 # Linear Programming complementation and its application to fractional graph theory

In this paper, we introduce a new kind of duality for Linear Programming (LP), that we call LP complementation. We prove that the optimal values of an LP and of its complement are in bijection (provided that either the original LP or its complement has an optimal value greater than one). The main consequence of the LP complementation theorem is for hypergraphs. We introduce the complement of a hypergraph and we show that the fractional packing numbers of a hypergraph and of its complement are in bijection; similar results hold for fractional matching, covering and transversal numbers. This hypergraph complementation theorem has several consequences for fractional graph theory. In particular, we relate the fractional dominating number of a graph to the fractional total dominating number of its complement. We also show that the edge toughness of a graph is equal to the fractional transversal number of its cycle matroid. We then consider the following particular problem: let G be a graph and b be a positive integer, then how many vertex covers of G, say S_1, ..., S_t_b, can we construct such that every vertex appears at most b times in total? The integer b can be viewed as a budget we can spend on each vertex, and given this budget we aim to cover all edges for as long as possible (up to time t_b). We then prove that t_b ∼χ_f/χ_f - 1 b, where χ_f is the fractional chromatic number of G.

## Authors

##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 Linear Programming complementation

### 1.1 The complementation theorem

For any linear program (LP) , we denote its optimal value (if it exists) as

. We denote the all-zero vector or matrix as

, regardless its dimension; similarly, the all-ones vector or matrix is denoted as .

We define the complement of an LP , which we denote , as follows. Let , , , then for the following maximisation LP , we have

 P: max c⊤x s.t. Ax ≤b x ≥0, ¯¯¯¯P: min c⊤x s.t. (bc⊤−A)x ≥b x ≥0.

Similarly, let , , , then for the following minimisation LP , we have

 Q: min v⊤x s.t. Mx ≥u x ≥0, ¯¯¯¯Q: max v⊤x s.t. (uv⊤−M)x ≤u x ≥0.

Complementation is indeed an involution: for any (minimisation or maximisation) LP , we have . Moreover, complementation commutes with duality: Indeed, if denotes the dual of , then we have

 ¯¯¯¯¯¯R∗=¯¯¯¯R∗.

The main result is that, provided or , then the optimal values of an LP and that of its complement are in bijection.

###### Theorem 1.1.

Let be a minimisation or maximisation LP. Then if and only if , in which case

 1\textscOpt(P)+1\textscOpt(¯¯¯¯P)=1.
###### Proof.

Without loss of generality, let be a maximisation problem. Suppose , say for some . Let be an optimal solution of , and let . We then have and

 (bc⊤−A)¯¯¯x=1+aab−1aAx≥b,

and hence is a feasible solution of , with value .

We have just shown that has a feasible solution of value greater than one. We now prove that . For the sake of contradiction, suppose that has a feasible solution with value at most , then for any , has a feasible solution with value . Let , then by the same reasoning as above, is a feasible solution of with value ; we conclude that is unbounded, which is the desired contradiction.

Having established that , we find that the first paragraph showed that

 1\textscOpt(P)+1\textscOpt(¯¯¯¯P)≥1a+1+aa+1=1.

We now prove the reverse inequality. Let with and be an optimal solution of . Then is a feasible solution of with value , and we obtain

 1\textscOpt(P)+1\textscOpt(¯¯¯¯P)≤¯¯¯a¯¯¯a+1+1¯¯¯a+1=1.

The case where we suppose instead is similar and hence omitted. ∎

### 1.2 Feasibility and boundedness

As we can see, the complementation theorem only considers LPs with an optimal value greater than one. This condition is met by many natural LPs, as we shall see in the next section. Nonetheless, different scenarios can occur when one of the LPs is infeasible, unbounded, or with an optimal value less than or equal to one, as seen below.

###### Lemma 1.2.

If and , then is infeasible or unbounded. Moreover, both cases can occur.

###### Proof.

Assume that is feasible. We will show that is unbounded in this case. It follows from Theorem 1.1 that there exists a feasible solution of such that .

First, notice that implies that is a feasible solution of for every . Furthermore, since , we have , and therefore for any , is a feasible solution of . The latter together with the boundedness of imply that . Consequently, is unbounded.

Now, for all , the one-dimensional LP has , while is infeasible.

Finally, consider the one-dimensional LP , with complement . If , then while is unbounded. ∎

### 1.3 Game theoretic interpretation

The links between two-player zero-sum (matrix) games and linear progamming are well established; see [5, 1] for instance. We shall review these and then show that LP complementation can be interpreted using two complementary games.

Given any matrix , the matrix game with payoff matrix is played by two persons, Rose and Colin, as follows: Rose selects a row of , Colin a column. If the row and the column are chosen, then Rose’s payoff is . In particular, if

, then Rose earns money; otherwise, Rose loses money. A strategy for Rose is then a probability distribution on the rows:

such that and . Rose aims at maximising her expected payoff, while Colin aims at minimising it. The value of the game, denoted as , is the maximum expected payoff over all strategies for Rose (and is equal to the minimum expected payoff over all strategies for Colin).

Without loss of generality, suppose that . Then the value of the game is also between and ; let us omit the two extreme cases and suppose that . For any strategy for Rose, let , then we have , (by optimality of the value), and . We can then express , where

 P: max 1⊤x s.t. Ax ≤1 x ≥0.

LP duality then corresponds to taking Colin’s point of view: , with

 P∗: min 1⊤y s.t. A⊤y ≥1 y ≥0.

LP complementation, on the other hand, corresponds to taking the complementary payoff. Consider a second game, where the players change their roles (Rose chooses columns of the payoff matrix and Colin chooses rows), and the payoff is equal to minus the original payoff. Thus, the new payoff matrix is and the value of the new game is . But then, taking Rose’s point of view, we have , where

 Q=¯¯¯¯P: min 1⊤y s.t. (1−A)y ≥1 y ≥0.

We then have and and

 1\textscOpt(P)+1\textscOpt(¯¯¯¯P)=1.

### 1.4 Consequence for integer programming

The proof of Theorem 1.1 actually shows that, whenever , is an optimal solution of if and only if is an optimal solution of . This has a consequence for integer programming.

Firstly, for any linear program , adding the constraint that the variables be integral yields an integer program, which we denote . Suppose is , where , , and are all integral, and that is an optimal solution of with value . The vector is rational, say the components of the vector are rational numbers with denominator for some ; then the value is equal to for some , and is an optimal solution of . Consider the following two IPs

 PIs: max c⊤x s.t. Ax ≤sb x ≥0, x ∈Z, ¯¯¯¯PIt: min c⊤x s.t. (bc⊤−A)x ≥tb x ≥0, x ∈Z.

Then is an optimal solution to both and with value .

## 2 Application to fractional hypergraph theory

### 2.1 Fractional hypergraph parameters

Many important graph parameters, such as the clique number, chromatic number, matching number, etc. can be viewed as the optimal values of IPs defined on hypergraphs related to the original graph. Hypergraph fractional theory then lifts the integrality constraint and focuses on the fractional analogues of those parameters, which are then optimal values of the corresponding LP relaxations. In this subsection, we review four important fractional hypergraph parameters, and how they are related. A comprehensive account of those parameters can be found in .

A (finite) hypergraph is a pair , where is a set of vertices and is a multiset of edges, each being a subset of vertices. Recall the following concepts for a hypergraph . Its incidence matrix is such that, for all and ,

 Mve={1if v∈e0otherwise.

A vertex is universal if it belongs to all edges of . On the other hand, a vertex is isolated if it does not belong to any edge of . Say an edge is complete if and that it is empty if . We say is nontrivial if it has no empty edges and no universal vertices.

A covering of is a set of edges whose union is equal to . The covering number of is the minimum size of a covering of . The fractional covering number of is the optimal value of the following LP, which we give in two forms: a concise matrix form and a more explicit form.

 K(H): min 1⊤x min ∑e∈Exe s.t. MHx ≥1, s.t. ∑e∋vxe ≥1 ∀v∈V, x ≥0. xe ≥0 ∀e∈E.

It is easily seen that the covering number is actually the optimal value of . We remark that is feasible if and only if has no isolated vertices. Clearly, if is feasible, then it has an optimal solution. In that case, , with strict inequality if and only if has no complete edges.

A packing of is a set of vertices such that every edge contains at most one of those vertices. The packing number of is the maximum size of a packing of . The fractional packing number of is the optimal value of the LP dual to :

 P(H)=K(H)∗: max 1⊤y max ∑v∈Vyv s.t. M⊤Hy ≤1, s.t. ∑v∈eyv ≤1 ∀e∈E, y ≥0. yv ≥0 ∀v∈V.

Again, the maximum size of a packing of corresponds to the optimal value of the analogous IP. We remark that is always feasible. However, is bounded if and and only if has no isolated vertices. In that case, if and only if it has no complete edges. LP duality then yields , whenever has no isolated vertices.

For any hypergraph , its dual is , where . We then have and . We note that has no empty edge if and only if has no isolated vertex, and vice versa.

A matching of is a set of disjoint edges; it corresponds to a packing of . The fractional matching number is then , i.e. the optimal value of:

 M(H)=P(H∗): max 1⊤y max ∑e∈Eye s.t. MHy ≤1, s.t. ∑e∋vye ≤1 ∀v∈V, y ≥0. ye ≥0 ∀e∈E.

A transversal of is a set of vertices such that every edge contains a vertex from that set; it corresponds to a covering of . The fractional transversal number is then , i.e. the optimal value of:

 T(H)=K(H∗): min 1⊤x min ∑v∈Vxv s.t. M⊤Hx ≥1, s.t. ∑v∈exv ≥1 ∀e∈E, x ≥0. xv ≥0 ∀v∈V.

Again, LP duality yields , whenever has no empty edges.

In summary, for any nontrivial we have

 τf(H)=kf(H∗)=pf(H∗)=μf(H)>1.

### 2.2 Hypergraph complementation

We define the complement of as . We then have . We remark that hypergraph complementation is an involution: . Moreover, hypergraph complementation and duality commute:

 ¯¯¯¯¯¯¯H∗=¯¯¯¯¯H∗.

Again, if is nontrivial, then

 τf(¯¯¯¯¯H∗)=kf(¯¯¯¯¯H)=pf(¯¯¯¯¯H)=μf(¯¯¯¯¯H∗)>1.
###### Theorem 2.1 (Hypergraph complementation).

For any nontrivial hypergraph ,

 1kf(H∗)+1kf(¯¯¯¯¯H)=1.
###### Proof.

The fractional packing number of is the optimal value of

 P(H∗): max 1⊤x s.t. MHx ≤1, x ≥0.

The complement LP is

 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P(H∗)=K(¯¯¯¯¯H): min 1⊤x s.t. (1−MH)x ≥1, x ≥0.

Theorem 1.1 then applies. ∎

Obviously, the hypergraph complementation theorem holds for all four parameters reviewed above.

###### Corollary 2.2.

For any nontrivial hypergraph , we have

 1kf(H∗)+1kf(¯¯¯¯¯H)=1pf(H∗)+1pf(¯¯¯¯¯H)=1μf(H∗)+1μf(¯¯¯¯¯H)=1τf(H∗)+1τf(¯¯¯¯¯H)=1.

## 3 Applications to fractional graph theory

### 3.1 Fractional domination in graphs and digraphs

Let be a loopless digraph, i.e. . For any , the open in-neighbourhood of is ; the closed in-neighbourhood of is . We thus define two hypergraphs and , both with vertex set , and where the edges of are the open in-neighbourhoods of all vertices and the edges of are the closed in-neighbourhoods instead. Open and closed out-neighbourhoods are defined similarly, and hence we define and similarly as well.

We note that , the adjacency matrix of ; similarly, . We then have and . Moreover, we have , where is the (digraph) complement of , with and .

An in-dominating set of is a set of vertices such that for any , there exists ; in other words, it is a transversal of . Similarly, a total in-dominating set of is a transversal of . We note that always has an in-dominating set ( itself), while has a total dominating set if and only if it has no sources (vertices with empty in-neighbourhoods). Out-dominating and total out-dominating sets are defined similarly.

The fractional in-dominating number of and the fractional total out-dominating number of are then, respectively:

 γinf(D) :=τf(Hinc(D))=τf(Houtc(D)∗), Γoutf(¯¯¯¯¯D) :=τf(Houto(¯¯¯¯¯D))=τf(¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯Houtc(D)).

Let us call a vertex in-universal in if for all . We note that if and only if has no in-universal vertices; the latter is also equivalent to being well defined. We obtain the following

###### Theorem 3.1 (Fractional domination vs. fractional total domination).

For any loopless digraph without in-universal vertices,

 1γinf(D)+1Γoutf(¯¯¯¯¯D)=1.

We focus on three special cases of Theorem 3.1. Firstly, a graph can be viewed as a symmetric digraph. For a graph , in-neighbourhoods and out-neighbourhoods coincide. We then refer to as the fractional dominating number of ; the fractional total dominating number of is defined and denoted similarly.

###### Corollary 3.2.

Let be a graph without universal vertices, then

 1γf(G)+1Γf(¯¯¯¯G)=1.

Secondly, if is a tournament, then is obtained by reversing the direction of every arc in . Thus, and we obtain the following corollary.

###### Corollary 3.3.

If is a tournament without an in-universal vertex, then

 1γinf(T)+1Γinf(T)=1.

Thirdly, is -regular if for every vertex , . Clearly, if has vertices, then is -regular if and only if is -regular. The following result was proven in [4, Theorem 7.4.1] in the case of graphs.

###### Corollary 3.4.

If has vertices and is -regular, then and .

###### Proof.

The value is an obvious upper bound for (assign to each vertex); similarly, an upper bound for . By Theorem 3.1, these bounds must be tight. ∎

### 3.2 Application to edge toughness of matroids

The hypergraph complementation theorem yields two bounds on the sizes of the intersections of edges in hypergraphs. For any hypergraph , any and any , let

 ρH(S) :=max{|S∩e|:e∈E}, α(H) :=maxS⊆V,ρH(S)>0|S|ρH(S).

We similarly define

 ~ρH(Z) :=min{|{e∈Z:v∈e}|:v∈V}, β(H) :=minZ⊆E,~ρH(Z)>0|Z|~ρH(Z).
###### Proposition 3.5.

For any hypergraph , we have

 p(H)≤α(H)≤pf(H)=kf(H)≤β(H)≤k(H).
###### Proof.

We prove . Let be a maximum packing of , then , thus

 α(H)≥|S|ρH(S)≥p(H).

We now prove . Let be such that , then the vector given by for all and for all is a feasible fractional packing of value .

The proofs of are similar, and hence omitted. ∎

Let be a matroid, where is the collection of independent sets of . A basis of is a maximal independent set. We then denote the set of bases of as and we construct the hypergraph . The rank function of is then , i.e. . The dual matroid is then defined as 222The dual of a matroid is commonly denoted as , but in this paper, denoting it as better reflects that its definition is in terms of hypergraph complementation, instead of duality..

The edge toughness (or strength) of is 

 σ′(M):=minS⊆V,ρM(V)>ρM(S)|V∖S|ρM(V)−ρM(S).

The edge toughness is well defined unless for all . Moreover, if and only if has a coloop, i.e. an element that belongs to all bases. Say that is nontrivial if it falls in neither case mentioned above; then its edge toughness satisfies .

###### Theorem 3.6.

The fractional transversal number and fractional matching number of a matroid coincide with its edge toughness: for any nontrivial matroid , we have

 μf(HB(M))=τf(HB(M))=σ′(M).

The proof is based on the following lemma.

###### Lemma 3.7.

If is a nontrivial hypergraph, then

 1α(¯¯¯¯¯H)+1β(H∗)=1.
###### Proof.

We shall prove the equivalent statement:

 1α(H)+1β(¯¯¯¯¯¯¯H∗)=1.

We have

 1−1α(H)=1−minS⊆V,ρH(V)>ρH(S){ρH(S)|S|}=maxS⊆V,ρH(V)>ρH(S){|S|−ρH(S)|S|}=1γ(H),

where

 γ(H):=minT⊆V,|T|>ρH(T)|T||T|−ρH(T).

We only need to prove that . We denote the set of edges of as , and the set of edges of as . For any , we have

 ~ρH∗(T)=min{|T∩e|:e∈E}=|T|−max{|T∩¯¯¯e|:¯¯¯e∈¯¯¯¯E}=|T|−ρ¯¯¯¯H(T),

and hence

 β(H∗)=minT⊆V,~ρH∗(T)>0|T|~ρH∗(T)=minT⊆V,|T|>ρ¯¯¯¯H(T)>0|T||T|−ρ¯¯¯¯H(T)=γ(¯¯¯¯¯H).

###### Proof of Theorem 3.6.

Firstly, by [4, Theorem 5.4.1], the fractional covering number of a matroid reaches the bound in Proposition 3.5:

 kf(HB(M))=α(HB(M)).

The hypergraph complementation theorem then yields

 1−1α(HB(M))=1−1kf(HB(M))=1kf(¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯HB(M)∗)=1τf(HB(¯¯¯¯¯¯M)).

Secondly, we recognise that , where is defined in the proof of Lemma 3.7. Indeed, using the formula for the rank function of the dual matroid , we obtain

 σ′(M) =minT⊆V,ρM(V)>ρM(V∖T)|T|ρM(V)−ρM(V∖T) =minT⊆V,|T|>ρ¯¯¯¯M(T)|S||T|−ρ¯¯¯¯¯M(T) =γ(HB(¯¯¯¯¯¯M)).

We then have . Thus, by Lemma 3.7, we obtain

 1σ′(M)=1β(HB(M)∗)=1−1α(HB(¯¯¯¯¯¯M))=1τf(HB(M)).

The edge toughness of a matroid generalises the edge toughness of a graph. Indeed, let be the cycle matroid of a connected graph , where the elements of are the edges of and the bases of are all spanning trees of 333The case where is not connected is a straightforward generalisation.. Then the edge toughness of reduces to the edge toughness of , defined as follows. For any , let denote the graph obtained by removing the edges from , and let denote the number of its connected components. Then

 σ′(G)=minZ⊆E(G),c(G−Z)>1|Z|c(G−Z)−1.

We remark that is well defined if and only if is nonempty. Moreover, if and only if has a cut edge, i.e. is not -edge connected.

Denote . The matching number of is the maximum number of edge-disjoint spanning trees in . On the other hand, the transversal number of is the smallest size of an edge cut set of . In particular, these two quantities are equal to whenever has a cut edge. Their fractional analogues are then equal to the edge toughness of .

###### Corollary 3.8.

For any -edge connected graph ,

 μf(HST(G))=τf(HST(G))=σ′(G).

## 4 Vertex cover with budget

### 4.1 The vertex cover hypergraph

Let be a graph. A vertex cover is a set of vertices such that is independent. We define as the hypergraph whose edges are all the vertex covers of . Then its complement is , whose edges are the independent sets of . It immediately follows that , the fractional chromatic number of . We note that is nontrivial if and only if is not empty. Denoting , we obtain the following.

###### Theorem 4.1.

For any nonempty graph ,

 1κf(G)+1χf(G)=1.

Let us give some properties of the quantity. Firstly, we can give general bounds on . Let denote the independence number of and denote its chromatic number. Then since and , we obtain

 χ(G)χ(G)−1≤κf(G)≤nn−α(G).

Conversely, since

 χf(G)≥χ(G)1+lnα(G),

we obtain

 κf(G)≤χ(G)χ(G)−1−lnα(G).

Moreover, if denotes the clique number of , we have

 κf(G)≤ω(G)ω(G)−1.

Secondly, is a rational number in . Conversely, for any rational number , there is with (since for the Kneser graph with ).

Thirdly, for any , determining whether is NP-complete (an immediate consequence of ). On the other hand, if is a line graph, or is perfect, then can be computed in polynomial time. If is vertex-transitive, then , where