Local Algorithms for Finding Densely Connected Clusters
Authors: Peter Macgregor, He Sun
ICML 2021 | Conference PDF | Archive PDF | Plain Text | LLM Run Details
| Reproducibility Variable | Result | LLM Response |
|---|---|---|
| Research Type | Experimental | In this section we evaluate the performance of our proposed algorithms on both synthetic and real-world data sets. |
| Researcher Affiliation | Academia | School of Informatics, University of Edinburgh, Edinburgh, United Kingdom. Correspondence to: Peter Macgregor <peter.macgregor@ed.ac.uk>, He Sun <h.sun@ed.ac.uk>. |
| Pseudocode | Yes | Algorithm 1 Loc Bipart DC; Algorithm 2 Approximate Pagerank DC; Algorithm 3 dcpush; Algorithm 4 Evo Cut Directed (ECD) |
| Open Source Code | Yes | Our code can be downloaded from https://github.com/pmacg/local-denselyconnected-clusters. |
| Open Datasets | Yes | Real-world Dataset. We demonstrate the significance of our algorithm on the Dyadic Militarised Interstate Disputes Dataset (v3.1) (Maoz et al., 2019)... Finally, we evaluate the algorithms performance on the US Migration Dataset (U.S. Census Bureau, 2000). |
| Dataset Splits | No | The paper discusses using synthetic and real-world datasets but does not explicitly provide details about train/validation/test dataset splits, specific proportions, or cross-validation setups. It mentions averaging results over 10 runs with a random starting vertex, but this is not a data splitting strategy. |
| Hardware Specification | Yes | All experiments were performed on a Lenovo Yoga 2 Pro with an Intel(R) Core(TM) i7-4510U CPU @ 2.00GHz processor and 8GB of RAM. |
| Software Dependencies | No | The paper provides a link to the code repository but does not explicitly list specific software dependencies with their version numbers (e.g., Python, specific libraries, or specialized solvers with versions). |
| Experiment Setup | Yes | Algorithm 1 sets α = β2 / 378, and ϵ = 1 / 20γ. Algorithm 4 sets T = (100φ^2 / 3)^(-1). The synthetic dataset parameters are given, e.g., 'p1 = 0.001, q1 = 0.018'. |