{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-connected-component"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-connected-component-01","title":{"zh-Hant":"連通分量：以完整 BFS 求出有限簡單圖的全部分量","en":"Connected components: finding every component of a finite simple graph by complete BFS"},"summary":{"zh-Hant":"設 G=(V,E)，V={0,1,2,3,4,5,6}，E 恰為 {0,1}、{0,2}、{1,2}、{3,4}、{4,5} 五條邊。先證明「有 path 相連」是 V 上的等價關係，再以完整 BFS 求出每個等價類，證明 BFS 回傳的集合恰為該類，並得出 Q(G)={{0,1,2},{3,4,5},{6}}，其中孤立頂點 6 自成單點分量。","en":"Let G=(V,E) with V={0,1,2,3,4,5,6} and E consisting of exactly the five edges {0,1}, {0,2}, {1,2}, {3,4}, {4,5}. First prove that 'joined by a path' is an equivalence relation on V; then compute each equivalence class by complete BFS, prove that the set BFS returns is exactly that class, and conclude Q(G)={{0,1,2},{3,4,5},{6}}, where the isolated vertex 6 forms a singleton component."},"mko_ids":["mko-connected-component"]}],"title":"Concept derivation guides"}