Undirected Depth-First Search. CSE 421 Introduction to Algorithms. Undirected Depth-First Search. Directed Depth First Search
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1 CSE Intoduction to Algoithms Undiected Depth-Fist Seach It s not just fo tees Depth Fist Seach and Stongly Connected Components DFS() back edge tee edge if maked then etun; mak ; # := ++count; fo all edges (,w) do DFS(w); Main() count := 0; fo all unmaked do DFS(); Undiected Depth-Fist Seach Diected Depth Fist Seach Key Popeties:. No coss-edges ; only tee- o back-edges. Befoe etuning, DFS() isits all etices eachable fom ia paths though peiously unisited etices Algoithm: Unchanged Key Popeties:. Unchanged. Edge (,w) is: As befoe New Tee-edge if w unisited Back-edge if w isited, #w<#, on stack Coss-edge if w isited, #w<#, not on stack Fowad-edge if w isited, #w># Note: Coss edges only go Right to Left An Application: G has a cycle DFS finds a back edge Clea. Why can t we hae something like this?: Stongly Connected Components Defn: G is stongly connected if fo all u, thee is a (diected) path fom u to and fom to u. [Equialently: thee is a cycle though u and.] Defn: a stongly connected component of G is a maimal stongly connected subgaph.
2 Note: collapsed gaph is a DAG Uses fo SCC s Optimizing compiles need to find loops, which ae SCC s in the pogam flow gaph. If (u,) means pocess u is waiting fo pocess, SCC s show deadlocks. Two Simple SCC Algoithms u, in same SCC iff thee ae paths u & u Tansitie closue: O(n ) DFS fom eey u, : O(ne) = O(n ) Goal: Lemma Find all Stongly Connected Components in linea time, i.e., time O(n+e) (Tajan, ) Befoe etuning, dfs() isits all unisited etices eachable fom only unisited etices eachable fom All become descendants of in the tee. dfs follows all diect out-edges call dfs ecusiely at each by induction on path length, isits all
3 Definition Lemma The oot of an SCC is the fist ete in it isited by DFS. All membes of an SCC ae descendants of its oot. Equialently, the oot is the ete in the SCC with the smallest numbe. all membes ae eachable fom all othes so, all ae eachable fom its oot all ae unisited when oot is isited so, all ae descendants of its oot (Lemma ) Subgoal Definition Can we identify some oot? How about the oot of the fist SCC completely eploed by DFS? Key idea: no eit fom fist SCC (fist SCC is leftmost leaf in collapsed DAG) is an eit fom (fom s subtee) if is not a descendant of, but is the head of a (coss- o back-) edge fom a descendant of (including itself) NOTE: # < # Lemma # oot eits - - -,,,,, - If is not a oot, then has an eit. let be oot of s SCC is a pope ancesto of (Lemma ) let be the fist ete that is not a descendant of on a path. is an eit Co: If has no eit, then is a oot. NB: conese not tue; some oots do hae eits
4 Lemma z? How to Find Eits (in st component) If is the fist oot fom which dfs etuns, then has no eit Suppose is an eit let z be oot of s SCC not eachable fom z, else in same SCC #z # (z ancesto of ; Lemma ) # < # ( is an eit fom ) #z < #, no z path, so etun fom z fist Contadiction All eits fom hae # < # Suffices to find any of them, e.g. min # Defn: LOW() = min({ # an eit fom } {#}) Calculate inductiely: LOW() = min of: # { LOW(w) w a child of } { # (,) is a back- o coss-edge } w w w 0 # oot eits LOW - - -,,,,, - Finding Othe Components Key idea: No eit fom st SCC nd SCC, ecept maybe to st d SCC, ecept maybe to st and/o nd... Lemma If is not a oot, then has an eit. in s SCC let be oot of s SCC is a pope ancesto of (Lemma ) let be the fist ete that is not a descendant of on a path. is an eit in s SCC Co: If has no eit, then is a oot. Lemma If is the fist oot fom which dfs etuns, then has no eit Suppose is an eit let z be oot of s SCC ecept possibly to the fist (k-) components not eachable fom z, else in same SCC #z # (z ancesto of ; Lemma ) # < # ( is an eit fom ) #z < #, no z path, so etun fom z fist Contadiction in s SCC k th z? i.e., in fist (k-)
5 How to Find Eits (in st component) All eits fom hae # < # Suffices to find any of them, e.g. min # Defn: LOW() = min({ # an eit fom } {#}) Calculate inductiely: LOW() = min of: and not in fist # (k-) components { LOW(w) w a child of } { # (,) is a back- o coss-edge } k th SCC Algoithm # = DFS numbe.low = LOW().scc = component # SCC() # = ete_numbe++;.low = #; push() fo all edges (,w) if #w == 0 then SCC(w);.low = min(.low, w.low) // tee edge else if #w < # && w.scc == 0 then.low = min(.low, #w) // coss- o back-edge if # =.low then // is oot of new scc scc#++; epeat w = pop(); w.scc = scc#; // mak SCC membes until w== Compleity Look at eey edge once Look at eey ete (ecept ia inedge) at most once # oot eits LOW - - -,,,,, - Time = O(n+e)
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