Gomory–Hu tree

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 ${\displaystyle G=(V_{G},E_{G},c)}$ be an undirected graph with ${\displaystyle c(u,v)}$ being the capacity of the edge ${\displaystyle (u,v)}$ respectively.

Denote the minimum capacity of an s-t cut by ${\displaystyle \lambda _{st}}$ for each ${\displaystyle s,t\in V_{G}}$.

Let ${\displaystyle T=(V_{G},E_{T})}$ be a tree, and denote the set of edges in an s-t path by ${\displaystyle P_{st}}$ for each ${\displaystyle s,t\in V_{G}}$.

Then T is said to be a Gomory–Hu tree of G, if for each ${\displaystyle s,t\in V_{G}}$

${\displaystyle \lambda _{st}=\min _{e\in P_{st}}c(S_{e},T_{e}),}$

where

1. ${\displaystyle S_{e},T_{e}\subseteq V_{G}}$ are the two connected components of ${\displaystyle T\setminus \{e\}}$, and thus ${\displaystyle (S_{e},T_{e})}$ forms an s-t cut in G.
2. ${\displaystyle c(S_{e},T_{e})}$ is the capacity of the ${\displaystyle (S_{e},T_{e})}$ cut in G.

Algorithm

Gomory–Hu Algorithm

Input: A weighted undirected graph ${\displaystyle G=((V_{G},E_{G}),c)}$

Output: A Gomory–Hu Tree ${\displaystyle T=(V_{T},E_{T}).}$

1. Set ${\displaystyle V_{T}=\{V_{G}\},\ E_{T}=\emptyset .}$
2. Choose some ${\displaystyle X\in V_{T}}$ with |X| ≥ 2 if such X exists. Otherwise, go to step 6.
3. For each connected component ${\displaystyle C=(V_{C},E_{C})\in T\setminus X,}$ let ${\textstyle S_{C}=\bigcup _{v_{T}\in V_{C}}v_{T}.}$

   Let ${\displaystyle S=\{S_{C}\mid C{\text{ is a connected component in }}T\setminus X\}.}$

   Contract the components to form ${\displaystyle G'=((V_{G'},E_{G'}),c'),}$ where:$${\displaystyle {\begin{aligned}V_{G'}&=X\cup S\\[2pt]E_{G'}&=E_{G}|_{X\times X}\cup \{(u,S_{C})\in X\times S\mid (u,v)\in E_{G}{\text{ for some }}v\in S_{C}\}\\[2pt]&\qquad \qquad \quad \!\cup \{(S_{C1},S_{C2})\in S\times S\mid (u,v)\in E_{G}{\text{ for some }}u\in S_{C1}{\text{ and }}v\in S_{C2}\}\end{aligned}}}$$

   ${\displaystyle c':V_{G'}\times V_{G'}\to R^{+}}$ is the capacity function, defined as:$${\displaystyle {\begin{aligned}&{\text{if }}\ (u,S_{C})\in E_{G}|_{X\times S}:&&c'(u,S_{C})=\!\!\!\sum _{\begin{smallmatrix}v\in S_{C}:\\(u,v)\in E_{G}\end{smallmatrix}}\!\!\!c(u,v)\\[4pt]&{\text{if }}\ (S_{C1},S_{C2})\in E_{G}|_{S\times S}:&&c'(S_{C1},S_{C2})=\!\!\!\!\!\!\!\sum _{\begin{smallmatrix}(u,v)\in E_{G}:\\u\in S_{C1}\,\land \,v\in S_{C2}\end{smallmatrix}}\!\!\!\!\!c(u,v)\\[4pt]&{\text{otherwise}}:&&c'(u,v)=c(u,v)\end{aligned}}}$$

4. Choose two vertices s, t ∈ X and find a minimum s-t cut (A′, B′) in G'.

   Set ${\displaystyle A={\Biggl (}\bigcup _{S_{C}\in A'\cap S}\!\!\!S_{C}\!{\Biggr )}\cup (A'\cap X),\ }$${\displaystyle B={\Biggl (}\bigcup _{S_{C}\in B'\cap S}\!\!\!S_{C}\!{\Biggr )}\cup (B'\cap X).}$

5. Set ${\displaystyle V_{T}=(V_{T}\setminus X)\cup \{A\cap X,B\cap X\}.}$

   For each ${\displaystyle e=(X,Y)\in E_{T}}$ do:

   1. Set ${\displaystyle e'=(A\cap X,Y)}$ if ${\displaystyle Y\subset A,}$ otherwise set ${\displaystyle e'=(B\cap X,Y).}$
   2. Set ${\displaystyle E_{T}=(E_{T}\setminus \{e\})\cup \{e'\}.}$
   3. Set ${\displaystyle w(e')=w(e).}$

   Set ${\displaystyle E_{T}=E_{T}\cup \{(A\cap X,\ B\cap X)\}.}$

   Set ${\displaystyle w((A\cap X,B\cap X))=c'(A',B').}$

   Go to step 2.

6. Replace each ${\displaystyle \{v\}\in V_{T}}$ by v and each ${\displaystyle (\{u\},\{v\})\in E_{T}}$ by (u, v). Output T.

Analysis

Using the submodular property of the capacity function c, one has $${\displaystyle c(X)+c(Y)\geq c(X\cap Y)+c(X\cup Y).}$$ 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, t ∈ X.

To show that for all ${\displaystyle (P,Q)\in E_{T},}$ ${\displaystyle w(P,Q)=\lambda _{pq}}$ for some p ∈ P, q ∈ Q throughout the algorithm, one makes use of the following lemma,

For any i, j, k in V_G, ${\displaystyle \lambda _{ik}\geq \min(\lambda _{ij},\lambda _{jk}).}$

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) ∪ {A ∩ X, B ∩ X} = { {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) ∪ {A ∩ X, B ∩ X} = { {0, 1, 2, 4}, {3}, {5} }.     Since (X, {0, 1, 2, 4}) ∈ ET and {0, 1, 2, 4} ⊂ A, replace it with (A ∩ X, 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,v ∈ X. 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) ∪ {A ∩ X, B ∩ X} = { {3}, {5}, {1, 4}, {0, 2} }.     Since (X, {3}) ∈ ET and {3} ⊂ A, replace it with (A ∩ X, 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,v ∈ X. 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) ∪ {A ∩ X, B ∩ X} = { {3}, {5}, {0, 2}, {1}, {4} }.     Since (X, {3}) ∈ ET and {3} ⊂ A, replace it with (A ∩ X, Y) = ({1}, {3}).     Since (X, {0, 2}) ∈ ET and {0, 2} ⊂ B, replace it with (B ∩ X, 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) ∪ {A ∩ X, B ∩ X} = { {3}, {5}, {1}, {4}, {0}, {2} }.     Since (X, {4}) ∈ ET and {4} ⊂ B, replace it with (B ∩ X, 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 X∈VT 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

• Cut (graph theory)
• Max-flow min-cut theorem
• Maximum flow problem