S. Erfani, ECE Dept., University of Windsor 0688-590-18 Network Security
1- El Gamal and Digital Signature Algorithm
A public-key algorithm was devised in 1984 by T. El Gamal based on discrete logarithms. The scheme is closely related to the Diffie-Hellman technique. It is used in the Digital Signature Standard (DSS) by NIST.
1.1– El Gamal Algorithm
Step 1 – Global Public Elements – As with Diffie-Hellman, to generate a key pair, first choose a prime number q and , a primitive root of q.
Step 2 – Key Generation – Alice selects a private key XA<Q and calculate a public key YA as in Diffie_hellman
YA= XA
Independently, Bob also generates his public key YB and private key.
Step 3–User A Signs a Message– Alice encrypts a plaintext M<q intended for Bob as follows:
a) Choose a random integer k, 1kq
a)Compute:
K=(YB)k mod q
(C1, C2) where:
C1=k mod q
C2=KM mod q
These two numbers together make up the signature.
Step 4–User B Verifies the Signature– Bob verifies the signature by recovering the plaintext M as follows:
(a)Compute
K= mod q
Which is
K= mod q = mod q= mod q
(b)Compute
M=(C2K-1) mod q
Where K-1 is the multiplicative inverse of K.
Therefore:
(C2K-1) mod q=(KMK-1) mod q
=MKK-1 mod q
=M mod q
Note 1– This scheme is sometimes referred to as DSA stands for Digital Signature Algorithm.
Note 2– The plaintext M is usually a digest of a message. It is seen that DSS does not encrypt the digest. The input to the algorithm is the digest of the data to sign, M, the key, YB and a random number, k. The output is a pair of numbers C1, and C2, as shown in Fig. 1. There will be many ciphertexts that are encryptions of the same digest, since the output depends on both the digest M and on the random value k chosen by Alice.
Fig. 1 DSS takes in three inputs and gives two numbers as a result
Note 3– To defeat this scheme and infer the values of XBand k givenC1, C2 and M, the intrude, Oscar, could find a means of computing a discrete logarithm to solve
YB= and C1=k
Example 1– Consider an El Gamal scheme with a common prime q=71 and a primitive root =7.
(a)If Bob has public key YB=3 and Alice chose the random integer k=2, what is the ciphertext of M=30?
(b)If Alice now chooses a different value of k, so that the encoding of M=30 is C=(59, C2), what is the integer C2?
Solution
(a) K=(YB)k mod q=32 mod 71=9
C1=k mod q=72 mod 71=49
C2=KM mod q=9×30 mod 71
=270 mod 7157
(b) In this case we have
C1=59=k mod q=7k mod 71
We need to solve a discrete logarithm to find k. It can be shown that k=3 because:
73 mod 71=343 mod 71=59=C1
K= mod q=33 mod 71
=27
C2=KM mod 71=27×30 mod 71
=810 mod 7129
Note 4– Informally, this is how the El Gamal algorithm works: The plaintext M is “masked” by multiplying it by , yielding C2. The value C1=k is also transmitted as past of the ciphertext. Bob who knows the private key, XB, can compute from C1. Then he can “remove the mask” by dividing C2 by to obtain M.
Note 5– The El Gamal cryptosystem can be defined mathematically as follows:
El Gamal CryptosystemLet q be a prime such that the discrete logarithm problem in (Zq,.) is infeasible and let Zqbe a primitive element. Let P=Zq, C=Zq×Zq, and define
K={(q, , XB, YB): YB= mod q}.
The values q, and YB are the public key and XB is the private key.
For K=(q, , XB, YB), and for a (secret) random number kZq-1, define
ek(x, k)=(C1, C2)
where
C1=k mod q
C2=x mod q
For C1 and C2Zq, define
k(C1, C2)=C2 mod q
Example 2– Suppose q=2579 and =2. Is a primitive element modulo q. Let XB=7. Now, suppose that Alice wishes to send the message M=1299 to Bob. Say k=853 is the random integer she chooses. Show the steps in El Gamal algorithm.
Solution
XB=765, =2, q=2579
YB= mod q=2765 mod 2579 949
K=(YB)k mod q = (949)853 mod 2579
C1 = k mod q = 2853 mod 2579 435
C2=KM mod q =1299×949853 mod 2579 2396
When Bob receives the ciphertext (C1, C2)=(435, 2396), He computes:
M=(C2K-1) mod q = 2396×(435765)-1 mod 2579 =1299,
Which was the plaintext that Alice encrypted.
1.2- Digital Signature Algorithm(DSA)
The U.S. Digital Signature Algorithm is the El Gamal algorithm with a few restrictions:
(a)The size of q is specifically fixed at 2511<q<2512 (so that q is roughly 170 decimal digits long).
(b)The large prime factor of (q-1) is chosen, so that 2519<p<2160.
(c)The algorithm uses a hash value instead of the full message plaintext M.
(d)The computations of C1 and C2 are taken mod p instead of mod q.
2- Elliptic Curve Cryptography
The vast majority of the products and standards that use public key cryptography for encryption and digital signatures use RSA. Recently, a competing system has begun to challenge RSA: elliptic curve cryptography (ECC, or EC for short). Already, ECC is showing up in standardization efforts, including the IEEE P1363 standards for public-key cryptography.
This scheme is based on solutions to elliptic curves modulo a prime p. We begin looking briefly at elliptic curves defined over the real numbers and move on to operations defined for elliptic curve modulo a prime p.
2.1- Elliptic Curves over the Reals
Elliptic curves are described by the set of solutions to certain equations in two variables. In general, cubic equations for elliptic curves take the form:
y2 + axy + by = x3 + cx2 + dx + e
where a, b, c, d, and e are real numbers that satisfy some simple conditions. They get the name because they used for calculating the circumference of an ellipse.
Note 1: It can be shown that the cubic equation
x3 + cx2 + dx + e = 0
can be reduced to the canonical form:
x3 + px + q = 0
to be solved.
Proof:
Let X = x + c/3 in the above equation we get:
X3 + (d – c2/3)X + (e – cd/3 + 2c3/27) = 0
Solutions of the original cubic are then in terms of the canonical cubic roots. The three roots of the canonical cubic are:
X1 = (A)1/3 + B1/3
X2= W (A)1/3 + W2 (B)1/3
X3= W2 (A)1/3 + W(B)1/3
Where
Where 4p3 + 27q2 0, A is complex.
Note 2: For ECC, we are concerned with a restricted form of elliptic curve that is defined over a finite field. More specifically:
y
P
R’
Q
x
-20 2
R
-P
Figure 2 – Adding Points on an Elliptic Curve
Note 3 – Elliptic curves are not ellipses. They receive their name from their relation to elliptic integrals such as and and arise in the computation of the arc length of ellipses.
2- Adding Points on an Elliptic Curve
Let us find the addition rules over E for all points P, QE, where P= (x1, y1) and Q = (x2, y2). The rules for addition over E correspond to the geometric technique illustrated in Figure 2. Given two points P and Q on E, to obtain a third point R(x3, y3) on E draw the line L through P and Q, the line L intersects E in a third point R’, reflect R’ through the x-axis (i.e., change y3 to –y3) to get R. Define the law of addition by P+Q=R. We consider three cases:
- x1 x2
- x1 = x2and y1 = -y2
- x1 = x2and y1 = y2
Case 1:
Let L to be the line through P and Q. L intersects E in the two points P and Q, and it is easy to see that L will intersect E in one further point, which we call R’. We reflect R’ in the x-axis, then we get a point which we call R. We define P+Q=R to compute coordinates of R, i.e., (x3, y3) note that equation for line L is
L:y = x +
Where = (y2-y1)/(x2-x1)
= y1-x1 = y2 – x2
To find the coordinates of R’, (x3, y3), which are the intersection of line L and curve E, we substitute equation for line L into the equation for E:
L: y = x +
E:y2 = x3 + ax + b
(x + )2 = x3 + ax + b
x3 - 2x2 + (a - 2)x + (b - 2) = 0
The roots are x-coordinates of points in L E, i.e., P, Q, and R’. Therefore
x1 + x2 + x3 = 2
x3 = 2 - x1 - x2
To find y3, note that slope of line L, i.e. can be determined by any two points on this line. We will denote the y-coordinate of R’ by –y3, so the y-coordinate of R will be y3. If we use the points (x1, y1) and (x3, -y3) to compute this slope, we get
= (-y3-y1)/(x3-x1)
y3 =(x1-x3) – y1
Thus, we have derived a formula for P+Q in case 1, when x1x2, for
(x1, y1) + (x2, y2) = (x3, y3) as:
Case 2:
x1 = x2 and y1 = -y2
If we try to add , we get a line through and P, which is vertical. It intersects E in P(x1, y1) and also in –P in (x1, -y1), (see Fig. 2). When we reflect (x1,-y1) across the x-axis, we get back P (x1,y1). Therefore, . Now, try to add P and –P. The line through (x1,y1) and (x1,-y1) is vertical, so the third point of intersection with E is . The reflection across the x-axis is still . Therefore, in this case, we define
(x, y) + (x, -y) = (Point at infinity) for all (x, y) E Therefore,
(x, y) and (x, -y) are inverses with respect to the elliptic curve addition operation.
The point at infinity, , will be defined as the identity element, so
P+ = +P =PP E
+ =
P + (-P) =
Case 3:
x1 = x2, y1 = y2
That is adding a point P to itself. In this case, the line L in case 1 is to be tangent to E at the point E. The slope L can be computed using implicit differentiation of equation of E:
2y dy/dx = 3x2 + a
Substituting x = x1, y=y1, we get the slope of the line L as:
= (3x12 + a) / 2y1
The rest is identical to case 1.
Def. Addition Law – Let E be given by and Let P(x1,y1) and let Q(x2,y2) be on E. Then:
where
(y2-y1)/(x2-x1) if P Q
=
(3x12+a)/(2y1) if P = Q
If the slop is infinite, then (point at the infinity). There is one additional law:
Note 4 -
Associative Law
Commutative Law
3- Elliptic Curves Modulo a Prime
Let p3 be a prime. The elliptic curve
y2 = x3 + ax + b
over Zp is the set of solutions (x, y)ZpX
to the congruence
y2 = x3 + ax + b (mod p)
where a, b, Zp are constants such that 4a3 + 27b2/ 0 (mod p), together with a special point called the point at infinity.
Example 1 – Let’s take the following elliptic curve and apply to it the modulus 11.
y2 mod 11 = (x3 + x + 2) mod 11
check to see if point P = (9, 5) is on the curve :
25 mod 11 = (729 + 9 + 2) mod 11
25 mod 11 = 740 mod 11
3 mod 11 (67 x 11 + 3) = 3 mod 11
We can find other points on the curve. For instance:
(1, 2)(6, 2)
(1, 9)(6, 9)
(2, 1)(8, 4)
(4, 2)(9, 5)
(4, 9)(9, 6)
infinity
Let’s multiply point P1 = ( 4, 2 ) by k = 3 :
kP1 = 3 x ( 4, 2) = P1 + P1 + P1
P1 + P1 P2 [(4, 2) + (4, 2) ] mod 11
We can easily find P2 using the following addition rule :
x22 – 2x1 (mod p)
y2 (x1 – x2) – y1 (mod p)
(3x12 + a) / 2y1 (mod p)
That is :
3x(42+1) / (2 x 2) mod 11 49/4 mod 11
5 x 4-1 mod 11 = 5x3 mod 11 4
x2 (42 – 2 x 4) mod 11 = 8
y2 [4 (4 – 8) – 2 ] mod 11 -7 mod 11 = 4
P1 + P2 = P2 = (8, 4)
P2 + P1 = P3 = (8, 4) + (4, 2) = (x3, y3)
Where
x3 = 2 – x1 - x2 62 – 8 – 4 2
y3 = (x1 – x3) – y1 = 6(8-2)-410
= (y2-y1)/(x2-x1) = (2-4)/(4-8) 1x2-1 mod 11 6
Thus, kP1 = 3 x (4, 2) = (2, 10). The multiplier k (i.e., 3 in the example) is known as a scalar.
Note 5 –The addition of points on an elliptic curve over Zp does not have the nice geometric interpretation that it does on an elliptic curve over the reals. However, the same formulas are used to define addition.
Note 6 –The addition operation in ECC is the counterpart of modular multiplication in RSA, and multiple addition is the counterpart of modular exponentiation.
Note 7 –To form a cryptographic system using elliptic curves, we need to have a “hard problem” corresponding to factoring the product of two primes or taking the discrete logarithm. For example consider the equation Q = kP, where Q, P E and k < p. It is relatively easy to calculate Q given k and P, but it is relatively hard to determine k given Q and P.
4- The ECC Diffie–Hellman Algorithm
Step 1 – Pick a prime number p and elliptic curve parameters a and b for equation
y23 + a + b ( mod p )
Pick a generator point P = (1, y1) in E.
The integers a and b, prime number p, and generator point P are parameters of the cryptosystem known to all participants.
Step 2 – Alice selects an integer dA and generates a public key QA = dA x P. The key QA is a point in E. dA is Alice’s private key.
Step 3 – Similarly, Bob selects a private key dB and computes a public key QB = dB x P.
Step 4 – Alice generates the secret key K as
K = dA x QB = dA x dB x P
Independently, Bob also generates the secret key K as
K = dB x QA = dB x dA x P
That is the same thing Alice computed.
Note 8 – Since that secret key K is another point on the elliptic curve, and we need just a number,. Alice and Bob need to decide beforehand which coordinates of or y to use. The most common way is to use the x-coordinate, and ignore the y-coordinate.
Example 2 – Take p = 211, a = 0, b = -4, and P = ( 2, 2 ). Alice selects dA = 121 as her private key. So her public key is :
QA = dA x P 121 x ( 2, 2 ) mod 211 = ( 115 , 48 )
Bob picks dB = 203 as his private key. His public key can be computed as:
QB = dB x P = 203 x (2, 2) mod 211 ( 130, 203 ).
Therefore the shared secret key K is
K = dA x QB = dB x QA = 121 x ( 130 , 203 ) = 203 x ( 115, 48 ) = (161, 169).
Note 9 – The security of ECC depends on how difficult it is to determine k given kP and P. This is referred to as the elliptic curve logarithm problem. It can be shown that a considerably smaller key size can be used for ECC compared to RSA. Furthermore, for equal key lengths, the computational efforts required for ECC and RSA is comparable. Thus it appears that there is a computational advantage to using ECC with a shorter key length than a comparably secure RSA.
Oct. 15, 031