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CSC 413/513: Intro to Algorithms. Graph Algorithms DFS. 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

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depth first search
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 search1
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
depth first search the code1
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

What does u->d represent?

depth first search the code2
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

What does u->f represent?

depth first search the code3
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

Will all vertices eventually be colored black?

dfs example
DFS Example

sourcevertex

dfs example1
DFS Example

sourcevertex

d f

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dfs example2
DFS Example

sourcevertex

d f

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dfs example3
DFS Example

sourcevertex

d f

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dfs example4
DFS Example

sourcevertex

d f

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3 | 4

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dfs example5
DFS Example

sourcevertex

d f

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3 | 4

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dfs example6
DFS Example

sourcevertex

d f

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3 | 4

5 | 6

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dfs example7
DFS Example

sourcevertex

d f

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2 | 7

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3 | 4

5 | 6

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dfs example8
DFS Example

sourcevertex

d f

1 |

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2 | 7

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3 | 4

5 | 6

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dfs example9
DFS Example

sourcevertex

d f

1 |

8 |

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2 | 7

9 |

3 | 4

5 | 6

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What is the structure of the grey vertices? What do they represent?

dfs example10
DFS Example

sourcevertex

d f

1 |

8 |

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2 | 7

9 |10

3 | 4

5 | 6

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dfs example11
DFS Example

sourcevertex

d f

1 |

8 |11

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2 | 7

9 |10

3 | 4

5 | 6

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dfs example12
DFS Example

sourcevertex

d f

1 |12

8 |11

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2 | 7

9 |10

3 | 4

5 | 6

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dfs example13
DFS Example

sourcevertex

d f

1 |12

8 |11

13|

2 | 7

9 |10

3 | 4

5 | 6

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dfs example14
DFS Example

sourcevertex

d f

1 |12

8 |11

13|

2 | 7

9 |10

3 | 4

5 | 6

14|

dfs example15
DFS Example

sourcevertex

d f

1 |12

8 |11

13|

2 | 7

9 |10

3 | 4

5 | 6

14|15

dfs example16
DFS Example

sourcevertex

d f

1 |12

8 |11

13|16

2 | 7

9 |10

3 | 4

5 | 6

14|15

depth first search the code4
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

What will be the running time?

depth first search the code5
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

Running time: O(n2) because call DFS_Visit on each vertex, and the loop over Adj[] can run as many as |V| times

depth first search the code6
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

BUT, there is actually a tighter bound. How many times will DFS_Visit() actually be called?

depth first search analysis
Depth-First Search Analysis
  • For a node u
    • DFS_Visit(u) cannot be called more than once
    • DFS_Visit(u) cannot be called less than once
    • Therefore, exactly once
  • DFS_Visit(u) requires c + ϴ(|Adj(u)|) time
    • The fact that recursive calls are made to v’s is accounted for by DFS_Visit(v)
  • Total= ∑u ( c + ϴ(|Adj(u)|) ) = ϴ(V+E)
    • Considered linear in input size, since adjacency list is a ϴ(V+E) representation
dfs kinds of edges
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 example17
DFS Example

sourcevertex

d f

1 |12

8 |11

13|16

2 | 7

9 |10

3 | 4

5 | 6

14|15

Tree edges

dfs kinds of edges1
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 example18
DFS Example

sourcevertex

d f

1 |12

8 |11

13|16

2 | 7

9 |10

3 | 4

5 | 6

14|15

Tree edges

Back edges

dfs kinds of edges2
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
      • If forward edge, then grey node to black node
        • Converse not true; could be a cross edge (see next)
dfs example19
DFS Example

sourcevertex

d f

1 |12

8 |11

13|16

2 | 7

9 |10

3 | 4

5 | 6

14|15

Tree edges

Back edges

Forward edges

dfs kinds of edges3
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 or within trees
      • (Also) from a grey node to a black node
      • No ancestor-descendent relation, if in the same DFS tree
dfs example20
DFS Example

sourcevertex

d f

1 |12

8 |11

13|16

2 | 7

9 |10

3 | 4

5 | 6

14|15

Tree edges

Back edges

Forward edges

Cross edges

dfs kinds of edges4
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 edges5
DFS: Kinds Of Edges
  • Thm 22.10: 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 edges6
DFS: Kinds Of Edges
  • Thm 22.10: 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 treevertex, before other vertex is explored
      • So in fact the picture is wrong…bothlower tree edges cannot in fact betree edges

source

C?

dfs and graph cycles
DFS And Graph Cycles
  • Thm: An undirected graph is acyclic iff a DFS yields no back edges
    • If acyclic, then no back edges (because a back edge implies a cycle)
    • If no back edges, then 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
DFS And Cycles
  • How would you modify the code to detect cycles?

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(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);

}

}

If v->color == GREY and Pred(u)≠v, …

dfs and cycles1
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 cycles2
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 if no cycle, then O(V+E) is O(V)
    • If cycle, count the edges: if ever see |V| distinct edges, must have seen a back edge along the way
      • Can halt and output “cycle exists”
      • O(V) time
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