Jump to content

Gomory–Hu tree

From Wikipedia, the free encyclopedia

In combinatorial optimization, the Gomory–Hu tree[1] of an undirected graph with capacities is a weighted tree that represents the minimum s-t cuts for all s-t pairs in the graph. The Gomory–Hu tree can be constructed in |V| − 1 maximum flow computations. It is named for Ralph E. Gomory and T. C. Hu.

Definition

[edit]

Let be an undirected graph with being the capacity of the edge respectively.

Denote the minimum capacity of an s-t cut by for each .
Let be a tree, and denote the set of edges in an s-t path by for each .

Then T is said to be a Gomory–Hu tree of G, if for each

where

  1. are the two connected components of , and thus forms an s-t cut in G.
  2. is the capacity of the cut in G.

Algorithm

[edit]

Gomory–Hu Algorithm

Input: A weighted undirected graph
Output: A Gomory–Hu Tree
  1. Set
  2. Choose some with |X| ≥ 2 if such X exists. Otherwise, go to step 6.
  3. For each connected component let
    Let
    Contract the components to form where:
    is the capacity function, defined as:
  4. Choose two vertices s, tX and find a minimum s-t cut (A′, B′) in G'.
    Set
  5. Set
    For each do:
    1. Set if otherwise set
    2. Set
    3. Set
    Set
    Set
    Go to step 2.
  6. Replace each by v and each by (u, v). Output T.

Analysis

[edit]

Using the submodular property of the capacity function c, one has Then it can be shown that the minimum s-t cut in G' is also a minimum s-t cut in G for any s, tX.

To show that for all for some pP, qQ throughout the algorithm, one makes use of the following lemma,

For any i, j, k in VG,

The lemma can be used again repeatedly to show that the output T satisfies the properties of a Gomory–Hu Tree.

Example

[edit]

The following is a simulation of the Gomory–Hu's algorithm, where

  1. green circles are vertices of T.
  2. red and blue circles are the vertices in G'.
  3. grey vertices are the chosen s and t.
  4. red and blue coloring represents the s-t cut.
  5. dashed edges are the s-t cut-set.
  6. A is the set of vertices circled in red and B is the set of vertices circled in blue.
G' T
1. Set VT = {VG} = { {0, 1, 2, 3, 4, 5} } and ET = ∅.
2. Since VT has only one vertex, choose X = VG = {0, 1, 2, 3, 4, 5}. Note that | X | = 6 ≥ 2.
1.
3. Since T\X = ∅, there is no contraction and therefore G' = G.
4. Choose s = 1 and t = 5. The minimum s-t cut (A', B') is ({0, 1, 2, 4}, {3, 5}) with c'(A', B') = 6.
    Set A = {0, 1, 2, 4} and B = {3, 5}.
5. Set VT = (VT\X) ∪ {AX, BX} = { {0, 1, 2, 4}, {3, 5} }.
    Set ET = { ({0, 1, 2, 4}, {3, 5}) }.
    Set w( ({0, 1, 2, 4}, {3, 5}) ) = c'(A', B') = 6.
    Go to step 2.
2. Choose X = {3, 5}. Note that | X | = 2 ≥ 2.
2.
3. {0, 1, 2, 4} is the connected component in T\X and thus S = { {0, 1, 2, 4} }.
    Contract {0, 1, 2, 4} to form G', where
c'( (3, {0, 1, 2 ,4}) ) = c( (3, 1) ) c( (3, 4) ) = 4.
c'( (5, {0, 1, 2, 4}) ) = c( (5, 4) ) = 2.
c'( (3, 5)) = c( (3, 5) ) = 6.
4. Choose s = 3, t = 5. The minimum s-t cut (A', B') in G' is ( {{0, 1, 2, 4}, 3}, {5} ) with c'(A', B') = 8.
    Set A = {0, 1, 2, 3, 4} and B = {5}.
5. Set VT = (VT\X) ∪ {AX, BX} = { {0, 1, 2, 4}, {3}, {5} }.
    Since (X, {0, 1, 2, 4}) ∈ ET and {0, 1, 2, 4} ⊂ A, replace it with (AX, Y) = ({3}, {0, 1, 2 ,4}).
    Set ET = { ({3}, {0, 1, 2 ,4}), ({3}, {5}) } with
w(({3}, {0, 1, 2 ,4})) = w((X, {0, 1, 2, 4})) = 6.
w(({3}, {5})) = c'(A', B') = 8.
    Go to step 2.
2. Choose X = {0, 1, 2, 4}. Note that | X | = 4 ≥ 2.
3.
3. { {3}, {5} } is the connected component in T\X and thus S = { {3, 5} }.
    Contract {3, 5} to form G', where
c'( (1, {3, 5}) ) = c( (1, 3) ) = 3.
c'( (4, {3, 5}) ) = c( (4, 3) ) c( (4, 5) ) = 3.
c'(u,v) = c(u,v) for all u,vX.
4. Choose s = 1, t = 2. The minimum s-t cut (A', B') in G' is ( {1, {3, 5}, 4}, {0, 2} ) with c'(A', B') = 6.
    Set A = {1, 3, 4, 5} and B = {0, 2}.
5. Set VT = (VT\X) ∪ {AX, BX} = { {3}, {5}, {1, 4}, {0, 2} }.
    Since (X, {3}) ∈ ET and {3} ⊂ A, replace it with (AX, Y) = ({1, 4}, {3}).
    Set ET = { ({1, 4}, {3}), ({3}, {5}), ({0, 2}, {1, 4}) } with
w(({1, 4}, {3})) = w((X, {3})) = 6.
w(({0, 2}, {1, 4})) = c'(A', B') = 6.
    Go to step 2.
2. Choose X = {1, 4}. Note that | X | = 2 ≥ 2.
4.
3. { {3}, {5} }, { {0, 2} } are the connected components in T\X and thus S = { {0, 2}, {3, 5} }
    Contract {0, 2} and {3, 5} to form G', where
c'( (1, {3, 5}) ) = c( (1, 3) ) = 3.
c'( (4, {3, 5}) ) = c( (4, 3) ) c( (4, 5) ) = 3.
c'( (1, {0, 2}) ) = c( (1, 0) ) c( (1, 2) ) = 2.
c'( (4, {0, 2}) ) = c( (4, 2) ) = 4.
c'(u,v) = c(u,v) for all u,vX.
4. Choose s = 1, t = 4. The minimum s-t cut (A', B') in G' is ( {1, {3, 5}}, {{0, 2}, 4} ) with c'(A', B') = 7.
    Set A = {1, 3, 5} and B = {0, 2, 4}.
5. Set VT = (VT\X) ∪ {AX, BX} = { {3}, {5}, {0, 2}, {1}, {4} }.
    Since (X, {3}) ∈ ET and {3} ⊂ A, replace it with (AX, Y) = ({1}, {3}).
    Since (X, {0, 2}) ∈ ET and {0, 2} ⊂ B, replace it with (BX, Y) = ({4}, {0, 2}).
    Set ET = { ({1}, {3}), ({3}, {5}), ({4}, {0, 2}), ({1}, {4}) } with
w(({1}, {3})) = w((X, {3})) = 6.
w(({4}, {0, 2})) = w((X, {0, 2})) = 6.
w(({1}, {4})) = c'(A', B') = 7.
    Go to step 2.
2. Choose X = {0, 2}. Note that | X | = 2 ≥ 2.
5.
3. { {1}, {3}, {4}, {5} } is the connected component in T\X and thus S = { {1, 3, 4, 5} }.
    Contract {1, 3, 4, 5} to form G', where
c'( (0, {1, 3, 4, 5}) ) = c( (0, 1) ) = 1.
c'( (2, {1, 3, 4, 5}) ) = c( (2, 1) ) c( (2, 4) ) = 5.
c'( (0, 2) ) = c( (0, 2) ) = 7.
4. Choose s = 0, t = 2. The minimum s-t cut (A', B') in G' is ( {0}, {2, {1, 3, 4, 5}} ) with c'(A', B') = 8.
    Set A = {0} and B = {1, 2, 3 ,4 ,5}.
5. Set VT = (VT\X) ∪ {AX, BX} = { {3}, {5}, {1}, {4}, {0}, {2} }.
    Since (X, {4}) ∈ ET and {4} ⊂ B, replace it with (BX, Y) = ({2}, {4}).
    Set ET = { ({1}, {3}), ({3}, {5}), ({2}, {4}), ({1}, {4}), ({0}, {2}) } with
w(({2}, {4})) = w((X, {4})) = 6.
w(({0}, {2})) = c'(A', B') = 8.
    Go to step 2.
2. There does not exist XVT with | X | ≥ 2. Hence, go to step 6.
6.

6. Replace VT = { {3}, {5}, {1}, {4}, {0}, {2} } by {3, 5, 1, 4, 0, 2}.
    Replace ET = { ({1}, {3}), ({3}, {5}), ({2}, {4}), ({1}, {4}), ({0}, {2}) } by { (1, 3), (3, 5), (2, 4), (1, 4), (0, 2) }.
    Output T. Note that exactly | V | − 1 = 6 − 1 = 5 times min-cut computation is performed.

Implementations: Sequential and Parallel

[edit]

Gusfield's algorithm can be used to find a Gomory–Hu tree without any vertex contraction in the same running time-complexity, which simplifies the implementation of constructing a Gomory–Hu Tree.[2]

Andrew V. Goldberg and K. Tsioutsiouliklis implemented the Gomory-Hu algorithm and Gusfield algorithm, and performed an experimental evaluation and comparison.[3]

Cohen et al. report results on two parallel implementations of Gusfield's algorithm using OpenMP and MPI, respectively.[4]

[edit]

In planar graphs, the Gomory–Hu tree is dual to the minimum weight cycle basis, in the sense that the cuts of the Gomory–Hu tree are dual to a collection of cycles in the dual graph that form a minimum-weight cycle basis.[5]

See also

[edit]

References

[edit]
  1. ^ Gomory, R. E.; Hu, T. C. (1961). "Multi-terminal network flows". Journal of the Society for Industrial and Applied Mathematics. 9 (4): 551–570. doi:10.1137/0109047.
  2. ^ Gusfield, Dan (1990). "Very Simple Methods for All Pairs Network Flow Analysis". SIAM J. Comput. 19 (1): 143–155. doi:10.1137/0219009.
  3. ^ Goldberg, A. V.; Tsioutsiouliklis, K. (2001). "Cut Tree Algorithms: An Experimental Study". Journal of Algorithms. 38 (1): 51–83. doi:10.1006/jagm.2000.1136.
  4. ^ Cohen, Jaime; Rodrigues, Luiz A.; Silva, Fabiano; Carmo, Renato; Guedes, André Luiz Pires; Duarte, Jr., Elias P. (2011). "Parallel implementations of Gusfield's cut tree algorithm". In Xiang, Yang; Cuzzocrea, Alfredo; Hobbs, Michael; Zhou, Wanlei (eds.). Algorithms and Architectures for Parallel Processing – 11th International Conference, ICA3PP, Melbourne, Australia, October 24–26, 2011, Proceedings, Part I. Lecture Notes in Computer Science. Vol. 7016. Springer. pp. 258–269. doi:10.1007/978-3-642-24650-0_22. ISBN 978-3-642-24649-4.{{cite conference}}: CS1 maint: multiple names: authors list (link)
  5. ^ Hartvigsen, D.; Mardon, R. (1994). "The all-pairs min cut problem and the minimum cycle basis problem on planar graphs". SIAM J. Discrete Math. 7 (3): 403–418. doi:10.1137/S0895480190177042..