# Tag Info

19

The question seems somewhat under-specified in the sense that it did not specify the desired error probability of the procedure. Assuming one means constant error probability, then the above is indeed the best I know. For a detailed discussion see Sec 2.5.2.4 in my book "The Foundations of Cryptography - Volume 1" available at http://www.wisdom.weizmann.ac....

14

In his CCC'17 paper [1], Avishay Tal improved the bound to $$\left(\log\frac{m}{\varepsilon}\right)^{O(d)}\,. \tag{1}$$ You may want to check p.15:4 for a discussion. It also refers to (see Footnote 30 to a paper of Harsha and Srinivasan, which improves on (1)) and answers Tal's conjecture: $k$-wise independent, for $$k = \left(\log m \right)^{O(d)}\cdot\... 11 LMN theorem shows that if f is a boolean function(f:\{-1,1\}^n \rightarrow \{-1,1\}) computable by an \text{AC}^0 circuit of size M,$$\sum_{S:|S|> k} \hat f(S)^{2} \leq 2^{-\Omega(k/(\log M)^{d-1})}$$\Rightarrow \hat f([n])^{2} \leq 2^{-\Omega(n/(\log M)^{d-1})} \Rightarrow |\hat f([n])| \leq 2^{-\Omega(n/(\log M)^{d-1})} |\hat f([n])| is ... 9 This paper came up on the arXiv today and it improves on the upper bound on bs(f) in terms of s(f). They prove the following bound:$$ bs(f) \leq 2^{s(f)-1}s(f). $$This along with the connection that Marcos mentioned in his comment should give better bounds than previously known. 8 I don't think parities are the only orthonormal Boolean basis, for example Paley basis provides a Boolean orthonormal basis in some cases. It's natural to ask if such bases, interpreted as Boolean functions, have any interesting applications. 8 Below I show how to explicity construct an average-case \varepsilon-biased space on n bits of size O(1/\varepsilon). In contrast, the best worst-case \varepsilon-biased spaces on n bits have size \tilde\Theta(n/\varepsilon^2). So you save a factor of about n/\varepsilon by going from worst-case to average-case. Unfortunately, as you will see,... 6 Yes. Let f : \{-1,1\}^n \to \mathbb{R}, and let F : [-1,1]^n \to \mathbb{R} be its multilinear extension. If f is monotone, then so is F. proof: Fix a variable index i; we'll show that \frac{\partial F}{\partial x_i} \ge 0 at all x \in [-1,1]^n. If this holds for all i, we're done. Since F is multilinear, we can write this partial ... 6 The representation theory of the symmetric group plays a key role in the Geometric Complexity Theory approach to lower bounds on the determinant or on matrix multiplication. Bürgisser and Ikenmeyer prove a lower bound on the border-rank of matrix multiplication using the representation theory of S_n. For how the representation theory of S_n relates to ... 6 Here might be another take on this question. Assuming the pseudo Boolean function is k-bounded, the Walsh polynomial representation of the function can be decomposed into k subfunctions. All of the linear terms are collected into one subfunction, the all of the pairwise interactions into one subfunction, then the 3-way interactions, etc. Each one of ... 5 Here's one way of attacking the problem. Any boolean function f: [2]^n \rightarrow [2] can be written as $$f(\alpha) = \sum_{\beta \in [2]^n} \chi_{\beta}\delta_{\beta} (\alpha)$$ where \chi_{\beta} \in [2] and \delta_{\beta} is defined as \delta_{\beta} (\alpha) = \begin{cases} 1 & \text{if } \alpha = \... 5 From D.W.'s comment, we see that your problem is equivalent to the same problem with s=0 (and with d only increased by 1). The resulting problem is precisely the Linear Code Equivalence Problem over \mathbb{F}_2, just phrased slightly differently. (Usually CodeEq is phrased as "you have two d \times n generator matrices of linear codes, is there a ... 5 A quantum Fourier transform is a unitary operation, so the number of basis states of the input and output must be the same. The number of basis states before the Fourier transform is 120, the number of group elements. The number of basis states after the Fourier transform is 120, in this case broken up according to the identity$$ 120=1^2+1^2+4^2+4^2+5^2+...

5

This is Corollary 3.22 of Analysis of Boolean Functions, by Ryan O'Donnell (2014). You may want to consult the proof in the book, or look directly at the online version (which has a different numbering: this corresponds to Corollary 22 there). Note that you can obtain a version of the book following this link.

5

Andrew(the asker) and I had discussed this over email, and we have shown the conjecture is false. The polytope is not integral for Abelian groups, not even for cyclic groups. On the positive side. Theorem: For cyclic groups with order $p^kq$, where $p$ and $q$ are primes and $k\in \mathbb{N}$, the incidence matrix of elements and subgroups is totally ...

4

Discarding in this context can be carried out by performing a Partial Trace. The system you're interested in is that of the first register. If you want to know only its state, as opposed to the combined state of the two registers, then you take the density matrix of the whole system and trace out the second register. The resultant state will, in this case, ...

4

We were eventually able to analyze hypercontractive properties of $R_{p_1, p_2}$ (http://arxiv.org/abs/1404.1191), building off of the main Fourier analysis of $R_{p,0}$ by Ahlberg, Broman, Griffiths and Morris (http://arxiv.org/abs/1108.0310). To summarize, the effect of a biased operator $R_{p,0}$ on a function $f$ can be analyzed as a symmetric noise ...

3

The question is about the spectral norm of the symmetric function $h: \{0,1\}^n \to [-1,+1]$ defined as $h(x) = f(1-2|x|/d)$. (I have assumed range $[-1,+1]$ instead of $[0,1]$ which is wlog by appropriate rescaling.) Ada, Fawzi and H. Hatami show that for any boolean $g: \{0,1\}^n \to \{-1, +1\}$, $$\log \|\hat{g}\|_1 = \Theta\left(r(g) \log\left(\frac{n}{... 3 I have figured out that the answer to this question is yes. The proof goes via sandwiching polynomials. It's a simple modification of a proof in [GMRTV12] \S 4. (Instead of keeping track of \mathrm{L}_1, we keep track of degree.) 2 Just posting an answer in order to close the question. See the helpful comments to the question which contain all the info; Yes it is vacuous, but if you use the stronger bound you get something nontrivial for large \epsilon, of course for fixed k you also have interesting things, it's just the estimates in the book lose too much. Better bounds are ... 2 You have the lower bound of \mathbf{Inf}[f] \geq \operatorname{Var}[ f ] for f\colon \{-1,1\}^n\to\mathbb{R} (Poincaré Inequality), so that for an (almost) balanced f\colon \{-1,1\}^n\to\{-1,1\} one must have$$ \mathbf{Inf}[f] \gtrsim 1 \tag{1} $$(say, 1-\epsilon, depending on the near-unbalanced assumption). Further, for a dictator function, ... 2 You shouldn't have a square root. Namely, for every \delta-biased distribution Z (using your notation), we have$$ \delta^2+2^{-n} \geq \lVert Z\rVert^2_2 \geq \frac{1}{\lvert\operatorname{supp} Z\rvert}\tag{1} $$since the squared \ell_2 norm of a distribution over support of size N is at least that of the uniform distribution on N elements, which ... 2 It is a standard fact that if f:\{-,1,1\}^n \to \{-1,1\} is a function of Fourier degree d, then its Fourier coefficients are multiples of 2^{-d+1}. In particular, every non-zero coefficient must be at least 2^{-d+1} in absolute value. Therefore, by Parseval, there are at most 2^{2(d-1)} non-zero coefficients, and so the spectral norm of f is at ... 2 As pointed out in the comments if u\in \{\pm 1\} then x=x(u) \in \{0,1\} where$$x(u)=\frac{1-u}{2},$$with x(-1)=1, and x(1)=0. This will then yield$$f(x)=2^{n-1}f(0)-\frac{1}{2} \sum_{S \in \{1,\ldots,n\}} \hat{f}(S) \prod_{i \in S} x_i,$$and if we denote the \{0,1\} valued version of f as \tilde{f},$$\frac{1-\tilde{f}(x)}{2}=2^{n-1}f(0)-\...

1

The Fourier transform is a linear operation. In particular, for $f:\{-1,1\}\to\mathbb{R}$ and $S\subseteq[n]$, the Fourier coefficient $\hat f(S)$ is a linear functional of $f$. If $\hat f(S)\neq 0$, its magnitude can be made arbitrarily large or small by multiplying $f$ by an appropriate scalar -- without affecting $||f||_0$. So the answer to your question ...

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