Learning Complex Heterogeneous Multimodal Fake News via Social Latent Network Inference

Authors: Mingxin Li, Yuchen Zhang, Haowei Xu, Xianghua Li, Chao Gao, Zhen Wang

AAAI 2025 | Venue PDF | Archive PDF | Plain Text | LLM Run Details

Reproducibility Variable Result LLM Response
Research Type Experimental Extensive experiments demonstrate that the proposed method outperforms the SOTA in real social media news datasets. [...] In this section, we conduct some extensive experiments to evaluate the proposed framework, HML.
Researcher Affiliation Academia Mingxin Li, Yuchen Zhang, Haowei Xu, Xianghua Li*, Chao Gao, Zhen Wang Northwestern Polytechnical University EMAIL, EMAIL
Pseudocode No The paper describes the proposed method HML in detail using figures, mathematical formulas, and descriptive text. However, it does not include a clearly labeled pseudocode or algorithm block.
Open Source Code Yes 1https://github.com/cgao-comp/HML.
Open Datasets Yes This paper applies two complex heterogeneous multimodal fake news detection datasets, Fake SV (Qi et al. 2023a) and FVC (Papadopoulou et al. 2019). Additionally, to prove the social latent graph inference method can be widely applied to multimodal fake news datasets, this paper additionally uses Twitter (Boididou et al. 2018) and Weibo (Wang et al. 2018) datasets to evaluate the inference task as a plugin.
Dataset Splits Yes To prevent the model from capturing features of news evolution, this paper divides the dataset 80:20 following the Fake SV benchmark.
Hardware Specification Yes For parameters, this paper uses 4 NVIDIA RTX A6000 as GPU.
Software Dependencies No The paper mentions various models and tools used (e.g., BERT, VGGish, VGG19, C3D, Doubao, GPT-4o) but does not provide specific version numbers for software libraries, programming languages, or development environments used for the implementation of HML.
Experiment Setup Yes For parameters, this paper uses 4 NVIDIA RTX A6000 as GPU. Besides, the method achieves optimal performance with λ = 0.5 and pruning ratio e = 0.6.