1. A few motivations for Graph Search:
a) check if a network is connected ( can get to anywhere from anywhere else)
b) driving direction
c) formulate a plan ( the nodes are the steps and arcs are the choose of the step sequences)
d) compute the "pieces" or "components" of a graph
2. Generic Graph Search:
Goals: a) find everything findable from a given start vertex
b) don't explore anything twice
Algorithm: Given Graph G , vertex s :
a) initially s is explored and all the other vertices are unexplored
b) while possible :
-- choose an edge (u, v) with u explored and v unexplored
-- mark v explored
3. Claim : at then end of Generic Graph Search, v is explored <==> G has a path from s to v.
Proof : (==>) by induction on number of iterations
(<==) by contradiction: if G has path from s to v and v is unexplored, there must be an edge (u, w) in the path , that u is explored and w is unexplored, in that case, the algorithm should not terminate.
4. BFS V.S. DFS: how to choose among the possible many "frontier" edges.
Breadth-FIrst-Search :
-- explore nodes in "layers"
-- can compute "shortest path"
-- can compute connected components of an undirected graph
Depth-First-Search:
-- explored aggressively like a maze, backtrack only when necessary.
-- can compute topological ordering of directed acyclic graph
-- can compute connected components of a directed graph
5. Breadth-First-Search Implementation:BFS(Graph G, start vertex s)
a) all nodes initially unexplored
b) mark s as explored
c) let Q = queue data structure (FIFO) , initialized with s
d) while Q is not empty :
-- remove the first node of Q , call it v
-- for each edge (v, w)
-- if w is unexplored, mark w as explored and add w to Q at end
Running Time is O ( m + n)
6. Application of BFS : Shortest Path:
Goal : compute dist(v) , the fewest number of edges on a path from s to v.
Extra code than BFS :
a) initialize dist(v) = 0 if v=s or plus inifitive if v != s
b) when considering an edge (v,w) :
if w is unexplored , then set dist(w) = dist(v) + 1
Claim : at termination , dist(v) = i <==> v in ith layer <==> shortest s-v path has i edges
Proof : every node w in layer i is added to Q by a node v in layer i - 1 via the edge (v, w)
7. Application of BFS : Undirected Connectivity:
Let G = (V, E) be an undirected graph.
Equivalent relation u ~ v <==> there is a path from u to v.
Definition of connected components: equivalence classes of relation ~
Application : check if network is disconnected, graph visualization ( clustering)
Algorithm to compute all connected components of an undirected graph:
a) initally all nodes unexplored
b) for i = 1 to n :
-- if node i not yet explored , BFS( G, i )
Running time: O (m + n)
8. Non-Recursive implementation of DFS(Graph G, start vertex s)
a) all nodes initially unexplored
b) mark s as explored
c) let Stack = stack data structure (FILO) , initialized with s
d) while Stack is not empty :
-- remove the last node of Stack , call it v
-- for the next edge (v, w) that is not explored through v
-- if w is unexplored, mark w as explored and add v to Stack at end
Running Time is O ( m + n)
9. Recursive implementation of DFS(Graph G, start vertex s):
-- initially all nodes unexplored
-- mark s as explroed
-- for every edge (s, v) :
-- if v unexplored : DFS(G, v)
10. Application of DFS : Topological Sort :
Definition : A topological ordering of a directed graph G is a labelling f of G's nodes such that :
a) for different v , f(v) is different and f(v) is in {1, 2, .., n}
b) If there is an edge (u, v) in G => f(u) < f(v)
Motivation : Sequence tasks while respecting all precedence constraints.
Claim : G has directed cycle => no topological ordering.
no directed cycle => can compute topological ordering in O( m + n ) time.
11. Claim : Every directed acyclic graph has a sink vertex ( vertex that doesn't has outgoing arcs)
Proof : if not can keep following outgoing arcs to produce a path that has more than n-1 arcs, then it must contain duplicated vertex.
12. To compute topological ordering :
-- Let v be a sink vertex of G
-- Set f(v) = n
-- recurse on G-{v} // A DAG removing a vertex is still a DAG
13. Topological Sort via DFS:
DFS-Loop ( graph G ) :
-- mark all nodes unexplored.
-- current-label = n
-- for each vertex v in G
-- if v not yet explored , DFS (G, v)
DFS (G, s)
-- mark s as explroed
-- for every edge (s, v) :
-- if v unexplored : DFS(G, v)
-- label s as current-label
-- current-label --
14. Proof of correctness of DFS for computing topological ordering:
Goal : need to show that if ( u, v ) is an edge in G, then f(u) < f(v);
case 1: if u is explored before v => recursive call on v finishes before that on u, so f(u) < f(v)
case 2: if u isexplored after v => because there is no cycle in the graph , so there is not path from v to u, so after v finish the recursive call , u is still not explored so , f(u) < f(v)
15. The strong connected components (scc) of a directed graph G are the equivalence classes of the relation ~ :
u ~ v <==> there is a path from u to v and a path from v to u.
16. Kosaraju’s Two-Pass Algorithm for computing SCC:
a) Let G(rev) = G with all arcs reversed
b) run DFS-Loop on G(rev), Let f(v) = "finishing time" of each v in G
(Goal : compute magical ordering of notes. )
c) run DFS-Loop on G , proccessing nodes in decreasing order of finishing times
(Goal : discover the SCCs one by one , SCCs = nodes with the same "leader" )
DFS-Loop ( graph G ) :
-- mark all nodes unexplored.
-- global variable t = 0 // # of nodes finished so far, for finishing time in 1st DFS loop
-- global variable l = NULL // current source vertex , for leaders in 2nd DFS loop
-- for i = n downto 1
-- v = ith node in G
-- if v not yet explored , l = v , DFS (G, v)
DFS (G, s)
-- mark s as explroed
-- for every edge (s, v) :
-- if v unexplored : DFS(G, v)
-- f(s) = ++ t
-- leader(s) = l
17. SCC in G and G(rev) are exactly the same.
18. Claim: Consider two ajacent SCCs in G, if there is a path from SCC1 to SCC2 ( no path from SCC2 to SCC1 , otherwise they are in the same SCC), Let f(v) = finishing times of DFS-Loop in G(rev), then:
max ( v in SCC1){f(v)} < max (v in SCC2) {f(v)}
Proof: Let v = 1st node of SCC1 U SCC2 explored by 1st pass of DFS-Loop ( on G(rev) ).
Case 1 : v is in SCC1, in G(rev) no path from SCC1 to SCC2, so all nodes in SCC1 get explored before nodes in SCC2. So finishing times of nodes SCC2 are greater than all those in SCC1. Case 2 : v is in SCC2, there is a path from SCC2 to SCC1 in G(rev) , so DFS on v won't finish untill all nodes in SCC1 U SCC2 are explored. so f(v) has the greater finishing time.
19. Correctness of DFS for computing SCC :
a) max value of f(v) in G must lie in a "Sink SCC" SCC*
b) First call to DFS of 2nd pass DFS-Loop on G discoveres SCC*
c) rest of DFS-Loop on G like recursing on G with SCC* deleted.
d) successive calls to DFS "peal off" the SCCs one by one.(in the reverse topological order of "meta-Graph")
20. The Web Graph:
a) vertices = web pages
b) directed edges = hyperlinks
c) the graph forms a Bow Tie :
Main Findings:
a) all 4 parts have roughly the same size
b) within core , very well connected
c) outside, surprisingly poorly connected
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