chap7 2 ecc intro

Elliptic Curve Cryptography

Introduction

Additi f P i t i Elli ti CAddition of Points in Elliptic Curve

Key Exchangey g

Encryption

Elliptic Curve CryptographyElliptic Curve Cryptography

• Majority of public‐key crypto (RSA, D‐H) use either integer or polynomial arithmetic with g p yvery large numbers/polynomials.

• This imposes a significant load in storing and• This imposes a significant load in storing and processing keys and messages.

• An alternative is to use elliptic curves (ECC).

• ECC offers the same security with smaller bit• ECC offers the same security with smaller bit sizes

Key SizeKey Size

• Symmetric vs public‐key vs ECC key sizes

Symmetric Public‐key ECC

(40 bits) — —

56 bits (400 bits) —

64 bits 512 bits —

80 bits 768 bits —80 bits 768 bits —

90 bits 1024 bits 160 bits




Elliptic Curves over Real NumbersElliptic Curves over Real Numbers• An elliptic curve is defined by an equation in two variables x & y with• An elliptic curve is defined by an equation in two variables x & y, with

coefficients

• The general equation for an elliptic curve is

Elli ti l b i l l f lli ti• Elliptic curves over real numbers use a special class of elliptic curves of the form– y2 = x3 + ax + b (equation 1)– where x, y, a, b are all real numbers– also define zero point O (also known as infinity point)– If the curve has no repeated factors, then the following condition is true:

4a3 +27b2 <> 0

• The set of points E(a,b) which consists of all points (x,y) satisfy the (equation 1) together with the element O.( q ) g

Examples of Real Elliptic CurvesExamples of Real Elliptic Curves

b l h ll h 2 3 d 2• Figures below shows two elliptic curves with equations y2 = x3 − 4x and y2

= x3 − 1. Both are nonsingular. However, the first has three real roots (x = −2, x = 0, and x = 2), but the second has only one real root (x = 1) and two imaginary ones.

Checking a Point on the CurveChecking a Point on the Curve

• Is (4,7) a point on the elliptic curve y2= x3‐ 5x + 5 over real numbers?

• Yes since the equation holds true for x = 4 and• Yes, since the equation holds true for x = 4 and y = 7: ( )2 ( )3 ( )(7)2= (4)3‐ 5(4) + 5 49 = 64 ‐ 20 + 5

49 = 49

Elliptic Curve Groups over FElliptic Curve Groups over FpE i l f h i h h fi i• Essential property for cryptography is that a group has a finite number of points.

• Calculations over the real numbers are slow and inaccurateCalculations over the real numbers are slow and inaccuratedue to round‐off error.

• Cryptographic applications require fast and precise arithmetic; thus elliptic curve groups over the finite fields of Fp and F2mare used in practice.

• Field F uses the numbers from 0 to p 1 and computations• Field Fp uses the numbers from 0 to p ‐ 1, and computations end by taking the modulo of p. – For example, in F23 the field is composed of integers from 0 to p , 23 p g

22, and any operation within this field will result in an integer also between 0 and 22.

Finite Elliptic CurvesFinite Elliptic Curves

• Elliptic curve cryptography uses curves whose variables & coefficients are finite

• have two families commonly used:i E ( b) d fi d Z– prime curves Ep(a,b) defined over Zp

• use integers modulo a prime

• best in software

– binary curves E2m(a,b) defined over GF(2m)• use polynomials with binary coefficients

• best in hardware

FFpA lli ti ith th d l i fi ld f F b f d b h i• An elliptic curve with the underlying field of Fp can be formed by choosing the variables a and b within the field of Fp. The elliptic curve includes all points (x, y) which satisfy the elliptic curve equation modulo p (where xand y are numbers in F )and y are numbers in Fp).

• For example: y2 mod p = x3 + ax + bmod p has an underlying field of Fp if aand b are in Fp. The curve can be noted as Ep(a,b).

• If x3 + ax + b contains no repeating factors (or, equivalently, if 4a3 + 27b2mod p <> 0), then the elliptic curve can be used to form a group.

• An elliptic curve group over Fp consists of the points on the corresponding pelliptic curve, together with a special point O called the point at infinity. There are finitely many points on such an elliptic curve.

Checking Points on Elliptic CurveChecking Points on Elliptic Curve• Consider an elliptic curve E13(1,1)over the field F23. • Since p = 13, a = 1 and b = 1, the elliptic curve equation is y2 = (x3 + x + 1)

mod 13. The point (4, 2) satisfies this equation since: y2 mod p = x3 + x + 1mod p22 mod 23 = (43 + 4 + 1) mod 13 = 17 mod 134 mod 23 = 4 mod 23

• There are 16 points which satisfy the above equation:

Inverse of a PointInverse of a Point

N t th t th i t• Note that there is two points for every x value. Even though the graph

• The additive inverse of a point (x,y) is (x, ‐y). It is also known as a negative point

seems random, there is still symmetry about y =

6 5

negative point

• The inverse of (1,4) is (1, 9)

6.5.

ExerciseExercise

• Does the elliptic curve equation y2 = x3 +10x +5 define a group over F17?g p 17

• a = 10, b = 5, p = 17

4a3 +27b2 mod 17 = 4(10)3 +27(5)2 mod 17

= 4675 mod 17

= 0

• Therefore the curve does not define a group.

Activity 1Activity 1

• Does the elliptic curve equationDoes the elliptic curve equation

y2 = x3 +x +6

define a group over F11?

• Write the equation in terms of Ep(a,b).

Addition of Points in EC

Addition Rules over an Elliptic CurveAddition Rules over an Elliptic Curve

P Q R O if P Q d R th i t lli ti th t li• P+Q+R = O if P, Q and R are three points on an elliptic curve that lie on a straight line.

• O = ‐O; P+O =P: then O is an additive identity

• The negative of a point P is the point with the same x coordinate but theThe negative of a point P is the point with the same x coordinate but the negative of the y coordinate (i.e. if P1 = (x,y) and P2 = (x, ‐y), then P1 = ‐P2.

T dd 2 i t P d Q ith diff t di t• To add 2 point P and Q with different coordinates:– Q+R+P1 = O, Therefore Q+R = ‐P1

• To double a point Q, draw the tangent line and find the other point of intersection S. Then Q+Q = 2Q = ‐S (Note: Q+R+S = O, If R=Q, then 2Q =‐S).

Adding 2 points: P+QAdding 2 points: P+Q

R ll th t i f th t h ith th lli ti• Recall that a is one of the parameters chosen with the elliptic curve and that λ is the tangent on the point P.

• The negative of the point P = (xP, yP) is the point ‐P = (xP, ‐yP mod p).

• If P = (x1, y1) and Q = (x2, y2) are distinct points such that P is not ‐Q, then P + Q = R where (meaning have different coordinates)

λ = (y2 – y1) / (x2 – x1) mod px3 = λ 2 – x1 – x2 mod py = y + λ(x x ) mod py3 = ‐y1 + λ(x1 – x3) mod p

• Note that λ is the slope of the line through P and Q.

Doubling the point PDoubling the point P

• Provided that y1 is not 0, 2P = R where

λ = (3x12 + a) / (2y1 ) mod p x3 = λ 2 ‐ 2x1 mod px3 = λ 2x1 mod py3 = ‐y1 + λ(x1 – x3) mod p

Example: The Negative PointsExample: The Negative Points

• What are the negatives of the following elliptic curve points over F17? p p 17P=(5,8) Q=(3,0) R=(0,6)

• The negative of a point P = (xP, yP) is the point ( )‐P = (xP, ‐yP mod p).

• Thus ‐P=(5,9) ‐Q=(3,0) ‐R=(0,11)

Example: Adding 2 PointsExample: Adding 2 Points

I h lli i d fi d b 2 3 7 F• In the elliptic curve group defined by y2 = x3 + x + 7 over F17, what is P + Q if P = (2,0) and Q = (1,3)?

• λ = (y – y ) / (x – x ) mod pλ = (y2 y1) / (x2 x1) mod p= (‐3) / 1 mod 17 = ‐3 mod 17 = 14 λ 2 dx3 = λ 2 – x1 – x2 mod p= 196 ‐ 2 ‐ 1 mod 17 = 193 mod 17 = 6

y3 = ‐y1 + λ(x1 – x3) mod p = 0 + 14*(2 ‐ 6) mod 17 = 56 mod 17 = 12= ‐56 mod 17 = 12

• Thus P + Q = (6,12) Q ( , )

Example: Doubling the PointsExample: Doubling the Points

I h lli i d fi d b 2 3 7 F• In the elliptic curve group defined by y2 = x3 + x + 7 over F17, what is 2P if P = (1, 3)?

• λ = (3x12 + a) / (2y1 ) mod p = (3 + 1) * 6‐1 mod 17 4 * 3 d 17 12= 4 * 3 mod 17 = 12

x3 = λ 2 ‐ 2x1 mod p= 144 ‐ 2 mod 17 = 142 mod 17 = 6

y3 = ‐y1 + λ(x1 – x3) mod p = 3 + 12 * (1 6) mod 17 = 63 mod 17 = 5= ‐3 + 12 * (1 ‐ 6) mod 17 = ‐63 mod 17 = 5

• Thus 2P = (6 5)Thus 2P = (6,5)

Elliptic Curve: y2 = x3+x+6(mod11)Elliptic Curve: y2 = x3+x+6(mod11)

• All the points on the curve are:

(2, 4) (7, 2)(2, 4) (7, 2)

(2, 7) (7, 9)

(3, 5) (8, 3)

(3 6) (8 8)(3, 6) (8, 8)

(5, 2) (10, 2)

(5, 9) (10, 9)


• Given P = (2, 4) and Q = (3, 5)

• Calculate :Calculate :– P+Q

– 2P

– 3P

– 4P

D 4P 3P P 2P 2P?– Does 4P = 3P + P = 2P + 2P?

Generating Public and Private KeysGenerating Public and Private Keys

• can do key exchange analogous to D‐H• users select a suitable curve E (a,b)users select a suitable curve Ep(a,b)

• select base point G=(x1,y1) with large order n h hsuch that nG=O

• A & B select private keys nA<n, nB<np y A , B

• compute public keys: PA=nA×G, PB=nB×G

Determining nDetermining n

Starting with P = (2, 7), 2P = P + P

λ = (3 * 22 + 1) / (2 * 7) 3-1 mod 11

= (13 mod 11)/ (14 mod 11)

= (2 mod 11) / (3 mod 11)= 39 mod 1132 mod 11=934 mod 11=81 mod 11=4

= 2 *4 = 8

x3 = 82 ‐ 2 ‐ 2 (mod 11)

38 mod 11=16 mod 11=539 mod 11= (38 × 3) mod 11

= (5 × 3) mod 11= 5

y3 = 8 (2 ‐ 5) ‐ 7 (mod 11)

= 15 mod 11= 4

= 2

Therefore, P + P = (2, 7) + (2, 7) = (5, 2).

Multiples of PMultiples of P

2P = (5, 2) 3P = (8, 3)

4P = (10, 2) 5P = (3, 6)4P (10, 2) 5P (3, 6)

6P = (7, 9) 7P = (7, 2)

8P = (3, 5) 9P = (10, 9)

10P = (8 8) 11P = (5 9)10P = (8, 8) 11P = (5, 9)

12P = (2, 4) 13P = O

RECALLING KEY GENERATIONRECALLING KEY GENERATION

Gl b l P bli El tGlobal Public Elements:Ep(a,b) – elliptic curve G – point on elliptic curve

Alice Bob• Key Generation

– Alice selects a private no.

• Key Generation

– Bob selects a private no.p• nA < n

– Alice calculates her public k

p• nB < n

– Bob calculates his public kkey

• PA = nA × Gkey

• PB = nB × G

Key GenerationKey Generation

• Recall equation y2 = x3+x+6(mod11)• Select G(x1, y1) = (2, 7)

and Ep(a,b) = E11(1, 6) where they are known to both parties. • Alice chooses a Private key :Alice chooses a Private key :

– Select integer nA = 7 < n, where nG = O. – In this case n = 13

• Alice’s Public key :– PA = nA * G ∈ E11(1,6)

= 7 * (2 7)= 7 (2, 7) = (7, 2)


• You are Bob:

• Choose a private keyChoose a private key

• Calculate your public key– PB = nB × G

ECC Encryption/DecryptionECC Encryption/Decryption

• several alternatives, will consider simplest

• must first encode any message M as a point on the elliptic Pcurve Pm

• select suitable curve & point G as in D‐Hh h i t k Message M• each user chooses private key nA<n

• and computes public key PA=nA×Gk d

Message M

Bob’s public key

• to encrypt Pm : Cm={kG, Pm+k PB}, k random

• decrypt Cm compute: Pm+kPB–nB(kG) = Pm+k(nBG)–nB(kG) = Pm

Bob’s private key

Example: EncryptionExample: Encryption

• Encrypt pm = (10, 9), k = 3, G(2, 7) Private key, KBpr = 7 ( < 13)y Bpr ( )Public key KBpu = (7, 2)

• c = {kG p + kK }• cm = {kG, pm + kKBpu }= {3(2, 7), (10, 9) + 3(7, 2)}{( ) ( ) ( )}= {(8, 3), (10, 9) + (3, 5)}

= {(8, 3), (10, 2)}

Example: DecryptionExample: Decryption

• Procedure:– Receiver multiplies the first point in the pair by p p p ythe receiver’s secret keyKBpr(kG) = 7*(8, 3) = (3, 5)Bpr( ) ( , ) ( , )

– Subtracts the result from the second point.

(p + kK ) K (kG)(pm + kKBpu) ‐ KBpr(kG)= Pm +k(KBprG) – KBpr(kG) = Pm

(10, 2) – (3, 5) = (10,2) + (– (3, 5)) = ?What is the +ve value of this?What is the ve value of this?

y2 = x3+x+6(mod11)

Security StrengthSecurity Strength

h f d d h d k k d• The security of ECC depends on how to determine k given kP and P. This is referred to as elliptic curve logarithm problem.

• The fastest method for taking the elliptic curve logarithm is “Pollard g p grho method”.

• Compared to factoring, can use much smaller key sizes than with RSA etcetc

• Eg. To solve logarithm problem for ECC with key size = 150 bit takes more time than factoring problem of 512 bit key size of RSA.

• For equivalent key lengths, computations are roughly equivalent.

• Hence for similar security ECC offers significant computational advantagesadvantages.

Discrete Log Problem Over ECC (ECDLP)Discrete Log Problem Over ECC (ECDLP)

Th ECDLP i i l t d f th f d• The ECDLP is, given elements r and q of the group formed over elliptic curve find a number k such that r = kq.

• Example:p– In the elliptic curve defined by y2 = x3 + 9x + 17, what is the

discrete log k of Q(4,5) to the base P(16,5) ?– One way to find k is to compute multiples of P until Q is foundOne way to find k is to compute multiples of P until Q is found.– P = (16,5) 2P = (20,20) 3P = (14,14)....8P = (12,17) 9P = (4,5)– Since, 9P = (4,5) = Q, the discrete log of Q is k = 9.

• In a real application, k would be large enough such that it would be infeasible to determine k in this manner.

Advantages of ECC over other Cryptosystem

• For RSA which depends on Integer Factorization, DSA which depends upon p pFermat’s little Theorem and Diffie‐Hellman protocol which depends upon DLPprotocol which depends upon DLP– time required to crack them grows subexponentially (a√n) with the length of keysubexponentially (a√n) with the length of key.

• For Elliptic Curve Discrete Log Problem (ECDLP) the time required to crack it grows fully exponentially an.y p y

Industrial AdvantagesIndustrial Advantages

• ECC offers equivalent security with much smaller key size.y

• Practical advantages of ECC :F t– Faster

– Low power consumption

– Low memory usage

– Low CPU utilizationLow CPU utilization

– Benefits of over its competitors increases with increase in the security needsincrease in the security needs.

chap7 2 ecc intro

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