text:
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Interview: A. T. Greenblatt
by Caroline M. Yoachim
A. T. Greenblatt is a mechanical engineer by day and a writer by night. She lives in Philadelphia where she's well acquainted with all four seasons and is known to frequently subject her friends to various cooking and home-brewing experiments. Her recent publications include "A Place to Grow" and "A Non-Hero's Guide to the Road of Monsters." She has a knack for creating protagonists that readers can genuinely root for, and puts those characters into situations where finding the right path is not always easy or clear. "And Yet" is Greenblatt's first appearance in Uncanny.
Uncanny Magazine: One of the themes in "And Yet" is about facing your fears. I loved the way the haunted house shifted over time. The things that frighten us as children are often different from the things that frighten us as adults, and our reactions to that fear can also change. How do you deal with the things that scare you (or, if you prefer: ideally how would you like to)?
A. T. Greenblatt: As someone who is almost always battling anxiety, facing my fears is a two-step process. First, I try to think of what I'd do in the situations I'm worried about. (What if I get stranded at the airport in the middle of a terrible storm? Pack spare clothes in your carry-on. Have the number of a local hotel written down.) I found having backup plans helps. Sometimes it's even enough that I can skip the second part.
But if it isn't, the second step is to use my anxiety against itself, which I'm not sure is exactly healthy, but it's effective. I tell myself that if I don't do this scary thing, something scarier will happen. (If you don't go on this trip, you will never get another chance to travel here and you will always regret it.)
Though it might not seem like it from this example, I really do love traveling. But no matter how many miles of experience I have, every time I visit a new place, I have to stare down the same fears.
Uncanny Magazine: You recently attended the Clarion West Writers Workshop, during which you lived in a big older house with all of your classmates for six weeks. Was that house a source of inspiration for this story? More generally, what impact do you think the workshop has had on you as a writer?
A. T. Greenblatt: "And Yet" was the first official story I wrote in Clarion West and it was the first piece of work my classmates saw from me. That initial draft was a complete wreck though—I wrote it in five days. I don't think the sorority house itself inspired this story, at least not consciously because I scribbled down the opening lines a week or two earlier. But it did take shape there. I remember there was this very precise moment when I was sitting in my dorm room, staring out the window, and I realized that the protagonist's point of view was all wrong. I had started writing the piece in first person and I was struggling, but in that moment, the lines and voice started to write themselves in second person, and the character's voice and the tone just clicked. It was one of those magical little moments in writing.
I'm still trying to sort out how Clarion West has altered me as a writer. I suspect it's the summation of a hundred small lessons. I do think it's made me a more efficient critiquer and I can also work through story drafts faster. Which is not to say drafting has become easier, because it hasn't. It's more like going up a difficult trail that you've been hiking more consistently. There's no new short cuts and just as many rocks, but you're more familiar with the route and you know you can make it to the end, because you've done it plenty of times before.
Uncanny Magazine: This story features parallel universes. If it was possible to go and visit a parallel universe to see the other possibilities of what could have happened differently at some juncture in your life, would you want to do it?
A. T. Greenblatt: A small part of me is fiercely curious about what would have happened if I'd made different choices at different points in my life. The rest of me realizes that having that knowledge would probably be a terrible, terrible idea. I mean, I already have hindsight and self-doubt constantly following me around.
I imagine if people could reference parallel universes, it'd be a rabbit hole. Once you began exploring the different scenarios, you won't be able to make a decision without checking out other parallel universes first. But then, the more you look, the more possibilities there are until the simplest decision becomes a chore. It'd be like trying to pick cereal in the grocery store. Too many options, so you just give up and go with oatmeal instead.
Uncanny Magazine: In your Kickstarter essay for Disabled People Destroy Science Fiction, you talk about having cerebral palsy and the need to find a balance between your abilities and your goals. You describe the kinds of disabled characters you'd like to see more of in fiction—characters who have found balance, who solve problems and have adventures. "And Yet" has a protagonist who fits this description. In addition to the need for better representations of disabled characters, what changes would you like to see in fictional representations of societies or worlds?
A. T. Greenblatt: Yeah, I realized after I wrote that essay I needed to start creating more of the characters I wanted to see. With "And Yet," I based the protagonist's disability on cerebral palsy, partially because I haven't read too many stories featuring characters with CP and partially because I wrote the first draft in less than a week and decided to go with the old and flawed advice of "write what you know."
In terms of other fictional representations, I was actually complaining to friends last weekend that we don't see enough healthy friendships between women in fiction. Which is sort of strange to me because it's not like it's a rarity in real life. Actually, I would love to see more equal platonic relationships in stories in general.
I guess I know what I need to write next.
Uncanny Magazine: There is a description of a terrible meal that the protagonist has to eat before leaving the table. What was your least favorite meal (either from childhood or as an adult)?
A. T. Greenblatt: Unlike the protagonist's mom, my mom is actually a really good cook, so I don't remember too many meals I didn't like as a kid.
With one notable exception: Lasagna.
It was something about the combination of textures of the cooked spinach and cottage cheese. I was so adamantly against it when I was a kid that I wrote a story about the horror of being the only person in the family who didn't like lasagna for a school project once. I think it was one of the first stories I ever wrote because my parents still have it a copy of it and it still made my dad chuckle when I asked him about it over the weekend.
I eat my mom's lasagna now, but the texture of cottage cheese still weirds me out and even though I love to cook, I've never made it for myself.
Uncanny Magazine: What are you working on next?
A. T. Greenblatt: I always have a short story or five to revise. My goal this year is to learn how to write some longer works, so I'm currently working on a second world fantasy novelette and hopefully (hopefully) outlining the basis for a novella. Usually when I'm writing, I make things up as I go along, so outlining is like a strange, oddly shaped tool that I'm trying to cram into my writing toolbox. We'll see how it goes.
Uncanny Magazine: Thank you so much for sharing your thoughts with us!
© 2018 by Uncanny Magazine
Caroline M. Yoachim
Caroline M. Yoachim is the author of the 2017 Hugo and Nebula finalist short story "Carnival Nine." Her fiction has appeared in Lightspeed, Beneath Ceaseless Skies, Uncanny, Fantasy & Science Fiction, Clarkesworld, and Asimov's, among other places. Her debut short story collection, Seven Wonders of a Once and Future World & Other Stories, came out with Fairwood Press in 2016. For more about Caroline, check out her website at carolineyoachim.com
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text:
Girl Named Tom moves forward to semis
By Drew Stambaugh - dstambaugh@aimmediamidwest.com
Girl Named Tom performs Monday on the "The Voice."
Trae Patton | NBC
Girl Named Tom learns their fate on Tuesday's episode of "The Voice" on NBC.
Girl Named Tom again rose to the challenge Monday, staying alive for another week in the NBC singing competition, "The Voice." For challenge week on the show, the trio performed Coldplay's "Viva la Vida."
"These three have been taking risks and challenging themselves their whole lives, and so far, it has definitely been paying off," said host Carson Daly, prior to their performance.
Last week, they were challenged with the country song "More Hearts Than Mine," and their fans responded.
"It was crazy," said Josh Liechty. "We got up to number two on the country charts, on iTunes, this last week."
For challenge week, coach Kelly Clarkson said they "really make people go fish out of water" for their performances. "I love having to push an artist to do something maybe they wouldn't do, because it's really cool to know other genres of music and not just be one lane."
To do that, she chose the Coldplay hit "Viva la Vida" for Girl Named Tom.
"What's cool about Girl Named Tom is they started out like folky, singer-songwriter, and now I have them on challenge week doing this big, huge stadium kind of vibe song," said Clarkson. "And it's very different from their sound so, pretty stoked about it."
Caleb Liechty said challenging themselves to grow is a risk. "This whole band thing for us was a risk in that, we were gonna be doctors."
Their grandfather was a doctor in Indiana and they had planned on following in his footsteps.
"Caleb and I had graduated college with biochemistry majors and we were looking at med schools. Bekah was looking at the same track, when our mom suggested, 'Why don't you form a band.' It didn't take long for us to take her up on it," said Josh Liechty.
And that decision led them onto national television and a chance for stardom on "The Voice."
Following their performance on Monday, Blake Shelton said it was probably the best performance they have had so far.
"You're such a cool family group and seeing you together, having fun stage, being musicians on stage together, is so much fun for us to watch," said John Legend. "I love Girl Named Tom everybody."
On Tuesday, they were the third of eight acts saved and move on to the semifinal round Monday. From there four acts will move on to the final round.
In a message to the Pettisville community, school superintendeent Josh Clark spoke of the season of gift giving, and what the trio has brought.
"We've been given the gift of community," he said. "This journey we are on with Girl Named Tom has allowed us all to dream a bit, and to unify behind something good. In a world with so much division, it's been a blessing to be drawn into an incredible journey together."
https://www.fcnews.org/wp-content/uploads/sites/45/2021/12/web1_GNT-Nov-29.jpgGirl Named Tom performs Monday on the "The Voice." Trae Patton | NBC
https://www.fcnews.org/wp-content/uploads/sites/45/2021/12/web1_NUP_195980_01721.jpgGirl Named Tom learns their fate on Tuesday's episode of "The Voice" on NBC. Trae Patton | NBC
By Drew Stambaugh
dstambaugh@aimmediamidwest.com
Reach Drew Stambaugh at 419-335-2010
Hi! A visitor to our site felt the following article might be of interest to you: Girl Named Tom moves forward to semis. Here is a link to that story: https://www.fcnews.org/news/34133/girl-named-tom-moves-forward-to-semis

redpajama_set_name: RedPajamaCommonCrawl


Nature declares the glory of God. Knowledge pours forth from the heavens. But of course there are no words. The stars are silent. The psalmist reminds us that truth is communicated in different ways. We are said to be becoming a knowledge society. By that we usually mean the written, spoken and digital word. We communicate through reason and language. The psalmist calls us back to other ways of finding knowledge and truth.
As the sun travels across the sky, we are reminded that we are not in control. We did not put the sun in its tracks; someone else did. With eternal regularity it rises and sets. Who has determined its path? Who wrote the laws of nature that let nature support life? If it's only brute fact, why does the sun bring warmth to everyone?
Creation points to an all-powerful, all-loving Creator. But general revelation raises more questions than it answers. What does it mean that the world is the way it is? How then should we live? The heavens are silent on many questions, but God is not. He has spoken in his Word to address many of our questions. When we study his words carefully and appropriately, we find truth, wisdom and beauty.
The psalmist declares that truth is more than accurate and logical. Scientists talk about a beautiful theory. When everything comes together, truth is found in beautiful simplicity and profundity. The beauty of the sun reflects the creativity of its maker. The truth of the Bible reflects the trustworthiness of its author. Its words not only bring rational truth, they give joy to the heart and light to the eyes. They are sweeter than honey.


text:
\section{Introduction}
Compressive sensing is a powerful signal acquisition approach  with  which one  can  recover signals beyond bandlimitedness from noisy under-determined measurements whose number  is closer to the intrinsic complexity of the target signals
than the Nyquist rate \cite{CandesRombergTao:2006,Donoho:2006,FazelCandesRcht:2008,FoucartRauhut:2013}.
Quantization that transforms the infinite-precision measurements into discrete ones is necessary for storage and transmission \cite{Sayood:2017}.
The binary  quantizer, an extreme case of scalar quantization,
that   codes the measurements  into binary values with  a single bit has been introduced into compressed sensing
\cite{BoufounosBaraniuk:2008}. The 1-bit compressed sensing (1-bit CS) has drawn much attention because of its low cost in hardware mentation and storage  and its robustness in the low signal-to-noise ratio scenario  \cite{Laska:2012}.
\subsection{Related  work}
A lot of  efforts  have been devoted to studying the theoretical and computational issues  in  the 1-bit CS under the sparsity assumption, i.e., $\|x^*\|_0\leq s \ll m$.
 Support recovery  can be achieved  in both  noiseless and noisy setting provided that
$m > \mathcal{O}(s\log n)$ \cite{GopiJain:2013,JacquesDegraux:2013,PlanVershynincpam:2013,BauptBaraniuk:2011,JacquesLaska:2013,GuptaNowakRech:2010,BauptBaraniuk:2011,PlanVershynin:2013,ZhangYiJin:2014,Ahsen:2019}. Greedy methods \cite{LiuGongXu:2016,Boufounos:2009,JacquesLaska:2013} and first order methods \cite{BoufounosBaraniuk:2008, LaskaWenYinBaraniuk:2011,YanYangOsher:2012,DaiShenXuZhang:2016}  are developed  to minimize the sparsity promoting nonconvex objective function caused by    the unit sphere constraint or the  nonconvex regularizers. Convex  relaxation models are also proposed \cite{ZhangYiJin:2014,PlanVershynin:2013,PlanVershynincpam:2013, ZymnisBoydCandes:2010,PlanVershynin:2017,Vershynin:2015,HuangShiYan:2015}  to address the nonconvex optimization  problem.
Using least squares to estimate parameters in the scenario of   model  misspecification    goes back to
\cite{Brillinger:2012}, and see also  \cite{LiDuan:1989} and the references therein  for related  development  in the setting $m \gg n$.

Recently, with this idea,  \cite{PlanVershynin:2016,Neykov:2016,HuangJiao:2018,ding2020robust} proposed least square with $\ell_1$/$\ell_0$ regularized or generalized lasso to estimate parameters from   general under-determined  nonlinear measurements.
In addition to the sparse structure  of the  signals/images under certain linear transform \cite{Mallat:2008},  the natural signals/images data have been verified having low intrinsic dimension, i.e., they can be represented  by a generator  $G$,  such as pretrained  neural network, that maps from $\mathbb{R}^k$ to $\mathbb{R}^{n}$ with $k\ll n$. Such a  $G$ can be obtained via GAN    \cite{goodfellow14},  VAE \cite{kingma14} or flow based method \cite {rezende2015variational}. In these models, the generative part learns a mapping from a low dimensional representation space $z\in \mathbb{R}^k$ to the high dimensional sample space $G(z)\in \mathbb{R}^n$. While training, this mapping is encouraged to produce vectors that resemble the vectors in the training dataset. With this generative prior, several tasks have been studied such as image restoration  \cite{ulyanov2018deep}, phase retrieval \cite{hand2018phase} and  compressed sensing \cite{wu2019deep,bora2017compressed,huang2018provably,liu2020information} and  nonlinear single index models under certain measurement and noise models  \cite{wei2019statistical,liu2020generalized}.

 In \cite{bora2017compressed}, the authors propose the least squares estimator (\ref{ls1})-(\ref{ls2}) to recover signals in standard compressed sensing  with generative prior and prove  sharp sample complexities \cite{liu2020information}. Surprisingly, the sharp  sample complexity for the squares decoder (\ref{ls1})-(\ref{ls2}) can be derived in this paper even if the measurements are highly quantized and corrupted by noise and sign flips.
Very recently, under generative prior, \cite{liu2020sample} and \cite{qiu2020robust} derived     sample complexity results for 1-bit CS.
The sample complexity obtained  in  \cite{liu2020sample} is $O(k\log L)$ under the assumption that the generator $G$ is $L$- Lipschitz continuous and  the rows of $A$   are   i.i.d.   sampled from   $\mathcal{N}(\textbf{0},\mathbf{I})$.  However, the estimator proposed in  \cite{liu2020sample},  $\hat{x} = G(\hat{z})$ with $\hat{z} \in \{z: y=\mathrm{sign}(AG(z))\}$,  is quite different from our least squares decoder and the analysis technique used there are also not applicable to our decoder.
 \cite{qiu2020robust} proposed unconstrained empirical
risk minimization to recovery in 1-bit CS and  derived  the sample complexity to be  $\mathcal{O}(kd\log n)$ for $d$-layer ReLU network $G$ via assuming  the rows of $A$ are i.i.d. sampled from   subexponential distributions. However, the 1-bit CS model considered in \cite{qiu2020robust} is without sign flips and require additional   quantization threshold before sampling  to measure and the empirical
risk minimization decoder used there  is also different from our least squares  (\ref{ls1})-(\ref{ls2}). The results  in \cite{liu2020generalized}
 can be applied to  1-bit CS model,  however,  it  requires that the target signals are exact contained  in the range of the generator. In contrast, we only need a more realistic assumption that the target signals can be approximated by  a generator.
\subsection{Notation and Setup}\label{setting}
We use $[n]$  to denote the set $\{1,...,n\}$, use $A_i\in \mathbb{R}^{m\times1}, i \in [n]$ and $a_j \in \mathbb{R}^{n\times 1},j\in [m]$ to denote the $i$th column and $j$th row  of $A$, respectively.
The multivariate normal distribution is denoted by $\mathcal{N}(\textbf{0},\Sigma)$ with a symmetric and positive definite matrix $\Sigma$.
Let $\|x\|_{\Sigma} = (x^{t}\Sigma x)^{\frac{1}{2}}$, and $\|x\|_p = (\sum_{i=1}^{n}|x_{i}|^p)^{1/p}, p\in [1,\infty)$  be the $\ell_p$-norm of  $x$. Without causing confusion, $\|\cdot\|$ defaults to $\|\cdot\|_2$.
Sign function $\textrm{sign}(\cdot)$ is defined componentwise as $\textrm{sign}(z) =1$ for $z \geq 0$ and $\textrm{sign}(z) = -1$ for $z<0$. We use $\odot $ to denote the Hadamard product.
For any set $B$, $|B|$ is defined as the number of elements contained in $B$.

Following \cite{PlanVershynincpam:2013,HuangJiao:2018},  we consider the following 1-bit CS model
\begin{equation}\label{setup}
 y = \eta \odot\textrm{sign} (A x^* + \epsilon),
\end{equation}
where $y\in \mathbb{R}^{m}$ are the binary  measurements, $x^{*}\in \mathbb{R}^{n}$ is an unknown signal.
The measurement matrix $A \in \mathbb{R}^{m\times n}$ is a random  matrix whose rows $a_i, i \in [m]$  are   i.i.d.  random vectors sampled from   $\mathcal{N}(\textbf{0}, \Sigma)$ with  an unknown
covariance matrix $\Sigma $, $\eta\in \mathbb{R}^{m}$ is a random vector modeling the sign flips of $y$ whose  coordinates $\eta_is$ are i.i.d.  satisfying
 $\mathbb{P}[\eta_i = 1] = 1- \mathbb{P}[\eta_i = -1] = q \neq \frac{1}{2},$ and $\epsilon \in \mathbb{R}^{m}$ is a random vector sampled from  $\mathcal{N}(\textbf{0},\sigma^2\textbf{I}_m)$ with an
unknown noise level $\sigma$ modeling errors before quantization.   We assume  $\eta_i, \epsilon_i$ and $a_i$ are independent.

Model \eqref{setup} is  unidentifiable under positive scaling, the best one can do is to recover $x^*$ up to a constant, $c=(2q-1)\sqrt{\frac{2}{\pi(\sigma^2+1)}}$ which has been proved in \cite{HuangJiao:2018}. Without loss of generality
we may assume $\| x^*\|_{\Sigma} = 1$.

Let $\ell$-dimensional unit sphere and a ball in the $\ell^p$ norm to be
$$\mathcal{S}_{p}^{\ell-1} = \{x\in\mathbb{R}^{\ell}: \|x\|_p=1\}, \quad \mathcal{B}_p^{\ell}(r)=\{z\in\mathbb{R}^{\ell}: \|z\|_p\leq r\}.$$
For an $L$-Lipschitz generator $G: \mathbb{R}^k\rightarrow\mathbb{R}^{n}$, denote by
$$\mathcal{G}_{k,\tau,p}(r) = \{x\in\mathbb{R}^{n}: \exists  z\in \mathcal{B}_{2}^{k}(r), \ \ s. t. \ \ \|cx-G(z)\|_p \leq \tau\},$$
where signals that can generated by $G$ with tolerance $\tau$. When $p=2$, we denote $\mathcal{G}_{k,\tau,2}(r)$ by $\mathcal{G}_{k,\tau}(r)$ for simplicity. The target signal $x^*$ is assumed with
low generative intrinsic dimension, i.e., $x^*\in \mathcal{G}_{k,\tau,p}(r)$ for some $p$ and $r$.

\subsection{Contributions}
It is a  challenging task  to decode from nonlinear, noisy, sign-flipped and under determined ($m\ll n$)  binary  measurements.
For a given Lipschtiz generator $G$, we use $\hat{x} =G(\hat{z})$ to estimate $x^*$  in  the 1-bit CS model \eqref{setup} via exploring the intrinsic low dimensional structure of the target signals,
 where the latent code $\hat{z}$ is solved by the least square problem \eqref{ls2}.
 \begin{itemize}
 \item[(1)] We prove  that, with high probability  the estimation error
   $\|\hat{x}- c x^*\| \leq \mathcal{O} (\sqrt{\frac{k\log (Ln)}{m}})$
   is  sharp provided that   the sample complexity satisfies
   $m\geq  \mathcal{O}( k  \log (Ln))$, if the target signal $x^*$ can  be approximated well by generate $G$.
    \item[(2)] By constructing a ReLU network with  properly chosen depth and width,
we verify  the desired approximation in (1) holds if the target signals have low intrinsic dimensions.
 \item[(3)] Extensive numerical simulations    and comparisons with state-of-the-art methods  show that
the proposed least square decoder is the robust to noise and sign flips,  as demonstrated  by our theory.
\end{itemize}

The rest of the paper is organized as follows. In Section 2  we  consider the  least squares decoder  and prove several  bounds on   $\|\hat{x} -cx^* \|$.
In Section 3 we conduct numerical simulation  and compare  with  existing  state-of-the-art 1-bit CS  methods.
We conclude  in Section 4.
\section{Analysis of the Least Square Decoder}
We first propose the least square decoder in details. Consider the following least square problem for the latent code $z$:
 \begin{equation}\label{ls2}
 \hat{z} \in \arg \min_{z\in \mathcal{B}_2^{k}(r)} \frac{1}{2m}\|y - AG(z)\|^2.
  \end{equation}
Then for a given $L$-Lipschtiz generator $G$, the signal is approximated by
   \begin{equation}\label{ls1}
 \hat{x} =G(\hat{z}).
 \end{equation}
In this section, we will prove under proper assumption on generator and sample complexity, the error between the decoder $\hat{x}$ and the underlying signal $x^*$ can be estimated, i.e., Theorem \ref{thel1} and \ref{theorem2}. Moreover we also provide the construction of a ReLU network such that the approximation to the target signals are satisfied, see Theorem \ref{thapp}.
\begin{theorem}\label{thel1}
Given a Lipschitz generator satisfying  $G(\mathcal{B}_2^k(1))\subset \mathcal{B}_1^n(1)$.
Assume the 1-bit CS model \eqref{setup} holds with  $x^* \in \mathcal{G}_{k,\tau,1}(1)$, and  $m \geq \mathcal{O} \left(\max \{\log n, k\log {\frac{L}{\tau}}\}\right)$, then with  probability at least $1-O(\frac{1}{n^2})-e^{-O(m/k)}$, the least squares decoder defined in (\ref{ls2})-(\ref{ls1}) (for $r=1$) satisfies
\begin{equation*}
\|\hat{x}-c{x^*}\|\leq \mathcal{O}\left(\sqrt{\tau} + \left(\frac{\log n}{m}\right)^{1/4}\right).
\end{equation*}
\end{theorem}

To prove Theorem \ref{thel1}, we need some technical Lemmas. Firstly we introduce the concept of S-REC with some minor changes and $\epsilon-$net, which is defined in \cite{bora2017compressed}.
\begin{definition}\cite{bora2017compressed}.
Let $S\subseteq\mathbb{R}^n$ and two positive parameters $\gamma>0, \delta>0$. The matrix $A\in \mathbb
{R}^{m\times n}$ is said to satisfy the S-REC$(S, \gamma, \delta)$, if  $\forall x_1, x_2\in S$,
\begin{equation}
\frac{1}{m}\|A(x_1-x_2)\|^2\geq\gamma\|x_1 - x_2\|^2 - \delta.
\end{equation}
\end{definition}
\begin{definition}
Let $N \subseteq S\subseteq\mathbb{R}^n$ and $\epsilon > 0$. We say that $N$ is an $\epsilon-$net of $S$, if
$\forall s \in S$, there exist an $\tilde{s} \in N$ such that $\|s-\tilde{s}\|\leq \epsilon$.
\end{definition}

\begin{lemma}\label{cov}\cite{boucheron2013concentration}
$\forall \epsilon > 0$, there exists an $\epsilon-$net $N_{\epsilon}$ of $\mathcal{B}^k_2(r)$ with finite many points in $N_\epsilon$, such that
 $$\log|N_\epsilon|\leq k\log(\frac{4r}{\epsilon}).$$
\end{lemma}
The proof follows directly from the standard volume arguments, see \cite{boucheron2013concentration}.

\begin{lemma}\label{A_delta}
Let $G: \mathbb{R}^k\rightarrow\mathbb{R}^n$ be an $L$-Lipschitz function.
If $N$ is a $\frac{\delta}{L}-$net on $\mathcal{B}^k_2(r)$, then, $G(N)$ is a $\delta-$net on $G(\mathcal{B}^k_2(r))$, i.e.,
\begin{equation}\label{eq:1}
\forall z\in \mathcal{B}^k_2(r), \quad \exists z_1 \in N, \; s.t. \;\|G(z)-G(z_1)\|\leq \delta.
\end{equation}
Furthermore, let
$A\in\mathbb{R}^{m\times n}$ be a random matrix and the rows are i.i.d. random vectors sampled from the multivariate normal distribution $\mathcal{N}(\mathbf{0},\Sigma)$, then,
\begin{equation}\label{eq:2}
\frac{1}{\sqrt{m}}\|AG(z)-AG(z_1)\|\leq\mathcal{O}(\delta)
\end{equation}
holds with probability $1-e^{- O(m)}$ as long as $m = \mathcal{O}\left(k\log\frac{L}{\delta}\right)$.
\end{lemma}
\begin{proof}
Let $N$ be $\frac{\delta}{L}-$ net on $\mathcal{B}_2^k(r)$ satisfying $$\log|N|\leq k\log(\frac{4Lr}{\delta}).$$ Since G is $L$-Lipschitz function, then by definition  we can check that  $ G(N)$ is $\delta-$net on $G(\mathcal{B}_2^k(r))$.

For fixed $\delta >0$, let $N_i$ be a $\frac{\delta_i}{L}-$net on $\mathcal{B}^k_2(r)$ satisfying  $\log|N_i|\leq k\log\frac{4Lr}{\delta_i}$  with $\delta_i = \frac{\delta}{2^i}$, and  $$N = N_0\subset N_1\subset\ldots\subset N_l,$$  with $2^l>\sqrt{n}.$  \\
$\forall x \in G(\mathcal{B}^k_2(r))$, $\exists x_i\in G(N_i)$, such that $$\|x - x_l\|\leq\frac{\delta}{2^l} \text{ and } \|x_{i+1} - x_i\|\leq\frac{\delta}{2^i}, i= 1, \ldots, l-1.$$\\
By triangle inequality we get that,
\begin{equation}\label{eqlem1}
\begin{array}{l}
\frac{1}{\sqrt{m}}\|Ax-Ax_0\|\\
= \|\frac{1}{\sqrt{m}}A\sum_{i=0}^{l-1}(x_{i+1}-x_i) + \frac{1}{\sqrt{m}}A(x-x_l)\|\\
\leq\sum_{i=0}^{l-1}\frac{1}{\sqrt{m}}\|A(x_{i+1}-x_i)\|+\|\frac{1}{\sqrt{m}}A(x-x_l)\|.
\end{array}
\end{equation}
By construction, the last term
\begin{equation}\label{eqlem2}
\begin{array}{l}
\|\frac{1}{\sqrt{m}}A(x-x_l)\|\leq(2+\sqrt{\frac{n}{m}})\frac{\delta}{2^l}=\mathcal{O}({\delta}).
\end{array}
\end{equation}
Let $\widetilde{A} = A\Sigma^{-1/2}$,
by Lemma 1.3 in \cite{vempala2005random}, with probability at least $1-\exp(-\mathcal{O}(\epsilon_i^2m))$, the following holds
\begin{equation*}
\|\frac{1}{\sqrt{m}}\widetilde{A}(x_{i+1}-x_i)\|^2\leq(1+\epsilon_i)\|x_{i+1}-x_i\|^2,
\end{equation*}
equivalently,
\begin{equation*}
\begin{array}{l}
\|\frac{1}{\sqrt{m}}A\Sigma^{-\frac{1}{2}}\Sigma^{\frac{1}{2}}(x_{i+1}-x_i)\|^2\\
\leq(1+\epsilon_i)\|\Sigma^{\frac{1}{2}}\|^2\|x_{i+1}-x_i\|^2,
\end{array}
\end{equation*}
i.e.,
\begin{equation}\label{ineq_A_x_i}
\|\frac{1}{\sqrt{m}}A(x_{i+1}-x_i)\|
\leq(1+\frac{\epsilon_i}{2})\|\Sigma^{\frac{1}{2}}\|\|x_{i+1}-x_i\|,
\end{equation}
the last inequality is derived from  $\sqrt{1+\epsilon_i}\leq 1+\frac{\epsilon_i}{2}, \text{ }\epsilon_i\in(0,1)$.
Set $\epsilon_i^2 =\epsilon +\frac{ik}{m}$, and use union bound and \eqref{ineq_A_x_i}, we have
$\forall i\in[l]$,
\begin{equation}\label{eqlem3}
\|\frac{1}{\sqrt{m}}A(x_{i+1}-x_i)\|
\leq(1+\frac{\epsilon_i}{2})\|\Sigma^{1/2}\|\|x_{i+1}-x_i\|,
\end{equation}
with probability at least $1-\exp{(-\mathcal{O}(\epsilon m))}$.
Then, it follow from \eqref{eqlem1}, \eqref{eqlem2} and \eqref{eqlem3} that,
\begin{equation*}
\begin{array}{l}
\frac{1}{\sqrt{m}}\|Ax-Ax_0\|\\
\leq\|\frac{1}{\sqrt{m}}A\sum_{i=0}^{l-1}(x_{i+1}-x_i)\| + \mathcal{O}(\delta)\\
\leq\sum_{i=0}^{l-1}(1+\frac{\epsilon_i}{2})(\sigma_{max}(\Sigma))^{\frac{1}{4}}\frac{\delta}{2^i} + \mathcal{O}(\delta)\\
\leq \delta(\sigma_{max}(\Sigma))^{\frac{1}{4}}\sum_{i=0}^{l-1}\frac{\sqrt{\epsilon}}{2^{i+1}}(1+\frac{ik}{m\epsilon})+\mathcal{O}(\delta)\\
=\mathcal{O}(\delta).
\end{array}
\end{equation*}
\end{proof}

\begin{lemma}\label{A_REC}
Let $G: \mathbb{R}^k\rightarrow\mathbb{R}^n$ be $L$-Lipschitz generator, $S = G(\mathcal{B}^k_2(r))$,
and $A\in\mathbb{R}^{m\times n}$ be a random matrix and the rows are i.i.d. random vectors sampled from the multivariate normal distribution $\mathcal{N}(\mathbf{0},\Sigma)$. if $m = \mathcal{O}\left(k\log\frac{Lr}{\delta}\right),$ $A$ satisfy the S-REC$(S, \frac{1}{2}\sqrt{\sigma_{min}(\Sigma)}, O(\delta))$, with probability $1 - e^{-O(m/k)}$.
\end{lemma}

\begin{proof}
We construct a $\frac{\delta}{L}-$net on $\mathcal{B}_2^k(r)$, which is denoted as  $N$  and satisfy $\log|N|\leq k\log(\frac{4Lr}{\delta}).$ Since G is $L$-Lipschitz function, then by Lemma \ref{A_delta},   $ G(N)$ is $\delta-$net on $G(\mathcal{B}_2^k(r))$, i.e.,\\
$\forall z, z'\in \mathcal{B}_2^k(r), \exists z_1, z_2\in N$ s.t.
\begin{equation}\label{lem2eq1}
\begin{aligned}
\|z-z_1\|\leq\frac{\delta}{L}, \ \  \|G(z)-G(z_1)\|\leq\delta,\\
\|z'-z_2\|\leq\frac{\delta}{L}, \ \  \|G(z')-G(z_2)\|\leq\delta.
\end{aligned}
\end{equation}

 By triangle inequality, Lemma \ref{A_delta} and \eqref{lem2eq1}, we get
\begin{equation}\label{ineq_G}
\begin{array}{ll}
\|G(z)-G(z')\|
&\leq\|G(z)-G(z_1)\|+\|G(z_1)-G(z_2)\| +\|G(z_2)-G(z')\| \\[1.5ex]
& \leq 2\delta + \|G(z_1)-G(z_2)\|
\end{array}
\end{equation}
and
\begin{equation}\label{A_z1_z2}
\begin{array}{ll}
\frac{1}{\sqrt{m}}\|AG(z_1)-AG(z_2)\|
&\leq\frac{1}{\sqrt{m}}\left(\|AG(z_1)-AG(z)\|+\|AG(z)-AG(z')\| +\|AG(z')-AG(z_2)\|\right)\\[1.5ex]
&\leq \mathcal{O}(\delta) + \frac{1}{\sqrt{m}}\|AG(z)-AG(z')\|.
\end{array}
\end{equation}
Recall $N$ is a $\frac{\delta}{L}-$net on $\mathcal{B}_2^k(r)$, consider
$$G(N) = \{G(z): z\in N\}, \quad T= \Sigma^{\frac{1}{2}}G(N) = \{t: t=\Sigma^{\frac{1}{2}} G(z), z\in N\}, $$
then $|T| \leq  |G(N)| \leq |N|\leq (\frac{4Lr}{\delta})^k$. Similar as Lemma \ref{A_delta}, let $\widetilde{A}=A\Sigma^{-\frac{1}{2}}$, then the rows of $\widetilde{A}$ are i.i.d standard Gaussian vectors. By the Johnson-Lindenstrauss Lemma, the projection $F: \mathbb{R}^n\rightarrow \mathbb{R}^m$ with $F(t) = \frac{1}{\sqrt{m}}A\Sigma^{-\frac{1}{2}}t$ preserves distances in the sense that, given any $\epsilon\in (0,1)$, with probability at least $1-e^{-\mathcal{O}(\epsilon^2 m/k)}$, for all $t_1, t_2\in T$,
\begin{equation*}
(1-\epsilon)\|t_1 - t_2\|\leq\|F(t_1)-F(t_2)\|\leq(1+\epsilon)\|t_1-t_2\|
\end{equation*}
provided that $m \geq \mathcal{O}(\frac{k}{\epsilon^2}\log\frac{Lr}{\delta}).$
We may choose $\epsilon = 0.5$ and hence
\begin{equation}\label{J-L_A}
\frac{1}{\sqrt{m}}\|AG(z_1)-AG(z_2)\|
\geq 0.5\|\Sigma^{\frac{1}{2}}(G(z_1)-G(z_2))\| \geq 0.5 \sqrt{\sigma_{min}(\Sigma)} \|G(z_1)-G(z_2)\|,
\end{equation}
holds with probability at least $1-e^{-\mathcal{O}(m/k)}.$
 It follows from   \eqref{ineq_G}-\eqref{J-L_A} that
\begin{equation*}
\begin{array}{ll}
\frac{1}{\sqrt{m}}\|AG(z)-AG(z')\|
&\geq  \frac{1}{\sqrt{m}}\|AG(z_1)-AG(z_2)\| - \mathcal{O}(\delta) \\ [1.5ex]
 &\geq 0.5 \sqrt{\sigma_{min}(\Sigma)} \|G(z_1)-G(z_2)\| - \mathcal{O}(\delta) \\ [1.5ex]
&\geq  0.5 \sqrt{\sigma_{min}(\Sigma)} \|G(z)-G(z')\| - \mathcal{O}(\delta).
\end{array}
\end{equation*}
The above inequality implies that  $A$ satisfy the S-REC$(G(\mathcal{B}^k_2(r)),0.5\sqrt{\sigma_{min}(\Sigma)}, O(\delta))$ with probability at least $1-e^{-\Omega(m/k)}$, for $m \geq \mathcal{O}(k\log\frac{Lr}{\delta}).$
\end{proof}

Next Lemma shows that least square decoder can be good in the subgaussian setting.
\begin{lemma}\label{linf} \cite{HuangJiao:2018}
Let $A\in\mathbb{R}^{m\times n}$, whose rows  $a_i\in\mathbb{R}^n$, are independent subgaussian vectors with mean $\mathbf{0}$ and covariance matrix $\Sigma$. If  $m\geq \mathcal{O}(\log n)$,  then
\begin{equation}
\left\|\sum_{i=1}^{m}\left(\mathbb{E}\left[a_{i} y_{i}\right]-a_{i} y_{i}\right) / m\right\|_{\infty} \leq  \mathcal{O}(\sqrt{\frac{\log n}{m}})
\end{equation}
 holds with probability at least $1-\frac{2}{n^3}$,
and
\begin{equation}
\left\|A^T A / m-\Sigma\right\|_{\infty} \leq  \mathcal{O}(\sqrt{\frac{\log n}{m}})
\end{equation}
holds with probability at least $1-\frac{1}{n^2}$, where $\|\Psi\|_\infty$ is the maximum pointwise absolute value of $\Psi$.
\end{lemma}

Now we are ready to prove Theorem \ref{thel1}.

\begin{proof}
Recall that
$$y = \eta\odot sign(Ax^*+\epsilon)$$ and
\begin{equation}
 \widehat{z} = \arg\min_{z\in \mathcal{B}_2^k(1)}\frac{1}{2m}\|y - AG(z)\|^2.
\end{equation}
Our goal is to  bound  $\|G(\widehat{z})-\widetilde{x^*}\|_2$ with $\widetilde{x^*}=cx^*$.
By triangle inequality,
\begin{equation*}
\begin{aligned}
\|G(\widehat{z})-\widetilde{x^*}\|
&=\|G(\widehat{z})- G(\overline{z}) + G(\overline{z})-\widetilde{x^*}\|\\
&\leq \|G(\widehat{z})- G(\overline{z})\| + \|G(\overline{z})-\widetilde{x^*}\|,
\end{aligned}
\end{equation*}
where $\overline{z}\in \mathcal{B}_{2}^{k}(1)$ is chosen such that  $\|G(\overline{z})-\widetilde{x^*}\|_1 \leq \tau$ by the assumption $x^*\in \mathcal{G}_{k,\tau,1}(1)$,
we have
\begin{equation}\label{the1}
\|G(\widehat{z})-\widetilde{x^*}\| \leq \|G(\widehat{z})- G(\overline{z})\|+ \tau.
\end{equation}
From the definition of $\widehat{z}$ we have
$$\|AG(\widehat{z}) - y\|^2 \leq \|AG(\overline{z}) - y\|^2.$$
Direct computation shows that
\begin{equation*}
\begin{aligned}
0 &\geq \|AG(\widehat{z}) - y\|^2  -\|AG(\overline{z}) - y\|^2 = \|AG(\widehat{z}) - AG(\overline{z})+ AG(\overline{z})- y\|^2 -\|AG(\overline{z}) - y\|^2 \\%  [1.5ex]
& = \|AG(\widehat{z}) - AG(\overline{z})\|^2  + 2\langle G(\widehat{z}) - G(\overline{z}), A^T(AG(\overline{z})- y) \rangle,
\end{aligned}
\end{equation*}
which hence
\begin{equation}\label{eql}
\begin{array}{ll}
\frac{1}{m}\|AG(\widehat{z}) - AG(\overline{z})\|^2
&\leq 2\langle G(\widehat{z}) - G(\overline{z}), \frac{1}{m}A^T(y -AG(\overline{z})) \rangle\\ [1.5ex]
&\leq 2\|G(\widehat{z}) - G(\overline{z})\|_1\|\frac{1}{m}A^T(y -AG(\overline{z}))\|_\infty \\[1.5ex]
&\leq 4 \|\frac{1}{m}A^T(y -AG(\overline{z}))\|_\infty,
\end{array}
\end{equation}
where the last step is from the assumption $G(\mathcal{B}_2^k(1))\subset \mathcal{B}_1^{n} \Rightarrow \|G(\widehat{z}) - G(\overline{z})\|_1\leq 2$. Next we bound $\frac{1}{m}\|A^T(y - AG(\overline{z}))\|_\infty$.
By triangle inequality,
\begin{equation}\label{the5}
\begin{array}{ll}
\frac{1}{m}\|A^T(y - AG(\overline{z}))\|_\infty
&= \frac{1}{m}\|A^T(y - A\widetilde{x^*} + A\widetilde{x^*} - AG(\overline{z}))\|_\infty \\[1.5ex]
&\leq\frac{1}{m}\|A^T(y - A\widetilde{x^*})\|_\infty + \frac{1}{m}\|A^T(A\widetilde{x^*} - AG(\overline{z}))\|_\infty.
\end{array}
\end{equation}
The first term in \eqref{the5} can be estimated by
\begin{equation}\label{theq7}
\begin{array}{ll}
\frac{1}{m}\|A^T(y - A\widetilde{x^*})\|_\infty
&=\|\frac{1}{m}A^Ty -\Sigma\widetilde{x^*} + \Sigma\widetilde{x^*} - \frac{1}{m}A^TA\widetilde{x^*}\|_\infty\\[1.5ex]
&\leq\|\frac{1}{m}A^Ty -\Sigma\widetilde{x^*}\|_\infty + \|\Sigma\widetilde{x^*} - \frac{1}{m}A^TA\widetilde{x^*}\|_\infty\\[1.5ex]
&\leq\frac{1}{m}\|A^Ty - \mathbb{E}[A^Ty]\|_\infty + \|\Sigma - \frac{1}{m}A^TA\|_\infty\|\widetilde{x^*}\|_1\\[1.5ex]
&= \frac{1}{m}\|\sum_{i=1}^m (A_iy_i-\mathbb{E}[A_iy_i])\|_\infty + \|\Sigma - \frac{1}{m}A^TA\|_\infty\|\widetilde{x^*}\|_1\\[1.5ex]
& \leq \mathcal{O}\left(\sqrt{\frac{\log n}{m}}\right),
\end{array}
\end{equation}
where the last inequality is from Lemma \ref{linf}. To estimate the second term in \eqref{the5}, denote by $\widetilde{\Delta} = \widetilde{x^*} - G(\overline{z})$,
we then have
\begin{equation}\label{the8}
\begin{array}{ll}
\frac{1}{m}\|A^T(A\widetilde{x^*} - AG(\overline{z}))\|_\infty
&= \frac{1}{m}\|A^TA\widetilde{\Delta}\|_\infty\\[1.5ex]
&= \|\frac{1}{m}A^TA\widetilde{\Delta} - \Sigma\widetilde{\Delta} + \Sigma\widetilde{\Delta}\|_\infty\\[1.5ex]
&\leq \|(\frac{1}{m}A^TA - \Sigma)\widetilde{\Delta}\|_\infty + \|\Sigma\widetilde{\Delta}\|_\infty\\[1.5ex]
&\leq \|\widetilde{\Delta}\|_1(\|\frac{1}{m}A^TA - \Sigma\|_\infty + \|\Sigma\|_\infty)\\[1.5ex]
&\leq \mathcal{O}\left(\sqrt{\frac{\log n}{m}}+1\right)\tau.
\end{array}
\end{equation}

From lemma \ref{A_REC}, $A$ satisfies the S-REC$(G(\mathcal{B}_2^k(1)), 0.5\sqrt{\sigma_{min}(\Sigma)}, O(\delta)),$ with probability $1 - e^{-O(m/k)}$ as long as
 $m = O(k\log\frac{L}{\delta})$, i.e.,
 \begin{equation}\label{the3}
   \frac{1}{m}\|AG(\widehat{z}) - AG(\overline{z})\|^2
   \geq 0.5\sqrt{\sigma_{min}(\Sigma)}\|G(\widehat{z}) - G(\overline{z})\|^2 - \mathcal{O}(\delta).
   \end{equation}
Substituting \eqref{the5} - \eqref{the3} into \eqref{eql} we obtain

\begin{equation}\label{eq2}
0.5\sqrt{\sigma_{min}(\Sigma)}\|G(\widehat{z}) - G(\overline{z})\|^2 - \mathcal{O}(\delta) \leq \mathcal{O}\left(\sqrt{\frac{\log n}{m}}+\tau + \sqrt{\frac{\log n}{m}}\tau\right).
\end{equation}
We may choose $\delta = O(\tau)$ in \eqref{eq2} and substituted it into \eqref{the1}, we conclude
\begin{equation*}
\|G(\widehat{z}) - \widetilde{x}^*\| \leq \mathcal{O}\left(\left(\frac{\log n}{m}\right)^{1/4} + \sqrt{\tau}\right).
\end{equation*}
\end{proof}

Obviously, $\tau$ measures the approximation error between the target $x^*$ and the generator $G$. If we assume that $\tau$ is smaller than $\mathcal{O}((\frac{\log n}{m})^{1/2})$,
Theorem \ref{thel1} shows that under that approximate low generative dimension prior, our proposed least decoder  (\ref{ls1})-(\ref{ls2})
can achieve an estimation error  $\mathcal{O}((\log n/m)^{1/4})$  provide that the number of samples $m \geq \mathcal{O}(\max \{\log n, k\log \frac{L}{\tau}\})$.
 Similar results has been established for 1-bit CS under the sparsity prior $\|x^*\|_0\leq s$ in the literatures. For example,
\cite{PlanVershynincpam:2013} proposed  a linear programming  decoder
 \begin{equation*}
 x_{\mathrm{lp}}\in \arg\min_{x\in \mathbb{R}^n} \|x\|_1 \quad \mathrm{s.t.} \quad  y \odot A x\geq 0  \quad \|A x\|_1  =  m.
 \end{equation*}
 in the   noiseless setting  without sign flips. It has been proved in \cite{PlanVershynincpam:2013} that
 $$\|\frac{x_{\mathrm{lp}}}{\|x_{\mathrm{lp}}\|} - x^*\|\leq \mathcal{{O}}((\frac{s\log n}{m})^{1/5}),$$
 provided that $m=\mathcal{O}(s\log^2(n/s))$.
Later,  in  \cite{PlanVershynin:2013}, another convex decoder
  \begin{equation*}
 x_{\mathrm{cv}} \in \arg\min_{x\in \mathbb{R}^n} -\langle y, A x\rangle/m \quad \mathrm{s.t.} \quad \|x\|_1 \leq s,  \quad \|x\| \leq 1,
 \end{equation*}
 is shown to achieve a estimation error bound
$$\|\frac{x_{\mathrm{cv}}}{\|x_{\mathrm{cv}}\|} - x^*\|\leq \mathcal{{O}}((\frac{s\log n}{m})^{1/4}).$$

Although the order of estimation error   proved in Theorem  \ref{thel1} does  not depend on the Lipschtiz constant of the generator $G$  which is usually  exponential order of the depth of   the  neural networks  \cite{bora2017compressed},  it  is  sub-optimal.  Next we  improve the estimation error bound  by using the tool of local (Gaussian) mean width. The definition of local mean width is given below, it can also be found in \cite{PlanVershynin:2016,PlanVershynin:2017}.
\begin{definition}\label{gmw}
Let $S\subseteq\mathbb{R}^n$. The local mean width of $S$ is a function of scale $t \geq 0$ defined as
$$
\omega_{t}(S)=\mathbb{E}_{g\sim \mathcal{N}(\mathrm{0},\mathbf{I})} \left[\sup _{x \in S \cap t \mathcal{B}_2^{n}(1)}\langle x, g\rangle\right].
$$
\end{definition}

\begin{theorem}\label{theorem2}
Given an $L$-Lipschitz generator satisfying  $G(\mathcal{B}_2^k(r))\subset \mathcal{S}_2^{n-1}$.
Assume the 1-bit CS model \eqref{setup} holds with  $x^* \in \mathcal{G}_{k,\tau}(r)$,
and  $m \geq \mathcal{O}\left(\max\{k\log\frac{Lrn}{k},\log n\}\right)$,
 then with high  probability,
 the least square decoder defined in (\ref{ls2})-(\ref{ls1}) satisfies
\begin{equation*}
\|\hat{x}-c{x^*}\|_2\leq \mathcal{O}\left(\sqrt{\frac{k}{m}\log\frac{rLn}{k\gamma}}\right) +\mathcal{O}(\frac{\tau n}{m}),
\end{equation*}
for $\gamma=\max\{\tau,\frac{k}{m}\log \frac{Lrn}{k}+\sqrt{\frac{\log n}{m}}\}$. If the approximation error satisfies $\tau = \mathcal{O} (\frac{\sqrt{mk\log (Ln)}}{n})$ and $r = \mathcal{O}(1)$, then we have
\begin{equation*}
\|\hat{x}-c{x^*}\|_2\leq \mathcal{O}\left(\sqrt{\frac{k}{m}\log(Ln)}\right).
\end{equation*}
\end{theorem}

First we do some preparing work before the proof. Similar as the proof to Theorem \ref{thel1}, let $\overline{z}\in \mathcal{B}_{2}^{k}(r)$ satisfying  $\|G(\overline{z})-\widetilde{x^*}\| \leq \tau$. By triangle inequality \eqref{the1}, we have
\begin{equation}\label{eqth21}
\|G(\widehat{z})-\widetilde{x^*}\| \leq \|G(\widehat{z})- G(\overline{z})\|+ \tau.
\end{equation}
Let $h = G(\hat{z})-G(\overline{z})$, $\gamma=\max\{\tau,\frac{k}{m}\log \frac{Lrn}{k}+\sqrt{\frac{\log n}{m}}\}$, $S = G(\mathcal{B}_2^k(r))$, and $D_{\gamma}(S,G(\overline{z}))$ be the  tangent cone which is defined by
$$ D_{\gamma}(S,G(\overline{z})) = \{t u:  t>0, u = G(z)-G(\overline{z}), \|u\|>\gamma\}.$$
If $\|h\|\leq \gamma $ this Theorem is trivial by (\ref{eqth21}), otherwise $\|h\|> \gamma $, then $h\in D_{\gamma}(S,G(\overline{z}))$.
Let $$\mathcal{D} = D_{\gamma}(S,G(\overline{z})) \cap \mathcal{S}_{2}^{n-1}.$$
We need the following two lemmas to proceed the proof.
\begin{lemma}\label{pv}
 With probability at least $0.99$, both
 \begin{equation}\label{eqth25}
 \inf _{v \in \mathcal{D}} \frac{1}{\sqrt{m}}\|A v\|_{2} \geq C_0
 \end{equation}
 \begin{equation}\label{eqth26}
 \sup _{v \in \mathcal{D}}\frac{1}{m}\left\langle v, A^{T} (y -A\widetilde{x^*})\right\rangle \leq C\frac{\omega_1(\mathcal{D})}{\sqrt{m}}.
 \end{equation}
 hold, where $\omega_1(\mathcal{D})$ is the local (Gaussian) mean width of $\mathcal{D}$ given  in Definition \ref{gmw}.
 \end{lemma}
\begin{proof}
The results can be found in the proof to Theorem 1.4 in \cite{PlanVershynin:2016}.
\end{proof}

 \begin{lemma}\label{gw}
$$\omega_1(\mathcal{D})=  \mathcal{O}\left(\sqrt{k\log(\frac{rLn}{k\gamma})}\right).$$
 \end{lemma}
\begin{proof}
Recall that $$D_{\gamma}(S,G(\overline{z})) = \{t u:  t>0, u = G(z)-G(\overline{z}), \|u\|>\gamma\}.$$
  and  $$\mathcal{D} = D_{\gamma}(S,\overline{z}) \cap \mathcal{S}_{2}^{n-1}.$$
  Then $\mathcal{D}=\{\frac{ G(z)-G(\overline{z})}{\| G(z)-G(\overline{z})\|}: z\in \mathcal{B}_2^k(r), \| G(z)-G(\overline{z})\|>\gamma\}.$
Let $\mathcal{U}$ be $\frac{\epsilon\gamma}{2L}-$ net on $\mathcal{B}_2^k(r)$ satisfying $$\log|\mathcal{U}|\leq k\log(\frac{8Lr}{\gamma\epsilon}),$$ which can be obtained by Lemma \ref{cov}.
Then, $\mathcal{C} = \{\frac{G(u)-G(\overline{z})}{G(u)-G(\overline{z})}:u \in \mathcal{U} \ \ \mathrm{and} \ \ \|G(u)-G(\overline{z})\|\geq \gamma\}$ is a $\epsilon-$ net of $\mathcal{D}$.
Indeed,
let $\frac{a}{\|a\|}$ with $a =  G(z)-G(\overline{z})$ be an arbitrary element in $\mathcal{D}$, and $u\in \mathcal{U}$ such that
$\|u-z\|\leq \frac{\epsilon\gamma}{2L}$. Let   $b=  G(u)-G(\overline{z})$,  then $\frac{b}{\|b\|}\in \mathcal{C}$ and satisfies that

\begin{align*}
& \left\| \frac{a}{\|a\|}-\frac{b}{\|b\|}\right\| = \left\|\frac{a\|b\|-b\|a\|}{\|a\|\|b\|}\right\|  \\
& \leq \frac{\|(a-b)\|b\|\|+\|b(\|b\|-\|a\|)\|}{\|a\|\|b\|}\\
& \leq 2\frac{\|a-b\|}{\|a\|} \leq 2L\frac{\frac{\epsilon\gamma}{2L}}{\gamma}=\epsilon,
\end{align*}
where in last inequality we use the facts that $G$ is $L$-Lipschitz and $\|a\|\geq \gamma.$

By Massart's finite class Lemma in \cite{boucheron2013concentration}, the  local Gaussian  width of $\mathcal{C}$  satisfies
\begin{equation}\label{gf}
\omega_1(\mathcal{C})\leq \sqrt{2k\log(\frac{16Lr}{\gamma\epsilon})}.
\end{equation}
Since $\forall x \in \mathcal{D}$, there exist $\tilde{x} \in \mathcal{C}$ such that
$\|x-\tilde{x}\|\leq \epsilon$. We then have $\forall g \sim \mathcal{N}(\mathbf{0}, \mathbf{I})$
\begin{equation*}
\langle g, x\rangle  \leq \langle g, \tilde{x}\rangle + \langle g, x-\tilde{x}\rangle \leq  \langle g, \tilde{x}\rangle + \epsilon \|g\|.
\end{equation*}
The above display and the definition of local Gaussian mean width and the fact $\mathcal{D}\subset \mathcal{B}_2^{n}(1)$
implies
\begin{align*}
\omega_1(\mathcal{D})&= \mathbb{E}[\sup_{x\in \mathcal{D}}\langle g, x\rangle]\leq \mathbb{E} [\sup_{\tilde{x} \in \mathcal{C}}\langle g, \tilde{x}\rangle + \epsilon \|g\|]\\
& \leq \omega_1(\mathcal{C})+\sqrt{n}\epsilon \\
&\leq  \sqrt{2k\log(\frac{16Lr}{\gamma\epsilon})} + \sqrt{n}\epsilon,
\end{align*}
where the second equality follow from $\mathbb{E}[\|g\|] = \sqrt{n}$, and in the third inequality  we use \eqref{gf}.
The proof will be finished by  setting $\epsilon = \sqrt{\frac{k}{n}}$.

\end{proof}

 \begin{lemma}\label{spectralnorm}
Let $a_i\in \mathbb{R}^{n}, i = 1,...n$ are i.i.d samples with mean $0$ and covariance matrix $\Sigma$. Denote $\Sigma_m = \sum_{i=1}^{m} a_ia_i^{T}/m$. Then for any $u \geq 0$
$$
\left\|\Sigma_{m}-\Sigma\right\| \leq \mathcal{O}\left(\sqrt{\frac{n+u}{m}}+\frac{n+u}{m}\right)\|\Sigma\|
$$
with probability at least $1-2 e^{-u}$.
 \end{lemma}
 \begin{proof}
 See  exercise 4.7.3 in \cite{vershynin2018high}.
 \end{proof}

Now we can move to the proof to Theorem \ref{theorem2}.

\begin{proof}
Similar as \eqref{eql} in the proof to Theorem \ref{thel1}, by \eqref{eqth25} in Lemma \ref{pv} and triangle inequality, we have
with probability  at least $0.99$ that
\begin{equation}\label{eq23}
\begin{array}{ll}
C_0 \|h\|^2 \leq \frac{1}{m}\|Ah\|^2
&\leq 2\langle h, \frac{1}{m}A^T(y -AG(\overline{z}))\rangle.\\[1.5ex]
&\leq 2|\langle h, \frac{1}{m}A^T(y -A\widetilde{x^*}) \rangle|
+ 2|\langle h, \frac{1}{m}A^T A(\widetilde{x^*}-G(\overline{z})) \rangle|.
\end{array}
\end{equation}
 We have to bound the two terms in \eqref{eq23}. For the first term, let $v = \frac{h}{\|h\|} = \frac{G(\widehat{z}) - G(\overline{z})}{\|G(\widehat{z}) - G(\overline{z})\|}$, hence $v\in \mathcal{D}$. Then by \eqref{eqth26} in Lemma \ref{pv} and Lemma \ref{gw}, we obtain that with probability  at least $0.99$
\begin{equation}\label{eq30}
|\langle h, \frac{1}{m}A^T(y -A\widetilde{x^*}) \rangle|  \leq \mathcal{O}\left(\sqrt{\frac{k\log(rLn/(k\gamma))}{m}}\right)  \|h\|.
\end{equation}
For the second term in \eqref{eq23}, we apply Cauchy-Schwarz inequality and using spectral norm estimation for random matrix in Lemma \ref{spectralnorm}  , we get with high probability at least $1-e^{-n}$ that
\begin{align}\label{eq31}
 |\langle h, \frac{1}{m}A^T A(\widetilde{x^*}-G(\overline{z})) \rangle|

 &\leq \|h\|  \frac{1}{m}\|A^T(A\widetilde{x^*} - AG(\overline{z}))\| \nonumber\\
 &\leq (\|A^TA/m-\Sigma\|_2+\|\Sigma\|) \|\widetilde{x^*}-G(\overline{z})\| \|h\| \nonumber \\
 &\leq \mathcal{O}(\sqrt{\frac{2n}{m}}+\frac{2n}{m}+1)\|\Sigma\|\tau\|h\|\\
 & \leq \mathcal{O}(\tau \frac{n}{m})\|h\|.
 \end{align}

Combining \eqref{eq23}, \eqref{eq30} and \eqref{eq31} we get
$$\|h\|\leq \mathcal{O}(\sqrt{\frac{k\log(rLn/(k\gamma))}{m}}) +\mathcal{O}(\frac{\tau n}{m}).$$
Moreover, if the approximation error $\tau = \mathcal{O} (\frac{\sqrt{mk\log (Ln)}}{n})$ and $r = \mathcal{O}(1)$, the above inequality can be reduced to
\begin{equation*}
\|\hat{x}-c{x^*}\|_2\leq \mathcal{O}\left(\sqrt{\frac{k}{m}\log(Ln)}\right),
\end{equation*}
which completes the proof.
\end{proof}

By assuming the   Lipschitz constant $L$ is  larger than $n$ (this usually  holds in deep neural network generators), the estimation
error $\mathcal{O}(\sqrt{\frac{k\log L}{m}})$ and the sample complexity $m \geq \mathcal{O}(k\log L) $ proved in Theorem \ref{theorem2} are  sharp
even in  the standard compressed sensing with generative prior \cite{liu2020information}.
 Under generative prior,   \cite{liu2020sample} proposed
the estimator   $\hat{x} = G(\hat{z})$ with $\hat{z} \in \{z: y=\mathrm{sign}(AG(z))\}$ in the setting the rows of $A$   are   i.i.d.   sampled from   $\mathcal{N}(\textbf{0},\mathbf{I})$.   The sample complexity obtained  in \cite{liu2020sample}
 is also $\mathcal{O}(k\log L)$.
 \cite{qiu2020robust} proposed unconstrained empirical
risk minimization to recover the  1-bit CS in the scenario that   the rows of $A$   are   i.i.d.   sampled from   subexponential distributions, and the generator $G$ is restricted to be  a  $d$-layer ReLU network. The sample complexity derived in \cite{qiu2020robust} is $\mathcal{O}(kd\log n)$.

There are some works on generative priors which assumes that the target signals can be exactly generated by a generator $G$, i.e.,  $x^* \in \mathcal{G}_{k,\tau}(r)$ with $\tau=0$, see e.g.
\cite{liu2020generalized,qiu2020robust}.
As mentioned in Theorem \ref{theorem2}, we can relax this assumption by requiring
the  target signal $x^*$ can  be generated  by $G$  approximately, i.e.,
$x^* \in \mathcal{G}_{k,\tau}(r)$  with
\begin{equation}\label{eq:tau}
\tau = \mathcal{O} (\frac{\sqrt{mk\log (Ln)}}{n}).
\end{equation}

Since natural signals/images data  with low intrinsic dimension can be represented   approximately by   neural networks is empirically verified in  \cite{goodfellow14,kingma14,rezende2015variational}.
 Next, we verify the assumption \eqref{eq:tau} by construct   a generator  ${G}$ with  properly chosen depth and width  based on  the recent approximation ideas of  deep neural networks \cite{shen2019nonlinear,vershynin2020memory,vardi2021optimal,huang2021error} by  utilizing the bit extraction techniques \cite{bartlett1999almost,bartlett2019nearly}. To this end, we recall the
definition of  Minkowski dimension  which is used to measures the intrinsic dimension of the target signals living in a large ambient dimension.

\begin{definition}\label{Minkowski dimensions}
\textnormal{
The upper and the lower Minkowski dimensions of a set $A \subseteq \mathbb{R}^n$ are defined respectively as
\begin{align*}
\overline{\dim}_M(A) := \limsup_{\epsilon\to 0} \frac{\log |N_{\epsilon}|}{-\log \epsilon}, \\
\underline{\dim}_M(A) := \liminf_{\epsilon\to 0} \frac{\log |N_{\epsilon}|}{-\log \epsilon},
\end{align*}
where $N_{\epsilon}$ is the $\epsilon$-net of $A$.
If $\overline{\dim}_M(A) = \underline{\dim}_M(A) = \dim_M(A)$, then $\dim_M(A)$ is called the \emph{Minkowski dimension} of the set $A$.
}
\end{definition}

The Minkowski dimension measures how the  number of elements in the $\epsilon$-net $N_{\epsilon}$ of  $A$ decays when the radius of covering balls converges to zero. We collect the useful properties from  \cite{falconer2004fractal} of Minkowski dimension.
\begin{proposition}\cite{falconer2004fractal}\label{pmd}
$\dim_M(A) < \bar{n}$ if and only if $\forall \gamma >0$, $$\epsilon^{-(\bar{n}-\gamma)}\leq |N_{\epsilon}| \leq \epsilon^{-(\bar{n}+\gamma)}$$ holds when $\epsilon$ small enough.
Furthermore,    $\dim_M(A) \leq \bar{n}$ implies that
$\forall \epsilon>0$, there exist an $\epsilon$-net $N_{\epsilon}$ of $A$  such that $|N_{\epsilon}| \leq c\epsilon^{-\bar{n}}$, where $c$ is a  finite number.
\end{proposition}

The next three Lemmas present the approximation ability of the deep neural networks.

 \begin{lemma}\label{fit1}
 For any $\mathcal{W},\ell\in \mathbb{N}$, given $\mathcal{W}^{2} \ell$ samples $\left(z_{i}, y_{i}\right), i=1, \ldots, \mathcal{W}^{2} \ell$, with distinct $z_{i} \in \mathbb{R}^{k}$ and $y_{i} = \sum_{j=1}^{\ell}2^{-j}b_{i,j}$, $b_{i,j}\in\{0,1\}$. There exists a ReLU network $G_1$ with width $4 \mathcal{W}+4$ and depth $\mathcal{\ell}+2$ such that $G_1\left(z_{i}\right)=y_{i}$ for $i=1, \ldots, \mathcal{W}^{2} \ell.$
 \begin{proof}
 This lemma   follows directly from  of Lemma $2.1$ and $2.2$ in
\cite{shen2019nonlinear}.
 \end{proof}
 \end{lemma}
 \begin{lemma}\label{bitext}
 For any $\ell \in \mathbb{N}$, there exists a ReLU network $G_2$ with width $8$ and depth $2\ell$  such that $G_2(x, j)=b_{j}$ for any $x=\sum_{j=1}^{\ell}2^{-j}b_j$  with $b_{j} \in\{0,1\}$ and $j=1,2, \ldots, \ell$.
 \end{lemma}
 \begin{proof}
 This lemma follow from  Lemma 5.7 in \cite{huang2021error}.
 \end{proof}
 \begin{lemma}\label{intmul}
  Let $\mathcal{W}\in \mathbb{N}$. Given any $\mathcal{W}^{2} \ell^2$ points  $\{\left(z_{i},  b_{i,j}\right) i=1, \ldots, \mathcal{W}^{2} \ell,  j=1,...,\ell\}$, where $z_{i} \in \mathbb{R}^{k}$ are distinct and $b_{i,j} \in \{0,1\}$. There exists a ReLU network $G_3$ with width $4 \mathcal{W}+6$ and depth $3 \ell+1$ such that $G_3\left(z_{i},j\right)=b_{i,j}$, $i=1, \ldots, \mathcal{W}^{2} \ell,   j=1,...,\ell$.
  \end{lemma}
  \begin{proof}
    $\forall i=1, \ldots, \mathcal{W}^{2} \ell$, let $y_{i}= \sum_{j=1}^{\ell}2^{-j}b_{i,j} \in[0,1].$  By  Lemma \ref{fit1} there exists a network $G_{1}$ with width $4 \mathcal{W}+4$ and depth $\ell+2$ such that $G_{1}\left(z_{i}\right)=y_{i}$ for $i=1, \ldots, \mathcal{W}^{2} \ell$.
By Lemma 6.4, there exists a network $G_{2}$ with width 8 and depth $2\ell$ such that $G_{2}\left(y_{i}, \ell\right)=b_{i, \ell}$ for any $i=1, \ldots, \mathcal{W}^{2} \ell$ and $j=1, \ldots, \ell$. Therefore, the function $G_3(\cdot, j)=G_{2}\left(G_{1}(\cdot), j\right)$  implemented by a ReLU network with width $ 4\mathcal{W}+6$ and depth $3 \ell+1$ satisfies our requirement.
  \end{proof}

\begin{theorem}\label{thapp}
Assume the target signals $x^*\in A^*\subseteq [0,1]^n$ with $\dim_M(A^*) = k$. Then $\forall \tau \in (0,1)$ there exist a generator network $G:\mathbb{R}^{k}\rightarrow \mathbb{R}^n$
with depth $3\ell+2$  and width  $(4\lceil \sqrt{sn/\ell}\rceil+6)n$ such that $x^*\in \mathcal{G}_{k,\tau,2}(1), \forall x^*\in A^*$, where
$\ell = \lceil\log_2(\frac{2n}{\tau})\rceil+1, \quad s = \mathcal{O}(\tau^{-k}).$
\end{theorem}
\begin{proof}
Let $\epsilon = \tau/2$. Since the target signals $x^*$ are contained in $A^*$ with   $\dim_M(A^*) = k$, then there exist an $\epsilon$-net $N_{\epsilon} = \{o^{*}_i\}_{i=1}^{s}$ of $A^*$ with $s\leq c\epsilon^{-k}$ by Proposition \ref{pmd}.
For any $o^{*}_i \in N_{\epsilon}$, let the binary representation of $ o^{*}_i$ be $o^{*}_i=\sum_{j=1}^{\infty} 2^{-j} \tilde{o}^{*}_{i,j}$ with $\tilde{o}^{*}_{i,j}\in\mathbb{R}^n$ whose entries $\in \{0,1\}$.
 Let $\ell = \lceil\log_2(\frac{n}{\epsilon})\rceil+1$, the truncation of $o^{*}_i$ be $T_{\ell}o^{*}_i = \sum_{j=1}^{\ell} 2^{-j} \tilde{o}^{*}_{i,j}$,
then it implies that \begin{equation}\label{tc}
\|o^{*}_i-T_{\ell}o^{*}_i\|\leq \epsilon, \forall i =1,...,s.
\end{equation}
By construction of $N_{\epsilon}$,  (\ref{tc}) and triangle inequality, we have $\{T_{\ell}o^{*}_i\}_{i=1}^s$ is an $\tau$-net of $A^*$.
  Let $e= (1,0,0,...0)^T\in\mathbb{R}^s$, $\mathcal{W} = \lceil \sqrt{sn/\ell}\rceil$,  and  $z_i$ be the $i$-th element of $\{e,  e/2,...,e/(ns)\}$,  $b_{i,j} = G_2(T_{\ell}o^{*}_i,j)$,

   $i=1,...,sn, j=1,...\ell$.
  By Lemma \ref{intmul}, we have $G_3\left(z_{i},j\right)=b_{i,j}$, $i=1, \ldots, sn,   j=1,...,\ell$.
  $\forall x\in \mathbb{R}^k$, define  $$G(x)=(\sum_{j=1}^{\ell} 2^{-j}G_3(a_1 x, j),\sum_{j=1}^{\ell} 2^{-j}G_3(a_2x, j),...\sum_{j=1}^{\ell} 2^{-j}G_3(a_n x,j))^T:\mathbb{R}^k\rightarrow \mathbb{R}^n$$ with $a_1>0, a_2>0,...,a_n>0.$
  Let $\theta_1$ and $\theta_2$ be the parameters of the ReLU network $G_1$ and $G_2$, respectively.
   Denote $a =(a_1,...,a_n)\in\mathbb{R}^n$ and $\theta=(a,\theta_1,\theta_2)$.
  Then $G(x)$ is a ReLU network with free parameter $\theta$ and depth $3\ell+2$  and width  $(4\lceil \sqrt{sn/\ell}\rceil+6)n$. We use $G_{\theta}(x)$ to emphasize the dependence of $G$ on $\theta$.
 For $i=1,2,...s$, let  $a^{(i)} = (\frac{1}{(i-1)n+1},\frac{1}{(i-1)n+2}, ..., \frac{1}{(i-1)n+n})^T \in\mathbb{R}^n,$
 $\theta^{(i)}  = (a^{(i)},\theta_1,\theta_2)$,   by construction, we have
    $G_{\theta^{(i)}} (e) = T_{\ell}o^{*}_{i}$.

\end{proof}

\section{Numerical Experiments}

\subsection{Experiments setting}
The rows of the matrix $A$ are i.i.d. random vectors sampled from the multivariate normal distribution $\mathcal{N}(\mathbf{0},\Sigma)$ with $\Sigma_{jk}=\nu^{|j-k|}$, $1\leq j,k\leq n$, $\nu =0.3$ in our tests. The elements of $\epsilon$ are generated from $\mathcal{N}(\mathbf{0},\sigma^2\mathbf{I}_m)$ with $\sigma=0.1$ in our examples. $\eta$ has independent coordinates with $\mathbb{P}\{\eta_i=1\}=1-\mathbb{P}\{\eta_i=-1\}=q\neq\frac{1}{2}$, with different $q$ which will be clarified in each example.
The generative model $G$ in our experiments is a pretrained variational autoencoder (VAE) model\footnote{We use the pre-trained generative model of (Bora et., 2017) available at https://github.com/
AshishBora/csgm.}. The MNIST dataset \cite{lecun1998gradient} consisting $60000$ handwritten images of size 28$\times$28 is applied in our tests. For this dataset, we set the VAE model with a latent dimension $k=20$. Input to the VAE is a vectorized binary image of input dimension $784$. Encoder and decoder are both fully connected network with two hidden layers, i.e., encoder and decoder are with size $784-500-500-20$ and $20-500-500-784$, respectively.

To avoid the  norm constraint $\|z\|_2 \leq r$ in  the least square decoder (\ref{ls1})-(\ref{ls2}), we use its Lagrangian form as following:
\begin{equation}\label{LS generative model}
\min_{z} \frac{1}{2m}\|y - AG(z)\|^2 + \lambda\|z\|^2,
\end{equation}
where the regularization parameter $\lambda$ is chosen as $0.001$ for all the experiments.
We do $10$ random restarts with $1000$ steps per restart and choose the best estimation. The reconstruction error is calculated over $10$ images by averaging the per-pixel error in terms of the $l_2$ norm.
\subsection{Experiment Results}
We compare our results with two SOTA algorithms: BIHT \cite{JacquesLaska:2013}
and generative prior based algorithm  VAE \cite{liu2020sample}.
The least square decoder with VAE in our paper is named by LS-VAE

Figures $1-4$ indicate that with or without sign flip in measurements,
generative prior based methods attain more accurate reconstruction than BIHT.
Additionally, if sign flips are added, Figures $3$ and $4$ show that LS-VAE
attain the higher accurate reconstruction.

In Figure $5$, we plot the reconstruction error for different measurements (from $50$ measurements to $300$ measurements).
VAE and LS-VAE both have smaller reconstruction errors, but LS-VAE is slightly better.
Moreover, after $200$ measurements, the reconstruction error emerges saturation for generative prior based methods, due to
its output is constrained to the presentation error \cite{bora2017compressed}.
\begin{figure}[ht!]
\begin{center}
\includegraphics[scale=0.4]{100-03-biht.pdf}
\setlength{\abovecaptionskip}{1pt}
\caption{original images, reconstructions by BIHT, VAE and LS-VAE (from top to bottom row)
with 100 measurements}
\end{center}
\end{figure}
\begin{figure}[ht!]
\begin{center}
\includegraphics[scale=0.4]{300-03-biht.pdf}
\setlength{\abovecaptionskip}{1pt}
\caption{original images, reconstructions by BIHT, VAE and LS-VAE (from top to bottom row) with 300 measurements}
\end{center}
\end{figure}
\begin{figure}[ht!]
\begin{center}
\includegraphics[scale=0.42]{100-003-03-biht.pdf}
\setlength{\abovecaptionskip}{1pt}
\caption{original images, reconstructions by BIHT, VAE and LS-VAE (from top to bottom row)  with 100 measurements
and 3\% sign flips}
\end{center}
\end{figure}
\begin{figure}[ht!]
\begin{center}
\includegraphics[scale=0.4]{300-003-03-biht.pdf}
\setlength{\abovecaptionskip}{1pt}
\caption{original images, reconstructions by BIHT, VAE and LS-VAE (from top to bottom row)  with 300 measurements
and 3\% sign flips}
\end{center}
\end{figure}
\begin{figure}[ht!]
\begin{center}
\includegraphics[scale=0.5]{mnist_00001_00_0001_to300_nolasso_l2.pdf}
\includegraphics[scale=0.5]{mnist_00001_003_0001_to300_nolasso_l2.pdf}
\setlength{\abovecaptionskip}{1pt}
\caption{pixel-wise reconstruction error as the number of measurements varies. Error bars indicate 95\% confidence intervals.
The  result with  no sign flips and  with 3\% sign flips are shown in the left and right, respectively.}
\end{center}
\end{figure}
\section{Conclusion}
We present a least square decoder by exploring the low generative intrinsic dimension structure of the target for the
1-bit compressive sensing with possible sign-flips.
Under the assumption  that the target signals can be approximately generated via
$L$-Lipschitz generator $G: \mathbb{R}^k\rightarrow\mathbb{R}^{n}, k\ll n$,
we prove that, up to a constant $c$, with high probability, the least square decoder
achieves a sharp  estimation error  $\mathcal{O} (\sqrt{\frac{k\log (Ln)}{m}})$ as long as  $m \geq O( k \log (Ln))$.
We verify  the (approximately) deep generative prior holds if the target signals have low intrinsic dimensions by constructing a ReLU network with  properly chosen depth and width.
Extensive numerical simulations and comparisons with state-of-the-art methods  demonstrated
the least square decoder is the robust to noise and sign flips, which verifies our theory. We only consider the analysis of the  least squares decoders, we will leave the analysis of   the regularized least squares decoder  in the future work.

\section*{Acknowledgements}
Y. Jiao is supported in part
by the National Science Foundation of China under Grant 11871474 and by the research fund of KLATASDSMOE of China.
X. Lu is partially supported by the National Key Research and Development Program of China (No.2018YFC1314600), the National Science Foundation of China (No. 11871385), and the Natural Science Foundation of Hubei Province (No. 2019CFA007).
\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
App Growth Programme
App Launch Strategy
Paid User Acquisition
2023 iOS and Android Updates: What to Expect and Keep an Eye On
It's fair to say that 2022 saw a lot of changes, and most of the world found it somewhat difficult to keep up. But we're not here to talk politics or economics: we're talking operating systems. There was the launch of Apple's iOS16 and Google's Android 13; a renewed focus on privacy and security; and something for the UA team in terms of new App Store advertising options. We're here to give you the details on iOS and Android updates so that you can go into 2023 fully optimised for the year ahead.
Android updates to keep an eye on in 2023
Privacy labels
TL;DR: Providing compelling reasons for why your app collects data can be the difference between a user downloading your app or not.
Since July, apps on the Google Play Store are required to ensure more transparency on data tracking. App developers are now required to disclose:
the data their app collects,
if and how it's shared with third parties,
the app's security practices and more.
The focus is on whether data collection follows best practices, as well as the 'why' data is being collected – whether for functionality, personalisation, etc. – to give users the information they need before they download the app.
Source blog.google
TL;DR: It's worth spelling out to users what value your Push notifications will provide if you want to maximise opt-ins.
In another Apple-like move, Android 13 requires apps to ask for Push notification permission rather than granting permission by default. This means that if a user doesn't opt-in, you can't send anything. (For apps that have already been downloaded, users will be prompted to decide whether to allow notifications when they next use the app.)
Developers who target Android 13 have control over when the permission dialogue appears. It could be after a specific user action that demonstrates your app's value, or after enough time has elapsed for them to appreciate why notifications would enrich their experience. And while the default permission dialogue is simply "allow" or "don't allow", you can build a review prompt proactively asking for permission and include a compelling argument as to why. Following best practices for requesting push notification permission can improve opt-in rates by up to 50%, so it's worth investing time in your push strategy.
Android updates on app standby bucket rules
TL;DR: Android's new app standby bucket rules requires a different approach to key metrics.
App standby buckets is a feature that helps manage battery usage by categorizing apps into different "buckets" based on how frequently they are used. Apps in the "active" bucket are allowed to run normally, while apps in the "working set" and "frequent" buckets may have certain restrictions placed on them, such as reduced access to network resources, to help conserve battery life.
Android updates have introduced new rules that would relegate an app to that last restricted bucket. These are:
no app interaction in eight days
app solicits too many broadcasts in a 24-hour time interval
app drains too much device's battery life within 24 hours via several background app activities
As a result, it's worth renewing focus on interactions as a key metric to monitor, given that low interaction will mean a downward move through those all-important app buckets.
iOS updates to look out for in 2023
Better SKAdNetwork conversion tracking
TL;DR: Advertisers will now be able to better understand user engagement over time.
Apple's SKAdNetwork can now return up to three conversion values for each download, each with a different attribution window (0-2 days, 3-7 days, and 8-35 days). This means multiple in-app purchases can be tracked, as opposed to just the one as previously set. It's going to make it significantly easier to evaluate in-app behaviour and calculate ROAS, tracking the full impact of UA campaigns.
The takeaway: There are certain apps that can benefit from showing off Apple's new features.
We spoke about Apple's love of highlighting new features by promotion in the app store in our article, How to get featured in the App Store. The launch of iOS16 has provided that opportunity for apps that can show off Apple's new features and capabilities, including:
Cross-generational apps – the new Parental Control setting in Family Sharing provides new functionality to apps used by the whole family.
Content apps – Apple's new SharePlay feature allows users to share media from within FaceTime.
News, sports, delivery apps – "Live Activities" notifications streamline any thread of notifications, such as game updates, into one dynamically updated notification.
Widget-friendly apps – the complete redesign of the lock screen allows for prominent placement of widgets, enriching the user experience even when not using an app.
Source: apple.com
TL;DR: Focus your efforts where there is still room for improvement, rather than spending more time in areas you're already excelling.
App Store Connect now allows you to compare your app's performance to industry standards, across acquisition, usage and monetisation metrics. Apple is keen to point out that this is done in a privacy-compliant way, which means you're plotted against apps that are similar to your own in terms of Category and Monetisation, with your own performance illustrated along a continuum. While not providing enough information to replicate what leading apps are doing right, it should provide additional insight into where resources should be directed, into areas where there is still ground to make up.
New ad placements
TL;DR: Competitors will start appearing on your product page with new app page ad placements, just in time for the holidays.
Apple is now offering ad spots on the Today page and on individual app product pages, in addition to the existing Search tab and results placements. The former isn't entirely surprising, but it's the individual app pages that may set developers' hearts racing. It won't be possible to bid directly on specific competitors, but ads will be relevant to the app in question, indicating that you can bid on apps similar to yours or within the same category.
In conclusion, the updates to iOS and Android operating systems in 2022 bring a range of changes and new features that will impact app marketers and users in 2023. With 2023's Google I/O likely to take place in May and Apple's WWDC at the beginning of June, there are five glorious months of the new year to make sure you're up-to-date – before the next round of iOS and Android innovations hit. But get this done now, and you'll be able to kick back and relax (or at least take a quick breath) in the new year.
By Agata Brown | 22nd December 2022 | App Growth, Mobile trends, WWDC
About the Author: Agata Brown
Agata is the Marketing Manager at Yodel Mobile, a leading mobile app marketing company. Assisting the agency growth efforts, Agata regularly shares insights on the latest app marketing strategies, promoting sustainable and long-term growth.
App Marketing Trends for 2023: The Yodel Crystal Ball
The What, Whys and Hows of Personalised App Marketing
26 Tips to Get your App Ready for the Holiday Season
Smokehouse Yard, 44-46 St John Street, London, EC1M 4DF
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text:
Q: regsvr32 error: the module was loaded but the entry point was not found People here use an old compiled VB program, the last version is 16 years old. It works on some machines but not on others, all with the same 1709 version and updates of Windows 10.
Where it does not work a message says either that msdatlst.ocx or msdatgrd.ocx is not registered. Then when I try to register with regsvr32 'MSDATGRD.OCX' there is a message "the module was loaded but the entry-point was not found". The same command is successful in the machines where the program works.
I did read some questions here but none of the answers solved it for me. Any ideas?

A: Thanks to @Mark's comment and all the others. The problem was indeed what version of regsvr32 was executed.
In the machines where it was possible to register the ocx the default version is at \windows\syswow64 and in those where that was not possible at \windows\system32. Then I did run the syswow64 one and the registration succeeded and the program now works.

A: Check if the .def file is specified, otherwise, the required DllRegisterServer function will not be exported.

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BEN VOLIN | ON FOOTBALL
Projecting the Patriots roster: Are there actually 53 players worthy of spots?
By Ben Volin Globe Staff,August 6, 2018, 11:42 a.m.
Bill Belichick (center) will have some tough decisions when picking the 53-man roster.John Tlumacki/Globe Staff/Globe Staff
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Projecting a 53-man roster for the Patriots is not easy. I can usually whittle the 90-man training camp roster down to about 60 spots, but the last six or seven cuts are tough. The Patriots led the league last year with four of their cut players claimed by other teams after roster cutdowns.
This year's projection is equally difficult, but for a different reason: Are there actually 53 players worthy of a roster spot?
The competition is wide open, and getting up to 53 players wasn't easy.
I count 39 players who are guaranteed of a roster spot, assuming they can make it through training camp in good health. This includes Cordarrelle Patterson, who will make the team more for his special teams prowess (kickoff returner and punt team gunner); core special teamers such as Brandon Bolden and Brandon King; and of course, three spots for the kicker, punter, and long snapper. It doesn't include Julian Edelman, who will be suspended for the first four games.
But unlike in previous years, there isn't much separating players 40 to 70 on the roster. Several positions are wide open, such as receiver, tight end, and linebacker.
And injuries will play a significant factor.
Will Nate Ebner, who tore an ACL last November and is one of the 39 roster locks, start the season on the physically unable to perform list, knocking him out for the first six-plus games? What about Jonathan Jones, who hurt his ankle in the playoffs, and Cyrus Jones, a 2016 second-round pick who tore an ACL last preseason? What's going on with Kenny Britt? How badly are Sony Michel and Marcus Cannon hurt?
The answers to these questions can send the Patriots roster in 100 different ways.
Let's take our first attempt at projecting a 53-man roster in advance of the Patriots' preseason opener Thursday against Washington:
Offense — 23 players
■ Quarterback (2): Tom Brady, Brian Hoyer.
Practice squad: Danny Etling.
Analysis: The only real question is what to do with Etling, the seventh-round pick out of LSU. The Patriots kept three quarterbacks in 2016, but that was because of Brady's suspension. They kept two last year, and probably only need two this year. Etling has a strong arm, good mechanics, and experience in a pro-style offense, but the Patriots don't need to use a roster spot on him.
James White (left) is entering his fifth NFL season.John Tlumacki/Globe Staff
■ Running back (6): James White, Rex Burkhead, Sony Michel, Mike Gillislee, Brandon Bolden, James Develin.
Analysis: The first three are obvious, though the fact that Michel reportedly had his knee drained is concerning. Develin is in, and Bolden is a lock as a core special-teamer. The only real question is Gillislee. My first inclination was to leave him off, given his struggles last year, his lack of special teams contributions, the depth at running back, and the fact that the Patriots can save $2.4 million with no dead salary-cap money.
But Michel is already hurt, and Burkhead was banged up a lot last year, missing seven games and playing just 208 snaps total. So Gillislee makes the roster for now as a first-down and short-yardage back, beating out newcomer Jeremy Hill and rookie Ralph Webb.
■ Tight end (3): Rob Gronkowski, Dwayne Allen, Jacob Hollister.
Practice squad: Ryan Izzo.
Analysis: Other than Gronkowski, this position is wide open. Coach Bill Belichick raves about Allen's approach and loves his blocking ability, and Allen appears to be back for another season as the No. 2 tight end, even though the Patriots can save $5 million cash and cap with him. Hollister has had a nice start to camp and might be making that second-year jump Belichick talks so much about.
But the one to watch is Izzo, the seventh-round pick out of Florida State. He has made several nice catches in training camp, and comes from a pro-style system in college that required him to be an in-line blocker. If Izzo keeps up his solid camp, he could make Allen or Hollister expendable.
Will Tye has been noticeable in the passing game so far, but he's the same type of player as Hollister, who looks like he is beating out Tye so far.
■ Wide receiver (4): Chris Hogan, Phillip Dorsett, Eric Decker, Cordarrelle Patterson.
Suspended: Julian Edelman.
Practice squad: Riley McCarron.
Analysis: This position is also wide open, and the depth is not very good beyond Hogan and Dorsett, who has looked much better in his second season in the offense and could be used to replace Brandin Cooks's production. Jordan Matthews already was released. Malcolm Mitchell and his chronically injured knee were finally released Monday. Britt hasn't practiced yet because of injury. Braxton Berrios hasn't taken any reps with the starting offense yet. Devin Lucien and Paul Turner have made a few plays, but they are camp bodies.
Cordarrelle Patterson will probably only be used as a gadget player and occasional receiver.Matthew J. Lee/Globe staff
So for now we're going with four receivers until Edelman returns, and really it's only three, because Patterson probably will be used only as a gadget player and occasional receiver.
Decker gets one of the spots based on his ability to play the slot or outside, his size (6 feet 3 inches, 214 pounds), and his veteran savvy. But he reportedly received only $75,000 guaranteed on his one-year deal, meaning he absolutely is fighting for a job. If Britt can't get on the field soon, his time with the Patriots might end swiftly.
McCarron is intriguing enough of a slot receiver to keep in Foxborough, but I'm not sure if he's worthy of a roster spot.
■ Offensive line (8): Trent Brown, Joe Thuney, David Andrews, Shaq Mason, Marcus Cannon, Isaiah Wynn, LaAdrian Waddle, Luke Bowanko.
Practice squad: James Ferentz, Cole Croston.
Analysis: The top seven spots are easy to account for, though I'm not quite sure how all the pieces will fit, and Cannon's injury could complicate matters. Will Thuney hold onto the left guard spot? Is Brown the starting left tackle? Is Wynn, the 23rd overall pick, just going to be a super sub this year?
Waddle was brought back on a one-year deal, and I'm counting him as a roster lock as the backup tackle. For the eighth spot, I'm going with Bowanko, a fifth-year veteran who has started NFL games at center and left guard. With Wynn able to play left tackle or left guard, and Thuney and Bowanko able to fill in at left guard or center, this group has a ton of versatility.
Ted Karras is also in the mix for that guard/center spot, but the fact that the Patriots brought in so much competition at that spot doesn't look good for him.
Defense — 24 players
■ Defensive line (9): Trey Flowers, Deatrich Wise, Adrian Clayborn, Derek Rivers, Lawrence Guy, Danny Shelton, Malcom Brown, Vincent Valentine, Adam Butler.
Practice squad: Eric Lee, John Atkins.
Analysis: I count six locks in Flowers, Wise, Clayborn, Rivers, Guy, and Shelton. Brown is an interesting case, a 2015 first-round pick who hasn't quite lived up to the billing. The Patriots declined his fifth-year option, and now he's in the final year of his contract.
The Patriots also acquired Shelton in a trade. But he looks to be Alan Branch's replacement, and I believe that Brown, who played 54 percent of snaps last year, still should have a significant role in the rotation, particularly on rushing downs.
The returns of Rivers and Dont'a Hightower and addition of Clayborn means the Patriots don't have a need for Lee anymore, but they can try to stash him on the practice squad. Butler was solid as a rookie and can play end or tackle, and Valentine, great as a rookie in 2016 and injured for 2017, offers good depth up the middle and should regain his roster spot.
Dont'a Hightower (54) is returning from injury, and Danny Shelton (71) is in his first season in New England.Matthew J. Lee/Globe staff
■ Linebackers (6): Dont'a Hightower, Kyle Van Noy, Marquis Flowers, Ja'Whaun Bentley, Nicholas Grigsby, Harvey Langi.
Practice squad: Christian Sam.
Analysis: We'll quote Dennis Eckersley to describe this group: "Yuck." Hightower is back, though he's more of a defensive end and pass rusher at this point. Van Noy will be the middle linebacker and play communicator. But it's basically a toss-up after those two.
One name you'll notice I don't include is Elandon Roberts, the third-year linebacker who played 676 snaps last year including the postseason (52.9 percent). I have Roberts as one of my surprise cuts because (A) he's terrible in pass coverage, (B) he consistently shoots the wrong gap or overruns a play, and (C) when given the chance to play significant snaps last year, he struggled.
The Patriots also re-signed Flowers and drafted two linebackers in Bentley and Sam, which tells me they are trying to upgrade the position. Grigsby and Langi hit hard and can take Roberts's place on special teams.
But this position is wide open, and I wouldn't be surprised to see the Patriots try to upgrade it around roster cutdowns.
■ Defensive back (9): Stephon Gilmore, Eric Rowe, Duke Dawson, Jonathan Jones, J.C. Jackson, Devin McCourty, Patrick Chung, Duron Harmon, Jordan Richards.
PUP: Cyrus Jones.
Practice squad: Ryan Lewis, Keion Crossen.
Analysis: One prominent name that's missing: Jason McCourty. Even though McCourty is listed as a starter on the initial depth chart, the 10-year veteran and first-time Patriot is my No. 1 surprise cut candidate.
Despite what the depth chart says, Gilmore and Rowe have been working almost exclusively as the top outside cornerbacks. Dawson, the second-round pick, is the slot cornerback. Jonathan Jones played 35 percent of snaps last year and could see an uptick after Malcolm Butler's departure.
Jason McCourty (30) has 16 career interceptions.Charles Krupa/AP
That leaves McCourty battling with a host of youngsters for the fifth cornerback spot. It's not that McCourty can't still play, he just has a lot working against him — a $3 million salary with no dead money and age. (He turns 31 next week.)
Young guys Jackson, Lewis, and Crossen are all making league minimum, have fresher legs, and offer Butler-like upside. Jackson has had a really nice start to camp and has NFL talent, but went undrafted because of off-field issues at Florida and Maryland.
It's worth keeping McCourty until the end of camp for depth purposes. Jonathan Jones starts the season on PUP with an ankle injury, and his availability is unclear. But if he can come back and everyone else stays healthy, I have McCourty getting squeezed out.
I also think the Patriots should keep Cyrus Jones as long as possible and give him another chance. He was talented enough to be a second-round pick in 2016, and while his rookie year was disappointing, his chance at redemption was ruined last year by a torn ACL in August. They can start him on PUP for the first six weeks and see how he develops before deciding on his future.
At safety, the only question is Richards, and I have him cracking the 53 one last time because of his special teams contributions.
Special teams – 6 players
■ Core special teamers (3): Matthew Slater, Nate Ebner, Brandon King.
■ Specialists (3): K Stephen Gostkowski, P Corey Bojorquez, LS Joe Cardona.
Analysis: If Ebner remains on PUP for the start of the season, it opens a roster spot elsewhere — another receiver or linebacker, for example. The other competition is at punter, where we have the rookie, Bojorquez, unseating the incumbent, Ryan Allen, who is in the last year of his contract.
Bojorquez has the stronger leg, and booms his punts higher and farther than Allen, with more hang time. Bojorquez is also a lot cheaper, set to make $465,000 compared with $1.5 million for Allen. Of course, there's more to punting than just bombing away, especially for the Patriots, who often punt around midfield and require more precision than distance.
So while Borjorquez may have a leg up, let's see how he does with his situational punting in preseason games.
Ben Volin can be reached at ben.volin@globe.com. Follow him on Twitter @BenVolin.


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Radazz Hearns' mother wants release of 14-year-old's record probed
Updated: Jan. 17, 2019, 2:27 a.m. | Published: Sep. 09, 2015, 4:07 p.m.
By Keith Brown | For NJ.com
TRENTON - Angered by the release of her son's juvenile police record to a local newspaper, the mother of Radazz Hearns on Wednesday called for an investigation into how Hearns' juvenile history was released by a Mercer County police department.
"Just because you are the law doesn't mean you're above the law,'' Jackson said.
Flanked by members of the United Mercer Interfaith Organization and the state chapter of the American Civil Liberties Union, the group announced that letters to seven different police agencies were being delivered Wednesday, demanding that an investigation into the "illegal and immoral'' release of Hearns' juvenile record.
"The law is clear,'' said Alexander Shalom, of the New Jersey ACLU. "It's a punishable, criminal offense for law enforcement officers to disclose that information.''
Last month, the Hopewell Township Police department released to a Trenton newspaper copies of Hearns' juvenile record showing that the teen had been arrested three times on simple assault and minor drug possession charges prior to his encounter with police on Aug. 7, when he was shot seven times by officers investigating a report of gunfire in the are of Louise Lane near Calhoun Street.
RELATED: Trenton 14-year-old shot by police among 5 injured in gunfire
Hearns was 14 years old when he was wounded last month, and has since turned 15.
The shooting is under investigation by the state Office of the Attorney General because it involved a State Police trooper as well as a Mercer County sheriff's officer, who were in the city as part of to supplement the Trenton Police Department.
The two officers and another State Police trooper saw Hearns and two other males as they were walking along a street. The three officers got out of their vehicle and confronted the trio, and Hearns fled around a townhouse.
The State Police trooper and Mercer sheriff's officer opened fire as they gave chase, according to the Attorney General's Office. Witnesses told authorities the teen reached for his waistband before the officers opened fire, the Attorney General's office has said.
Hearns spent a week at Capital Health Regional Medical Center in Trenton before he was released. He was charged with aggravated assault on a police officer and weapons offenses.
Hearns' lawyer, Samuel A. Anyan Jr., has said Hearns was unarmed when officers opened fire. The Attorney General's office has said Hearns had a .22 caliber handgun that he pointed at police officers before he was shot.
The Attorney General's Office has not named the two officers involved in the shooting.
5 unanswered questions in police shooting of 14-year-old
Rev. Lukata Mjumbe, of the United Mercer Interfaith Organization, said the letters were being hand-delivered Wednesday to the Attorney General's Office, the U.S. Attorney's Office, the State Police, Mercer County Prosecutor's Office, the Mercer County Sheriff's Office, the Trenton Police and the Hopewell Police Department.
Hopewell Township police arrested Hearns three times in town, where he attended an alternative high school. The newspaper also published three separate mug shots, which the paper said were received by an anonymous police source.
"We believe that a wrong has been done," Mjumbe said. "We believe the law has been broken. We are calling for an internal affairs investigation of all the departments that potentially may have provided information."
Hopewell Township Police Chief Lance Maloney declined to comment Wednesday. Acting Mercer County Prosecutor Angelo Onofri acknowledged receiving the letter on Wednesday, but he declined to comment on whether his office is investigating.
"It's our office policy not to comment on any investigation or potential investigation,'' Onofri said.
Keith Brown may be reached at kbrown@njadvancemedia.com. Follow him on Twitter @KBrownTrenton. Find The Times of Trenton on Facebook.


There was no outlook written for Cam Phillips in 2018. Check out the latest news below for more on his current fantasy value.
Phillips was signed to a reserve futures contract Monday, Chris Brown of the Bills' official site reports.
Phillips, an undrafted free agent, bounced around with the Bills in 2018 between the practice squad, the active roster (he played in two games) and being waived from the team, but the Virginia Tech product will be given a shot to make the team this summer. The Bills don't have many proven options at wide receiver, but they have a large amount of cap money to spend and will address the position with a lot more than Phillips this offseason.
How do Cam Phillips' 2018 advanced stats compare to other wide receivers?
See where Cam Phillips lined up on the field and how he performed at each spot.
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Phillips was waived by the Bills on Monday, Mike Rodak of ESPN.com reports.
Phillips (coach's decision) is inactive for Sunday's game versus the Bears.
Phillips was signed off the Bills' practice squad and promoted to the active roster Thursday, Chris Brown of the Bills' official site reports.
Phillips caught one pass for five yards during Friday's preseason game against the Browns.
Phillips exited Friday's practice with a groin injury, Mike Rodak of ESPN.com reports.


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