# Tag Info

## Hot answers tagged lower-bounds

32

Here's a lower bound from sorting. Given an input set $S$ of length $n$ to be sorted, create an input to your running median problem consisting of $n-1$ copies of a number smaller than the minimum of $S$, then $S$ itself, then $n-1$ copies of a number larger than the maximum of $S$, and set $k=2n-1$. The running medians of this input are the same as the ...

30

Let $f\colon \{0,1\}^n \to \{0,1\}$ be a boolean function. If it has a polynomial representation $P$ then it has a multilinear polynomial representation $Q$ of degree $\deg Q \leq \deg P$: just replace any power $x_i^k$, where $k \geq 2$, by $x_i$. So we can restrict our attention to multilinear polynomials. Claim: The polynomials $\{ \prod_{i \in S} x_i : ... 26 Let$p$be a polynomial such that for all$x\in \{0,1\}^n$,$p(x) = \sf{OR}(x)$. Consider the symmetrization of the polynomial$p$: $$q(k) = \frac{1}{\binom{n}{k}} \sum_{x: |x| = k} p(x).$$ Note that, since the OR function is a symmetric boolean function, we have that for$k = 1, 2, \ldots, n$,$q(k) = 1$, and$q(0) = 0$. Since$q-1$is a non-zero ... 21 Markov proved that any function of$n$inputs can be computed with only$\lceil \log (n+1)\rceil$negations. An efficient constructive version was described by Fisher. See also an exposition of the result from the GLL blog. More precisely: Theorem: Suppose$f : \{0,1\}^n \to \{0,1\}^m$is computed by a circuit$C$with$g$gates, then it is also computed by ... 19 A family of bitvectors is the class of solutions to a 2-SAT problem if and only if it has the median property: if you apply the bitwise majority function to any three solutions you get another solution. See e.g. https://en.wikipedia.org/wiki/Median_graph#2-satisfiability and its references. So if you can find three solutions for which this is not true, then ... 18 According to the paper A$5n − o(n)$Lower Bound on the Circuit Size over$U_2$of a Linear Boolean Function by Kulikov, Melanich, and Mihajlin, when$m=o(n)$there are no lower bounds known better than$3n - o(n)$. It also outlines a method for obtaining functions for which a$4n - o(n)$lower bound holds, when$m=n$, based on a result of Lamagne and ... 17 This question has been completely resolved (up to constant factors) by a recent result of Benjamin Rossman (http://eccc.hpi-web.de/report/2013/169/). As Kaveh points out above, a depth$d$, size$S$, circuit can be converted to a depth$d$, size$S^d$formula. Rossman shows that this is essentially tight. For any depth$d$, he exhibits a function that can ... 16 On techniques for proving poly-log circuit-depth lower bounds, all current approaches work under restricted settings. Like, in the work leading to GCT that you mention, the lower bound applies to a restricted PRAM model without bit operations. Under another restriction, which is the monotone restriction for monotone boolean functions, there is a Fourier-... 15 The lower bound is correct (2) - you can not do this better than$\Omega(n^2 \log n)$and (1) is of course wrong. Lets us first define what is a sorted matrix - it is a matrix where the elements in each row and column are sorted in increasing order. It is now easy to verify that each diagonal might contains elements that are in any arbitrary order - you ... 15 There are "natural" problems hard for Ackermannian time; in fact there is a growing body of literature on the subject. A place to start is the survey in Section 6 of Complexity Hierarchies Beyond Elementary. What is even more interesting is that there are complete problems for Ackermannian time, under say primitive-recursive many-one reductions: define for ... 15 Hmm, turns out there's actually an matching upper bound that isn't too hard: To produce the truth table$HALT_n$in a finite amount of time, the only information that is needed is the number of machines of description length at most$n$which halt. This number is not more than$2^{n}$, so it can be represented with about$n$bits. Then we can start all such ... 14 Following the suggestion of Kaveh, I am putting my comment as an (expanded) answer. Concerning$Q1$, a word of caution is in order: even logarithmic depth if far from being understood, not speaking about poly-logarithmic. So, in the non-monotone world, the real problem is much less ambitious: Beating Log-depth Problem: Prove a super-linear(!) lower bound ... 13 According to Theorem 3.1 in Alexis Maciel and Denis Therien Threshold Circuits of Small Majority-Depth there is indeed a depth-3 circuit for computing the addition of two numbers. The precise bound is$\Delta_2 \cdot \mathsf{NC}^0_1$where$\Delta_2 = \Sigma_2 \cap \Pi_2$are problems which have depth-2$\mathsf{AC}^0$circuits with both$\vee,\wedge$... 13 One approach on this question, related to complexity-theoretic questions, is due to Shub and Smale  who proved that if$n!$is ultimately hard to compute, then$\mathsf{VP}\neq\mathsf{VNP}$over some field. Their model of computation is the straight-line programs: The goal is to compute$n!$from the constant$1$, using only additions, subtractions and ... 13 The paper Counting Quantifiers, Successor Relations and Logarithmic Space, by Kousha Etessami proves that the problem$\mathbf{ORD}$(which is essentially checking if a vertex$s$precedes a vertex$t$in an outdegree one graph$G$, that is promised to be a path) is$\mathsf{L}$-hard under quantifier free projections. The problem$\mathbf{ORD}$can be seen ... 12 The bounds... We have in fact$NFA(L) \ge Cov(M) + Cov(N)$, see Theorem 4 in (Gruber & Holzer 2006). For an upper bound, we have$2^{Cov(M)+Cov(N)} \ge DFA(L) \ge NFA(L)$, see Theorem 11 in the same paper. ...cannot be substantially improved There can be a subexponential gap between$Cov(M)+Cov(N)$and$NFA(L)$. The following example, and the proof ... 11 It is very possible that the determinant is, in a way, harder than the permanent. They are both polynomials, the Waring Rank(sums of n powers of linear forms) of the permanent is roughly 4^n, Chow Rank(sums of products of linear forms) is roughly 2^n. Clearly, Waring Rank \leq 2^{n-1} Chow Rank. For the determinant, those numbers are just lower bounds. On ... 11 Permanent is complete for VNP under p-projections over any field not of characteristic 2. This provides a positive answer to your second question. If this reduction were linear, it would give a positive answer to your first question, but I believe that remains open. In more detail: there is some polynomial$q(n)$such that$det_n(X)$is a projection of$...

11

The result can be found in Beigel, R., Gasarch, W. I., Gill, J., and Owings, J. Terse, superterse, and verbose sets. Inform. and Comput. 103 (1993), no. 1, 68–85. Their proof in fact shows that your problem cannot be solved with fewer than $\log_2 n$ queries to any set $X$, let alone to the halting problem itself. For some notation, your problem is ...

11

here are new results on this said to be the 1st in ~3 decades and some brief commentary A better-than-3n lower bound for the circuit complexity of an explicit function / Find, Golovnev, Hirsch, Kulikov We consider Boolean circuits over the full binary basis. We prove a $(3+\frac{1}{86})n−o(n)$ lower bound on the size of such a circuit for an explicitly ...

11

It depends what you count as "the GCT program." Consider the specific suggestion (GCT I, GCT II) to use the vanishing/nonvanishing of certain multiplicities in the orbit closures of the determinant and permanent to resolve the strong permanent versus determinant conjecture (i.e., that the permanent is not in the orbit closure of any polynomially larger ...

11

It depends on the precise definition of RAM being used, but (for most reasonable definitions of RAMs) this would also imply that SAT is not solvable in $O(n^{2-e})$ time by multitape TMs, a longstanding open problem. The reason is that there is a very efficient reduction from linear time on RAMs to SAT (in general, nondeterministic quasi linear time on RAMs ...

10

EDIT: The argument that I had answered with was not wrong, but it was a bit misleading, in that it only showed that the upper bound had to be tight for some $n$ (which is actually trivial, since it has to be tight when $n=2$ and the bound is 1). Here is a more precise argument. It shows that if the upper bound of $\log_2 n$ is loose for any particular $n$,...

10

Let $P(x_1,\ldots,x_n)$ be a property on $n$ variables. Suppose that there is a 2CNF formula $\varphi(x_1,\ldots,x_n,y_1,\ldots,y_m)$ such that $$P(x_1,\ldots,x_n) \Leftrightarrow \exists y_1 \cdots \exists y_m \varphi(x_1,\ldots,x_n,y_1,\ldots,y_m).$$ We claim that $\varphi$ is equivalent to a 2CNF formula $\psi$ involving only $x_1,\ldots,x_n$. To prove ...

9

It made my day when my friend James told me that this thread from long ago was rekindled. Thank you for that. Also, I had an urge to share some interesting references that are relevant to L vs Log(DCFL) vs Log(CFL). Have a great day! http://link.springer.com/chapter/10.1007%2F978-3-642-14031-0_35#page-1 http://link.springer.com/chapter/10.1007/3-540-...

9

Alon, Matias, and Szegedy proved that finding the frequency of the most frequent of $n$ items requires $\Omega(n)$ space in the worst case in the streaming model, even if you allow a constant number of passes over the input. Since finding the most frequent item gives you a two-pass algorithm for computing its frequency (find in the first pass, track its ...

9

The tight bound is $n(n^n -(n-1)^n)$, which was given in Removing Bidirectionality from Nondeterministic Finite Automata by Christos Kapoutsis (2005). I am also putting a figure from this paper giving a clear picture: To get more, I think this paper is a good starting point. To check the related recent developments, I can also recommend to check the ...

9

This question has just been resolved! As I mentioned, it was known that $Pn(f) \leq D(f) \leq (Pn(f))^2$, but it was a major open problem to show that either $Pn(f) = \Theta(D(f))$ or that there exists a function for which $Pn(f) = o(D(f))$. A few days ago this was resolved by Mika Göös, Toniann Pitassi, Thomas Watson (http://eccc.hpi-web.de/report/...

9

Edit: This algorithm is now presented here: http://arxiv.org/abs/1406.1717 Yes, to solve this problem it is sufficient to perform the following operations: Sort $n/k$ vectors, each with $k$ elements. Do linear-time post-processing. Very roughly, the idea is this: Consider two adjacent blocks of input, $a$ and $b$, both with $k$ elements; let the elements ...

9

I don't know any nontrivial lower bounds other than the $\Omega(n\log n)$ algebraic-decision-tree lower bound for two dimensions, which also extends to larger $d$. As for upper bounds: In 3d, the whole hull can be found in $O(n\log n)$; in four or higher dimensions that doesn't work because the hull complexity grows exponentially with $d$, but nevertheless ...

Only top voted, non community-wiki answers of a minimum length are eligible