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Introdution
本文是簡(jiǎn)單介紹橢圓曲線密碼學(xué)的基本內(nèi)容，目標(biāo)是直觀、簡(jiǎn)化，忽略嚴(yán)謹(jǐn)性。切記，簡(jiǎn)單是為了導(dǎo)向嚴(yán)肅而不是庸俗。
閱讀以下內(nèi)容，在有群論與密碼學(xué)的基礎(chǔ)上，大概需要1個(gè)小時(shí)。如果沒(méi)有密碼學(xué)基礎(chǔ)，可忽略密碼學(xué)，而重點(diǎn)關(guān)注橢圓曲線的定義與相關(guān)計(jì)算。
Why we need ECC?
- Efficiency
- More properties
What exactly is an elliptic curve?
Definition
Let $$a ∈ ?$$, $$b ∈ ?$$, be constants such that $$4a^3 + 27b^2 ≠ 0$$. A non-singular elliptic curve is the set $$E$$ of solutions $$(x, y) ∈ ? \times ?$$ to the equation: $$y^2 = x^3 + ax + b$$ together with a special point $O$ called the point at infinity.
The solution set $$E$$ forms an Abelian group.
Graphical Representation
Addition of the Group E
Three cases:
Case1 x1 ≠ x2
Addition is defined as follows:
Let $$(x_1, y_1) + (x_2, y_2) = (x_3, y_3) ∈ E$$ where
$$x_3 = λ^2 - x_1 - x_2$$
$$y_3 = λ(x_1 – x_3) - y_1$$,
and $$λ = (y_2 – y_1) / (x_2 – x_1)$$
**Case2 x1 = x2 and y1 = - y2 **
Addition is defined as follows:
$$(x_1, y_1) + (x_2, y_2) = (x_3, y_3) ∈ E$$ where
$$(x, y) + (x, -y) = O$$, the point at infinity.
Case3 x1 = x2 and y1 = y2
Addition is defined as follows:
$$(x_1, y_1) + (x_2,y_2) = (x_3, y_3) ∈ E$$ where
$$x_3 = λ^2 - x_1 - x_2$$
$$y_3 = λ(x_1 – x_3) - y_1$$, and
$$λ = (3x_1^2 + a) / 2y_1$$
If you want to know why the computations are correct, please read Silverman's text: Rational Points on Elliptic Curves.
ECC mod p
Definition. Let $$p > 3$$ be prime. The elliptic curve $$y^2 = x^3 + ax + b$$ over $$?_p$$ is the set of solutions $$(x, y) \in ?_p \times ?_p$$ to the congruence: $$y^2 ≡ x^3 + ax + b ; (mod ; p) $$ where $$a \in ?_p$$, $$b \in ?_p$$, are constants such that $$4a^3 + 27b^2 ? 0 \quad (mod ; p)$$, together with a special point $$O$$ called the point at infinity.
Solutions still form an Abelian group.
Example
Let’s examine the following elliptic curve as an example:
$$y^2 = x^3 + x + 6$$ over $$?_{11}$$
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| $$y^2 = x^3 + x + 6$$ | 6 | 8 | 5 | 3 | 8 | 4 | 8 | 4 | 9 | 7 | 4 |
| QR? | N | N | Y | Y | N | Y | N | Y | Y | N | Y |
| Y | 4, 7 | 5, 6 | 2, 9 | 2, 9 | 3, 8 | 2, 9 |
Generating the group
Since the $$O(E)$$ is prime, the group is cyclic. We can generate the group by choosing any point other than the point at infinity. Let our generator be $$ g = (2, 7)$$
Generate the group by using the rules of addition we defined earlier where $$2g = g + g$$:
| g | 2g | 3g | 4g | 5g | 6g | 7g | 8g | 9g | 10g | 11g | 12g | 13g |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| (2, 7) | (5, 2) | (8, 3) | (10, 2) | (3, 6) | (7, 9) | (7, 2) | (3, 5) | (10, 9) | (8, 8) | (5, 9) | (2, 4) | (0, 6) |
Where $$\lambda = 8$$. You could check that.
Discrete logarithm problem on elliptic curve groups
Definition. (Informal!) Suppose $$E$$ is an elliptic curve and $$P \in E $$ . Given a multiple $$Q$$ of $$P$$, the elliptic curve discrete log problem(ECDLP) is to find $$n$$ such that $$nP = Q$$.
Example. Given $$g$$ and $$Q=(3, 5)$$, find $$n=8$$, s.t., $$ng = Q$$.
We believe ECDLP problem is hard, more rigorous results may be found here.
A real example from the NSA
Curve P-192
- p = 62771017353866807638578942320766641608390870039024961279
- r = 627710173538668076385789423176059013767194773182842284081
- a = 3099d2bb bfcb2538 542dcd5f b078b6ef 5f3d6fe2 c745de65
- b = 64210519 e59c80e7 0fa7e9ab 72243049 feb8deec c146b9b1
- Gx = 188da89e b03090f6 7cbf20eb 43a18800 f4ff0afd 82ff1012
- Gy = 07192b95 ffc8da78 631011ed 6b24cdd5 73f977a1 1e794811
ECC in Sage
Play with following simple code:
sage: p = 137
sage: F = FiniteField(p)
sage: E = EllipticCurve(F, [F.random_element(), F.random_element()])
sage: print E
Elliptic Curve defined by y^2 = x^3 + 112*x + 106 over Finite Field of size 137
sage: E.points()
[(0 : 1 : 0), (2 : 8 : 1), (2 : 129 : 1),......
Now, to visualize a EC, actually, it looks like that ...
sage: show(plot(E, hue=.9))
Diffie-Hellman in ECC
sage: p = next_prime(randrange(10^40))
sage: print p
sage: F = FiniteField(p)
sage: E = EllipticCurve(F, [F.random_element(), F.random_element()])
sage: P = E.random_element()
sage: print P
sage: b = randrange(1000); b
sage: B = b*P
sage: a = randrange(1000); a
sage: A = a*P
sage: if(a*B == b*A): print "We share a common secret."
Show me more...
2017年8月24日
