Security of Using Special Integers in Elliptic Scalar Multiplication

Security of Using Special Integers in Elliptic Scalar Multiplication

Mun-Kyu Leeo Jin Wook Kim Kunsoo Park

School of CSE, Seoul National University

1. Preliminaries

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Elliptic Curve

A curve of the formy2 + xy = x3 + ax2 + b ory2 = x3 + ax + b

There are many cryptosystems that use elliptic curve operations.

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Elliptic Curve Operations [1]

Point Addition: R = P +Q– First, draw the

line through P and Q.– Then this line intersects

the elliptic curvein a third point.

– Define R = P +Q(the sum of P and Q)as the reflection ofthis point in the x-axis.

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Elliptic Curve Operations [2]

Point Doubling: R = 2P– First, draw the tangent line

to the curve at P.– Then this line intersects

the curve ina second point.

– Define R = 2P (the double of P)as the reflection ofthis point in the x-axis.

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Elliptic Curve Operations [3]

Scalar Multiplication kP– For a nonnegative integer k and a point P,

scalar multiplication kP is defined as• 0P = O, for k = 0,

where O is the “point at infinity”which is the additive identity element.

• kP = (k-1)P + P for k > 0.

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ECDLP

Elliptic Curve Discrete Log Problem – Given two points P and Q on an elliptic curve,– ECDLP is to find k such that kP = Q

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Scalar Mult. vs. ECDLP

Scalar Multiplicationk, P Q = kP

Efficient

ECDLPP, Q k s.t. Q = kP

- Computationally infeasible- Hence, security of elliptic curve based cryptosystems is based on this problem.

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The purpose of this paper

is to develop a technique to find harder instances of ECDLP,while keeping the efficiency of a scalar multiplication as the same level.

2. Previous Results:Efficient Scalar Mult. Algos.

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[1] Binary Method

To compute Q = kP,– represent k as a binary form.– scan each bit of k from left to right.– if the bit is 1, do a doubling and an addition.

if the bit is 0, do a doubling only. Example: 61P = (1, 1, 1, 1, 0, 1)(2)P

PDBL

2P

1

ADD P

3P

1

6P

1

DBL

ADD P

7P

14PDBL

ADD P

15P

01 1

30PDBL

DBL

60P

ADD P

Q = 61P

101

11

110

111

1110

1111

11110 111100

111101

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[1] Binary Method

Complexity– log k doublings and– HW(k)-1 additions,

where HW(k) is the Hamming weight of k,i.e., the number of 1’s in the binary representation of k.

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[2] Signed Binary Method

[Morain, Olivos 90] Use the following facts.

– For a point P on an elliptic curve, computation of an additive inverse –P is almost free.

• For example,on y2 = x3 + ax + b,–P is the reflection of Pin the x-axis.

– Hence, a subtraction P - Q has the same complexity as that of an addition P +Q.

P = (x, y)

-P = (x, -y)

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[2] Signed Binary Method

To compute Q = kP,– convert k to a signed binary representation k’ with smaller

number of nonzero digits than k.– if a digit is 1, do a doubling and an addition.

if a digit is –1, do a doubling and a subtraction.if a digit is 0, do a doubling only.

Example: 61P = (26 - 22 +1)P = (1, 0, 0, 0,-1, 0, 1)P

PDBL

2P

01

4P

0

8P

-10 1101 100 1000

0DBL DBL DBL

16P10000

SUB

15P10001

DBL

30P100010

60P1000100DBL

ADD

Q = 61P1000101

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[2] Signed Binary Method

Complexity– log k doublings and– SHW(k)-1 additions/subtractions,

where SHW(k) is the signed Hamming weight of k, i.e., the number of nonzeros in the signed binary representation of k.

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[3] AMV method

In many elliptic curve based systems, we compute kP for a randomly chosen k.

[Agnew, Mullin, Vanstone 93]– Choose special k’s that have small HW(k) to reduce the

number of additions.– Specifically, generate random k’s of length m in a

binary form s.t. HW(k) = w for a fixed small w.– One can control the Hamming weight, and thus the

number of additions.

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[3] AMV method

Example: m = 8, w = 30. Initially, there are 8 empty bits.

1. Choose 3 random positions for ‘1’.

2. Set them as ‘1’ and others as ‘0’.

For kP, we need 7 doublings and 2 additions.k = (1, 0, 1, 0, 0, 0, 0, 1)

3. Proposed Method

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Our Method

Use special k’s– Generate random k’s that have small SHW(k).– Specifically, generate random k’s of length m in a signed

binary form s.t. SHW(k) = w for a fixed small w. More secure than the AMV selection method, i.e.,

random selection of k’s with HW(k)=w.– (Recall that an ECDLP is to find k such that kP = Q.)– The number of possible k’s in our method is much

larger,– while the amount of computation is the same,

i.e., m-1 doublings and w-1 additions/subtractions,in both of the methods.

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Naïve Approach

In order to generate a random k of length m s.t. SHW(k) = w,– randomly select w locations for nonzero digits out of m

possible digits of k,– and then assign ‘1’ or ‘-1’ to each of these digits

randomly. Problem

– k’s are not unique.• Hence, the search space for k is much smaller than what

we have intended.– Example: m = 6, w = 3

• (1,0,0,1,0,-1) = (1,0,0,0,1,1) = 35

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Solution

select k’s in the nonadjacent form (NAF).– NAF is a signed binary representation with the

property that no two consecutive digits are nonzero.

– A number’s NAF is unique.

35

(1, 0, 0, 1, 0,-1) (1, 0, 0, 0, 1, 1)

possible representations

in NAF not in NAF

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Selection Algorithm

Now, we want to generate a random k of length m in NAF s.t. SHW(k) = w to guarantee the uniqueness of k.

To satisfy the NAF property, we use ‘10’ and ‘-10’ as single nonzero units instead of ‘1’ and ‘-1’.

The algorithm has six steps.

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Step 1

Initially there is an array of m-w+1 consecutive empty slots.

Example: m = 8, w = 3 (m -w +1 = 6)

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Step 2

Assign two-digit binary number 10 to the first slot to guarantee that k > 0 and that k has exactly m digits.

Example: m = 8, w = 3

10

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Step 3

Choose w - 1 random slots out of the remaining m – w slots and assign10 or –10 randomly to each of them.

Example: m = 8, w = 3 (w -1 = 2, m -w = 5)

10 10-10

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Step 4

Assign 0 to each remaining slot.

Example: m = 8, w = 3

10 10-10 0 0 0

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Step 5

Concatenate all slots to get a number k with m +1 signed binary digits.

Note that, for now, k is always even.

Example: m = 8, w = 3

10 10-10 0 0 0

(1, 0, 0,-1, 0, 0, 0, 1, 0)

9 digits

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Step 6

Set k = k / 2

Example: m = 8, w = 3

k = (1, 0, 0,-1, 0, 0, 0, 1)

8 digits

(1, 0, 0,-1, 0, 0, 0, 1, 0)

For kP, we need 7 DBLs and 2 ADD/SUBs.

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Distribution of k’s

k’s generated by this algorithm are unique.

k’s generated by this algorithm form a uniform distribution of k’s thathave m digits and satisfy SHW(k) = w.

4. Security Analysis

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Security Against BSGS

With k’s of special forms, the best possible attack algorithm against the ECDLP is the baby-step giant-step algorithm, which is a time-memory trade-off version of the exhaustive search.

Hence, k’s with larger search space is more secure against this attack.

Now we compare the size of the search space of our method with that of the AMV method.

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Comparison of the Sizes of Search Spaces for k

AMV Our Method

# digits

# nonzeros

m

w

m

w

sizes ofsearch spaces

1

1

w

m 121

w

w

wm

complexity ofa scalar mult.

m-1 DBLsw-1 ADDs

m-1 DBLsw-1 ADD/SUBs

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Comparison in Typical Settings

The size of the search space of our method is much larger.

Our method is expected to be more secure.