Gomory–Hu tree

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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

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

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

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

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

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]

Related concepts

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

References

  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.{{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..