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 [1].
Latent Domains Modeling for Visual Domain Adaptation
Authors: Caiming Xiong, Scott McCloskey, Shao-Hang Hsieh, Jason Corso
AAAI 2014 | Venue PDF | LLM Run Details
| Reproducibility Variable | Result | LLM Response |
|---|---|---|
| Research Type | Experimental | In experiment, we test our approach on two common image datasets, the results show that our method outperforms the existing state-of-the-art methods, and also show the superiority of multiple latent domain discovery. |
| Researcher Affiliation | Collaboration | Caiming Xiong SUNY at Buffalo EMAIL Scott Mc Closkey Honeywell ACS Scott.Mc EMAIL Shao-Hang Hsieh SUNY at Buffalo EMAIL Jason J. Corso SUNY at Buffalo EMAIL |
| Pseudocode | No | The paper presents mathematical formulations and derivations, but no pseudocode blocks or algorithms are visually presented. |
| Open Source Code | No | The paper does not explicitly state that its own source code is released or provide a link to it. |
| Open Datasets | Yes | Our experiments use two different datasets: of๏ฌce dataset (Saenko et al. 2010) and bing-caltech dataset (Alessandro Bergamo 2010). |
| Dataset Splits | No | The paper does not provide specific percentages, sample counts, or explicit citations to predefined splits for training, validation, and testing. |
| Hardware Specification | No | The paper does not provide any specific details about the hardware (e.g., GPU models, CPU types, memory) used to run the experiments. |
| Software Dependencies | No | The paper mentions software components like SURF, Random Forest, and SVM, but does not provide specific version numbers for any of these or other relevant software dependencies. |
| Experiment Setup | Yes | In the experiment, we set ฮป = 5 and ฮฒ = 10. |