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- 1. Algorithm Design and Complexity Course 8
- 2. Overview Applications for DFS Strongly Connected Components Articulation Points Bridges Biconnected Components
- 3. Applications for DFS Directed graphs We have seen topological sorting Strongly connected components Undirected graphs Articulation points Bridges Biconnected components
- 4. Strongly Connected Components Applications: Definitions: Given a directed graph G(V, E), G is strongly connected if ∀ u, v ∈ V, there exists a path u..v and a path v..u Compiler analysis, data mining, scientific computing u∈R(v) and v∈R(u) ∀ u, v ∈ V Not all directed graphs are strongly connected Given a directed graph G(V, E), a strongly connected component (SCC) is a maximal subset of vertices C⊆V such that ∀ u, v ∈ C, there exists a path u..v and a path v..u
- 5. SCC - Example For any graph, we have a single way to partition it into SCCs I I A A J G G B B K C H D E J K C L H D E F F L
- 6. SCCs - Properties Let C⊆V – a SCC of G and ∀ u, v ∈ C. Any vertex z on a path u..z..v is also in C. Let x be any vertex in C => x∈R(v), but v∈R(z) => x∈R(z) => u∈R(x), but z∈R(u) => z∈R(x) Therefore z ∈ C Let C⊆V – a SCC of G. In any DFS traversal, all the vertices in C are always in the same DFS tree! Let u be the first vertex in C that is discovered by the DFS At d[u], all the other vertices in C are WHITE There exists a path from u to all the other vertices in C that consists of intermediate vertices that are all in C (def. of a SCC) Therefore, there exists a white path from u to all the other vertices in C => all the others vertices in C become descendants of u in the same DFS tree
- 7. Run DFS on the Example Graph I A J G B K C H L D E F
- 8. Transposing a Graph Given G(V, E) GT – transpose of G: a graph that has the same vertices, but the direction of the edges is reversed GT (V, ET) (u, v) ∈ ET <=> (v, u) ∈ E We can compute the transpose of a graph in: Θ(n+m) for adjacency lists (need to recompute the lists) Θ(1) for adjacency matrix (just use A[j, i] = AT[i, j])
- 9. SCCs – Property of G and GT G and GT have the same SCCs Let C be a SCC in G and ∀ u, v ∈ C: u∈R(v) in G => u∈R(v) in GT v∈R(u) in G => v∈R(u) in GT => C is also a SCC in GT
- 10. The SCCs Graph GSCC = (VSCC, ESCC) VSCC contains a vertex for each SCC in G ESCC contains an edge between two different SCC C1, C2 if there exists in G at least an edge between two vertices u ∈ C1 and v ∈ C2 ((u, v) ∈ E[G]) Property: GSCC is a DAG Let C1, C2 be two distinct SCC in G. Let u, v ∈ C1 and u’, v’ ∈ C2. If there exists a path u .. u’ in G, then there cannot be any path v’ .. v in G Why? If it was such a path, then there is also a path u .. u’ .. v’ .. v .. u , therefore all the vertices would be in the same SCC
- 11. SCC – Algorithm We can find the SCCs of a graph using two DFS traversals: Complexity: A DFS for G, regardless of the order for considering the roots of the DFS trees GT = Transpose G A second DFS for GT, taking the roots of the DFS trees in an order provided by the first DFS for G Θ(n+m) – adjacency lists Θ(n2) – adjacency matrix Name: Kosaraju’s algorithm
- 12. SCC – Algorithm SCC(G) f[1..n] = DFS(G) // call DFS for G in order to compute the finish times // f[u] for all u Gt = transpose(G) DFS_modified(Gt, f[1..n]) // call the modified DFS for G that picks the // roots of the DFS trees in decreasing // order of f[u] (plus u being WHITE as // usual) PRINT the vertices in each separate DFS tree computed by the call to DFS_modified. Each DFS tree contains the nodes of a SCC of G
- 13. Example – DFS(G) I 17/24 A 1/16 J 18/23 G 11/14 B 2/15 K 19/22 C 3/10 H 12/13 D 5/6 E 4/9 F 7/8 L 20/21
- 14. Example – DFS(GT) I 17/24 A 1/16 I 1/6 A 9/16 J 18/23 G 11/14 B 2/15 G 11/14 K 19/22 C 3/10 H 12/13 D 5/6 E 4/9 L 20/21 B 12/13 H 10/15 C 17/22 D 18/21 E 19/20 F 7/8 F 23/24 J 3/4 K 2/5
- 15. Idea for the Algorithm If GSCC is a DAG => (GSCC )T is also a DAG How many DFS trees do we find when running DFS for GSCC? It depends on how we pick the roots of the trees: A single tree that contains all SCCs |VSCC| trees, each containing a single node We want to find how to walk GSCC in order to have a tree for each node => an order for choosing the roots of the DFS trees
- 16. Idea for the Algorithm (2) Knowing this order, we can also walk the original graph and choose each time as a new root any node in the SCC that is next in order. This order is the topological sorting order for GSCC We can compute a topological sorting of GSCC regardless of how we choose the roots of the DFS trees This is done by running the first DFS on G We run DFS on (GSCC )T considering the vertices in the topological sorting order we have computed for GSCC This is done by running the first DFS on G T by choosing the root vertices ordered descending on the finish time of the first DFS
- 17. Topological Sorting of GSCC By choosing the vertices in the topological ordering of GSCC: The first node is the one with the highest finish time The last node is the one with the lowest finish time We want to show that for any edge of GSCC: the source vertex has a higher finish time than the one of the destination vertex Therefore, for any edge of (GSCC )T: the source vertex has a lower finish time than the one of the destination vertex
- 18. Demonstration Consider the results of the first DFS (for G): d[u] and f[u] Extend the notation for d and f to sets of vertices Let C⊆V: d(C) = min(d[u]) for all u∈C Earliest discovery time of any vertex in C f(C) = max(f[u]) for all u∈C Latest finish time of any vertex in C
- 19. Lemma Let C and C’ be 2 distinct SCCs in G. Suppose there exists an edge (u, v)∈E[G], u∈C, v∈C’ (then there also exists an edge in GSCC between the vertices corresponding to C and C’) => f(C) > f(C’) Two possibilities: d(C) < d(C’) Let x∈C be the first vertex discovered in C and C’ White path theorem: all nodes in C and C’ are reachable from x using only intermediate vertices that are WHITE f(C) = f[x] > f(C’) d(C) > d(C’) Let y∈C’ be the first vertex discovered in C and C’ f(C’) = f[y] At f[y], all nodes in C are WHITE => f(C) > d(C) > f(C’)
- 20. Lemma – Conclusion The first DFS on G provides a topological sorting order for GSCC
- 21. Corollary Let C and C’ be 2 distinct SCCs in G. Suppose there exists an edge (u, v)∈ET, u∈C, v∈C’ (then there also exists an edge in (GSCC)T between the vertices corresponding to C and C’) => f(C) < f(C’) (u, v)∈ET => (v, u)∈E, u∈C, v∈C’ =>f(C’) > f(C) Using previous lemma
- 22. Corollary – Conclusion Let C and C’ be 2 distinct SCCs in G. If f(C) > f(C’) there cannot be an edge from C to C’ in (GSCC)T If f(C) > f(C’) there cannot be an edge from C to C’ in GT
- 23. SCC Algorithm – Conclusion When running the second DFS on GT , we pick the vertex with the highest finish time, x∈V such that f(x)=maximum Therefore, if C is the SCC that contains x, f(C)>f(C’) for any other C’ a SCC of G. Thus, by choosing x as the root of the first DFS tree we can only explore the nodes that are in C And we explore all of them => we color them in BLACK The second vertex that we pick as a root is x’ that has the maximum finish time out of the remaining white vertices. Let C’ be the SCC that x’ is part of. Then: There can be an edge only from C’ to C, but not to any other SCC And we repeat this process until no vertices are left unvisited! Thus, we are visiting (GSCC)T in reverse topological sorting order!
- 24. Articulation Points Given a connected undirected graph G(V, E) An articulation point (or cut vertex) is any u∈V such that by removing it from the graph (and all the adjacent edges), the remaining graph becomes disconnected The definition is similar for graphs that are not connected. An articulation point is any vertex that after removal increases the number of connected components of the resulting graph
- 25. Articulation Points (2) Formal definition: Given G(V, E), u is an articulation point if there exist two vertices x, y ∈ V, x != y != u such that any path from x..y contains u u – articulation point u – not an articulation point
- 26. Articulation Points (3) Lots of applications for detecting critical points in any kind of networks for communication, airline traffic, electric circuits, etc.
- 27. Simple Algorithm Given G(V, E) that is connected Foreach (u∈V) G’ = remove u from G if (G’ is not connected) // can use BFS or DFS u is an articulation point Complexity: θ(n * (n+m)) = θ(n2 + n*m) We want a better algorithm!
- 28. Theorem Let G(V, E) undirected graph Let any DFS of G u∈V is an articulation point of G if and only if p[u] == null and u has at least two children p[u] != null and u has at least a child v such that no vertex in Subtree(v) is connected to an ancestor of u using a back edge u is a root of a DFS tree u is not a root of a DFS tree Demonstration: on the whiteboard
- 29. Implementation The first case of the previous theorem is very simple to detect Count the number of children for each node children[u] The second case is trickier We need to remember for each vertex u the earliest node that can be reached from Subtree(u) Earliest means minimum discovery time, therefore we want to discover if we can reach the ancestors of a node u from Subtree(u) low[u]
- 30. low[u] Can be defined recursively: low[u] = min( d[u], d[x], for all x that can be reached from any node in Subtree(x) by a back edge ) We can improve the definition in order to integrate it efficiently into a DFS
- 31. low[u] Revisited low[u] = min( d[u], d[w], for all back edges (u, w) visited from u low[v], for all v children of u in the DFS tree ) Thus, we can take advantage that low[v] has been computed and will never change when we compute low[u], for all v – children of u in the DFS tree Because color[v] is BLACK and we have already explored everything reachable from v !
- 32. Articulation Points - Algorithm We can use a simple DFS algorithm and make small changes We do not need to compute f[u] We need color[u] to detect back edges, d[u], children[u] and low[u] This algorithm is called Tarjan’s algorithm
- 33. Tarjan’s Algorithm DFS(G) FOREACH (u ∈ V) color[u] = WHITE; p[u] = NULL; articulation[u] = FALSE ; time = 0 FOREACH (u ∈ V) IF (color[u] == WHITE) DFS_Visit (u) IF (children[u] > 1) articulation[u] = TRUE // the first case of the theorem
- 34. Tarjan’s Algorithm DFS_Visit(u) d[u] = ++time low[u] = d[u] // low[u] = min(d[u], …) color[u] = GREY children[u] = 0 FOREACH (v ∈ Adj(u)) // look for undiscovered neighbors IF (color[v] == WHITE) children[u]++ p[v] = u DFS_Visit(v) low[u] = min(low[v], low[u]) // finished a child… look for new minimum IF (p[u] != NULL AND low[v] >= d[u]) articulation[u] = TRUE // the second case of the theorem ELSE IF (color[v] == GREY AND p[u] != v) // back edge… look for new minimum low[u] = min(low[u], d[v]) color[u] = BLACK Complexity: θ(n + m) if using adjacency lists, just like normal DFS
- 35. Example
- 36. Example (2)
- 37. Bridges Given a connected undirected graph G(V, E) A bridge (or cut edge) is any edge (u, v)∈E such that by removing it from the graph, the remaining graph becomes disconnected The definition is similar for graphs that are not connected. A brdige is any edge that after removal increases the number of connected components of the resulting graph
- 38. Simple Algorithm Given G(V, E) that is connected Foreach ((u, v)∈E) G’ = remove (u, v) from G if (G’ is not connected) // can use BFS or DFS (u, v) is a bridge Complexity: θ(m * (n+m)) = θ(m2 + n*m) We want a better algorithm! Use an adapted version of Tarjan’s algorithm for articulation points
- 39. Tarjan’s Algorithm for Finding Bridges For any edge in the DFS search (v, u = p[v]) It is a bridge if and only if low[v] > d[u] Only tree edges can be bridges! There does not exist any other alternative to reach u or an ancestor of u from Subtree(v)! Why? bridge[u] = TRUE if (u, p[u]) is a bridge
- 40. Tarjan’s Algorithm for Finding Bridges DFS(G) FOREACH (u ∈ V) color[u] = WHITE; p[u] = NULL; bridge[u] = FALSE ; time = 0 FOREACH (u ∈ V) IF (color[u] == WHITE) DFS_Visit (u)
- 41. Tarjan’s Algorithm for Finding Bridges DFS_Visit(u) d[u] = ++time low[u] = d[u] color[u] = GREY FOREACH (v ∈ Adj(u)) IF (color[v] == WHITE) children[u]++ p[v] = u DFS_Visit(v) low[u] = min(low[v], low[u]) IF (low[v] > d[u]) bridge[v] = TRUE ELSE IF (color[v] == GREY AND p[u] != v) low[u] = min(low[u], d[v]) color[u] = BLACK // look for undiscovered neighbors // finished a child… look for new minimum // check if the explored edge is a bridge // back edge… look for new minimum Complexity: θ(n + m) if using adjacency lists, just like normal DFS
- 42. Biconnected Graph A biconnected graph is a graph that has no articulation points Between any two vertices in the graph, there are at least two different paths The graph is also called 2-connected!
- 43. Biconnected components As not all the graphs are biconnected, an interesting problem is to find the biconnected components of a graph Biconnected component = maximal subgraph that is biconnected Any two edges in the same biconnected component lie on a common simple cycle Can also use Tarjan’s algorithm for computing the biconnected components of a graph
- 44. Example Image source: Wikipedia
- 45. Tarjan’s Algorithm for SCCs Tarjan’s algorithm can also be adapted for computing the SCCs of a directed graph Use only a single DFS walk instead of two walks that Kosaraju’s algorithm uses However, the two algorithms have the same order of growth How would you adapt Tarjan’s algortihm for computing the SCCs ?
- 46. References CLRS – Chapter 23 www.cse.ohio-state.edu/~lai/780/9.graph.pdf

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please add definition for DFS - otherwise it's used too much without clear sense.