Алгоритмы быстрого вычисления разреженного преобразования Фурье (Sparse FFT), осень 2015: БПФ разреженных

Sparse Fourier Transform(lecture 2)

Michael Kapralov1

1IBM Watson → EPFL

St. Petersburg CS ClubNovember 2015

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Given x ∈Cn, compute the Discrete Fourier Transform of x :

xf =1n

∑j∈[n]

xjω−f ·j ,

where ω= e2πi/n is the n-th root of unity.

Goal: find the top k coefficients of x approximately

In last lecture:

Ï 1-sparse noiseless case: two-point sampling

Ï 1-sparse noisy case: O(logn loglogn) time and samples

Ï reduction from k -sparse to 1-sparse case, via filtering

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xf =1n

∑j∈[n]

xjω−f ·j ,



In last lecture:




2 / 72


xf =1n

∑j∈[n]

xjω−f ·j ,



In last lecture:




2 / 72


xf =1n

∑j∈[n]

xjω−f ·j ,



In last lecture:




2 / 72

Partition frequency domain into B ≈ k buckets

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For each j = 0, . . . ,B−1 let

ujf =

xf , if f ∈ j-th bucket0 o.w.

Restricted to a bucket, signal is likely approximately 1-sparse!

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ujf =



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ujf =



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ujf =



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ujf =



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ujf =



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ujf =



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ujf =



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ujf =



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ujf =



6 / 72

We want time domain access to u0: for any a= 0, . . . ,n−1,compute

u0a = ∑

− n2B ≤f≤ n

2B

xf ·ωf ·a.

Let

Gf =

1, if f ∈ [− n2B : n

2B]

0 o.w.

Thenu0

a = (x·+a ∗ G)(0)

For any j = 0, . . . ,B−1

uja = (x·+a ∗ G)(j · n

B)

7 / 72

We want time domain access to u0: for any a= 0, . . . ,n−1,compute

u0a = ∑

− n2B ≤f≤ n

2B

xf ·ωf ·a.

Let

Gf =

1, if f ∈ [− n2B : n

2B]

0 o.w.

Thenu0

a = (x·+a ∗ G)(0)

For any j = 0, . . . ,B−1

uja = (x·+a ∗ G)(j · n

B)

7 / 72

Reducing k -sparse recovery to 1-sparse recovery

For any j = 0, . . . ,B−1

uja = (x·+a ∗ G)(j · n

B)

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magnitu

de

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For any j = 0, . . . ,B−1

uja = (x·+a ∗ G)(j · n

B)

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−1

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magnitu

de

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For any j = 0, . . . ,B−1

uja = (x·+a ∗ G)(j · n

B)

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−1

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de

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frequency

magnitu

de

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Need to evaluate(x·+a ∗ G)

(j · n

B

)for j = 0, . . . ,B−1.

We have access to x , not x ...

By the convolution identity

x·+a ∗ G = á(x·+a ·G)

Suffices to compute

àx·+a ·Gj · nB

, j = 0, . . . ,B−1

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(j · n

B

)for j = 0, . . . ,B−1.



x·+a ∗ G = á(x·+a ·G)

Suffices to compute

àx·+a ·Gj · nB

, j = 0, . . . ,B−1

11 / 72


(j · n

B

)for j = 0, . . . ,B−1.



x·+a ∗ G = á(x·+a ·G)

Suffices to compute

àx·+a ·Gj · nB

, j = 0, . . . ,B−1

11 / 72

Suffices to compute

àx·+a ·Gj · nB

, j = 0, . . . ,B−1

Sample complexity? Runtime?

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Computing x ·G takes Ω(N) time and samples!

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Suffices to compute

x ·Gj · nB

, j = 0, . . . ,B−1


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Suffices to compute

x ·Gj · nB

, j = 0, . . . ,B−1


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Suffices to compute

x ·Gj · nB

, j = 0, . . . ,B−1


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To sample all signals uj , j = 0, . . . ,B−1 in time domain, it sufficesto compute x ·Gj · n

B, j = 0, . . . ,B−1

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Computing x ·G takes supp(G) samples.

Design G with supp(G)≈ k that approximates rectangular filter?

11 / 72

In this lecture:

Ï permuting frequencies

Ï filter construction

12 / 72

1. Pseudorandom spectrum permutations

2. Filter construction

13 / 72



14 / 72

Pseudorandom spectrum permutationsPermutation in time domain plus phase shift =⇒ permutation infrequency domain

ClaimLet σ,b ∈ [n], σ invertible modulo n. Let yj = xσjω

−jb. Then

yf = xσ−1(f+b).

(proof on next slide; a close relative of time shift theorem)

Pseudorandom permutation:

Ï select b uniformly at random from [n]

Ï select σ uniformly at random from 1,3,5, . . . ,n−1(invertible numbers modulo n)

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−jb. Then

yf = xσ−1(f+b).





15 / 72



−jb. Then

yf = xσ−1(f+b).





15 / 72

Pseudorandom spectrum permutationsClaimLet yj = xσjω

−jb. Then yf = xσ−1(f+b).

Proof.

yf =1n

∑j∈[n]

yjω−f ·j

= 1n

∑j∈[n]

xσjω−(f+b)·j

= 1n

∑i∈[n]

xiω−(f+b)·σ−1i (change of variables i =σj)

= 1n

∑i∈[n]

xiω−σ−1(f+b)·i

= xσ−1(f+b)

16 / 72

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Design G with supp(G)≈ k that approximates rectangular filter?

Our filter G will approximate the boxcar. Bound collisionprobability now.

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Partition frequency domain into buckets, permute spectrum

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Frequency i collides with frequency j only if |σi −σj | ≤ nB .

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collision


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Collision probabilityLemmaLet σ be a uniformly random odd number in 1,2, . . . ,n. Then forany i , j ∈ [n], i 6= j one has

Prσ[|σ · i −σj | ≤ n

B

]=O(1/B)

Proof.Let ∆ := i − j = d2s for some odd d .

The orbit of σ ·∆ is 2s ·d ′ for all odd d ′.

[ ]σi − n

Bσi σi + n

B

There are O( n

B2s

)values of d ′ that make σ ·∆ fall into

[− nB , n

B],

out of n/2s+1.

21 / 72



B

]=O(1/B)



[ ]σi − n

Bσi σi + n

B

There are O( n

B2s


[− nB , n

B],

out of n/2s+1.21 / 72



B

]=O(1/B)



[ ]σi − n

Bσi σi + n

B

There are O( n

B2s


[− nB , n

B],

out of n/2s+1.21 / 72



B

]=O(1/B)



[ ]σi − n

Bσi σi + n

B

There are O( n

B2s


[− nB , n

B],

out of n/2s+1.21 / 72



B

]=O(1/B)



[ ]σi − n

Bσi σi + n

B

There are O( n

B2s


[− nB , n

B],

out of n/2s+1.21 / 72



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Rectangular buckets G have full support in time domain...

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Approximate rectangular filter with a filter G with small support?

Need supp(G)≈ k , so perhaps turn the filter around?

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Let

Gj :=

1/(B+1) if j ∈ [−B/2,B/2]0 o.w.

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Have supp(G)=B ≈ k , but buckets leak

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=⇒

In what follows: reduce leakage at the expense of increasingsupp(G)

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Window functionsDefinitionA symmetric filter G is a (B,δ)-standard window function if

1. G0 = 12. Gf ≥ 03. |Gf | ≤ δ for f 6∈ [− n

2B , n2B

]

− n2B 0 n

2B

ideal bucket

leakage to other buckets

bounded by δ¿ 1

26 / 72

Window functions

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

Start with the sinc function:

Gf :=sin(π(B+1)f/n)(B+1) ·πf/n

For all |f | > n2B we have

|Gf | ≤1

(B+1)πf/n≤ 1π/2

≤ 2/π≤ 0.9

27 / 72

Window functions

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

Start with the sinc function:

Gf :=sin(π(B+1)f/n)(B+1) ·πf/n


|Gf | ≤1

(B+1)πf/n≤ 1π/2

≤ 2/π≤ 0.9

27 / 72

Window functions

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

Consider powers of the sinc function:

Grf :=

(sin(π(B+1)f/n)(B+1) ·πf/n

)r


|Gf |r ≤ (0.9)r

28 / 72

Window functions

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

Consider powers of the sinc function:

Grf :=

(sin(π(B+1)f/n)(B+1) ·πf/n

)r


|Gf |r ≤ (0.9)r

29 / 72

Window functions

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

Consider powers of the sinc function: Grf


|Gf |r ≤ (0.9)r

So setting r =O(log(1/δ)) is sufficient!

30 / 72

Window functions

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

Consider powers of the sinc function: Grf


|Gf |r ≤ (0.9)r

So setting r =O(log(1/δ)) is sufficient!31 / 72


1. G0 = 12. Gf ≥ 03. |Gf | ≤ δ for f 6∈ [− n

2B , n2B

]

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

32 / 72


1. G0 = 12. Gf ≥ 03. |Gf | ≤ δ for f 6∈ [− n

2B , n2B

]

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

How large is supp(G)⊆ [−T ,T ]? 33 / 72

Let

Gj :=

1/(B+1) if j ∈ [−B/2,B/2]0 o.w.

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Let Gr := (G0)r . How large is the support of Gr ?

By the convolution identity Gr =G0 ∗G0 ∗ . . .∗G0

Support of G0 is in [−B/2,B/2], so

supp(G∗ . . .∗G)⊆ [−r ·B/2,r ·B/2]

33 / 72

Let

Gj :=

1/(B+1) if j ∈ [−B/2,B/2]0 o.w.

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supp(G∗ . . .∗G)⊆ [−r ·B/2,r ·B/2]

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Let

Gj :=

1/(B+1) if j ∈ [−B/2,B/2]0 o.w.

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supp(G∗ . . .∗G)⊆ [−r ·B/2,r ·B/2]

31 / 72

Let

Gj :=

1/(B+1) if j ∈ [−B/2,B/2]0 o.w.

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supp(G∗ . . .∗G)⊆ [−r ·B/2,r ·B/2]

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Let

Gj :=

1/(B+1) if j ∈ [−B/2,B/2]0 o.w.

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supp(G∗ . . .∗G)⊆ [−r ·B/2,r ·B/2]

31 / 72

Let

Gj :=

1/(B+1) if j ∈ [−B/2,B/2]0 o.w.

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supp(G∗ . . .∗G)⊆ [−r ·B/2,r ·B/2]

31 / 72

Let

Gj :=

1/(B+1) if j ∈ [−B/2,B/2]0 o.w.

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supp(G∗ . . .∗G)⊆ [−r ·B/2,r ·B/2]

31 / 72

Let

Gj :=

1/(B+1) if j ∈ [−B/2,B/2]0 o.w.

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supp(G∗ . . .∗G)⊆ [−r ·B/2,r ·B/2]

31 / 72

Let

Gj :=

1/(B+1) if j ∈ [−B/2,B/2]0 o.w.

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supp(G∗ . . .∗G)⊆ [−r ·B/2,r ·B/2]

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Flat window functionDefinitionA symmetric filter G is a (B,δ,γ)-flat window function if

1. Gj ≥ 1−δ for all j ∈ [−(1−γ) n2B ,(1−γ) n

2B]

2. Gj ∈ [0,1] for all j

3. |Gf | ≤ δ for f 6∈ [− n2B , n

2B]

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

32 / 72



2B]

2. Gj ∈ [0,1] for all j

3. |Gf | ≤ δ for f 6∈ [− n2B , n

2B]

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

1−γ fraction of bucket

33 / 72



2B]

2. Gj ∈ [0,1] for all j

3. |Gf | ≤ δ for f 6∈ [− n2B , n

2B]

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1

0.99 fraction of bucket

34 / 72

Flat window function – construction

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


Let H be a (2B/γ,δ/n)-standard window function. Note that

|Hf | ≤ δ/n

for all f outside of [−γ n

4B,γ

n4B

].

35 / 72


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


1−γ/2 fraction of bucket


|Hf | ≤ δ/n


4B,γ

n4B

].

36 / 72


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1



|Hf | ≤ δ/n


4B,γ

n4B

].

37 / 72


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1



|Hf | ≤ δ/n


4B,γ

n4B

].

38 / 72

Flat window function – constructionTo construct G:

1. sum up shifts H ·−∆ over all ∆ ∈ [−U ,U], where

U = (1−γ/2)n

2B2. normalize so that G0 = 1±δ

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


39 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


39 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


39 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


39 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


40 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


40 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


41 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


41 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


41 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


41 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


41 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


41 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


41 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


41 / 72



U = (1−γ/2)n


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


41 / 72

To construct G:1. sum up shifts H ·−∆ over all ∆ ∈ [−U ,U], where

U = (1−γ/2)n

2B

2. normalize so that G0 = 1±δ

Formally:

Gf :=1Z

(H f−U + H f+1−U + . . .+ H f+U

)where Z is a normalization factor.

Upon inspection, Z =∑f∈[n] Hf works.

42 / 72


U = (1−γ/2)n

2B


Formally:

Gf :=1Z

(H f−U + H f+1−U + . . .+ H f+U



42 / 72


U = (1−γ/2)n

2B


Formally:

Gf :=1Z

(H f−U + H f+1−U + . . .+ H f+U



43 / 72

Formally:

Gf :=1Z

(H f−U + H f+1−U + . . .+ H f+U



(Flat region) For any f ∈ [−(1−γ) n2B ,(1−γ) n

2B ] (flat region) onehas

H f−U + H f+1−U + . . .+ H f+U ≥ ∑f∈[−γ n

4B ,γ n4B ]

Hf

≥Z − tail of H≥Z − (δ/n)n ≥Z −δ

44 / 72

Formally:

Gf :=1Z

(H f−U + H f+1−U + . . .+ H f+U



Indeed, for any f 6∈ [− n2B , n

2B ] (zero region) one has

H f−U + H f+1−U + . . .+ H f+U ≤ ∑f>γ n

4B

Hf

≤ tail of H ≤ (δ/n)n ≤ δ

45 / 72

Flat window function

− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1


How large is support of G := 1Z

(H ·−U + . . .+ H ·+U

)?

By time shift theorem for every q ∈ [n]

Gq :=Hq · 1Z

U∑j=−U

ωqj

Support of G a subset of support of H!

46 / 72


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1



(H ·−U + . . .+ H ·+U

)?


Gq :=Hq · 1Z

U∑j=−U

ωqj

Support of G a subset of support of H!

46 / 72


− n2B 0 n

2B

ideal bucket


bounded by δ¿ 1



(H ·−U + . . .+ H ·+U

)?


Gq :=Hq · 1Z

U∑j=−U

ωqj

Support of G a subset of support of H!46 / 72

Flat window functions – construction

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