Wikipedia:Reference desk/Archives/Mathematics/2022 November 10

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November 10

Locality-preserving functions

Having a problem with locality-preserving functions. In particular, a locality-preserving hash function $f:{\mathcal {M}}\rightarrow \mathbb {R}$ (where ${\mathcal {M}}=(M,d)$ is a metric space) is a hash function satisfying $\forall p,q,r\in {\mathcal {M}},d(p,q)<d(q,r)\Rightarrow |f(p)-f(q)|<|f(q)-f(r)|$, and more generally, we can say that a locality-preserving function $f:{\mathcal {M}}\rightarrow {\mathcal {K}}$ for metric spaces ${\mathcal {M}}=(M,d_{M}),{\mathcal {K}}=(K,d_{K})$ is a function satisfying $\forall p,q,r\in {\mathcal {M}},d_{M}(p,q)<d_{M}(q,r)\Rightarrow d_{K}(f(p),f(q))<d_{K}(f(q),f(r))$.

It can be seen that any such function must be injective. If $f(p)=f(q)$ but $p\neq q$, then for any $r\in {\mathcal {M}}$ with $d_{M}(p,r)\neq d_{M}(q,r)$ (for example, $p$ or $q$ itself), we must have either $d_{M}(p,r)<d_{M}(q,r)$ or $d_{M}(p,r)>d_{M}(q,r)$, but $d_{K}(f(p),f(r))=d_{K}(f(q),f(r))$, which is contradictory.

The area I'm having trouble with is that I have no idea if locality-preserving functions have to be continuous. I'm pretty sure I have it narrowed down to one problem which I am completely stuck on: if $f:{\mathcal {M}}\rightarrow {\mathcal {K}}$ is a locality-preserving function, is it possible for there to be some point $x\in {\mathcal {M}}$ such that $x$ is an accumulation point but $f(x)$ is an isolated point? Similar problem in establishing whether $f$ is an open mapping, where the problem becomes whether it is possible for $x$ to be an isolated point while $f(x)$ is an accumulation point.

As a quick aside, I'm fairly sure that if ${\mathcal {K}}$ (resp. ${\mathcal {M}}$) is compact, then the answer is no for continuity (resp. open mapping) GalacticShoe (talk) 04:22, 10 November 2022 (UTC)

Neither one needs to hold. For instance, let ${\mathcal {M}}=\{0\}\cup \left\{{\frac {1}{n}}\colon n\in \mathbb {Z} ^{+}\right\}$, ${\mathcal {K}}=\ell ^{\infty }(\mathbb {R} )$ under the supremum norm, and $f(0)=e_{1}$, $f\left({\frac {1}{n}}\right)={\frac {n+1}{n}}e_{n+1}$ (where $e_{i}$ is the sequence with 1 in the $i$-th position and 0 elsewhere). $f$ is a locality-preserving function and 0 is an accumulation point of ${\mathcal {M}}$, but $f(0)=e_{1}$ is an isolated point of $f({\mathcal {M}})$. The inverse of $f$ is also a locality-preserving function from $f({\mathcal {M}})$ back to ${\mathcal {M}}$. (Wrong, see below for a correct example.) BentSm (talk) 05:58, 10 November 2022 (UTC)

Writing $f:{\mathcal {M}}\to {\mathcal {S}},$ is the codomain ${\mathcal {S}}$ in this proposed counterexample one of scalar values?  --Lambiam 10:27, 10 November 2022 (UTC)

No. (I was basing my answer from his (extended) definition of a locality-preserving function.) BentSm (talk) 01:08, 11 November 2022 (UTC)

Hey BentSM, thanks for the response! Just wanted to clarify a point of confusion; is the distance under the supremum norm defined as $d_{sup}(x,y)=sup_{n}|x_{n}-y_{n}|$, and if so, would $|0-{\frac {1}{3}}|={\frac {1}{3}}>{\frac {1}{6}}=|{\frac {1}{3}}-{\frac {1}{2}}|,d_{sup}(f(0),f({\frac {1}{3}}))={\frac {4}{3}}<{\frac {3}{2}}=d_{sup}(f({\frac {1}{3}}),f({\frac {1}{2}}))$ not render $f$ non-locality preserving? Thanks again! GalacticShoe (talk) 18:07, 11 November 2022 (UTC)

You're quite right. This does work, though:

${\mathcal {M}}=\{0\}\cup \left\{{\frac {1}{n}}\colon n\in \mathbb {Z} ^{+}\right\}$, ${\mathcal {K}}=\ell ^{\infty }(\mathbb {R} )$ under the $L^{1}$ norm, and $f(0)=e_{2}$, $f\left({\frac {1}{n}}\right)={\frac {1}{n}}e_{1}+e_{n+2}$. If $x\neq y$, $d_{1}(f(x),f(y))=2+|x-y|$ ($d_{1}(f(x),f(x))=0$, of course). BentSm (talk) 00:32, 12 November 2022 (UTC)

Suppose we want to hash (one-dimensional) real values into integer-valued buckets. So the desired hash function has signature $f:\mathbb {R} \to \mathbb {Z} .$ A prime candidate is the rounding function defined by

$f(x)=\lfloor x+0.5\rfloor .$

If any function from the reals to the integers is a locality-preserving hash, this should be one. But (under the usual metric in which $d(x,y)=|x-y|$ ) we have

$d(3.4,3.6)=0.2<d(3.6,3.9)=0.3,$

whereas

$|f(3.4)-f(3.6)|=|3-4|=1>|f(3.6)-f(3.9)|=|4-4|=0.$

This suggest strongly that the given definition is off. Perhaps it should be the other way around: values mapped to the same bucket should be close. That would explain the suggested relation to space-filling curves: themselves being continuous, they are surjective and so have a (not unique) inverse. Such an inverse can be viewed as a hash, and pairs of points whose hashes are close are themselves close. Naively reversing the implication does not work, however:

$|f(2.8)-f(3.4)|=|3-3|=0<|f(3.4)-f(3.6)|=|3-4|=1,$

whereas

$d(2.8,3.4)=0.6>d(3.4,3.6)=0.2\,.$

So we may want a condition in the shape of

$|f(x)-f(y)|<\varepsilon \,\Rightarrow \,d(x,y)<\delta .$

 --Lambiam 12:11, 10 November 2022 (UTC)

I do agree that the definition is rather unusual. It is certainly is not that of the two references listed in the corresponding section of the Wikipedia article referenced. I tried doing some searching, which turned up more references to "locality-preserving" than direct definitions; see, e.g., [1], [2], [3], [4]. For what it's worth, none of what I found matched up with the definition given in the article (except for one site that cites Wikipedia). BentSm (talk) 03:38, 11 November 2022 (UTC)

For example, the Hilbert curve satisfies this "inverse continuity" condition if we put $\delta ={\sqrt {c\varepsilon }},$ where I think we can take $c={\tfrac {27}{5}}.$  --Lambiam 16:11, 10 November 2022 (UTC)

Slopes and angles

If a line had a slope of S (assuming the line goes through the origin and S cannot be negative), how can I use that to find out the angle that the line makes with the X or Y-axis? Primal Groudon (talk) 14:18, 10 November 2022 (UTC)

By consulting our article Slope: it gives the relation between the slope of a line and its angle of inclination. It is not relevant whether the line goes through the origin, and the relation is valid regardless of the sign of the slope.  --Lambiam 16:35, 10 November 2022 (UTC)

• The angle of the line to the X axis will be the arctan of the slope. That's because t the line will always be the hypotenuse of a right triangle where the X axis is one leg and a line parallel to the Y axis is the other leg. Since tangent Θ = Y/X = slope, to find the angle Θ you take the inverse tangent (arctan) of the slope. --2600:1004:B024:84EE:F02:CC26:4A9A:E65 (talk) 21:54, 10 November 2022 (UTC)

There's two angles corresponding to a slope differing by 180°. Very often you'll want the one between -90° and +90°. I'm not sure why you specified that the slope is not negative, that doesn't matter much. If anything the slope that is a bit peculiar is the vertical slope going straight up and down where the ratio is infinity. NadVolum (talk) 01:19, 11 November 2022 (UTC)