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❗️ Debian has patched 1,313 Linux kernel CVEs in a single security update for Debian 13 "Trixie", possibly its largest ever. The flaws allow privilege escalation, denial of service and information leaks. Fixed in version 6.12.111-1. Kernel CVEs per release have tripled as AI tools scan the codebase.
6 条评论
NaoX | Post-Quantum Chain3 天前
1313 in one update says more about what automated review can now reach than about Trixie itself. Plenty of those bugs had been sitting in kernel code for years, just waiting for something patient enough to walk every path
Krzysztof Hajdamowicz3 天前
I had a shower thought: at that pace of AI, distros are becoming less and less relevant. Current infra is going to hit a trajectory called "hourly builds straight from main branch"
John Noble3 天前
@grok what LLM was used?
alias2 天前
This is wild.
Cybermerge3 天前
1,313 CVEs in one DSA is a backlog clear, not 1,313 fires. Don't triage them one by one; the fix is one package. Get Trixie onto 6.12.111-1 and book the reboot, because a kernel fix isn't live until you boot it. Shared and container hosts first, where local priv-esc bites.
International Cyber Digest3 天前
Link:
![SVM by hand ✍️ ~ 19 steps walkthrough below (Linear vs RBF) Support Vector Machines reigned supreme in machine learning before the deep learning revolution. An SVM predicts with dot products, the same matrix multiplication every model uses. What it does not do is train by backpropagation: it is fitted by convex optimization, so there is no matrix-multiplication backward pass for a GPU to accelerate. I drew and calculated two SVMs by hand: a linear one (top) and an RBF one (bottom), classifying the same two test vectors. Goal: turn six training vectors and their learned coefficients into a prediction, and see what changing the kernel actually changes. = 1. Given = Six training vectors, their labels, and the coefficients and bias already learned. A coefficient of zero means that vector is not a support vector: too far from the boundary to matter. = 2. Linear kernel, test vector 1 = Let us take the dot product of the test vector with every training vector. The dot product stands in for cosine similarity, and the column of results is the first column of the kernel matrix K. = 3. Linear kernel, test vector 2 = We do the same for the second, and K is complete. = 4. Signed weights = Let us multiply each coefficient by its label. The second training vector drops out here, because its coefficient is 0. = 5. Weighted combination = We multiply the signed weights through K and add the bias b. The result is a signed distance to the decision boundary: 17 and 5. = 6. Classify = Let us take the sign. Both are positive. = 7 to 11. RBF kernel, test vector 1 = Now the same picture with a different kernel, in five moves: square the differences, sum them, take the square root for the L2 distance, multiply by minus gamma, and raise e to that power. The negation is what turns a distance into a similarity, and gamma controls how far a single training vector's influence reaches. = 12 to 16. RBF kernel, test vector 2 = We repeat all five. The numbers change, the moves do not. = 17 to 19. Decision boundary, again = Signed weights, weighted combination, sign. Identical arithmetic to steps 4 through 6, on a K that was built a completely different way. The outputs: Linear K, first column = [13, 25, 12, 15, 19, 27] Linear decision values = 17 and 5, both positive RBF decision values = -2 and 1, so negative and positive The takeaway: the kernel is the only thing that changed, and it changed the answer. The linear SVM calls both test vectors positive; the RBF one splits them. Everything after the kernel matrix, the signed weights and the weighted combination and the sign, is the same page of arithmetic twice. 💾 Save this post!](https://image.24vids.com/tw-2080658072842703031/ext_tw_video_thumb/2080658053083394048/pu/img/Lrl4Xyr6GSkbbbrN.jpg)
