Notice: The reproducibility variables underlying each score are classified using an automated LLM-based pipeline, validated against a manually labeled dataset. LLM-based classification introduces uncertainty and potential bias; scores should be interpreted as estimates. Full accuracy metrics and methodology are described in Coakley et alK. L. Coakley, T. Snelleman, H. Hoos, and O. E. Gundersen, "The embrace of open science: An analysis of a decade of AI research and 56 800 conference papers," Under Review, 2026..
Linear Contextual Bandits With Interference
Authors: Yang Xu, Wenbin Lu, Rui Song
ICML 2025 | Venue PDF | LLM Run Details
| Reproducibility Variable | Result | LLM Response |
|---|---|---|
| Research Type | Experimental | The effectiveness of our approach is demonstrated through simulations and synthetic data generated based on Movie Lens data. |
| Researcher Affiliation | Academia | 1Department of Statistics, North Carolina State University, USA. Correspondence to: Rui Song <EMAIL>. |
| Pseudocode | Yes | Algorithm 1 Linear Contextual Bandits with Interference |
| Open Source Code | Yes | All supplementary code is available at our Github repository. |
| Open Datasets | Yes | The Movie Lens 1M dataset (Harper & Konstan, 2015) contains over 1 million movie ratings from 6k users, aiding movie recommendations based on historical ratings. |
| Dataset Splits | No | Based on the timestamps of each rating and the relative user density, we divided the dataset into T = 200 rounds. |
| Hardware Specification | No | All experiments were conducted on a local computer with 16 GB of memory. |
| Software Dependencies | No | No specific software dependencies with version numbers are mentioned in the paper. |
| Experiment Setup | Yes | The simulation setup of testing coverage probability is as follows. In the estimation of β, the entire process is replicated for B = 1000 times to calculate the empirical coverage. For each replication, we assume there are a total of T = 500 rounds, and we randomly sample the true β from β0 = (2, 3, 1) and β1 = (1, 1, 3) . |