Model: cristiano-sartori/bge-base-en-v1.5_finetuned
Author: cristiano-sartori
Pipeline: sentence-similarity
Library: sentence-transformers
Tags: sentence-transformers, safetensors, bert, sentence-similarity, feature-extraction, generated_from_trainer, dataset_size:35129, loss:MatryoshkaLoss, loss:MultipleNegativesRankingLoss, en, arxiv:1908.10084, arxiv:2205.13147, arxiv:1705.00652, base_model:BAAI/bge-base-en-v1.5, base_model:finetune:BAAI/bge-base-en-v1.5, license:apache-2.0, model-index, text-embeddings-inference, endpoints_compatible, region:us
Base models (finetune): id: BAAI/bge-base-en-v1.5
Card metadata: language: en
license: apache-2.0
tags: sentence-transformers, sentence-similarity, feature-extraction, generated_from_trainer, dataset_size:35129, loss:MatryoshkaLoss, loss:MultipleNegativesRankingLoss
base_model: BAAI/bge-base-en-v1.5
widget: source_sentence: I know the familiar differences between tight and overtwisted contact structures. For example,  each homotopy class of plane-bundles on a three-manifold has an overtwisted representative but tight contact structures are less common. Also a tight contact structure gives a genus bound on an embedded closed surface and therefore can tell things about the topology of the ambient manifold. My question is what were the motivations behind distinguishing between the two originally? Is there possibly an intuition which makes the two fundamentally different and then proving the above differences comes as a consequence or it was these differences which led to distinguishing between the two? In other words I am trying to see the difference between the two without appealing to the statements like the above.
sentences: Yes there are other hard problems you can base asymmetric cryptography on. 

Lattices. The NTRU systems is based on the shortest vector problem in ideal lattices. Lattice-based cryptography is of much interest these days for two reasons: (1) unlike factorization and discrete logarithms, there isn't an efficient algorithm for breaking these problems on a (still theoretic but should happen within our lifetime) quantum computer; (2) you can use lattice-based systems to do arbitrary operations on data while it is still encrypted (not with NTRU though but other lattice-based systems). 

Coding Theory. The McEliece system is based on decoding linear codes. It is a very old system but recently has seen renewed interest for the same post-quantum reasons as lattice-systems.

Misc. There are other more obscure (research-level) cryptosystems based on things like learning parity with noise, learning with error (related to SVP in lattices), subset sum problem (including a famous broken example), solving multivariate or diophantine equations, etc. 

In response to your second question, is it that hard to find alternatives, the answer is it is reasonably hard. For most cryptographic primitives, you need to find examples of functions that are hard to invert, and for asymmetric encryption specifically, you are generally interested in functions that have a specific value that allow them to be inverted (but only if you know that value). See trapdoor functions. It is difficult to find new candidates for trapdoor functions and to establish they are suitably hard. 
, First of all: French effacer means "to erase", and it is of course a composite of the prefix  e- (latin ex-) [away from] face. I'm not aware of a different meaning than erase (maybe wipe up or annihilate are also viable translations in some circumstances), and what you're describing is probably more or less what Grothendieck had in mind. As far as I know Grothendieck is the inventor of this terminology.

Let me focus on the simplest case:

Let us assume that the abelian category has enough injectives and let $F$ be a (left exact) functor to another abelian category. As you know, the associated right derived functors $R^nF$ can be constructed by choosing an injective resolution of each object, applying $F$ to the resolution and taking homology.


  The right derived functors are a universal $\delta$-functor due to the fact that each $A$ admits an injection $A \to I$ such that $R^{n}F(I) = 0$ for $n \geq 1$.


Let us look at a specific instance of the arguments involved: The first step in proving that given any other $\delta$-functor $T^n$ and a natural transformation $\phi : F \Rightarrow T^0$ there is a unique sequence of natural transformations $\phi^n: R^nF \Rightarrow T^n$ extending $\phi$ and compatible with the connecting morphisms of both $R^nF$ and $T^n$.

This sequence of natural transformations is built by induction. For each $A$ choose an injective effacement $0 \to A \to I \to C \to 0$ and apply both $R^nF$ and $T^n$ and the induction hypotheses. We get to the following diagram:

$$\begin{array}{ccccccc}
    R^nF(I) & \to & R^{n}F(C) & \to & R^{n+1}F(A) & \to & \mathbf{0} \\ 
\downarrow & & \downarrow & & \downarrow \exists? & & \\
T^n(I) & \to & T^n(C) & \to & T^{n+1}(A)
\end{array}$$
The bold zero is due to effaceability of $R^{n+1}F$. The two left hand vertical maps are already constructed and the map marked with a question mark exists and is unique by an easy diagram chase because the bold zero implies that the connecting morphism $R^{n}F(C) \to R^{n+1}F(A)$ is an epimorphism. The map just constructed is the $A$-component $\phi_{A}^{n+1}$ of $\phi^{n+1}$. Such arguments are applied some further times when checking that this construction actually defines a natural transformation $\phi^{n+1} : R^{n+1} F \Rightarrow T^{n+1}$, that the $\phi^n$'s are compatible with the connecting morphisms and that they are unique in that they don't depend on the chosen effacement.

In summary and conclusion: 


  Universality follows from effaceability due to the fact that the connecting morphisms $R^{n}F(C) \to R^{n+1}F(A)$ associated to the sequence $0 \to A \to I \to C \to 0$ are forced to be epimorphisms, as $R^{n+1}F(I) = 0$. In other words, because $R^{n+1}F$ gets erased at $I$.


I hope that this is clear enough, if not, I strongly recommend to go through the proof of universality of derived functors once again, e.g. in Theorem 2.4.7 of Weibel. (Don't miss the exercises at the end of the section!)

The effaceability condition can be used to show more generally: If $S^n$ is a $\delta$-functor and $S^{n}$ is effaceable for $n \geq 1$, then the $\delta$-functor $S^n$ is universal.

This is used when one applies dimension shifting: Suppose we have an effacement $0 
\to A \to I \to C \to 0$ and $S^{n}(I) = 0 = S^{n+1}(I)$. Then the sequence $0 \to S^n(C) \to S^{n+1}(A) \to 0$ shows that the connecting morphism $S^n(C) \to S^{n+1}(A)$ is an isomorphism because of this effacement.

For instance, the definition of flatness implies that $Tor_{n}(A,F)$ is zero for a flat module $F$ and $n \geq 1$. A variant of the previous argument can be used to show that $Tor_{n+1}(A,B)$ can be constructed from $Tor_{n}(A,-)$ using a flat (co-)effacement of $B$, and the resulting $\delta$-functor is universal due to the above.

Sometimes a slightly more general variant of the effaceability is called Buchsbaum's criterion (or weak effaceability) due to Proposition 4.2 in Buchsbaum's Annals paper Satellites and Universal Functors.



Added:

Here is, to the best of my knowledge, the most useful general criterion for $F$ to admit a (classical) right derived functor, which I learned from Bernhard Keller. It covers all the "oodles of applications" explained by Akhil in his answer:

Suppose $\mathcal{C}$ is an abelian (or more generally an exact) category and assume that $F:\mathcal{C} \to \mathcal{D}$ is a functor to an abelian category $\mathcal{D}$.

Theorem. If there exists a full subcategory $\mathcal{A}$ of $\mathcal{C}$ such that


$\mathcal{A}$ is closed under extensions: if $0 \to A' \to C \to A'' \to 0$ is exact in $\mathcal{C}$ then $C$ is an object of $\mathcal{A}$.
The restriction of $F$ to $\mathcal{A}$ is exact.
For every object of $\mathcal{C}$ there exists a monomorphism $C \to A$ with $A$ in $\mathcal{A}$.
For every short exact sequence $0 \to A' \to C \to C''$ with $A'$ in $\mathcal{A}$ there exists a commutative diagram
$$\begin{array}{ccccccccc}
0 & \to & A' & \to & C & \to & C'' & \to & 0 \\
 & & \parallel & & \downarrow & & \downarrow \\
0 & \to & A' & \to & A & \to & A'' & \to & 0
\end{array}$$
where the second row is exact and $A, A''$ are in $\mathcal{A}$.


Then $F$ admits a right derived functor.

Notes:


The technical condition 4. is satisfied in presence of 3. if $\mathcal{A}$ is supposed to be closed under quotients in $\mathcal{C}$. That is: Whenever $0 \to A' \to A \to C \to 0$ is short exact then $C$ is an object of $\mathcal{A}$.
No left exactness of $F$ on $\mathcal{C}$ is imposed: this is only needed to guarantee that $R^0F = F$. In general, $R^{0}F$ is the best left exact approximation to $F$. In highbrow terms: $R^{0}$ can be seen as reflection of the inclusion of the left exact functors into all additive functors.
The right derived functors of $F$ can be computed by choosing for each $C$ an $\mathcal{A}$-resolution — which exists due to the effaceability conditions 2 and 3.
The objects of $\mathcal{A}$ are called $F$-acyclic or adapted to $F$ if $\mathcal{A}$ satisfies the hypotheses of the theorem.


Final remarks (coming back to effaceability):


The above theorem admits a partial converse: if $F$ admits a right derived functor, the category $\mathcal{A}(F)$ of $F$-acyclic objects (a notion I don't want to define in full generality here), satisfies conditions 1., 2. and 4. above. The effaceability of the higher right derived functors is thus seen as the part missing from a complete characterization of derived functors.
The usefulness of injective effacements is of course that injective objects are $F$-acyclic for every additive functor. This is because short exact sequences of injectives split, by definition of injectivity. In particular, every functor on an abelian category with enough injectives can be derived (not only left exact ones).
I'm not aware of a complete proof of the theorem as stated in the literature without making a detour via derived categories, which certainly looks like overkill given its rather elementary nature. See sections 12ff in Keller's article or sections 10.5, 10.6 and 12 in my notes on exact categories.

, As mentioned in the previous answer, the first use of the overtwisted/tight dichotomy is most certainly Bennequin's Theorem that there are non-isomorphic contact structures on $\mathbb{R}^3$ and $\mathbb{S}^3$, a landmark result. 

However, the relevance of this dichotomy goes now far beyond this. As you probably know, contact topology has a Darboux theorem: locally, all contact structures are the same, isomorphic to the standard contact structure on $\mathbb{R}^3$. So, all of them are "locally tight", and having an overtwisted disc must be a global condition. A global way of distinguishing objects which are locally the same is tremendously important, here it can be thought of as some analogue of Gromov's non-squeezing theorem in symplectic geometry.

Moreover, while the classification of overtwisted structure has been achieved quite early by Eliashberg (Inventiones 1989), the tight contact structures happened to be very rich (see e.g. Giroux, Inventiones 1999 and its introduction -- in French) : certain manifold have only one of them, but infinitely many are shown to bear infinitely many non-isomorphic tight contact structure in the paper cited. The relation with symplectic fillings is one more indication of the relevance of this dichotomy. One can now consider that the tight contact structure are the one to study, as overtwisted ones are pretty well understood.

Note that the relevance of this dichotomy has recently been more firmly established in higher dimensions too by Niederkrüger, Massot and Wendl (Inventiones 2013).

About Bennequin's theorem, I can not explain its proof, although I learned one in my graduate years. I cannot give a complete account of it any more, but I can give some flavor. Consider the usual cylindrically symmetric version of the standard contact structure, and decompose $\mathbb{R}^3$ as a trivial open book (the vertical line through the origin is the binding, and each half plane it bounds is a page). Assume there is a closed horizontal curve(in the sense of the contact structure) which bounds a disc tangent to the structure, which you assume in general position, and look at the way this disc intersects the pages. In each page the non-transversal points are vertices of a graph, so you get combinatorial objects to work with.
That's pretty much what I remember, but there are several proofs in the literature (at least one by Giroux, but likely to be written in French; it is probably not a waste of time to learn enough French to read mathematics if you are planning to work in contact topology). In fact, a friend of mine working in this area once told me that it was almost a duty for anyone working in this area to find its own proof of Bennequin's theorem. You should be ale to locate several proofs through the literature, but I am too far from this field to help you there.
, source_sentence: I read somewhere in a book that mostly desert found in the western part of a continent.  So my question is that why is it so? I searched on the internet but I didn't get a reasonable answer for that.
sentences: Some of the driest deserts on Earth occur in the western side of continents and they are called Coastal Deserts. Examples of such deserts are the Atacama desert (Chile, the driest desert on Earth), the Baja California desert (USA/Mexico), the Namibia desert (southwestern Africa), and the Atlantic coastal desert of Morocco/western Sahara/Mauritania. In mid-latitudes, the western side of continents is most frequently occupied by deserts, while the eastern side tends to be covered by forests (e.g., Amazon rainforest).

. Distribution of non-polar arid regions. Source: USGS. 

Coastal deserts occur east of the strongest coastal upwelling regions in the world. The upwelling is associated with eastern boundary currents flowing equatorward (due to Coriolis): the Canary Current (off Northwest Africa), the Benguela Current (off Southwest Africa), the Humboldt Current (off Chile and Peru), and the California Current (western North America). The prevailing winds in these regions are the trade winds that flow parallel to the coast and generate upwelling dynamics: surface Ekman balance is setup (in deep enough waters) with water transport being to the right (left) of the wind in the northern (southern) hemisphere; this causes a deficit along the coast that needs a compensating flow in the deeper part of the water column bringing cold, nutrient rich waters (some of the most productive regions in the world) to the surface.

 Source Commons Wikipedia.

The cold waters near the ocean surface results in a cool, stable coastal atmosphere. In this region, evaporation from the ocean is reduced and produces extremely low rainfall over land. Precipitation is limited to morning fog and produces some of the driest ecosystems on Earth. The Atacama desert is the best example of such environment with average rainfalls of 15 mm/year (the driest non-polar region). In some areas, they are trying to take advantage of the little moisture the fog (Camanchaca) brings to establish some agricultural zones. The fog droplets are too small (1-40 micrometers) to form water drops and precipitate, so they use fog-catchers to collect moisture from the fog.

 Source: newatlas.com
, Actually, it is also true for squares (and for regular polygons in general!). The problem you ran into is what the equivalent of "r" is. The side length of a square is actually more comparable to the circle's diameter.

Instead, the correct analogue of the circle's radius is the distance from the center of the square to the midpoint of one side, which is only half as long as the square's side.

  

Here, we have $A = (2r)^2 = 4 r^2$ and $P = 4 (2r) = 8 r$. The perimeter is the derivative of the area with respect to $r$, just as in the case of a circle.
, Waves always travel. Even standing waves can always be interpreted as two traveling waves that are moving in opposite directions (more on that below).

Keeping the idea that waves must travel in mind, here's what happens whenever you figure out a way to build a region in which the energy of such a moving wave cancels out fully: If you look closely, you will find that you have created a mirror, and that the missing energy has simply bounced off the region you created.

Examples include opals, peacock feathers, and ordinary light mirrors. The first two reflect specific frequencies of light because repeating internal structures create a physical regions in which that frequency of light cannot travel - that is, a region in which near-total energy cancellation occurs. An optical mirror uses electrons at the top of their Fermi seas to cancel out light over a much broader range of frequencies. In all three examples the light bounces off the region, with only a little of its energy being absorbed (converted to heat).

A skip rope (or perhaps a garden hose) provides a more accessible example. First, lay out the rope or hose along its length, then give it quick, sharp clockwise motion. You get a helical wave that travels quickly away from you like a moving corkscrew. No standing wave, that!

You put a friend at the other end, but she does not want your wave hitting her. So what does she do? First she tries sending a clockwise wave at you too, but that seems to backfire. Your wave if anything seems to hit harder and faster. So she tries a counterclockwise motion instead. That seems to work much better. It halts the forward progress of the wave you launched at her, converting it instead to a loop. That loop still has lots of energy, but at least now it stays in one place. It has become a standing wave, in this case a classic skip-rope loop, or maybe two or more loops if you are good at skip rope.

What happened is that she used a canceling motion to keep your wave from hitting her. But curiously, her cancelling motion also created a wave, one that is twisted in the opposite way (counterclockwise) and moving towards you, just as your clockwise wave moved towards her. As it turns out, the motion you are already doing cancels her wave too, sending it right back at her. The wave is now trapped between your two cancelling actions. The sum of the two waves, which now looks sinusoidal instead of helical, has the same energy as your two individual helical waves added together.

I should note that you really only need one person driving the wave, since any sufficiently solid anchor for one end of the rope will also prevent the wave from entering it, and so end up reflecting that wave just as your friend did using a more active approach. Physical media such as peacock features and Fermi sea electrons also use a passive approach to reflection, with the same result: The energy is forbidden by cancellation from entering into some region of space.

So, while this is by no means a complete explanation, I hope it provides some "feel" for what complete energy cancellation really means: It's more about keeping waves out. Thinking of cancellation as the art of building wave mirrors provides a different and less paradoxical-sounding perspective on a wide variety of phenomena that alter, cancel, or redirect waves.
, source_sentence: For some of us, refereeing papers is part of the job. When refereeing statistical methodology papers, I think advice from other subject areas is fairly useful, i.e. computer science and Maths.

This question concerns reviewing more applied statistical papers. By this I mean, the paper is submitted to a non-statistical/mathematical journal and statistics is just mentioned in the "methods" section.

Some particular questions:


How much effort should we put in to understand the application area?
How much time should I spend on a report?
How picky are you when looking at figures/tables.
How do you cope with the data not being available.
Do you try and rerun the analysis used.
What's the maximum number of papers your would review in a year?


Have a missed any questions? Feel free to edit or add a comment.

Edit

I coming to this question as a statistician reviewing a biology paper, but I'm interested in the statistical review of any non-mathematical discipline.



I'm not sure if this should be a CW. On one hand it's a bit open, but on the other I can see myself accepting an answer. Also, answers will probably be fairly long.

sentences: The tangent complex to $\operatorname{Bun}_G(C)$ can be identified with $T_{\operatorname{Bun}_G(C)}=\mathbf{R}\pi_*{\operatorname{ad} P[1]}$, where $\pi:\operatorname{Bun}_G(C)\times C\rightarrow \operatorname{Bun}_G(C)$ is the natural projection and $P$ is the universal bundle.
Then the cotangent stack is $T^*{\operatorname{Bun}_G(C)} = \operatorname{Spec} \operatorname{Sym} (T_{\operatorname{Bun}_G(C)})$. Maps from $U$ into the total space of the bundle $T^*{\operatorname{Bun}_G(C)}\rightarrow \operatorname{Bun}_G(C)$ are the same as maps $U\rightarrow \operatorname{Bun}_G(C)$ together with a section of the dual sheaf of $T_{\operatorname{Bun}_G(C)}$. Relative Serre duality identifies $\mathcal{Hom}(T_{\operatorname{Bun}_G(C)}, \mathcal{O})$ with $\mathbf{R}\pi_*\mathcal{Hom}(\operatorname{ad} P, \omega_C)\cong\mathbf{R}\pi_*(\operatorname{ad} P\otimes\omega_C)$ using the Killing form.
So, maps $U\rightarrow H^0(T^*{\operatorname{Bun}_G(C)})$ to the underlying ordinary stack are identified with $G$-bundles over $U\times C$ and a section $\phi\in H^0(U\times C, \operatorname{ad} P\otimes \omega_C)$.
, You can delete databases with DBCA which takes care of most of it.

Or you can do as below, but this will do the same as removing the datafiles, redo logs, controlfiles manually.

sqlplus / as sysdba
startup mount exclusive restrict
exit
rman target /
drop database including backups noprompt;
exit


After this, you still have to remove the entry that belongs to the database from /etc/oratab, remove init.ora/spfile, password file from $ORACLE_HOME/dbs, and clean log directories (adump, bdump, cdump, udump).
, I am not sure about which area of science you are referring to (I'm sure the answer would be really different if dealing with biology vs physics for instance...)

Anyway, as a biologist, I will answer from a "biological" point of view:


  How much effort should we put in to understand the application area?


I tend at least to read the previous papers from the same authors and look for a few review on the subject if I am not too familiar with it. This is especially true when dealing with new techniques I don't know, because I need to understand if they did all the proper controls etc. 


  How much time should I spend on a report?


As much as needed (OK, dumb answer, I know! :P)
In general I would not like someone reviewing my paper to do an approximative job just because he/she has other things to do, so I try not to do it myself.


  How picky are you when looking at figures/tables.


Quite picky. Figures are the first thing you look at when browsing through a paper. They need to be consistent (e.g. right titles on the axes, correct legend etc.). On occasion I have suggested to use a different kind of plot to show data when I thought the one used was not the best. This happens a lot in biology, a field that is dominated by the "barplot +/- SEM" type of graph.
I'm also quite picky on the "materials and methods" section: a perfect statistical analysis on a inherently wrong biological model is completely useless.


  How do you cope with the data not being available.


You just do and trust the Authors, I guess. In many cases in biology there's not much you can do, especially when dealing with things like imaging or animal behaviour and similar. Unless you want people to publish tons of images, videos etc (that you most likely would not go through anyways), but that may be very unpractical. If you think the data are really necessary ask for the authors to provide them as supplementary data/figures.


  Do you try and rerun the analysis used.


Only if I have serious doubts on the conclusions drawn by the authors.
In biology there's often a difference between what is (or not) "statistically significant" and what is "biologically significant". I prefer a thinner statistical analysis with good biological reasoning then the other way around. But again, in the very unlikely event that I were to review a bio-statistics paper (ahah, that would be some fun!!!) I would probably pay much more attention to the stats than to the biology in there. 
, source_sentence: Is it true that for all proofs $\forall x \exists y : R(x, y)$, then $y = y(x)$?

A while back I remember reading a book on functional programming that was leading into some questions about what exactly mathematical proofs were and how they were related to functions. In fact, I just found the book: link to PDF, page 16 (marked page 3). 

The author mentions that if we have a proof that  $\forall x \exists y : R(x, y)$, then, in a sense, because $x$ is bound before $y$, then $y$ is dependent on $x$, and therefore $y = y(x)$. There may be counterexamples. Perhaps $y$ is a constant itself, but then $y(x)$ can also just be a constant function. I'm trying to understand under what circumstances this is not true.

Example:

Let $f : \mathbb {R}^n \to \mathbb  R$ be continuous at a point $\vec{v_0}$ and $f(\vec{v_0}) = A < 0$. Prove that there exists a number $\delta > 0$ such that if $\| \vec{v} - \vec{v_0} \| < \delta $ then $f(\vec{v}) < 0$. 

My argument was simply to use the statement of continuity at $\vec{v_0}$ to prove this. That is, since $f$ is continuous, the following holds:

$$ \forall \epsilon > 0 \quad \exists \delta > 0 : \quad \| \vec{v} - \vec{v_0} \| < \delta \Rightarrow \| f(\vec{v}) - f(\vec{v_0}) \| < \epsilon .$$

So, as referred to above, we must have $\delta = \delta(\epsilon)$. Then we can make the replacement $\delta = \delta(|A|)$, and we have the necessary $\delta$ to prove that $f(\vec{v}) < 0$.

I have one main question (the one that closes this), and one side question.


Under what conditions can we assume that $\forall x \exists y : R(x, y)$, then $y = y(x)$?

Are there indirect ways to prove a proposition of the form $\forall x \exists y : R(x, y)$ where $y$ is somehow not dependent on $x$ and also not a constant? Perhaps it varies on $R$, but that is given, I suppose. Certainly all $\epsilon$-$\delta$ proofs that I've come across have been constructive in this way.
Is there a canonical way to prove the proposition above?

I actually struggled with a possible proof for a while before I came up with the technique above. I used this technique in a HW assignment and on my latest Advanced Calculus test. The HW was marked as correct although with the note that it was "indirect", while the test was marked as wrong. Granted, the HW was done by the TA, the test by my prof. I personally don't see a more "direct" way to prove this result.


sentences: This is not a bug.
As we have explored (extensively) in the comments, there are two things happening. The first is that the columns of $U$ are constrained to meet the SVD requirements: each must have unit length and be orthogonal to all the others.  Viewing $U$ as a random variable created from a random matrix $X$ via a particular SVD algorithm, we thereby note that these $k(k+1)/2$ functionally independent constraints create statistical dependencies among the columns of $U$.
These dependencies might be revealed to a greater or lesser extent by studying the correlations among the components of $U$, but a second phenomenon emerges: the SVD solution is not unique.  At a minimum, each column of $U$ can be independently negated, giving at least $2^k$ distinct solutions with $k$ columns.  Strong correlations (exceeding $1/2$) can be induced by changing the signs of the columns appropriately.  (One way to do this is given in my first comment to Amoeba's answer in this thread: I force all the $u_{ii},i=1,\ldots, k$ to have the same sign, making them all negative or all positive with equal probability.)  On the other hand, all correlations can be made to vanish by choosing the signs randomly, independently, with equal probabilities.  (I give an example below in the "Edit" section.)
With care, we can partially discern both these phenomena when reading scatterplot matrices of the components of $U$.  Certain characteristics--such as the appearance of points nearly uniformly distributed within well-defined circular regions--belie a lack of independence.  Others, such as scatterplots showing clear nonzero correlations, obviously depend on choices made in the algorithm--but such choices are possible only because of the lack of independence in the first place.
The ultimate test of a decomposition algorithm like SVD (or Cholesky, LR, LU, etc.) is whether it does what it claims.  In this circumstance it suffices to check that when SVD returns the triple of matrices $(U, D, V)$, that $X$ is recovered, up to anticipated floating point error, by the product $UDV^\prime$; that the columns of $U$ and of $V$ are orthonormal; and that $D$ is diagonal, its diagonal elements are non-negative, and are arranged in descending order.  I have applied such tests to the svd algorithm in R and have never found it to be in error.  Although that is no assurance it is perfectly correct, such experience--which I believe is shared by a great many people--suggests that any bug would require some extraordinary kind of input in order to be manifest.
What follows is a more detailed analysis of specific points raised in the question.

Using R's svd procedure, first you can check that as $k$ increases, the correlations among the coefficients of $U$ grow weaker, but they are still nonzero.  If you simply were to perform a larger simulation, you would find they are significant. (When $k=3$, 50000 iterations ought to suffice.) Contrary to the assertion in the question, the correlations do not "disappear entirely."
Second, a better way to study this phenomenon is to go back to the basic question of independence of the coefficients.  Although the correlations tend to be near zero in most cases, the lack of independence is clearly evident.  This is made most apparent by studying the full multivariate distribution of the coefficients of $U$.  The nature of the distribution emerges even in small simulations in which the nonzero correlations cannot (yet) be detected.  For instance, examine a scatterplot matrix of the coefficients.  To make this practicable, I set the size of each simulated dataset to $4$ and kept $k=2$, thereby drawing $1000$ realizations of the $4\times 2$ matrix $U$, creating a $1000\times 8$ matrix.  Here is its full scatterplot matrix, with the variables listed by their positions within $U$:

Scanning down the first column reveals an interesting lack of independence between $u_{11}$ and the other $u_{ij}$: look at how the upper quadrant of the scatterplot with $u_{21}$ is nearly vacant, for instance; or examine the elliptical upward-sloping cloud describing the $(u_{11}, u_{22})$ relationship and the downward-sloping cloud for the $(u_{21}, u_{12})$ pair.  A close look reveals a clear lack of independence among almost all of these coefficients: very few of them look remotely independent, even though most of them exhibit near-zero correlation.
(NB: Most of the circular clouds are projections from a hypersphere created by the normalization condition forcing the sum of squares of all components of each column to be unity.)
Scatterplot matrices with $k=3$ and $k=4$ exhibit similar patterns: these phenomena are not confined to $k=2$, nor do they depend on the size of each simulated dataset: they just get more difficult to generate and examine.
The explanations for these patterns go to the algorithm used to obtain $U$ in the singular value decomposition, but we know such patterns of non-independence must exist by the very defining properties of $U$: since each successive column is (geometrically) orthogonal to the preceding ones, these orthogonality conditions impose functional dependencies among the coefficients, which thereby translate to statistical dependencies among the corresponding random variables.

Edit
In response to comments, it may be worth remarking on the extent to which these dependence phenomena reflect the underlying algorithm (to compute an SVD) and how much they are inherent in the nature of the process.
The specific patterns of correlations among coefficients depend a great deal on arbitrary choices made by the SVD algorithm, because the solution is not unique: the columns of $U$ may always independently be multiplied by $-1$ or $1$.  There is no intrinsic way to choose the sign.  Thus, when two SVD algorithms make different (arbitrary or perhaps even random) choices of sign, they can result in different patterns of scatterplots of the $(u_{ij}, u_{i^\prime j^\prime})$ values.  If you would like to see this, replace the stat function in the code below by

stat <- function(x) {
  i <- sample.int(dim(x)[1]) # Make a random permutation of the rows of x
  u <- svd(x[i, ])$u         # Perform SVD
  as.vector(u[order(i), ])   # Unpermute the rows of u
}

This first randomly re-orders the observations x, performs SVD, then applies the inverse ordering to u to match the original observation sequence.  Because the effect is to form mixtures of reflected and rotated versions of the original scatterplots, the scatterplots in the matrix will look much more uniform. All sample correlations will be extremely close to zero (by construction: the underlying correlations are exactly zero).  Nevertheless, the lack of independence will still be obvious (in the uniform circular shapes that appear, particularly between $u_{i,j}$ and $u_{i,j^\prime}$).
The lack of data in some quadrants of some of the original scatterplots (shown in the figure above) arises from how the R SVD algorithm selects signs for the columns.
Nothing changes about the conclusions.  Because the second column of $U$ is orthogonal to the first, it (considered as a multivariate random variable) is dependent on the first (also considered as a multivariate random variable).  You cannot have all the components of one column be independent of all the components of the other; all you can do is to look at the data in ways that obscure the dependencies--but the dependence will persist.

Here is updated R code to handle the cases $k\gt 2$ and draw a portion of the scatterplot matrix.
k <- 2    # Number of variables
p <- 4    # Number of observations
n <- 1e3  # Number of iterations
stat <- function(x) as.vector(svd(x)$u)
Sigma <- diag(1, k, k); Mu <- rep(0, k)
set.seed(17)
sim <- t(replicate(n, stat(MASS::mvrnorm(p, Mu, Sigma))))
colnames(sim) <- as.vector(outer(1:p, 1:k, function(i,j) paste0(i,",",j)))
pairs(sim[, 1:min(11, p*k)], pch=".")

, As others have pointed out, you should probably either try to change your boss' mind or find employment where you don't have to put up with such a backwards mentality.

However, in the mean time, you could make your job a bit more interesting by trying to move whatever functionality you can client side and use async calls to trigger stuff to happen on the server.  Think of this as an HTML/JavaScript front end with Webservices (implemented in classic ASP) on the back end.  Developing a RESTful API could be an interesting challenge, there are some tools like JSON parsers for Classic ASP to move data back and forth in a more standard manner and client side templating would let you format data you get from your web services for nicer presentation.  LinkedIn did something similar to unify different back-end technologies.

Once you have a RESTful API, you could try to write some managed web services to emulate functionality of existing Classic ASP stuff.
, You seem to be asking: if for every $x$ there is a $y$ for which $R(x,y)$ holds, can we then define a function $f$ (in the usual sense) such that for every $x$, $R(x,f(x))$ holds?

From a set-theoretic point of view, that statement is in fact none other than the Axiom of Choice.

The Axiom of Choice has many forms, but one form is the following: given a nonempty family of nonempty sets, $S_i$ with $i\in I$ ("nonempty family" means that $I$ is not empty, so we have some sets; "nonempty sets" means that each $S_i$ is not empty), there exists a "choice function": a way to select for each $i\in I$ an element of the corresponding $S_i$. Formally, there exists a function $f\colon I\to \cup S_i$ from the index set to the union of the sets, with the property that $f(i)\in S_i$ for each $i$.

Suppose that by assumption, for every $x$ in our universe $X$, the set $S_x = \{y\mid R(x,y)\}$ is nonempty, and you have a function $f$ with the desried property. The function $f$ will be a "choice function": a function  $f:X \to \cup_{x\in U}S_x$ such that for each $x\in X$, $f(x)\in S_x$. Conversely, a choice function for the family $\{S_x\}_{x\in X}$ will yield and $f$ with the desired property. The existence of such functions for every nonempty family of nonempty sets is precisely the Axiom of Choice. Conversely, suppose that the Axiom of Choice fails, and let $\{S_x\}_{x\in X}$ be a nonempty family of nonempty sets for which no choice function exists. Then let $R(x,y)$ be "$y$ is in $S_x$". For every $x$ there exists a $y$ such that $R(x,y)$, since $S_x\neq\emptyset$. But since we are assuming that the family has no choice function, there does not exist any function $f$ such that for every $x$, $f(x)$ is in $S_x$. Thus, the question not only depends on the Axiom of Choice, but is equivalent to the Axiom of Choice. 

In some circumstances, the exact nature of the sets $S_x$ mean that you do not need to invoke the Axiom of Choice, because we have some "natural way" of defining the choice. For instance, if every set $S_x$ is a subset of the natural numbers, then you can define $f(x)$ to be "the least element of $S_x$".  Another standard example is when $S_x$ is a singleton for every $x$; that is, for each $x$ there exists one and only one $y$ for which $R(x,y)$ holds. In that case, you can define your function $f$ to be "$f(x)$ is the unique $y$ for which $R(x,y)$ holds". 

But in general, the Axiom of Choice is, in some sense, "inherently non-constructive".  And the thing is, the Axiom of Choice is independent of the usual axioms of set theory. Like the parallel postulate in geometry, you can accept it or not at your taste (the consequences of accepting and of not accepting it are different; for example, if you accept the Axiom of Choice, then that means that there is a way to slice a ball of radius 1 into a finite number of parts, and then rearrange the parts without changing their shape or size, into two balls of radius of 1; this is called the Banach-Tarski paradox; on the other hand, if you want all your vector spaces to have bases, then you must accept the Axiom of Choice. In short, it has good consequences and weird/bad consequences). 

Edit: In the situation you have, it is possible to describe a function that will work; however, the "description" is a bit like cheating, in the sense that you can give an explicit rule for what the value of the function will be, but in practice it would be difficult to actually find the value. Namely, for every $\epsilon\gt 0$ you know that the set of $\delta\gt 0$ for which the implication $||v-v_0||\lt\delta\Rightarrow ||f(v)-f(v_0)||\lt\epsilon$ holds is nonempty. In particular, by the Archimedean property, the set $A_{\epsilon}=\{n\in\mathbb{N}| ||v-v_0||\lt\frac{1}{n} \Rightarrow ||f(v)-f(v_0)||\lt\epsilon\}$ is nonempty. This is a nonempty set of positive integers, so it has a least element. You can then define your function $\mathbf{f}\colon (0,\infty)\to(0,\infty)$ by the "formula" $\mathbf{f}(\epsilon)=1/\min(A_{\epsilon})$, the reciprocal of the smallest $n$ in the set. This definition would not require you to invoke the Axiom of Choice.

As to your question 2: you almost had it, I think, but you should not invoke (implicitly) the Axiom of Choice: pick a specific $\epsilon$ (for example, $|A|/2$). You know that there exists a $\delta\gt 0$ that works for that epsilon, so you pick any $\delta_0$ that works and use that. That would be the standard way of proving the proposition at hand. If you assume you have some kind of "black-box function" that will hand you a $\delta$ whenever you hand it an $\epsilon$, you are invoking the Axiom of Choice and you are making your proof somewhat non-constructive. 
, source_sentence: I have the following code in my microcontroler program:

int analogValue = ADCH;        // ADC Data Register

//
// Simple analog comparator. 
// If analogValue lower than threshold then toggle output high,
// Otherwise toggle it low.
//
if ( analogValue > 128 ) {
    PORTB = 0;                 // Port B Data Register
} else {
    PORTB = _BS( outputPin );  // Port B Data Register
}


Where: 


ADCH is the register that contains the value from the ADC
PORTB is a dital output port that toggles an LED


Looking at the resulting assembly code, I noticed that it is doing a 16 bit compare (lines 40-44) where strictly speaking only 8 bits would have been sufficient:

40:   90 e0           ldi     r25, 0x00       ; 0
42:   81 38           cpi     r24, 0x81       ; 129
44:   91 05           cpc     r25, r1
46:   14 f0           brlt    .+4             ; 0x4c <__SREG__+0xd>
48:   18 ba           out     0x18, r1        ; PORTB
4a:   f5 cf           rjmp    .-22            ; 0x36 <__CCP__+0x2>
4c:   28 bb           out     0x18, r18       ; PORTB
4e:   f3 cf           rjmp    .-26            ; 0x36 <__CCP__+0x2>


I realize I declared analogValue as int, which indeed is 16 bit on AVR, but ...

How can I instruct the compiler to use 8 bit comparison? The Arduino IDE allows me to use byte, but avr-gcc by default doesn't.

Check this page for the complete program and its disassembled resulting code.

EDIT1:

Changing int to char changes the assembly code to:

14:   11 24           eor     r1, r1          ; r1 = 0
3e:   18 ba           out     0x18, r1        ; PORTB


Basically skipping the test entirely.

EDIT2: (Thnx: Wouter van Ooijen)

Changing int to unsigned char changes the assembly code to:

3c:   85 b1           in      r24, 0x05       ; ADCH
3e:   ...
40:   87 fd           sbrc    r24, 7          ; compare < 128 (well optimized)
42:   02 c0           rjmp    .+4             ; 0x48 <__SREG__+0x9>
44:   18 ba           out     0x18, r1        ; 24
46:   f7 cf           rjmp    .-18            ; 0x36 <__CCP__+0x2>
48:   98 bb           out     0x18, r25       ; 24
4a:   f5 cf           rjmp    .-22            ; 0x36 <__CCP__+0x2>


sentences: Entropy is a function of the distribution. That is, the process used to generate a byte stream is what has entropy, not the byte stream itself. If I give you the bits 1011, that could have anywhere from 0 to 4 bits of entropy; you have no way of knowing that value.

Here is the definition of Shannon entropy. Let $X$ be a random variable that takes on the values $x_1,x_2,x_3,\dots,x_n$. Then the Shannon entropy is defined as

$$H(X) = -\sum_{i=1}^{n} \operatorname{Pr}[x_i] \cdot \log_2\left(\operatorname{Pr}[x_i]\right)$$

where $\operatorname{Pr}[\cdot]$ represents probability. Note that the definition is a function of a random variable (i.e., a distribution), not a particular value!

So what is the entropy in a single flip of a coin? Let $F$ be a random variable representing such. There are two events, heads and tails, each with probability $0.5$. So, the Shannon entropy of $F$ is:

$$H(F) = -(0.5\cdot\log_2 0.5 + 0.5\cdot\log_2 0.5) = -(-0.5 + -0.5) = 1.$$

Thus, $F$ has exactly one bit of entropy, what we expected. 

So, to find how much entropy is present in a byte stream, you need to know how the byte stream is generated and the entropy of any inputs (in the case of PRNGs). Recall that a deterministic algorithm cannot add entropy to an input, only take it away, so the entropy of all inputs to a deterministic algorithm is the maximum entropy possible in the output. 

If you're using a hardware RNG, then you need to know the probabilities associated with the data it gives you, else you cannot formally find the Shannon entropy (though you could give it a lower bound if you know the probabilities of some, but not all, events). 

But note that in any case, you are dependent on the knowledge of the distribution associated with the byte stream. You can do statistical tests, like you mention, to verify that the output "looks random" (from a certain perspective). But you'll never be able to say any more than "it looks pretty uniformly distributed to me!". You'll never be able to look at a bitstream without knowing the distribution and say "there are X bits of entropy here."
, I actually think a better practice that avoids this architectural ambiguity is to include <stdint.h> then use declarative types like:


uint8_t for unsigned 8-bit integers
int8_t for signed 8-bit integers
uint16_t for unsigned 16-bit integers
uint32_t for unsigned 32-bit integers


and so on...
, Give them the chance to shine

I've actually had a very similar position for some time but now I think I'm making some progress with the developer. I think in the end it will only be a case of commit shyness but I just told him "I need you to commit and push to the server so I can help you better if you get stuck, and you can help me better to oversee the project", and then he started committing.

Give them responsibilities they can deal with (i.e. below their level) to improve morale.

Give them the chance to fail

It is also important to let them take decisions and fail, it's called learning. Give them the power to take little architectural or design decisions and see how they perform. Once they fail, tell them how they can do better and then sit with them.

Don't get impatient because you know you will do in 5 minutes the job they will do in five hours. Let them do it.

Smaller assignments

Having smaller assignments and clearer objectives will make them feel less lost and attack better the problems at hand; baby steps if you will.

Not all people have enough initiative, sadly, but initiative can be taught.

Be there by not being there

Partly continuing the "let them fail" argument, but always stay in reach (e.g. IM) but not physically. If you stay connected with them by IM, you can sometimes answer them with the correct answer, but other times you can just leave them hanging and pretend you are busy (cruel, I know *evil grin*), and just tell them "brb, google XHR and ActiveX" or "google the compiler error", or "google "... by doing this you are partly making them feel stupid in a good way and progressively loose the fear of researching stuff by themselves.

Provide them with the tools

Do they know about Firebug or Chrome Dev Tools?. How about common js pitfalls?. Are they using a good javascript framework? (e.g. MooTools). You really need to give them the weapons to make his work as fun and productive as possible.

Think of them as your past self

You weren't always a pro, right?. How did you learn?, remember that and try to put them on the same path.

If they are not actually cut for the job...

Ask them if they are having trouble with anything, and let them know their performance is not good enough. Personally I like to be direct with people, and I do tell them that beforehand so there is the least resentment.

They might not know there is a problem if you don't tell them.

pipeline_tag: sentence-similarity
library_name: sentence-transformers
metrics: cosine_accuracy@1, cosine_accuracy@3, cosine_accuracy@5, cosine_accuracy@10, cosine_precision@1, cosine_precision@3, cosine_precision@5, cosine_precision@10, cosine_recall@1, cosine_recall@3, cosine_recall@5, cosine_recall@10, cosine_ndcg@10, cosine_mrr@10, cosine_map@100
model-index: name: BGE base Financial Matryoshka
results: task: type: information-retrieval
name: Information Retrieval
dataset: name: dim 768
type: dim_768
metrics: type: cosine_accuracy@1
value: 0.45952408061026984
name: Cosine Accuracy@1
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name: Cosine Precision@5
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name: Cosine Precision@10
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name: Cosine Ndcg@10
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name: Cosine Mrr@10
verified: no, type: cosine_map@100
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name: Cosine Map@100
verified: no, task: type: information-retrieval
name: Information Retrieval
dataset: name: dim 512
type: dim_512
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name: Cosine Mrr@10
verified: no, type: cosine_map@100
value: 0.5550461100265889
name: Cosine Map@100
verified: no, task: type: information-retrieval
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verified: no, type: cosine_map@100
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name: Cosine Map@100
verified: no, task: type: information-retrieval
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name: Cosine Mrr@10
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name: Cosine Map@100
verified: no
Likes: 0 | All-time downloads: 9
Created: 2025-06-03 17:30:40