This article presents a refined notion of incompatible JPEG images for a quality factor of 100. It can be used to detect the presence of steganographic schemes embedding in DCT coefficients. We show that, within the JPEG pipeline, the combination of the DCT transform with the quantization function can map several distinct blocks in the pixel domain to the same block in the DCT domain. However, not every DCT block can be obtained: we call those blocks incompatible. In particular, incompatibility can happen when DCT coefficients are manually modified to embed a message. We show that the problem of distinguishing compatible blocks from incompatible ones is an inverse problem with or without solution and we propose two different methods to solve it. The first one is heuristic-based, fast to find a solution if it exists. The second is formulated as an Integer Linear Programming problem and can detect incompatible blocks only for a specific DCT transform in a reasonable amount of time. We show that the probability for a block to become incompatible only relies on the number of modifications. Finally, using the heuristic algorithm we can derive a Likelihood Ratio Test depending on the number of compatible blocks per image to perform steganalysis. We simulate the result of this test and show that it outperforms a deep learning detector e-SRNet for every payload between 0.001 and 0.01 bpp by using only 10% of the blocks from 256x256 images. A Selection-Channel-Aware version of the test is even more powerful and outperforms e-SRNet while using only 1% of the blocks.

翻译：本文提出了一种改进的JPEG不兼容图像概念，适用于质量因子为100的情况，可用于检测嵌入到DCT系数中的隐写方案。我们证明，在JPEG处理流程中，DCT变换与量化函数的组合可以将像素域中的多个不同块映射到DCT域中的同一块。然而，并非所有DCT块都能被获取：我们将那些无法获取的块称为不兼容块。特别地，当人为修改DCT系数以嵌入消息时，不兼容性可能发生。我们展示了区分兼容块与不兼容块的问题是一个有无解的逆问题，并提出了两种解决方法。第一种方法基于启发式规则，能快速找到解（若存在）；第二种方法被表述为整数线性规划问题，可在合理时间内仅针对特定DCT变换检测不兼容块。我们证明，块变为不兼容的概率仅与修改次数有关。最后，利用启发式算法，我们可根据每幅图像中兼容块的数量推导出似然比检验以进行隐写分析。我们模拟了该检验的结果，并证明在256x256图像中仅使用10%的块时，该检验在0.001至0.01 bpp的每个负载下均优于深度学习检测器e-SRNet。考虑选择信道感知的检验版本性能更强，仅使用1%的块即可超越e-SRNet。