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3. One minor subtlety is that while each sampled [MATH] comes from [MATH] as stated, our algorithm Find-invariant-structure is adaptive. Consequently, the above two items do not imply that the marginal distribution of all queries is coming from [MATH] The cause of non-adaptivity is that in the routine Find-candidate-di... |
A lower bound in terms of surface area The query complexity of our testing algorithm depends on the surface area of the set being tested. In this section, we prove that a polynomial dependence on surface area is necessary for non-adaptive tester, by proving a lower bound for distinguishing 1-juntas and 2-juntas in two ... |
Theorem 42 Any non-adaptive algorithm which can distinguish between a [MATH] -junta with surface area at most [MATH] versus [MATH] -far from a linear [MATH] -junta makes at least [MATH] queries. |
To prove this theorem, as is standard, we will use the Yao’s minimax lemma. More specifically, we will describe a distribution [MATH] |
over 1-juntas with surface area at most [MATH] and a distribution [MATH] over functions that are far from 1-juntas and have surface area [MATH] , such that for any choice of [MATH] with [MATH] if [MATH] and [MATH] then [MATH] |
and [MATH] have almost the same distribution. We begin with the description of [MATH] : let [MATH] be a uniformly random unit vector. Choose [MATH] |
uniformly from [MATH] , and then put them in increasing order. We also set [MATH] and [MATH] Then choose independent random bits [MATH] |
and define [MATH] by [EQUATION] Clearly, such a function [MATH] is a 1-junta, and its surface area is at most [MATH] because the boundary of [MATH] is a collection of at most [MATH] |
lines, and each line has surface area at most [MATH] To describe the construction of [MATH] , we begin with the same collection of random variables as before (i.e., [MATH] |
[MATH] [MATH] ). Let [MATH] be a [MATH] clockwise rotation of [MATH] choose [MATH] independent of the other random variables, and define [MATH] by |
[EQUATION] Note that the boundary of [MATH] is contained in at most [MATH] lines, and so it has surface area at most [MATH] We will prove below that (with high probability) functions drawn from [MATH] are far from 1-juntas. Then the following Theorem will demonstrate that testing 1-juntas with surface area [MATH] |
requires [MATH] queries. Theorem 43 For any query set [MATH] with [MATH] , if [MATH] and [MATH] then the distributions of [MATH] and [MATH] are [MATH] -close in total variation distance. |
In order to study the distinguishability of [MATH] and [MATH] , we give a slightly different description of [MATH] and [MATH] for [MATH] set |
[EQUATION] and note that [MATH] was defined by independently assigning a random [MATH] value on each set [MATH] while [MATH] was defined by independently assigning opposite random [MATH] values on each pair |
[MATH] [MATH] Also, [MATH] and [MATH] are both identically one on [MATH] Let [MATH] be the set of query points, and consider the event that for every [MATH] , at least one of [MATH] or [MATH] contains no point in [MATH] call this event [MATH] Then [MATH] depends on [MATH] [MATH] , and [MATH] , but not on [MATH] Thanks ... |
[MATH] the random variables [MATH] and [MATH] have the same distribution. In particular, we can couple [MATH] and [MATH] so that |
[MATH] with probability at least [MATH] and so we will prove Theorem 43 by showing that for any choice of [MATH] with [MATH] [MATH] To do this, we will divide the pairs [MATH] into “close” pairs and “far” pairs: we say that [MATH] and [MATH] are [MATH] -close if |
[MATH] , and [MATH] -far otherwise. The following lemma will complete the proof of Theorem 43 , because it implies that with high probability no pair of points lies in the same strips [MATH] but on different sides of the line [MATH] |
Lemma 44 Suppose that [MATH] and set [MATH] . For any set [MATH] , with probability at least [MATH] 1. every pair of points [MATH] that are [MATH] -far do not belong to the same set [MATH] |
for any [MATH] 2. every pair of points [MATH] that are [MATH] -close lie on the same side of the line [MATH] The first step of Lemma 44 is the simple observation that far points remain reasonably far even after projecting them in the direction [MATH] |
Lemma 45 For all sufficiently small [MATH] and any [MATH] [MATH] Proof. If [MATH] is the angle between [MATH] and [MATH] then [MATH] |
exactly when [MATH] , which has probability [MATH] Since [MATH] has derivative 1 at zero, this is approximately [MATH] for small [MATH] In particular, if [MATH] is sufficiently small then this probability is at most [MATH] |
Proof of Lemma 44 Let [MATH] be the line [MATH] By Lemma 45 applied to [MATH] , if [MATH] and [MATH] are [MATH] -far then with probability at least [MATH] [MATH] By a union bound, with probability at least [MATH] |
[MATH] for every [MATH] -far pair [MATH] ; from now on, we will condition on this event (call it [MATH] ) occurring. Now, if either [MATH] |
or [MATH] lies outside of the interval [MATH] then [MATH] and [MATH] do not both lie in any single [MATH] . On the other hand, if both [MATH] |
and [MATH] lie in [MATH] , then each line [MATH] has (independently) probability [MATH] to “split” [MATH] from [MATH] Hence, with probability at least [MATH] , there will be a line [MATH] that splits [MATH] from [MATH] , and so they will not belong to any single set [MATH] Taking a union bound over all pairs [MATH] , w... |
[MATH] , which with our choice of parameters is at least [MATH] If [MATH] and [MATH] are [MATH] -close then [MATH] , and hence the probability that they land on opposite sides of the line |
[MATH] is at most [MATH] . By a union bound over all pairs, with probability at least [MATH] every pair of [MATH] -close [MATH] land on the same side of that line. |
5.1 [MATH] is far from a 1-junta So far, we have shown that one cannot distinguish [MATH] from [MATH] from few samples. It remains to show that functions from [MATH] are far (with high probability) from 1-juntas, it will follow that one cannot 1-juntas with [MATH] surface area with fewer than [MATH] queries. |
Theorem 46 There is a constant [MATH] such that with probability at least [MATH] over [MATH] [MATH] is [MATH] -far from every 1-junta. |
Now recall that the construction of [MATH] and [MATH] involved dividing up the strip [MATH] into [MATH] strips [MATH] and assigning random values on each strip. Since both the construction of [MATH] and the notion of distance to a 1-junta are rotationally invariant, we will assume from now on that [MATH] , which means ... |
[MATH] are vertically oriented. Let [MATH] and let [MATH] Definition 47 Let [MATH] be an interval. We say that [MATH] is [MATH] -balanced if of both [MATH] and [MATH] are at least [MATH] , where [MATH] denotes the one-dimensional Lebesgue measure. We say that [MATH] is wide if [MATH] |
We extend these definitions to strips in two dimensions: say that [MATH] is [MATH] -balanced (resp. wide) if [MATH] is [MATH] -balanced (resp. wide). |
Definition 48 For any line [MATH] , we say that [MATH] is [MATH] -balanced if both [EQUATION] are at least [EQUATION] We will now describe the outline of Theorem 46 ’s proof: note that if [MATH] is a [MATH] -junta then [MATH] for some [MATH] Now, Fubini’s theorem implies that |
[EQUATION] Now, whenever the line [MATH] is [MATH] -balanced, the inner integral is at least [MATH] . Therefore, in order to prove Theorem 46 it suffices to show that there is a constant [MATH] such that at least a constant fraction of the lines |
[MATH] are [MATH] -balanced. To be precise, let [MATH] be the set of lines of the form [MATH] for [MATH] . Since [MATH] is bounded from below on [MATH] it suffices to show that there is a constant [MATH] such that with high probability, for every [MATH] , a constant fraction of [MATH] are [MATH] -balanced. For the rema... |
We will consider two cases depending on [MATH] : if the lines in [MATH] are “steep,” then these lines will be balanced because a constant fraction of them will cross the horizontal line [MATH] near the middle of a strip. Since the value of [MATH] on a strip changes sign at that horizontal line, this will imply that suc... |
We will first deal with the case of steep lines. In this case, it is deterministically the case that [MATH] is far from [MATH] Lemma 49 |
At least half of the points on the line segment from [MATH] to [MATH] are in a wide strip [MATH] Proof. There are [MATH] strips in total, and so the narrow ones can take up at most a total width of 1, which is only half of the width of the line segment in question. |
Lemma 50 There is a constant [MATH] such that if the absolute value of the slope of [MATH] is at least [MATH] then a [MATH] -fraction of [MATH] are [MATH] -balanced. |
Proof. We may assume without loss of generality that [MATH] By Lemma 49 , at least a constant fraction of [MATH] intersect the line [MATH] in the middle third of a wide strip [MATH] . In this case, |
[MATH] belongs to [MATH] for a distance of at least 1/3, and to [MATH] for a distance of at least 1/3, and it follows that [MATH] is [MATH] -balanced for a constant [MATH] |
depending on the minimum and maximum values of [MATH] for [MATH] For the remainder of the section we will deal with lines that are not steep. For [MATH] |
with [MATH] , consider an interval of the form [MATH] let [MATH] be the set of all such intervals. Lemma 51 There is a constant [MATH] such that with probability at least [MATH] for every [MATH] for which [MATH] , at least a [MATH] -fraction of the intervals [MATH] are [MATH] -balanced. |
Proof. For technical convenience, we will consider a slightly different way of generating the strips [MATH] Instead of dividing [MATH] using exactly [MATH] vertical lines, we will take a Poisson number (with mean [MATH] ) of vertical lines. We will prove the claim for this modified model, with a probability estimate of... |
Our first claim is that for [MATH] , each interval in [MATH] has a constant probability of being [MATH] -balanced. First, consider the largest [MATH] for which [MATH] . In this case, the width of each [MATH] |
is within a factor 2 of [MATH] (we will call such an interval a primitive interval. It is easy to verify that for each [MATH] , there is a constant probability that [MATH] will intersect exactly two strips, each taking up at least 1/3 of the width of [MATH] , and that these two strips will receive different labels [MAT... |
Now consider [MATH] for which [MATH] . Every [MATH] is made up of [MATH] primitive intervals, each of which has a constant probability of being balanced. Moreover (thanks to our Poissonized model) the events that different primitive intervals are balanced are independent. By Chebyshev’s inequality, there is a constant ... |
[MATH] -balanced. This proves our first claim (that for each [MATH] , each interval in [MATH] has a constant probability of being [MATH] -balanced). Now, for each such [MATH] there are at least [MATH] |
such intervals, and so a Chernoff bound implies that with probability at least [MATH] at least a constant fraction of these intervals are balanced. Taking a union bound over [MATH] |
proves the claim whenever [MATH] For smaller [MATH] , we claim that with high probability, every [MATH] is balanced. Indeed, such [MATH] contain at least [MATH] primitive intervals, and so a Chernoff bound implies that with probability [MATH] , at least a constant fraction of those primitive intervals are balanced, and... |
and all [MATH] To complete the proof of Theorem 46 , it remains to show that with high probability, every non-steep line is balanced. |
Lemma 52 There is a constant [MATH] such that if the absolute value of the slope of [MATH] is at most [MATH] then with probability at least [MATH] , a [MATH] -fraction of [MATH] are [MATH] -balanced. |
Proof. Choose [MATH] so that the slope of all lines in [MATH] are between [MATH] and [MATH] Consider a rectangle of the form [MATH] , where the interval [MATH] is balanced. Since the slope of [MATH] |
is at most [MATH] , if the line [MATH] intersects the rectangle [MATH] then it crosses the entire vertical strip [MATH] within the horizontal strip [MATH] Since the interval [MATH] is balanced, it follows that the line [MATH] is also balanced. (We’re assuming here, without loss of generality, that [MATH] ). |
Finally, it is easy to verify that if a constant fraction of the intervals [MATH] are balanced then a constant fraction of [MATH] intersect with some rectangle of the form above. By Lemma 51 , this completes the proof. |
Appendix A Small net for noise attenuated linear juntas In this section, we are going to prove the following theorem which essentially shows the existence of a small cover for noise stable linear juntas. To state this theorem, we will require one crucial fact about noise attenuated functions (due to Bakry and Ledoux Ba... |
Lemma 53 Let [MATH] . Then, [MATH] is [MATH] -Lipschitz for [MATH] For the rest of this section, we are going to use [MATH] to denote this quantity. |
We can now state the main theorem of this section. Theorem 54 For any error parameter [MATH] , noise parameter [MATH] and [MATH] , there is a set of functions [MATH] (mapping [MATH] to [MATH] ) such that the following holds: |
1. Let [MATH] and [MATH] be a [MATH] -dimensional space such that [MATH] is [MATH] -close to a [MATH] -junta. Further, [MATH] be any orthonormal basis of [MATH] . Then, [MATH] is [MATH] -close to [MATH] for some [MATH] |
2. Every function in [MATH] is [MATH] -Lipschitz. 3. [MATH] The proof of this theorem relies on the following two lemmas. Lemma 55 |
For any [MATH] , error parameter [MATH] and [MATH] , there is a set [MATH] consisting of functions mapping [MATH] such that the following holds: |
1. For every [MATH] which is [MATH] -Lipschitz, there is a function [MATH] such that [MATH] 2. Every function in [MATH] is [MATH] -Lipschitz. |
3. [MATH] Proof. Let [MATH] . Let [MATH] be a maximal [MATH] -packing of [MATH] (that is, a maximal subset of [MATH] such that any two distinct points in [MATH] are at least [MATH] apart. It is well-known (see, e.g. LT91 ) that [MATH] is a [MATH] -net of [MATH] and that [MATH] |
(the [MATH] term comes from the diameter of [MATH] For [MATH] , we now define [MATH] by simply rounding [MATH] to the nearest integer multiple of [MATH] . To check the Lipschitz constant of [MATH] , note that if [MATH] then |
[EQUATION] where the last inequality used the fact that [MATH] is [MATH] -Lipschitz and that every pair of points in [MATH] is [MATH] -separated. In particular, [MATH] is [MATH] -Lipschitz. Let [MATH] be the set of all functions [MATH] obtained in this way. Then the size of [MATH] is at most [MATH] because there are at... |
to a function [MATH] . McShane’s Lemma McS34 implies that this extension can be done without increasing its Lipschitz constant. Hence, properties 2 and 3 hold. |
To check property 1, note that if [MATH] and [MATH] is the closest point to [MATH] then [EQUATION] It then follows that [EQUATION] |
The last inequality just follows from the fact that a [MATH] -dimensional standard Gaussian is in a ball of radius [MATH] with probability [MATH] . This proves property 1 modulo the constant [MATH] , which can be dropped by redefining [MATH] |
Lemma 56 Let [MATH] be a [MATH] -Lipschitz function. Further, for [MATH] , let [MATH] be a [MATH] -junta such that [MATH] is [MATH] -close to [MATH] . Then, there is a function [MATH] which is [MATH] -Lipschitz and [MATH] -junta which is [MATH] -close to [MATH] |
Proof. Reorient the axes so that [MATH] is the space spanned by the first [MATH] -axes. Let us define the [MATH] -junta [MATH] defined as |
[EQUATION] For any fixed choice of [MATH] , we have [EQUATION] However, the second term can be bounded as [EQUATION] The last inequality is simply Jensen’s inequality. Combining these two, we get |
[EQUATION] This in turn implies that [EQUATION] Finally, we see that [EQUATION] This finishes the proof. With these two lemmas, we can now finish the proof of Theorem 54 |
Proof of Theorem 54 First, we apply Lemma 53 to obtain that [MATH] is [MATH] -Lipschitz. Since [MATH] is [MATH] -close to a [MATH] -junta, we obtain that [MATH] is [MATH] close to a [MATH] -junta [MATH] which is [MATH] -Lipschitz (follows from Lemma 56 ). Let [MATH] (constructed in Lemma 55 ). By a rotation of the coor... |
Appendix B Some useful results from linear algebra The next lemma states for any [MATH] which are [MATH] -linearly independent, we can find a set of vectors [MATH] (expressed as linear combination of [MATH] ) which is close to being an orthonormal basis of the [MATH] provided we have sufficiently good approximations of... |
Lemma 57 Let [MATH] be a [MATH] -linearly independent vectors. Then, for any error parameter [MATH] and [MATH] defined as [EQUATION] |
given numbers [MATH] such that [MATH] , we can compute numbers [MATH] such that: 1. For [MATH] defined as [EQUATION] we have [MATH] |
2. There is an orthonormal basis [MATH] of [MATH] such that for [MATH] Proof. Consider the symmetric matrix [MATH] defined as [MATH] . By Proposition 58 [MATH] is non-singular. Define the matrix [MATH] . It is easy to see that the columns of |
[MATH] form an orthonormal basis of [MATH] . Here [MATH] . Of course, we cannot compute the matrix [MATH] exactly and consequently, we cannot compute the matrix [MATH] either. Instead, if we define the matrix [MATH] as [MATH] , then observe that [MATH] is symmetric. Next, observe that Proposition 58 , we have that |
[EQUATION] Define a parameter [MATH] as [EQUATION] Now, with this setting, observe that [EQUATION] Further, since entrywise, [MATH] and [MATH] differ by at most [MATH] , hence [MATH] . First, by Weyl’s inequality (Lemma 17 ), we have that |
[EQUATION] Thus, [MATH] is also psd. Now, we apply the matrix perturbation bound to matrices [MATH] and [MATH] (Corollary 20 with parameter [MATH] to obtain that |
[EQUATION] We now define [MATH] . We also define [MATH] . Note that the vectors [MATH] forms an orthonormal basis. As the matrices [MATH] and [MATH] are [MATH] close in operator norm, this immediately implies item 2. To get item 1, we recall the following basic inequality for Frobenius norm of an inverse matrix. In par... |
[MATH] . Thus, [EQUATION] The last inequality uses ( ) and the fact that [MATH] . This immediately implies the first item. Proposition 58 |
Let [MATH] be a [MATH] -linearly independent vectors. Let [MATH] . Then, the smallest singular value of [MATH] is at least [MATH] |
Proof. Let us set a parameter [MATH] . Recall that if [MATH] is the smallest singular value of [MATH] , then [EQUATION] Let us try to lower bound the right hand side. To do this, let [MATH] be any unit vector and note that [MATH] Now, let [MATH] be the largest coordinate such that [MATH] (note that there has to be such... |
# Source: arxiv 1806.10389 # Title: Computing the metric dimension by decomposing graphs into extended biconnected components # Sections: all # Downloaded: 2026-03-03T01:44:37.639623+00:00 |
Computing the metric dimension by decomposing graphs into extended biconnected components Abstract A vertex set [MATH] of an undirected graph [MATH] is a resolving set for [MATH] , if for every two distinct vertices [MATH] there is a vertex [MATH] such that the distances between [MATH] and [MATH] and the distance betwe... |
keywords: Graph algorithm, Complexity, Metric Dimension, Resolving Set, Biconnected Component Introduction An undirected graph [MATH] has metric dimension at most [MATH] if there is a vertex set [MATH] such that [MATH] and [MATH] [MATH] , there is a vertex [MATH] such that [MATH] , where [MATH] is the distance (the len... |
and Slater If for three vertices [MATH] , we have [MATH] , then we say that [MATH] and [MATH] are resolved by vertex [MATH] . If every pair of vertices is resolved by at least one vertex of a vertex set [MATH] , then [MATH] is a resolving set for [MATH] . The metric dimension of [MATH] is the size of a minimum resolvin... |
Determining the metric dimension of a graph is a problem that has an impact on multiple research fields such as chemistry , robotics |
, combinatorial optimization and sensor networks . Deciding whether a given graph [MATH] has metric dimension at most [MATH] for a given integer [MATH] is known to be NP-complete for general graphs |
, planar graphs , even for those with maximum degree 6 and Gabriel unit disk graphs . Epstein et al. showed the NP-completeness for split graphs, bipartite graphs, co-bipartite graphs and line graphs of bipartite graphs |
and Foucaud et al. for permutation and interval graphs There are several algorithms for computing the metric dimension in polynomial time for special classes of graphs, as for example for trees |
, wheels , grid graphs [MATH] -regular bipartite graphs , amalgamation of cycles , outerplanar graphs , cactus block graphs and chain graphs |
. The approximability of the metric dimension has been studied for bounded degree, dense, and general graphs in . Upper and lower bounds on the metric dimension are considered in |
for further classes of graphs. There are many variants of the Metric Dimension problem. For the weighted version Epstein et al. gave a polynomial-time algorithm on paths, trees and cographs |
. Hernando et al. investigated the fault-tolerant Metric Dimension in , Estrada-Moreno et al. the [MATH] -metric Dimension in and Oellermann et al. the strong metric Dimension in |
The parameterized complexity was investigated by Hartung and Nichterlein. They showed that for the standard parameter the problem is [MATH] -complete on general graphs, even for those with maximum degree at most three |
. Foucaud et al. showed that for interval graphs the problem is FPT for the standard parameter . Afterwards Belmonte et al. extended this result to the class of graphs with bounded treelength, which is a superclass of interval graphs and also includes chordal, permutation and AT-free graphs |
In this paper, we introduce a concept that allows us to compute the metric dimension based on a tree structure given by the decomposition of a graph [MATH] into components like bridges legs , and so-called extended biconnected components . An extended biconnected component |
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