Algorithm 9 Energy estimation procedure for individual frequencies
1: procedure EstimateValues(X,χ, L, n, kb 0, k1, δ, p)
2: B ← 1200δ k0k1 3: F ← 10 log1δ
4: G ← (n, B, F )-flat filter . See Definition 2.1
5: F ← {f ∈ [n] : round(kf
1) ∈ L}
6: T ← 10 log2p
7: fort ∈ {1, . . . , T } do
8: ∆ ← uniform random sample from [kn
1]
9: σ ← uniform random sample from odd numbers in [kn
1]
10: U ← HashToBins(X,b χ, G, n, B, σ, ∆)b . of(f ), h(f ) in Definition 4.2
11: Wf(t) ← bG−1o
f(f )Ubh(f )ω−σ∆f for eachf ∈ F
12: Wf ← Mediant(x(t)f ) for each f ∈ F . Separately for the real and imaginary parts
13: returnW
Once we have located the blocks, we need to estimate the frequency values with them. The function EstimateValues in Algorithm 9 performs this task for us via basic hashing techniques.
The following lemma characterizes the guarantee on the output.
Lemma 5.2 (Re-stated from Section 5.1) For any integers(n, k0, k1), list of block indices L, param-etersδ ∈ n1,201
and p ∈ (0, 1/2), and signals X ∈ Cn andχ ∈ Cb n with k bX −χkb 2≥ poly(n)1 kχkb 2, the outputW of the function EstimateValues(X,χ, L, n, kb 0, k1, δ, p) has the following property:
X
f ∈S
j∈LIj
|Wf − ( bX −χ)b f|2 ≤ δ|L|
3k0k bX −χkb 22
with probability at least 1 − p, where Ij is the j-th block. Moreover, the sample complexity is O(k0δk1 log1plog1δ), and if kχkb 0 = O(k0k1), then the runtime is O(k0δk1log1plog1δlog n+k1·|L| log1p).
Proof. LetX0 = X − χ, and let bU be the output of HashToBins and bU∗ its exact counterpart. We start by calculatingWf(t) for an arbitraryf ∈ F:
Wf(t) = bG−1o (σ, ∆) implicitly depending on t.
Bounding err(t)f and errf(t)f : Using Lemma G.1 in Appendix G, we have wherec is used in HashToBins, and c0 is the exponent in the poly(n) notation of the assumption k bX −χkb 2 ≥ poly(n)1 kχkb 2.
In order to characterize |Wf(t)− bXf0|2, we use the following:
|errff|2+ 2|err(t)f | · |errf(t)f | ≤ 4n2(−c+c0)k bX0k22+ 4n−c+c0k bX0k2· |err(t)f |. (117) which follows directly from (116).
Characterizing |Wf(t)− bXf0|2: We have from (114), (115), and (117) that where the last line follows by writing E
|err(t)f |
≤ q
E
|err(t)f |2
via Jensen’s inequality, and then applying (115).
Taking the median: Recall that Wf is the median of T independent random variables, with the median taken separately for the real and imaginary parts. Since |W |2 = |Re(W )|2+ |Im(W )|2, we find that (119) holds true whenWf(t)− bXf0 is replaced by its real or imaginary part. Hence, with probability at least1 − T /2T 1
6t
T /2
, we have |Re(Wf(t)− bXf0)|2 < 160tB k bX0k22, and analogously for the imaginary part. Combining these and applying the union bound, we obtain
P
where we first applied T /2T
≤ 2T, and then the choice T = 10 log2p from Algorithm 9 and the choice ofp ≤ 1/2.
We now bound the error as follows:
|Wf − bXf0|2 ≤ 320
We write the expected value of the second term as Eh |Wf− bXf0|2−320 where we applied the change of variablev = 1 + 320 u
B k bX0k22. By incorporating (120) into this integral,
where the second line is by explicitly evaluating the integral (withT > 2), and the third by T /2−1 ≥ 4 (cf., Algorithm 9).
Summing (121) over F = ∪j∈LIj, we find that the total error is upper bounded as follows:
X
with probability at least1 − p, by Markov’s inequality. The lemma now follows by the choice B =
1200
δ k0k1 in Algorithm 9.
Sample complexity and runtime: To calculate the sample complexity, note that the only opera-tion in the algorithm that takes samples is the call to HashToBins. By Lemma 4.6, and the choices B = 1200k0δk1 andF = 10 log1δ, the sample complexity isO(k0δk1 log1δ) per hashing performed. Since we run the hashing in a loop10 log2p times, this amounts to a total ofO(k0δk1log1δlogp1).
The runtime depends on two operations. The first one is calling HashToBins, whose analysis follows similarly to the aforementioned sample complexity analysis using the assumption kχkb 0 = O(k0k1), but with an extra log n factor compared to the sample complexity, as per Lemma 4.6.
The other operation is computation ofWf(t), which takes unit time for eachf ∈ F. Since the size of |F |= k1· |L|, running it in a loop costs O(k1· |L| logp1). Computing the median is done in linear time which consequently results in |F |T = O(k1· |L| log1p).
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