1

L ATTICE E RASURE C ODES OF L OW R ANK
WITH

N OISE M ARGINS

Vinay A. Vaishampayan

arXiv:1801.04466v1 [] 13 Jan 2018

Dept. of Engineering Science and Physics
City University of New York-College of Staten Island
Staten Island, NY USA

Abstract
We consider the following generalization of an (n, k) MDS code for application to an erasure
channel with additive noise. Like an MDS code, our code is required to be decodable from any k
received symbols, in the absence of noise. In addition, we require that the noise margin for every
allowable erasure pattern be as large as possible and that the code satisfy a power constraint. In this
paper we derive performance bounds and present a few designs for low rank lattice codes for an additive
noise channel with erasures.
Index terms Lattices, Erasure Codes, MDS Codes, Compound Channel.

I. I NTRODUCTION
YS1

Z
EncoderModulator

X

Y
+

Erasure
Network

YS2
YSi
YSm

Fig. 1. Coding and Modulation for the Erasure Network.

We consider low rank lattice codes for transmission over a noisy erasure channel as illustrated
in Fig. 1. In this figure k information symbols are mapped by an encoder/modulator to a vector
x = (x1 , x2 , . . . , xn ) ∈ Λ, where Λ is a rank-k lattice in Rn . The output of the additive noise
channel is y = x + z, where z = (z1 , z2 , . . . , zn ) is a noise vector independent of x and with
independent components. Components of y are then erased by an erasure network, whose outputs

2

are obtained by retaining only those symbols of y indexed by subsets S ⊂ {1, 2, . . . , n} in a
given sub-collection of subsets; thus yS coincides with y is the positions identified by S. As
an example, with n = 4, S = {2, 4} and y = (a, b, c, d), yS = (b, d). A decoder estimates
the source symbols based on yS with a probability of error denoted Pe (S). The objective is to
minimize Pe (S) for each S by designing a single codebook which satisfies a power constraint
E[X t X] ≤ nP , where E denotes expectation with respect to a uniform distribution on the
codebook. Here we consider as our sub-collection S, all k-subsets of {1, 2, . . . , n}. Our paper is
organized as follows. Prior work and lattice background is in Sec. II. Two performance bounds
are presented in Sec. III, constructions for codes in dimension n = 4 are presented and compared
to the derived bounds in Sec. IV. A summary is in Sec. V.
We use the acronym w.l.o.g to mean ‘without loss of generality’.
II. P RIOR W ORK AND R EVIEW OF L ATTICE T ERMINOLOGY
The problem considered here may be viewed as a code design problem for a special case of
the compound channel, see e.g. [1], [5]. This work was motivated by a study on cross layer
coding that appeared in [4]. For prior contributions on the Gaussian erasure channel, please refer
to [8] and the references therein.We now develop notation and some basic definitions for low
rank lattices in Rn . Let {φi , i = 1, 2, . . . , k} be a collection of k ≤ n orthonormal column
vectors in Rn and let Φ = (φi , i = 1, 2, . . . , k) denote the associated n × k orthonormal matrix.
Let V denote a k × k generator matrix of full rank for a lattice ΛV = V Zk := {V u, u ∈ Zk }.
We will refer to ΛV as the mother lattice. Let
Λ = ΦΛV := ΦV Zk .

(1)

Λ is a rank-k lattice in Rn . Let G(ΛV ) = V t V denote the Gram matrix of ΛV (t is the transpose
operator).
The determinant of a lattice det Λ is defined in terms of the determinant of its Gram matrix by
det Λ := det(G(Λ)). Let ρ(Λ) denote the radius of the largest inscribed sphere in a Voronoi cell
of Λ. The packing density of Λ is defined in terms of Vk the volume of a unit-radius Euclidean
ball in Rk by

√
∆k (Λ) = Vk ρk / det Λ.

(2)

3

We denote by ∆k (opt), the largest packing density that can be acheived by any lattice in Rk . The
problem of finding lattices that maximize the packing density is a classical problem in number
theory and geometry, with several excellent references [3], [6].
The following definitions are from [6]. A body captures the notion of a solid subset of Rn ,
specifically, B ⊂ Rn is a body if it has nonempty interior and is contained in the closure
of its interior [6]. A body B ⊂ Rn is said to be centrally symmetric if B = −B, where
−B = {−x : x ∈ B}. A closed body B with the property that for any x ∈ B, the point
λx ∈ B for every 0 ≤ λ < 1 is called a star body. While convex bodies are star bodies, the
converse is not true. A simple example, and one directly relevant to us is the star body formed by
the union of centrally symmetric ellipsoids in Rn . A lattice Λ is said to be admissible for B ⊂ Rn ,
p
or B-admissible, if no non-zero point in Λ lies in B. The greatest lower bound of ( det Λ)
over all B−admissible lattices is called the lattice constant of B, denoted ∆(B) (which is set
to ∞ is there are no B− admissible lattices). A B-admissible lattice Λ with det Λ = ∆(B)2 is
said to be a critical lattice for B.
A lattice Λ is said to be a packing lattice for a body B if the sets B and B + λ are disjoint for
all non-zero λ ∈ Λ. It is known, Thm.1, Ch. 3, Sec 20 [6], that Λ is a packing lattice for centrally
symmetric, convex body B if and only if it is admissible for 2B. Thus, for a convex body, the
problem of finding a packing lattice for B is equivalent to that of finding an admissible lattice
for 2B. The connection between packings and admissibility for non-convex bodies is messier.
The distinction arises because for a centrally symmetric body B, Λ is a lattice packing of B
if and only if it is admissible for B + B, where + denotes the set sum or Minkowski sum. If
the centrally symmetric body is also convex, then B + B = 2B and thus packing problems and
admissibility problems are closely related. On the other hand, if B is centrally symmetric but
non-convex, in order to solve a packing problem for B one must solve an admissibility problem
for B + B, and this set may not be as easily described as B.
In our application, we need to index body B by subset S in a given sub-collection of subsets
S
and our problem is one of packing S∈S B(S), which is non-convex. While admissibility for
non convex centrally symmetric body 2C says nothing in general about packings for C, it turns
S
out that our design problem is equivalent to finding a critical lattice for 2 S∈S B(S) because the
decoder knows S. Thus it is possible to draw on the theory of admissible lattices for star bodies.
This theory provides several key ingredients to help find a solution to this problem. Most notably,
in the chapter on Mahler’s compactness theorem [2], Theorem VII states that every critical lattice

4

for a bounded star body S has n linearly independent points on the boundary of S.
III. B OUNDS
Let In = {1, 2, . . . , n}. For S ⊂ In , |S| = k let ΛS be the lattice obtained be retaining
only those coordinates that are in S or equivalently ΛS is the projection of Λ into the subspace
CS := Span{ei , i ∈ S} where ei = (0, ..., 0, 1, 0, ..., 0)t is the ith unit vector in Rn . For any
k-subset S ⊂ {1, 2, . . . , n}, we denote by ΦS the k × k submatrix obtained by extracting from Φ
the k rows identified by S. The generator matrix for ΛS is ΦS V and its Gram matrix G(ΛS ) =
V t ΦtS ΦS V .
Define the (packing volume) contraction ratio
βS = (ρ(ΛS )/ρ(ΛV ))k

(3)

let ρmin = minS ρ(ΛS ) and let βmin = minS βS .
A. Determinant Upper Bound
We will use symbols x̄, x# to denote the arithmetic mean and geometric mean, respectively,
of the real numbers xi over some index set I. When a k-dim mother lattice ΛV is set in IRn

using a basis Φ, the projections on the nk subsets S, cannot all be simultaneously good. There
are two important factors that measure the ‘goodness’ of the projections—the packing density
and the scale of the lattices ΛS . The following theorem develops one of two bounds presented
in this paper.
Theorem 1. (Determinant Bound) Given a mother lattice ΛV and orthonormal basis Φ, let
β # and ∆# be respectively, the geometric mean of the volume contraction ratios and packing
densities of the child lattices ΛS , taken over all k-subsets of {1, 2, . . . , n}. Then
(β # ∆(ΛV ))2 ≤

(∆# )2
 .
n

(4)

k

Equality holds if and only if all child lattices have equal determinants.
Proof. The packing densities of the mother lattice ΛV and child lattice ΛS are related by the
following identity
∆2 (ΛV )βS2 = ∆2 (ΛS )

det ΛS
.
det ΛV

(5)

5

Compute the geometric mean of both sides over the collection of k-subsets S to get
! 1
Y det ΛS (nk)
∆2 (ΛV )(β # )2 = (∆# )2
.
det ΛV
S

(6)

From the arithmetic-geometric mean inequality it follows that
∆2 (ΛV )(β # )2 ≤ (∆# )2

1

n
k

X det ΛS
det ΛV
S

!
(7)

and equality holds if and only if det ΛS is a constant with respect to S. However
X
X
det ΛS =
det(G(ΛS ))
S

S

=

X

det((ΦS V )t ) det(ΦS V )

S
(a)

= det((ΦV )t (ΦV ))
= det ΛV ,

(8)

where in (a) we have used the Cauchy-Binet formula, see e.g. [7]. The remainder of the proof
follows directly.
The following corollary is immediate.
Corollary 1. Given mother lattice ΛV and orthonormal basis Φ, let βmin be the minimum volume
contraction ratio of the child lattices ΛS , taken over all k-subsets of {1, 2, . . . , n}, and ∆# the
geometric mean of the packing densities. Then
(βmin ∆(ΛV ))2 ≤

(∆# )2
∆k (opt)2

 .
≤
n
n
k

(9)

k

Equality holds in the left inequality iff all the contraction ratios are equal and all the child lattices
have equal determinants. Equality holds in the right inequality iff all child lattices achieve the
optimal packing density in dimension k.
B. Trace Upper Bound
Theorem 2. (Trace Bound) For an (n, k) code, the compaction ratio is bounded as
2/k

2/k

βmin ≤ βS

≤

k
.
n

(10)

Equality holds if the shortest vector of each lattice ΛS is the image of the shortest vector in ΛV .

6

Proof. Upon summing over all k-subsets S we obtain
X
X
G(ΛS ) =
V t ΦtS ΦS V
S

S

!
X

= Vt

ΦtS ΦS

V

S


n−1 t
Φ ΦV
= V
k−1


n−1
=
V t V.
k−1
t



(11)

By definition the smallest packing radius of any child lattice ρmin satisfies
ρmin 2 ≤ ρ(ΛS )2 ≤ (1/2)ut G(ΛS )u

(12)

for any non-zero u ∈ Zk and any k-subset S. Upon averaging over subsets S we obtain the
upper bound
ρ(ΛS )2 ≤
=

2

1 X t

u G(ΛS )u
n
k

S

1
2

 t
n u
k

X

G(ΛS )u

S

n−1
k−1

2 nk



=

ut G(ΛV )u.

(13)

Equality holds if ΦS V u is the shortest vector in ΛS for all S. Thus
ρ(ΛS )2 ≤

k 2
ρ (ΛV )
n

(14)

and (10) follows immediately.
IV. A NALYSIS OF S OME (4, k) C ODES
We construct Φ for n = 4 for various values of k and various mother lattices ΛV . Numerical
2/k

results for the (4, k), k = 2, 3 are presented in Fig. 2, in which βmin is plotted as a function of
the packing density of the mother lattice. We have plotted the determinant bound using both the
optimal and the cubic lattice for the child lattices. We have also plotted the trace bound.
Observe that in the (n, k) = (4, 2) case there is a significant gap between the best possible
construction and the upper bounds. In the (4, 3) case performance close to the determinant bound
is achieved by setting the mother lattice to be the cubic lattice. Also with the cubic lattice as
the mother lattice, since the trace bound is lower than the determinant bound, this is proof that

7

it is impossible to simultaneously achieve the packing density of D3 when the mother lattice is
the cubic lattice.
(n,k)=(4,3)

(n,k)=(4,2)
0.9

0.6

0.8
0.5
0.7

0.6

- 2/k
min

- 2/k
min

0.4

0.3

0.5

0.4

0.3

0.2
Det. Bound (A 2)
Trace Bound
Det Bound (Z 2)
$ M=Z2

0.1

$ M=A2
Lower Bound

0
0.75

0.8

0.85

0.9

0.2

Det. Bound: D 3
Trace Bound

0.1

Det Bound: Z 3
Z3
FCC
BCC

0
0.4

0.95

0.5

0.6

"($ M)

0.7

0.8

"($ M)

Fig. 2. Bounds on β 2/k and values obtained from the construction.

4

4

4

2

2

2

0

0

0

-2

-2

-2

-4
-5

0

5

-4
-5

0

5

-4
-5

4

4

4

2

2

2

0

0

0

-2

-2

-2

-4
-5

0

5

-4
-5

0

5

0

0.5

0

-4
-5

5

0

5

(a)

2

1.5

1

0.5

0

-0.5

-1

-1.5

-2
-2

-1.5

-1

-0.5

1

1.5

2

(b)

Fig. 3. Packings (a) and the star body (b) derived from noise spheres in each of the six subspaces for the (4, 2) code with
ΛV = Z2 as described in the text.

Geometrically, the ability to decode correctly, post-erasure, with iid Gaussian noise is determined by the largest noise sphere which can be packed by the projected (or child) lattice ΛS

in each of the nk subspaces CS . When each noise sphere is projected back onto the subspace
spanned by the columns of Φ, a noise sphere is transformed into an ellipsoid. To see this
consider the noise sphere kxk2 ≤ r2 in subspace CS . Setting x = ΦS y, this leads to the noise

8

ellipsoid BS (r) = {y : kΦS yk2 ≤ r2 }. The packing of the noise ellipsoids BS (ρmin ), (recall
that ρmin is the half the length of the shortest non-zero vector in ΛS ), by the mother lattice
ΛV is shown in the six panels in Fig. 3(a), one for each subspace. Fig. 3(b) shows the star
S
body B(2ρmin ) = s∈S BS (2ρmin ). This illustrates the star body, which is the union of the six
ellipses, two of which are circles. Also shown are points of the lattice ΛV , which in this case
is an admissible lattice for this star body and illustrates also the interpretation as a code design
problem for the compound channel. Observe that Z2 is simultaneously good as a packing for
all six erasure configurations, i.e. for each of the bodies BS (ρmin ). Also, ΛV is simultaneously
critical for five of the six bodies BS (2ρmin ) (notice that one circle does not touch any of the
lattice points).
A. (4, 1)
t





2
= 1/4. The trace and determinant upper
Let Φ = a a a a , a = 1/2. We obtain βmin

bounds yield β 2 ≤ 1/4. Hence this construction is optimal.
B. (4, 2)
√
With ΛV = Z2 , a = 1/ 3 and

Φt = 

a

a

a 0

a −a 0 a

we obtain six child lattices with Gram matrices




2a2 0
a2 a2
,

, 
2
2
2
0 2a
a 2a

 

2
2
2
2
a
−a
2a −a

, 
,
2
2
2
2
−a 2a
−a
a









2a2 a2
a

2

a2
0

2


,

a

0
.
2
a

(15)

All six child lattices are similar to Z2 . The first one has shortest vector of square length 2a2 and all
2/k

the others have square length a2 . This code achieves βmin = βmin = 1/3, β # = 21/6 /3 = 0.374.
The trace upper bound is βmin ≤ βS ≤ 1/2, regardless of the mother lattice, while the
√
√
determinant upper bound depends on the mother lattice, and is 2/3 and 1/ 6 = 0.408 for
ΛV = A2 (hexagonal lattice) and Z2 , resp. Thus the determinant bound is tighter than the trace
bound but greater than 1/3. This construction does not meet the trace bound or the determinant

9

bound with equality. However with ΛV = Z2 , the following theorem shows this to be the best
βmin possible. A computer-based search has also failed to reveal any improvements in β # .
Theorem 3. Let Φ be a 4 × 2 matrix with orthonormal columns, so that Φt Φ = I. Let ΛV = Z2 .
Let ΛS = ΦS ΛV where S is a 2-subset of {1, 2, 3, 4}. Let rS be the length of the shortest non-zero
vector in ΛS . Then rS2 ≤ 1/3 for at least one of the six 2-subsets of {1, 2, 3, 4}.
Proof. For this proof we will write


x 1 x2 x3 x4
.
Φt = 
y1 y2 y3 y4

(16)

Our proof is by contradiction. Suppose that the shortest non-zero vector in all ΛS has square
length r2 > 1/3. Now x2i < r2 /2, in at most one position i, else the shortest would be smaller
than r2 . The same is true for yi2 , (xi − yi )2 and (xi + yi )2 . Hence there exists one position i,
w.l.o.g. i = 1, such that x2i ≥ r2 /2 , yi2 ≥ r2 /2 and (xi − yi )2 ≥ r2 /2. It follows that (i) x1 and
y1 must be of opposite sign and (ii) (x1 + y1 )2 < r2 /2. (i) is true because if x1 , y1 are of the
√
√
√
same sign and (x1 − y1 )2 ≥ r2 /2 then |y1 | ≥ 2r or |x1 | ≥ 2r. Assuming x1 ≥ 2r, we have
r2 ≤ x23 + x24 = 1 − x21 − x22 ≤ 1 − 2r2 which contradicts the hypothesis that r2 > 1/3. (ii) is
√
√
true for if not |x1 | ≥ 2r or |y1 | ≥ 2r, and by the same argument as in (i) r2 cannot exceed
1/3.
Since (x1 + y1 )2 < r2 /2, it follows that (xi + yi )2 ≥ r2 /2 for positions i = 2, 3, 4. In at least
one of these positions, say i = 2, x2i ≥ r2 /2 and yi2 ≥ r2 /2. Now x2 and y2 must be of the
same sign and (x2 − y2 )2 < r2 /2, by a proof similar to that used before. Thus for i = 3, 4,
(xi + yi )2 ≥ r2 /2 and (xi − yi )2 ≥ r2 /2. Again for i = 3, 4, both x2i ≥ r2 /2 and yi2 ≥ r2 /2
cannot hold for the same i, hence, either x23 < r2 /2, y32 ≥ r2 /2 and x24 ≥ r2 /2, y42 < r2 /2 or
x24 < r2 /2, y42 ≥ r2 /2 and x23 ≥ r2 /2, y32 < r2 /2. We assume the first case. The proof for the
other case is similar.
We have already proved that x2 and y2 are of the same sign. Assume they are both positive
(if not reverse signs of all elements of Φ). Further, assume that y2 > x2 and consider positions
i = 2, 3 (if x2 > y2 , then the same proof applies but for positions i = 2, 4). We now break up
the proof into two cases:
Case 1: (y3 and x3 of the same sign): Either (a) (y3 − x3 )2 ≥ r2 or (b) (y3 − x3 )2 < r2 . If (a)
then |y3 | − |x3 | ≥ r and y22 + y32 ≥ x22 + (|x3 | + r)2 ≥ x22 + x23 + r2 ≥ 2r2 . Thus r2 ≤ y12 + y42 =

10

1 − (y22 + y33 ) ≤ 1 − 2r2 , hence r2 ≤ 1/3.
If (b) then let (y3 − x3 )2 = r2 − 2 (0 < 2 ≤ r2 /2). Thus |y3 | − |x3 | =

√
r2 − 2 and it follows

that y32 ≥ x23 + r2 − 2 . Since (y2 − x2 )2 + (y3 − x3 )2 ≥ r2 it follows that (y2 − x2 )2 ≥ 2 and
that |y2 | ≥ |x2 | +  and thus y22 ≥ x22 + 2 . Thus y22 + y32 ≥ x22 + 2 + x23 + r2 − 2
= x22 + x23 + r2 ≥ 2r2 . But then r2 ≤ y12 + y42 = 1 − (y22 + y32 ) ≤ 1 − 2r2 and it follows that
r2 ≤ 1/3.
Case 2: (y3 and x3 are of opposite signs): Either (a) (y3 + x3 )2 ≥ r2 or (b) (y3 + x3 )2 < r2 .
If (a) then |y3 + x3 | ≥ r and due to opposite signs |y3 | − |x3 | ≥ r from which y32 ≥ x23 + r2 . It
follows that y22 + y32 ≥ x22 + x23 + r2 ≥ 2r2 . This implies r2 ≤ 1/3. If (b) then let
(y3 + x3 )2 = r2 − 2 . Since y3 and x3 have opposite signs (|y3 | − |x3 |)2 = r2 − 2 and thus
y32 ≥ x23 + r2 − 2 . Since (y2 + x2 )2 + (y3 + x3 )2 ≥ r2 it follows that (y2 + x2 )2 + r2 − 2 ≥ r2
and hence (y2 + x2 )2 ≥ 2 which implies that y22 ≥ x22 + 2 . Thus
y22 + y32 ≥ x22 + 2 + x23 + r2 − 2 ≥ 2r2 . Once again this means r2 ≤ 1/3.

C. (4, 3)
It is checked by direct evaluation that for

a

t
Φ = a

a
the four child lattices have Gram matrices


1/4 
3/4 1/4


1/4 3/4 −1/4 ,


1/4 −1/4 3/4


 3/4 1/4 −1/4


 1/4 3/4 1/4 


−1/4 1/4 3/4

a = 1/2, ΛV = Z3 and

−a −a −a

a −a a  ,

−a a
a

(17)




 3/4 −1/4 1/4


−1/4 3/4 1/4


1/4
1/4 3/4


 3/4 −1/4 −1/4


−1/4 3/4 −1/4 .


−1/4 −1/4 3/4

√
All child lattices are congruent to the body-centered cubic lattice with packing density π 3/8.
This construction achieves the trace bound r2 = 3/4 and is optimal.

11

Set ΛV to be the face-centered cubic lattice, the densest lattice packing in R3 , packing density
√
∆V = π/ 18 = 0.7408, so that det ΛV = 4,




0
−1 1
2 0 1




V = −1 −1 1  , G(ΛV ) = 0 2 1 .
(18)




0
0 −1
1 1 2
With Φ such that



1

0 0





 2/3 1 1


Φ∗V =
,
−2/3 1 0


1/3 0 1

(19)

we get β 2/k = (ρmin /ρ(ΛV ))2 = 0.5. Each child lattice has unit determinant and shortest vector
of length equal to unity. All four child lattices have an identical packing density of π/6, which
is the packing density of the cubic lattice Z3 . Also observe that the determinant bound (4) holds
with equality, which implies that the selected Φ is optimal among rotations for which all child
lattices achieve the packing density of the cubic lattice.
√
Let ΛV the body-centered cubic lattice, ∆(ΛV ) = π 3/8, ρ2 = 3/4,




 3 −1 −1
 1 −1 1




V = −1 1 1 , G(ΛV ) = −1 3 −1 .




−1 −1 3
−1 −1 1
√
√
Upon setting a = 1/2, b = a + a, c = a − a and
√



a a
0
−c c
b



√ 
 0
−c b
a − a
c




Φ=
,Φ ∗ V = 

√

 0

−b c
a
a
b




√
a −a
0
b −b −c
we obtain child lattices with Gram matrices given in terms of d = 2c2 + b2 , e = 2b2 + c2 ,
f = −c2 − 2bc, g = c2 + 2bc, h = 3bc, i = −b2 − 2bc,

 

 d f −g   d −g f 

 

 f d h  , −g e −h ,

 

−g h e
f −h d

12




e
−g
i




−g d −h ,


i −h e




e
−i
−g




 −i e −h .


−g −h d

(20)

√
Each child lattice has determinant 4 and shortest vector with square length d = (9/4 − 1/ 2) =
1.5429 and packing density ∆ = πd3/2 /12 = 0.5017. This results in β 2/k = d/3 = 0.5143.
V. S UMMARY
We considered the design of power-constrained rank-k lattice codes in Rn , with the property
that error free recovery is possible for any k code symbols and the minimum distance for each of

the nk lattices obtained by projection onto k-dim subspaces spanned by k coordinate vectors are
bounded from below. The potential application is to erasure channels with additive noise. Bounds
on the performance are derived and the performance of specific constructions are investigated
for n = 4.
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[4] T. A. Courtade and R. D. Wesel. Optimal allocation of redundancy between packet-level erasure coding and physical-layer
channel coding in fading channels. IEEE Transactions on Communications, 59(8):2101–2109, 2011.
[5] I. Csiszár and P. Narayan. Capacity of the gaussian arbitrarily varying channel. IEEE Transactions on Information Theory,
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