D Omitted Proofs from Section 5

D.1 Proof of Lemma 5.3

Note on _poly(n)¹ assumptions in lemmas: Throughout the proof, we apply Lemmas 3.5, 5.1, and 5.2. The first of these assumes that χb₀ uniformly distributed over an arbitrarily length-Ω _poly(n)^{kχkb 2}
interval, and the latter two use the assumption k bX −χkb 2 ≥ _poly(n)¹ kχkb ²₂.

We argue that these assumptions are trivial and can be ignored. To see this, we apply a minor

technical modification to the algorithm as follows. Suppose the implied exponent to the poly(n) notation isc⁰. By adding a noise term toχb₀ on each iteration uniform in[−n^−c0+10kχkb ², n^c0+10kχkb ²], we immediately satisfy the first assumption above, and we also find that the probability of k bX −χkb ₂ <

1

poly(n)kχkb ²₂ is at most O(n⁻¹⁰), and the additional error in the estimate is O(n^c0+10kχkb ²). Since we only doO(log SNR⁰) = O(n) iterations (by the assumption SNR⁰= O(poly(n))), this does not affect the result because the accumulated noise added to χb0 which we denote by err(χb0), does not exceed

k bXk²₂

poly(n) which by the final assumption of the lemma implies that err(χb0) ≤ ν².

Overview of the proof: We introduce the approximate support set of the input signal bX, given by the union of the top k0 blocks of the signal and the blocks whose energy is more than the tail noise level:

S0 :=n j ∈h n

k1

i

: k bXIjk²₂ ≥ µ²o

∪



arg min

S⊂[_k1ⁿ]

|S|=k0

X

j∈[ⁿ

k1]\S

k bXIjk²₂



. (77)

From the definition of µ² in Definition 1.2, we readily obtain |S₀| ≤ 2k₀. For each t = 1, 2, ..., T , define the setSt as

St= St−1∪ L⁰_t,

whereL⁰_tis the output of PruneLocation at iteration t of ReduceSNR. Stcontains the set of the head elements of bX, plus every element that is modified by the algorithm so far.

We prove by induction on the iteration number t = 1, . . . , T that there exist events E₀ ⊇ E₁ ⊇ ... ⊇ ET such that conditioned on E_t, the following conditions hold true:

a. |St| ≤ 2k0+^tk_T⁰; b. kχb^(t)_I

jk²₂ = 0 for all j ∈_n

k1

\S_t;

c. k bX −χb^(t)k²₂ ≤ 99 · SNR⁰(k0ν²)/2^t;

and for eacht ≤ T , we have P[Et+1|E_t] ≥ 1 −_10T¹ .

Base case of the induction: We have already deduced that |S₀| ≤ 2k₀ and definedχb⁽⁰⁾ = 0, and we have k bX −χb⁽⁰⁾k²₂ = k bXk²₂ ≤ SNR⁰· (k0µ²)/2⁰ by the two assumptions of the lemma. Hence, we can let E₀ be the trivial event satisfying P[E0] = 1.

Inductive step: We seek to define an event E_t+1 that occurs with probability at least 1 − _10T¹ conditioned on E_t, and such that the induction hypotheses a, b, and c are satisfied fort+1 conditioned on E_t+1. To do this, we will introduce three events E_loc,t, E_prune,t, and E_est,t, and set E_t+1 = Eloc,t∩ E_prune,t∩ E_est,t∩ E_t. Throughout the following, we let δ, θ, and p be chosen as in Algorithm 6 Success event associated with MultiBlockLocate: Let Eloc,t be the event of having a successful run of MultiBlockLocate(X,χb^(t), n, k₁, k₀, δ, p) at iteration t + 1 of the algorithm,

meaning the following conditions on the outputL:

|L| ≤ C · T ·k0

δ logk0

δ log³ 1

δp (78)

X

j∈St\L

k( bX −χb^(t))Ijk²₂ ≤ 0.1k bX −χb^(t)k²₂, (79) whereC is a constant to be specified shortly, for a small enough δ. To show this, we invoke Lemma 3.5 with S^∗ = St. Note that inductive hypothesis (a) implies |St| ≤ (2 + _{log SNR}^t 0)k0 ≤ 3k0. By the first part of Lemma 3.5, we have E

|L|

≤ C^{0 k}_δ⁰ log^k_δ⁰ log¹_plog^{2 1}_δp for an absolute constantC⁰, and hence (78) follows withC = 100C⁰ and probability at least 1 −_100T¹ , by Markov’s inequality.

By the second part of Lemma 3.5, (79) holds with probability at least 1 − p provided that δ ≤

0.1 200

2

, so by the union bound, the event E_loc,toccurs with probability at least P[Eloc,t|E_t] ≥ 1−p−_100T¹ . Success event associated with PruneLocation: Let Eprune,tbe the event of having a success-ful run of PruneLocation(X,χb^(t), L, k0, k1, δ, p, n, θ) at iteration t + 1 of the algorithm, meaning the following conditions on the outputL⁰:

|L⁰\St| ≤ k0

T (80)

X

j∈[ⁿ

k1]\L⁰

k( bX −χb^(t))_Ijk²₂ ≤ 0.2k bX −χb^(t)k²₂+ k₀(µ²+ 33ν²SNR⁰/2^t+1). (81)

The probability of (80) holding: In order to bound |L⁰\St|, first recall that the set Stail, defined in Lemma 5.1 part (a), has the following form:

S_tail =n j ∈h n

k1

i : k( bX −χb^(t))_Ijk₂≤√ θ −

r δ k0

k bX −χb^(t)k₂o .

By substitutingθ = 10 · 2^−(t+1)· ν²(SNR⁰) and using k bX −χb^(t)k²₂ ≤ 99 · SNR⁰(k0ν²)/2^t from part c of the inductive hypothesis, we have

√ θ −

r δ k0

k bX −χb^(t)k₂

≥ q

10 · 2^−(t+1)· ν²(SNR⁰) − r δ

k0

q

99 · SNR⁰(k0ν²)/2^t

≥ q

9 · ν²(SNR⁰)/2^t+1,

where the last inequality holds whenδ is sufficiently small. Hence, Stail ⊇n

j ∈h n k1

i

: k( bX −χb^(t))Ijk²₂ ≤ 9 · ν²(SNR⁰)/2^t+1o

. (82)

Now, to prove that (80) holds with high probability, we write

|L⁰\S_t| = |(L⁰∩ S_tail)\S_t| + |L⁰\(S_tail∪ S_t)|. (83) To upper bound the first term, note that by the first part of Lemma 5.1, we have

E

hL⁰∩ S_taili

≤ δp · |L|,

and hence by Markov’s inequality, the following holds with probability at least1 −_100T¹ : (L⁰∩ Stail)\St

 ≤

L⁰∩ Stail



≤ 100T δp · |L|

≤ 100T δp · CTk0

δ logk0

δ log³ 1 δp

= 100CT²p · k₀logk₀ δ log³ 1

δp

= 100Cδ · k₀log^{3 1}_δp log^k_δ⁰ · log²SNR⁰,

where the third line follows from (78) (we condition on E_loc,t), and the fifth line follows from T = log SNR⁰ and the choice p = ^δ

log^{2 k0}_δ log⁴SNR⁰ in Algorithm 6. Again using this choice of p, we claim that ^{100Cδ log}

3 1 δp

log^k0_δ ·log SNR⁰ ≤ 1 for sufficiently small δ regardless of the values (k0, SNR⁰); this is because the dependence of 1/p on k₀ and SNR⁰ is logarithmic, so in the numerator contains log³log k₀ and log³log SNR⁰ while the denominator contains log k0 andlog²SNR⁰. Hence,

(L⁰∩ S_tail)\S_t

 ≤ k0

log SNR⁰ with probability at least1 −_100T¹ .

We now show that the second term in (83) is zero, by showing that S_tail∪ S_t= [_kⁿ

1]. To see this, note that the termν²(SNR⁰)/2^t+1 in the bound on Stail in (82) satisfies

ν²(SNR⁰)/2^t+1≥ 1 2ν²≥ 1

2µ², (84)

by applyingt ≤ T = log SNR⁰, followed by the first assumption of the lemma. Hence, Stail\St⊇

 j ∈h n

k1

iSt : k( bX −χb^(t))Ijk²₂≤ 4µ²

 .

By part b of the inductive hypothesis, we have k( bX −χb^(t))_Ijk²₂ = k bX_Ijk²₂ for all j /∈ St, and hence Stail\St⊇

 j ∈h n

k1

iSt : k bXIjk²₂ ≤ 4µ²

 .

But from (77), we know that S0 (and hence St) contains all j with k bXIjk²₂ > 4µ², so we obtain Stail\St⊃_n

k1

\St, and henceStail∪ St= (Stail\St) ∪ St=_n

k1

, as required.

Therefore, the probability of (80) holding conditioned on E_t and E_loc,t is at least 1 −_100T¹ . The probability of (81) holding: To show (81), we use the second part of Lemma 5.1. The set S_head therein is defined as

S_head=n j ∈h n

k₁ i

: k( bX −χb^(t))_Ijk₂≥√ θ +

r δ

k₀k bX −χb^(t)k₂o .

By substitutingθ = 10 · 2^−(t+1)· ν²(SNR⁰) and using k bX −χb^(t)k²₂ ≤ 99 · SNR⁰(k0ν²)/2^t from part c of the inductive hypothesis, we have

√ θ +

r δ k0

k bX −χkb 2

= q

10 · 2^−(t+1)· ν²(SNR⁰) + r δ

k0

q

99 · SNR⁰(k₀ν²)/2^t

≤ q

11 · ν²(SNR⁰)/2^t+1 for sufficiently small δ, and hence

S_head⊇n j ∈h n

k1

i : k( bX −χ)b _Ijk²₂ ≥ 11 · ν²(SNR⁰)/2^t+1o

. (85)

Next, we write X

j∈[_k1ⁿ]\L⁰

k( bX −χb^(t))Ijk²₂ = X

j∈(St∩S_head∩L)\L⁰

k( bX −χb^(t))Ijk²₂+ X

j∈(St∩S_head)\(L⁰∪L)

k( bX −χb^(t))Ijk²₂

+ X

j∈St\(S_head∪L⁰)

k( bX −χb^(t))Ijk²₂+ X

j∈[_k1ⁿ]\(L⁰∪St)

k( bX −χb^(t))Ijk²₂, (86)

and we proceed by upper bounding the four terms.

Bounding the first term in (86): By the second part of Lemma 5.1 and the use of Markov, we

have X

j∈(St∩S_head∩L)\L⁰

k( bX −χb^(t))_Ijk²₂ ≤ δ X

j∈L∩S_head

k( bX −χb^(t))_Ijk²₂ ≤ δk bX −χb^(t)k²₂

with probability at least1 − p.

Bounding the second term in (86): Conditioned on E_loc,t, we have X

j∈(St∩S_head)\(L∪L⁰)

k( bX −χb^(t))Ijk²₂≤ X

j∈St\L

k( bX −χb^(t))Ijk²₂≤ 0.1k bX −χb^(t)k²₂, where we have applied (79).

Bounding the third term in (86): We have X

j∈St\(S_head∪L⁰)

k( bX −χb^(t))Ijk²₂≤ X

j∈St\S_head

k( bX −χb^(t))Ijk²₂

≤ |S_t\S_head| · max

j∈St\S_headk( bX −χb^(t))_Ijk²₂

≤ |St|(11 · ν²SNR⁰/2^t+1)

by (85). Part a of the inductive hypothesis implies that |S_t| ≤ 3k₀, and hence X

j∈(L∩St)\(Shead∪L⁰)

k( bX −χb^(t))_Ijk²₂≤ 33k₀· ν²SNR⁰/2^t+1.

Bounding the fourth term in (86):

X

j∈[_k1ⁿ]\(L⁰∪St)

k( bX −χb^(t))Ijk²₂≤ X

[_k1ⁿ]\St

k( bX −χb^(t))Ijk²₂

= X

[ⁿ

k1]\St

k bXIjk²₂ ≤ k0µ²,

where the equality follows from part b of the inductive hypothesis, and the final step holds sinceSt

contains all elements with k bX_Ijk²₂≥ µ² (cf., (77)).

Adding the above four contributions and applying the union bound, we find that conditioned on E_t and E_loc,t, (81) holds with probability at least P[Eprune,t|E_t∩ E_loc,t] ≥ 1 − p −_100T¹ , provided that δ is a sufficiently small constant (δ ≤ 0.1).

Success event associated with EstimateValues: Let Eest,tbe the event of having a successful run of EstimateValues(X,χb^(t), L⁰, k₀, k₁, δ, p) at iteration t + 1 of the algorithm conditioned on E_t, meaning the following conditions on the output signalW :

W_f = 0 for all f /∈ F X

j∈L⁰

k( bX −χb^(t)− W )_Ijk²₂ ≤ δk bX −χb^(t)k²₂, (87) where F contains the frequencies within the blocks indexed by L⁰. By Lemma 5.2 and the fact that |L⁰| ≤ 3k₀ conditioned on E_prune,t, E_loc,t, and E_t, it immediately follows that E_est,t occurs with probability at least P[Eest,t|E_prune,t∩ E_loc,t∩ E_t] ≥ 1 − p.

Combining the events: We can now wrap everything up as follows:

P E_t+1

E_t

= P

E_loc,t∩ E_prune,t∩ E_est,t∩ E_t E_t

= P E_est,t

E_loc,t∩ E_prune,t∩ E_t P

E_prune,t

E_loc,t∩ E_t P

E_loc,t E_t

. Substituting the probability bounds into the above equation, we have

P E_t+1

E_t

≥ 1 − 3p − 2

100T ≥ 1 − 1 20T, by the choice ofp in Algorithm 6 along with T = log SNR.

Now we show that the event E_t+1 = E_loc,t∩ E_prune,t∩ E_est,t∩ E_t implies the induction hypothesis.

Conditioned on E_prune,t∩E_t, we have (80), which immediately gives part a. Conditioned on E_loc,t∩E_est,t, from the definitionS_t+1= S_t∪ L⁰, part b of the inductive hypothesis follows from the fact that only elements inL⁰ are updated. Finally, conditioned on E_t∩ E_prune,t∩ E_est,t, we have

k bX −χb^(t+1)k²₂ = X

j∈L⁰

k( bX −χb^(t+1))Ijk²₂+ X

j∈[_k1ⁿ]\L⁰

k( bX −χb^(t+1))Ijk²₂

= X

j∈L⁰

k( bX −χb^(t)− W )Ijk²₂+ X

j∈[_k1ⁿ]\L⁰

k( bX −χb^(t+1))Ijk²₂

≤ (0.2 + δ)k bX −χb^(t)k²₂+ k0(µ²+ 33ν²SNR⁰/2^t+1)

≤ 99ν²k0SNR⁰/2^t+1,

where the second line holds since W is non-zero only for the blocks indexed by L⁰, the third line follows from (81) and (87), and the last line holds for sufficiently smallδ from part c of the induction hypothesis, and the upper boundµ²≤ 2ν²(SNR⁰)/2^t+1 given in (84).

The first part of the lemma now follows from a union bound over the T iterations and the fact thaterr(χb₀) ≤ ν², and by noting that the three parts of the induction hypothesis immediately yield the two claims therein. We conclude by analyzing the sample complexity and runtime.

Sample complexity: We have from Lemma 3.5 that the expected sample complexity of Multi-BlockLocate in a given iteration is O^{∗ k}_δ⁰ log(1 + k₀) log n + ^k0_δ^k2¹

, and multiplying by the number T = O(log SNR⁰) of iterations gives a total of O^∗ log SNR^{0 k}_δ⁰ log(1 + k0) log n + ^k0_δ^k2¹

. Hence, by Markov’s inequality, this is also the total sample complexity across all calls to Multi-BlockLocate with probability at least 1 − ₁₀₀¹ ; this probability can be combined with the union bound that we applied above. Since δ = Ω(1), the above sample complexity simplifies to O^∗(k₀log(1 + k₀) log SNR⁰log n + k₀k₁log SNR⁰).

By Lemma 5.1, the sample complexity of PruneLocation is O(^k0_δ^k1log_δp¹ log¹_δ), and by Lemma 5.2, the sample complexity of EstimateValues is O(^k0_δ^k1log¹_plog¹_δ). Substituting the choices of δ andp, these behave as O^∗(k₀k₁) per iteration, or O^∗(k₀k₁log SNR⁰) overall.

Runtime: By Lemma 3.5, the expected runtime of MultiBlockLocate in a given iteration is O^{∗ k}_δ⁰ log(1 + k0) log²n + ^k0_δ^k2¹ log²n + ^k0_δ^k1log³n

. Moreover, by Lemma 5.1, the runtime of PruneLocation as a function of |L| is O(^k0_δ^k1 log_δp¹ log¹_δlog n + k₁· |L| log_δp¹), and substituting E

|L|

= O ^k_δ⁰ log^k_δ⁰ log¹_plog^{2 1}_δp

from Lemma 3.5, this becomes O^{∗ k0}_δ^k1 log n

in expectation, by absorbing thelog¹_δ and log¹_p factors into theO^∗(·) notation.

Summing the preceding per-iteration expected runtimes, multiplying by the number of itera-tions T , and substituting the choices of T , p and δ, we find that their combined expectation is O^∗(k0log k0log SNR⁰log²n + k0k1log SNR⁰log³n). Hence, by Markov’s inequality, this is also the total sample complexity across all calls to MultiBlockLocate and PruneLocation with prob-ability at least 1 − ₁₀₀¹ . Applying the union bound over this failure event, ¯E_T, and the 1/poly(n) probability failure event arising from random perturbations of χb0, we obtain the required bound of 0.9 on the success probability.

By Lemma 5.2, the runtime of EstimateValues is O(^k0_δ^k1 log¹_plog¹_δlog n + k1· |L⁰| log¹_p), which behaves asO(^k0_δ^k1log¹_plog¹_δlog n) conditioned on E_prune,t∩ E_t (see (80) and recall that |S_t| ≤ 3k₀).

By our choices of p and δ, this simplifies to O^∗(k₀k₁log n) per iteration, or O^∗(k₀k₁log SNR⁰log n) overall.

D.2 Proof of Lemma 5.4

The proof resembles that of Lemma 5.3, but is generally simpler, and has some differing details. We provide the details for completeness.

Note on _poly(n)¹ assumptions in lemmas: We use an analogous argument to the start of Section D.1 to handle the assumptions | bX0−χb0|₂ ≥ _poly(n)¹ kχkb ²₂ and k bX −χkb 2 ≥ _poly(n)¹ kχkb ²₂ in Lemmas 3.5, 5.1, and 5.2. Specifically, we add a noise term to χb0 uniform in [−n^−c0+10kχkb ², n^c0+10kχkb ²]. This does not affect the result, since the noise added toχb₀ which we denote by err(χb₀), does not exceed

k bXk²₂

poly(n) which by the assumptions of the lemma implies that err(χb0) ≤ ν².

Overview of the proof: We first introduce the approximate support set of the input signal bX −χ,b given by the top10k0 blocks of the signal:

S0:= arg min

S⊂[_k1ⁿ]

|S|=10k0

X

j /∈S

k( bX −χ)b Ijk²₂ (88)

We also introduce another set indexing blocks whose energy is sufficiently large:

S_ =n j ∈h n

k₁

i : k( bX −χ)b _Ijk²₂≥ Err²( bX −χ, 10kb ₀, k₁) k₀

o∪ S₀. (89)

It readily follows from this definition and Definition 1.2 that |S\S0| ≤ k0/.

The function calls three other primitives, and below, we show that each of them succeeds with high probability by introducing suitable success events. Throughout, we letθ, p, and η be as chosen in Algorithm 6

Success event of the location primitive: Let E_loc be the event of having a successful run of MultiBlockLocate(X,χ, kb 1, k0, n, ², p), meaning the following conditions on the output L:

|L| ≤ Ck₀

² logk₀

² log³ 1

²p (90)

X

j∈S0\L

k( bX −χ)b Ijk²₂≤ 200k bX −χkb ²₂, (91) whereC is a constant to be specified shortly. To verify these conditions, we invoke Lemma 3.5 with S^∗= S0. By the first part of Lemma 3.5, we have E

|L|

≤ C^{0 k}_2⁰ log^k_⁰ log^{3 1}_pfor an absolute constant C⁰, and hence (90) follows withC = 100C⁰ and probability at least 1 −₁₀₀¹ , by Markov’s inequality.

By the second part of Lemma 3.5 with δ = ², (91) holds with probability at least 1 − p, so by the union bound, the event E_loc occurs with probability at least 1 − p −₁₀₀¹ .

Success event of the pruning primitive: Let E_prune be the event of having a successful run of PruneLocation(X,χ, L, n, kb ₀, k₁, , p, θ), meaning the following conditions on the output L⁰:

|L⁰\S₀| ≤ 2k₀

 (92)

X

j∈[ⁿ

k1]\L⁰

k( bX −χ)b Ijk²₂≤ 300k bX −χkb ²₂+ 6000ν²k0+ Err²( bX −χ, 10kb 0, k1). (93)

Bounding the probability of (92): In order to bound |L⁰\S0|, first note that the set Stail, defined in Lemma 5.1 part (a), has the following form:

S_tail =n Now, to prove that (92) holds with high probability, we write

|L⁰\S0| = |(L⁰∩ S)\S0| + |L⁰\(S0∪ S)|

≤ |(L⁰∩ S)\S0| + |(L⁰∩ Stail)\S| + |L⁰\(Stail∪ S)|.

(95) We first upper bound the first term as follows:

|(L⁰∩ S_)\S₀| ≤ |S_\S₀| ≤ k₀/,

which follows directly from the definition ofS. To upper bound the second term in (95), note that by Lemma 5.1 part (a) withδ = ,

Eh

L⁰∩ Stail

i

≤ p · |L|,

and hence by Markov’s inequality, the following holds with probability at least1 −₁₀₀¹ : (L⁰∩ S_tail)\S

where the third line follows from (90) (we condition on E_loc), and the fifth line follows from and the choicep = ^η

log^{2 k0}_ in Algorithm 6. Again using this choice ofp, we claim that ^{100Cη log}

3 1

p

log^k0_δ ≤ 1 for sufficiently small η regardless of the value of k0; this is because the dependence of 1/p on k0

is logarithmic, so the numerator contains log³log k₀, while the denominator contains log k₀ which means that the ratio is upper bounded and can be made arbitrarily small by choosing a small enough constantη. Hence

(L⁰∩ S_tail)\S_  ≤ k0

with probability at least1 −₁₀₀¹ .

We now show that the third term in (95) is zero, by showing that S_tail∪ S_ = [_kⁿ

1]. To see this, note that the termν² in the definition of S_tail is more than ^{Err2( bX−}_k^{χ,10kb 0,k1)}

0 by the first assumption of the lemma, and hence

Stail\S ⊃n

Bounding the probability of (93): To show (93), we use the second part of Lemma 5.1. The set Shead therein is defined as

S_head =n

and we proceed by upper bounding the four terms.

Bounding the first term in (97): By part b of Lemma 5.1, the choice δ = , and the use of with probability at least1 − p.

Bounding the second term in (97): Conditioned on E_loc, we have X where we have applied (91).

Bounding the third term in (97): We have X

j∈S0\(S_head∪L⁰)

k( bX −χ)b Ijk²₂ ≤ X

j∈S0\S_head

k( bX −χ)b Ijk²₂

≤ |S₀\S_head| · max

j∈S0\S_headk( bX −χ)b _Ijk²₂

≤ |S0|(600 · ν²)

by (96). We have by definition that |S₀| = 10k₀ (cf., (88)), and hence X

j∈(L∩S0)\(Shead∪L⁰)

k( bX −χ)b _Ijk²₂ ≤ 6000k₀ · ν².

Bounding the fourth term in (97): We have X

j∈[ⁿ

k1]\(L⁰∪S0)

k( bX −χ)b Ijk²₂ ≤ X

j∈[ⁿ

k1]\S0

k( bX −χ)b Ijk²₂

= Err²( bX −χ, 10kb ₀, k₁), which follows from the definition ofS0 in (88), along with Definition 1.2.

Hence by the union bound, it follows that E_prune holds with probability at least 1 − p − ₁₀₀₀¹ conditioned on E_loc.

Success event of estimation primitive: Let E_est be the event of having a successful run of EstimateValues(X,χ, L, n, 3kb ₀/, k₁, , p), meaning the following conditions on the output, W :

W_f = 0 for all f /∈ F X

j∈L⁰

k( bX −χ − W )b _Ijk²₂≤ k bX −χkb ²₂, (98) where F contains the frequencies within the blocks indexed by L⁰. Since the assumption of the theorem implies that kχkb ₀ = O(k₀k₁), by Lemma 5.2 (with δ =  and 3k₀/ in place of k₀) and the fact that conditioned on E_prune we have |L⁰| ≤ 3k₀/ (cf., (92)), it follows that E_est occurs with probability at least1 − p.

Combining the events: We can now can wrap everything up.

Letting E denote the overall success event corresponding to the claim of the lemma, we have P[E ] = P

E_loc∩ E_prune∩ E_est

= P

E_estE^loc∩ E_prune P

E_prune E_loc

P E_loc

. By the results that we have above, along with the union bound, it follows that

P E

≥ 1 − 2/100 − 3p ≥ 0.95

for sufficiently smallη in Algorithm 6. Applying the union bound over ¯E and the 1/poly(n) probability failure event arising from random perturbations of χb₀, we obtain the required bound of 0.9 on the success probability.

What remains is to first show that the statement of the lemma follows from E_loc∩ E_prune∩ E_est.

To do this, we observe that, conditioned on these events, k bX −χb⁰k²₂ = X

j∈L⁰

k( bX −χb⁰)Ijk²₂+ X

j∈[_k1ⁿ]\L⁰

k( bX −χb⁰)Ijk²₂+ err(χb0)

= X

j∈L⁰

k( bX −χ − W )b Ijk²₂+ X

j∈[_k1ⁿ]\L⁰

k( bX −χ)b Ijk²₂+ err(χb0)

≤ k bX −χkb ²₂+ 300k bX −χkb ²₂+ 6000ν²k0+ Err²( bX −χ, 10kb 0, k1) + ν²

≤ (4 · 10⁵)ν²k0+ Err²( bX −χ, 10kb 0, k1),

where the second line follows since χb⁰ = χ + W and W is non-zero only within the blocks indexedb by L⁰, the third line follows from (93) and (98), and the final line follows from the assumption k bX −χkb ²₂ ≤ 100k0ν² in the lemma.

Sample complexity: We condition on E_loc∩E_prune∩E_estand calculate the number of samples used.

By Lemma 3.5 withδ = ², the sample complexity of MultiBlockLocate is O^∗(|L|·log n+^k0_^k4¹) which behavesO^∗(^k_2⁰ log(1+k0) log n+^k0_^k4¹ log¹_p) conditioned on Eloc(see (91)). Moreover, by Lemma 5.1 with δ = , the sample complexity of PruneLocation is O(^k0_^k1 log_p¹ log¹_). Moreover, by Lemma 5.2 withδ = , the sample complexity of EstimateValues is O(^k0_2^k1log¹_plog¹_).

The claim follows by summing the three terms and noting from the choice of p in Algorithm 6 that, up tolog log^k_⁰ factors, we can replace p by  in the above calculations.

Runtime: As in sample complexity analysis, we condition on E_loc∩ E_prune∩ E_est.

First, by Lemma 3.5 with δ = ², the runtime of MultiBlockLocate is O^∗ |L| · log²n +

k0k1

⁴ log²n + ^k0_2^k1 log³n

, which behaves as O^{∗ k}_2⁰ log^k_⁰ · log²n + ^k0_^k4¹ log²n + ^k0_2^k1log³n

condi-tioned on E_loc (see (90)). Second, by Lemma 5.1 with δ = , the runtime of PruneLocation is O(^k0_^k1 log_p¹ log¹_log n+k1·|L| log_p¹), which behaves as O(^k0_^k1 log_p¹ log¹_log n+k1·^k_2⁰ log^k_⁰ log^{4 1}_p) conditioned on E_loc (see (91)). Finally, by Lemma 5.2 with δ = , the runtime of EstimateValues is O(^k0_^k1log¹_plog¹_ log n + k1 · |L⁰| log¹_p), which behaves as O(^k0_^k1 log¹_plog¹_ log n) conditioned on E_prune (see (92) and recall that |S0| = 10k0).

The claim follows by summing the above terms and replacingp by , with the log log n, log log SNR⁰ andlog¹_ terms absorbed into the O^∗(·) notation.

In document An Adaptive Sublinear-Time Block Sparse Fourier Transform (Page 50-61)