lecture 18

CS 332: Algorithms Graph Algorithms

Administrative Test postponed to Friday Homework: Turned in last night by midnight: full credit Turned in tonight by midnight: 1 day late, 10% off Turned in tomorrow night: 2 days late, 30% off Extra credit lateness measured separately

Review: Graphs A graph G = (V, E) V = set of vertices, E = set of edges Dense graph: |E|  |V| 2 ; Sparse graph: |E|  |V| Undirected graph: Edge (u,v) = edge (v,u) No self-loops Directed graph: Edge (u,v) goes from vertex u to vertex v, notated u  v A weighted graph associates weights with either the edges or the vertices

Review: Representing Graphs Assume V = {1, 2, …, n } An adjacency matrix represents the graph as a n x n matrix A: A[ i , j ] = 1 if edge ( i , j )  E (or weight of edge) = 0 if edge ( i , j )  E Storage requirements: O(V 2 ) A dense representation But, can be very efficient for small graphs Especially if store just one bit/edge Undirected graph: only need one diagonal of matrix

Review: Graph Searching Given: a graph G = (V, E), directed or undirected Goal: methodically explore every vertex and every edge Ultimately: build a tree on the graph Pick a vertex as the root Choose certain edges to produce a tree Note: might also build a forest if graph is not connected

Review: Breadth-First Search “Explore” a graph, turning it into a tree One vertex at a time Expand frontier of explored vertices across the breadth of the frontier Builds a tree over the graph Pick a source vertex to be the root Find (“discover”) its children, then their children, etc.

Review: Breadth-First Search Again will associate vertex “colors” to guide the algorithm White vertices have not been discovered All vertices start out white Grey vertices are discovered but not fully explored They may be adjacent to white vertices Black vertices are discovered and fully explored They are adjacent only to black and gray vertices Explore vertices by scanning adjacency list of grey vertices

Review: Breadth-First Search BFS(G, s) { initialize vertices; Q = {s}; // Q is a queue (duh); initialize to s while (Q not empty) { u = RemoveTop(Q); for each v  u->adj { if (v->color == WHITE) v->color = GREY; v->d = u->d + 1; v->p = u; Enqueue(Q, v); } u->color = BLACK; } } What does v->p represent? What does v->d represent?

Breadth-First Search: Example         r s t u v w x y

Breadth-First Search: Example   0      r s t u v w x y s Q:

Breadth-First Search: Example 1  0 1     r s t u v w x y w Q: r

Breadth-First Search: Example 1  0 1 2 2   r s t u v w x y r Q: t x

Breadth-First Search: Example 1 2 0 1 2 2   r s t u v w x y Q: t x v

Breadth-First Search: Example 1 2 0 1 2 2 3  r s t u v w x y Q: x v u

Breadth-First Search: Example 1 2 0 1 2 2 3 3 r s t u v w x y Q: v u y

Breadth-First Search: Example 1 2 0 1 2 2 3 3 r s t u v w x y Q: u y

Breadth-First Search: Example 1 2 0 1 2 2 3 3 r s t u v w x y Q: y

Breadth-First Search: Example 1 2 0 1 2 2 3 3 r s t u v w x y Q: Ø

BFS: The Code Again BFS(G, s) { initialize vertices; Q = {s}; while (Q not empty) { u = RemoveTop(Q); for each v  u->adj { if (v->color == WHITE) v->color = GREY; v->d = u->d + 1; v->p = u; Enqueue(Q, v); } u->color = BLACK; } } What will be the running time? Total running time: O(V+E) Touch every vertex: O(V) u = every vertex, but only once ( Why? ) So v = every vertex that appears in some other vert’s adjacency list

BFS: The Code Again BFS(G, s) { initialize vertices; Q = {s}; while (Q not empty) { u = RemoveTop(Q); for each v  u->adj { if (v->color == WHITE) v->color = GREY; v->d = u->d + 1; v->p = u; Enqueue(Q, v); } u->color = BLACK; } } What will be the storage cost in addition to storing the graph? Total space used: O(max(degree(v))) = O(E)

Breadth-First Search: Properties BFS calculates the shortest-path distance to the source node Shortest-path distance  (s,v) = minimum number of edges from s to v, or  if v not reachable from s Proof given in the book (p. 472-5) BFS builds breadth-first tree , in which paths to root represent shortest paths in G Thus can use BFS to calculate shortest path from one vertex to another in O(V+E) time

Depth-First Search Depth-first search is another strategy for exploring a graph Explore “deeper” in the graph whenever possible Edges are explored out of the most recently discovered vertex v that still has unexplored edges When all of v ’s edges have been explored, backtrack to the vertex from which v was discovered

Depth-First Search Vertices initially colored white Then colored gray when discovered Then black when finished

Depth-First Search: The Code DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; }

Depth-First Search: The Code DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } What does u->d represent?

Depth-First Search: The Code DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } What does u->f represent?

Depth-First Search: The Code DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } Will all vertices eventually be colored black?

Depth-First Search: The Code DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } What will be the running time?

Depth-First Search: The Code DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } Running time: O(n 2 ) because call DFS_Visit on each vertex, and the loop over Adj[] can run as many as |V| times

Depth-First Search: The Code DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } BUT, there is actually a tighter bound. How many times will DFS_Visit() actually be called?

Depth-First Search: The Code DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } So, running time of DFS = O(V+E)

Depth-First Sort Analysis This running time argument is an informal example of amortized analysis “ Charge” the exploration of edge to the edge: Each loop in DFS_Visit can be attributed to an edge in the graph Runs once/edge if directed graph, twice if undirected Thus loop will run in O(E) time, algorithm O(V+E) Considered linear for graph, b/c adj list requires O(V+E) storage Important to be comfortable with this kind of reasoning and analysis

DFS Example 1 | | | | | 3 | 2 | | source vertex d f

DFS Example 1 | | | | | 3 | 4 2 | | source vertex d f

DFS Example 1 | | | | 5 | 3 | 4 2 | | source vertex d f

DFS Example 1 | | | | 5 | 6 3 | 4 2 | | source vertex d f

DFS Example 1 | 8 | | | 5 | 6 3 | 4 2 | 7 | source vertex d f

DFS Example 1 | 8 | | | 5 | 6 3 | 4 2 | 7 9 | source vertex d f What is the structure of the grey vertices? What do they represent?

DFS Example 1 | 8 | | | 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f

DFS Example 1 | 8 |11 | | 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f

DFS Example 1 |12 8 |11 | | 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f

DFS Example 1 |12 8 |11 13| | 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f

DFS Example 1 |12 8 |11 13| 14| 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f

DFS Example 1 |12 8 |11 13| 14|15 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f

DFS Example 1 |12 8 |11 13|16 14|15 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f

DFS: Kinds of edges DFS introduces an important distinction among edges in the original graph: Tree edge : encounter new (white) vertex The tree edges form a spanning forest Can tree edges form cycles? Why or why not?

DFS Example 1 |12 8 |11 13|16 14|15 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f Tree edges

DFS: Kinds of edges DFS introduces an important distinction among edges in the original graph: Tree edge : encounter new (white) vertex Back edge : from descendent to ancestor Encounter a grey vertex (grey to grey)

DFS Example 1 |12 8 |11 13|16 14|15 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f Tree edges Back edges

DFS: Kinds of edges DFS introduces an important distinction among edges in the original graph: Tree edge : encounter new (white) vertex Back edge : from descendent to ancestor Forward edge : from ancestor to descendent Not a tree edge, though From grey node to black node

DFS Example 1 |12 8 |11 13|16 14|15 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f Tree edges Back edges Forward edges

DFS: Kinds of edges DFS introduces an important distinction among edges in the original graph: Tree edge : encounter new (white) vertex Back edge : from descendent to ancestor Forward edge : from ancestor to descendent Cross edge : between a tree or subtrees From a grey node to a black node

DFS Example 1 |12 8 |11 13|16 14|15 5 | 6 3 | 4 2 | 7 9 |10 source vertex d f Tree edges Back edges Forward edges Cross edges

DFS: Kinds of edges DFS introduces an important distinction among edges in the original graph: Tree edge : encounter new (white) vertex Back edge : from descendent to ancestor Forward edge : from ancestor to descendent Cross edge : between a tree or subtrees Note: tree & back edges are important; most algorithms don’t distinguish forward & cross

DFS: Kinds Of Edges Thm 23.9: If G is undirected, a DFS produces only tree and back edges Proof by contradiction: Assume there’s a forward edge But F? edge must actually be a back edge ( why? ) source F?

DFS: Kinds Of Edges Thm 23.9: If G is undirected, a DFS produces only tree and back edges Proof by contradiction: Assume there’s a cross edge But C? edge cannot be cross: must be explored from one of the vertices it connects, becoming a tree vertex, before other vertex is explored So in fact the picture is wrong…both lower tree edges cannot in fact be tree edges source C?

DFS And Graph Cycles Thm: An undirected graph is acyclic iff a DFS yields no back edges If acyclic, no back edges (because a back edge implies a cycle If no back edges, acyclic No back edges implies only tree edges ( Why? ) Only tree edges implies we have a tree or a forest Which by definition is acyclic Thus, can run DFS to find whether a graph has a cycle

DFS And Cycles How would you modify the code to detect cycles? DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; }

DFS And Cycles What will be the running time? DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; }

DFS And Cycles What will be the running time? A: O(V+E) We can actually determine if cycles exist in O(V) time: In an undirected acyclic forest, |E|  |V| - 1 So count the edges: if ever see |V| distinct edges, must have seen a back edge along the way

lecture 18

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lecture 18