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arXiv 2609.38161cs.LGcs.DM

LLM图重构中失真的谱理论:尖锐界与实证刻画

A Spectral Theory of Distortion in LLM Graph Reconstruction: Sharp Bounds and Empirical Characterization

  • Columbia University(哥伦比亚大学)

机构由 AI 辅助整理,请以论文原文为准。

Jianru Shen

中文总结 AI 辅助

本文提出LLM图重构失真的谱理论,证明Wasserstein距离被边数净变化和对称差所夹住,并实证刻画了135个重构中的单侧与混合编辑模式。

中文摘要 AI 辅助

语言模型对图重构的评估通常报告原始图与重构图之间的单一聚合距离。我们证明,对于拉普拉斯谱之间的Wasserstein距离,这样的摘要被两个边数所夹住:下界为边数的净变化,上界为对称差,每个都乘以$2/n$,其中$n$是顶点数。该界限是尖锐的:当重构仅添加边或仅删除边时,其两端恰好重合,并且在该类上,距离是重新缩放的边数,它不说明哪些边发生了变化。当两端不同时,距离与下界之间的残差仅在重构既发明又丢失边时为正,这使其成为仅从报告的摘要中可计算的混合编辑的证书。我们在45个合成图上由三个开放权重模型产生的135个重构中刻画了这些区域。七十七个输出是单侧的,29个混合输出的$X > 0$,包括边数恰好保持不变而同时发明和丢失十九条边的情况。这三个模型在编辑策略上有所不同,从复制输入到以大量幻觉体积为代价尝试完成,这种区别是聚合失真所不能揭示的。

英文摘要

Evaluations of graph reconstruction by language models typically report a single aggregate distance between the original and the reconstructed graph. We prove that for the Wasserstein distance between Laplacian spectra such a summary is bracketed by two edge counts, the net change in edge number from below and the symmetric difference from above, each scaled by $2/n$ where $n$ is the number of vertices. The bracket is sharp: its two ends coincide exactly when the reconstruction only adds edges or only deletes them, and on that class the distance is a rescaled edge count that says nothing about which edges changed. When the ends differ, the residual between the distance and the lower end is positive only if the reconstruction both invented and lost edges, which turns it into a certificate of mixed editing computable from the reported summaries alone. We characterize these regimes in 135 reconstructions produced by three open-weight models over 45 synthetic graphs. Seventy-seven outputs are one-sided and 29 mixed outputs have $X > 0$, including cases where edge count is exactly preserved while nineteen edges were simultaneously invented and lost. The three models differ in editing policy, ranging from copying the input to attempting completion at the cost of large hallucination volume, a distinction that aggregate distortion does not reveal.

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