Elliptical curve cryptography

Elliptic Curve Cryptography
Presented by
M. Baranitharan
Kings College of Engineering
Tanjore

Outline of the Talk…
• Introduction to Elliptic Curves
• Elliptic Curve Cryptosystems (ECC)
• Implementation of ECC in Binary Fields

Introduction to Elliptic Curves

Lets start with a puzzle…
• What is the number of balls that may be
piled as a square pyramid and also
rearranged into a square array?
• Soln: Let x be the height of the pyramid…
Thus,
We also want this to be a square:
Hence,
2 2 2 2 ( 1)(2 1)
1 2 3 ...
6
x x x
x
+ +
+ + + + =
2 ( 1)(2 1)
6
x x x
y
+ +
=

Graphical Representation
X axis
Y axis
Curves of this nature
are called ELLIPTIC
CURVES

Method of Diophantus
• Uses a set of known points to produce new points
• (0,0) and (1,1) are two trivial solutions
• Equation of line through these points is y=x.
• Intersecting with the curve and rearranging terms:
• We know that 1 + 0 + x = 3/2 =>
x = ½ and y = ½
• Using symmetry of the curve we also have (1/2,-1/2)
as another solution
3 23 1
0
2 2
x x x− + =

Diophantus’ Method
• Consider the line through (1/2,-1/2) and (1,1) =>
y=3x-2
• Intersecting with the curve we have:
• Thus ½ + 1 + x = 51/2 or x = 24 and y=70
• Thus if we have 4900 balls we may arrange
them in either way
3 251
... 0
2
x x− + =

Elliptic curves in Cryptography
• Elliptic Curve (EC) systems as applied to
cryptography were first proposed in 1985
independently by Neal Koblitz and Victor
Miller.
• The discrete logarithm problem on elliptic
curve groups is believed to be more
difficult than the corresponding problem in
(the multiplicative group of nonzero
elements of) the underlying finite field.

Discrete Logarithms
in Finite Fields
Alice Bob
Pick secret, random
X from F
Pick secret, random
Y from F
gy
mod p
gx
mod p
Compute k=(gy
)x
=gxy
mod p
Compute k=(gx
)y
=gxy
mod p
Eve has to compute gxy
from gx
and gy
without knowing x and y…
She faces the Discrete Logarithm Problem in finite fields
F={1,2,3,…,p-1}

Elliptic Curve on a finite set of
Integers
• Consider y2
= x3
+ 2x + 3 (mod 5)
x = 0 ⇒ y2
= 3 ⇒ no solution (mod 5)
x = 1 ⇒ y2
= 6 = 1 ⇒ y = 1,4 (mod 5)
x = 2 ⇒ y2
= 15 = 0 ⇒ y = 0 (mod 5)
x = 3 ⇒ y2
= 36 = 1 ⇒ y = 1,4 (mod 5)
x = 4 ⇒ y2
= 75 = 0 ⇒ y = 0 (mod 5)
• Then points on the elliptic curve are
(1,1) (1,4) (2,0) (3,1) (3,4) (4,0)
and the point at infinity: ∞
Using the finite fields we can form an Elliptic Curve Group
where we also have a DLP problem which is harder to solve…

Definition of Elliptic curves
• An elliptic curve over a field K is a nonsingular
cubic curve in two variables, f(x,y) =0 with a
rational point (which may be a point at infinity).
• The field K is usually taken to be the complex
numbers, reals, rationals, algebraic extensions
of rationals, p-adic numbers, or a finite field.
• Elliptic curves groups for cryptography are
examined with the underlying fields of Fp (where
p>3 is a prime) and F2
m
(a binary representation
with 2m
elements).

General form of a EC
• An elliptic curve is a plane curve defined by an
equation of the form
baxxy ++= 32
Examples

Weierstrass Equation
• A two variable equation F(x,y)=0, forms a curve
in the plane. We are seeking geometric
arithmetic methods to find solutions
• Generalized Weierstrass Equation of elliptic
curves:
2 2 2
1 3 2 4 6y a xy a y x a x a x a+ + = + + +
Here, A, B, x and y all belong to a field of say rational
numbers, complex numbers, finite fields (Fp) or
Galois Fields (GF(2n
)).

• If Characteristic field is not 2:
• If Characteristics of field is neither 2 nor 3:
22
2 3 23 31 1
2 4 6
2 3 ' 2 ' '
1 2 4 6
( ) ( ) ( )
2 2 4 4
a aa x a
y x a x a x a
y x a x a x a
+ + = + + + + +
⇒ = + + +
'
1 2
2 3
1 1 1
/3x x a
y x Ax B
= +
⇒ = + +

Points on the Elliptic Curve (EC)
• Elliptic Curve over field L
• It is useful to add the point at infinity
• The point is sitting at the top of the y-axis
and any line is said to pass through the
point when it is vertical
• It is both the top and at the bottom of the
y-axis
2 3
( ) { } {( , ) | ... ...}E L x y L L y x= ∞ ∪ ∈ × + = +

The Abelian Group
• P + Q = Q + P (commutativity)
• (P + Q) + R = P + (Q + R) (associativity)
• P + O = O + P = P (existence of an identity element)
• there exists ( − P) such that − P + P = P + ( − P)
= O (existence of inverses)
Given two points P,Q in E(Fp), there is a third
point, denoted by P+Q on E(Fp), and the
following relations hold for all P,Q,R in E(Fp)

Elliptic Curve Picture
• Consider elliptic curve
E: y2
= x3
- x + 1
• If P1 and P2 are on E, we can
define
P3 = P1 + P2
as shown in picture
• Addition is all we need
P1
P2
P3
x
y

Addition in Affine Co-ordinates
x
y
1 1 2 2
3 3
( , ), ( , )
( ) ( , )
P x y Q x y
R P Q x y
= =
= + =
y=m(x-x1)+y1
2 1
2 1
2
1 1
3 2 2
2
3 1
3 1 2
;
To find the inters
( ( ) )
, 0
,
( )
y y
m
x x
m x x y
or x m x
So x m x
y m x x
−
=
−
− + =
= − +
= − −
⇒= −
Let, P≠Q,
y2
=x3
+Ax+B

Doubling of a point
• Let, P=Q
• What happens when P2=∞?
2
2
1
1
1 1 2
3 2 2
2
3 1 3 1 3 1
2 3
3
2
, 0 (since then P +P = ):
0 ...
2 , ( )
dy
y x A
dx
dy x A
m
dx y
If y
x m x
x m x y m x x y
= +
+
⇒ = =
≠ ∞
∴ = − +
⇒ = − = − −

Why do we need the reflection?
P2=O=∞
P1
y
P1=P1+ O=P1

Sum of two points






=
+
≠
−
−
=
21
1
2
1
21
12
12
_
2
3
_
xxfor
y
ax
xxfor
xx
yy
λ
Define for two points P (x1,y1) and
Q (x2,y2) in the Elliptic curve
Then P+Q is given by R(x3,y3) :
1133
213
)( yxxy
xxx
+−=
−−=
λ
λ

P+P = 2P
Point at infinity O
As a result of the above case P=O+P
O is called the additive identity of
the elliptic curve group.
Hence all elliptic curves have an
additive identity O.

Projective Co-ordinates
• Two-dimensional projective space over K
is given by the equivalence classes of triples
(x,y,z) with x,y z in K and at least one of x, y,
z nonzero.
• Two triples (x1,y1,z1) and (x2,y2,z2) are said to
be equivalent if there exists a non-zero
element λ in K, st:
– (x1,y1,z1) = (λx2, λy2, λz2)
– The equivalence class depends only the ratios
and hence is denoted by (x:y:z)
2
KP

Projective Co-ordinates
• If z≠0, (x:y:z)=(x/z:y/z:1)
• What is z=0? We obtain the point at infinity.
• The two dimensional affine plane over K:
2
2 2
{( , ) }
Hence using,
( , ) ( : :1)
K
K K
A x y K K
x y X Y
A P
= ∈ ×
→
⇒ =
There are advantages with projective co-ordinates
from the implementation point of view

Singularity
• For an elliptic curve y2
=f(x), define
F(x,y)=y2
-F(x). A singularity of the EC is a pt
(x0,y0) such that:
0 0 0 0
0 0
0 0
( , ) ( , ) 0
,2 '( ) 0
, ( ) '( )
f has a double root
F F
x y x y
x y
or y f x
or f x f x
∂ ∂
= =
∂ ∂
= − =
=
∴
It is usual to assume the EC has no singular points

If Characteristics of
field is not 3:
1. Hence condition
for no singularity is
4A3
+27B2
≠0
2. Generally, EC
curves have no
singularity
0 0 0 0
0 0
0 0
2 3
3 2
2
4 2
2 2
2
2
2
3 2
( , ) ( , ) 0
,2 '( ) 0
, ( ) '( )
f has a double root
For double roots,
3 0
/3.
Also, +Bx=0,
0
9 3
2
9
2
3( ) 0
9
4 27 0
F F
x y x y
x y
or y f x
or f x f x
y x Ax B
x Ax B x A
x A
x Ax
A A
Bx
A
x
B
A
A
B
A B
∂ ∂
= =
∂ ∂
= − =
=
∴
= + +
+ + = + =
⇒ = −
+
⇒ − + =
⇒ =
⇒ + =
⇒ + =
2 3
( )y f x x Ax B= = + +

Elliptic Curves in Characteristic 2
• Generalized Equation:
• If a1 is not 0, this reduces to the form:
• If a1 is 0, the reduced form is:
• Note that the form cannot be:
2 3 2
y xy x Ax B+ = + +
2 3 2
1 3 2 4 6y a xy a y x a x a x a+ + = + + +
2 3
y Ay x Bx C+ = + +
2 3
y x Ax B= + +

Outline of the Talk…
• Introduction to Elliptic Curves
• Elliptic Curve Cryptosystems
• Implementation of ECC in Binary Fields

Elliptic Curve Cryptosystems
(ECC)

Public-Key Cryptosystems
Secrecy: Only B can Decrypt
the message
Authentication: Only A can
generate the encrypted message

What Is Elliptic Curve
Cryptography (ECC)?
• Elliptic curve cryptography [ECC] is a public-
key cryptosystem just like RSA, Rabin, and El
Gamal.
• Every user has a public and a private key.
– Public key is used for encryption/signature
verification.
– Private key is used for decryption/signature
generation.
• Elliptic curves are used as an extension to other
current cryptosystems.
– Elliptic Curve Diffie-Hellman Key Exchange
– Elliptic Curve Digital Signature Algorithm

Using Elliptic Curves In
Cryptography
• The central part of any cryptosystem involving
elliptic curves is the elliptic group.
• All public-key cryptosystems have some
underlying mathematical operation.
– RSA has exponentiation (raising the message or
ciphertext to the public or private values)
– ECC has point multiplication (repeated addition of two
points).

Generic Procedures of ECC
• Both parties agree to some publicly-known data items
– The elliptic curve equation
• values of a and b
• prime, p
– The elliptic group computed from the elliptic curve equation
– A base point, B, taken from the elliptic group
• Similar to the generator used in current cryptosystems
• Each user generates their public/private key pair
– Private Key = an integer, x, selected from the interval [1, p-1]
– Public Key = product, Q, of private key and base point
• (Q = x*B)

Example – Elliptic Curve
Cryptosystem Analog to El Gamal
• Suppose Alice wants to send to Bob an
encrypted message.
– Both agree on a base point, B.
– Alice and Bob create public/private keys.
• Alice
– Private Key = a
– Public Key = PA = a* B
• Bob
– Private Key = b
– Public Key = PB = b * B
– Alice takes plaintext message, M, and encodes it onto
a point, PM, from the elliptic group

Cryptosystem Analog to El Gamal
– Alice chooses another random integer, k from the
interval [1, p-1]
– The ciphertext is a pair of points
• PC = [ (kB), (PM + kPB) ]
– To decrypt, Bob computes the product of the first
point from PC and his private key, b
• b * (kB)
– Bob then takes this product and subtracts it from the
second point from PC
• (PM + kPB) – [b(kB)] = PM + k(bB) – b(kB) = PM
– Bob then decodes PM to get the message, M.

Example – Compare to El Gamal
– The ciphertext is a pair of points
• PC = [ (kB), (PM + kPB) ]
– The ciphertext in El Gamal is also a pair.
• C = (gk
mod p, mPB
k
mod p)
--------------------------------------------------------------------------
– Bob then takes this product and subtracts it from the
second point from PC
• (PM + kPB) – [b(kB)] = PM + k(bB) – b(kB) = PM
– In El Gamal, Bob takes the quotient of the second
value and the first value raised to Bob’s private value
• m = mPB
k
/ (gk
)b
= mgk*b
/ gk*b
= m

Diffie-Hellman (DH) Key Exchange

ECC Diffie-Hellman
• Public: Elliptic curve and point B=(x,y) on curve
• Secret: Alice’s a and Bob’s b
Alice, A Bob, B
a(x,y)
b(x,y)
• Alice computes a(b(x,y))
• Bob computes b(a(x,y))
• These are the same since ab = ba

Diffie-Hellman Exchange
• Alice and Bob want to agree on a shared key.
– Alice and Bob compute their public and private keys.
• Alice
» Private Key = a
» Public Key = PA = a* B
• Bob
» Private Key = b
» Public Key = PB = b * B
– Alice and Bob send each other their public keys.
– Both take the product of their private key and the other user’s
public key.
• Alice  KAB = a(bB)
• Bob  KAB = b(aB)
• Shared Secret Key = KAB = abB

Why use ECC?
• How do we analyze Cryptosystems?
– How difficult is the underlying problem that it
is based upon
• RSA – Integer Factorization
• DH – Discrete Logarithms
• ECC - Elliptic Curve Discrete Logarithm problem
– How do we measure difficulty?
• We examine the algorithms used to solve these
problems

Security of ECC
• To protect a 128 bit
AES key it would take
a:
– RSA Key Size: 3072
bits
– ECC Key Size: 256
bits
• How do we
strengthen RSA?
– Increase the key
length
• Impractical?

Applications of ECC
• Many devices are small and have limited
storage and computational power
• Where can we apply ECC?
– Wireless communication devices
– Smart cards
– Web servers that need to handle many encryption
sessions
– Any application where security is needed but
lacks the power, storage and computational
power that is necessary for our current
cryptosystems

Benefits of ECC
• Same benefits of the other cryptosystems:
confidentiality, integrity, authentication and
non-repudiation but…
• Shorter key lengths
– Encryption, Decryption and Signature
Verification speed up
– Storage and bandwidth savings

Summary of ECC
• “Hard problem” analogous to discrete log
– Q=kP, where Q,P belong to a prime curve
given k,P  “easy” to compute Q
given Q,P  “hard” to find k
– known as the elliptic curve logarithm problem
• k must be large enough
• ECC security relies on elliptic curve
logarithm problem
– compared to factoring, can use much smaller key sizes than with RSA
etc
 for similar security ECC offers significant
computational advantages

Implementation of ECC in
Binary Fields

Sub-Topics
1. Scalar Multiplication: LSB first vs MSB first
2. Montgomery Technique of Scalar Multiplication
3. Fast Scalar Multiplication without pre-
computation.
4. Lopez and Dahab Projective Transformation to
Reduce Inverters
5. Mixed Coordinates
6. Parallelization Techniques
7. Half and Add Technique for Scalar
Multiplication

ECC operations: Hierarchy
ECC
Point
multiplication:
kP
Group operation:
point add/double
Finite field arithmetic: multiplication,
addition, subtraction, inversion, …
Parallelize
the
architectures
Level 0
Level 1
Level 2
Level 3

Scalar Multiplication: MSB first
• Require k=(km-1,km-2,…,k0)2, km=1
• Compute Q=kP
– Q=P
– For i=m-2 to 0
• Q=2Q
• If ki=1 then
– Q=Q+P
• End if
– End for
– Return Q
Sequential Algorithm
Requires m point doublings and
(m-1)/2 point additions on the
average

Example
• Compute 7P:
– 7=(111)2
– 7P=2(2(P)+P)+P=> 2 iterations are required
– Principle: First double and then add
(accumulate)
• Compute 6P:
– 6=(110)2
– 6P=2(2(P)+P)

Scalar Multiplication: LSB first
• Require k=(km-1,km-2,…,k0)2, km=1
• Compute Q=kP
– Q=0, R=P
– For i=0 to m-1
• If ki=1 then
– Q=Q+R
• End if
• R=2R
– End for
– Return Q
Can Parallelize…
What you are doubling and what
you are accumulating are
different…
On the average m/2 point
Additions and m/2 point
doublings

Example
• Compute 7P, 7=(111)2, Q=0, R=P
– Q=Q+R=0+P=P, R=2R=2P
– Q=P+2P=3P, R=4P
– Q=7P, R=8P
• Compute 6P, 6=(110)2, Q=0, R=P
– Q=0, R=2R=2P
– Q=0+2P=2P, R=4P
– Q=2P+4P=6P, R=8P

Compute 31P…
1. Q=2P
2. Q=3P
3. Q=6P
4. Q=7P
5. Q=14P
6. Q=15P
7. Q=30P
8. Q=31P
1. Q=P, R=2P
2. Q=3P, R=4P
3. Q=7P, R=8P
4. Q=15P, R=16P
5. Q=31P, R=32P
31=(11111)2
MSB First LSB First

Weierstrass Point Addition
• Let, P=(x1,y1) be a point on the curve.
• -P=(x1,x1+y1)
• Let, R=P+Q=(x3,y3)
2 3 2
, ( , ) (2 ) (2 )m m
y xy x ax b x y GF GF+ = + + ∈ ×
2
1 2 1 2
1 2
1 2 1 2
3 2
1 2
1
1 2
1 3 3 1
1 2
3 2 1
1 1 3 3
1
;
;
( ) ;
( ) ;
y y y y
x x a P Q
x x x x
x b
x P Q
x
y y
x x x y P Q
x x
y y
x x x x P Q
x
 + +
 + + + + ≠ ÷
+ + 

= 
+ =


 +
+ + + ≠ ÷
+ 
= 
+ + + =


1. Point addition and doubling
each require 1 inversion
& 2 multiplications
2. We neglect the costs of
squaring and addition
3. Montgomery noticed that the
x-coordinate of 2P does not
depend on the y-coordinate of
P

Montgomery’s method to perform scalar
multiplication
• Input: k>0, P
• Output: Q=kP
1. Set k<-(kl-1,…,k1,k0)2
2. Set P1=P, P2=2P
3. For i from l-2 to 0
If ki=1,
Set P1=P1+P2, P2=2P2
else
Set P2=P2+P1, P1=2P1
1. Return Q=P1
Invariant Property:
P=P2-P1
Question: How to implement the
Operation efficiently?

Example
Compute 7P
• 7=(111)2
• Initialization:
P1=P; P2=2P
• Steps:
– P1=3P, P2=4P
– P1=7P, P2=8P
Compute 6P
• 7=(110)2
• Initialization:
P1=P; P2=2P
• Steps:
– P1=3P, P2=4P
– P2=7P, P1=6P

Fast Multiplication on EC
without pre-computation

Result-1
• Let P1 = (x1,y1) and P2=(x2,y2) be elliptic points.
Then the x-coordinate of P1+P2, x3 can be
computed as:
2 2
1 2 2 1 1 2 2 1
3 2
1 2( )
x y x y x x x x
x
x x
+ + +
=
+
Hint: Remember that the field has a characteristic 2
and that P1 and P2 are points on the curve

Result-2
• Let P=(x,y), P1 = (x1,y1) and P2=(x2,y2) be elliptic
points. Let P=P2-P1 be an invariant.
Then the x-coordinate of P1+P2, x3 can be
computed in terms of the x-coordinates
as: 2
1 1
1 2
1 2 1 2
3 2
1 1 22
1
;
;
x x
x P P
x x x x
x b
x P P
x
  
 + + ≠ ÷
+ +  

= 
+ =



Result-3
Let P=(x,y), P1=(x1,y1) and P2=(x2,y2) be elliptic
points. Assume that P2-P1=P and x is not 0.
Then the y-coordinates of P1 can be
expressed in terms of P, and the
x-coordinates of P1 and P2 as follows:
2
1 1 1 2( ){( )( ) }/y x x x x x x x y x y= + + + + + +

Final Algorithm
Input: k>0, P=(x,y)
Output: Q=kP
1. If k=0 or x=0 then output(0,0)
2. Set k = (kl-1,kl-2,…,k0)2
3. Set x1=x, x2=x2
+b/x2
1. Set t=x1/(x1+x2)
2. If ki=1,
x1=x+t2
+t, x2=x2
2
+b/x2
2
else
x1=x1
2
+b/x1
2
, x2=x+t2
+t
1. r1=x1+x, r2=x2+x
2. y1=r1(r1r2+x2
+y)/x+y
3. Return Q=(x1,y1)
• #INV:2(l-2)+1;
• #MULT: 2(l-2)+4
• #ADD: 4(l-2)+6
• #SQR: 2(l-2)+2

How to reduce inversions?
1. In affine coordinates Inverses are very
expensive
2. For each inversion requires around 7
multipliers (in hardware designs)
3. Lopez Dahab Projective coordinates:
• (X,Y,Z), Z≠0, maps to (X/Z,Y/Z2
)
• Motivation is to replace inversions by the
multiplication operations and then perform one
inversion at the end (to obtain back the affine
coordinates)
128n ≥

Doubling
• Remember:
• In Projective Coordinates:
2
1 1
1 2
1 2 1 2
3 2
1 1 22
1
;
;
x x
x P P
x x x x
x b
x P P
x
  
 + + ≠ ÷
+ +  

= 
+ =


4 4
1 2 3 1 1
2 2
3 1 1
2
1 2 3 1 2 2 1
3 3 1 2 2 1
, .
.
, ( . . )
. ( . ).( . )
P P X X b Z
Z Z X
P P Z X Z X Z
X x Z X Z X Z
= = +
=
≠ = +
= +
• 2 inverses
• 1 general field
multiplication
• 4 additions
• 2 squarings
• 0 inverses
• 4 general field
multiplications
• 3 additions
• 5 squarings

Montgomery Algorithm
• Input: k>0, P=(x,y)
• Output: Q=kP
• Set k<-(kl-1,…,k1,k0)2
• Set X1=x, Z1=1; X2=x4
+b, Z2=x2
• For i from l-2 to 0
– If ki =1,
Madd(X1,Z1,X2,Z2), Mdouble(X2,Z2)
else
Madd(X2,Z2,X1,Z1), Mdouble(X1,Z1)
• Return Q=(Mxy(X1,Y1,X2,Y2))

Mxy: Projective to Affine
3 1 1
2 1
3 1 1 1 1 2 2 1 2 1 2
/
( / )[( )( ) ( )( )]( )
x X Z
y x X Z X xZ X xZ x y Z Z xZ Z y−
=
= + + + + + +
Requires 10 multiplications and one inverse operation

Final Comparison
Affine Coordinates
Inv: 2logk + 1
Mult: 2logk + 4
Add: 4logk + 6
Sqr: 2logk + 2
Projective Coordinates
Inv: 1
Mult: 6logk + 10
Add: 3logk + 7
Sqr: 5logk + 3
Hence, final decision depends upon the I:M ratio of the finite field operators

Addition in Mixed Coordinates
• Theorem: Let P1=(X1/Z1,Y1/Z1
2
) and
P2=(X2/Z2,Y2/Z2
2
) be two points on the curve. If
Z1=1, then P1+P2=(X3/Z3,Y3/Z3
2
) st.
Number of multiplications are further reduced.
Squaring is increased a bit, but they are cheap in GF(2n
)
Improvement by 10 % if a≠0, otherwise 12 %...
2 2
2 1 2 2 1 2 2 3
2 2
3 1 3
2
3 3 3 3
, , , ,
, ( ),
( )( )
U Z Y Y S Z X X T Z S Z T
V Z X X U T U S Ta
Y V X TU Z Z C
= + = + = =
= = + + +
= + + +

Parallel Strategies for Scalar Point
Multiplication
• Point Doubling
– Cycle 1: T=X1
2
, M=cZ1
2
, Z2=T.Z1
2
– Cycle 1a: X2=T2
+M2
• Point Addition
– Cycle 1: t1=(X1.Z2); t2=(Z1.X2)
– Cycle 1a: M=(t1+t2), Z1=M2
– Cycle 2: N=t1.t2, M=xZ1
– Cycle 2a: X1=M+N
1 multiplier
2 multipliers
We assume that squarings and multiplications with constants can be
performed without multipliers…

Parallelizing Montgomery Algorithm
1. Input: k>0, P=(x,y)
2. Output: Q=kP
3. Set k<-(kl-1,…,k1,k0)2
4. Set X1=x, Z1=1; X2=x4
+b, Z2=x2
If ki =1,
5a) Madd(X1,Z1,X2,Z2), Mdouble(X2,Z2)
else
5b) Madd(X2,Z2,X1,Z1), Mdouble(X1,Z1)
1. Return Q=(Mxy(X1,Y1,X2,Y2))

Looking back at our
Design Hierarchy
ECC
Point
multiplication:
kP
Group operation:
point add/double
Finite field arithmetic: multiplication,
addition, subtraction, inversion, …
Parallelize
the
architectures
Level 0
Level 1
Level 2
Level 3

Parallelizing Strategies
• Parallelize level 1: If we allocate one multiplier
to each of Madd and Mdouble, then we can
parallelize steps 5a and 5b. Thus 4 clock cycles
are required for each iteration. Total time is
nearly 4l.
• Parallelize level 2: If we can parallelize the
underlying Madd and Mdouble, then we cannot
parallelize level 1, if we have constraint of 2
multipliers. So, we have a sequential step 5a
and 5b. Total time is 3l.

Parallelizing Strategies
• Parallelize both the levels: Total time is
2l clock cycles. Require 3 multipliers.
• Thus Montgomery algorithm is highly
parallelizable
• Helpful in high performance designs (low
power, high thoughput etc)

Point Halving
• In 1999 Scroeppel and Knudsen proposed
further speed up
• Idea is to replace point doubling by halving
• Point Halving is three times as fast than
doubling
• The scalar k, has to be expressed in the
negative powers of 2

Computing the Half
• Problem: Let E be the Elliptic Curve, defined
by the equation:
• Let Q=(u,v)=2P
• Compute P=(x,y)
• Remember :
2 3 2
, 0y xy x ax b b+ = + + ≠
2
2
2
( )
b
u x
x
y
v x x u u
x
= +
= + + +

Halving (contd.)
• Thus, we have to solve the above
equations
• λ-representation: (x, λx)
2
2
,
( 1) ( 1)
:
y
Let x
x
v x u x v u
Note u a
λ
λ λ
λ λ
= +
∴ = + + ⇒ = + +
+ = +
Square
Root
Solving
Quadratics

Trace of a point
• Define:
• Properties of Trace:
– Tr(c)=Tr(c2
)=Tr(c)2
, Tr(c) can be 0 or 1
– Tr(c+d)=Tr(c)+Tr(d)
– NIST Curves : Tr(a)=1
– If x,y belongs to the Elliptic Curve, Tr(x)=Tr(a)
1
2 2
( ) ...
m
Tr C C C C
−
= + + +

Computing λ
• The roots of are λ1= λ or λ+1
• Theorem:
2
u aλ λ+ = +
2
Let, ( , ), ( , ) , . 2
ˆand denote / . Let be a solution
ˆto and . Suppose that
ˆ( ) 1. Then if and only if ( ) 0.
P x y Q u v G st Q P
x y x
u a t v u
Tr a Tr t
λ λ
λ λ λ
λ λ
= = ∈ =
= +
+ = + = +
= = =

Halving Algorithm
• Input: (u,v) , Output: (x,y)
1. Solve for λ. Let the root be
2. Compute
3. If Tr(t)=0, then λP= , x=(t+u)1/2
else λP= +1,x=(t)1/2
4. Return (x,λP)
2
u aλ λ+ = + ˆλ
ˆt v uλ= +
ˆλ
ˆλ

Implementation of Trace
• Trace :
• Can be evaluated in O(1) time
• Example: GF(2163
), with reduction polynomial
p(x)=x163
+x7
+x6
+x3
+1, Tr(xi
)=1, iff i=0 or 159.
• Thus, the implementation is only one xor gate
to add the 0th
and the 159th
bits of the register
storing C.
1 1
0 0
( ) ( ) ( )
m m
i i
i i
i i
Tr C Tr c x cTr x
− −
= =
= =∑ ∑

Solving a Quadratic over GF(2m
)
• Solve x2
+x=c+Tr(c), c is an element of GF(2m
)
• Define Half Trace:
2
( 1)/2
2
0
2
2
( )
1. ( ) ( ) ( )
2. ( ) is a root for ( ), as
( ) ( ) ( )
i
m
i
H C C
H C D H C H D
H C x x C Tr C
H C H C C Tr C
−
=
=
+ = +
+ = +
= + +
∑
H(C) gives a root for the quadratic equation. A simple method to find
H(C) requires storage for m elements and m/2 field additions on an average

Obtaining Square Root
• Field squaring in binary field is linear
• Hence squaring can be rephrased as:
– C=MA=A2
• We require to compute D st. D2
=A
• Let, D=M-1
A=> A=MD
• D2
=MD (as M is the squaring matrix)
=M(M-1
A)=A
• Hence, D=(A)1/2

An Example
7 7
10 1
2
9 8 6 2
7
7 7
7 7 7
Compute: 763R , where order of R 1013
10
2 (763) 651(mod 1013) (1010001011)
1 1 1 1
763 ( 1)mod(1013)
2 2 2 2
763 may be computed using the following steps:
1
Step 1:
2
1 1
Step 2: ( )
2 2
m
R
R R
R R R
−
=
⇒ =
= =
∴ = + + + +
∴
+
+ +
Step 3: Similarly continue...

Half and Add Algorithm
1. Input: 0<k<n, P=(x,y)
2. Output: Q=kP
3. Compute: , k1=(2t-1
k)mod n
4. Q=O
5. for i=0 to m-1 do
1. Q=[1/2]Q
2. If, k1
i
=1, then Q=Q+P
6. return Q
2t= log 1n +  
No method is currently known to perform point halving in projective
Coordinates. Keep Q in affine coordinates and P in Projective
Coordinates. Then step 5.2 is a mixed operation, giving further
efficiency.

Key References
• Papers:
– J. Lopez and R. Dahab, “Fast Multiplication on Elliptic Curves
over GF(2m
) without pre-computation”, CHES 1999
– K. Fong etal, “Field Inversion and Point Halving Revisited”, IEEE
Trans on Comp, 2004
– G. Orlando and C. Paar, “A High Performance Reconfigurable
Elliptic Curve Processor for GF(2m
)”, CHES 2000
– N. A. Saqib etal, “A Parallel Architecture for Fast Computation of
Elliptic Curve Scalar Multiplication over GF(2m
)”, Elsevier Journal
of Microprocessors and Microsystems, 2004
– Sabiel Mercurio etal, “ An FPGA Arithmetic Logic Unit for
Computing Scalar Multiplication using the Half-and-Add
Method”, IEEE ReConfig 2005

Key References
• Books:
– Elliptic Curves: Number Theory and
Cryptography, by Lawrence C. Washington
– Guide to Elliptic Curve Cryptography, Alfred J.
Menezes
– Guide to Elliptic Curve Cryptography, Darrel
R. Hankerson, A. Menezes and A. Vanstone
– http://cr.yp.to/ecdh.html ( Daniel Bernstein)

Elliptical curve cryptography

More Related Content

What's hot

Similar to Elliptical curve cryptography

More from Barani Tharan

Recently uploaded

Elliptical curve cryptography

Editor's Notes