The Sparse Fourier Transform. Haitham Hassanieh
Читать онлайн книгу.href="#fb3_img_img_c1f269be-a7e7-51e8-899f-6e50b7dc5aed.png" alt="image"/> time. Location loops thus take
For estimation loops, we get the following guarantee.
Lemma 3.1 Let S be the support of the largest k coefficients of x̂, and x̂−S contain the rest. Then for any ∊ ≤ 1,
Proof The proof can be found in Appendix A.1. ■
Furthermore, since
Algorithm 3.1 SFT 1.0: Non-iterative Sparse Fourier Transform for
Figure 3.1 Example inner loop of the algorithm on sparse input. This run has parameters n = 256, k = 4, G being the (0.11, 0.06, 2 × 10−9, 133) flat window function in Figure 2.1, and selecting the top 4 of B = 16 samples. In part (a), the algorithm begins with time domain access to Pσ, τ, bx given by (Pσ, τ, bx)i = xσ(i−τ)ωσbi, which permutes the spectrum of x by permuting the samples in the time domain. In part (b), the algorithm computes the time domain signal y = G · Pσ, τ, bx. The spectrum of y (pictured) is large around the large coordinates of Pσ, τ, bx. The algorithm then computes ẑ, which is the rate B subsampling of ŷ as pictured in part (c). During estimation loops, the algorithm estimates x̂i based on the value of the nearest coordinate in ẑ, namely
Lemma 3.2 Define
Proof The proof can be found in Appendix A.2 ■
3.2.2 Non-Iterative Sparse Fourier Transform
Our SFT 1.0 algorithm shown in Algorithm 3.1 is parameterized by ∊ and δ. It runs L = O(log n) iterations of the inner loop, with parameters
Lemma 3.3 The algorithm runs in time
Proof To analyze this algorithm, note that
or |I′| ≤ 2dkn/B. Therefore, the running time of both the location and estimation inner loops is
Theorem 3.1 Running the algorithm with parameters ∊, δ < 1 gives x̂′ satisfying
with probability 1 − 1/n and running time
Proof Define
Lemma 3.2 says that in each location iteration r, for any i with |x̂i| ≥ 4E,
Thus, E[si] ≥ 3L/4, and each iteration is an independent trial, so by a Chernoff bound the chance that si < L/2 is at most 1/2Ω(L) < 1/n3. Therefore by a union bound, with probability at least 1 − 1/n2, i ∈ I′ for all i with |x̂i| ≥ 4E.
Next, Lemma 3.1 says that for each estimation iteration r and index i,
Therefore, with probability
Since real
By a union bound over I′, with probability at least 1 − 1/n2 we have