Probability Bounds for Two-Dimensional Algebraic...

Probability Bounds for Two-Dimensional Algebraic LatticeCodes

David Karpuk

Aalto University

April 16, 2013

(Joint work with C. Hollanti and E. Viterbo)

David Karpuk (Aalto University) Probability Bounds for Two-Dimensional Algebraic Lattice CodesApril 16, 2013 1 / 20

Alice, Bob, and Eve

Suppose that Alice wants to transmit information to Bob over a potentiallynoisy wireless channel, while an eavesdropper, (St)Eve, listens in.

−−−− ∗ ∗ ∗ ∗ − −−− >>|*|

Alice, Bob, and Eve

This wireless channel can be modeled by the equations

yb = Hbx + zb (1)ye = Hex + ze (2)

wherex ∈ Rn is the vector intended for transmission.

Hb,He ∈ Mn(R) are Bob’s and Eve’s fading matrices, respectively.zb, ze ∈ Rn are the corresponding noise vectors, whose entries areGaussian random variables with variance σ2

b, σ2e .

yb, ye ∈ Rn are the vectors received by Bob and Eve.We assume that σ2

e >> σ2b, i.e. that Eve’s channel is much noisier than

Bob’s.

Alice, Bob, and Eve

wherex ∈ Rn is the vector intended for transmission.Hb,He ∈ Mn(R) are Bob’s and Eve’s fading matrices, respectively.

zb, ze ∈ Rn are the corresponding noise vectors, whose entries areGaussian random variables with variance σ2

b, σ2e .

Bob’s.

Alice, Bob, and Eve

wherex ∈ Rn is the vector intended for transmission.Hb,He ∈ Mn(R) are Bob’s and Eve’s fading matrices, respectively.zb, ze ∈ Rn are the corresponding noise vectors, whose entries areGaussian random variables with variance σ2

b, σ2e .

Bob’s.

Alice, Bob, and Eve

b, σ2e .

yb, ye ∈ Rn are the vectors received by Bob and Eve.

We assume that σ2e >> σ2

b, i.e. that Eve’s channel is much noisier thanBob’s.

Alice, Bob, and Eve

b, σ2e .

Bob’s.

Coset Coding

Alice uses coset coding, a variant of lattice coding, to confuse Eve.

Alice selects a “fine” lattice Λb whose elements encode data intended forBob. At the same time, Alice chooses a “coarse” sublattice

Λe ⊂ Λb, (3)

containing random bits intended to confuse Eve.

Coset Coding

Alice now sends codewords of the form

x = r + c (4)

where r is a random element of Λe intended to confuse Eve, and c is acoset representative of Λe in Λb.

Alice’s strategy ensures that Eve can easily recover the “random” data r ,but not the actual data c .

Coset Coding

In practice, we construct Eve’s codebook from a finite subset CR of Λe ,which we’ll define as

CR := {x ∈ Λe : ||x ||∞ ≤ R} (5)

for some positive R > 0. In this picture, the blue dots represent elementsof Λe , and R = 5:

Probability of Eve’s Correct Decision

Given the above scheme to be employed by Alice, what is the probabilitythat Eve correctly decodes the data c? It is known that this probability canbe estimated by

Pe ≤ C (σ2e ,Λb)

∑x∈CR

∏xi 6=0

1|xi |3

This bound motivates the following design criteria for Eve’s lattice. For afixed dimension n, find the lattice Λ which minimizes the inverse norm sum

SΛ(R, s) =∑x∈CR

∏xi 6=0

1|xi |s

Probability of Eve’s Correct Decision

Given the above scheme to be employed by Alice, what is the probabilitythat Eve correctly decodes the data c? It is known that this probability canbe estimated by

Pe ≤ C (σ2e ,Λb)

∑x∈CR

∏xi 6=0

1|xi |3

This bound motivates the following design criteria for Eve’s lattice. For afixed dimension n, find the lattice Λ which minimizes the inverse norm sum

∏xi 6=0

1|xi |s

Algebraic Lattices

From now on, we’ll only deal with the case of n = 2. For algebraic lattices,the inverse norm sum takes a particularly interesting form.

Let K = Q(√d) be a totally real quadratic number field with ring of

integers OK , and Gal(K/Q) = 〈σ〉.

For example, one could take K = Q(√5), so that OK = Z[1+

2 ] andσ(√5) = −

Algebraic Lattices

We can embed OK ↪→ R2 as a lattice Λ via the canonical embedding

Λ := {(x , σ(x)) : x ∈ OK}. (8)

In this case, the inverse norm sum becomes

∏xi 6=0

1|xi |s

=∑x∈CR

1|N(x)|s

where N : K → Q is the field norm, defined by N(x) = x · σ(x).

The Inverse Norm Sum

From now on, we identify OK with the lattice Λ it determines in R2. Howdo we estimate

SΛ(R, s) =∑

x∈OK||x ||∞≤R

1|N(x)|s

, (10)

and study how it grows as R →∞?

For any x ∈ OK , we have N(x) ∈ Z. Thus any x ∈ OK lives on one of thehyperbolas XY = ±k for some integer k , allowing for a convenientgeometrical grouping of the codewords.

Estimating the Inverse Norm Sum

Now letbk,R = #{x ∈ OK : |N(x)| = k , ||x ||∞ ≤ R} (11)

so that, for example, b1,R is the number of units inside the bounding box.

We have the following bounds for SΛ(R, s):

b1,R ≤ SΛ(R, s) ≤ ζ1K (s)b1,R , (12)

ζ1K (s) =

∑a⊆OK

a principal

1N(a)s =

∑k≥1

a1kks (13)

is the partial zeta function of K , so that a1k is the number of principal

ideals of norm k in OK .

Now letbk,R = #{x ∈ OK : |N(x)| = k , ||x ||∞ ≤ R} (11)

so that, for example, b1,R is the number of units inside the bounding box.

We have the following bounds for SΛ(R, s):

b1,R ≤ SΛ(R, s) ≤ ζ1K (s)b1,R , (12)

ζ1K (s) =

∑a⊆OK

a principal

1N(a)s =

∑k≥1

a1kks (13)

is the partial zeta function of K , so that a1k is the number of principal

ideals of norm k in OK .

Proof: (See also paper by Vehkalahti et al) Rewrite the inverse norm sumas

SΛ(R, s) =∑

x∈OK||x ||∞≤R

1|N(x)|s

=∑k≥1

ks . (14)

Taking log | · | of each coordinate, one sees that bk,R ≤ a1kb1,R for all k > 0:

Experimental Data

How good are these estimates? Let’s take K = Q(√5):

blog(R)c b1,R SΛ(R, 3) b1,Rζ1K (3)

1 10 10.0472 10.27552 18 18.2576 18.49593 26 26.4809 26.71624 34 34.7068 34.93665 42 42.9276 43.15706 50 51.2105 51.3774

In order for these estimates to be practically useful, we have to have a wayof calculating ζ1

K (s), which is equivalent to calculating a1k for k = 1, . . . ,N.

Evaluating the Partial Zeta Function

First, let us suppose that k = p is prime, and we wish to calculate thenumber a1

p of principal ideals of norm p in OK .

The only ideals, principal or otherwise, in K which have norm p, mustappear in the prime factorization of the ideal (p) in OK .

Let D be the discriminant of K . The ideal (p) factors in OK as

(p) is prime iff (p,D) = 1,D 6≡ y2 (mod p), for any y ∈ Zpq, p 6= q iff (p,D) = 1,D ≡ y2 (mod p), for some y ∈ Z

p2 iff p|D(15)

and we say that p is inert, split, or ramified in K , respectively.

First, let us suppose that k = p is prime, and we wish to calculate thenumber a1

p of principal ideals of norm p in OK .

The only ideals, principal or otherwise, in K which have norm p, mustappear in the prime factorization of the ideal (p) in OK .

Let D be the discriminant of K . The ideal (p) factors in OK as

(p) is prime iff (p,D) = 1,D 6≡ y2 (mod p), for any y ∈ Zpq, p 6= q iff (p,D) = 1,D ≡ y2 (mod p), for some y ∈ Z

p2 iff p|D(15)

and we say that p is inert, split, or ramified in K , respectively.

If p is inert, so that (p) is prime, then the only prime ideal appearing in thefactorization of (p) is (p) itself. But this ideal has norm p2, so in this casea1p = 0.

If p is split, so that (p) = pq, then p and q are Galois conjugate andtherefore simultaneously principal or non-prinicipal. Hence a1

p = 0 or 2,accordingly.

If p is ramified, so that (p) = p2, then p is the only ideal of norm p. Soa1p = 0 or 1, depending on whether p is principal.

Algorithms for determining whether or not an ideal in a ring of integers isprincipal are implemented in SAGE.

What to do if k = pe11 · · · pem

m is not prime?

If k is composite, one can use the prime factorization of k , and how the pifactor in K , to list all of the ideals of norm k . It’s easier to see this byexample.

Example: Let K = Q(√229), and let k = 225 = 32 · 52. Let us calculate

a1225. In K the ideals (3) and (5) both split, and we have factorizations

(3) = p3q3, p3 =(3, (1−

√229)/2

), q3 =

(3, (1 +

√229)/2

)(5) = p5q5, p5 =

(5, (7−

√229)/2

), q5 =

(5, (7 +

√229)/2

thus the list of all ideals of norm k is

25, p3q3p

23p5q5, p3q3p5q5, q

23p5q5, p

25, p3q3q

Exactly three of these ideals are principal, so that a1225 = 3. Specifically,

25 = (2−

√229), p3q3q

25 = (2 +

√229), p3q3p5q5 = (15).

Example: Let K = Q(√229), and let k = 225 = 32 · 52. Let us calculate

a1225. In K the ideals (3) and (5) both split, and we have factorizations

(3) = p3q3, p3 =(3, (1−

√229)/2

), q3 =

(3, (1 +

√229)/2

)(5) = p5q5, p5 =

(5, (7−

√229)/2

), q5 =

(5, (7 +

√229)/2

)thus the list of all ideals of norm k is

25, p3q3p

23p5q5, p3q3p5q5, q

23p5q5, p

25, p3q3q

Exactly three of these ideals are principal, so that a1225 = 3. Specifically,

25 = (2−

√229), p3q3q

25 = (2 +

√229), p3q3p5q5 = (15).

Conclusion

Design criteria for coset coding using algebraic lattices over fading wiretapchannels consists of studying the inverse norm sum,

SΛ(R, s) =∑

x∈OK||x ||∞≤R

1|N(x)|s

, (16)

which itself is inversely proportional to the regulator of K , and directlyproportional to the values of the partial zeta function of K .

Further work consists of studying for which number fields both of thesequantities are optimal, as well as extending results to MIMO systems.

The End! Thanks!

References

1. F. Oggier, J.C. Belfiore, and E. Viterbo, Cyclic Division Algebras: ATool for Space-Time Coding, Foundations and Trends in Communicationsand Information Theory. 2007. Vol. 4, No 1, pp 1-95.

2. J.C. Belfiore and F. Oggier, An Error Probability Approach to MIMOWiretap Channels, January 2013, http://arxiv.org/abs/1109.6437.

3. R. Vehkalahti, F. Lu, and L. Luzzi, Inverse Determinant Sums andConnections Between Fading Channel Information Theory and Algebra,December 2012, http://arxiv.org/abs/1111.6289.

4. C. Hollanti, E. Viterbo, and D. Karpuk, Nonasymptotic ProbabilityBounds for Fading Channels Exploiting Dedekind Zeta Functions, January2013, http://arxiv.org/abs/1303.3475.

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