lecture 19
TRANSCRIPT
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David Luebke 1 04/14/23
CS 332: Algorithms
Graph Algorithms
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David Luebke 2 04/14/23
Review: 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
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David Luebke 3 04/14/23
Review: DFS 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 = YELLOW;
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;
}
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David Luebke 4 04/14/23
DFS Example
sourcevertex
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David Luebke 5 04/14/23
DFS Example
1 | | |
| | |
| |
sourcevertex
d f
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David Luebke 6 04/14/23
DFS Example
1 | | |
| | |
2 | |
sourcevertex
d f
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David Luebke 7 04/14/23
DFS Example
1 | | |
| | 3 |
2 | |
sourcevertex
d f
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David Luebke 8 04/14/23
DFS Example
1 | | |
| | 3 | 4
2 | |
sourcevertex
d f
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David Luebke 9 04/14/23
DFS Example
1 | | |
| 5 | 3 | 4
2 | |
sourcevertex
d f
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David Luebke 10 04/14/23
DFS Example
1 | | |
| 5 | 63 | 4
2 | |
sourcevertex
d f
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David Luebke 11 04/14/23
DFS Example
1 | 8 | |
| 5 | 63 | 4
2 | 7 |
sourcevertex
d f
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David Luebke 12 04/14/23
DFS Example
1 | 8 | |
| 5 | 63 | 4
2 | 7 |
sourcevertex
d f
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David Luebke 13 04/14/23
DFS Example
1 | 8 | |
| 5 | 63 | 4
2 | 7 9 |
sourcevertex
d f
What is the structure of the yellow vertices? What do they represent?
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David Luebke 14 04/14/23
DFS Example
1 | 8 | |
| 5 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
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David Luebke 15 04/14/23
DFS Example
1 | 8 |11 |
| 5 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
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David Luebke 16 04/14/23
DFS Example
1 |12 8 |11 |
| 5 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
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David Luebke 17 04/14/23
DFS Example
1 |12 8 |11 13|
| 5 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
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David Luebke 18 04/14/23
DFS Example
1 |12 8 |11 13|
14| 5 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
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David Luebke 19 04/14/23
DFS Example
1 |12 8 |11 13|
14|155 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
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David Luebke 20 04/14/23
DFS Example
1 |12 8 |11 13|16
14|155 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
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David Luebke 21 04/14/23
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?
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David Luebke 22 04/14/23
DFS Example
1 |12 8 |11 13|16
14|155 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
Tree edges
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David Luebke 23 04/14/23
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 yellow vertex (yellow to yellow)
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David Luebke 24 04/14/23
DFS Example
1 |12 8 |11 13|16
14|155 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
Tree edges Back edges
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David Luebke 25 04/14/23
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 yellow node to black node
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David Luebke 26 04/14/23
DFS Example
1 |12 8 |11 13|16
14|155 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
Tree edges Back edges Forward edges
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David Luebke 27 04/14/23
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 yellow node to a black node
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David Luebke 28 04/14/23
DFS Example
1 |12 8 |11 13|16
14|155 | 63 | 4
2 | 7 9 |10
sourcevertex
d f
Tree edges Back edges Forward edges Cross edges
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David Luebke 29 04/14/23
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
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David Luebke 30 04/14/23
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?)
sourceF?
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David Luebke 31 04/14/23
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 treevertex, before other vertex is explored
○ So in fact the picture is wrong…bothlower tree edges cannot in fact betree edges
source
C?
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David Luebke 32 04/14/23
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
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David Luebke 33 04/14/23
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;
}
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David Luebke 34 04/14/23
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;
}
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David Luebke 35 04/14/23
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