Strongly Connected Components. Uses for SCC s. Two Simple SCC Algorithms. Directed Acyclic Graphs

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1 CSE Intoduction to Algoithms Depth Fist Seach and Stongly Connected Components W.L. Ruzzo, Summe 00 Undiected Depth-Fist Seach It s not just fo tees DFS() back edge tee edge if maked then etun; mak ; # := ++count; fo all edges (,w) do DFS(w); Main() count := 0; NB: book uses deceasing fo all unmaked do DFS(); Undiected 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 Diected Depth Fist Seach 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 Easy - back edge (,y) plus tee edges y,, fom a cycle. Why can t we hae something like this?: Lemma Befoe etuning, dfs() isits w iff w is unisited w is eachable fom ia a path though unisited etices dfs follows all diect out-edges call dfs ecusiely at each unisited one by induction on path length, isits all

2 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 cicuit though u and.] Defn: a stongly connected component of G is a maimal stongly connected (ete-induced) subgaph. 0 0 Uses fo SCC s Optimizing compiles: SCC s in pogam flow gaph = loops SCC s in call gaph = mutual ecusion Opeating Systems: If (u,) means pocess u is waiting fo pocess, SCC s show deadlocks. Econometics: SCC s might show highly intedependent sectos of the economy. Etc. Note: collapsed gaph is a DAG 0 Diected Acyclic Gaphs Two Simple SCC Algoithms If we collapse each SCC to a single ete we get a diected gaph with no cycles a diected acyclic gaph o DAG Many poblems on diected gaphs can be soled as follows: Compute SCC s and esulting DAG Do one computation on each SCC Do anothe on the oeall DAG Eample: Speadsheet ealuation u, in same SCC iff thee ae paths u & u Tansitie closue: O(n ) DFS fom eey u, : O(ne) = O(n )

3 Goal: Definition Find all Stongly Connected Components in linea time, i.e., time O(n+e) (Tajan, ) The oot of an SCC is the fist ete in it isited by DFS. Equialently, the oot is the ete in the SCC with the smallest DFS numbe. Lemma Eecise: show that each SCC is a contiguous subtee. Subgoal All membes of an SCC ae descendants of its oot. 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 ) Can we identify some oot? How about the oot of the fist SCC completely eploed (etuned fom) by DFS? Key idea: no eit fom fist SCC (fist SCC is leftmost leaf in collapsed DAG) Definition 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: # < # E: node # cannot hae an eit. 0 # oot eits - - -,,, 0 0,, 0 0 -

4 Lemma : Nonoots hae eits Lemma : No Escaping st Root z? Idea: Follow cycle to oot 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 (contapositie): If has no eit, then is a oot. NB: conese not tue; some oots do hae eits Idea: Eit Bigge Cycle If is the fist oot fom which dfs etuns, then has no eit Poof (by contadiction): 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 0 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: # { LOW(w) w a child of } { # (,) is a back- o coss-edge } st oot : LOW()= w w w 0 # oot eits LOW - - -,,, 0 0,, st oot: LOW()= Finding Othe Components Lemma Key idea: No eit fom st SCC nd SCC, ecept maybe to st d SCC, ecept maybe to st and/o nd... 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. in s SCC

5 Lemma k th z? 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 i.e., in fist (k-) 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== 0 # oot eits LOW - - -,,, 0 0, 0, Compleity Look at eey edge once Look at eey ete (ecept ia inedge) at most once Time = O(n+e) Whee to stat Unlike undiected DFS, stat ete mattes Add oute loop : mak all etices unisited while thee is unisited ete do scc() Eecise: edo eample stating fom anothe ete, e.g. # o # (which become #) 0

6 Low() Low() Eample A B C D dfs# oot eits low() Eample E F