Introduction
Background
Given an n×d matrix A and p∈[1,∞], let
For simplicity, we will use κp, σpmin, and σpmax when the underlying matrix is clear.
Given an n×d matrix A and p∈[1,∞], we always have
For a discussion of ellipsoidal rounding, we refer readers to Clarkson et al. . In this paper, we simply cite the following lemma, which is based on ellipsoidal rounding.
Stable distributions.
where Xi∼iidD and X∼D. By “X≃Y”, we mean X and Y have the same distribution.
By a result due to Lévy , it is known that p-stable distributions exist for p∈(0,2]; and from Chambers et al. , it is known that p-stable random variables can be generated efficiently, thus allowing their practical use. Let us use Dp to denote the “standard” p-stable distribution, for p∈, specified by its characteristic function ψ(t)=e−∣t∣p. It is known that D1 is the standard Cauchy distribution, and that D2 is the Gaussian distribution with mean and variance 2.
Tail inequalities.
We note two inequalities from Clarkson et al. regarding the tails of the Cauchy distribution.
For i=1,…,m, let Ci be m (not necessarily independent) standard Cauchy variables, and γi>0 with γ=∑iγi. Let X=∑iγi∣Ci∣. For any t>1,
For simplicity, we assume that m≥3 and t≥1, and then we have Pr[X>tγ]≤2log(mt)/t.
For i=1,…,m, let Ci be independent standard Cauchy random variables, and γi≥0 with γ=∑iγi. Let X=∑iγi∣Ci∣. Then, for any t>0,
We also note the following result about Gaussian variables. This is a direct consequence of Maurer’s inequality (), and we will use it to derive lower tail inequalities for p-stable distributions.
For i=1,…,m, let Gi be independent standard Gaussian random variables, and γi≥0 with γ=∑iγi. Let X=∑iγi∣Gi∣2. Then, for any t>0,
In addition, ΠA can be computed in O(nnz(A)) time.
The construction of Π in this theorem is the same as the construction in Clarkson and Woodruff . For them, s=O((d/ϵ)4log2(d/ϵ)) in order to achieve (1±ϵ) distortion with a constant probability. Theorem 1 shows that it actually suffices to set s=O((d2+d)/ϵ2). Surprisingly, the proof is rather simple. Let X=UTΠTΠU, where U is an orthonormal basis for A2. Compute E[∥X−I∥F2] and apply Markov’s inequality to ∥X−I∥F2≤ϵ2, which implies ∥X−I∥2≤ϵ and hence the embedding result. See Appendix A.1 for a complete proof.
Remark. The O(nnz(A)) running time is indeed optimal, up to constant factors, for general inputs. Consider the case when A has an important row aj such that A becomes rank-deficient without it. Thus, we have to observe aj in order to compute a low-distortion embedding. However, without any prior knowledge, we have to scan at least a constant portion of the input to guarantee that aj is observed with a constant probability, which takes O(nnz(A)) time. Note that this optimality result applies to general p.
In addition, ΠA can be computed in O(nnz(A)) time.
Instead of analyzing Dp directly, for any p∈(1,2), we establish an order among the Cauchy distribution, the p-stable distribution, and the Gaussian distribution, and then we derive upper and lower tail inequalities for the p-stable distribution similar to the ones we used to prove Theorem 2. We state these technical results here since they are of independent interest. We start with the following lemma, which is proved in Appendix A.4 and which establishes this order.
For any p∈(1,2), there exist constants αp>0 and βp>0 such that
Our numerical results suggest that the constants αp and βp are not too far away from 1. See Figure 1, which plots of the CDFs of ∣Xp/2∣p for p=1,0,1.1,…,2.0, based on which we conjecture ∣Xp1/2∣p1⪰∣Xp2/2∣p2, for all 1≤p1≤p2≤2. This implies that 2p−1∣C∣⪰∣Xp∣p and ∣Xp∣p⪰2p−2∣X2∣2≃2p−1∣G∣2, which therefore provides a value for the constants αp and βp.
Lemma 8 suggests that we can use Lemma 5 (regarding Cauchy random variables) to derive upper tail inequalities for general p-stable distributions and that we can use Lemma 7 (regarding Gaussian variables) to derive lower tail inequalities for general p-stable distributions. The following two lemmas establish these results; the proofs of these lemmas are provided in Appendix A.5 and Appendix A.6, respectively.
Given p∈(1,2), for i=1,…,m, let Xi be m (not necessarily independent) random variables sampled from Dp, and γi>0 with γ=∑iγi. Let X=∑iγi∣Xi∣p. Assume that m≥3. Then for any t≥1,
For i=1,…,m, let Xi be independent random variables sampled from Dp, and γi≥0 with γ=∑iγi. Let X=∑iγi∣ci∣. Then,
In addition, ΠA can be computed in O(nnz(A)) time.
In addition, ΠA can be computed in O(nnz(A)⋅dlogd) time.
Improving the Embedding Dimension
time. The second term comes from solving a subsampled problem of size O(d3+p/2log(1/ϵ)/ϵ2)×d.
Remark. We have stated our results in the previous sections as poly(d) without stating the value of the polynomial because there are numerous trade-offs between the conditioning quality and the running time. For example, let p=1. We can use a rounding algorithm instead of QR to compute the R matrix. If we use the input-sparsity time embedding with the O(d)-rounding algorithm of , then the running time to compute the (1±ϵ)-distortion embedding is O(nnz(A)⋅logn+d8/ϵ2) and the embedding dimension is O(d6.5/ϵ2) (ignoring log factors). If, on the other hand, we use QR to compute R, then the running time is O(nnz(A)⋅logn+d7/ϵ2) and the embedding dimension is O(d8/ϵ2). However, with the result from this section, the running time is simply O(nnz(A)⋅logn+poly(d)+Tp(ϵ;d3+p/2/ϵ2,d)) and the poly(d) term can be absorbed by the nnz(A) term.
Acknowledgments
References
Appendix A Appendix
Let the n×d matrix U be an orthonormal basis for the range of the n×d matrix A. Rather than proving the theorem by establishing that
where sij is the (i,j)-th element of S, dj is the j-th diagonal element of D, and ujk is the (j,k)-th element of U. We will use the following facts in the proof:
Given these results, it is easy to obtain that
For any δ∈(0,1), set s=(d2+d)/(ϵ2δ). Then, by Markov’s inequality,
Therefore, with probability at least 1−δ, we have ∥X−I∥2≤∥X−I∥F≤ϵ, which implies
We start with the following result, which establishes the existence of the so-called Auerbach’s basis of a d-dimensional normed vector space. For our proof, we will only need its existence and not an algorithm to construct it.
(Auerbach ) Let (A,∥⋅∥) be a d-dimensional normed vector space. There exists a basis {e1,…,ed} of A, called Auerbach basis, such that ∥ek∥=1 and ∥ek∥∗=1 for k=1,…,d, where {e1,…,en} is a basis of A∗ dual to {e1,…,en}.
For any y=Ux∈Y, we have ∥y∥1=∥Ux∥1≥∥x∥∞=1,
and thus ∥y∥1≤∥v∥1=d. Define YL={y∈Y∣∥yL∥1≥21∥y∥1} and YH=Y\YL. Given S, define a mapping ϕ:{1,…,n}→{1,…,s} such that sϕ(j),j=1, j=1,…,n, and split L into two subsets: L^={j∈L∣ϕ(j)∈ϕ(H)} and Lˉ=L\L^. Consider these events:
EU: ∣ΠU∣1≤ω1dlogd for some ω1>0.
EL: ∥SvL∥∞≤ω2/(dlogd) for some ω2>0.
EH: ϕ(j1)=ϕ(j2), ∀j1=j2, j1,j2∈H.
EC: minj∈∣H∣∣cj∣≥ω3/(d2log2d) for some ω3>0.
EL^: ∣ΠUL^∣1≤ω4/(d2log2d) for some ω4>0.
Recall that we set s=ωd5log5d in Theorem 2. We will show that, with ω sufficiently large and proper choices of ω1, ω2, ω3, and ω4, the event EU leads to an upper bound of ∥Πy∥1 for all y∈range(A), EU and EL lead to a lower bound of ∥Πy∥1 for all y∈YL with probability at least 0.9, and EH, EL^, and EC together imply an lower bound of ∥Πy∥1 for all y∈YH.
Provided EU, we have
For any y∈range(A), we can find an x such that y=Ux. Then,
Provided EL, for any fixed y∈YL, we have
By assumption EL and ∥yL∥1≥21∥y∥1≥21, we obtain the result. ∎
Assume both EU and EL. If ω1 and ω2 satisfy
for some δ∈(0,1) regardless of d, then, with probability at least 1−δ, we have
Set ϵ=1/(2+8ω1dlogd) and create an ϵ-net YϵL⊆YL such that for any y∈YL, we can find a yϵ∈YϵL such that ∥y−yϵ∥1≤ϵ. Since ∥y∥1≤d for all y∈YL, there exist such an ϵ-net with at most (3d/ϵ)d elements (Bourgain et al. ). By Lemma 13, we can apply a union bound for all the elements in YϵL:
For any y∈YL, we have, noting that y−yϵ∈range(A),
So we establish a lower bound for all y∈YL. ∎
Provided EH and EL^, if ω3>4ω4, we have
which creates a lower bound for all y∈YH. ∎
We continue to show that, with ω sufficiently large, by setting τ=ω1/4/(dlog2d) and choosing ω1, ω2, ω3, and ω4 properly, we have each event with probability at least 1−0.08=0.92 and thus
Moreover, the condition in Lemma 14 holds with δ=0.1, and the condition in Lemma 15 holds. Therefore, Π=SC has the desired property with probability at least 0.5, which would conclude the proof of Theorem 2.
With probability at least 0.92, EU holds with ω1=500(1+logω).
Setting ω1=500(1+logω) and t=ω1logd, we have
We assume that logd≥1 and logω≥1. ∎
For any δ∈(0,0.1), if s≥d/τ, we have,
Let Xij=sijvjL. We have E[Xij]=vjL/s, E[Xij2]=(vjL)2/s, and 0≤Xij≤vjL≤τ. Fixed i, Xij are independent, j=1,…,n. By Bernstein’s inequality,
where we use Holder’s inequality: ∥vL∥22≤∥vL∥1∥vL∥∞≤dτ. To obtain a union bound for all i with probability 1−δ, we need
Given δ<0.1, it suffices to choose s=d/τ and t=2log(d/(δτ))τ. Note that ∥vL∥1/s≤∥v∥1/s=τ. We have
Increasing s will decrease the failure rate, so it holds for all s≥d/τ. ∎
With probability at least 0.92, EL holds with ω2=(15+logω)/ω1/4.
By Lemma 17, with probability at least 0.92, EL holds with
With the above choices of ω1 and ω2, the condition in Lemma 13 holds with δ=0.1 for sufficiently large ω.
With ω1=500(1+logω), and ω2=(15+logω)/ω1/4, the first term in
increases much slower than the second term as ω increases, while both are at the order of dlogd. Therefore, if ω is sufficiently large, the condition hold with δ=0.1. ∎
If ω≥160, event EH holds with probability at least 0.92.
Given j1,j2∈H and j1=j2, let Xj1j2=1 if ϕ(j1)=ϕ(j2) and Xj1j2=0 otherwise. It is easy to see that Pr[Xj1j2=1]=s1. Therefore,
With probability at least 0.92, event EC holds with ω3=1/(8ω1/4).
∣H∣ is at most d/τ=ω1/4d2log2d. Then
Therefore, ω3=1/(8ω1/4) would suffice. ∎
With probability at least 0.92, event EL^ holds with ω4=25000(1+logω)/ω3/4. Thus with ω sufficiently large and the above choice of ω3, the condition in Lemma 15 ω3>4ω4 holds.
Assume that ∣UL^∣1≤ω3/4d2log3d25. Similar to the proof of Lemma 16, we have
It suffices to choose t=1000(1+logω)logd to make the RHS less than 0.04. So with probability at least 0.92, we have EL^ holds with ω4=25000(1+logω)/ω3/4. ∎
By Theorem 2 and Lemma 3, we know that Steps 2 and 4 of Algorithm 1 succeed with a constant probability. Conditioning on this event, we have
where the last inequality is due to ϵ<1/2. By Theorem 2, Step 2 takes O(nnz(A)) time, and Step 3 takes O(poly(d)) time because ΠA has O(poly(d) rows. Then, by Lemma 3, Step 4 takes O(nnz(A)⋅logn) time, and Step 5 takes T1(ϵ/4;O(poly(d)log(1/ϵ)/ϵ2),d) time. Therefore, the total running time of Algorithm 1 is as stated.
A.4 Proof of Lemma 8
Next, we state the following lemma, which is due to Nolan .
(Nolan [27, Thm. 1.12]) Let X∼Dp with p∈[1,2). Then as x→∞,
where cp=sin2πp⋅Γ(p)/π.
By Lemma 23, it follows that, as t→∞,
Hence, there exist αp′>0 and t1>0 such that for all t>t1,
Note that all the p-stable distributions with p∈ have finite and positive density at x=0. Therefore, there exists αp′′>0 such that for all 0≤t≤t1,
Let αp=max{αp′,αp′′}. We get αp∣C∣⪰∣Xp∣p. For the Gaussian distribution, we have, as t→∞,
which converges to zero much faster than t−1, so we can apply similar arguments to obtain βp.
A.5 Proof of Lemma 9 (Upper Tail Inequality for p𝑝p-stable Distributions)
Let Ci=Fc−1(Fp(Xi)), i=1,…,m, where Fc is the CDF of the standard Cauchy distribution and Fp is the CDF of Dp. Ci follows the standard Cauchy distribution, and, by Lemma 8, we have αp∣Ci∣≥∣Xi∣p. Therefore, for any t≥1,
A.6 Proof of Lemma 10 (Lower Tail Inequality for p𝑝p-stable Distributions)
Let Gi be independent random variables sampled from the standard Gaussian distribution, i=1,…,m. By Lemma 8, we have
The lower tail inequality from Lemma 7 concludes the proof.
A.8 Proof of Theorem 7 (Improving the Embedding Dimension)
Each of Steps 1, 3, and 5 of Algorithm 2 succeeds with a constant probability. We can control the success rate of each by adjusting the constant factor in the embedding dimension, such that all steps succeed with a constant probability. Conditioning on this event, we have κp(AR−1)=6d because
By Lemma 1, κˉp(AR−1)≤6d1/p+1, and then by Lemma 3, the embedding dimension of S is O(κˉpp(AR−1)d∣p/2−1∣dlog(1/ϵ)/ϵ2)=O(d3+p/2log(1/ϵ)/ϵ2).