverse, and hopefully we will get an answer of unity. Let’s do:
现在让我们提供这是真的。 为此，我们可以将原始数与逆数相乘，并希望我们能得到统一的答案。 让我们做：
r=f×f′=(modp)
r = f × f '=(模p )
This gives
这给
r=f×f′=(−1+x²+x³)×(-19+7x+13x²+26x³)(mod 23)
R = F×F'=( - 1 + X 2 + X³)×(-19 + 7×13 X 2 + 26׳)(MOD 23)
r=f×f′=(−19−7x−13x²−26x³+19x²+7x³+13x⁴+26x⁵+19x³+7x⁴+13x⁵+26x⁶) (mod p)
R = F×F'=( - 19-7 X -13 X²-26׳+ 19×2 + 7׳+ 13×⁴+ 26×⁵+ 19׳+ 7×⁴+ 13×⁵+ 26×⁶)(mod p)的
We then get the result of:
然后我们得到以下结果：
r=(−19−7x−13x²+19x²−26x³+19x³+7x³+13x⁴+7x⁴+26x⁵+13x⁵+26x⁶)
R =( - 19-7 X -13 X 2 + 19ײ-26׳+ 19׳+ 7׳+ 13×⁴+ 7×⁴+ 26×⁵+ 13×⁵+ 26×⁶)
r=−19−7x+6x²+20x⁴+39x⁵+26x⁶
R = -19-7 X 6 X 2 20 X⁴+ 39 X⁵+ 26×⁶
With polynomial operations in cryptography we then limit the highest power of the polynomial by dividing by D=x⁴−1 and where 3 is the highest polynomial power we can have:
在密码学多项式的操作，我们然后通过d除以限制多项式的最高功率= X⁴-1，并且其中图3是可以具有最高多项式功率：
r=(−19−7x+6x²+20x⁴+39x⁵+26x⁶)/(x⁴−1)
R =( - 19-7 X 6 X 2 + 20×⁴+ 39 X⁵+ 26×⁶)/(X⁴-1)
and which gives a result of 1. The code is here:
结果为1。代码在这里：
演示地址
If you are interested, here is NTRU:
如果您有兴趣，这里是NTRU：
and which is one of the finalists for the NIST Post Quantum standard for Key Exchange Mechanisms (KEMs):
并且是NIST密钥交换机制(KEM)的Post Quantum标准的决赛入围者之一：
翻译自: https://medium.com/asecuritysite-when-bob-met-alice/the-trap-door-with-polynomials-towards-a-post-quantum-crypto-era-dcc0e18686d1
后量子密码 区块链
