Model: AShi846/all-MiniLM-L6-v2_rag_ft_e-7
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: Suppose we use the Simplex method to solve the following linear program: \begin{align*} \textbf{maximize} \hspace{0.8cm} & \hspace{0.4cm}4x_1 - 6x_2 + 4x_3 \\ \textbf{subject to}\hspace{0.6cm} & x_1 - 3x_2 + x_3  + s_1  = 1 \\ \hspace{0.8cm} & \hspace{1.90cm}x_1  + s_2  = 8 \\ \hspace{0.8cm} & \hspace{0.65cm} 3x_2  + 2x_3 + s_3 = 6 \\ \hspace{0.8cm} &\hspace{-0.35cm}  x_1,\: x_2, \: x_3, \:s_1, \:s_2, \:s_3 \geq 0 \end{align*} At the current step, we have the following Simplex tableau: \begin{align*} \hspace{1cm} x_1 &= 1 + 3x_2 - x_3 - s_1 \\ s_2 &= 7 -3x_2 + x_3  + s_1  \\ s_3 &= 6 - 3x_2  - 2x_3 \\ \cline{1-2} z &= 4 + 6 x_2   -  4s_1 \end{align*} Write the tableau obtained by executing one iteration (pivot) of the Simplex method starting from the above tableau.
sentences: We want to assign each job to a machine. For any job $j$ which the LP assigns to a single machine $i$, i.e., $x^*_{ij} = 1$, it clearly makes sense to listen to the LP and assign $j$ to $i$. To assign the other jobs, we will use the graph $H$, which has the following properties: \begin{itemize} \item $H$ is acyclic, i.e., a forest. \item Vertices corresponding to the already-assigned jobs are isolated. \item Vertices corresponding to the yet-unassigned jobs have degree at least two. \end{itemize} While $H$ contains an edge, we iteratively do the following: \begin{itemize} \item Choose a machine $i$ whose vertex $a_i$ has degree one in $H$, i.e., there is only a single edge $\{a_i, b_j\}$ incident to $a_i$. (Such a machine exists because $H$ is nonempty, and a nonempty forest contains a degree-one vertex; since job-vertices have degrees either zero or at least two, a degree-one vertex must be a machine-vertex.) \item Assign job $j$ to machine $i$. \item Remove all edges incident to $j$ from $H$. (Note that this preserves all the above properties of $H$.) \end{itemize} Obviously, this procedure will terminate. Note that it will assign all jobs to machines, because as long as there is an unassigned job $j$, the vertex $b_j$ has at least two incident edges in $H$. It remains to reason that for every machine $i$, the sum of processing times of jobs assigned to $i$ (this is called the \textit{makespan} of $i$) is at most $T + \max_{j \in J} p_j$. This is because there are two kinds of jobs assigned to $i$: \begin{itemize} \item Assigned in the beginning: the jobs $j$ which had $x^*_{ij} = 1$. The total processing time of these jobs is \[ \sum_{j\in J: x^*_{ij} = 1} p_j = \sum_{j\in J: x^*_{ij} = 1} x^*_{ij} p_j \le \sum_{j\in J: i \in N(j)} x^*_{ij} p_j \le T. \] \item Assigned later: at most one other job. For note that we only assign a job to $i$ when $a_i$ is of degree-one, and once we do, $a_i$ becomes isolated and no more jobs can be assigned to $i$. This job has processing time at most $\max_{j \in J} p_j$. \end{itemize} Another perspective on the solution is the following. By looking at the makespan bound $T + \max_j p_j$ that we should satisfy, we can notice that it will be fine if we assign the integral jobs to the machines that the LP has selected and the non-integral jobs in such a way that each machine gets at most one. In other words, we just need to prove that the graph $H$ has a matching which matches all non-integral jobs (any such matching will be fine). Our iterative procedure is one way to prove that such a matching exists., 1, It breaks backward compatibility, because the signature changes and "Document" is not a special kind of "String" thus callers will have to be updated, source_sentence: A service is an application component that performs long-running operations, usually in the background. A service doesn't provide a UI.
While reviewing the pull request of a friend you notice that he periodically fetches data from the cloud in his activity? What potential problem this could lead to, and how can you fix it?
sentences: The + operation is not associative.  Notice that due to the nature of our definition, wherein we maintain a finite precision (just 4 bits!), adding a small number to a large one may not produce a change at all. We use this property to construct a counterexample:  val a = Float8(3, 8) // a: Float8 = Float8(3,8) val b = Float8(1, 8) // b: Float8 = Float8(1,8) val c = Float8(1, 10) // c: Float8 = Float8(1,10)  // adding this number does not change c! c + Float8(3, 8) // res2: Float8 = Float8(1,10)  // but a slightly larger one will c + Float8(4, 8) // res3: Float8 = Float8(2,10)  a + (b + c) // res4: Float8 = Float8(1,10) (a + b) + c // res5: Float8 = Float8(2,10), ... a list of strings, each string corresponding to the 8-bit representation of an element if and only
if that element is between 0 and 255 included. The most significant bit is farthest to the left in the
characters sequence.
Hint: The most significant bit represents the largest value in a multiple-bit binary number., The benefit of depending on the latest available minor version means that we will always be up-to-date with the latest bugfixes and security/performance improvements as well. But the problem with this approach is that it could lead to unexpected behaviour which could cause bugs, because two compatible versions can still have slightly different behaviour., source_sentence: Up to which linguistic processing level can each of the following sentences be considered as correct? The glass broke its leg, I no go rain, The cook put cherry stones in the cake, Cars flow beautifully; syntactic, pragmatic, syntactic, semantic, lexical
sentences: The provided 9 rules could be expanded as follows to take simple number agreement into account:
R1.1: S --> NPs VPs
R1.2: S --> NPp VPp
R2.1: NPs --> NNs
R2.2: NPp --> NNp
R3.1: NPs --> Dets NNs
R3.2: NPp --> Detp NNp
R4.1: NNs --> Ns
R4.2: NNp --> Np
R5.1: NNs --> NNs NNs
R5.2: NNp --> NNs NNp
R5.3: NNs --> NNp NNs
R5.4: NNp --> NNp NNp
R6.1: NNs --> NNs PNP
R6.2: NNp --> NNp PNP
R7.1: PNP --> Prep NPs
R7.2: PNP --> Prep NPp
R8.1: VPs --> Vs
R8.2: VPp --> Vp
R9: VPs --> Adv Vs
R9: VPs --> Adv Vs
thus resulting in a set of 20 syntactic rules. 
Note that rule R5 may be expanded in only 2 rules instead of 4 if the assumption that in nominal compounds corresponding to a sequence of several (e.g. "satellite antenna frames"), all the nouns but the last one must be singular:
 R5.1: NNs --> NNs NNs
 R5.2: NNp --> NNs NNp, When we want to prove something by contradiction, we start by assuming that the negation (of whatever we are trying to prove) is true. We said in the classroom that p → q is equivalent to ¬p ∨ q. Therefore the negation of p → q is p ∧¬q.
With that said, let's assume that a^2 is even and a is odd. Since a is odd, a can be written as a=2k+1. Therefore, a^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1. Thus, a^2 is odd, a contradiction!, a (DET)
computer (N)
process (V, N)
programs (V, N)
accurately (ADV)
which leads to 4 solutions., source_sentence: Give some concrete examples of NLP applications that might benefit from the semantic vectorial representations.
sentences: Modify the “candeliver” function.
Function candeliver(m) returns Boolean is return #(ack[m]) > N / 2

Suppose that a correct process delivers m. That means that at least one correct process p “acknowledged” m (rebroadcast m using BestEffortBroadcast). Consequently, all correct processes eventually deliver m from BestEffortBroadcast broadcast by p and rebroadcast m themselves (if they have not done that yet). Hence, every correct process eventually collects at least N/2 acknowledgements and delivers m., No it does not, we have no clue to which separating hyperplane the Perceptron algorithm is converging to.
, Transformers
Transformers don’t have a recurrent computation, so the representations at each time step can directly attend to the representations at other time steps. As a result, it is more effective for modeling long-term dependencies because there is no vanishing gradients effect across time.
Because there is no recurrence, the representations at each time step can be computed in parallel, source_sentence: Show a code snippet which represents the kernel of a Spectre
  attack (use any convenient programming language or assembly).

sentences: Use a step of all-to-all communication.
In particular, very process that gets a message relays it immediately.

upon initialization do delivered := {}
upon RB-broadcast(m) do 
  send(m) to Π \ {p}
  RB-deliver(m)

upon BEB-receive(m) from q do 
  if not m ∈ delivered
    send (m) to Π \ {p, q} 
    RB-deliver(m)
    delivered := delivered ∪ m
  
Agreement: Before RB-delivering m, a correct process p forwards m to all processes. By the properties of perfect channels and the fact that p is correct, all correct processes will eventually receive m and RB-deliver it., Whatever your personal convictions, you have to adapt to the convention used by the project., No, this will force other members to sit through an explanation that's not relevant to them, the colleague was right.
pipeline_tag: sentence-similarity
library_name: sentence-transformers
Likes: 0 | All-time downloads: 11
Created: 2025-06-10 20:32:39