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SentenceTransformer(
(0): Transformer({'max_seq_length': 512, 'do_lower_case': False}) with Transformer model: NewModel
(1): Pooling({'word_embedding_dimension': 1024, '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})
(2): Dense({'in_features': 1024, 'out_features': 1024, 'bias': True, 'activation_function': 'torch.nn.modules.linear.Identity'})
)pip install -U sentence-transformers1from sentence_transformers import SentenceTransformer
2
3# Download from the 🤗 Hub
4model = SentenceTransformer("cristiano-sartori/stella_finetuned")
5# Run inference
6sentences = [
7 'Suppose we use the Simplex method to solve the following linear program: \\begin{align*} \\textbf{maximize} \\hspace{0.8cm} & 2x_1 - x_2 \\\\ \\textbf{subject to}\\hspace{0.8cm} & x_1 - x_2 + s_1 = 1 \\\\ \\hspace{0.8cm} & \\hspace{0.85cm}x_1 + s_2 = 4 \\\\ \\hspace{0.8cm} & \\hspace{0.85cm} x_2 + s_3 = 2 \\\\ \\hspace{0.8cm} &\\hspace{-0.8cm} x_1,\\: x_2, \\: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 + x_2 - s_1 \\\\ s_2 &= 3 -x_2 + s_1 \\\\ s_3 &= 2 -x_2 \\\\ \\cline{1-2} z &= 2 + x_2 - 2s_1 \\end{align*} Write the tableau obtained by executing one iteration (pivot) of the Simplex method starting from the above tableau.',
8 "To execute one iteration (pivot) of the Simplex method from the given tableau, we need to follow these key steps:\n\n1. **Identify the entering variable:** This is the variable that will increase to improve the objective function. In our case, we look at the coefficients of the objective function row (z-row). We want to maximize z, so we will select the variable with the highest positive coefficient. In this tableau, the coefficients for \\(x_1\\), \\(x_2\\), and the slack variables \\(s_1\\), \\(s_2\\), and \\(s_3\\) in the objective function are \\(2\\), \\(1\\), \\(-2\\), \\(0\\), and \\(0\\), respectively. The variable \\(x_1\\) has the highest positive coefficient of \\(2\\), so it will be our entering variable.\n\n2. **Identify the leaving variable:** Next, we need to determine which variable will leave the basis. This is done using the minimum ratio test. We look at the constraints and find the ratios of the current solution values to the coefficients of the entering variable in each constraint that has a positive coefficient for that variable.\n\n From the current tableau:\n - For the first equation, \\( x_1 = 1 + x_2 - s_1 \\): The coefficient of \\(x_1\\) is \\(1\\). If we set \\(x_2 = 0\\) and \\(s_1 = 0\\), we can solve for the right-hand side (RHS), which is \\(1\\). The ratio is \\( \\frac{1}{1} = 1\\).\n - For the second equation, \\(s_2 = 3 - x_2 + s_1\\): The coefficient of \\(x_1\\) is \\(1\\). The RHS is \\(3\\), so the ratio is again \\( \\frac{3}{1} = 3\\).\n - For the third equation, \\(s_3 = 2 - x_2\\): The coefficient of \\(x_1\\) is \\(0\\), so we can't use this equation.\n\n The minimum ratio is \\(1\\) from the first equation, so \\(s_1\\) will leave the basis.\n\n3. **Perform the pivot operation:** We pivot on the intersection of the entering variable \\(x_1\\) and the leaving variable \\(s_1\\). We want to express all variables in terms of the new basic variable \\(x_1\\).\n\n The tableau before pivoting is:\n\n \\[\n \\begin{array}{c|c|c|c|c|c}\n & x_1 & x_2 & s_1 & s_2 & s_3 \\\\\n \\hline\n x_1 & 1 & 1 & -1 & 0 & 0 \\\\\n s_2 & 0 & 1 & 1 & 1 & 0 \\\\\n s_3 & 0 & 0 & 0 & 0 & 2 \\\\\n \\hline\n z & 0 & 1 & 2 & 0 & 0 \\\\\n \\end{array}\n \\]\n\n Now we perform row operations to update the tableau:\n\n - **Row for \\(x_1\\)**: This will remain unchanged as it becomes our new basic variable.\n - **Row for \\(s_2\\)**: We will subtract the \\(0\\) multiplied by row \\(x_1\\).\n - **Row for \\(s_3\\)**: This row will also not change as \\(x_1\\) does not appear.\n - **Row for \\(z\\)**: We need to replace the \\(z\\) row to reflect the new basis. The new coefficient of \\(z\\) will be \\(2\\) because we have removed \\(s_1\\) from the basis.\n\nAfter performing these operations, we can represent the new tableau:\n\n\\[\n\\begin{array}{c|c|c|c|c|c}\n & x_1 & x_2 & s_2 & s_3 \\\\\n \\hline\n x_1 & 1 & 1 & -1 & 0 \\\\\n s_2 & 0 & 1 & 1 & 1 \\\\\n s_3 & 0 & 0 & 0 & 2 \\\\\n \\hline\n z & 0 & 1 & 2 & 0 \\\\\n\\end{array}\n\\]\n\nHere, we have successfully updated the tableau after one pivot step of the Simplex method. The new solution reflects a state where \\(x_1\\) is now a basic variable. We will repeat this process until we find the optimal solution.",
9 "Certainly! Let's reevaluate the proof that \\( w(S_\\ell) = \\max_{T \\in \\mathcal{I}: |T| = \\ell} w(T) \\) for all \\( \\ell = 1, \\ldots, k \\) in a more rigorous manner.\n\n### Overview of the Proof Structure\n\nWe need to show that for each \\( \\ell \\), the greedy algorithm produces a set \\( S_\\ell \\) of elements with the maximum weight among all independent sets of size \\( \\ell \\) in a matroid \\( \\mathcal{M} = (E, \\mathcal{I}) \\) with a weight function \\( w: E \\rightarrow \\mathbb{R} \\).\n\n### Step 1: Base Case\n\nFor \\( \\ell = 1 \\):\n- \\( S_1 = \\{s_1\\} \\) is the element with the highest weight.\n- The maximum weight of any independent set of size 1 is indeed \\( \\max_{e \\in E} w(e) \\), and since \\( s_1 \\) is chosen to be the element with the maximum weight, we have:\n\n\\[\nw(S_1) = w(s_1) = \\max_{T \\in \\mathcal{I}: |T| = 1} w(T)\n\\]\n\nThis base case holds true.\n\n### Step 2: Induction Hypothesis\n\nNow we assume that for some \\( \\ell \\) where \\( 1 \\leq \\ell < k \\):\n\n\\[\nw(S_\\ell) = \\max_{T \\in \\mathcal{I}: |T| = \\ell} w(T)\n\\]\n\nThis means that the greedy choice up to \\( \\ell \\) produces a maximum weight independent set of size \\( \\ell \\).\n\n### Step 3: Induction Step\n\nWe need to prove that:\n\n\\[\nw(S_{\\ell + 1}) = \\max_{T \\in \\mathcal{I}: |T| = \\ell + 1} w(T)\n\\]\n\nLet \\( S_{\\ell + 1} = \\{s_1, s_2, \\ldots, s_{\\ell + 1}\\} \\).\n\n#### Constructing the Set \\( S_{\\ell + 1} \\)\n\nTo show this equality, we will analyze any independent set \\( T \\) of size \\( \\ell + 1 \\).\n\n1. **Case 1: \\( s_{\\ell + 1} \\notin T \\)**\n\n If \\( s_{\\ell + 1} \\) is not in \\( T \\), then \\( T \\) must consist of elements among \\( \\{s_1, s_2, \\ldots, s_\\ell\\} \\) and possibly other elements from \\( E \\). Since \\( S_\\ell \\) consists of the elements with the highest weights, we can state:\n\n \\[\n w(T) \\leq w(S_\\ell)\n \\]\n\n By the induction hypothesis, this means:\n\n \\[\n w(T) \\leq \\max_{T' \\in \\mathcal{I}: |T'| = \\ell} w(T') = w(S_\\ell)\n \\]\n\n2. **Case 2: \\( s_{\\ell + 1} \\in T \\)**\n\n In this case, we can remove \\( s_{\\ell + 1} \\) from \\( T \\) to obtain an independent set \\( T' \\) of size \\( \\ell \\). Thus, we have:\n\n \\[\n w(T) = w(T') + w(s_{\\ell + 1})\n \\]\n\n Since \\( T' \\) is an independent set, we can apply the induction hypothesis, which gives us:\n\n \\[\n w(T') \\leq w(S_\\ell)\n \\]\n\n Therefore:\n\n \\[\n w(T) = w(T') + w(s_{\\ell + 1}) \\leq w(S_\\ell) + w(s_{\\ell + 1})\n \\]\n\n### Conclusion from the Induction Step\n\nTo compare \\( w(S_{\\ell + 1}) \\) with \\( w(T) \\), we note:\n\n- \\( S_{\\ell + 1} \\) contains the highest weight elements, and thus:\n\n\\[\nw(S_{\\ell + 1}) = w(S_\\ell) + w(s_{\\ell + 1})\n\\]\n\nSince \\( s_{\\ell + 1} \\) is the highest weight among the remaining elements, we establish the following:\n\n- If \\( s_{\\ell + 1} \\) is included in \\( T \\), we have \\( w(T) \\leq w(S_{\\ell + 1}) \\).\n- If \\( s_{\\ell + 1} \\) is not included, \\( w(T) \\leq w(S_\\ell) \\) which is already less than or equal to \\( w(S_{\\ell + 1}) \\).\n\nThus, we conclude that:\n\n\\[\nw(S_{\\ell + 1}) = \\max_{T \\in \\mathcal{I}: |T| = \\ell + 1} w(T)\n\\]\n\n### Final Remarks\n\nThe proof holds without any assumptions being violated. Each step has been justified based on the properties of the matroid and the greedy choice made at each step, ensuring that the maximum weight independent set is correctly identified for all sizes from 1 to \\( k \\).\n\nThis completes the proof with careful consideration of all cases and steps involved. Thank you for your patience!",
10]
11embeddings = model.encode(sentences)
12print(embeddings.shape)
13# [3, 1024]
14
15# Get the similarity scores for the embeddings
16similarities = model.similarity(embeddings, embeddings)
17print(similarities.shape)
18# [3, 3]dim_768InformationRetrievalEvaluator with these parameters:
1{
2 "truncate_dim": 768
3}| Metric | Value |
|---|---|
| cosine_accuracy@1 | 0.2772 |
| cosine_accuracy@3 | 0.8807 |
| cosine_accuracy@5 | 0.9298 |
| cosine_accuracy@10 | 0.9825 |
| cosine_precision@1 | 0.2772 |
| cosine_precision@3 | 0.2936 |
| cosine_precision@5 | 0.186 |
| cosine_precision@10 | 0.0982 |
| cosine_recall@1 | 0.2772 |
| cosine_recall@3 | 0.8807 |
| cosine_recall@5 | 0.9298 |
| cosine_recall@10 | 0.9825 |
| cosine_ndcg@10 | 0.6591 |
| cosine_mrr@10 | 0.5512 |
| cosine_map@100 | 0.5524 |
dim_512InformationRetrievalEvaluator with these parameters:
1{
2 "truncate_dim": 512
3}| Metric | Value |
|---|---|
| cosine_accuracy@1 | 0.2807 |
| cosine_accuracy@3 | 0.8772 |
| cosine_accuracy@5 | 0.9298 |
| cosine_accuracy@10 | 0.9825 |
| cosine_precision@1 | 0.2807 |
| cosine_precision@3 | 0.2924 |
| cosine_precision@5 | 0.186 |
| cosine_precision@10 | 0.0982 |
| cosine_recall@1 | 0.2807 |
| cosine_recall@3 | 0.8772 |
| cosine_recall@5 | 0.9298 |
| cosine_recall@10 | 0.9825 |
| cosine_ndcg@10 | 0.659 |
| cosine_mrr@10 | 0.5511 |
| cosine_map@100 | 0.5523 |
dim_256InformationRetrievalEvaluator with these parameters:
1{
2 "truncate_dim": 256
3}| Metric | Value |
|---|---|
| cosine_accuracy@1 | 0.2772 |
| cosine_accuracy@3 | 0.8737 |
| cosine_accuracy@5 | 0.9368 |
| cosine_accuracy@10 | 0.9895 |
| cosine_precision@1 | 0.2772 |
| cosine_precision@3 | 0.2912 |
| cosine_precision@5 | 0.1874 |
| cosine_precision@10 | 0.0989 |
| cosine_recall@1 | 0.2772 |
| cosine_recall@3 | 0.8737 |
| cosine_recall@5 | 0.9368 |
| cosine_recall@10 | 0.9895 |
| cosine_ndcg@10 | 0.6595 |
| cosine_mrr@10 | 0.5497 |
| cosine_map@100 | 0.5504 |
dim_128InformationRetrievalEvaluator with these parameters:
1{
2 "truncate_dim": 128
3}| Metric | Value |
|---|---|
| cosine_accuracy@1 | 0.2947 |
| cosine_accuracy@3 | 0.8702 |
| cosine_accuracy@5 | 0.9333 |
| cosine_accuracy@10 | 0.986 |
| cosine_precision@1 | 0.2947 |
| cosine_precision@3 | 0.2901 |
| cosine_precision@5 | 0.1867 |
| cosine_precision@10 | 0.0986 |
| cosine_recall@1 | 0.2947 |
| cosine_recall@3 | 0.8702 |
| cosine_recall@5 | 0.9333 |
| cosine_recall@10 | 0.986 |
| cosine_ndcg@10 | 0.6667 |
| cosine_mrr@10 | 0.5604 |
| cosine_map@100 | 0.5613 |
dim_64InformationRetrievalEvaluator with these parameters:
1{
2 "truncate_dim": 64
3}| Metric | Value |
|---|---|
| cosine_accuracy@1 | 0.2702 |
| cosine_accuracy@3 | 0.8456 |
| cosine_accuracy@5 | 0.9333 |
| cosine_accuracy@10 | 0.9789 |
| cosine_precision@1 | 0.2702 |
| cosine_precision@3 | 0.2819 |
| cosine_precision@5 | 0.1867 |
| cosine_precision@10 | 0.0979 |
| cosine_recall@1 | 0.2702 |
| cosine_recall@3 | 0.8456 |
| cosine_recall@5 | 0.9333 |
| cosine_recall@10 | 0.9789 |
| cosine_ndcg@10 | 0.6493 |
| cosine_mrr@10 | 0.5396 |
| cosine_map@100 | 0.5411 |
anchor and positive| anchor | positive | |
|---|---|---|
| type | string | string |
| details |
|
|
| anchor | positive |
|---|---|
In the following let $\kappa_{1}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)$ and $\kappa_{2}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)$ be two valid kernels. Show that the following is also valid kernel: $\kappa\left(\mathbf{x}, \mathbf{x}^{\prime}\right)=\kappa_{1}\left(\mathbf{x}, \mathbf{x}^{\prime}\right) \kappa_{2}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)$. | To show that the product of two kernels, (\kappa(\mathbf{x}, \mathbf{x}^{\prime}) = \kappa_1(\mathbf{x}, \mathbf{x}^{\prime}) \kappa_2(\mathbf{x}, \mathbf{x}^{\prime})), is also a valid kernel, we can use the property of positive semi-definiteness.[object Object][object Object]### Step 1: Understanding Kernels[object Object]A function (\kappa(\mathbf{x}, \mathbf{x}^{\prime})) is a kernel if it is symmetric (i.e., (\kappa(\mathbf{x}, \mathbf{x}^{\prime}) = \kappa(\mathbf{x}^{\prime}, \mathbf{x}))) and positive semi-definite (PSD). A function is PSD if for any finite set of points ({\mathbf{x}_1, \mathbf{x}[object Object]{j=1}^n c_i c_j \kappa(\mathbf{x}_i, \mathbf{x}_j} \geq 0.[object Object]][object Object][object Object]### Step 2: Properties of (\kappa_1) and (\kappa_2)[object Object]Given that (\kappa_1) and (\kappa_2) are kernels, we know:[object Object]1. They are symmetric: (\kappa_1(\mathbf{x}, \mathbf{x}^{\prime}) = \kappa_1(\mathbf{x}^{\prime}, \mathbf{x})) and... |
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 understand the components required for the calculation. Cohen's Kappa (( \kappa )) is a statistic that measures inter-rater agreement for categorical items. The formula for Cohen's Kappa 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 (the proportion of times both annotators agreed).[object Object]- ( P_e ) is the expected agreement by chance.[object Object][object Object]Let's break down the problem step by step:[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 and ( N - a ) as negative.[object Object] - The second annotator rated ( b ) reviews as positive and ( N - b ) as negative.[object Object] - We know that both annotators agreed on 80 reviews being positive.[object Object][object Object]2. [object Object]:[object Object] - The total number of reviews is ( N ).[object Object] - The observed agreement ( P_o ) is the proportion of reviews where both annotators agreed on the rating. [object Object] - Therefore... |
Consider the (toy) grammar $G$ consisting of the following rules:[object Object]R1: S --> NP VP[object Object]R2: NP --> NN[object Object]R3: NP --> Det NN[object Object]R4: NN --> N[object Object]R5: NN --> NN NN[object Object]R6: NN --> NN PNP[object Object]R7: PNP --> Prep NP[object Object]R8: VP --> V[object Object]R9: VP --> Adv V[object Object]Precisely define the type of grammar G is corresponding to (for that, consider at least the following aspects: dependency-based vs. constituency-based, position in the Chomsky hierarchy, and CNF).[object Object]Justify your answer for each of the aspects you will be mentioning. | To analyze the grammar ( G ) you've provided, we need to consider several aspects: whether it is dependency-based or constituency-based, its position in the Chomsky hierarchy, and whether it is in Chomsky Normal Form (CNF).[object Object][object Object]### 1. Dependency-based vs. Constituency-based[object Object][object Object][object Object]: [object Object]- [object Object] focuses on the relationships between words in a sentence, where a word is dependent on another word (its head).[object Object]- [object Object] organizes sentences into nested structures (constituents) that can be replaced with single units.[object Object][object Object][object Object]:[object Object]The grammar ( G ) is [object Object] because it uses hierarchical structures (Non-terminal symbols like ( S, NP, VP, NN )) to define the grammatical relationships. The rules show that phrases (like Noun Phrases and Verb Phrases) are made up of smaller constituents. For instance, ( S ) is defined as consisting of a noun phrase ( NP ) and a verb phrase ( VP ). This clearly indicates a constituency str... |
MatryoshkaLoss with these parameters:
1{
2 "loss": "MultipleNegativesRankingLoss",
3 "matryoshka_dims": [
4 768,
5 512,
6 256,
7 128,
8 64
9 ],
10 "matryoshka_weights": [
11 1,
12 1,
13 1,
14 1,
15 1
16 ],
17 "n_dims_per_step": -1
18}eval_strategy: epochper_device_train_batch_size: 2per_device_eval_batch_size: 16gradient_accumulation_steps: 16learning_rate: 2e-05num_train_epochs: 4lr_scheduler_type: cosinewarmup_ratio: 0.1bf16: Truetf32: Falseload_best_model_at_end: Trueoptim: adamw_torch_fusedbatch_sampler: no_duplicatesoverwrite_output_dir: Falsedo_predict: Falseeval_strategy: epochprediction_loss_only: Trueper_device_train_batch_size: 2per_device_eval_batch_size: 16per_gpu_train_batch_size: Noneper_gpu_eval_batch_size: Nonegradient_accumulation_steps: 16eval_accumulation_steps: Nonetorch_empty_cache_steps: Nonelearning_rate: 2e-05weight_decay: 0.0adam_beta1: 0.9adam_beta2: 0.999adam_epsilon: 1e-08max_grad_norm: 1.0num_train_epochs: 4max_steps: -1lr_scheduler_type: cosinelr_scheduler_kwargs: {}warmup_ratio: 0.1warmup_steps: 0log_level: passivelog_level_replica: warninglog_on_each_node: Truelogging_nan_inf_filter: Truesave_safetensors: Truesave_on_each_node: Falsesave_only_model: Falserestore_callback_states_from_checkpoint: Falseno_cuda: Falseuse_cpu: Falseuse_mps_device: Falseseed: 42data_seed: Nonejit_mode_eval: Falseuse_ipex: Falsebf16: Truefp16: Falsefp16_opt_level: O1half_precision_backend: autobf16_full_eval: Falsefp16_full_eval: Falsetf32: Falselocal_rank: 0ddp_backend: Nonetpu_num_cores: Nonetpu_metrics_debug: Falsedebug: []dataloader_drop_last: Falsedataloader_num_workers: 0dataloader_prefetch_factor: Nonepast_index: -1disable_tqdm: Falseremove_unused_columns: Truelabel_names: Noneload_best_model_at_end: Trueignore_data_skip: Falsefsdp: []fsdp_min_num_params: 0fsdp_config: {'min_num_params': 0, 'xla': False, 'xla_fsdp_v2': False, 'xla_fsdp_grad_ckpt': False}fsdp_transformer_layer_cls_to_wrap: Noneaccelerator_config: {'split_batches': False, 'dispatch_batches': None, 'even_batches': True, 'use_seedable_sampler': True, 'non_blocking': False, 'gradient_accumulation_kwargs': None}deepspeed: Nonelabel_smoothing_factor: 0.0optim: adamw_torch_fusedoptim_args: Noneadafactor: Falsegroup_by_length: Falselength_column_name: lengthddp_find_unused_parameters: Noneddp_bucket_cap_mb: Noneddp_broadcast_buffers: Falsedataloader_pin_memory: Truedataloader_persistent_workers: Falseskip_memory_metrics: Trueuse_legacy_prediction_loop: Falsepush_to_hub: Falseresume_from_checkpoint: Nonehub_model_id: Nonehub_strategy: every_savehub_private_repo: Nonehub_always_push: Falsegradient_checkpointing: Falsegradient_checkpointing_kwargs: Noneinclude_inputs_for_metrics: Falseinclude_for_metrics: []eval_do_concat_batches: Truefp16_backend: autopush_to_hub_model_id: Nonepush_to_hub_organization: Nonemp_parameters:auto_find_batch_size: Falsefull_determinism: Falsetorchdynamo: Noneray_scope: lastddp_timeout: 1800torch_compile: Falsetorch_compile_backend: Nonetorch_compile_mode: Noneinclude_tokens_per_second: Falseinclude_num_input_tokens_seen: Falseneftune_noise_alpha: Noneoptim_target_modules: Nonebatch_eval_metrics: Falseeval_on_start: Falseuse_liger_kernel: Falseeval_use_gather_object: Falseaverage_tokens_across_devices: Falseprompts: Nonebatch_sampler: no_duplicatesmulti_dataset_batch_sampler: proportional| Epoch | Step | Training Loss | dim_768_cosine_ndcg@10 | dim_512_cosine_ndcg@10 | dim_256_cosine_ndcg@10 | dim_128_cosine_ndcg@10 | dim_64_cosine_ndcg@10 |
|---|---|---|---|---|---|---|---|
| 0.2807 | 10 | 0.1249 | - | - | - | - | - |
| 0.5614 | 20 | 0.8091 | - | - | - | - | - |
| 0.8421 | 30 | 0.0235 | - | - | - | - | - |
| 1.0 | 36 | - | 0.6463 | 0.6516 | 0.6468 | 0.6425 | 0.6363 |
| 1.1123 | 40 | 0.0293 | - | - | - | - | - |
| 1.3930 | 50 | 0.0474 | - | - | - | - | - |
| 1.6737 | 60 | 0.0062 | - | - | - | - | - |
| 1.9544 | 70 | 0.0022 | - | - | - | - | - |
| 2.0 | 72 | - | 0.6535 | 0.6501 | 0.6488 | 0.6488 | 0.6454 |
| 2.2246 | 80 | 0.9144 | - | - | - | - | - |
| 2.5053 | 90 | 0.0139 | - | - | - | - | - |
| 2.7860 | 100 | 0.0019 | - | - | - | - | - |
| 3.0 | 108 | - | 0.6533 | 0.6582 | 0.6523 | 0.6651 | 0.6499 |
| 3.0561 | 110 | 0.3805 | - | - | - | - | - |
| 3.3368 | 120 | 0.0075 | - | - | - | - | - |
| 3.6175 | 130 | 0.0035 | - | - | - | - | - |
| 3.8982 | 140 | 0.0012 | - | - | - | - | - |
| 4.0 | 144 | - | 0.6591 | 0.659 | 0.6595 | 0.6667 | 0.6493 |
1@inproceedings{reimers-2019-sentence-bert,
2 title = "Sentence-BERT: Sentence Embeddings using Siamese BERT-Networks",
3 author = "Reimers, Nils and Gurevych, Iryna",
4 booktitle = "Proceedings of the 2019 Conference on Empirical Methods in Natural Language Processing",
5 month = "11",
6 year = "2019",
7 publisher = "Association for Computational Linguistics",
8 url = "https://arxiv.org/abs/1908.10084",
9}1@misc{kusupati2024matryoshka,
2 title={Matryoshka Representation Learning},
3 author={Aditya Kusupati and Gantavya Bhatt and Aniket Rege and Matthew Wallingford and Aditya Sinha and Vivek Ramanujan and William Howard-Snyder and Kaifeng Chen and Sham Kakade and Prateek Jain and Ali Farhadi},
4 year={2024},
5 eprint={2205.13147},
6 archivePrefix={arXiv},
7 primaryClass={cs.LG}
8}1@misc{henderson2017efficient,
2 title={Efficient Natural Language Response Suggestion for Smart Reply},
3 author={Matthew Henderson and Rami Al-Rfou and Brian Strope and Yun-hsuan Sung and Laszlo Lukacs and Ruiqi Guo and Sanjiv Kumar and Balint Miklos and Ray Kurzweil},
4 year={2017},
5 eprint={1705.00652},
6 archivePrefix={arXiv},
7 primaryClass={cs.CL}
8}