Internally Disjoint Paths - PowerPoint PPT Presentation

1 / 28
About This Presentation
Title:

Internally Disjoint Paths

Description:

Title: Corollary 3.3.8 Author: grace Last modified by: MJTsai Created Date: 3/13/2005 4:09:19 AM Document presentation format: Company – PowerPoint PPT presentation

Number of Views:88
Avg rating:3.0/5.0
Slides: 29
Provided by: Grace94
Category:

less

Transcript and Presenter's Notes

Title: Internally Disjoint Paths


1
Internally Disjoint Paths
  • Internally Disjoint Paths Two paths u to v are
    internally disjoint if they have no common
    internal vertex.

Internally disjoint paths
u
v
u
v
2
Theorem 4.2.2
  • A graph G having at least three vertices is
    2-connected if and only if for each pair u,v?V(G)
    there exists internally disjoint u,v-paths in G.
  • (Proof of if part)
  • It suffices to show for each pair u,v?V(G),
  • deletion of any vertex in V(G) cannot separate u
    from v.
  • 2. This is clearly true because
  • G has internally disjoint u,v-paths.

3
Theorem 4.2.2
  • (Proof of only if part)
  • G is 2-connected. (Premise)
  • That G has internally disjoint u,v-paths is
    proved by induction on d(u,v).
  • 3. Basis Step d(u,v)1.
  • 3.1. The graph G-uv is connected
  • since ?(G)gt?(G)gt2.
  • 3.2. There exists a u,v-path in G-uv, which is
    internally disjoint in G from the u,v-path formed
    by the edge uv itself.

4
Theorem 4.2.2
  • 4. Induction Step d(u,v)gt1.
  • 4.1. Let kd(u,v).
  • 4.2. Let w be the vertex before v on a shortest
    u,v-path.
  • d(u,w)k-1.
  • 4.3 G has internally disjoint u,w-paths P and Q.
  • (Induction Hypothesis)

5
Theorem 4.2.2
4.4. If v?V(P)?V(Q), then we find the desired
paths in the cycle P?Q.
6
Theorem 4.2.2
since G is 2-connected.
  • 4.5. Otherwise, G-w is connected
  • 4.6. G-w contains a u,v-path R.
  • 4.7. If R avoids P or Q, we are done.

Q
v
w
P
u
R
7
Theorem 4.2.2
  • 4.8. Otherwise, let z be the last vertex of R
    (before v) belonging to P?Q. We assume that z?P
    by symmetry.
  • 4.9. We combine the u,z-subpath of P with the
    z,v-subpath of R to obtain a u,v-path internally
    disjoint from Q?wv.

R
z
P
w
v
u
Q
8
Lemma 4.2.3 (Expansion Lemma)
  • If G is a k-connected graph, and G is obtained
    from G by adding a new vertex y with at least k
    neighbors in G, then G is k-connected.

9
Theorem 4.2.4
  • For a graph G with at least three vertices, the
    following conditions are equivalent (and
    characterize 2-connected graphs).
  • G is connected and has no cut-vertex.
  • For all x,y?V(G), there are internally disjoint
    x,y-paths.
  • For all x,y?V(G), there is a cycle through x and
    y.
  • ?(G)gt1, and every pair of edges in G lies on a
    common cycle.

10
Theorem 4.2.4
Proof. 1. Theorem 4.2.2 proves A?B. 2. For B?C,
the cycles containing x and y corresponds to
pairs of internally disjoint x,y-paths. 3. For
D?C, ?(G)gt1 implies that vertices x and y are
not isolated.
11
Theorem 4.2.4
  • 4. Consider edges incident to x and y.
  • 5. Case 1 there are at least two such edges e
    and f.
  • Then e and f lies on a common cycle.
  • There is a cycle through x and y.
  • 6. Case 2 only one such edge e.
  • Let f be an edge incident to the third vertex.
  • e and f lies on a common cycle.
  • ? There is a cycle through x and y.

12
Theorem 4.2.4
  • 7. For C?D. G satisfies condition C.
  • G satisfies condition A.
  • G is connected.
  • ?(G)gt1.
  • 8. We need to show any two edges, uv and xy, lie
    on a common cycle.
  • 9. Add to G the vertices w with neighborhood
    u,v and z with neighborhood x,y to form G.

13
Theorem 4.2.4
10. Since G is 2-connected, Lemma 4.2.3 implies
G is 2-connected. 11. w and z lie on a cycle C
in G. 12. Since w,z each have degree 2, C must
contain the paths u,w,v and x,z,y but not the
edges uv or xy. 13. Replacing the path u,w,v and
x,z,y in C with the edges uv and xy yields the
desired cycle through uv and xy in G.
14
x,y-cut
  • x,y-cut Given x,y?V(G), a set S?V(G)-x,y is an
    x,y-separator or x,y-cut if G-S has no x,y-path.
  • ?(x,y) the minimum size of x,y-cut.
  • ?(x,y) the maximum size of a set of pairwise
    internally disjoint x,y-paths.

15
Example 4.2.16
  • b,c,z,d is an x,y-cut of size 4. ? ?(x,y)lt4.
  • G has four internally disjoint x,y-paths. ?
    ?(x,y)gt4.
  • b,c,x is an w,z-cut of size 3. ? ?(w,z)lt3.
  • G has three internally disjoint w,z-paths. ?
    ?(w,z)gt3.

16
Theorem 4.2.17 (Menger Theorem)
  • If x,y are vertices of a graph G and xy?E(G),
    then ?(x,y) ?(x,y).
  • Proof. 1. An x,y-cut must contain an internal
    vertex of every internally disjoint x,y-paths,
    and no vertex can cut two internally disjoint
    x,y-paths.
  • ?(x,y)gt ?(x,y).
  • 2. We prove equality by induction on n(G).

17
Theorem 4.2.17 (Menger Theorem)
  • Basis Step n(G)2.
  • xy?E(G) yields ?(x,y) ?(x,y)0.
  • Induction Step n(G)gt2.
  • Let k ?G(x,y).
  • No minimum cut properly contains N(x) or N(y)
    since N(x) and N(y) are x,y-cuts.
  • 3. Case 1 G has a minimum x,y-cut S other than
    N(x) or N(y).
  • 4. Case 2 Every minimum x,y-cut is N(x) or N(y).

18
Theorem 4.2.17
5. For Case 1, let V1 be the set of vertices on
x,S-path, and let V2 be the set of vertices on
S,y-path. 6. S?V1 and S?V2 ? S?V1?V2. 7. If there
exists v such that v?V1?V2S, then combing
x,v-portion of some x,S-path and v,y-portion of
some S,y-path yields an x,y-path that avoids the
x,y-cut S. It contradicts that S is a minimum
x,y-cut. 8. This implies SV1?V2.
S
V1
V2
x
y
G
v
19
Theorem 4.2.17
9. Form H1 by adding to GV1 a vertex y with
edges from S, and form H2 by adding to GV2 a
vertex x with edges from S.
20
Theorem 4.2.17
  • 10. Every x,y-path in G starts with an x,S-path
    (contained in H1).
  • Every x,ycut in H1 is an x,y-cut in G.
  • ?H1(x,y) k.
  • 11. ?H2(x,y) k by the same argument in 10.
  • 12. H1 and H2 are smaller than G since
  • N(y) ? S and N(x) ? S.
  • ? ?H1(x,y)k ?H2(x,y).

21
Theorem 4.2.17
13. SV1?V2. ? Deleting y from the k paths in H1
and x from the k paths in H2 yields the desired
x,S-paths and S,y-paths in G that combine to form
k internally disjoint x,y-paths in G.
?
22
Theorem 4.2.17
  • 14. For Case 2, if there exists node u?N(x)?N(y),
    then S-u is x,y-cut in G-u.
  • ?G-u(x,y)k-1.
  • G-u has k-1 internally disjoint x,y-paths by
    induction hypothesis.
  • 15. Combining these k-1 x,y-path and the path
    x,u,y yields k internally disjoint x,y-paths in
    G.

23
Theorem 4.2.17
  • 16. If there exists node v?x?N(x)?N(y)?y,
    then S is minimum x,y-cut in G-v.
  • (If there exists a x,y-cut, S, in G-v whose size
    is smaller than S, then S?v is a x,y-cut in
    G. It is a contradiction.)
  • ?G-v(x,y)k.
  • G-v has k internally disjoint x,y-paths by
    induction hypothesis.
  • These are k internally disjoint x,y-paths in G.

24
Theorem 4.2.17
17. We may assume that N(x) and N(y) partition
V(G)-x,y. 18. Let G be the bipartite graph
with bipartition N(x), N(y) and edge set
N(x),N(y).
  • 19. Every x,y-path in G uses some edge from N(x)
    to N(y).
  • x,y-cuts in G are the vertex covers of G.
  • ?(G)k.
  • G has a matching of size k by Theorem 3.1.16.
  • These k edges yield k internally disjoint
    x,y-paths of length 3.

25
Line Graph (Digraph)
Line Graph (Digraph) The line graph (digraph) of
a graph (digraph) G, written L(G), is the graph
(digraph) whose vertices are the edges of G, with
ef?E(L(G)) when euv and fvw in G.
26
Theorem 4.2.19
If x and y are distinct vertices of a graph or
digraph G, then the minimum size of an
x,y-disconnecting set of edges equals the maximum
number of pairwise edge-disjoint x,y-paths.
Proof. 1. Modify G to obtain G by adding two new
vertices s, t and two new edges sx and yt. 2.
Cleary, ?G(x,y) ?G(x,y) and ?G(x,y)
?G(x,y).
a
d
f
y
x
c
t
s
b
g
e
27
Theorem 4.2.19
3. A x,y-path exists in G that traverses edges
p, q, r if and only if a sx,yt-path exists in
L(G) that traverses vertices p, q, r .
28
Theorem 4.2.19
4. Edge-disjoint x,y-paths in G become
internally disjoint sx,yt-paths in L(G), and
vice versa. ? ?G(x,y)? L(G)(sx,yt). 5. A
set of edges disconnects y from x in G if and
only if the corresponding vertices of L(G) form
an sx,yt-cut. ? ?G(x,y)?L(G)(sx,yt). 6.
?L(G)(sx,yt)? L(G)(sx,yt). ? ?G(x,y)
?G(x,y). ? ?G(x,y) ?G(x,y).
Write a Comment
User Comments (0)
About PowerShow.com