Depth First Search (DFS) Lecture 16 Thursday, October 26, 2017 - PowerPoint PPT Presentation

pre and post numbers Node u is active in time interval [ pre ( u ) , post ( u )] Proposition For any two nodes u and v , the two intervals [ pre ( u ) , post ( u )] and [ pre ( v ) , post ( v )] are disjoint or one is contained in the other. Proof. Assume without loss of generality that pre ( u ) < pre ( v ) . Then v visited after u . If DFS( v ) invoked before DFS( u ) finished, post ( v ) < post ( u ) . If DFS( v ) invoked after DFS( u ) finished, pre ( v ) > post ( u ) . pre and post numbers useful in several applications of DFS Sariel Har-Peled (UIUC) CS374 10 Fall 2017 10 / 60

pre and post numbers Node u is active in time interval [ pre ( u ) , post ( u )] Proposition For any two nodes u and v , the two intervals [ pre ( u ) , post ( u )] and [ pre ( v ) , post ( v )] are disjoint or one is contained in the other. Proof. Assume without loss of generality that pre ( u ) < pre ( v ) . Then v visited after u . If DFS( v ) invoked before DFS( u ) finished, post ( v ) < post ( u ) . If DFS( v ) invoked after DFS( u ) finished, pre ( v ) > post ( u ) . pre and post numbers useful in several applications of DFS Sariel Har-Peled (UIUC) CS374 10 Fall 2017 10 / 60

DFS in Directed Graphs DFS( G ) Mark all nodes u as unvisited T is set to ∅ time = 0 while there is an unvisited node u do DFS( u ) Output T DFS( u ) Mark u as visited pre ( u ) = ++ time for each edge ( u , v ) in Out ( u ) do if v is not visited add edge ( u , v ) to T DFS( v ) post ( u ) = ++ time Sariel Har-Peled (UIUC) CS374 11 Fall 2017 11 / 60

Example B A C E F D G H Sariel Har-Peled (UIUC) CS374 12 Fall 2017 12 / 60

Example [1, 16] B A C [2, 11] [12, 15] B A C E F D [3, 10] [6, 7] [13, 14] E F D [4, 5] G H [8, 9] G H Sariel Har-Peled (UIUC) CS374 12 Fall 2017 12 / 60

DFS Properties Generalizing ideas from undirected graphs: DFS( G ) takes O ( m + n ) time. 1 Edges added form a branching : a forest of out-trees. Output of 2 DFS ( G ) depends on the order in which vertices are considered. If u is the first vertex considered by DFS ( G ) then DFS ( u ) 3 outputs a directed out-tree T rooted at u and a vertex v is in T if and only if v ∈ rch ( u ) For any two vertices x , y the intervals [ pre ( x ) , post ( x )] and 4 [ pre ( y ) , post ( y )] are either disjoint or one is contained in the other. Note: Not obvious whether DFS( G ) is useful in directed graphs but it is. Sariel Har-Peled (UIUC) CS374 13 Fall 2017 13 / 60

DFS Properties Generalizing ideas from undirected graphs: DFS( G ) takes O ( m + n ) time. 1 Edges added form a branching : a forest of out-trees. Output of 2 DFS ( G ) depends on the order in which vertices are considered. If u is the first vertex considered by DFS ( G ) then DFS ( u ) 3 outputs a directed out-tree T rooted at u and a vertex v is in T if and only if v ∈ rch ( u ) For any two vertices x , y the intervals [ pre ( x ) , post ( x )] and 4 [ pre ( y ) , post ( y )] are either disjoint or one is contained in the other. Note: Not obvious whether DFS( G ) is useful in directed graphs but it is. Sariel Har-Peled (UIUC) CS374 13 Fall 2017 13 / 60

DFS Properties Generalizing ideas from undirected graphs: DFS( G ) takes O ( m + n ) time. 1 Edges added form a branching : a forest of out-trees. Output of 2 DFS ( G ) depends on the order in which vertices are considered. If u is the first vertex considered by DFS ( G ) then DFS ( u ) 3 outputs a directed out-tree T rooted at u and a vertex v is in T if and only if v ∈ rch ( u ) For any two vertices x , y the intervals [ pre ( x ) , post ( x )] and 4 [ pre ( y ) , post ( y )] are either disjoint or one is contained in the other. Note: Not obvious whether DFS( G ) is useful in directed graphs but it is. Sariel Har-Peled (UIUC) CS374 13 Fall 2017 13 / 60

DFS Properties Generalizing ideas from undirected graphs: DFS( G ) takes O ( m + n ) time. 1 Edges added form a branching : a forest of out-trees. Output of 2 DFS ( G ) depends on the order in which vertices are considered. If u is the first vertex considered by DFS ( G ) then DFS ( u ) 3 outputs a directed out-tree T rooted at u and a vertex v is in T if and only if v ∈ rch ( u ) For any two vertices x , y the intervals [ pre ( x ) , post ( x )] and 4 [ pre ( y ) , post ( y )] are either disjoint or one is contained in the other. Note: Not obvious whether DFS( G ) is useful in directed graphs but it is. Sariel Har-Peled (UIUC) CS374 13 Fall 2017 13 / 60

DFS Properties Generalizing ideas from undirected graphs: DFS( G ) takes O ( m + n ) time. 1 Edges added form a branching : a forest of out-trees. Output of 2 DFS ( G ) depends on the order in which vertices are considered. If u is the first vertex considered by DFS ( G ) then DFS ( u ) 3 outputs a directed out-tree T rooted at u and a vertex v is in T if and only if v ∈ rch ( u ) For any two vertices x , y the intervals [ pre ( x ) , post ( x )] and 4 [ pre ( y ) , post ( y )] are either disjoint or one is contained in the other. Note: Not obvious whether DFS( G ) is useful in directed graphs but it is. Sariel Har-Peled (UIUC) CS374 13 Fall 2017 13 / 60

DFS Tree Edges of G can be classified with respect to the DFS tree T as: Tree edges that belong to T 1 A forward edge is a non-tree edges ( x , y ) such that 2 pre ( x ) < pre ( y ) < post ( y ) < post ( x ) . A backward edge is a non-tree edge ( y , x ) such that 3 pre ( x ) < pre ( y ) < post ( y ) < post ( x ) . A cross edge is a non-tree edges ( x , y ) such that the intervals 4 [ pre ( x ) , post ( x )] and [ pre ( y ) , post ( y )] are disjoint. Sariel Har-Peled (UIUC) CS374 14 Fall 2017 14 / 60

Types of Edges A Forward Backward B C D Cross Sariel Har-Peled (UIUC) CS374 15 Fall 2017 15 / 60

Cycles in graphs Question: Given an undirected graph how do we check whether it has a cycle and output one if it has one? Question: Given an directed graph how do we check whether it has a cycle and output one if it has one? Sariel Har-Peled (UIUC) CS374 16 Fall 2017 16 / 60

Using DFS ... ... to check for Acylicity and compute Topological Ordering Question Given G, is it a DAG ? If it is, generate a topological sort. Else output a cycle C . DFS based algorithm: Compute DFS( G ) 1 If there is a back edge e = ( v , u ) then G is not a DAG . Output 2 cyclce C formed by path from u to v in T plus edge ( v , u ) . Otherwise output nodes in decreasing post-visit order. Note: no 3 need to sort, DFS ( G ) can output nodes in this order. Algorithm runs in O ( n + m ) time. Correctness is not so obvious. See next two propositions. Sariel Har-Peled (UIUC) CS374 17 Fall 2017 17 / 60

Using DFS ... ... to check for Acylicity and compute Topological Ordering Question Given G, is it a DAG ? If it is, generate a topological sort. Else output a cycle C . DFS based algorithm: Compute DFS( G ) 1 If there is a back edge e = ( v , u ) then G is not a DAG . Output 2 cyclce C formed by path from u to v in T plus edge ( v , u ) . Otherwise output nodes in decreasing post-visit order. Note: no 3 need to sort, DFS ( G ) can output nodes in this order. Algorithm runs in O ( n + m ) time. Correctness is not so obvious. See next two propositions. Sariel Har-Peled (UIUC) CS374 17 Fall 2017 17 / 60

Using DFS ... ... to check for Acylicity and compute Topological Ordering Question Given G, is it a DAG ? If it is, generate a topological sort. Else output a cycle C . DFS based algorithm: Compute DFS( G ) 1 If there is a back edge e = ( v , u ) then G is not a DAG . Output 2 cyclce C formed by path from u to v in T plus edge ( v , u ) . Otherwise output nodes in decreasing post-visit order. Note: no 3 need to sort, DFS ( G ) can output nodes in this order. Algorithm runs in O ( n + m ) time. Correctness is not so obvious. See next two propositions. Sariel Har-Peled (UIUC) CS374 17 Fall 2017 17 / 60

Back edge and Cycles Proposition G has a cycle iff there is a back-edge in DFS( G ) . Proof. If: ( u , v ) is a back edge implies there is a cycle C consisting of the path from v to u in DFS search tree and the edge ( u , v ) . Only if: Suppose there is a cycle C = v 1 → v 2 → . . . → v k → v 1 . Let v i be first node in C visited in DFS . All other nodes in C are descendants of v i since they are reachable from v i . Therefore, ( v i − 1 , v i ) (or ( v k , v 1 ) if i = 1 ) is a back edge. Sariel Har-Peled (UIUC) CS374 18 Fall 2017 18 / 60

Proof Proposition If G is a DAG and post ( v ) > post ( u ) , then ( u , v ) is not in G. Proof. Assume post ( v ) > post ( u ) and ( u , v ) is an edge in G . We derive a contradiction. One of two cases holds from DFS property. Case 1: [ pre ( u ) , post ( u )] is contained in [ pre ( v ) , post ( v )] . Implies that u is explored during DFS( v ) and hence is a descendent of v . Edge ( u , v ) implies a cycle in G but G is assumed to be DAG! Case 2: [ pre ( u ) , post ( u )] is disjoint from [ pre ( v ) , post ( v )] . This cannot happen since v would be explored from u . Sariel Har-Peled (UIUC) CS374 19 Fall 2017 19 / 60

Example c a b e d g f h Sariel Har-Peled (UIUC) CS374 20 Fall 2017 20 / 60

Part II Strong connected components Sariel Har-Peled (UIUC) CS374 21 Fall 2017 21 / 60

Strong Connected Components ( SCC s) B A C Algorithmic Problem Find all SCC s of a given directed graph. E F D Previous lecture: Saw an O ( n · ( n + m )) time algorithm. This lecture: sketch of a O ( n + m ) time algorithm. G H Sariel Har-Peled (UIUC) CS374 22 Fall 2017 22 / 60

Graph of SCC s B A C B, E, F A, C, D E F D G H Graph of SCC s G SCC G H G: Meta-graph of SCC s Let S 1 , S 2 , . . . S k be the strong connected components (i.e., SCC s) of G. The graph of SCC s is G SCC Vertices are S 1 , S 2 , . . . S k 1 There is an edge ( S i , S j ) if there is some u ∈ S i and v ∈ S j 2 such that ( u , v ) is an edge in G. Sariel Har-Peled (UIUC) CS374 23 Fall 2017 23 / 60

Reversal and SCC s Proposition For any graph G, the graph of SCC s of G rev is the same as the reversal of G SCC . Proof. Exercise. Sariel Har-Peled (UIUC) CS374 24 Fall 2017 24 / 60

SCC s and DAG s Proposition For any graph G, the graph G SCC has no directed cycle. Proof. If G SCC has a cycle S 1 , S 2 , . . . , S k then S 1 ∪ S 2 ∪ · · · ∪ S k should be in the same SCC in G. Formal details: exercise. Sariel Har-Peled (UIUC) CS374 25 Fall 2017 25 / 60

Part III Directed Acyclic Graphs Sariel Har-Peled (UIUC) CS374 26 Fall 2017 26 / 60

Directed Acyclic Graphs Definition 2 3 A directed graph G is a directed acyclic graph ( DAG ) if there is no directed 1 4 cycle in G. Sariel Har-Peled (UIUC) CS374 27 Fall 2017 27 / 60

Is this a DAG? a c b b a k e d c d g f e n f i h k j g p h m l i r j n o p q m t s l r s o v u w u t w q v Sariel Har-Peled (UIUC) CS374 28 Fall 2017 28 / 60

Sources and Sinks 2 3 source sink 1 4 Definition A vertex u is a source if it has no in-coming edges. 1 A vertex u is a sink if it has no out-going edges. 2 Sariel Har-Peled (UIUC) CS374 29 Fall 2017 29 / 60

Simple DAG Properties Proposition Every DAG G has at least one source and at least one sink. Proof. Let P = v 1 , v 2 , . . . , v k be a longest path in G . Claim that v 1 is a source and v k is a sink. Suppose not. Then v 1 has an incoming edge which either creates a cycle or a longer path both of which are contradictions. Similarly if v k has an outgoing edge. G is a DAG if and only if G rev is a DAG . 1 G is a DAG if and only each node is in its own strong 2 connected component. Formal proofs: exercise. Sariel Har-Peled (UIUC) CS374 30 Fall 2017 30 / 60

Simple DAG Properties Proposition Every DAG G has at least one source and at least one sink. Proof. Let P = v 1 , v 2 , . . . , v k be a longest path in G . Claim that v 1 is a source and v k is a sink. Suppose not. Then v 1 has an incoming edge which either creates a cycle or a longer path both of which are contradictions. Similarly if v k has an outgoing edge. G is a DAG if and only if G rev is a DAG . 1 G is a DAG if and only each node is in its own strong 2 connected component. Formal proofs: exercise. Sariel Har-Peled (UIUC) CS374 30 Fall 2017 30 / 60

Simple DAG Properties Proposition Every DAG G has at least one source and at least one sink. Proof. Let P = v 1 , v 2 , . . . , v k be a longest path in G . Claim that v 1 is a source and v k is a sink. Suppose not. Then v 1 has an incoming edge which either creates a cycle or a longer path both of which are contradictions. Similarly if v k has an outgoing edge. G is a DAG if and only if G rev is a DAG . 1 G is a DAG if and only each node is in its own strong 2 connected component. Formal proofs: exercise. Sariel Har-Peled (UIUC) CS374 30 Fall 2017 30 / 60

Topological Ordering/Sorting 2 3 1 2 3 4 Topological Ordering of G 1 4 Graph G Definition A topological ordering / topological sorting of G = ( V , E ) is an ordering ≺ on V such that if ( u , v ) ∈ E then u ≺ v . Informal equivalent definition: One can order the vertices of the graph along a line (say the x -axis) such that all edges are from left to right. Sariel Har-Peled (UIUC) CS374 31 Fall 2017 31 / 60

DAG s and Topological Sort Lemma A directed graph G can be topologically ordered iff it is a DAG . Need to show both directions. Sariel Har-Peled (UIUC) CS374 32 Fall 2017 32 / 60

DAG s and Topological Sort Lemma A directed graph G can be topologically ordered if it is a DAG . Proof. Consider the following algorithm: Pick a source u , output it. 1 Remove u and all edges out of u . 2 Repeat until graph is empty. 3 Exercise: prove this gives topological sort. Exercise: show algorithm can be implemented in O ( m + n ) time. Sariel Har-Peled (UIUC) CS374 33 Fall 2017 33 / 60

Topological Sort: Example c a b e d g f h Sariel Har-Peled (UIUC) CS374 34 Fall 2017 34 / 60

DAG s and Topological Sort Lemma A directed graph G can be topologically ordered only if it is a DAG . Proof. Suppose G is not a DAG and has a topological ordering ≺ . G has a cycle C = u 1 , u 2 , . . . , u k , u 1 . Then u 1 ≺ u 2 ≺ . . . ≺ u k ≺ u 1 ! That is... u 1 ≺ u 1 . A contradiction (to ≺ being an order). Not possible to topologically order the vertices. Sariel Har-Peled (UIUC) CS374 35 Fall 2017 35 / 60

DAG s and Topological Sort Note: A DAG G may have many different topological sorts. Question: What is a DAG with the most number of distinct topological sorts for a given number n of vertices? Question: What is a DAG with the least number of distinct topological sorts for a given number n of vertices? Sariel Har-Peled (UIUC) CS374 36 Fall 2017 36 / 60

Cycles in graphs Question: Given an undirected graph how do we check whether it has a cycle and output one if it has one? Question: Given an directed graph how do we check whether it has a cycle and output one if it has one? Sariel Har-Peled (UIUC) CS374 37 Fall 2017 37 / 60

To Remember: Structure of Graphs Undirected graph: connected components of G = ( V , E ) partition V and can be computed in O ( m + n ) time. Directed graph: the meta-graph G SCC of G can be computed in O ( m + n ) time. G SCC gives information on the partition of V into strong connected components and how they form a DAG structure. Above structural decomposition will be useful in several algorithms Sariel Har-Peled (UIUC) CS374 38 Fall 2017 38 / 60

Part IV Linear time algorithm for finding all strong connected components of a directed graph Sariel Har-Peled (UIUC) CS374 39 Fall 2017 39 / 60

Finding all SCC s of a Directed Graph Problem Given a directed graph G = ( V , E ) , output all its strong connected components. Straightforward algorithm: Mark all vertices in V as not visited. for each vertex u ∈ V not visited yet do find SCC ( G , u ) the strong component of u : Compute rch( G , u ) using DFS ( G , u ) Compute rch( G rev , u ) using DFS ( G rev , u ) SCC ( G , u ) ⇐ rch( G , u ) ∩ rch( G rev , u ) ∀ u ∈ SCC ( G , u ) : Mark u as visited. Running time: O ( n ( n + m )) Is there an O ( n + m ) time algorithm? Sariel Har-Peled (UIUC) CS374 40 Fall 2017 40 / 60

Finding all SCC s of a Directed Graph Problem Given a directed graph G = ( V , E ) , output all its strong connected components. Straightforward algorithm: Mark all vertices in V as not visited. for each vertex u ∈ V not visited yet do find SCC ( G , u ) the strong component of u : Compute rch( G , u ) using DFS ( G , u ) Compute rch( G rev , u ) using DFS ( G rev , u ) SCC ( G , u ) ⇐ rch( G , u ) ∩ rch( G rev , u ) ∀ u ∈ SCC ( G , u ) : Mark u as visited. Running time: O ( n ( n + m )) Is there an O ( n + m ) time algorithm? Sariel Har-Peled (UIUC) CS374 40 Fall 2017 40 / 60

Finding all SCC s of a Directed Graph Problem Given a directed graph G = ( V , E ) , output all its strong connected components. Straightforward algorithm: Mark all vertices in V as not visited. for each vertex u ∈ V not visited yet do find SCC ( G , u ) the strong component of u : Compute rch( G , u ) using DFS ( G , u ) Compute rch( G rev , u ) using DFS ( G rev , u ) SCC ( G , u ) ⇐ rch( G , u ) ∩ rch( G rev , u ) ∀ u ∈ SCC ( G , u ) : Mark u as visited. Running time: O ( n ( n + m )) Is there an O ( n + m ) time algorithm? Sariel Har-Peled (UIUC) CS374 40 Fall 2017 40 / 60

Structure of a Directed Graph B A C B , E , F A , C , D E F D G H G H Graph of SCC s G SCC Graph G Reminder G SCC is created by collapsing every strong connected component to a single vertex. Proposition For a directed graph G, its meta-graph G SCC is a DAG . Sariel Har-Peled (UIUC) CS374 41 Fall 2017 41 / 60

Linear-time Algorithm for SCC s: Ideas Exploit structure of meta-graph... Wishful Thinking Algorithm Let u be a vertex in a sink SCC of G SCC 1 Do DFS( u ) to compute SCC ( u ) 2 Remove SCC ( u ) and repeat 3 Justification DFS( u ) only visits vertices (and edges) in SCC ( u ) 1 2 3 4 Sariel Har-Peled (UIUC) CS374 42 Fall 2017 42 / 60

Linear-time Algorithm for SCC s: Ideas Exploit structure of meta-graph... Wishful Thinking Algorithm Let u be a vertex in a sink SCC of G SCC 1 Do DFS( u ) to compute SCC ( u ) 2 Remove SCC ( u ) and repeat 3 Justification DFS( u ) only visits vertices (and edges) in SCC ( u ) 1 2 3 4 Sariel Har-Peled (UIUC) CS374 42 Fall 2017 42 / 60

Linear-time Algorithm for SCC s: Ideas Exploit structure of meta-graph... Wishful Thinking Algorithm Let u be a vertex in a sink SCC of G SCC 1 Do DFS( u ) to compute SCC ( u ) 2 Remove SCC ( u ) and repeat 3 Justification DFS( u ) only visits vertices (and edges) in SCC ( u ) 1 ... since there are no edges coming out a sink! 2 3 4 Sariel Har-Peled (UIUC) CS374 42 Fall 2017 42 / 60

Linear-time Algorithm for SCC s: Ideas Exploit structure of meta-graph... Wishful Thinking Algorithm Let u be a vertex in a sink SCC of G SCC 1 Do DFS( u ) to compute SCC ( u ) 2 Remove SCC ( u ) and repeat 3 Justification DFS( u ) only visits vertices (and edges) in SCC ( u ) 1 ... since there are no edges coming out a sink! 2 DFS( u ) takes time proportional to size of SCC ( u ) 3 4 Sariel Har-Peled (UIUC) CS374 42 Fall 2017 42 / 60

Linear-time Algorithm for SCC s: Ideas Exploit structure of meta-graph... Wishful Thinking Algorithm Let u be a vertex in a sink SCC of G SCC 1 Do DFS( u ) to compute SCC ( u ) 2 Remove SCC ( u ) and repeat 3 Justification DFS( u ) only visits vertices (and edges) in SCC ( u ) 1 ... since there are no edges coming out a sink! 2 DFS( u ) takes time proportional to size of SCC ( u ) 3 Therefore, total time O ( n + m ) ! 4 Sariel Har-Peled (UIUC) CS374 42 Fall 2017 42 / 60

Big Challenge(s) How do we find a vertex in a sink SCC of G SCC ? Can we obtain an implicit topological sort of G SCC without computing G SCC ? Answer: DFS( G ) gives some information! Sariel Har-Peled (UIUC) CS374 43 Fall 2017 43 / 60

Big Challenge(s) How do we find a vertex in a sink SCC of G SCC ? Can we obtain an implicit topological sort of G SCC without computing G SCC ? Answer: DFS( G ) gives some information! Sariel Har-Peled (UIUC) CS374 43 Fall 2017 43 / 60

Big Challenge(s) How do we find a vertex in a sink SCC of G SCC ? Can we obtain an implicit topological sort of G SCC without computing G SCC ? Answer: DFS( G ) gives some information! Sariel Har-Peled (UIUC) CS374 43 Fall 2017 43 / 60

Linear Time Algorithm ...for computing the strong connected components in G do DFS( G rev ) and output vertices in decreasing post order. Mark all nodes as unvisited for each u in the computed order do if u is not visited then DFS( u ) Let S u be the nodes reached by u Output S u as a strong connected component Remove S u from G Theorem Algorithm runs in time O ( m + n ) and correctly outputs all the SCCs of G . Sariel Har-Peled (UIUC) CS374 44 Fall 2017 44 / 60

Linear Time Algorithm: An Example - Initial steps Graph G: Reverse graph G rev : B A C B A C = ⇒ E F D E F D G H G H Pre/Post DFS numbering of reverse graph: DFS of reverse graph: [1, 6] [7, 12] B A C [3, 4] B A C ⇒ ⇒ = = [9, 10] [8, 11] [2, 5] E F D E F D [13 , 16] [14 , 15] G H G H Sariel Har-Peled (UIUC) CS374 45 Fall 2017 45 / 60

Linear Time Algorithm: An Example Removing connected components: 1 Do DFS from vertex G Original graph G with rev post remove it. numbers: 6 12 4 B A C 6 12 11 5 10 E F D 4 B A C = ⇒ 15 H 11 5 10 E F D 16 SCC computed: 15 G H { G } Sariel Har-Peled (UIUC) CS374 46 Fall 2017 46 / 60

Linear Time Algorithm: An Example Removing connected components: 2 Do DFS from vertex G Do DFS from vertex H , remove it. remove it. 6 6 12 12 4 4 B A C B A C 11 5 11 5 10 E F D 10 E F D = ⇒ 15 H SCC computed: SCC computed: { G } { G } , { H } Sariel Har-Peled (UIUC) CS374 47 Fall 2017 47 / 60

Linear Time Algorithm: An Example Removing connected components: 3 Do DFS from vertex B Do DFS from vertex H , Remove visited vertices: remove it. { F , B , E } . 6 12 6 4 B A C 4 A C 11 5 10 E F D = ⇒ 5 D SCC computed: SCC computed: { G } , { H } { G } , { H } , { F , B , E } Sariel Har-Peled (UIUC) CS374 48 Fall 2017 48 / 60

Linear Time Algorithm: An Example Removing connected components: 4 Do DFS from vertex A Do DFS from vertex F Remove visited vertices: Remove visited vertices: { A , C , D } . { F , B , E } . 6 4 A C 5 D = ⇒ SCC computed: SCC computed: { G } , { H } , { F , B , E } { G } , { H } , { F , B , E } , { A , C , D } Sariel Har-Peled (UIUC) CS374 49 Fall 2017 49 / 60

Linear Time Algorithm: An Example Final result B A C E F D G H SCC computed: { G } , { H } , { F , B , E } , { A , C , D } Which is the correct answer! Sariel Har-Peled (UIUC) CS374 50 Fall 2017 50 / 60

Obtaining the meta-graph... Once the strong connected components are computed. Exercise: Given all the strong connected components of a directed graph G = ( V , E ) show that the meta-graph G SCC can be obtained in O ( m + n ) time. Sariel Har-Peled (UIUC) CS374 51 Fall 2017 51 / 60

Solving Problems on Directed Graphs A template for a class of problems on directed graphs: Is the problem solvable when G is strongly connected? Is the problem solvable when G is a DAG? If the above two are feasible then is the problem solvable in a general directed graph G by considering the meta graph G SCC ? Sariel Har-Peled (UIUC) CS374 52 Fall 2017 52 / 60

Part V An Application to make Sariel Har-Peled (UIUC) CS374 53 Fall 2017 53 / 60

Make/Makefile (A) I know what make/makefile is. (B) I do NOT know what make/makefile is. Sariel Har-Peled (UIUC) CS374 54 Fall 2017 54 / 60

make Utility [Feldman] Unix utility for automatically building large software applications 1 A makefile specifies 2 Object files to be created, 1 Source/object files to be used in creation, and 2 How to create them 3 Sariel Har-Peled (UIUC) CS374 55 Fall 2017 55 / 60

An Example makefile project: main.o utils.o command.o cc -o project main.o utils.o command.o main.o: main.c defs.h cc -c main.c utils.o: utils.c defs.h command.h cc -c utils.c command.o: command.c defs.h command.h cc -c command.c Sariel Har-Peled (UIUC) CS374 56 Fall 2017 56 / 60

makefile as a Digraph main.c main.o utils.c project defs.h utils.o command.h command.o command.c Sariel Har-Peled (UIUC) CS374 57 Fall 2017 57 / 60