# Tag Info

12

As far as I know (and can interpret your question) no such result is known. There are two reasons: 1) Generally unique games hardness results (as well as NP hardness result) do not yield "instance based" hardness. That is, the UG-hardness result have the following flavor - "if the unique games conjecture is true then for problem $P$ no algorithm can get a ...

11

I think the following should answer your questions, even though it's not exactly in the same order. The original formulation of the small set expansion conjecture states that, analogously to the Unique Games Conjecture, for every $\epsilon >0$ there exists $\delta>0$ so that it is NP-hard to determine whether in a graph $G$ it's the "YES" case where ...

10

Let me see if I can clarify this, on a high level. Assume the UG instance is a bipartite graph $G = (V \cup W, E)$, bijections $\{\pi_e\}_{e \in E}$, where $\pi_e\colon \Sigma \to \Sigma$, and $|\Sigma| = m$. You want to construct a new graph $H$ so that if the UG instance is $1-\delta$ satisfiable, then $H$ has a large cut, and if the UG instance is not ...

8

The powering step fails. After the powering, each vertex is labeled with a neighborhood of the original graph. each edge checks that its endpoints agree on the intersection of their neighborhoods, and that this labeling satisfies the edges in this intersection. However, the edge cannot check anything about part of the labeling that lies outside the ...

7

Looking at the Goemans–Williamson algorithm in the SOS framework yields no technical advantages: it is exactly the same algorithm and the same ideas are used in the analysis. The only advantages in doing so are: Arguably the algorithm seems less "magical" in that viewpoint, though of course that's a matter of taste. It's a good basic case to get intuition ...

4

"So, applying MIS on $g$" To apply the Majority is Stablest theorem, you need to apply it to a non-negative parameter $\rho'\in[0,1)$ (read the statement of the theorem). Since in Proposition 7.3 the parameter $\rho\in(-1,0]$ is non-positive, this means you here apply it to $\rho' \stackrel{\rm def}{=} -\rho\in[0,1)$, giving \mathbb{S}_{-\rho}(g) \leq 1-\...

3

The degree $r$ pseudo-expectation operator operates on polynomials of at most degree $r$. Since the pseudo-expectation operator is positive semidefinite, we're guaranteed that the square of a polynomial (or sum of squares of polynomials) always has nonnegative pseudo-expectation. Also, if we have a system of polynomial equations $\{p_i = 0\}$ of degree at ...

2

Yes, that's exactly what it is: optimizing over higher degree pseudoexpectations. The results I think you refer to prove that the hypercontractive inequality holds when $f$ is replaced by a degree $4$ pseudo distribution over $\mathbb{R}^{\{-1,1\}^n}$, which is stronger than the classical theorem (see Lemma 2.10 in Boaz's notes). Specifically, the stronger ...

1

Suppose that $A$ is an $m\times n$ matrix, and let $\mathcal{A}$ be the corresponding linear operator whose matrix (w.r.t. the standard basis) is $A$. Let $W\subseteq \mathbb{R}^m$ be the span of the columns of $A$, or, equivalently, the range of $\cal A$. I am going to treat $\mathcal{A}$ as an operator from $\mathbb{R}^n$ to $W$. As long as $A$ has full ...

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