Introduction to Algorithms
Elementary Graph Algorithms
My T. Thai @ UF
Outline
Graph representation Graph-searching algorithms
Breadth-first search Depth-first search
Topological sort Strongly connected components
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Graphs
Given a graph G=(V, E) V: set of vertices E: set of edges
Types of graphs: Undirected: Directed: Weighted: each edge e has a weight w(e) Dense: Sparse: |E| = o(V2)
|E| = O(|V|2) My T. [email protected]
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Representations of graphs
Two common ways to represent for algorithms: Adjacency lists
Adjacency matrix
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Adjacency lists
Array Adj of |V| lists, one per vertex List of vertex u, Adj[u], include all vertices v
such that If edges have weights, can put the weights in
the lists
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Adjacency matrix
Store adjacent matrix |V| × |V| matrix A = (aij )
If edges have weights, aij is the weight of edge (i, j)
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Adjacency lists vs Adjacency matrix
Adjacency lists Adjacency matrix Space:
Time: to list all vertices adjacent to u:
Time: to determine if (u, v) ∈ E:
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Graph searching algorithms
Visiting all the nodes in a graph in a particular manner following the edges of the graph
Used to discover the structure of the graph Two common searching algorithms:
Breadth-first search (BFS): visit the sibling nodes before visiting the child nodes (visit node layer by layer based on the distance to the root)
Depth-first search (DFS): visit the child nodes before visiting the sibling nodes (follow the “depth” of the graph)
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Breadth-first search
Input: Graph G = (V, E), either directed or undirected, and source vertex s ∈ V
Output: d[v] = distance (smallest # of edges, or shortest path)
from s to v, for all v V such that (u, v) is last edge on shortest path s
to v u is v’s predecessor
Builds breadth-first tree with root s that contains all reachable vertices
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Breadth-first search Discover vertices in a wave manner from the
source s First hit all vertices 1 edge from s, then vertices 2 edges
from s … A vertex is discovered at the first time that it is encountered during
the search. A vertex is finished if all vertices adjacent to it have
been discovered. Colors the vertices to keep track of progress.
White: Undiscovered. Gray: Discovered but not finished. Black: Finished.
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Queue Q contains all discovered but not finished vertices – gray vertices
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Analysis of BFS Initialization takes O(V)
In while loop Each vertex is enqueued and dequeued at most once =>
total time for queuing is O(V) The adjacency list of each vertex is scanned at most
once. The sum of lengths of all adjacency lists is (E)
=>Total running time: O(V+E) linear in the size of the adjacency list representation of the graph
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Correctness of BFS
Where is the shortest distance from s to v
Proof (sketched): : If t is the sequence of vertices in the
queue, then If v is reached from u where d.u =
=>My T. Thai
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Depth-first search
Input: G = (V, E), directed or undirected. No source vertex given.
Output: 2 timestamps on each vertex: d[v] = discovery time f [v] = finishing time [v] : predecessor of v = u, such that v was
discovered during the scan of u’s adjacency list Build depth-first forest comprising several
depth-first trees
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Depth-first search
From a vertex v Follow an outgoing edge to explore Keep searching as deep as possible When all vertices reachable through the edge is
explored, follow another outgoing edge of v
When all vertices reachable from the original source are discovered, choose an undiscovered vertex as a new source then repeat the process
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Analysis of DFS
In DFS: loops on lines 1-2 & 5-7 take (V) time, excluding time to execute DFS-Visit
DFS-Visit is called once for each white vertex vV when it’s colored gray the first time. Lines 3-6 of DFS-Visit is executed |Adj[v]| times. The total cost of executing all DFS-Visit is vV|Adj[v]| = (E)
Total running time of DFS is (V+E)My T. Thai
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Parenthesis TheoremTheorem 22.7 For all u, v, exactly one of the following holds:
1. d[u] < f [u] < d[v] < f [v] or d[v] < f [v] < d[u] < f [u] and neither u nor v is a descendant of the other.
2. d[u] < d[v] < f [v] < f [u] and v is a descendant of u.
3. d[v] < d[u] < f [u] < f [v] and u is a descendant of v d[u] < d[v] < f [u] < f [v] cannot happen Like parentheses:
OK: ( ) [ ] ( [ ] ) [ ( ) ] Not OK: ( [ ) ] [ ( ] )
Corollary 22.8
v is a proper descendant of u if and only if d[u]<d[v]<f [v]<f [u]My T. Thai
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Proof of Parenthesis Theorem
W.l.g suppose d[u] < d[v], there are two cases: d[v] < f[u]
v is discovered while u is gray v is descendant of u v is discovered more recently than u v is finished
before the algorithm comes back and finishes u Interval [d[v], f[v]] is included in [d[u], f[u]]
d[v] > f[u] d[v] < f[v] Two intervals [d[u], f[u]] and [d[v], f[v]] are disjoint
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White-path theoremTheorem 22.9 In a depth-first forest of a (directed or
undirected) graph G =(V,E), vertex v is a descendant of vertex u if and only if at the time d[u] that the search discovers u, there is a path from u to v consisting entirely of white vertices.
Proof:
=> at time d[u], all vertices on the path from u to v, including v are descendant of u Apply corollary 22.8, d[u] < d[v], these vertices are white at time d[u]
<= Suppose there is a white path from u to v but v is not a descendant of u Let v’ the nearest vertex to u on the path s.t. v’ is not descendant of u w is the prior vertex of v’ on the path w is descendant of u d[u]
< d[w] < d[v’] < f[v’] < f[w]< f[u] v’ is the descendant of uMy T. [email protected]
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Classification of edges
Tree edge: in the depth-first forest. Found by exploring (u, v)
Back edge: (u, v), where u is a descendant of v Forward edge: (u, v), where v is a descendant of u, but
not a tree edge Cross edge: any other edge. Can go between vertices in
same depth-first tree or in different depth-first trees
Theorem 22.10 In a depth-first search of an undirected graph G, every edge of G is either a tree edge or a back edge.
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Topological sort
Directed acyclic graph (DAG) A directed graph with no cycles Good for modeling processes and structures that
have a partial order, edge (u, v) represent order u>v
Topology sort: make a total order of a given DAG such that vertex u appears before v if there is edge (u, v)
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Characterizing DAGs
Lemma 2.11. A directed graph G is acyclic if and only if a depth-first search of G yields no back edges.
Proof: ⇒: Show that back edge cycle⇒
Suppose there is a back edge (u, v) => v is ancestor of u in depth-first forest => there is a path from v to u => there is a cycle
⇐: Show that cycle back edge⇒ Suppose G contains cycle c. Let v be the first vertex discovered in
c, and let (u, v) be the preceding edge in c. At time d[v], vertices of c form a white path v to u By white-path theorem, u is descendant of v in depth-first forest
=>(u, v) is a back edge.
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Strongly connected components
Given directed graph G = (V, E), a strongly connected component (SCC) of G is a maximal set of vertices C V ⊆ such that for all u, v C∈ , there are paths both from u to v and from v to u
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Component Graph
GSCC = (VSCC, ESCC). VSCC has one vertex for each SCC in G ESCC has an edge if there’s an edge between the
corresponding SCC’s in G
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GSCC is a DAG
Proof Suppose there is a path v v in G there are paths u u v and v v u in G u and v are reachable from each other, so they
are not in separate SCC’s.
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Transpose of a Directed Graph
GT = transpose of directed graph G GT = (V, ET), ET = {(u, v) : (v, u) E}. GT is G with all edges reversed.
Can create GT in Θ(V + E) time if using adjacency lists.
Observation: G and GT have the same SCC’s. (u and v are reachable from each other in G if and only if reachable from each other in GT.)
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How does it work? Idea:
By considering vertices in second DFS in decreasing order of finishing times from first DFS, we are visiting vertices of the component graph in topologically sorted order.
Because we are running DFS on GT, we will not be visiting any v from u, where v and u are in different components.
Notation: d[u] and f [u] always refer to first DFS. Extend notation for d and f to sets of vertices U V: d(U) = minuU{d[u]} (earliest discovery time)
f (U) = maxuU{ f [u]} (latest finishing time)
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SCCs and DFS finishing times
Proof: Case 1: d(C) < d(C)
Let x be the first vertex discovered in C. At time d[x], all vertices in C and C are white.
=> there exist paths of white vertices from x to all vertices in C and C.
By the white-path theorem, all vertices in C and C are descendants of x in depth-first tree.
By the parenthesis theorem, f [x] = f (C) > f(C).
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SCCs and DFS finishing times
Proof: Case 2: d(C) > d(C)
Let y be the first vertex discovered in C At time d[y], all vertices in C are white and
there is a white path from y to each vertex in C all vertices in C become descendants of y Again, f [y] = f (C).
At time d[y], all vertices in C are also white. By earlier lemma, since there is an edge (u, v),
we cannot have a path from C to C => no vertex in C is reachable from y => at time f [y], all vertices in C are still white
=> for all w C, f [w] > f [y] => f (C) > f (C)
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SCCs and DFS finishing times
Proof: (u, v) ET (v, u) E Since SCC’s of G and GT are the same, f(C) >
f (C), by Lemma 22.14.
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Correctness
When we do the second DFS, on GT, start with SCC C such that f(C) is maximum. The second DFS starts from some x C, and it
visits all vertices in C. Corollary 22.15 says that since f(C) > f (C) for all
C C, there are no edges from C to C in GT.
=> DFS will visit only vertices in C.
=> the depth-first tree rooted at x contains exactly the vertices of C.
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Correctness The next root chosen in the second DFS is in SCC C such
that f (C) is maximum over all SCC’s other than C. DFS visits all vertices in C, but the only edges out of C go to C,
which we’ve already visited. Therefore, the only tree edges will be to vertices in C.
Each time we choose a root for the second DFS, it can reach only vertices in its SCC—get tree edges to these, vertices in SCC’s already visited in second DFS—get no tree edges
to these.
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Summary
Two common ways to represent a graph: adjacency lists and adjacency matrix. Choosing representation depends on: Type of graph: dense or sparse Type of operation that is used frequently in the problem adjacency lists saves more space than adjacency matrix
BFS starts from a given source and computes the shortest paths from the source to other vertices. Some vertices may be not discovered
BFS starts from an arbitrary vertex and discovers the whole graph to capture the graph’s structure
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Both BFS and DFS can provide a set of reachable vertices from a given vertex
Topology sort is used to form the total order from partial orders
Strongly connected components list all strongly connected components in their topology order
BFS, DFS, Topology Sort, Strongly Connected Component run in linear time in term of the graph’s size
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