Division Property: a New Attack Against Block Ciphers

Christina Boura

(joint on-going work with Anne Canteaut)

Séminaire du groupe Algèbre et Géometrie, LMVNovember 24, 2015

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Symmetric-key encryption

Alice and Bob exchange the secret key through a secure channel.

DecryptionEncryption

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Symmetric-key encryption

Alice and Bob exchange the secret key through a secure channel.

Key-exchange problem ⇒ birth of the public-key cryptography.

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Public-key encryption

Decryption

%gTi2z*

Encryption

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Advantages and disadvantages of each system

Advantages Disadvantages

Fast systemsSecret-key Need secure key-exchange

Relatively short-keysn users: n2 keys

No key-exchange neededPublic-key Slow systems

n users: 2n keysRelatively long-keys

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Hybrid encryption

Idea: Use a combination of asymmetric and symmetric encryption tobenefit from the strengths of every system.

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Hybrid encryption

Use a public-key cryptosystem to exchange a key (session key).

Use the exchanged key to encrypt data by using a symmetric-keycryptosystem.

Advantages:

Slow public-cryptosystem is used to encrypt a short string only.

Fast symmetric-key cryptosystem is used to encrypt the longercommunication session.

Used for example in the SSL protocol.

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Block ciphers

Outline

1 Block ciphers

2 Division property

3 Propagation through an Sbox

4 Extending the division property

5 Understanding Dnk for some specific values of k

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Block ciphers

Encrypt a block of message m into a block of ciphertext c under theaction of the key k.

ENC : {0, 1}n × {0, 1}κ → {0, 1}n

(m,k) 7→ ENC(m,k) = c

ENCm c

Given k, it must be easy to compute c from m.

Given m, c it must be hard to compute k such that ENC(m,k) = c.

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Block ciphers

Two important parameters:

block size, n

key size, κ

A block cipher generates a family of permutations indexedby a key k.

2κsubset

permutations

Ideal design: 2κ permutationschosen uniformly at random fromall 2n! ≈ 2(n−1)2n permutations.

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Block ciphers

Iterated block ciphers

Idea: Iterate a round function f several times. The function f r is waitedto be strong for large r.

Advantages:

Compact implementation.

Easier analysis.

f f fm c

k1 k2 kr

Key schedule

master key k

Use a key schedule to extend the user-supplied (or master) key to asequence of r subkeys.

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Block ciphers

How to build the round function?

Two major approaches:

Feistel network.

Substitution-Permutation Network (SPN).

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Block ciphers

How to build the round function?

Two major approaches:

Feistel network.

Substitution-Permutation Network (SPN).

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Block ciphers

Substitution Permutation Network (SPN)

Substitution

Permutation

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Block ciphers

Substitution Permutation Network (SPN)

S S S S

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Block ciphers

Cryptanalysis of block ciphers

Problem: Design block ciphers that are fast and secure at the same time.

In symmetric key cryptography, security proofs are partial and insufficient.

Only mean of proving that a design is secure:

cryptanalysis.

An algorithm is secure as long there is no attack against it.

The more an algorithm is analysed without being broken, the morereliable it is.

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Block ciphers

What does "broken" mean?

No attack faster than exhaustive search should exist.

If a block cipher encrypts messages with a k-bit key, no attack with timecomplexity less than 2k should be known.

Otherwise, the cipher is considered as broken (even if the complexity ofthe attack is not practical).

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Division property

Outline

1 Block ciphers

2 Division property

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Division property

A new property for block ciphers

In Eurocrypt 2015, Yosuke Todo introduces a new property, called thedivision property.

Combination (in some sense) of higher-order differential andsaturation attacks.

Construction of more powerful generic distinguishers for both SPN andFeistel constructions.

Use of this new property for breaking full MISTY-1 (best paper awardat CRYPTO 2015).

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Division property

Notation

If x, u ∈ Fn2 , we denote

Example: (n = 4)

x = (x1, x2, x3, x4) = (1, 1, 0, 1),u = (u1, u2, u3, u4) = (1, 0, 1, 0)

xu = x1u1x2

u2x3u3x4

u4 = 11100110 = 0.

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Division property

Let X be a multiset of elements in Fn2 .

For 0 ≤ k ≤ n, we say that X has the division property Dnk if

xu = 0,

for all u ∈ Fn2 such that wt(u) < k.

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Division property

Division property - Example

X = {0x0, 0x3, 0x3, 0x3, 0x5, 0x6, 0x8, 0xB, 0xD, 0xE}.

Compute⊕

xu for all u ∈ F42.

xu = 1,

for u = 1011, u = 1101 and u = 1110.

So,⊕

xu = 0 for all u with wt(u) < 3.

X has the division property D43 .

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Division property

Using the division property in practice

Prepare a set of plaintexts and evaluate its division property.

Propagate the input texts and evaluate the division property of theoutput set after one round.

Use rules to propagate the property through the different ciphercomponents (Sboxes, XOR, etc..)

Repeat the procedure and compute the division property of the set oftexts after several rounds.

If after several rounds some exploitable information is found, then weget a distinguisher.

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Division property

Distinguisher and key recovery attack

Distinguisher: A property that permits to distinguish the target blockcipher from an ideal permutation.

Division property:

EK(y) has the division property Dnk for k ≥ 1.

Key recovery: Exploit this property to recover the key by targeting firstthe subkey of the last round.

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Propagation through an Sbox

Outline

1 Block ciphers

2 Division property

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What is an Sbox?

Main component for providing non-linearity.

Can be seen as a vectorial Boolean function S : Fn2 → F

m2 (usually

m = n).

Algebraic Normal Form (ANF) of an Sbox

y0 = x0x2 + x1 + x2 + x3

y1 = x0x1x2 + x0x1x3 + x0x2x3 + x1x2 + x0x3 + x2x3 + x0 + x2

y2 = x0x1x3 + x0x2x3 + x1x2 + x1x3 + x2x3 + x0 + x1 + x3

y3 = x0x1x2 + x1x3 + x0 + x1 + x2 + 1.

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Algebraic degree of an Sbox

(y0, y1, y2, y3) = S(x0, x1, x2, x3)

y0 = x0x2 + x1 + x2 + x3

y1 = x0x1x2 + x0x1x3 + x0x2x3 + x1x2 + x0x3 + x2x3 + x0 + x2

y2 = x0x1x3 + x0x2x3 + x1x2 + x1x3 + x2x3 + x0 + x1 + x3

y3 = x0x1x2 + x1x3 + x0 + x1 + x2 + 1.

The algebraic degree of S is 3.

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Propagation of the division property through an Sbox

Let S be a permutation of Fn2 of algebraic degree d.

Let X be a multiset having the division property Dnk .

Question: What is the division property of Y = S(X)?

If k = n, then Y has the division property Dnn.

Proposition (Todo):

Y has the division property Dn⌈k

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Example - MISTY S7

MISTY’s Sbox S7 is a 7-bit Sbox of degree 3.

The input set X has the property D7k.

The output set Y has the property D7k′ , with k′ = ⌈k3⌉.

k 0 1 2 3 4 5 6 7

k′ 0 1 1 1 2 2 2 7

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Proof Sketch

Let the input set X have the division property Dnk . Then,

xu = 0, for all u ∈ Fn2 with wt(u) < k.

Goal: Evaluate for which v ∈ Fn2 ,

S(x)v vanishes.

If deg(Sv) < k then⊕

S(x)v = 0.

If deg(Sv) ≥ k,⊕

S(x)v is undetermined.

Obviously, deg(Sv) ≤ wt(v)× d, so the sum becomes unknown if

wt(v)× d ≥ k.

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An improvement idea

In the previous proof, the degree was bounded by

deg(Sv) ≤ wt(v) × d

This bound is not tight!

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The inverse permutation influences the degree

Let S be a permutation on Fn2 .

Denote by δk(S) the max. degree of the product of k coordinates of S.

Theorem [B.–Canteaut 2013]. For any k and ℓ,

δℓ(S) < n− k if and only if δk(S−1) < n− ℓ.

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Getting a tighter result

Use the previous theorem to better estimate deg(Sv):

deg(Sv) ≤ δwt(v)(S).

Then,δwt(v)(S) < k iff δn−k(S

−1) < n− wt(v).

By re-writing the second inequality we get

δwt(v)(S) < k iff wt(v) < n− δn−k(S−1).

The quantity⊕

x∈X(Sv)(x) becomes unknown when

wt(v) ≥ n− δn−k(S−1).

So Y has the division property Dnn−δn−k(S−1).

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Example - Back to MISTY S7

MISTY’s inverse Sbox S−17 is a 7-bit Sbox of degree 3.

k 1 2 3 4 5 6 7

δk(S−17 ) 3 5 5 6 6 6 7

The input set X has the property D7k.

The output set Y has the property D7k′ , with

k′ = ⌈k

3⌉ (Todo’s estimation)

k′ = 7− δ7−k(S−1

7) (our estimation)

k 0 1 2 3 4 5 6 7

k′ (Todo’s) 0 1 1 1 2 2 2 7

k′ (our) 0 1 1 1 2 2 4 7

For k = 6: k′ = 7− δ7−6(S−17 ) = 7− 3 = 4

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Extending the division property

Outline

1 Block ciphers

2 Division property

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Reduced set of a multiset

The corresponding reduced set X̃ is the set composed of all elements in X

having an odd multiplicity.

Example: If X = {0x0, 0x3, 0x3, 0x3, 0x5, 0x7, 0x7, 0xB, 0xC} then

X̃ = {0x0, 0x3, 0x5, 0xB, 0xC}.

A multiset X fulfills Dnk if and only if X̃ fulfills Dn

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Parity set of a multiset

Let X be a multiset of elements in Fn2 . The set U(X) is the subset of Fn

defined by

U(X) = {u ∈ Fn2 :

xu = 1},

is called the parity set of X.

Obviously U(X) = U(X̃).

The parity set provides a complete characterization of the reducedset of a multiset.

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Incidence vector of U(X)

Lemma. Let G be the 2n × 2n binary matrix whose entries areindexed by n-bit vectors and defined by

Gu,a = au, a, u ∈ Fn2 .

For any subset X of Fn2 , the incidence vector of U(X) is equal to

the product of G by the incidence vector of X.

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An example (n = 3)

00 10 20 30 40 50 60 70

01 11 21 31 41 51 61 71

02 12 22 32 42 52 62 72

03 13 23 33 43 53 63 73

04 14 24 34 44 54 64 74

05 15 25 35 45 55 65 75

06 16 26 36 46 56 66 76

07 17 27 37 47 57 67 77

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An example (n = 3)

1 1 1 1 1 1 1 10 1 0 1 0 1 0 10 0 1 1 0 0 1 10 0 0 1 0 0 0 10 0 0 0 1 1 1 10 0 0 0 0 1 0 10 0 0 0 0 0 1 10 0 0 0 0 0 0 1

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An example (n = 3)

X = {1, 3, 4}

U(X) =

1 1 1 1 1 1 1 10 1 0 1 0 1 0 10 0 1 1 0 0 1 10 0 0 1 0 0 0 10 0 0 0 1 1 1 10 0 0 0 0 1 0 10 0 0 0 0 0 1 10 0 0 0 0 0 0 1

01011000

10111000

U(X) = {0, 2, 3, 4}.

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Reed-Muller codes

Fm2 = {z0, . . . , z2m−1}. Let f : Fm

2 → F2. We define

cf = (f(z0), . . . , f(z2m−1)).

The Reed-Muller code RM(d,m) of order d and length 2m is defined as

RM(d,m) := {cf : deg(f) ≤ d}.

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Correspondance of X and U(X)

Corollary. For any subset U of Fn2 , there exists a unique set X ⊂ F

such that U(X) = U .

Proof.

The matrix G is a generator matrix of the Reed-Muller code of length2n and order n.

Dimension of the code : 2n ⇒ G is invertible.

The mapping matching the incidence vector of a set X, vX to theincidence vector of U(X) is an isomorphism of the set of 2n vectors.

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Parity set and the division property

Let Ek be a keyed permutation.

The division property is a distinguishing property of the multisetEk(X) for a given choice of the input multiset X.

We can now reformulate the division property Dnk of Ek(X) by a

simple property of U(Ek(X)). Indeed, Dnk characterizes a multiset X

by a lower bound on the weight of all elements in U(X).

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Proposition. Let X be a multiset of elements in Fn2 and k be an integer

0 ≤ k ≤ n. Then,the following assertions are equivalent:

(i) X fulfills the division property Dnk .

(ii)U(X) ⊆ {u ∈ F

n2 : wt(u) ≥ k} .

(iii) The incidence vector of the corresponding reduced set X̃belongs to the Reed-Muller code of length 2n andorder (n− k).

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Understanding Dn

kfor some specific values of k

Outline

1 Block ciphers

2 Division property

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Understanding Dn

Some specific values of k

Question: What can be said about a multiset X that verifies a propertyDn

k , for some value of k?

The cases Dn1 , Dn

2 , Dnn, have been characterized.

[Todo 2015], [Sun et al. 2015]

The cases Dnk , for k 6= {1, 2, n} had not been exploited before.

We provide some insight on these cases here by using the aboveintroduced new vision and some well known properties of Reed-Mullercodes.

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Understanding Dn

The property Dn1

X fulfills Dn1 if and only if its cardinality is even.

Indeed,

X has the property Dn1 : For u = (0, . . . , 0) :

⊕x∈X xu = 0

⇔⊕

x01 . . . x0n =

1 = #X mod 2 = 0

The inverse can be easily deduced.

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Understanding Dn

The property Dn2

X fulfills Dn2 if and only if its cardinality is even and it has the

Balance property.

Balance property: For any i, 1 ≤ i ≤ n⊕

xi = 0.

Indeed, if X has the property Dn2 :

⊕x∈X x01 . . . x

0n = 0 ⇒ X has even cardinality.

For all u with wt(u) = 1:⊕x∈X xu =

⊕x∈X x01 . . . x

0i−1x

1i . . . x

⊕x∈X xi = 0

⇒ X has the Balance property.

The inverse is proven easily.

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Understanding Dn

The property Dnn

X fulfills Dnn if and only if its reduced set X̃ is

either empty or equal to Fn2 .

Let v be the incidence vector of X̃.

Proof. X satisfies Dnn iff v ∈ R(0, n). Thus, either v is the all-zero vector,

i.e., X̃ is empty or v is the all-one vector i.e. X̃ = Fn2 .

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Understanding Dn

The property Dnn−1

Proposition. X satisfies Dnn−1 if and only if X̃ is an (affine) subspace of

dimension (n− 1).

Proof. X satisfies Dnn−1 iff v ∈ R(1, n).

R(1, n) consists of the incidence vectors of all (affine) hyperplanes of F2.Then, this equivalently means that X̃ is an (affine) hyperplane.

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Understanding Dn

Example [Todo, Eurocrypt 2015]

For the multiset of elements of F42

X = {0x0, 0x3, 0x3, 0x3, 0x5, 0x6, 0x8, 0xB, 0xD, 0xE},

the corresponding reduced set

X̃ = {0x0, 0x3, 0x5, 0x6, 0x8, 0xB, 0xD, 0xE}

is a linear subspace of dimension 3 spanned by {0x3, 0x5, 0x8}.

So, it can be directly deduced (without computation) that

X has the property D43.

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Understanding Dn

The property Dnk

Proposition. Let X be a multiset of elements in Fn2 satisfying Dn

such that X̃ is not empty. Then

|X̃ | ≥ 2k ,

and equality holds iff X̃ is an affine subspace of dimension k.

Proof. X satisfies Dnk iff v

X̃belongs to R(n− k, n). The minimum

distance of R(n− k, n) is 2k and that the minimum-weight codewords inthis code are the incidence vectors of the affine subspaces of dimension k.

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Understanding Dn

Conclusion

We have reformulated the division property to captivate moreinformation.

Complete characterisation of the property Dnk for different values of k.

More powerful distinguishers for a high number of block ciphers.

Work in progress. . .

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Understanding Dn

Conclusion

We have reformulated the division property to captivate moreinformation.

Complete characterisation of the property Dnk for different values of k.

More powerful distinguishers for a high number of block ciphers.

Work in progress. . .

Thanks for your attention!

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