Overview. CSE 548: (Design and) Analysis of Algorithms. Definition and Representations

Transcription

1 Overview Overview S 548: (esign and) nalysis of lgorithms raphs R. Sekar raphs provide a concise representation of a range problems Map coloring more generally, resource contention problems Figure Networks 3.1 (a) mapcommunication, and (b) its graph. traffic, social, biological,... (a) (b) / / 30 Overview efinition and Representations graph = (V, ), where V is a set of vertices, and a set of edges. n edge e of the form (v1, v2) is said to span vertices v1 and v2. The edges in a directed graph are directed. = (V, ) is called a subgraph of if V V and includes every edge in between vertices in V. djacency matrix graph (V = {v1,..., vn}, ) can be represented by an n n matrix a, where aij = 1 iff (vi, vj) djacency list ach vertex v is associated with a linked list consisting of all vertices u such that (v, u). Note that adjacency matrix uses O(n 2 ) storage, while adjacency list uses O( V + ) storage. oth can represent directed as well as undirected graphs. 3 / 30 necessary to use edges with directions on them. There can be directed edges e from x to y (written e = (x, y)), or from y to x (written (y, x)), or both. particularly enormous example of a directed graph is the graph of all links in the World Wide Web. It has a vertex for each Introsite FSon iconnectivity the Internet, s and Sa directed FS Paths edge and Matrices (u, v) whenever site u has a link to site v: in total, billions of nodes and edges! Understanding even the most basic connectivity properties of the Web is of great economic and social interest. lthough the size of this problem is daunting, epth-first we will soon see that asearch lot of valuable information (FS) about the structure of a graph can, happily, be determined in just linear time technique How is afor graph traversing represented? all vertices in the graph We Very canversatile, represent a graph forms bythe an adjacency linchpin matrix; of many if theregraph are n = algorithms V vertices v1,..., vn, this is an n n array whose (i, j)th entry is 1 if there is an edge from vi to vj dfs(v, ) aij = 0 otherwise. foreach For undirected v graphs, V dothe visited[v] matrix symmetric = falsesince an edge {u, v} can be taken in either foreach direction. v V do The biggest convenience of this format is that the presence of a particular edge can be checked if not invisited[v] constant time, then with just explore(v one memory,, access. v) On the other hand the matrix takes up O(n 2 ) space, which is wasteful if the graph does not have very many edges. n alternative representation, with size proportional to the number of edges, is the adjacency list., It, consists v) of V linked lists, one per vertex. The linked list for vertex u holds the explore(v names of vertices to which u has an outgoing edge that is, vertices v for which (u, v). visited[v] = true 88 previsit(v) /* placeholder for now*/ foreach (v, u) do if not visited[u] then explore(, V, u) postvisit(v) /*nother placeholder*/ 4 / 30

2 raphs, Mazes and FS Figure 3.2 xploring a graph is rather like navigating a maze. graph and its FS tree Figure 3.4 The result of explore() on the graph of Figure 3.2. Figure 3.2 xploring a graph is rather like navigating a maze. F H F H F I J J H F I J I J F H I Figure If a maze 3.3 Finding is represented all nodes reachable as a from graph, a particular then FS node. of the graph amounts procedure explore(, v) to an exploration and mapping of the maze. Input: = (V, ) is a graph; v V Output: visited(u) is set to true for all nodes u reachable from v visited(v) = true 5 / 30 previsit(v) for each edge (v, u) : if not visited(u): explore(u) postvisit(v) FS and onnected omponents and whenever you arrive at any junction (vertex) there are a variety of passages (edges) you Figure can follow. 3.6 (a) careless 12-node choice graph. of passages (b) FSmight searchlead forest. you around in circles or might cause you to overlook some accessible part of the maze. learly, you need to record some intermediate 1,10 11,22 23,24 information during exploration. (a) (b) This classic challenge has amused people for centuries. verybody knows that all youf need to explore a labyrinth is aball of string and a piece of chalk. The chalk prevents looping, by marking the junctions you have already visited. The string 4,9 always takes 12,21 you back to the starting place, enabling you to return to passages 2,3that you previously saw but did not yet investigate. F H I 5,8 13,20 H How can we simulate these two primitives, chalk and string, on a computer? The chalk marks are easy: for each vertex, maintain a oolean variable indicating whether it has been visited I already. J s forthe ball of string, the correct cyberanalog J is a14,17 stack. fter all, the exact role of the string is to offer two primitive operations unwind 6,7 to get to a new junction 18,19 (the stack equivalent is to push the new vertex) and rewind to return to the previous junction (pop the stack). 15,16 Instead of explicitly maintaining a stack, we will do so implicitly via recursion (which is implemented using a stack of activation records). The resulting algorithm is shown in connected component of a graph is a maximal subgraph where 90 These there regions is path are called between connected anycomponents: two vertices eachin of them subgraph, is a i.e., that itisisinternally a connected but has no edges to the remaining vertices. When explore is started at a particular vertex, maximal it identifies connected preciselysubgraph. the connected component containing that vertex. nd each time the FS outer loop calls explore, a new connected component is picked out. 7 / 30 Thus depth-first search is trivially adapted to check if a graph is connected and, more generally, to assign each node v an integer ccnum[v] identifying the connected component to J I FS uses O( V ) space Figure and 3.5O( epth-first + V search. ) time. procedure dfs() Figure 3.3 Finding all nodes reachable from a particular node. for all v V : 6 / 30 procedure explore(, visited(v) v) = false Input: = (V, ) is a graph; v V Output: visited(u) for allis v set V : to true for all nodes u reachable if not visited(v): explore(v) visited(v) = true previsit(v) FS Numbering for each edge (v, u) : if not visited(u): This loop takes explore(u) a different amount of time for each vertex, so let s co postvisit(v) gether. The total work done in step 1 is then O( V ). In step 2, over t ssociate post andfs, pre each numbers edge {x, with y} each is examined visited node exactly bytwice, defining once during expl ing explore(y). The overall time for step 2 is therefore O( ) and so previsit and postvisit has a running time of O( V + ), linear in the size of its input. Th could possibly hope for, since it takes this long even just to read the ad previsit(v) postvisit(v) Figure 3.6 shows the outcome of depth-first search on a 12-node gra and pre[v] whenever = clock you arrive ing tiesat alphabetically any junction post[v] (ignore (vertex) the = pairs clock there of numbers are a variety for theof time passages being). can follow. careless calls choice explore of passages three times, might on, lead, and you finally around F. ins circles a result, or there mig toclock++ overlook some accessible rooted at part one of of these starting clock++ maze. points. learly, Together you need theyto constitute record asome foresti information during exploration onnectivity in undirected graphs Property This classic challenge has amused people for centuries. verybody knows need to explore a labyrinth n undirected is a ball graph of string is connected and aif piece there of is achalk. path between The chalk anyprev pair For any two vertices u and v, the intervals [pre[u], post[u]] and by marking the junctions of Figure you 3.6have is notalready connectedvisited. because, The for instance, string always there is no takes pathyou fro starting [pre[v], post[v]] place, enabling aredoes either have three disjoint connected regions, corresponding to the follow youdisjoint, to return or one to passages is contained that entirely you previously within saw but investigate. another. {,,, I, J} {,,, H,, } {F } How can we simulate these two primitives, chalk and string, on a computer marks are easy: for each vertex, maintain a oolean variable 92 indicating whethe 8 / 30 visited already. s for the ball of string, the correct cyberanalog is a stack. fter role of the string is to offer two primitive operations unwind to get to a new j

3 FS of irected raph Tree edges are actually part of the FS forest. ure 3.7 FS on a directed graph. Forward edges lead from a node to a nonchild descendant in the FS tree. 1,16 ack edges lead to an ancestor in the FS tree. 2,11 12,15 ross edges lead to neither descendant nor ancestor; they therefore F lead to a node that has already 3,10 beencompletely 13,14 explored (that is, already postvisited). F H You can ross confirm each of these characterizations by consulting the diagram of 4,7 8,9 you see why no other orderings are possible? H No cross edges in undirected graphs! 5, irected acyclic graphs Figure 3.7 has two forward edges, two back edges, and two cross edges. an you ack spotand them? forward edges merge cycle in a directed graph is a circular path v0 v1 v2 vk v0. 9 / 30 quite a few of them, for example, F. graph without10 / cycles 30 is ncestor and descendant relationships, as well as edge types, can outbe we can read testoff fordirectly acyclicity in linear time, with a single depth-first search. procedure postvisit(v) post[v] = from clock pre and post numbers. ecause of the depth-first exploration strategy, vertex u is an clock = clock ancestor + 1 of vertex v exactly in those cases where u is discovered first and v is discovered 95 during explore(u). This is to say pre(u) < pre(v) < post(v) < post(u), which we can iconnectivity depict pictorially in Undirected as two nestedraphs intervals: Illustration of iconnected omponents se timings will soon take on larger significance. Meanwhile, you might have noticed from ure 3.4 that: perty For any nodes u and v, the two intervals [pre(u), upost(u)] v and [pre(v), v post(v)] u are er disjoint efinition or one is (iconnected contained within omponents) the other. The case of descendants is symmetric, since u is a descendant of v if and only if v is an ancestor Why? ecause n [pre(u), post(u)] is essentially the time during which vertex u was on the articulation of u. point nd in since a graph edge is any categories vertex a are that based lies on entirely every on ancestor-descendant relationships, k. The last-in, first-out behavior of a stack explains the rest. path it from follows between thatvertices they, too, v, w. can be read off from pre and post numbers. Here is a summary of epth-first biconnected the various search graph possibilities in contains directed nofor articulation an graphs edge points. (u, v): FS The andcase dge of descendants Types is symmetric, since u is a descendant of v if and on cestor of u. nd since edge categories are based entirely on ancestor-descendan it follows that they, too, can be read off from pre and post numbers. Here i FS tree the various possibilities for an edge (u, v): pre/post ordering for (u, v) dge type.1 biconnected component of a graph Types of edges pre/post is a maximal ordering subgraph for (u, that v) dge type is biconnected. depth-first search algorithm can be run verbatim on directed graphs, taking care Tree/forward to trase edges only in their prescribed directions. Figure u 3.7 vshows anv example u and the search that results when vertices are considered in lexicographic order. In further analyzing the directed case, it helps to have terminology for important relationsbetween nodes of a tree. istheroot of the search v tree; u everythingelseisitsdescendant. u v ack ach biconnected component given a different color rticulation points have multiple colors ilarly, has descendants F,, and H, and conversely, is an ancestor of these three 11 / 30 nodes. family analogy is carried further: is the parent of, which is its child. ross ack Tree Forward u v v u u v v u v u u v v v u u Tree/forward ack ross 12 / 30

4 iconnected raphs Finding articulation points during FS iconnected components are not disjoint. rticulation points are duplicated across them. rticulation points single-points of failure raph is disconnected if any of them go down. etween any two u, v V in a biconnected graph there are two node-disjoint paths (and hence a cycle). uring FS visit of vertex v, we want to decide if it is an articulation point or not. ivide V into to disjoint sets: Vi includes nodes inside the FS tree rooted at v, and Vo = V Vi of nodes outside the FS tree rooted at v. Observation (1) Suppose that every inside node vi (i.e., vi Vi {v}) can reach some outside node vo (i.e., vo Vo) without going through v. Then v is not an articulation point. 13 / / 30 Finding articulation points during FS (2) Finding articulation points during FS (3) Proof: Paths within Vo: y properties of FS tree, any two vertices in Vo can reach each other without going through v Paths between vi and Vo: Once vi can reach any node in Vo, it can reach all other nodes in Vo without having to go through v. Paths within Vi: onsider nodes vi and v i in Vi {v}. y our assumption there is a path from vi to some vo Vo, and another path from v i to some v o Vo, neither involving v. onsider the path vi to vo to v o to v i, which does not pass through v. Thus, for every pair of vertices, there is a path between them that avoids v, and hence v is not an articulation point. 15 / 30 Observation (2) If Observation (1) does not hold then v is an articulation point. This means that there exists some vi that cannot reach an outside node vo without going through v. y definition, this means that v is an articulation point. 16 / 30

5 Finding articulation points during FS (3) Finding articulation points during FS (4) We combine and slightly strengthen Observations (1), (2), while omitting the proof, as it is essentially the same as before. Observation (3) et Y denote the set of ancestors of v and X its immediate children in the FS tree. Now, v is not an articulation point iff each x X has a path to some y Y without going through v. y focusing on just the children and ancestors of v, this makes it easier to decide on articulation points during FS. Note than any such path from x to y will follow zero or more tree edges down, followed by a back edge. Note than any such path from x to y will follow zero or more tree edges down, followed by a back edge. If this path bypasses v, this is going to occur due to a single back-edge that goes to an ancestor of v. So there is no need to consider paths with multiple back-edges. Note: Pre-number increases while following tree edges down, while it decreases when a back edge is followed. ey Idea: uring FS, for each vertex x, maintain the highest ancestor that can be reached from x by following tree edges down and then a back edge. 17 / / 30 Finding articulation points during FS (5) ey Idea: uring FS, maintain the highest ancestor reachable from x by following tree edges down and then a back edge. This info is maintained in the array low. low[x] low[x ] for every child x of x in FS tree. This captures the fact that we can follow the tree edge down from x to x, then go to whichever ancestor is reachable from x. lgorithmically, let low[x] = min(low[x], low[x ]) when explore(x ) returns low[x] pre[x ] for x adjacent to x but not a parent or child in the FS tree. lgorithmically, set low[x] = min(low[x], pre[x ]) y properties of FS, x is either a descendant or ancestor of x. s we are taking min, statement effective only if x is an ancestor. 19 / 30 Finding articulation points during FS (6) ey Idea: Maintain low during FS when visiting x, initialize low[x] to pre[x]. when a FS call on a child x of x returns, check if low[x ] pre[x]. If so, the highest ancestor x can reach is not higher than x, i.e., x cannot go to ancestors of x without going through x. So, mark x as articulation point If an adjacent vertex x is already visited when x considers it, set low[x] = min(low[x], pre[x ]) unless x is parent of x ll these can be done during a FS, while increasing the cost of each step by only a constant amount so, overall complexity is O( + V ) 20 / 30

6 irected cyclic raphs (s) Strongly onnected omponents (S) directed graph that contains no cycles. Often used to represent (acyclic) dependencies, partial orders,... Property (s and FS) directed graph has a cycle iff its FS reveals a back edge. In a dag, every edge leads to a vertex with lower post number. very dag has at least one source and one sink. nalog of connected components for undirected graphs efinition (S) Two vertices u and v in a directed graph are connected if there is a path from u to v and vice-versa. directed graph is strongly connected if any pair of vertices in the graph are connected. subgraph of a directed graph is said to be an S if it is a maximal subgraph that is strongly connected. Ss are also similar to biconnected components! 21 / / 30 FigureS 3.9 (a) xample directed graph and its strongly connected components. (b) The meta-graph. readth-first Search (FS) (a) (b) I J F H,,F,H,I J,, Traverse the graph by levels FS(v) visits v first Then it visits all immediate children of v then it visits children of children of v, and so on. s compared to FS, FS uses a queue (rather than a stack) to remember vertices that still need to be explored The textbook describes an algorithm for computing S in linear-time n efficient using algorithm FS. 23 / 30 The decomposition of a directed graph into its strongly connected components is very informative and useful. It turns out, fortunately, that it can be found in linear time by making 24 / 30

7 FS lgorithm bfs(v,, s) foreach u V do visited[u] = false q = {s}; visited[s] = true while q is nonempty do u = deque(q) foreach edge (u, v) do if not visited[v] then queue(q, v); visited[v] = true Shortest Paths and FS FS automatically computes shortest paths! bfs(v,, s) foreach u V do dist[u] = q = {s}; dist[s] = 0 while q is nonempty do u = deque(q) foreach edge (u, v) do if dist[v] = then queue(q, v); dist[v] = dist[u] + 1 ut not all paths are created equal! We would like to compute shortest weighted path a topic of future lecture. Figure 4.2 physical model of a graph. FS lgorithm Illustration S Figure 4.4 The result of breadth-first search on the graph of Figure 4.1. Order Queue contents of visitation after processing node Figure 4.3 readth-first search. S [S] procedure bfs(, S s) [ ] Input: raph = (V, ), [ directed ] or undirected; vertex s V [ ] Output: For all vertices u reachable from s, dist(u) is set [ ] to the distance [] from s to u. [ ] for all u V : 25 / / 30 dist(u) = right thing. If S is the starting point and the nodes are ordered alphabetically, they get visited dist(s) = 0 in the sequence shown in Figure 4.4. The breadth-first search tree, on the right, contains the edges Intro FS through iconnectivity which seach S FS node Paths isand initially Matrices discovered. Unlike the FS tree we saw earlier, it Q = [s] (queue containing just s) has the property that all its paths from S are the shortest possible. It is therefore a shortestpath nottree. empty: while Q is raph paths and oolean Matrices u = eject(q) for all orrectness edges (u, v) and efficiency : if dist(v) We have= developed : the basic intuition behind breadth-first search. In order to check that inject(q, the algorithm v) works correctly, we need to make sure that it faithfully executes this intuition. dist(v) What graph we= expect, dist(u) and precisely, its boolean + 1 is that matrix representation For each d = 0, 1, 2,..., there is a moment at which (1) all nodes at distance d from s have their distances correctly set; (2) all other nodes have their distances set to ; and (3) the queue contains exactly the nodes at distance d readth-first This has been phrased search with an inductive argument in mind. We have already discussed both the base case and the inductive step. an you fill in the details? In Figure 4.2, the lifting of s partitions the graph into layers: = s itself, the nodes at dis The overall running time of this algorithm is linear, O( V + 0 ), 0 0for 1exactly the same 1 from it, the nodes reasons at as depth-first distancesearch. 2 fromach it, vertex and so is put on. on the convenient queue exactly 0 way 0once, 0 towhen 0 compute it is first dist encountered, vertices so thereis are to2 proceed V queue operations. layer by layer. The restonce of the work we have is donepicked in the algorithm s from s to the other 4 out the n at distance 0, 1, innermost 2,..., d, loop. theover onesthe atcourse d + 1 of are execution, easily determined: this loop looks at they eachare edgeprecisely once (in directed the a graphs) or twice (in undirected graphs), and therefore takes O( ) time. unseen nodes that are adjacent to the layer at distance d. This suggests an iterative algo in which two layers Noware thatactive we have atboth anyfs given and time: FS before some us: layer how do d, their which exploration has been styles fully compare? ident and d + 1, which epth-first is being search discovered makes deep byincursions scanning intothe a graph, neighbors retreating of only layer when d. it runs out of new 27 / 30 nodes to visit. This strategy gives it the wonderful, subtle, and extremely useful28 properties / 30 readth-first search (FS) directly implements this simple reasoning (Figure 4.3) we saw in the hapter 3. ut it also means that FS can end up taking a long and convoluted tially the queue Q consists only of s, the one node at distance 0. nd for each subseq S

8 raph paths and oolean Matrices et be the adjacency matrix for a graph, and =. Now, ij = 1 iff there is path in the graph of length 2 from vi to vj et = +. Then ij = 1 iff there is path of length 2 between vi and vj efine = If = then ij = 1 iff vj is reachable from vi. = = = Shortest paths and Matrix Operations Redefine operations on matrix elements so that + becomes min, and becomes integer addition. = then ij = k iff the shortest path from vj to vi is of length k 29 / / 30