You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
Copy file name to clipboardExpand all lines: .claude/skills/review-pipeline/SKILL.md
+1-1Lines changed: 1 addition & 1 deletion
Display the source diff
Display the rich diff
Original file line number
Diff line number
Diff line change
@@ -186,7 +186,7 @@ Invoke `/review-quality` (file: `.claude/skills/review-quality/SKILL.md`) with t
186
186
- Classify as: `confirmed` / `not reproducible in current worktree`
187
187
- For confirmed issues, note severity and recommended fix
188
188
189
-
**Do NOT fix any issues.** Only report them.
189
+
**Do NOT fix any issues.** Only report them. When dispatching the agentic-test subagent, explicitly instruct it: "This is a read-only review run. Do NOT offer to fix issues, do NOT select option (a) 'Review together and fix', and do NOT modify any files. Report findings only and stop after generating the report."
Given an undirected graph $G = (V, E)$ and an integer $k$, determine whether there exists a subset $K subset.eq V$ with $|K| >= k$ such that every pair of distinct vertices in $K$ is adjacent.
1447
+
][
1448
+
$k$-Clique is the classical decision version of Clique, one of Karp's original NP-complete problems @karp1972 and listed as GT19 in Garey and Johnson @garey1979. Unlike Maximum Clique, the threshold $k$ is part of the input, so this formulation is the natural target for decision-to-decision reductions such as $3$SAT $arrow.r$ Clique. The best known exact algorithm matches Maximum Clique via the complement reduction to Maximum Independent Set and runs in $O^*(1.1996^n)$@xiao2017.
1449
+
1450
+
*Example.* Consider the house graph $G$ with $n = #nv$ vertices, $|E| = #ne$ edges, and threshold $k = #k$. The set $K = {#K.map(i=>$v_#i$).join(", ")}$ is a valid witness because all three pairs #clique-edges.map(((u, v)) =>$(v_#u, v_#v)$).join(", ") are edges, so $|K| = 3 >= #k$ and this is a YES instance. This witness is unique, and no $4$-clique exists because every vertex outside $K$ misses at least one edge to the other selected vertices.
caption: [The house graph with satisfying witness $K = {#K.map(i=>$v_#i$).join(", ")}$ for $k = #k$. The selected vertices and their internal clique edges are highlighted in blue.],
0 commit comments