Model: AShi846/all-MiniLM-L6-v2_rag_ft_e-5
Author: AShi846
Pipeline: sentence-similarity
Library: sentence-transformers
Tags: sentence-transformers, safetensors, bert, sentence-similarity, feature-extraction, generated_from_trainer, dataset_size:475, loss:CosineSimilarityLoss, arxiv:1908.10084, base_model:sentence-transformers/all-MiniLM-L6-v2, base_model:finetune:sentence-transformers/all-MiniLM-L6-v2, text-embeddings-inference, endpoints_compatible, region:us
Base models (finetune): id: sentence-transformers/all-MiniLM-L6-v2
Card metadata: tags: sentence-transformers, sentence-similarity, feature-extraction, generated_from_trainer, dataset_size:475, loss:CosineSimilarityLoss
base_model: sentence-transformers/all-MiniLM-L6-v2
widget: source_sentence: Explain how precise exceptions are implemented in
  dynamically-scheduled out-of-order processors.
sentences: $ 	ext{Var}[\wv^	op \xx] = rac1N \sum_{n=1}^N (\wv^	op \xx_n)^2$ %
, Only $x_3$ has a positive coefficient in $z$, we will pivot $x_3$. We have $\nearrow x_3 \longrightarrow \ x_3 \leq \; \infty \ (1),\ x_3 \leq 3\ (2),\ x_3 \leq 2\ (3)$, Thus we use third equality to pivot $x_3$. Hence $x_3=\frac{1}{2}(4+3x_2-s_3)$. And we get \begin{align*} \hspace{1cm} x_1 &= 1 + \frac{1}{2}(4+3x_2-s_3) - s_1 \\ s_2 &= 3 -\frac{1}{2}(4+3x_2-s_3)  + s_1  \\ x_3&=\frac{1}{2}(4+3x_2-s_3) \\ \cline{1-2} z &= 4 - x_2  +  (4+3x_2-s_3) -  4s_1 \end{align*} That is \begin{align*} \hspace{1cm} x_1 &= 3 + \frac{3x_2}{2} -\frac{s_3}{2} - s_1 \\ s_2 &= 1 - \frac{3x_2}{2}  +\frac{s_3}{2} + s_1  \\ x_3&= 2+  \frac{3x_2}{2} -\frac{s_3}{2}  \\ \cline{1-2} z &= 8 + 2x_2  +  -s_3 -  4s_1 \\ x_1& :=3\text{  }x_2:=0\text{  }x_3:=2\text{  }s_1:=0\text{  }s_2:=1\text{  }s_3:=0 \end{align*}, This does not break compatibility, as the method is private so nobody else could call it., source_sentence: How is it possible to compute the average Precision/Recall curves? Explain in detail the
various steps of the computation.
sentences: there are 12 different bigrams (denoting here the whitespace with 'X' to better see it): Xc, Xh,
Xt, at, ca, cu, eX, ha, he, tX, th, ut,, 1. Dynamically scheduled processors have universally more
        physical registers than the typical 32 architectural ones and
        they are used for removing WARs and WAW (name
        dependencies). In VLIW processors, the same renaming must be
        done by the compiler and all registers must be architecturally
        visible.
        2. Also, various techniques essential to improve the
        performance of VLIW processors consume more registers (e.g.,
        loop unrolling or loop fusion). , G consists of syntactic rules.
G should be complemented with lexical rules with the following format:
T --> w, where T is a pre-terminal (i.e. a Part-of-Speech tag) and w is a terminal (i.e. a word)., source_sentence: You have $1$ Euro and your goal is to exchange it to Swiss francs during the next two consecutive days. The exchange rate is an arbitrary function from days to real numbers from the interval $[1,W^2]$, where $W\geq 1$ is known to the algorithm. More precisely, at  day $1$, you learn the exchange rate $x_1 \in [1,W^2]$, where $x_1$ is the amount of Swiss francs you can buy from $1$ Euro. You then need to decide between the following two options: \begin{enumerate}[label=(\roman*)] \item Trade the whole $1$ Euro at day $1$ and receive $x_1$ Swiss francs. \item Wait and trade the whole $1$ Euro at day $2$ at  exchange rate $x_2 \in [1,W^2]$. The exchange rate $x_2$ is known only at day 2, i.e., after you made your decision at day 1. \end{enumerate} In the following two subproblems, we will analyze the competitive ratio of optimal deterministic algorithms. Recall that we say that an online algorithm  is $c$-competitive  if, for any $x_1, x_2 \in [1,W^2]$, it exchanges the $1$ Euro into at least $c \cdot \max\{x_1, x_2\}$ Swiss francs. Show that any deterministic algorithm has  a competitive ratio of at most $1/W$. {\em (In this problem you are asked to prove that any deterministic algorithm has a competitive ratio of at most $1/W$ for the above problem.    Recall that you are allowed to refer to material covered in the lecture notes.)}
sentences: We use the idea of the AMS algorithm. We first describe how Alice Alice calculates the message $m$. Let $\Alg$ be the following procedure: \begin{itemize} \item Select a random $h: [n] \rightarrow \{\pm 1\}$ $4$-wise independent hash function. $h$ takes $O(\log n)$ bits to store. \item Calculate $A = \sum_{i=1}^n h(i) x_i$. \end{itemize} Let $t=6/\epsilon^2$. Alice runs $\Alg$ $t$ times. Let $h_i$ and $A_i$ be the hash function and the quantity calculated by $i$:th invokation of \Alg. Then Alice transmits the information $h_1, A_1, h_2, A_2, \ldots, h_t, A_t$ to Bob. Note that each $h_i$ takes $O(\log n)$ bits to store and each $A_i$ is an integer between $-n^2$ and $n^2$ and so it also takes $O(\log n)$ bits to store. Therefore the message Alice transmits to Bob is $O(\log(n)/\epsilon^2)$ bits. Now Bob calculates the estimate $Z$ as follows: \begin{itemize} \item For $\ell = 1, 2, \ldots, t$, let $Z_\ell = A_\ell +  \sum_{i=1}^n h_\ell(i) y_i$. \item Output $Z = \frac{\sum_{\ell=1}^t Z_\ell^2}{t}.$ \end{itemize} To prove that $Z$ satisfies~\eqref{eq:guaranteeStream}, we first analyze a single $Z_\ell$. First, note that $Z_\ell = A_\ell + \sum_{i=1}^n h_\ell(i)y_i = \sum_{i=1}^n h_\ell(i) (x_i + y_i) = \sum_{i=1}^n h_\ell(i) f_i$, where we let $f_i = x_i + y_i$. And so $Z_\ell = \sum_{i=1}^n h_\ell(i) f_i$ where $h_\ell$ is a random $4$-wise independent hash function. This is exactly the setting of the analysis of the AMS streaming algorithm seen in class. And so over the random selection of the hash function, we know that \begin{align*} \E[Z_\ell^2] = \sum_{i=1}^n f_i^2 = Q \end{align*} and \begin{align*} \Var[Z_\ell^2] \leq 2\left( \sum_{i=1}^n f_i^2 \right)^2 = 2Q^2\,. \end{align*} Therefore, we have that \begin{align*} \E[Z] =  Q \qquad \mbox{and} \qquad \Var[Z] \leq \frac{2 Q^2}{t}\,. \end{align*} So by Chebychev's inequality \begin{align*} \Pr[|Z- Q| \geq \epsilon Q] \leq \frac{2Q^2/t}{\epsilon^2 Q^2} \leq 1/3\,, \end{align*} by the selection of $t = 6/\epsilon^2$., False., Yes, it is possible.      d1>d2: without adding any document, it holds true     d2>d1: adding d3=”aaaa”, source_sentence: One of your colleagues has recently taken over responsibility for a legacy codebase, a library currently used by some of your customers. Before making functional changes, your colleague found a bug caused by incorrect use of the following method in the codebase:

public class User {
    /** Indicates whether the user’s browser, if any, has JavaScript enabled. */
    public boolean hasJavascriptEnabled() { … }

    // … other methods, such as getName(), getAge(), ...
}

Your colleague believes that this is a bad API. You are reviewing the pull request your colleague made to fix this bug. After some discussion and additional commits to address feedback, the pull request is ready. You can either "squash" the pull request into a single commit, or leave the multiple commits as they are. Explain in 1 sentence whether you should "squash" and why.
sentences: causal language modeling
learns to predict the next word, which you would need to generate a story., It is not suitable as the item is not specified properly ("doesn't render well" is not concrete). A bug item has to include details on what is wrong with the user experience., We name “central” the city that we can reach from every other city either directly or through exactly one intermediate city.
Base case (n=2): It obviously holds. Either one of the cities is “central”.
Inductive step: Suppose this property holds for n ≥ 2 cities. We will prove that it will still hold for n+1 cities.

Let n+1 cities, ci, i=0, ..., n, where for every pair of different cities ci, cj, there exists a direct route
(single direction) either from ci to cj or from cj to ci.
We consider only the first n cities, i.e. cities ci, i=0, ..., n-1. According to the inductive step, there
exists one central city among these n cities. Let cj be that city.
We now exclude city cj and consider the rest of the cities. Again, we have n cities, therefore there should exist one city among them that is central. Let ck be that city.
All cities apart from cj and ck can reach cj and ck either directly or through one intermediate city.
Furthermore, there exists a route between cj and ck:
● If the route is directed from cj to ck, then ck is the central city for the n+1 cities.
● If the route is directed from ck to cj, then cj is the central city for the n+1 cities., source_sentence: The data contains information about submissions to a prestigious machine learning conference called ICLR. Columns:
year, paper, authors, ratings, decisions, institution, csranking, categories, authors_citations, authors_publications, authors_hindex, arxiv. The data is stored in a pandas.DataFrame format. 

Create two fields called has_top_company and has_top_institution. The field has_top_company equals 1 if the article contains an author in the following list of companies ["Facebook", "Google", "Microsoft", "Deepmind"], and 0 otherwise. The field has_top_institution equals 1 if the article contains an author in the top 10 institutions according to CSRankings.
sentences: Let $S$ be a minimum $s,t$-cut; then the number of edges cut by $S$ is $\opt$. We shall exhibit a feasible solution $y$ to the linear program such that value of $y$ is $\opt$. This then implies that $\optlp \leq \opt$ as the minimum value of a solution to the linear program is at most the value of $y$. Define $y$ as follows: for each $e\in E$ \begin{align*} y_e = \begin{cases} 1 & \mbox{if $e$ is cut by $S$,}\\ 0 & \mbox{otherwise.} \end{cases} \end{align*} Notice that, by this definition, $\sum_{e\in E} y_e = \opt$. We proceed to show that $y$ is a feasible solution: \begin{itemize} \item for each $e\in E$, we have $y_e \geq 0$; \item for each $p\in P$, we have $\sum_{e\in p} y_e \geq 1$ since any path from $s$ to $t$ must exit the set $S$. Indeed, $S$ contains $s$ but it does not contain $t$, and these edges (that have one end point in $S$ and one end point outside of $S$) have $y$-value equal to $1$. \end{itemize}, 1, Recall that, in the Hedge algorithm we learned in class, the total loss over time is upper bounded by $\sum_{t = 1}^T m_i^t + \frac{\ln N}{\epsilon} + \epsilon T$. In the case of investments, we want to do almost as good as the best investment. Let $g_i^t$ be the fractional change of the value of $i$'th investment at time $t$. I.e., $g_i^t = (100 + change(i))/100$, and $p_i^{t+1} = p_i^{t} \cdot g_i^t$. Thus, after time $T$, $p_i^{T+1} = p_i^1 \prod_{t = 1}^T g_i^t$. To get an analogous bound to that of the Hedge algorithm, we take the logarithm. The logarithm of the total gain would be $\sum_{t=1}^T \ln g_i^t$. To convert this into a loss, we multiply this by $-1$, which gives a loss of $\sum_{t=1}^T (- \ln g_i^t)$. Hence, to do almost as good as the best investment, we make our cost vectors to be $m_i^t = - \ln g_i^t$. Now, from the analysis of Hedge algorithm in the lecture, it follows that for all $i \in [N]$, $$\sum_{t = 1}^T p^{(t)}_i \cdot m^{(t)} \leq \sum_{t = 1}^{T} m^{(t)}_i + \frac{\ln N}{\epsilon} + \epsilon T.$$ Taking the exponent in both sides, We have that \begin{align*} \exp \left( \sum_{t = 1}^T p^{(t)}_i \cdot m^{(t)} \right) &\leq \exp \left( \sum_{t = 1}^{T} m^{(t)}_i + \frac{\ln N}{\epsilon} + \epsilon T \right)\\ \prod_{t = 1}^T \exp( p^{(t)}_i \cdot m^{(t)} ) &\leq \exp( \ln N / \epsilon + \epsilon T) \prod_{t = 1}^T \exp(m^t_i) \\ \prod_{t = 1}^T \prod_{i \in [N]} (1 / g_i^t)^{p^{(t)}_i} &\leq \exp( \ln N / \epsilon + \epsilon T) \prod_{t = 1}^{T} (1/g^{(t)}_i) \end{align*} Taking the $T$-th root on both sides, \begin{align*} \left(\prod_{t = 1}^T \prod_{i \in [N]} (1 / g_i^t)^{p^{(t)}_i} \right)^{(1/T)} &\leq \exp( \ln N / \epsilon  T + \epsilon ) \left( \prod_{t = 1}^{T} (1/g^{(t)}_i) \right)^{(1/T)}. \end{align*} This can be interpreted as the weighted geometric mean of the loss is not much worse than the loss of the best performing investment.
pipeline_tag: sentence-similarity
library_name: sentence-transformers
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Created: 2025-06-10 20:31:53