Binary sparse signal recovery with binary least squares

Haifeng Li; Qi Chen

doi:10.1515/jiip-2023-0008

Article

Binary sparse signal recovery with binary least squares

Haifeng Li and Qi Chen

Published/Copyright: September 16, 2025

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From the journal Journal of Inverse and Ill-posed Problems

Abstract

The problem of obtaining a 𝐾-sparse binary signal 𝐱 from y = Φ ⁢ x + ν is often encountered in communications and signal processing. Based on orthogonal least squares (OLS), a greedy algorithm guaranteeing the support recovery of a 𝐾-sparse binary signal is presented in this paper. Using the mutual coherence and restricted isometric properties of Φ , sufficient conditions ensuring the exact recovery of the support of 𝐱 are given. Finally, compared with binary matching pursuit (BMP) and OLS, the simulation experiments imply the advantage of the proposed algorithm.

Keywords: Sparse recovery; binary sparse signal; greedy algorithm

MSC 2020: 94A12; 65F22; 65J22

Funding source: Natural Science Foundation of Henan Province

Award Identifier / Grant number: 242300420252

Funding source: Key Scientific Research Project of Colleges and Universities in Henan Province

Award Identifier / Grant number: 24A120007

Funding statement: This work was partially supported by the Natural Science Foundation of Henan Province (grant no. 242300420252) and in part by the Key Scientific Research Project of Colleges and Universities in Henan Province (grant no. 24A120007) and in part by Young Backbone Teachers in Henan Province (grant no. 2023GGJS037).

A Some lemmas

Lemma 1

Lemma 1 ([29])

Consider the ( k + 1 ) -th step in the BLS algorithm. Then BLS selects the index

s k + 1 = arg ⁢ max i ∈ T ∖ Λ k ⁡ | ⟨ r k , Φ i ⟩ | ∥ P Λ k ⟂ ⁢ Φ i ∥ 2 .

Lemma 2

Lemma 2 ([29])

Assuming that S 1 ∈ Ω , Φ is a column normalization matrix and satisfies RIP of order | S 1 | + 1 , then for any i ∈ Ω ∖ S 1 , we have that

∥ P S 1 ⟂ ⁢ Φ i ∥ 2 ≥ 1 − δ | S 1 | + 1 2 .

Lemma 3

Lemma 3 ([17])

Suppose that S 1 , S 2 satisfy | S 2 ∖ S 1 | > 1 , Φ is a column normalization matrix and satisfies RIP of order | S 1 ∪ S 2 | . Let w ∈ R n and supp ⁡ ( w ) = S 2 . One has

( 1 − δ S 1 ∪ S 2 ) ⁢ ∥ w ∥ 2 2 ≤ ∥ P S 1 ⟂ ⁢ Φ ⁢ w ∥ 2 2 ≤ ( 1 + δ S 1 ∪ S 2 ) ⁢ ∥ w ∥ 2 2 .

Lemma 4

Consider y = Φ ⁢ x . For any constant α > 0 and k ∈ { 0 , 1 , … , K − 1 } , one has

∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 > α ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 ⁢ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j ⟩ |

for all j ∈ Ω ∖ T , provided that Φ satisfies δ K − k + 1 < 1 / α + 1 .

Proof

The proof of Lemma 4 is similar to [29] except for the form of the residual. Define

β = 1 − α + 1 α .

Let

Ψ = 1 1 − β 4 ⁢ [ Φ T ∖ Λ k − 1 Φ j ] , u = [ x T ∖ Λ k − 1 0 ] ∈ R | T ∖ Λ k − 1 | + 1 , v = [ 0 | T ∖ Λ k − 1 | × 1 λ ⁢ β ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 ] ∈ R | T ∖ Λ k − 1 | + 1 ,

where

λ = { 1 if ⁢ ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j ⟩ ≥ 0 , − 1 if ⁢ ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j ⟩ < 0 .

Then one has

(A.1) ∥ Ψ ⁢ ( u + v ) ∥ 2 2 − ∥ Ψ ⁢ ( β 2 ⁢ u − v ) ∥ 2 2 = ∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 + 2 ⁢ ( 1 + β 2 ) ⁢ β 1 − β 4 ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 ⁢ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j ⟩ | = ∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 − α ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 ⁢ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j ⟩ | ,

where (A.1) follows from

2 ⁢ β 1 − β 2 = − α .

On the other hand, applying Lemma 3, we can have

(A.2) ∥ Ψ ⁢ ( u + v ) ∥ 2 2 − ∥ Ψ ⁢ ( β 2 ⁢ u − v ) ∥ 2 2 ≥ 1 − δ K − k + 1 1 − β 4 ⁢ ∥ u + v ∥ 2 2 − 1 + δ K − k + 1 1 − β 4 ⁢ ∥ β 2 ⁢ u − v ∥ 2 2 = (a) ⁢ ( 1 − δ K − k + 1 ⁢ α + 1 ) ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 2 > 0 ,

where (a) is from

1 + β 2 1 − β 2 = α + 1 .

By combining (A.1) and (A.2), we can get

∥ Ψ ⁢ ( u + v ) ∥ 2 2 − ∥ Ψ ⁢ ( β 2 ⁢ u − v ) ∥ 2 2 = ∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 − α ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 ⁢ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j ⟩ | > 0 .

So it is easy to see that ∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 > α ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 ⁢ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j ⟩ | . ∎

Lemma 5

Lemma 5 ([8, 25])

Let S 1 ⊂ T and w ∈ R | S 1 | . Suppose that Φ satisfies δ | S 1 | < 1 . One has δ | S 1 | ≤ ( | S 1 | − 1 ) ⁢ μ .

B Proof of Theorem 1

Proof

The proof of Theorem 1 is similar to [29] except for the form of the residual.

Our proof is based on mathematical induction. Assume that the BLS algorithm selects the correct index in the first k − 1 iterations, i.e., Λ k − 1 ⊂ T , so according to Algorithm 1, it is only necessary to prove s k ∈ T in the 𝑘-th iteration, i.e.,

(B.1) max i ∈ T ∖ Λ k − 1 ⁡ | ⟨ r k − 1 , Φ i ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ i ∥ 2 > max i ∈ Ω ∖ T ⁡ | ⟨ r k − 1 , Φ i ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ i ∥ 2 .

According to Algorithm 1, we can get

r k − 1 = y − ∑ i ∈ Λ k − 1 Φ i = Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 + ν .

Because Φ is the column normalization matrix, there is ∥ P Λ k − 1 ⟂ ⁢ Φ i ∥ 2 ≤ ∥ Φ i ∥ 2 = 1 . Then

max i ∈ T ∖ Λ k − 1 ⁡ | ⟨ r k − 1 , Φ i ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ i ∥ 2 ≥ max i ∈ T ∖ Λ k − 1 ⁡ | ⟨ r k − 1 , Φ i ⟩ | ≥ max i ∈ T ∖ Λ k − 1 ⁡ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ i ⟩ | − ∥ Φ T ∖ Λ k − 1 ′ ⁢ ν ∥ ∞ ≥ (a) ⁢ ∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 | T ∖ Λ k − 1 | ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 − ∥ ν ∥ 2 ,

where (a) follows from the Cauchy–Schwarz inequality.

On the right side of (B.1), one has

(B.2) max j ∈ Ω ∖ T ⁡ | ⟨ r k − 1 , Φ j ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ j ∥ 2 ≤ max j ∈ Ω ∖ T ⁡ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j ⟩ | + | ⟨ ν , Φ j ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ j ∥ 2 ⁢ ≤ (a) ⁢ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j 0 ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ j 0 ∥ 2 + ∥ ν ∥ 2 1 − δ K + 1 2 ,

where

j 0 = arg ⁢ max j ∈ Ω ∖ T ⁡ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ j ∥ 2 ,

(a) follows from Lemma 2 and the monotonicity of RIP.

Next, to prove (B.1), we should prove

(B.3) ∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 | T ∖ Λ k − 1 | ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 − | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j 0 ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ j 0 ∥ 2 > η + η 1 − δ K + 1 2 ≥ ∥ ν ∥ 2 + ∥ ν ∥ 2 1 − δ K + 1 2 .

According to Lemma 4, letting

α = | T ∖ Λ k − 1 | ∥ P Λ k − 1 ⟂ ⁢ Φ j 0 ∥ 2 2 ,

from (A.1) and (A.2), we can have

∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 − α ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 ⁢ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j 0 ⟩ | = ( 1 − δ K − k + 1 ⁢ α + 1 ) ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 2 .

(B.4) ∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 | T ∖ Λ k − 1 | ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 − α ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 ⁢ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j 0 ⟩ | | T ∖ Λ k − 1 | ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 ≥ ( 1 − δ K + 1 ⁢ α + 1 ) ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 | T ∖ Λ k − 1 | .

Note that

α = | T ∖ Λ k − 1 | ∥ P Λ k − 1 ⟂ ⁢ Φ j 0 ∥ 2 2 ;

then we have that

(B.5) δ K + 1 ⁢ α + 1 = | T ∖ Λ k − 1 | ∥ P Λ k − 1 ⟂ ⁢ Φ j 0 ∥ 2 + 1 ⁢ δ K + 1 ⁢ ≤ k = 1 ⁢ K + 1 ⁢ δ K + 1 < 1 ,

and

(B.6) δ K + 1 ⁢ α + 1 = | T ∖ Λ k − 1 | ∥ P Λ k − 1 ⟂ ⁢ Φ j 0 ∥ 2 + 1 ⁢ δ K + 1 ⁢ ≤ ( 1 < k ≤ K ) ⁢ K − 1 1 − δ | Λ k − 1 | + 1 2 + 1 ⁢ δ K + 1 ≤ K − 1 1 − δ K + 1 2 + 1 ⁢ δ K + 1 ≤ K + 1 ⁢ δ K + 1 < 1 .

(B.7) 1 − δ K + 1 ⁢ α + 1 > 0 .

Thus (B.4) can be changed into

(B.8) ∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 | T ∖ Λ k − 1 | ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 − | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j 0 ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ j 0 ∥ 2 ≥ ( 1 − δ K + 1 ⁢ α + 1 ) ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 | T ∖ Λ k − 1 | ⁢ > ( B.5 ), ( B.6 ) ⁢ ( 1 − δ K + 1 ⁢ K + 1 ) ⁢ min i ∈ T ∖ Λ k − 1 ⁡ | x i | .

In order to prove (B.3), by (B.8) and (B.7), we only need to show

(B.9) ( 1 − δ K + 1 ⁢ K + 1 ) ⁢ min i ∈ T ∖ Λ k − 1 ⁡ | x i | > 1 − δ K + 1 2 + 1 1 − δ K + 1 2 ⁢ η

Through (B.2)–(B.9), it can be easily obtained that, when

δ K + 1 < 1 K + 1 and η < 1 − δ K + 1 2 ⁢ ( 1 − δ K + 1 ⁢ K + 1 ) 2 ⁢ min i ∈ T ⁡ | x i | ,

(B.3) holds. So, for 1 ≤ k ≤ | T | , we have s k ∈ T .

In the following, we will prove that the BLS algorithm iterates 𝐾 steps. Firstly, for 1 ≤ k ≤ K − 1 , one has

∥ r k ∥ 2 = ∥ Φ T ∖ Λ k ⁢ x T ∖ Λ k + ν ∥ 2 ≥ ∥ Φ T ∖ Λ k ⁢ x T ∖ Λ k ∥ 2 − η ≥ 1 − δ | T | + 1 ⁢ ∥ x T ∖ Λ k ∥ 2 − η ≥ 1 − δ K + 1 − η ≥ (a) ⁢ 1 − δ K + 1 2 ⁢ ( 1 − δ K + 1 ⁢ K + 1 ) − η ⁢ > (b) ⁢ 2 ⁢ η − η = η ,

where (a) is from 1 − δ K + 1 2 ⁢ ( 1 − δ K + 1 ⁢ K + 1 ) ≤ 1 − K + 1 ⁢ δ K + 1 ≤ 1 − δ K + 1 ≤ 1 − δ K + 1 , and (b) is from (3.2). So the BLS algorithm iterates at least | T | steps. When k = K , ∥ r K ∥ 2 = ∥ Φ T ∖ T K ⁢ x T ∖ T K + ν ∥ 2 = ∥ ν ∥ 2 ≤ η . So, to sum up, the BLS algorithm iterates 𝐾 steps. ∎

C Proof of Theorem 2

Proof

We divide the proof into two steps. Firstly, it is proved that, for 1 ≤ k ≤ K , there is s k ∈ T ; then it is proved that the BLS algorithm iterates 𝐾 steps. Similarly, assuming that the correct index is selected in the first k − 1 steps, it is only necessary to show that (B.1) is valid in the 𝑘-th iteration.

For the left side of (B.1), we have

max i ∈ T ∖ Λ k − 1 ⁡ | ⟨ r k − 1 , Φ i ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ i ∥ 2 ≥ max i ∈ T ∖ Λ k − 1 ⁡ | ⟨ r k − 1 , Φ i ⟩ | ≥ ∥ Φ T ∖ Λ k − 1 ′ ⁢ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ ∞ − ∥ Φ T ∖ Λ k − 1 ′ ⁢ ν ∥ ∞ .

To the right of (B.1), we have

max i ∈ Ω ∖ T ⁡ | ⟨ r k − 1 , Φ i ⟩ | ∥ P Λ k − 1 ⟂ ⁢ Φ i ∥ 2 ≤ ∥ Φ Ω ∖ T ′ ⁢ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ ∞ min i ∈ Ω ∖ T ∥ P Λ k − 1 ⟂ Φ i ∥ 2 + ∥ Φ Ω ∖ T ′ ⁢ ν ∥ ∞ min i ∈ Ω ∖ T ∥ P Λ k − 1 ⟂ Φ i ∥ 2 .

At this point, we need to prove

(C.1) ∥ Φ T ∖ Λ k − 1 ′ ⁢ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ ∞ − ∥ Φ Ω ∖ T ′ ⁢ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ ∞ min i ∈ Ω ∖ T ∥ P Λ k − 1 ⟂ Φ i ∥ 2 ≥ ∥ Φ T ∖ Λ k − 1 ′ ⁢ ν ∥ ∞ + ∥ Φ Ω ∖ T ′ ⁢ ν ∥ ∞ min i ∈ Ω ∖ T ∥ P Λ k − 1 ⟂ Φ i ∥ 2 .

Let

p = ∥ Φ T ∖ Λ k − 1 ′ ⁢ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ ∞ and q = ∥ Φ Ω ∖ T ′ ⁢ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ ∞ min i ∈ Ω ∖ T ∥ P Λ k − 1 ⟂ Φ i ∥ 2 .

The lower bound of 𝑝 is

(C.2) p = ∥ Φ T ∖ Λ k − 1 ′ ⁢ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ ∞ ≥ ∥ Φ T ∖ Λ k − 1 ′ ⁢ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 | T ∖ Λ k − 1 | ⁢ ≥ (a) ⁢ ( 1 − ( | T ∖ Λ k − 1 | − 1 ) ⁢ μ ) ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 | T ∖ Λ k − 1 | ,

where (a) is from Lemma 5. The upper bound of 𝑞 is

(C.3) q = ∥ Φ Ω ∖ T ′ ⁢ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ ∞ min i ∈ Ω ∖ T ∥ P Λ k − 1 ⟂ Φ i ∥ 2 ≤ | T ∖ Λ k − 1 | ⁢ μ ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 min i ∈ Ω ∖ T ∥ P Λ k − 1 ⟂ Φ i ∥ 2 ⁢ ≤ (a) ⁢ | T ∖ Λ k − 1 | ⁢ μ ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 1 − δ | Λ k − 1 | + 1 2 ≤ | T ∖ Λ k − 1 | ⁢ μ ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 1 − δ | Λ k − 1 | + 1 ⁢ ≥ (b) ⁢ | T ∖ Λ k − 1 | ⁢ μ ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 1 − | Λ k − 1 | ⁢ μ ,

where (a) follows from Lemma 2, and (b) follows from Lemma 5.

Thus, according to (C.2) and (C.3), one has

(C.4) p − q ≥ ( 1 − ( | T ∖ Λ k − 1 | − 1 ) ⁢ μ | T ∖ Λ k − 1 | − | T ∖ Λ k − 1 | ⁢ μ 1 − | Λ k − 1 | ⁢ μ ) ⁢ | T ∖ Λ k − 1 | ⁢ min i ∈ T ∖ Λ k − 1 ⁡ | x i | = ( 1 − ( | T ∖ Λ k − 1 | − 1 ) ⁢ μ − | T ∖ Λ k − 1 | ⁢ μ 1 − | Λ k − 1 | ⁢ μ ) ⁢ min i ∈ T ∖ Λ k − 1 ⁡ | x i | = ( 1 − | Λ k − 1 | ⁢ μ − 2 ⁢ | T ∖ Λ k − 1 | ⁢ μ + μ + ( | Λ k − 1 | ⁢ | T ∖ Λ k − 1 | − | Λ k − 1 | ) ⁢ μ 2 1 − | Λ k − 1 | ⁢ μ ) ⁢ min i ∈ T ∖ Λ k − 1 ⁡ | x i | ≥ (a) ⁢ ( 1 − | Λ k − 1 | ⁢ μ − 2 ⁢ | T ∖ Λ k − 1 | ⁢ μ + μ 1 − | Λ k − 1 | ⁢ μ ) ⁢ min i ∈ T ∖ Λ k − 1 ⁡ | x i | = ( 1 − ( 2 ⁢ | T ∖ Λ k − 1 | − 1 ) ⁢ μ 1 − | Λ k − 1 | ⁢ μ ) ⁢ min i ∈ T ∖ Λ k − 1 ⁡ | x i | ≥ (b) ⁢ ( 1 − ( 2 ⁢ K − 1 ) ⁢ μ ) ⁢ min i ∈ T ∖ Λ k − 1 ⁡ | x i | ,

where (a) is from ( | Λ k − 1 | ⁢ | T ∖ Λ k − 1 | − | Λ k − 1 | ) ⁢ μ 2 > 0 , (b) is from

( 2 ⁢ | T ∖ Λ k − 1 | − 1 ) ⁢ μ 1 − | Λ k − 1 | ⁢ μ < 1

and the value of

( 2 ⁢ | T ∖ Λ k − 1 | − 1 ) ⁢ μ 1 − | Λ k − 1 | ⁢ μ

decreases as 𝑘 increases (i.e., the maximum value ( 2 ⁢ K − 1 ) ⁢ μ is obtained when k = 1 ).

On the other hand,

∥ Φ T ∖ Λ k − 1 ′ ⁢ ν ∥ ∞ + ∥ Φ Ω ∖ T ′ ⁢ ν ∥ 2 min i ∈ Ω ∖ T ∥ P Λ k − 1 ⟂ Φ i ∥ 2 ≤ η + η 1 − | Λ k − 1 | ⁢ μ ≤ η + η 1 − K ⁢ μ = 2 − K ⁢ μ 1 − K ⁢ μ ⁢ η .

In order to prove (C.1), according to (3.3) and (C.4), we need to show

(C.5) p − q ≥ ( 1 − ( 2 ⁢ K − 1 ) ⁢ μ ) ⁢ min i ∈ T ∖ Λ k − 1 ⁡ | x i | > 2 − K ⁢ μ 1 − K ⁢ μ ⁢ η .

Inequality (3.4) can guarantee that (C.5) holds.

Now, we will prove that BLS iterates 𝐾 steps under ∥ r k ∥ 2 ≤ η . For 1 ≤ k ≤ K − 1 , one has

∥ r k ∥ 2 = ∥ Φ T ∖ Λ k ⁢ x T ∖ Λ k + ν ∥ 2 ⁢ ≥ (a) ⁢ 1 − ( | T ∖ Λ k | − 1 ) ⁢ μ ⁢ min i ∈ T ⁡ | x i | − η ≥ (b) ⁢ 1 − ( | T ∖ Λ k | − 1 ) ⁢ μ ⁢ 2 − K ⁢ μ ( 1 − K ⁢ μ ) ⁢ ( 1 − ( 2 ⁢ K − 1 ) ⁢ μ ) − η ≥ 2 ⁢ η − η = η ,

where (a) is from Lemmas 3 and 5, (b) is from (3.4). That is, the BLS algorithm iterates at least | T | steps. In particular,

∥ r K ∥ 2 = ∥ Φ T ∖ T K ⁢ x T ∖ T K + ν ∥ 2 = ∥ ν ∥ 2 ≤ η ,

so BLS iterates 𝐾 steps. ∎

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Received: 2023-01-26

Revised: 2024-03-27

Accepted: 2025-08-29

Published Online: 2025-09-16

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https://doi.org/10.1515/jiip-2023-0008

Keywords for this article

Sparse recovery; binary sparse signal; greedy algorithm