Depth-First and Breadth-First Search - Combinatorial Computing | MAT 3770, Study notes of Mathematics

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Section 3.2 DepthñFirst and BreadthñFirst Search

I (^) Tree-based searches abound in applications, and are amenable to computerized solutions.

I (^) Depending upon the application, we may wish to nd

1. one solution
2. all solutions, or
3. an optimal solution

I (^) To organize the enumeration of possible solutions:

1. let the sequential choices be internal nodes in a rooted tree
2. the solutions and ìdead endsî be the leaves

I (^) The big challenge is to be sure to check that all possible ways to generate a solution are investigated: that the enumeration of possibilities is complete.

Example: DFS(G, f)

I (^) numbers will indicate visitation order

I (^) dotted lines will represent unmarked edges

I (^) use alphabetic order to choose among next nodes to visit

a

b

c

d f^

i

g

h

DFS Algorithm

Given : An undirected graph G = (V , E ), and a vertex v ∈ V Goal : Visit all vertices and edges of G starting with v.

Procedure DFS (G, v)

// PRE: G is connected

// POST: every vertex and edge of G

// is visited

begin

Mark (v)

Visit (v) // do application processing

for all unmarked edges <v, w> do

if w is unmarked then

MarkEdge (v, w)

DFS (G, w)

VisitEdge (v, w) // process if necessary

end

Proof:

I (^) Recall:

deg(v) = 2jE j, i.e., the sum of the degrees of all nodes is twice the number of edges.

I (^) For any v, DFS(G , v) is called exactly once ó when v is unmarked.

I (^) So, if < v, w > is an edge, it's visited and marked at that time by DFS(G , v) or DFS(G , w), whichever node is visited rst.

I Note also, this means the FOR loop is executed a total

of 2jE j times over all calls of DFS.

I (^) As for marked edges: Claim : DFS(G , v) marks edges which form a tree with root v. [Proof by induction on number of unmarked vertices in G .]

Lemma

If T is the tree formed by DFS (G , v), then for each edge e ∈ EG , either I (^) e ∈ T , or I (^) e connects a vertex in T to some ancestor or descendent in T. I.e., there are no cross–edges in T.

Proof. Let e = < u, W >. if u is marked rst, then w is marked during DFS (G , u). So, by the previous lemma (vertices are visited only once), w is a descendent of u.

DFS Drawback

I (^) Suppose we don't want to visit all the vertices ó we merely want to nd one with a certain property: i.e., we want to be able to stop.

I (^) Idea : nonñrecursive version of DFS using a stack.

I (^) We'll assume we're not interesting in visiting every edge.

Nonñrecursive DFS Algorithm

Procedure DFS2 (G, v)

// PRE: G is connected or we only visit

// subgraph in which v is found

// POST: every vertex of G is visited

begin

CreateStack(S)

Mark(v)

Push(S, v)

while not Empty(S)

x = Pop(S)

while x has unvisited neighbor (w)

Mark(w)

Push(S, w)

end

DFS in DiGraphs

Works the same as for undirected graphs, but there are differences with respect to the DFS tree ó it may not be a tree, but a forest.

There are four possible types of edges in the DFS forest:

I (^) the tree edges: of the DFS tree

I (^) back edges: edges to ancestors in the tree

I (^) forward edges: edges to descendents in the tree

I (^) cross edges: all others

DiGraph Example: DFS(G, f)

a

b

c

d f^

i

g

h

Lemma

Suppose there is a directed path v 1 ; v 2 ; : : : ; vn in G such that v 1 has lowest DFSnumber in fv 1 ; : : : ; vn g. Then, ∀ 1 < i  n, vi is a descendent of v 1.

Proof:

Note v 1 and all descendents have DFSnumbers in the interval [a..b], and DFSnumber(v 1 ) = a.

Proof will be by induction on the number of nodes in the path.

BC Let n = 2, and v 1 → v 2. By the last lemma, v 2 is a descendent of v 1.

IH Assume for directed paths with n 1 nodes where v 1 has lowest DFSnumber, that v 2 ; : : : ; vn 1 are descendents of v 1.

IS Show lemma is true for paths with n nodes. If vn is not a descendent of v 1 , and the DFSnumber(v 1 ) > DFSnumber(vn ), then DFSnumber(vn ) >DFSnumber(vn 1 ) since vn 1 is closer to v 1 in the path (and hence will be processed before vn ).

Application: Acyclic Digraphs

I (^) Suppose a program consists of many procedures.

I (^) Let us say procedure X depends on Procedure Y if there are procedures X = X 1 , X 2 ,... , Xn = Y such that Xi calls Xi + 1 ∀ i < n.

I (^) Question: Does any procedure depend on itself?

I (^) Why care?

I (^) As programmers, this is a sort of recursion I (^) As compiler writers, we can allocate one big activation record for all the procedures if none depend on themselves.

Directed Cycles and Backedges

I (^) Consider procedures as vertices of a directed graph, and X → X′^ if X calls X′

I (^) We want to know: does G contain any directed cycles?

I (^) Theorem. let G = (V, E) be a directed graph, and let T be a DFS tree (forest) of G. Then G contains a directed cycle IFF T contains a back edge.