Keys to Success: CAR Theorem CHAPTER 3 GRAPHS. Outline. Basics. Definitions and applications. Iris Hui-Ru Jiang Fall 2014.

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1 Keys to Success: CAR Theorem Prof. Chang s CAR Theorem Iris: Recap stable matching Criticality Extract the essence Identify the clean core Remoe extraneous detail CHAPTER GRAPHS Iris Hui-Ru Jiang Fall 0 Abstraction Restriction Represent in an abstract form First think at high-leel Deise the algorithm Then go down to low-leel Complete implementation Simplify unimportant things List the limitations Show how to extend Prof. Yao-Wen Chang@NTU Outline Content: Basic definitions and applications Graph connectiity and graph traersal Implementation Testing bipartiteness: an application of BFS Connectiity in directed graphs Directed acyclic graphs and ological ordering Reading: Chapter Basics Definitions and applications

2 Salute to Euler! Examples of (/) Our focus in this course is on problems with a discrete flaor. One of the most fundamental and expressie of combinatorial structures is the graph. Inented by L. Euler based on his proof on the Königsberg bridge problem (the seen bridge problem) in. Is it possible to walk across all the bridges exactly once and return to the starting land area? Abstraction! It s useful to digest the meaning of the nodes and the meaning of the edges in the following examples. It s not important to remember them. Transportation networks: No Eulerian walk China Airlines international route map (node: city; edge: non-s flight) London underground map (node: station; edge: adjacent stations) L. Euler, Solutioproblematis ad geometriamsituspertinentis, Commentarii Academiae Scientiarum Imperialis Petropolitanae, Vol., pp. 0, (published ). (H. Beck, ) Graph ( ) Examples of (/) A graph is simply a way of encoding pairwise relationships among a set of objects. A graphg = (V, E) consists of A collection V of nodes (a.k.a. ertices) A collection E of edges Each edge joins two nodes e = {u, } Î E for some u, Î V In an undirected graph: symmetric relationships Edges are undirected, i.e., {u, } == {, u} e.g., u and are family. In a directed graph: asymmetric relationships Edges are directed, i.e., (u, )!= (, u) u u tail head e.g., u knows (celebrity), while doesn t know u. is one of u sneighbor if there is an edge (u, ) Adjacency Communication networks Wireless sensor network (node: sensor; edge: signal broadcasting)

3 Examples of (/) Examples of (/) Information networks Dependency networks World Wide Web (node: webpage; edge: hyperlink) Food chain/web (node: species; edge: from prey to predator) Examples of (/) Examples of (/) 0 Social networks Technological Network Facebook (node: people; edge: friendship) Finite state machine (node: state; edge: state transition)

4 Paths and Connectiity (/) Trees One of the fundamental operations in a graph is that of traersing a sequence of nodes connected by edges. Browse Web pages by following hyperlinks Join a 0-day tour from Taipei to Europe on a sequence of flights Pass gossip by word of mouth (by message of mobile phone) from you to someone far away Hey upper east siders, Gossip Girl here! -- Gossip Girl An undirected graph is a tree if it is connected and does not contain a cycle. Trees are the simplest kind of connected graph: deleting any edge will disconnect it. Thm: Let G be an undirected graph on n nodes. Any two of the following statements imply the third. G is connected. G does not contain a cycle. G has n- edges. G is a tree if it satisfies any two of the three statements Tree? Grab and let the rest hang downward Yes! Paths and Connectiity (/) A path in an undirected graph G = (V, E) is a sequence P of nodes,,, k-, k with the property that each consecutie pair i, i+ is joined by an edge in E. A path is simple if all nodes are distinct. A cycle is a path,,, k-, k in which = k, k>, and the first k- nodes are all distinct. An undirected graph is connected if, for eery pair of nodes u and, there is a path from u to. The distance between nodes u and is the minimum number of edges in a u- path. ( for disconnected) Note: These definitions carry oer naturally to directed graphs with respect to the directionality of edges. Path P =,,,,,, Cycle C =,,,,, Rooted Trees A rooted tree is a tree with its root at r. Grab and let the rest hang downward root r s parent a tree the same tree, rooted at Rooted trees encode the notion of a hierarchy. e.g., sitemap of a Web site The tree-like structure facilitates naigation (root: entry page) Directed edges: asymmetric relationship s child leaf

5 Breadth-First-Search (BFS) Graph Connectiity and Graph Traersal BFS DFS Breadth-first search (BFS): propagate the waes Start at s and flood the graph with an expanding wae that grows to isit all nodes that it can reach. Layer L i : i is the time that a node is reached. Adjacent nodes Layer L 0 = {s}; layer L = all neighbors of L 0. Layer L j+ = all nodes that do not belong to an earlier layer and that are neighbors of L j. i = distance between s to the nodes that belong to layer L i. L L L L Node-to-Node Connectiity BFS Tree Q: Gien a graph G = (V, E) and two particular nodes s and t, is there a path from s to t in G? The s-t connectiity problem The maze-soling problem A: For small graphs, easy! (isual inspection) - connectiity? - connectiity? What if large graphs? How efficiently can we do? 0 room hallway 0 Let T be a BFS tree, let x and y be nodes in T belonging to layers L i and L j respectiely, and let (x, y) be an edge of G. Then i and j differ by at most. Pf: Without loss of generality, suppose j i >. By definition, xîl i, x s neighbors belongs to L i+ or earlier. Since (x, y) is an edge of G, y is x s neighbor, yî L j and j! i+. BFS tree L L L L 0 Nontree edge L 0 L L BFS tree Tree edge L

6 Connected Component Connected Component s R A connected component containing s is the set of nodes that are reachable from s. Connected component containing node is {,,,,,,, }. There are three connected components. The other two are {, 0} and {,, }. 0 Find all nodes reachable from s: Connected-Component(s) // R will consist of nodes to which s has a path. initialize R = {s}. while (there is an edge (u, ) where uîr andïr) do. R = R + {} it's safe to add Correctness: Upon termination, R is the connected component containing s. Pf: Q: How about any node ÎR? Q: How about a node wïr? Q: How to recoer the actual path from s to any node tîr? Q: How to explore a new edge in line? BFS: explore in order of distance from s. Any method else? u Color Fill Depth-First Search (DFS) Q: Gien lime green pixel in an image, how to change color of entire blob of neighboring lime pixels to blue? A: Model the image as a graph. Node: pixel. Edge: two neighboring lime pixels. Blob: connected component of lime pixels. recolor lime green blob to blue Depth-first search (DFS): Go as deeply as possible or retreat Start from s and try the first edge leading out, and so on, until reach a dead end. Backtrack and repeat. A mouse in a maze without the map. Another method for finding connected component DFS(u). mark u as explored and add u to R. foreach edge (u, ) incident to u do. if ( is not marked as explored) then. recursiely inoke DFS()

7 DFS Tree Let T be a DFS tree, let x and y be nodes in T, and let (x, y) be a nontree edge. Then one of x or y is an ancestor of the other. Pf: descendant WLOG, suppose x is reached first by DFS. When (x, y) is examined during DFS(x), it is not added to T because y is marked explored. Since y is not marked as explored when DFS(x) was first inoked, it is a node that was discoered between the inocation and end of the recursie call DFS(x). DFS tree DFS tree y is a descendant of x. Implementation Lists / arrays Queues / stacks Nontree edge Tree edge 0 Summary: BFS and DFS Representing Similarity: BFS/DFS builds the connected component containing s. Difference: BFS tree is flat/short; DFS tree is narrow/deep. What are the nontree edges in BFS/DFS? Q: How to produce all the connected components of a graph? A: A graph G = (V, E) V = the number of nodes = n E = the number of edges = m cardinality (size) of a set Dense or sparse? n For a connected graph, n! m!! n Linear time = O(m+n) Why? It takes O(m+n) to read the input 0

8 Adjacency Matrix What is a Queue? Two open ended container Consider a graph G = (V, E) with n nodes, V = {,, n}. The adjacency matrix of G is an nxn martix A where A[u, ] = if (u, ) Î E; A[u, ] = 0, otherwise. Time: Q() time for checking if (u, ) Î E. Q(n) time for finding out all neighbors of some u Î V. Visit many 0 s Space: Q(n ) What if sparse graphs? symmetric A queue is a set of elements from which we extract elements in first-in, first-out (FIFO) order. We select elements in the same order in which they were added. Q = (a,..., a n ) front rear Push (A) A Push (B) A B Push (C) A B C Push (D) front, rear front A B C D rear Pop () front front rear B C D rear front Push (E) front front rear Pop a a a n rear B C D E Push rear Adjacency List What is a Stack? One open ended and one close ended container 0 The adjacency list of G is an array Adj[]of n lists, one for each node represents its neighbors Adj[u] = a linked list of { (u, ) ÎE}. Time: degree of u: number of neighbors Q(deg(u)) time for checking one edge or all neighbors of a node. Space: O(n+m) [] [] 0 [] [] [] [] m 0 A stack is a set of elements from which we extract elements in last-in, first-out (LIFO) order. Each time we select an element, we choose the one that was added most recently. A S = (a,..., a n ) bottom B A C B A D C B A bottom a E D C B A a a n D C B A Pop Push [] [] 0 0 Push (A) Push (B) Push (C) Push (D) Push (E) Pop( ) = E

9 Linked Stacks and Queues Implementing DFS Implement queues and stacks by linked lists datalink... 0 Linked stack Pop Push Pop front data link Linked queue Adantage? - Variable sizes - Insert/delete in O() Push rear 0 We implement DFS by adjacency list Recursie procedure DFS(u). mark u as explored and add u to R. foreach edge (u, ) incident to u do. if ( is not marked as explored) then. recursiely inoke DFS() Alternatie implementation of DFS Process each adjacency list in the reerse order DFS(s) // S: a stack of nodes whose neighbors haen t been entirely explored. S = {s}. while (S is not empty) do. remoe a node u from S. if (Explored[u] = false) then. Explored[u] = true. for each (edge (u, ) incident to u) do. S = S + {} Q: Running time? O(n ) or O(n+m) Why? Implementing BFS Adjacency list is ideal for implementing BFS BFS(s) // T will be BFS tree rooted at s; layer counter i; layer list L[i]. Discoered[s] = true; Discoered[] = false for other. i = 0; L[0] = {s}; T = {};. while (L[i] is not empty) do. L[i+] = {};. for each (node uîl[i]) do. for each (edge (u, ) incident to u) do. if (Discoered[] = false) then. Discoered[] = true. T = T + {(u, )} 0. L[i+] = L[i+] + {}. i++ Q: How to manage each layer list L[i]? A: A queue/stack is fine Nodes in L[i] can be in any order Or, we can merge all layer lists into a single list L as a queue L L Q: Running time? O(n ) or O(n+m) Why? L L 0 Summary: Implementation Graph: Adjacency matrix s. Adjacency list Faster to find an edge? Faster to find degree? Faster to traerse the graph? Storage for sparse graph? Storage for dense graph? Edge insertion or deletion? Better for most applications? Graph traersal BFS: queue (or stack) DFS: stack O(n+m) time Winner Matrix List List List Matrix Matrix List

10 Is a Graph Bipartite? Testing Bipartiteness Q: Gien a graph G, is it bipartite? A: Color the nodes with blue and red (two-coloring) Application of BFS If a graph G is bipartite, then it cannot contain an odd cycle. Bipartite A bipartite graph (bigraph) is a graph whose nodes can be partitioned into sets X and Y in such a way that eery edge has one one end in X and the other end in Y. X and Y are two disjoint sets. No two nodes within the same set are adjacent. X Y G = (X, Y, E) 0 Testing Bipartiteness Color the nodes with blue and red Procedure: Arbitrarily chosen. Assume G = (V, E) is connected. Otherwise, we analyze connected components separately.. Pick any node sîv and color it red Anyway, s must receie some color.. Color all the neighbors of s blue.. Repeat coloring red/blue until the whole graph is colored.. Test bipartiteness: eery edge has ends of opposite colors. X Y

11 Implementation: Testing Bipartiteness Q: How to implement this procedure? Arbitrarily chosen A: BFS! O(m+n)-time We perform BFS from any s, coloring s red, all of layer L blue, Een/odd-numbered layers red/blue Insert the following statements after line 0 of BFS(s) (p. ) 0. L[i+] = L[i+] + {} 0a. if ((i+) is een) then 0b. Color[] = red 0c. else 0d. Color[] = blue X Y Proof: Correctness (/) Pf: (Case ) Suppose (x, y) is an edge with x, y in same layer L j. Let z = lca(x, y) = lowest common ancestor. Let L i be the layer containing z. Consider the cycle that takes edge from x to y, then path from y to z, then path from z to x. Its length is + (j-i) + (j-i), which is odd. x z = lca(x, y) (x, y) y ->z z ->x z Let xîl i, yîl j, and (x, y) ÎE. Then i and j differ by at most. s y L 0 L i L j Proof: Correctness (/) Let G be a connected graph and let L 0, L, be the layers produced by BFS starting at node s. Then exactly one of the following holds.. No edge of G joins two nodes of the same layer, and G is bipartite.. An edge of G joins two nodes of the same layer, and G contains an odd-length cycle (and hence is not bipartite). Pf: Case is triial. Connectiity in Directed L 0 L L L L 0 L L L Let xîl i, yîl j, and (x, y) ÎE. Then i and j differ by at most.

12 Recap: Directed Testing Strong Connectiity In a directed graph: asymmetric relationships u (u, ) Edges are directed, i.e., (u, )!= (, u) tail head e.g., u knows (celebrity), while doesn t know u. Directionality is crucial. Representation: Adjacency list Each node is associated with two lists, instead of one in an undirected graph. To which From which Graph search algorithms: BFS/DFS Almost the same as undirected graphs Again, directionality is crucial. Q: What can we reach from node? A: Q: Can we determine if a graph is strongly connected in linear time? A: Yes. How? Why? TestSC(G). pick any node s in G G re : reerse the direction of eery edge in G. R = BFS(s, G) (u, ) in G re if (, u)in G. R re = BFS(s, G re ). if (R = V = R re ) then return true else false Time: O(m+n) R = V = R Q: Correctness? re s Q: How to partition a directed graph into strong components? u Strong Connectiity Strong Component Nodes u and are mutually reachable if there is a path from u to and also a path from to u. u The strong component containing s in a directed graph is the maximal set of all s.t. s and are mutually reachable. a.k.a. strongly connected component A directed graph is strongly connected if eery pair of nodes are mutually reachable. Q: When are there no mutually reachable nodes? Lemma: If u and are mutually reachable, and and w are mutually reachable, then u and w are mutually reachable. Simple but important! Pf: Theorem: For any two nodes s and t in a directed graph, their strong components are either identical or disjoint. Q: When are they identical? When are they disjoint? Pf: Identical if s and t are mutually reachable s --, s -- t, -- t Disjoint if s and t are not mutually reachable Proof by contradiction u w u w

13 Topological Ordering DAGs and Topological Ordering Q: driers come to the junction simultaneously, who goes first? Deadlock! Dependencies form a cycle! The drier must come to a complete s at a s sign. Generally the drier who arries and ss first continues first. If two or three driers in different directions s simultaneously at a junction controlled by s signs, generally the driers on the left must yield the right-of-way to the drier on the far right. Gien a directed graph G, a ological ordering is an ordering of its nodes as,,, n so that for eery edge ( i, j ), we hae i<j. Precedence constraints: edge ( i, j ) means i must precede j. Lemma: If G has a ological ordering, then G is a DAG. Pf: Proof by contradiction! How? Consider a cycle, i,, j, i. i j Directed Acyclic Example 0 Q: If an undirected graph has no cycles, then what s it? A: A tree (or forest). At most n- edges. A directed acyclic graph (DAG) is a directed graph without cycles. A DAG may hae a rich structure. A DAG encodes dependency or precedence constraints e.g., prerequisite of Algorithms: Data structures Discrete math Programming C/C++ e.g., execution order of instructions in CPU Pipeline structures Example: a DAG Lemma: If G has a ological ordering, then G is a DAG. Q: If G is a DAG, then does G hae a ological ordering? Q: If so, how do we compute one? A: Key: find a way to get started! Q: How? the same DAG with ological ordering

14 Where to Start? w x u Topological Ordering A: A node that depends on no one, i.e., unconstrained. Lemma: In eery DAG G, there is a node with no incoming edges. Pf: Proof by contradiction! Suppose that G is a DAG where eery node has at least one incoming edge. Let's see how to find a cycle in G. Pick any node, and begin following edges backward from : Since has at least one incoming edge (u, ) we can walk backward to u. Then, since u has at least one incoming edge (x, u), we can walk backward to x; and so on. Repeat this process n+ times (the initial counts one). We will isit some node w twice, since G has only n nodes. Let C denote the sequence of nodes encountered between successie isits to w. Clearly, C is a cycle. Lemma: If G is a DAG, then G has a ological ordering. Pf: Proof by induction!. Base case: true if n =.. Inductie step: Induction hypothesis: true for DAGs with up to n nodes Gien a DAG on n+ nodes, find a node w/o incoming edges. G {} + <,,, n > = <,,,, n > G {} is a DAG, since deleting cannot create any cycles. G {} has n nodes. By induction hypothesis, G {} has a ological ordering. Place first in ological ordering. This is safe since all edges of point forward. Then append nodes of G {} in ological order after. Example: Topological Ordering A Linear-Time Algorithm In fact, the proof has already suggested an algorithm. TopologicalOrder(G). find a node without incoming edges. order. G = G {} // delete from G. if (G is not empty) then TopologicalOrder(G) Time: From O(n ) to O(m+n) O(n )-time: Total n iterations; line in O(n)-time. How? O(m+n)-time: How? Maintain the following information indeg(w) = # of incoming edges from undeleted nodes S = set of nodes without incoming edges from undeleted nodes Initialization: O(m+n) ia single scan through graph Update: line deletes Remoe from S Decrement indeg(w) for all edges from to w, and add w to S if indeg(w) hits 0; this is O() per edge