Optimal strategies in fractional games: vertex cover and domination

In a hypergraph with vertex set $V$ and edge set $E$, a real-valued function $f: V \to [0, 1]$ is a fractional transversal if $\sum_{v\in e} f(v) \ge 1$ for every edge $e \in E$. Its size is $|f| := \sum_{v \in V} f(v)$, and the fractional transversal number is the smallest possible $|f|$. We consider a game scenario where two players with opposite goals construct a fractional transversal incrementally, trying to minimize and maximize $|f|$, respectively. We prove that both players have strategies to achieve their common optimum, and they can reach their goals using rational weights.


Introduction
Let H = (V, E) be a finite hypergraph, where V is the finite vertex set and E is the edge set, a set system over the underlying set V .We assume that every edge contains at least one vertex; that is, E ⊆ 2 V \ {∅}.A hypergraph is k-uniform if |E| = k holds for all E ∈ E. A set T ⊆ V is a transversal1 of H if every edge is covered by a vertex of T , which formally means that T ∩ E = ∅ holds for all E ∈ E. Its real relaxation, called fractional transversal, is a function f : V → [0, 1] such that v∈E f (v) ≥ 1 holds for every E ∈ E. The size of f is defined as |f | := v∈V f (v).The transversal number τ (H) and the fractional transversal number τ * (H) of H are the minimum cardinality |T | of a transversal and minimum value |f | of a fractional transversal, respectively.
The transversal game is a competitive optimization version of hypergraph transversals, which was introduced in [8] and studied further in [9].It is played on a hypergraph H by two players called Edge-hitter and Staller.They take turns choosing a vertex.The game is over when all edges are covered, and the length of the game is the number of vertices chosen by the players.Edge-hitter wants to finish the game as soon as possible, while Staller wants to delay the end.To prevent Staller from making completely useless moves, we stipulate that the chosen vertex must be contained in at least one previously uncovered edge.
Assuming that both players play optimally and Edge-hitter starts, the length of the game on H is uniquely determined.It is called the game transversal number of H and is denoted by τ g (H).Analogously, the Staller-start game transversal number of H, denoted by τ g (H), is the length of the game under the same rules when Staller makes the first move.Among other results, it was proved in [8] that |τ g (H) − τ g (H)| ≤ 1 always holds.We further recall that, denoting by n(H) and m(H) the number of vertices and edges in H respectively, 4  11 (n(H) + m(H)) is a (sharp) upper bound on τ g (H) if H does not contain one-element edges and it is not isomorphic to the cycle C 4 .
Below we shall refer to this game as the integer game, as opposed to its fractional version which we will introduce in the next section.
Important motivation of this approach are the domination game [6] and the total domination game [15], where in fact the transversal game is played on the 'closed neighborhood hypergraph' and on the 'open neighborhood hypergraph' of a graph, respectively.Further variants studied so far include the disjoint domination [12], connected domination [2], and fractional domination [13] games on graphs, and the domination games on hypergraphs [11].Some of the most recent results can be found in [3,4,7,10,14,17,18,19,20,21].For a thorough survey and list of further references see the book [5].

Our results
In Section 2 we introduce the rules of the game, prove that its value is well-defined, and present some examples.The latter also show that it matters which player starts, moreover edges that are subsets of other edges of the hypergraph may influence the game value, which is not the case for the standard non-game version of transversal number.
In Section 3 we compare the game transversal number with other related parameters, and prove a monotonicity property, implying that changing the starting player can affect the value of the game by at most 1.
The rules of the game allow both players to split their moves into infinitely many submoves.In Section 4 we prove that any infinite move is equivalent to a finite move.
In Section 5 we prove that the game can be modeled in a way that leads to an optimization problem solvable via the theory of piecewise linear continuous rational functions.From this we derive that the game value is rational for every finite hypergraph, moreover both players can achieve their goals using rational submoves.
Consequences on domination games and several conjectures are given in the concluding section.

Fractional transversal game
Let H = (V, E) be a hypergraph.In the context of the fractional transversal game, we will consider a (partial) cover function t : V → [0, 1] that is updated after each move during the game.We denote by |t| the sum v∈V t(v).Given a cover function t, the corresponding load function is : E → [0, 1] defined by the rule for every E ∈ E. If ≡ 1, we say that H is fully covered.We shall write t i and i for the cover and load functions obtained after the i th move.
The game begins with t 0 ≡ 0 and therefore with 0 ≡ 0. It is finished when the hypergraph becomes fully covered.Edge-hitter and Staller take turns making moves under the following rules.As long as ≡ 1, the next player performs a move which is a sequence (v i 1 , w 1 ), (v i 2 , w 2 ), . . . of arbitrary length (possibly infinite).It consists of the submoves (v i k , w k ), k = 1, 2, . . ., where v i 1 , v i 2 , . . .are vertices of H with any number of repetitions allowed, and the weights w 1 , w 2 , . . .are real numbers from [0, 1].It is required that in each move except the last one, whereas it is enough to have k≥1 w k ≤ 1 in the last move.The i th move (v i 1 , w 1 ), (v i 2 , w 2 ), . . . is legal if it satisfies the following condition: ( * ) For every k ≥ 1 there exists an edge That is, for each legal submove there exists an edge E such that its load increases by exactly the weight in the submove.This rule forces that no part of w k be wasted during the submove.
The cover function t i can gradually be reached from t i−1 by adding the weight w k to t(v i k ) after each submove; this process converts also the corresponding load function from i−1 to i .
The value |G| of the game is defined as the value |t q | of the cover function obtained at the end, that is the sum of the weights that have been spent during the game.The goal of Edge-hitter is to achieve a value |G| as small as possible, while Staller wants a large |G|.
Assuming that Edge-hitter starts the fractional transversal game on H, we consider the set of upper bounds, U H = {a : Edge-hitter has a strategy that ensures |G| ≤ a} and the set of lower bounds, L H = {b : Staller has a strategy that ensures |G| ≥ b}.
Formally the game fractional transversal number τ * g (H) is defined as The Staller-start game fractional transversal number τ * g (H) is defined similarly, under the condition that the first move is made by Staller.
The following assertion shows that τ * g (H) is in fact the common optimum for Edgehitter and Staller, and so is τ * g (H) as well.The proof is essentially the same as the one for the game fractional domination number in [13].
Proposition 1.For every hypergraph H we have inf(U H ) = sup(L H ), and the analogous equality holds for the Staller-start game, too.
Proof.First, assume that inf(U H ) < sup(L H ) and consequently, there exist two reals x and y satisfying inf(U H ) < x < y < sup(L H ). By definition, x ∈ U H and, therefore, Edge-hitter can ensure that, under every strategy of Staller, the value of the game is at most x.Similarly, y ∈ L H and Staller has a strategy that ensures |G| ≥ y whatever strategy is followed by Edge-hitter.This is a contradiction that establishes inf(U H ) ≥ sup(L H ). Now, we prove the reverse inequality.By definition, z < inf(U H ) implies that Edgehitter does not have a strategy to achieve |G| ≤ z.That is, against each strategy of Edge-hitter there is a strategy of Staller which results in |G| > z.We may infer that z ∈ L H and therefore z ≤ sup(L H ). Since it holds for every z < inf(U H ), we conclude inf(U H ) ≤ sup(L H ).This completes the proof of the proposition.
Later, in Section 5, we will show that inf(U H ) = min(U H ) and sup(L H ) = max(L H ). Therefore, Edge-hitter and Staller have optimal strategies under which, respectively, |G| ≤ τ * g (H) and |G| ≥ τ * g (H) are achieved.
If Staller starts the fractional transversal game on C 4 with the move (v 1 , w 1 ), (v 2 , w 2 ), (v 3 , w 3 ), (v 4 , w 4 ) (or with any permutation of these submoves), then Edge-hitter can always ensure |G| = 2 by playing (v 1 , w 3 ), (v 2 , w 4 ), (v 3 , w 1 ), (v 4 , w 2 ).Indeed, 2 assigns w 1 + w 2 + w 3 + w 4 = 1 to every edge of the graph.Therefore, . Now we modify the previous example C 4 by adding four new vertices u 1 , . . ., u 4 and four new edges {v 1 , v 2 , u 1 }, . . ., {v 4 , v 1 , u 4 } to get the hypergraph H.When the (integer or) fractional transversal number is considered each edge that is a superset of another edge can be deleted.This gives τ * (H) = τ * (C 4 ) = 2.We show that the situation is different for the fractional transversal game on H. Again, an optimal first move for Edge-hitter is ), but here Staller may respond by playing and forces Edge-hitter to spend a weight of 1 in his next move.It can be proved that the value |G| = 3 equals τ * g (H) and therefore, τ * g (H) = τ * g (C 4 ).

Some basic facts and the Continuation Principle
In this section we first observe some simple inequalities which are analogous to the ones in other games concerning graph domination and hypergraph transversal, most notably to the fractional domination game [13].
(i) For every hypergraph H, it holds that , and not even with . The same holds true for τ * g (H), too.
Proof.No matter which player starts the game, at the end the cover function t q is a fractional transversal.This implies the lower bounds Concerning a fractional transversal game G on H and the upper bounds in (i), we can write the value of the game in the form |G| = W + W , where W and W denote the total sum of weights assigned by Edge-hitter and Staller, respectively.To keep the claimed bounds, first Edge-hitter can fix an optimal fractional transversal f , i.e. one with |f | = τ * (H).After that, in his moves he can apply the strategy to play submoves (v i j , w j ) with the largest possible weights w j which are not only allowed by ( * ) but also respect the inequalities t i−1 (v i j ) + w j ≤ f (v i j ).If such a legal submove with a positive weight does not exist anymore, then H is fully covered and the game is finished.This strategy yields W ≤ τ * (H), with strict inequality if the game is finished by Staller.We also have W ≤ W or W ≤ W + 1, depending on whether the first move is made by Edge-hitter or Staller, both with strict inequalities if the game is finished by Edge-hitter.Since only one of the players can make the last move, the claimed strict upper bounds follow.
For the proof of (ii) we apply the following result of Alon [1]: For every fixed > 0 there is an integer k and a k- On the other hand, a very simple fractional transversal f with |f | = |V |/k may be constructed by assigning k and we obtain due to the obvious fact τ ≤ τ g and the inequality τ * g < 2τ * from (i).For τ * g (H) the proof is similar, by the second part of (i).Proof.Consider the complete bipartite graph G = K k,k 2 on k + k 2 vertices as a 2-uniform hypergraph.Clearly, τ * (G) = k.In any submove of a fractional transversal game, while G is not fully covered, Staller can always select a vertex from the bigger partite class.Following this strategy, during k − 1 moves, Staller increases the sum of the loads by at most (k − 1)k.As ∆(G) = k 2 , k − 1 moves of Edge-hitter increase the loads by at most (k − 1)k 2 .Hence, no matter whether Edge-hitter or Staller starts the game, after 2k − 2 moves we have A monotone property of the game fractional transversal number is expressed in the following idea, which provides a useful tool in simplifying several arguments.Let a hypergraph H with a pre-defined load function be given, which we consider as a non-zero starting configuration.We ask about the value |G| of the game started by Edge-hitter, where the game is finished when is completed to a load function under which H is fully covered.The rules are the same as they were in the case of 0 ≡ 0, but here we have 0 = , while we still start with t 0 ≡ 0. Assuming that both players play optimally, the value of the game will be denoted by τ * g (H| ) and termed the game fractional -transversal number.The corresponding Staller-start value τ * g (H| ) is defined analogously.Proof.Assume for a contradiction that τ * g (H| ) < τ * g (H| ), and choose a real x with τ * (H| ) < x < τ * g (H| ).We use the imagination strategy between the following two games: Game 1: Edge-hitter plays on H| applying a strategy which ensures that the length is at most t 1 < x.
Game 2: Staller plays on H| applying a strategy which ensures that the length is at least t 2 > x.
Once this can be done, the assertion follows by the contradiction x > t 1 > t 2 > x.
The moves essentially are copied between Game 1 and Game 2 such that (E) ≤ (E) remains true for all E ∈ E after every move.If this inequality is valid before Staller's move, then her next move in H| is feasible in H| as well, so that Edgehitter can simply copy it and reply to it.On the other hand it may happen that one or more submoves (v i k , w k ) made by Edge-hitter in Game 1 are not feasible in Game 2. Observe, however, that if this happens, then already all loads on the edges incident with v i k reach 1 in H| .It follows that those loads will never become smaller than the corresponding ones in H| .Consequently, τ * g (H| ) ≥ τ * g (H| ) holds.The analogous conclusion can be reached in the Staller-start game as well, literally by the same argument, deriving a contradiction from the assumption τ * g (H| ) < τ * g (H| ).
We obtain the following immediate consequence.
Theorem 5.The game fractional transversal numbers for the Staller-start and for the Edge-hitter-start games on H may differ by at most 1.
Proof.Consider the Staller-start game.Whatever Staller moves first, she assigns total weight 1, and creates a situation which is at least as favorable for Edge-hitter as the all-zero load at the beginning of the original transversal game.Then, due to Theorem 4, Edge-hitter can ensure that the game ends using at most τ * g (H) further weight.This proves τ * g (H) ≤ τ * g (H) + 1.Similarly, if Edge-hitter starts, after his first move he is in at least as favorable position as with the all-zero load at the beginning of the Staller-start game.This proves the reverse inequality τ * g (H) ≤ τ * g (H) + 1.

Infinite moves are not necessary
The definition of a legal move in the transversal game admits the option that a player splits the value 1 into an infinite number of pieces; e.g., w k = 2 −k .It turns out, however, that each legal move on H = (V, E) is equivalent to a move which consists of at most |V | submoves.
Theorem 6.Every legal move in a fractional transversal game can be replaced with a legal move such that each vertex occurs in at most one submove of it and the two moves result in the same load function.
Proof.First, consider a vertex v which occurs in two different submoves (v i j , w j ) and (v i k , w k ) of a move.That is, v = v i j = v i k and we may assume j < k.By the condition ( * ), there exists an edge E ∈ E such that v ∈ E and the second submove (v i k , w k ) increases the load of E by exactly w k .If the submove (v i j , w j ) is deleted from the sequence and the weight w k is replaced by w j + w k in the k th submove, the submove and the whole move remain legal and result in the same load function as before.
Performing this modification repeatedly we can achieve that every vertex occurs in either zero or exactly one or many submoves of the move in question.This already proves the statement if the move contains only a finite number of submoves.Now, assume that the move is infinite.Then, the sequence of submoves can be split into two, such that the first part is finite, and in the second infinite part every vertex (which is present there) is repeated infinitely many times.Consider this infinite subsequence S = (v is , w s ), . . . .By renaming the vertices of H if necessary, we may assume that {v 1 , . . ., v } is the set of the vertices which are present in S. We prove that the finite sequence S = (v 1 , j: i j =1 w j ), . . ., (v , j: i j = w j ) is equivalent to S. Clearly, S and S yield the same load function after the move.So, it is enough to prove that S is legal.Assume for a contradiction that (v k , j: i j =k w j ) is not a legal submove in S , and let k be the smallest such index.Then, after the (k − 1) st submove of S , every edge E which contains v k has a load (E) > 1 − j: i j =k w j and, moreover, there is a positive constant such that min v k ∈E (E) + j: i j =k w j = 1 + .Now, consider S again.There is an index p = p( ) such that j≥p w j < and hence, before the p th submove of S, each edge containing v k is fully covered.As v k occurs infinitely often in S, and also the occurrences after the p th submove are legal, this is a contradiction.
We say that a legal finite move (v i 1 , w 1 ), . . ., (v i k , w k ) is transposable if for any permutation β(1), . . ., β(k) of 1, . . ., k, the move (v i β(1) , w β(1) ), . . ., (v i β (k) , w β(k) ) is also legal.Proposition 7. A legal move (v i 1 , w 1 ), . . ., (v i k , w k ), where w j > 0 for all 1 ≤ j ≤ k and no vertices are repeated, is transposable if and only if every vertex v i j is incident with an edge E i j such that v i ∈E i j t(v i ) ≤ 1 holds after performing the entire move.
Proof.If the inequality holds for E i j , then omitting the submove (v i j , w j ) from the move we obtain v i ∈E i j \{v i j } t(v i ) ≤ 1 − w j .Hence (v i j , w j ) is a legal submove no matter when it is performed during the move.This means that the move is transposable whenever the condition is satisfied for all j.
In the other direction assume that for some v i j every edge violates the inequality.Violation means that the corresponding sum is at least 1+ on every edge, for a certain fixed > 0. Consider now a permutation in which (v i j , w j ) is the last submove.Then (v i j , w j ) is not legal because at most w j − can be assigned to v i j legally.Consequently the move is not transposable.Theorem 8.If a legal move m = (v i 1 , w 1 ), . . ., (v i k , w k ) is not transposable in the fractional transversal game, then it can be replaced by a legal transposable move after which no edge gets smaller load than after m.
Proof.First, consider the legal move m = (v i 1 , w 1 ), . . ., (v i k , w k ) and the move m which is obtained by the permutation β = 2, . . ., k, 1.It is clear by definition that first k − 1 submoves of m remain legal.For the last (and not necessarily legal) submove, determine w * 1 as the maximum weight which results in a legal last submove with v i 1 .We have 0 ≤ w * 1 ≤ w 1 .If w 1 = w * 1 , then m is legal and gives exactly the same load function as m.If w 1 > w * 1 , then the same load function is obtained after the submove (v i 1 , w * 1 ) as after m, because in both cases every edge incident with v i 1 is fully covered and the loads of the other edges are unchanged.In this latter case, the sum of the weights used is 1 − (w 1 − w * 1 ) < 1.After this change, the submove (v i 1 , w * 1 ) will be legal in any permutation of (v i 1 , w 1 ), . . ., (v i k , w k ).That is, if a permutation is not feasible after this replacement, then infeasibility is due to another vertex.The same is true if some weights w s are replaced by smaller weights.
We repeat this step for the modified sequence with permutation β = 3, . . ., k, 1, 2, then with β = 4, . . ., k, 1, 2, 3, and so on, finally with β = k, 1, 2, . . ., k − 1, keeping all modifications incrementally.Decreasing the weight of the last submove in each step if necessary, at the end a legal transposable move m * is obtained, which yields the same load function as m and satisfies k j=1 w * j ≤ k j=1 w j .If the game is not over yet and if s * = k j=1 w j − k j=1 w * j is positive, we may use the weight s * to increase the loads of the non-fully covered edges.This may yield a move which is not transposable.If so, then we repeatedly perform the steps of modifiation as described above.
If the process terminates after a finite number of iterations, then the last version of the move is transposable, due to the way as the steps have been defined.Otherwise we obtain an infinite sequence s * 1 , s * 2 , . . . of re-distributions from the total unit weight of the move.Note that the total load of edges increases by at least i≥1 s * i , hence this sum is convergent because altogether the total load is at most |E|.On the other hand, the weight of each vertex changes by at most s * i in the i th iteration, hence the local changes in weight (one of which is negative in each iteration) sum up to at most i≥1 s * i , therefore they are also convergent.It means that for every > 0 there exists a threshold i( ) such that i≥i( ) s * i < /k.That is, the cummulative change of weight at each vertex obtained later than iteration i( ) is smaller than /k, and the change in the sum of weights inside each edge is smaller than .Let w * denote the limit sequence of the k weights of the k submoves under this infinite process, i.e. the submove at v j tends to weight w * j for j = 1, . . ., k (allowing that some of the w * j are zero).We claim that the move m * = (v 1 , w * 1 ), . . ., (v k , w * k ) is legal and transposable.This will complete the proof because the loads never decrease, hence under w * no edge gets smaller load than by move m.Let us also denote by t * the cover function obtained after m * , disregarding for the moment that the move is not yet proved to be legal.
Suppose for a contradiction that some permutation β yields a move which is not legal.We may assume without loss of generality that β = 1, 2, . . ., k and that the submove (v k , w * k ) is not legal.It means v j ∈E t * (v j ) ≥ 1+ for all edges E containing v k , for some constant > 0. We know, however, that if we subtract the distributed weight s * i( ) from the vertex loads in iteration i( ), we obtain a legal move; and com-pared to that situation, each vertex changed by no more than i≥i( ) s * i < /k.Consequently there exists an edge E ∈ E such that v k ∈ E and This contradiction completes the proof of the theorem.
Remark 9. Based on Theorem 8, Edge-hitter may restrict his strategy to transposable moves.On the other hand, the result suggests that Staller is advised to perform moves, if possible, which are 'very non-transposable' in a sense.

Algorithm for computing the value of the game
We consider an equivalent version of the game, the structured game, which is easier to analyze.
Each move consists of n + 1 rounds.Each round consists of n submoves, which are dedicated to the vertices v 1 , . . ., v n in succession.In each submove, the player whose turn it is can decide the amount w, the weight spent in the submove, by which the cover value of t(v i ) is increased, subject to the usual rules: The increase must be useful, i.e. each submove must satisfy the condition ( * ), and it must be within the budget constraint of total weight 1 to be spent per move.It is possible to skip a submove by simply choosing w to be zero.
The first n rounds are identical, but the last round is special: In each submove, the weight is greedily chosen as the largest possible legal weight, hence not allowing any freedom in choosing w for the player in those submoves.This ensures that the whole move spends a total weight of 1 unless a cover is obtained.
There are n moves, alternating between the two players.This is enough to ensure that a cover is constructed when the game terminates.Every move consists of n 2 + n submoves, and in total, the game consists of N = n 3 +n 2 submoves.We illustrate this for a small example with n = 4 vertices, where the Edge-hitter starts.The sequence of submoves is H i denotes a move of Edge-hitter for vertex i, and S i denotes a move of Staller for vertex i.The greedy moves are denoted by G i .
We do not stipulate as part of the rules that the whole budget of 1 unit must be spent during a move.This capacity is only an upper bound.It is still true that the whole budget is spent in each move if the game is played from the beginning.However, this arises as a consequence of the new setup, due to the greedy moves.
The duration of a play is defined as the total weight spent during all submoves.As soon as a cover is found, the rules imply that no more weight can be spent, and thus the game is effectively over.
Lemma 10.The structured game has the same value as the original game.
Proof.According to Theorem 6, we can assume that every vertex occurs at most once in a move.We can realize this in the structured game by selecting one vertex per round and leaving the weight at 0 otherwise.Thus, the structured game does not restrict the players' strategies, when compared to the original game.On the other hand, the structured game does not give the players more power: The greedy moves ensure that the total weight of 1 is used as long as it is possible.
In Remark 9 we have argued that, for the Edge-hitter, any permutation will do.Thus, a single round H 1 , H 2 , . . ., H n followed by n greedy submoves would be sufficient for Edge-hitter.For simplicity, we have however chosen to treat the two players uniformly.Note that the greedy submoves are necessary also in case of Edge-hitter.Otherwise, for example, he might pass the first move and transform the game to the Staller-start version which sometimes admits a smaller game value (cf. the example of C 4 ).
Consider the situation after the j th submove, 0 ≤ j ≤ N .Let x ∈ [0, 1] E be an arbitrary load vector, and let r ∈ [0, 1] be the budget for the current move that is still available.If j = k(n2 + n) + i, r is the total weight spent in the last i submoves.
We define T j ( x, r) as the remaining duration of the game if both players play optimally, starting from the current situation.If j is not very small then for a certain region of x it happens that a complete fractional cover is not reached because the game necessarily ends after the last submove of the n th round. 2 This convention has the advantage that T j is defined for arbitrary x and r.
The value of the whole game is T 0 ( 0, 1).w-axis.(In this hypothetical example, the resulting minimum is discontinuous; this cannot happen when F is continuous and its domain is the box Formally, we conduct the proof as follows.We know that the domain [0, 1] m+1 of F splits into finitely many rational convex (m + 1)-dimensional polytopes P on which F is linear: F (y, w) = a P y + b P w + c P , for (y, w) ∈ P for some rational coefficient vector a P and coefficients b P and c P .We can thus write T (y) as the minimum of finitely many functions T P (y) of the form T P (y) := min{ a P y + b P w + c P | 0 ≤ w ≤ 1, (y, w) ∈ P }, where the minimum of an empty set is taken as ∞.
For fixed y, the minimum in (8) depends on the sign of b P .If b P > 0, the minimum is achieved on a boundary point that lies on some facet P of P whose outer normal has negative w-coordinate.On such a facet, w can be expressed as a linear function of y, and thus, T P can be written as a linear function T P (y) = a P y + c P , for y ∈ P , where P is the projection of the facet P to [0, 1] m .Thus, T P (y) is the minimum of finitely many functions T P (y), with the understanding that T P (y) is taken as ∞ when y is outside its domain P .
The situation is similar for b P < 0. When b P = 0, then F does not depend on w and we can simply write T P (y) = a P y + c P , for y ∈ P , (10) where P is the projection of P .
summary, the function T (y) can be written as the minimum of finitely many pieces T P (y), each of which can in turn be written as the minimum of finitely many linear pieces (9) or (10).All these pieces have rational coefficients and rational domain boundaries, and since continuity of T has already been established, the PLCR property of T follows.
The proof of Theorem 11 is constructive and, in principle, it provides an algorithm for computing the value T 0 ( 0, 1) of the game.From this, we obtain the following important corollary.
Theorem 13.For every finite hypergraph H = (V, E), the game fractional transversal number τ * g (H) and its Staller-start version τ * g (H) are rational, and both players have optimal strategies in which they play only rational weights.Remark 14.It is not true in general that every optimal strategy uses only rational weights.A simple counterexample is the graph C 4 where Staller can start with placing x and 1−x on two vertices with any x ∈ [0, 1], no matter if x is rational or irrational.

Concluding remarks and open problems
Putting the fractional domination game [13] into a more general context, in this paper we introduced the fractional transversal game on hypergraphs.Among other results we proved that the game value is rational, and both players have optimal strategies using rational weights and with a finite number of submoves.Since a dominating set is a transversal of the closed neighborhood hypergraph, and total a dominating set is a transversal of the open neighborhood hypergraph, the following consequence is immediate.
Theorem 15.The fractional versions of both the domination game and the total domination game have rational game values on every hypergraph.Moreover, in either of these games, both players can achieve their common optimum using only rational weights and within finitely many submoves.
We conclude this paper with some conjectures.Conjecture 1.If each of the first 2k − 1 (k ≥ 1) moves was an integer move in the fractional transversal game, i.e. of the form (v i 1 , 1), then Staller has an integer move in the (2k) th turn, which is optimal in the fractional transversal game.
This means that fractional moves would be advantageous for Edge-hitter only.If true then this conjecture implies the following weaker one.Conjecture 2. For every hypergraph H, τ * g (H) ≤ τ g (H).
Perhaps the following stronger version of 1 is also true.
Conjecture 3. Starting from any partial covering function, there is an optimal strategy for Staller where, in every submove, she always spends the largest legal weight.

Proposition 3 .
The upper bounds in Proposition 2 (i) are tight apart from an additive constant at most 2.

Figure 1 :
Figure 1: The lower envelope of a polyhedral set in 2 dimensions (m = 1).