This is a sentence-transformers model finetuned from allenai/scibert_scivocab_uncased. It maps sentences & paragraphs to a 768-dimensional dense vector space and can be used for semantic textual similarity, semantic search, paraphrase mining, text classification, clustering, and more.

Model Details

Model Description

Model Type: Sentence Transformer
Base model: allenai/scibert_scivocab_uncased
Maximum Sequence Length: 512 tokens
Output Dimensionality: 768 dimensions
Similarity Function: Cosine Similarity

Model Sources

Documentation: Sentence Transformers Documentation
Repository: Sentence Transformers on GitHub
Hugging Face: Sentence Transformers on Hugging Face

Full Model Architecture

SentenceTransformer(
  (0): Transformer({'max_seq_length': 512, 'do_lower_case': False}) with Transformer model: BertModel 
  (1): Pooling({'word_embedding_dimension': 768, 'pooling_mode_cls_token': False, 'pooling_mode_mean_tokens': True, 'pooling_mode_max_tokens': False, 'pooling_mode_mean_sqrt_len_tokens': False, 'pooling_mode_weightedmean_tokens': False, 'pooling_mode_lasttoken': False, 'include_prompt': True})
)

Usage

Direct Usage (Sentence Transformers)

First install the Sentence Transformers library:

pip install -U sentence-transformers

Then you can load this model and run inference.

python

from sentence_transformers import SentenceTransformer

# Download from the 🤗 Hub
model = SentenceTransformer("ML5562/fine-tuned-scibert_scivocab_uncased-4json")
# Run inference
sentences = [
    'A binary prefix-free code $\\Gamma$ is made of four codewords. The first three codewords have codeword lengths $\\ell_1 = 2$, $\\ell_2 = 3$ and $\\ell_3 = 3$. What is the minimum possible length for the fourth codeword?',
    "To find the minimum possible length for the fourth codeword in a binary prefix-free code, we need to consider the properties of prefix-free codes and how they relate to the total number of codewords and their lengths.\n\nA binary prefix-free code is a type of code in which no codeword is a prefix of any other codeword. This property allows us to uniquely decode each codeword without ambiguity.\n\nGiven the three existing codeword lengths:\n- \\(\\ell_1 = 2\\)\n- \\(\\ell_2 = 3\\)\n- \\(\\ell_3 = 3\\)\n\nWe have three codewords of lengths 2, 3, and 3. \n\nNext, we can calculate the total number of binary codewords that can be generated with lengths up to \\(\\ell\\):\n- For length 1: \\(2^1 = 2\\)\n- For length 2: \\(2^2 = 4\\)\n- For length 3: \\(2^3 = 8\\)\n- For length 4: \\(2^4 = 16\\)\n\nThe first step is to determine how many possible codewords we have used with the existing lengths:\n- Length 2 can produce 4 codewords.\n- Length 3 can produce 8 codewords.\n\nHowever, the prefix-free condition restricts the total number of usable codewords. The maximum number of codewords that can be used without violating the prefix-free property is given by the formula:\n\n\\[\n\\sum_{i=1}^{n} 2^{-\\ell_i} \\leq 1\n\\]\n\nwhere \\(\\ell_i\\) are the lengths of codewords, and \\(n\\) is the number of codewords. \n\nFor our current codewords:\n\\[\n\\frac{1}{2^2} + \\frac{1}{2^3} + \\frac{1}{2^3} = \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{8} = \\frac{1}{4} + \\frac{2}{8} = \\frac{1}{4} + \\frac{1}{4} = \\frac{2}{4} = \\frac{1}{2}\n\\]\n\nThis means we have used up half of the allowed capacity for a prefix-free code with 4 codewords. For the total capacity of a 4 codeword prefix-free code, we can have a total of:\n\n\\[\n\\sum_{i=1}^{4} 2^{-\\ell_i} \\leq 1\n\\]\n\nThus, we can have:\n\\[\n\\frac{1}{2^2} + \\frac{1}{2^3} + \\frac{1}{2^3} + \\frac{1}{2^{\\ell_4}} \\leq 1\n\\]\n\nRearranging gives:\n\\[\n\\frac{1}{2^{\\ell_4}} \\leq \\frac{1}{2} \\implies 2^{-\\ell_4} \\leq \\frac{1}{2} \\implies \\ell_4 \\geq 1\n\\]\n\nHowever, we also need to ensure that the fourth codeword does not conflict with the existing codewords. The lengths of the existing codewords are 2 and 3, meaning that codewords of length 1 would not conflict with the existing codewords, since they are shorter.\n\nIf we choose \\(\\ell_4 = 1\\), we can have 2 codewords of length 1 (i.e., '0' and '1'). Thus, it is possible to have a valid prefix-free code with this configuration.\n\nLengths 2 and 3 are already being used, but if we chose \\(\\ell_4 = 2\\), we would still be legal, but the minimum possible length is what we are looking for.\n\nThus, the minimum possible length for the fourth codeword is:\n\n\\[\n\\boxed{1}\n\\]",
    'To determine the minimum length of the fourth codeword in a binary prefix-free code, we need to analyze the constraints imposed by the existing codewords.\n\nA binary prefix-free code ensures that no codeword is a prefix of any other codeword. The lengths of the first three codewords are given as follows:\n\n- Codeword 1: Length \\( \\ell_1 = 2 \\)\n- Codeword 2: Length \\( \\ell_2 = 3 \\)\n- Codeword 3: Length \\( \\ell_3 = 3 \\)\n\n### Understanding the Codeword Capacity\n\nThe total number of possible codewords of length \\( n \\) in a binary code is \\( 2^n \\):\n\n- For \\( \\ell = 2 \\): Possible combinations are \\( 2^2 = 4 \\).\n- For \\( \\ell = 3 \\): Possible combinations are \\( 2^3 = 8 \\).\n\nSince we have three codewords (two of length 3 and one of length 2), we must ensure that the fourth codeword does not conflict with these existing codewords and adheres to the prefix-free property.\n\n### Evaluating Possible Lengths for the Fourth Codeword\n\nLet’s examine the potential lengths for the fourth codeword \\( \\ell_4 \\):\n\n1. **If \\( \\ell_4 = 1 \\)**: There are only 2 possible codewords (0 and 1). These will serve as prefixes for all codewords of length 2 or greater, making this length infeasible.\n\n2. **If \\( \\ell_4 = 2 \\)**: There are 4 possible combinations. Depending on which codeword of length 2 is used, at least one of these could also act as a prefix for the codewords of length 3, hence this length is not valid.\n\n3. **If \\( \\ell_4 = 3 \\)**: There are 8 possible combinations. However, given that we already have two codewords of this length, we need to ensure that the new codeword does not share prefixes with the existing codewords. It might be possible to find suitable unused combinations, but this length is risky due to potential prefix conflicts.\n\n4. **If \\( \\ell_4 = 4 \\)**: There are 16 possible combinations. Since we are only constrained by the existing codewords of lengths 2 and 3, there will be sufficient combinations available that do not conflict with the existing codewords. This length is feasible and safe to use.\n\n### Conclusion\n\nTo maintain the prefix-free property while accommodating existing codewords, the minimum length for the fourth codeword must be at least 4.\n\n**Final Answer: 4.**',
]
embeddings = model.encode(sentences)
print(embeddings.shape)
# [3, 768]

# Get the similarity scores for the embeddings
similarities = model.similarity(embeddings, embeddings)
print(similarities.shape)
# [3, 3]

Evaluation

Metrics

Triplet

Dataset: val-eval
Evaluated with TripletEvaluator

Metric	Value
cosine_accuracy	0.7576

Training Details

Training Dataset

Unnamed Dataset

Size: 131 training samples
Columns: sentence_0, sentence_1, and sentence_2

Approximate statistics based on the first 131 samples:

	sentence_0	sentence_1	sentence_2
type	string	string	string
details	min: 5 tokens mean: 117.4 tokens max: 512 tokens	min: 16 tokens mean: 347.61 tokens max: 512 tokens	min: 11 tokens mean: 346.56 tokens max: 512 tokens

Samples:

sentence_0	sentence_1	sentence_2
`Consider an RSA encryption where the $(p, q)$ are determined as $(53, 61)$. True or false: $(e,d) = (7, 223)$ are valid encoding/decoding exponents.`	To determine if the pair ((e, d) = (7, 223)) are valid encoding/decoding exponents for RSA encryption with primes ( p = 53 ) and ( q = 61 ), we will perform the following steps:[object Object][object Object]1. [object Object]:[object Object] [[object Object] n = p \times q = 53 \times 61 = 3233[object Object] ][object Object][object Object]2. [object Object] (Euler's totient function):[object Object] [[object Object] \phi(n) = (p-1)(q-1) = (53-1)(61-1) = 52 \times 60 = 3120[object Object] ][object Object][object Object]3. [object Object]:[object Object] We check if (\gcd(e, \phi(n)) = 1):[object Object] [[object Object] \gcd(7, 3120) = 1[object Object] ][object Object] Since 7 is a prime number and does not divide 3120, this confirms that ( e = 7 ) is valid.[object Object][object Object]4. [object Object]:[object Object] We need to ensure that:[object Object] [[object Object] d \times e \equiv 1 \mod \phi(n)[object Object] ][object Object] This means calculating ( 223 \times 7 ):[object Object] [[object Object] 223 \times 7 = 1561[object Object] ][object Object] Now check if ( 1561 \equiv 1 \mod 3120):[object Object] [[object Object] 1561 \mod 3120 = 1561[object Object] ][object Object] Since ( 1561 \neq 1 ), it appears that ( d = 223 ) does not s...	To determine whether ((e, d) = (7, 223)) are valid encoding and decoding exponents for RSA encryption using the primes (p = 53) and (q = 61), we need to follow these steps:[object Object][object Object]1. [object Object]:[object Object] [[object Object] n = p \times q = 53 \times 61 = 3233[object Object] ][object Object][object Object]2. [object Object] (Euler's totient function):[object Object] [[object Object] \phi(n) = (p - 1)(q - 1) = (53 - 1)(61 - 1) = 52 \times 60 = 3120[object Object] ][object Object][object Object]3. [object Object]:[object Object] - (e) must be coprime to (\phi(n)) (i.e., (\gcd(e, \phi(n)) = 1)).[object Object] - Calculate (\gcd(7, 3120)):[object Object] - Since 7 is a prime number, we check if it divides 3120.[object Object] - (3120 \div 7 \approx 445.714) (not an integer).[object Object] - So, (\gcd(7, 3120) = 1).[object Object][object Object] Since (e = 7) is coprime to (\phi(n)), it is a valid choice.[object Object][object Object]4. [object Object]:[object Object] - (d) must satisfy the equation (d \cdot e \equiv 1 \mod \phi(n)).[object Object] - We need to check if (7 \cdot 223 \mod 3120 = 1).[object Object][object Object] First, calculate (7 \cdo...
`The first annotator rated {a} reviews as positive and the rest as negative. The second annotator rated {b} reviews as positive and the rest as negative. 80 reviews were rated as positive by both annotators. Compute the quality of the above reference using Cohen's Kappa.Give your answer as a numerical value to three decimal places.`	To compute Cohen's Kappa, we need to follow these steps:[object Object][object Object]1. [object Object][object Object] - Let ( P ) be the number of positive reviews rated by both annotators.[object Object] - Let ( N ) be the number of negative reviews rated by both annotators.[object Object] - Let ( a ) be the number of reviews rated positive by the first annotator.[object Object] - Let ( b ) be the number of reviews rated positive by the second annotator.[object Object] - Let ( C ) be the total number of reviews.[object Object][object Object]2. [object Object][object Object] - True Positive (TP): Reviews rated positive by both annotators = 80[object Object] - False Positive (FP): Reviews rated positive by the first annotator but negative by the second annotator = ( a - P )[object Object] - False Negative (FN): Reviews rated negative by the first annotator but positive by the second annotator = ( b - P )[object Object] - True Negative (TN): Reviews rated negative by both annotators = ( C - a - b + P )[object Object][object Object]3. [object Object][object Object] [[object Object] C = a + (C - a - b + P) + (b - P) + (a - P)[object Object] ...	To compute Cohen's Kappa, we need to assess the level of agreement between two annotators on a set of categorical items, which in this case are reviews. Cohen's Kappa (( \kappa )) quantifies how much agreement there is beyond what would be expected by chance. The formula is:[object Object][object Object][[object Object]\kappa = \frac{P_o - P_e}{1 - P_e}[object Object]][object Object][object Object]Where:[object Object]- ( P_o ) is the observed agreement.[object Object]- ( P_e ) is the expected agreement by chance.[object Object][object Object]To derive these values, let's break down the problem:[object Object][object Object]1. [object Object]:[object Object] - Let ( N ) be the total number of reviews.[object Object] - The first annotator rated ( a ) reviews as positive.[object Object] - The second annotator rated ( b ) reviews as positive.[object Object] - Both annotators agreed on 80 reviews, which were rated positively.[object Object][object Object]2. [object Object]:[object Object] - The first annotator rated ( N - a ) reviews negatively.[object Object] - The second annotator rated ( N - b ) reviews negatively.[object Object] - The number of reviews rated negatively by both annotators can be calculated as:[object Object] [[object Object] \text{Negative ...
`Assume we have $N$ training samples $(\xx_1, y_1), \dots, (\xx_N, y_N)$ where for each sample $i \in {1, \dots, N}$ we have that $\xx_i \in \R^d$ and $y_i \in \R$. For $\lambda \geq 0$, we consider the following loss:[object Object] L_{\lambda}(\ww) = rac{1}{N} \sum_{i = 1}^N (y_i - \xx_i^ op \ww)^2 + \lambda \Vert \ww \Vert_2, and let $C_\lambda = \min_{\ww \in \R^d} L_{\lambda}(\ww)$ denote the optimal loss value.[object Object] Which of the following statements is extbf{true}:`	The correct statement is: [object Object][object Object][object Object][object Object][object Object][object Object]1. [object Object]: The loss function is defined as:[object Object] [[object Object] L_{\lambda}(\ww) = \frac{1}{N} \sum_{i=1}^N (y_i - \xx_i^T \ww)^2 + \lambda \Vert \ww \Vert_2^2[object Object] ][object Object] This consists of two parts: the first part is the mean squared error term, which measures the fit of the model to the data, and the second part is the regularization term, which penalizes the magnitude of the weight vector ( \ww ).[object Object][object Object]2. [object Object]: When ( \lambda ) increases, the regularization term ( \lambda \Vert \ww \Vert_2^2 ) contributes more heavily to the overall loss. For any fixed ( \ww ), as ( \lambda ) increases, ( L_{\lambda}(\ww) ) becomes larger because the penalty for the weight vector ( \ww ) increases. [object Object][object Object]3. [object Object]: For two values ( \lambda_1 < \lambda_2 ), we can analyze the loss:[object Object] [[object Object] L_{\lambda_1}(\ww) = \frac{1}{N...	To analyze the given statements regarding the loss function ( L_{\lambda}(\ww) ) and the optimal loss value ( C_\lambda ), we will evaluate each claim based on our understanding of convexity, optimization, and regularization.[object Object][object Object]1. [object Object][object Object] - When ( \lambda = 0 ), the loss function simplifies to:[object Object] [[object Object] L_{0}(\ww) = \frac{1}{N} \sum_{i=1}^{N} (y_i - \xx_i^T \ww)^2[object Object] ][object Object] This is a quadratic function in ( \ww ) and is convex due to the squared terms. However, the uniqueness of the minimizer depends on whether the matrix ( \frac{1}{N} \sum_{i=1}^{N} \xx_i \xx_i^T ) is positive definite. If the features ( \xx_i ) are linearly independent, then there is a unique solution; otherwise, there could be infinitely many solutions. Therefore, this statement is not necessarily true for all cases.[object Object][object Object]2. [object Object][object Object] - The regularization term ( ...

Loss: TripletLoss with these parameters:
json
1{ 2 "distance_metric": "TripletDistanceMetric.EUCLIDEAN", 3 "triplet_margin": 5 4}

Training Hyperparameters

Non-Default Hyperparameters

eval_strategy: steps
num_train_epochs: 5
multi_dataset_batch_sampler: round_robin

All Hyperparameters

Click to expand

overwrite_output_dir: False
do_predict: False
eval_strategy: steps
prediction_loss_only: True
per_device_train_batch_size: 8
per_device_eval_batch_size: 8
per_gpu_train_batch_size: None
per_gpu_eval_batch_size: None
gradient_accumulation_steps: 1
eval_accumulation_steps: None
torch_empty_cache_steps: None
learning_rate: 5e-05
weight_decay: 0.0
adam_beta1: 0.9
adam_beta2: 0.999
adam_epsilon: 1e-08
max_grad_norm: 1
num_train_epochs: 5
max_steps: -1
lr_scheduler_type: linear
lr_scheduler_kwargs: {}
warmup_ratio: 0.0
warmup_steps: 0
log_level: passive
log_level_replica: warning
log_on_each_node: True
logging_nan_inf_filter: True
save_safetensors: True
save_on_each_node: False
save_only_model: False
restore_callback_states_from_checkpoint: False
no_cuda: False
use_cpu: False
use_mps_device: False
seed: 42
data_seed: None
jit_mode_eval: False
use_ipex: False
bf16: False
fp16: False
fp16_opt_level: O1
half_precision_backend: auto
bf16_full_eval: False
fp16_full_eval: False
tf32: None
local_rank: 0
ddp_backend: None
tpu_num_cores: None
tpu_metrics_debug: False
debug: []
dataloader_drop_last: False
dataloader_num_workers: 0
dataloader_prefetch_factor: None
past_index: -1
disable_tqdm: False
remove_unused_columns: True
label_names: None
load_best_model_at_end: False
ignore_data_skip: False
fsdp: []
fsdp_min_num_params: 0
fsdp_config: {'min_num_params': 0, 'xla': False, 'xla_fsdp_v2': False, 'xla_fsdp_grad_ckpt': False}
fsdp_transformer_layer_cls_to_wrap: None
accelerator_config: {'split_batches': False, 'dispatch_batches': None, 'even_batches': True, 'use_seedable_sampler': True, 'non_blocking': False, 'gradient_accumulation_kwargs': None}
deepspeed: None
label_smoothing_factor: 0.0
optim: adamw_torch
optim_args: None
adafactor: False
group_by_length: False
length_column_name: length
ddp_find_unused_parameters: None
ddp_bucket_cap_mb: None
ddp_broadcast_buffers: False
dataloader_pin_memory: True
dataloader_persistent_workers: False
skip_memory_metrics: True
use_legacy_prediction_loop: False
push_to_hub: False
resume_from_checkpoint: None
hub_model_id: None
hub_strategy: every_save
hub_private_repo: None
hub_always_push: False
gradient_checkpointing: False
gradient_checkpointing_kwargs: None
include_inputs_for_metrics: False
include_for_metrics: []
eval_do_concat_batches: True
fp16_backend: auto
push_to_hub_model_id: None
push_to_hub_organization: None
mp_parameters:
auto_find_batch_size: False
full_determinism: False
torchdynamo: None
ray_scope: last
ddp_timeout: 1800
torch_compile: False
torch_compile_backend: None
torch_compile_mode: None
dispatch_batches: None
split_batches: None
include_tokens_per_second: False
include_num_input_tokens_seen: False
neftune_noise_alpha: None
optim_target_modules: None
batch_eval_metrics: False
eval_on_start: False
use_liger_kernel: False
eval_use_gather_object: False
average_tokens_across_devices: False
prompts: None
batch_sampler: batch_sampler
multi_dataset_batch_sampler: round_robin

Training Logs

Epoch	Step	val-eval_cosine_accuracy
1.0	17	0.6061
2.0	34	0.5758
3.0	51	0.6364
4.0	68	0.6970
5.0	85	0.7576

Framework Versions

Python: 3.11.11
Sentence Transformers: 4.1.0
Transformers: 4.48.3
PyTorch: 2.7.0+cpu
Accelerate: 1.3.0
Datasets: 3.6.0
Tokenizers: 0.21.1

Citation

BibTeX

Sentence Transformers

bibtex

@inproceedings{reimers-2019-sentence-bert,
    title = "Sentence-BERT: Sentence Embeddings using Siamese BERT-Networks",
    author = "Reimers, Nils and Gurevych, Iryna",
    booktitle = "Proceedings of the 2019 Conference on Empirical Methods in Natural Language Processing",
    month = "11",
    year = "2019",
    publisher = "Association for Computational Linguistics",
    url = "https://arxiv.org/abs/1908.10084",
}

TripletLoss