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```Lower Bounds for Sparse Recovery∗
Khanh Do Ba
MIT CSAIL
Piotr Indyk
MIT CSAIL
Abstract
We consider the following k-sparse recovery problem:
design an m × n matrix A, such that for any signal
x, given Ax we can efficiently recover x̂ satisfying
kx − x̂k1 ≤ C mink-sparse x0 kx − x0 k1 . It is known
that there exist matrices A with this property that
have only O(k log(n/k)) rows.
In this paper we show that this bound is tight.
Our bound holds even for the more general randomized version of the problem, where A is a random
variable, and the recovery algorithm is required to
work for any fixed x with constant probability (over
A).
1
Introduction
In recent years, a new “linear” approach for obtaining a succinct approximate representation of ndimensional vectors (or signals) has been discovered.
For any signal x, the representation is equal to Ax,
where A is an m × n matrix. The vector Ax is often referred to as the measurement vector or sketch of
x. Although m is typically much smaller than n, the
sketch Ax contains plenty of useful information about
the signal x. A particularly useful and well-studied
problem is that of stable sparse recovery: given Ax,
recover a k-sparse vector x̂ (i.e., having at most k
non-zero components) such that
(1.1)
kx − x̂kp ≤ C
min
k-sparse
x0
kx − x0 kq
Eric Price
MIT CSAIL
David P. Woodruff
length m = O(k log(n/k)). In particular, a random
Gaussian matrix [CRT06]1 or a random sparse binary
matrix ([BGI+ 08], building on [CCFC04, CM05]) has
this property with overwhelming probability. In comparison, using a non-linear approach, one can obtain
a shorter sketch of length O(k): it suffices to store
the k coefficients with the largest absolute values, together with their indices.
Surprisingly, it was not known if the
O(k log(n/k)) bound for linear sketching could
be improved upon2 , although O(k) sketch length
was known to suffice if the signal vectors x are
required to be exactly k-sparse. This raised hope
that the O(k) bound might be achievable even for
general vectors x. Such a scheme would have been
of major practical interest, since the sketch length
determines the compression ratio, and for large n
any extra log n factor worsens that ratio tenfold.
In this paper we show that, unfortunately, such
an improvement is not possible. We address two
types of recovery schemes:
• A deterministic one, which involves a fixed matrix A and a recovery algorithm which work
for all signals x. The aforementioned results
of [CRT06] and others are examples of such
schemes.
• A randomized one, where the matrix A is chosen at random from some distribution, and for
each signal x the recovery procedure is correct
with constant probability. Some of the early
schemes proposed in the data stream literature
(e.g., [CCFC04, CM05]) belong to this category.
for some norm parameters p and q and an approximation factor C = C(k). Sparse recovery has applications to numerous areas such as data stream computing [Mut03, Ind07] and compressed sensing [CRT06,
Don06, DDT+ 08].
Our main result is that, even in the ranIt is known that there exist matrices A and as- domized case, the sketch length m must be at
sociated recovery algorithms that produce approxi- least Ω(k log(n/k)). By the aforementioned result
mations x̂ satisfying Equation (1.1) with p = q = 1 of [CRT06] this bound is tight.
(i.e., the ”`1 /`1 guarantee”), constant C and sketch
∗ This research has been supported in part by David and
Lucille Packard Fellowship, MADALGO (Center for Massive
Data Algorithmics, funded by the Danish National Research
Association) and NSF grant CCF-0728645. E. Price has been
supported in part by Cisco Fellowship.
1 In fact, they even achieve a somewhat stronger ` /`
2 1
guarantee, see Section 1.2.
2 The lower bound of Ω(k log(n/k)) was known to hold for
specific recovery algorithms, specific matrix types, or other
recovery scenarios. See Section 1.2 for an overview.
Thus, our results show that the linear compres- O(log n) bits. In this setting we show the followsion is inherently more costly than the simple non- ing: there is a method for encoding a sequence of
linear approach.
d = O(k log(n/k) log n) bits into a vector x, so that
any sparse recovery algorithm can recover that se1.1 Our techniques On a high level, our ap- quence given Ax. Since each entry of Ax conveys
proach is simple and natural, and utilizes the pack- only O(log n) bits, it follows that the number m of
ing approach: we show that any two “sufficiently” rows of A must be Ω(k log(n/k)).
different vectors x and x0 are mapped to images Ax
The encoding is performed by taking
and Ax0 that are “sufficiently” different themselves,
log
Xn
which requires that the image space is “sufficiently”
x=
D j xj ,
high-dimensional. However, the actual arguments are
j=1
somewhat subtle.
Consider first the (simpler) deterministic case. where D = O(1) and the xj ’s are chosen from the
We focus on signals x = y+z, where y can be thought error-correcting code Y defined as in the determinisof as the “head” of the signal and z as the “tail”. The tic case. The intuition behind this approach is that
“head” vectors y come from a set Y that is a binary a good `1 /`1 approximation to x reveals most of the
error-correcting code, with a minimum distance Ω(k), bits of xlog n . This enables us to identify xlog n exwhere each codeword has weight k. On the other actly using error correction. We could then comhand, the “tail” vectors z come from an `1 ball (say pute Ax − Axlog n = A(Plog n−1 Dj xj ), and identify
j=1
B) with a radius that is a small fraction of k. It xlog n−1 . . . x1 in a recursive manner. The only obcan be seen that for any two elements y, y 0 ∈ Y , the stacle to completing this argument is that we would
balls y + B and y 0 + B, as well as their images, must need the recovery algorithm to work for all xi , which
be disjoint. At the same time, since all vectors x would require lower probability of algorithm failure
live in a “large” `1 ball B 0 of radius O(k), all images (roughly 1/ log n). To overcome this problem, we
Ax must live in a set AB 0 . The key observation is replace the encoding argument by a reduction from
that the set AB 0 is a scaled version of A(y + B) and a related communication complexity problem called
therefore the ratios of their volumes can be bounded Augmented Indexing. This problem has been used in
by the scaling factor to the power of the dimension the data stream literature [CW09, KNW10] to prove
m. Since the number of elements of Y is large, this lower bounds for linear algebra and norm estimagives a lower bound on m.
tion problems. Since the problem has communication
Unfortunately, the aforementioned approach complexity of Ω(d), the conclusion follows.
does not seem to extend to the randomized case. A
We apply the argument to arbitrary matrices A
natural approach would be to use Yao’s principle, and by representing them as a sum A0 + A00 , where A0 has
focus on showing a lower bound for a scenario where O(log n) bits of precision and A00 has “small” entries.
the matrix A is fixed while the vectors x = y + z are We then show that A0 x = A(x + s) for some s with
“random”. However, this approach fails, in a very ksk < n−Ω(1) kxk . In the communication game,
1
1
strong sense. Specifically, we are able to show that this means we can transmit A0 x and recover xlog n
there is a distribution over matrices A with only O(k) from A0 (Plog n Dj x ) = A(Plog n Dj x + s). This
j
j
j=1
rows so that for a fixed y ∈ Y and z chosen uniformly means that j=1
the Augmented Indexing reduction applies
at random from the small ball B, we can recover y to arbitrary matrices as well.
from A(y + z) with high probability. In a nutshell,
the reason is that a random vector from B has an 1.2 Related Work There have been a number
`2 norm that is much smaller than the `2 norm of of earlier works that have, directly or indirectly,
elements of Y (even though the `1 norms are com- shown lower bounds for various models of sparse
parable). This means that the vector x is “almost” recovery and certain classes of matrices and algok-sparse (in the `2 norm), which enables us to achieve rithms. Specifically, one of the most well-known rethe O(k) measurement bound.
covery algorithms used in compressed sensing is `1 Instead, we resort to an altogether different ap- minimization, where a signal x ∈ Rn measured by
proach, via communication complexity [KN97]. We matrix A is reconstructed as
start by considering a “discrete” scenario where both
the matrix A and the vectors x have entries rex̂ := arg min kx0 k1 .
x0 : Ax0 =Ax
stricted to the polynomial range {−nc . . . nc } for
some c = O(1). In other words, we assume that the Kashin and Temlyakov [KT07] gave a characterizamatrix and vector entries can be represented using tion of matrices A for which the above recovery algo-
rithm yields the `2 /`1 guarantee, i.e.,
kx − x̂k2 ≤ Ck −1/2
min
k-sparse x0
kx − x0 k1
for some constant C, from which it can be shown that
such an A must have m = Ω(k log(n/k)) rows.
Note that the `2 /`1 guarantee is somewhat
stronger than the `1 /`1 guarantee investigated in this
paper. Specifically, it is easy to observe that if the
approximation x̂ itself is required to be O(k)-sparse,
then the `2 /`1 guarantee implies the `1 /`1 guarantee
(with a somewhat higher approximation constant).
For the sake of simplicity, in this paper we focus
mostly on the `1 /`1 guarantee. However, our lower
bounds apply to the `2 /`1 guarantee as well: see footnote on page 7.
On the other hand, instead of assuming a specific
recovery algorithm, Wainwright [Wai07] assumes a
specific (randomized) measurement matrix. More
specifically, the author assumes a k-sparse binary
signal x ∈ {0, α}n , for some α > 0, to which is added
i.i.d. standard Gaussian noise in each component.
The author then shows that with a random Gaussian
matrix A, with each entry also drawn i.i.d. from
the standard Gaussian, we cannot hope to recover
x from Ax with any sub-constant probability of
error unless A has m = Ω( α12 log nk ) rows. The
p
author also shows that for α = 1/k, this is tight,
i.e., that m = Θ(k log(n/k)) is both necessary and
sufficient. Although this is only a lower bound for a
specific (random) matrix, it is a fairly powerful one
and provides evidence that the often observed upper
bound of O(k log(n/k)) is likely tight.
More recently, Dai and Milenkovic [DM08], extending on [EG88] and [FR99], showed an upper
bound on superimposed codes that translates to a
lower bound on the number of rows in a compressed
sensing matrix that deals only with k-sparse signals
but can tolerate measurement noise. Specifically, if
we assume a k-sparse signal x ∈ ([−t, t] ∩ Z)n , and
that arbitrary noise µ ∈ Rn with kµk1 < d is added
to the measurement vector Ax, then if exact recovery
is still possible, A must have had m ≥ Ck log n/ log k
rows, for some constant C = C(t, d) and sufficiently
large n and k.3
(2.2)
kx − x̂k1 ≤ C
min
k-sparse x0
kx − x0 k1 .
We define a C-approximate deterministic `1 /`1
recovery algorithm to be a pair (A, A ) where A is
an m × n observation matrix and A is an algorithm
that, for any x, maps Ax (called the sketch of x) to
some x̂ that satisfies Equation (2.2).
We define a C-approximate randomized `1 /`1
recovery algorithm to be a pair (A, A ) where A is a
random variable chosen from some distribution over
m × n measurement matrices, and A is an algorithm
which, for any x, maps a pair (A, Ax) to some x̂ that
satisfies Equation (2.2) with probability at least 3/4.
We use Bpn (r) to denote the `p ball of radius r in
n
R ; we skip the superscript n if it is clear from the
context.
For any vector x, we use kxk0 to denote the “`0
norm of x”, i.e., the number of non-zero entries in x.
3
Deterministic Lower Bound
We will prove a lower bound on m for any Capproximate deterministic recovery algorithm. First
we use a discrete volume bound (Lemma 3.1) to find
a large set Y of points that are at least k apart
from each other. Then we use another volume bound
(Lemma 3.2) on the images of small `1 balls around
each point in Y . If m is too small, some two images
collide. But the recovery algorithm, applied to a
point in the collision, must yield an answer close to
two points in Y . This is impossible, so m must be
large.
Lemma 3.1. (Gilbert-Varshamov) For any q, k ∈
Z+ , ∈ R+ with < 1 − 1/q, there exists a set
Y ⊂ {0, 1}qk of binary vectors with exactly k ones,
such that Y has minimum Hamming distance 2k and
log |Y | > (1 − Hq ())k log q
where Hq is the q-ary entropy function Hq (x) =
x
− (1 − x) logq (1 − x).
−x logq q−1
See appendix for proof.
Lemma 3.2. Take an m × n real matrix A, positive
reals , p, λ, and Y ⊂ Bpn (λ). If |Y | > (1+1/)m , then
n
In this paper we focus on recovering sparse approx- there exist z, z ∈ Bp (λ) and y, y ∈ Y with y 6= y and
A(y
+
z)
=
A(y
+
z).
imations x̂ that satisfy the following C-approximate
`1 /`1 guarantee with sparsity parameter k:
Proof. If the statement is false, then the images of all
|Y | balls {y + Bpn (λ) | y ∈ Y } are disjoint. However,
3 Here A is assumed to have its columns normalized to have
n
`1 -norm 1. This is natural since otherwise we could simply those balls all lie within Bp ((1+)λ), by the bound on
scale A up to make the image points Ax arbitrarily far apart, the norm of Y . A volume argument gives the result,
effectively nullifying the noise.
as follows.
2
Preliminaries
Let S = ABpn (1) be the image of the ndimensional ball of radius 1 in m-dimensional space.
This is a polytope with some volume V . The image
of Bpn (λ) is a linearly scaled S with volume (λ)m V ,
and the volume of the image of Bpn ((1+)λ) is similar
with volume ((1 + )λ)m V . If the images of the former are all disjoint and lie inside the latter, we have
|Y | (λ)m V ≤ ((1 + )λ)m V , or |Y | ≤ (1 + 1/)m . If
Y has more elements than this, the images of some
two balls y + Bpn (λ) and y + Bpn (λ) must intersect,
implying the lemma.
a distribution Z, such that any algorithm given the
sketch of y + z must recover an incorrect y with nonnegligible probability.
Using our deterministic bound as inspiration,
we could take Y to be uniform over a set of ksparse binary vectors of minimum Hamming distance
k and Z to be uniform over the ball B1 (γk) for some
constant γ > 0. Unfortunately, as the following
theorem shows, one can actually perform a recovery
of such vectors using only O(k) measurements;
√ this is
because kzk2 is very small (namely, Õ(k/ n)) with
high probability.
Theorem 3.1. Any C-approximate deterministic recovery algorithm must have
Theorem 4.1. Let Y ⊂ Rn be a set of signals with
the property that for every distinct y1 , y2 ∈ Y , ky1 −
jnk
1 − Hbn/kc (1/2)
y2 k2 ≥ r, for some parameter r > 0. Consider “noisy
.
k log
m≥
log(4 + 2C)
k
signals” x = y + z, where y ∈ Y and z is a “noise
vector” chosen uniformly at random from B1 (s), for
Proof. Let Y be a maximal set of k-sparse n- another parameter s > 0. Then using an m × n
dimensional binary vectors with minimum Ham- Gaussian measurement matrix A = (1/√m)(g ),
ij
1
ming distance k, and let γ = 3+2C
.
By where g ’s are i.i.d. standard Gaussians, we can
ij
Lemma 3.1 with q = bn/kc we have log |Y | > (1 − recover y ∈ Y from A(y + z) with probability 1 − 1/n
Hbn/kc (1/2))k log bn/kc.
(where the probability is over both A and z), as long
Suppose that the theorem is not true; then as
!
m < log |Y | / log(4 + 2C) = log |Y | / log(1 + 1/γ),
rm1/2 n1/2−1/m
1 m
or |Y | > (1 + γ ) . Hence Lemma 3.2 gives us some
s≤O
.
1/m log3/2 n
|Y
|
y, y ∈ Y and z, z ∈ B1 (γk) with A(y + z) = A(y + z).
Let w be the result of running the recovery
To prove the theorem we will need the following
algorithm on A(y + z). By the definition of a
two lemmas.
deterministic recovery algorithm, we have
ky + z − wk1 ≤ C
min
k-sparse y 0
Lemma 4.1. For any δ > 0, y1 , y2 ∈ Y , y1 6= y2 , and
z ∈ Rn , each of the following holds with probability
at least 1 − δ:
ky + z − y 0 k1
ky − wk1 − kzk1 ≤ C kzk1
ky − wk1 ≤ (1 + C) kzk1 ≤ (1 + C)γk =
and similarly ky − wk1 ≤
1+C
3+2C
1+C
3+2C
k,
k, so
ky − yk1 ≤ ky − wk1 + ky − wk1 =
2 + 2C
k < k.
3 + 2C
But this contradicts the definition of Y , so m must
be large enough for the guarantee to hold.
Corollary 3.1. If C is a constant bounded away
from zero, then m = Ω(k log(n/k)).
4
Randomized Upper Bound for Uniform
Noise
The standard way to prove a randomized lower bound
is to find a distribution of hard inputs, and to
show that any deterministic algorithm is likely to
fail on that distribution. In our context, we would
like to define a “head” random variable y from a
distribution Y and a “tail” random variable z from
1/m
• kA(y1 − y2 )k2 ≥ δ 3 ky1 − y2 k2 , and
p
• kAzk2 ≤ ( (8/m) log(1/δ) + 1)kzk2 .
See the appendix for the proof.
Lemma 4.2. A random vector z chosen uniformly
from B1 (s) satisfies
√
Pr[kzk2 > αs log n/ n] < 1/nα−1 .
See the appendix for the proof.
Proof of theorem. In words, Lemma 4.1 says that A
cannot bring faraway signal points too close together,
and cannot blow up a small noise vector too much.
Now, we already assumed the signals to be far apart,
and Lemma 4.2 tells us that the noise is indeed small
(in `2 distance). The result is that in the image space,
the noise is not enough to confuse different signals.
Quantitatively, applying the second part of Lemma
4.1 with δ = 1/n2 , and Lemma 4.2 with α = 3, gives
us
(4.3)
!
!
s log3/2 n
log1/2 n
kzk2 ≤ O
kAzk2 ≤ O
m1/2
(mn)1/2
Proof. The
parameters in this case are r = k and
|Y | ≤ nk ≤ (ne/k)k , so by Theorem 4.1, it suffices
to have
3/2+k/m 1/2−(k+1)/m k
n
s≤O
.
log3/2 n
Choosing m = (k + 1)/ yields the corollary.
with probability ≥ 1 − 2/n2 . On the other hand,
given signal y1 ∈ Y , we know that every other signal
5 Randomized Lower Bound
y2 ∈ Y satisfies ky1 − y2 k2 ≥ r, so by the first part
of Lemma 4.1 with δ = 1/(2n|Y |), together with a Although it is possible to partially circumvent this
obstacle by focusing our noise distribution on “high”
union bound over every y2 ∈ Y ,
`2 norm, sparse vectors, we are able to obtain stronger
r
ky1 − y2 k2
results via a reduction from a communication game
≥
(4.4) kA(y1 − y2 )k2 ≥
and the corresponding lower bound.
3(2n|Y |)1/m
3(2n|Y |)1/m
The communication game will show that a mesholds for every y2 ∈ Y , y2 6= y1 , simultaneously with sage Ax must have a large number of bits. To show
probability 1 − 1/(2n).
that this implies a lower bound on the number of rows
Finally, observe that as long as kAzk2 < kA(y1 − of A, we will need A to be discrete. Hence we first
y2 )k2 /2 for every competing signal y2 ∈ Y , we are show that discretizing A does not change its recovery
guaranteed that
characteristics by much.
kA(y1 + z) − Ay1 k2
= kAzk2
5.1 Discretizing Matrices Before we discretize
by rounding, we need to ensure that the matrix is well
conditioned. We show that without loss of generality,
≤ kA(y1 + z) − Ay2 k2
the rows of A are orthonormal.
We can multiply A on the left by any invertible
for every y2 6= y1 , so we can recover y1 by simply
matrix
to get another measurement matrix with the
returning the signal whose image is closest to our
same
recovery
characteristics. If we consider the
measurement point A(y1 + z) in `2 distance. To
singular
value
decomposition
A = U ΣV ∗ , where U
achieve this, we can chain Equations (4.3) and (4.4)
and V are orthonormal and Σ is 0 off the diagonal,
together (with a factor of 2), to see that
this means that we can eliminate U and make the
!
entries of Σ be either 0 or 1. The result is a matrix
rm1/2 n1/2−1/m
consisting of m orthonormal rows. For such matrices,
s≤O
|Y |1/m log3/2 n
we prove the following:
< kA(y1 − y2 )k2 − kAzk2
Lemma 5.1. Consider any m × n matrix A with
orthonormal rows. Let A0 be the result of rounding
A to b bits per entry. Then for any v ∈ Rn there
The main consequence of this theorem is that for exists an s ∈ Rn with A0 v = A(v − s) and ksk1 <
the setup we used in Section 3 to prove a deterministic n2 2−b kvk1 .
lower bound of Ω(k log(n/k)), if we simply draw the
Proof. Let A00 = A − A0 be the roundoff error when
noise uniformly randomly from the same `1 ball (in
discretizing A to b bits, so each entry of A00 is less
fact, even one with a much larger radius, namely,
than 2−b . Then for any v and s = AT A00 v, we have
polynomial in n), this “hard distribution” can be
As = A00 v and
defeated with just O(k) measurements:
√
ksk1 = AT A00 v 1 ≤ n kA00 vk1
√
Corollary 4.1. If Y is a set of binary k-sparse vec≤ m n2−b kvk1 ≤ n2 2−b kvk1 .
tors, as in Section 3, and noise z is drawn uniformly
at random from B1 (s), then for any constant > 0, 5.2 Communication Complexity We use a few
m = O(k/) measurements suffice to recover any sig- definitions and results from two-party communicanal in Y with probability 1 − 1/n, as long as
tion complexity. For further background see the book
suffices. Our total probability of failure is at most
2/n2 + 1/(2n) < 1/n.
s≤O
k 3/2+ n1/2−
log3/2 n
.
by Kushilevitz and Nisan [KN97]. Consider the following communication game. There are two parties, Alice and Bob. Alice is given a string y ∈
{0, 1}d . Bob is given an index i ∈ [d], together with
yi+1 , yi+2 , . . . , yd . The parties also share an arbitrarily long common random string r. Alice sends
a single message M (y, r) to Bob, who must output yi
with probability at least 3/4, where the probability is
taken over r. We refer to this problem as Augmented
Indexing. The communication cost of Augmented Indexing is the minimum, over all correct protocols, of
the length of the message M (y, r) on the worst-case
choice of r and y.
The next theorem is well-known and follows
[BYJKK04]).
distance k. From Lemma 3.1 we have log |X| =
Ω(k log(n/k)). Let d = blog |X|c log n, and define
D = 2C + 3.
Alice is given a string y ∈ {0, 1}d , and Bob is
given i ∈ [d] together with yi+1 , yi+2 , . . . , yd , as in
the setup for Augmented Indexing.
Alice splits her string y into log n contiguous
chunks y 1 , y 2 , . . . , y log n , each containing blog |X|c
bits. She uses y j as an index into X to choose xj .
Alice defines
x = D1 x1 + D2 x2 + · · · + Dlog n xlog n .
Alice and Bob use the common randomness r to
Theorem 5.1. The communication cost of Augagree upon a random matrix A with orthonormal
mented Indexing is Ω(d).
rows. Both Alice and Bob round A to form A0 with
Proof. First, consider the private-coin version of the b = d2(1 + log D) log ne = O(log n) bits per entry.
0
problem, in which both parties can toss coins, but do Alice computes A x and transmits it to Bob.
From Bob’s input i, he can compute the value
not share a random string r (i.e., there is no public
j
coin). Consider any correct protocol for this problem. j = j(i) for which the bit yi occurs in y . Bob’s
We can assume the probability of error of the protocol input also contains yi+1 , . . . , yn , from which he can
is an arbitrarily small positive constant by increasing reconstruct xj+1 , . . . , xlog n , and in particular can
the length of Alice’s message by a constant factor compute
(e.g., by independent repetition and a majority vote).
z = Dj+1 xj+1 + Dj+2 xj+2 + · · · + Dlog n xlog n .
Applying Lemma 13 of [MNSW98] (with, in their
0
notation, t = 1 and a = c · d for a sufficiently
0
0
small constant c0 > 0), the communication cost of Bob then computes A z, and using A x and linearity,
0
such a protocol must be Ω(d). Indeed, otherwise A (x − z). Then
there would be a protocol in which Bob could output
j
X
yi with probability greater than 1/2 without any
D1+log n
< kD2 log n .
kx
−
zk
≤
kDi < k
1
interaction with Alice, contradicting that Pr[yi =
D
−
1
i=1
1/2] and that Bob has no information about yi . Our
theorem now follows from Newman’s theorem (see,
So from Lemma 5.1, there exists some s with A0 (x −
e.g., Theorem 2.4 of [KNR99]), which shows that the
z) = A(x − z − s) and
communication cost of the best public coin protocol
is at least that of the private coin protocol minus
ksk1 < n2 2−2 log n−2 log D log n kx − zk1 < k.
O(log d) (which also holds for one-round protocols).
Set w = x − z − s. Bob then runs the estimation
algorithm A on A and Aw, obtaining ŵ with the
Theorem 5.2. For any randomized `1 /`1 recovery property that with probability at least 3/4,
algorithm (A, A ), with approximation factor C =
kw − ŵk1 ≤ C
min
kw − w0 k1 .
O(1), A must have m = Ω(k log(n/k)) rows.
k-sparse w0
Proof. We shall assume, without loss of generality,
that n and k are powers of 2, that k divides n, and Now,
that the rows of A are orthonormal. The proof for
0
w − Dj xj min
kw
−
w
k
≤
1
the general case follows with minor modifications.
1
k-sparse w0
Let (A, A ) be such a recovery algorithm. We will
j−1
X
show how to solve the Augmented Indexing problem
Di xi ≤ ksk1 +
1
on instances of size d = Ω(k log(n/k) log n) with
i=1
communication cost O(m log n). The theorem will
< k(1 + D + D2 + · · · + Dj−1 )
Dj
Let X be the maximal set of k-sparse n.
<k·
D−1
dimensional binary vectors with minimum Hamming
5.3
Randomized Lower Bound Theorem
Hence
j
D xj − ŵ ≤ Dj xj − w + kw − ŵk
1
1
j1
≤ (1 + C) D xj − w
1
kDj
.
<
2
And since the minimum
Hamming
distance
in X is
k, this means Dj xj − ŵ1 < Dj x0 − ŵ1 for all
x0 ∈ X, x0 6= xj 4 . So Bob can correctly identify xj
with probability at least 3/4. From xj he can recover
y j , and hence the bit yi that occurs in y j .
Hence, Bob solves Augmented Indexing with probability at least 3/4 given the message A0 x. The entries in A0 and x are polynomially bounded integers
(up to scaling of A0 ), and so each entry of A0 x takes
O(log n) bits to describe. Hence, the communication
cost of this protocol is O(m log n). By Theorem 5.1,
m log n = Ω(k log(n/k) log n), or m = Ω(k log(n/k)).
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A
Proof of Lemma 3.1
Proof. We will construct a codebook T of block
length k, alphabet q, and minimum Hamming distance k. Replacing each character i with the qlong standard basis vector ei will create a binary
qk-dimensional codebook S with minimum Hamming
distance 2k of the same size as T , where each element of S has exactly k ones.
The Gilbert-Varshamov bound, based on volumes of Hamming balls, states that a codebook of
size L exists for some
L ≥ Pk−1
qk
k
i=0
i
(q − 1)i
.
Using the claim (analogous to [vL98], p. 21, proven
below) that for < 1 − 1/q
k X
k
i=0
i
(q − 1)i < q Hq ()k ,
In particular, for any α > 1,
we have that log L > (1 − Hq ())k log q, as desired.
i=0
i
Then
1 = ( + (1 − ))k
k X
k i
>
(1 − )k−i
i
i=0
i
k X
k
=
(q − 1)i
(1 − )k
i
(q
−
1)(1
−
)
i=0
k
k X
k
i
(q − 1)
(1 − )k
>
i
(q
−
1)(1
−
)
i=0
k X
k
−Hq ()k
=q
(q − 1)i
i
i=0
Proof of Lemma 4.1
Proof. By standard arguments (see, e.g., [IN07]), for
any D > 0 we have
m
3
ky1 − y2 k2
≤
Pr kA(y1 − y2 )k2 ≤
D
D
and
Pr[kAzk2 ≥ Dkzk2 ] ≤ e−m(D−1)
2
/8
.
Setting both right-hand sides to δ yields the lemma.
C
αs log n/n] = (1 − α log n/n)n < e−α log n
1/nα .
Now, by symmetry this holds for every other coordinate zi of z as well, so by the union bound
(q − 1)i < q Hq ()k .
Proof. Note that
−Hq ()
(1 − ) < (1 − ).
q
=
(q − 1)(1 − )
B
Pr[|z1 | >
=
Claim A.1. For 0 < < 1 − 1/q,
k X
k
proportional to (s − t)n−1 . Normalizing to ensure the
probability integrates to 1, we derive this probability
as
n
p(|z1 | = t) = n (s − t)n−1 .
s
It follows that, for any D ∈ [0, s],
Z s
n
Pr[|z1 | > D] =
(s − t)n−1 dt = (1 − D/s)n .
n
D s
Proof of Lemma 4.2
Proof. Consider the distribution of a single coordinate of z, say, z1 . The probability density of |z1 |
taking value t ∈ [0, s] is proportional to the (n − 1)(n−1)
dimensional volume of B1
(s − t), which in turn is
Pr[kzk∞ > αs log n/n] < 1/nα−1 ,
√
and since kzk2 ≤ n · kzk∞ for any vector z, the
lemma follows.
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