in Weighted Graphs of Netflix Games
Gregory Gutin1, Philip R. Neary2, and Anders Yeo3,4 1
Department of Computer Science, Royal Holloway, University of London, Egham, United [email protected]
2 Department of Economics, Royal Holloway, University of London, Egham, United
3 Department of Mathematics and Computer Science, University of Southern
Denmark, [email protected]
4 Department of Pure and Applied Mathematics, University of Johannesburg, South
Africa
Abstract. Gerke et al. (arXiv:1905.01693, 2019) introduced Netflix Games
and proved that every such game has a pure strategy Nash equilibrium. In this paper, we explore the uniqueness of pure strategy Nash equilib-ria in Netflix Games. LetG= (V, E) be a graph and κ: V →Z≥0 a
function, and call the pair (G, κ) aweighted graph. A spanning subgraph H of (G, κ) is called aDP-Nash subgraphifH is bipartite with partite sets D, P called the D-setand P-setof H, respectively, such that no vertex ofP is isolated and for everyx∈D, dH(x) = min{dG(x), κ(x)}.
We prove that whether (G, κ) has a unique DP-Nash subgraph can be decided in polynomial time. We also show that when κ(v) = k ∈ Z≥0
for everyv ∈ V, the problem of deciding whether (G, κ) has a unique D-set is polynomial time solvable for k= 0 and 1, andco-NP-complete fork≥2.
1
Introduction
In this paper, all graphs are undirected, finite, without loops or parallel edges. LetG= (V, E) be a graph andκ: V →Z≥0 a function. Forv∈V,we will call κ(v) the weightofv and the pair (G, κ) aweighted graph. A spanning subgraph H of (G, κ) is called aDP-Nash subgraphif H is bipartite with partite setsD and P called the D-set and P-set ofH, respectively, such that no vertex ofP is isolated and for every x ∈ D, dH(x) = min{dG(x), κ(x)}, where dH(x) and dG(x) are the degrees ofxinH andG,respectively. SinceHis a bipartite graph, we will write it as the triple (D, P;E0),where D, P are D-set andP-set ofH, respectively, andE0 is the edge set ofH.A vertex setB is aD-set of (G, κ) if (G, κ) has aDP-Nash subgraph in whichBis the D-set. Gerke et al. [6] proved the following:
Theorem 1 implies that every weighted graph has aD-set. We provide a proof of the theorem in Appendix A to make this paper self-contained.
Let us consider a few examples ofDP-Nash subgraphs andD-sets. Ifκ(x) = 0 for every x∈ V then there is only oneDP-Nash subgraph with D-set V and empty P-set. If κ(x) = 1 for every x∈ V then everyDP-Nash subgraph is a spanning vertex-disjoint collection of stars, each with at least two vertices. If κ(x) = dG(x) for every x ∈ V then the D-set of each DP-Nash subgraph of (G, κ) is a maximal independent set ofG. It is well-known that a vertex set is maximal independent if and only if it is independent dominating. Since finding both maximum size independent set and minimum size independent dominating set are bothNP-hard [2], so are the problems of finding aD-set of maximum and minimum size. For more information on complexity of independent domination, see [7].
The notion of aD-set is not directly related to theCapacitated
Domina-tion problem where the number of vertices which a vertex can dominate does
not exceed its weight (capacity) [3, 9]. D-sets provide what one can call exact
capacitated domination, not studied in the literature yet, as far as we know. Theorem 1 means that all Netflix Games introduced in [6] have pure strategy Nash equilibria; see Section 2 for a brief discussion of Netflix Games and their relation toDP-Nash subgraphs andD-sets in weighted graphs. As explained in Section 2, there are two natural problems of interest in economics.
DP-Nash Subgraph Uniqueness: decide whether a weighted graph has a
uniqueDP-Nash subgraph, and
D-set Uniqueness:decide whether a weighted graph has a uniqueD-set.
While the problems are clearly related, we show that their time complexities are not unlessP=co-NP:DP-Nash Subgraph Uniquenessis polynomial-time solvable andD-set Uniquenessisco-NP-complete. In fact, forD-set
Unique-nesswe prove the following complexity dichotomy whenκ(x) =kfor every
ver-tex x∈V,wherek is a non-negative integer. Ifk≥2 thenD-set Uniqueness is co-NP-complete and if k∈ {0,1} then D-set Uniquenessis in P. We note that the proof of Theorem 1 in Appendix A implies that constructing a DP -Nash subgraph and, thus, a D-set in every weighted graph is polynomial-time solvable.
Preliminaries are given in Section 3. To obtain the above polynomial-time complexity results forDP-Nash Subgraph UniquenessandD-set Unique-ness, we first prove in Section 4 a charaterization of weighted graphs with unique D-sets, which we believe is of interest in its own right. In Section 5, we show that DP-Nash Subgraph Uniqueness is in P. In Section 6 we prove the above-mentioned complexity dichotomy for D-set Uniqueness.1 We conclude the paper in Section 7.
1
2
Motivation
There are many economic situations that are collectively referred to as
combina-torial assignment problems. The first systematic approach to issues of this type
was by Gale and Shapley [4] who studied ‘matching’ in marriage markets. They imagined a group ofnwomen and another group ofnmen, where everyone wants to be matched with one member of the opposite sex. The problem of finding an assignment that leaves everyone ‘content’ is difficult since there aren! possible assignments and individuals have preferences. Gale and Shapley proposed a so-lution. They called an assignment between women and menstable if there does not exist a woman-man pair (call them Ann and Barry) such that: 1) Ann is not paired with Barry, 2) Ann prefers Barry to her match, and 3) Barry prefers Ann to his match. Gale and Shapley’s ‘deferred acceptance algorithm’ confirms that a stable match always exists. Variants and extensions of the algorithm have been applied to a wide variety of assignment problems in economics including college admissions, the market for kidney donors, and refugee resettlement (see [14] for a survey).2
The assignment problem that motivates our study arises in the provision of local public goods. The story is as follows. There is a society of individuals arranged in a social network modelled as a graph where vertices represent indi-viduals and edges capture friendships. There is a desirable product, say access to Netflix or Microsoft Office, that is available for purchase. While the product can be shared upon purchase, an owner may only share access with a limited number of friends. Individual preferences are such that it is always better to have access than not, but, since access is costly each individual prefers that a friend purchases and shares their access than vice versa. This describes the Net-flix Games of Gerke et al. [6].3For a given Netflix game, aD-set lists those who purchase the product in equilibrium, while aDP-Nash subgraph lists those who purchase (theD-set), those who free-ride (theP-set), and exactly who inDeach individual in P receives an offer of access from (the edge set).4 Netflix Games generalise the models of local public goods without weight constraints, see [1, 5], for which the stable outcomes correspond to maximal independent sets.
The rationale for a detailed focus on what weighted graphs (G, κ) admit a uniqueDP-Nash subgraph and/or a uniqueD-set is that economic models with
2 Lloyd Shapley and Alvin Roth received the 2012 Nobel Memorial Prize in Economics
for their work in this area. (David Gale died in 2007.)
3 Vertices being constrained in the number of neighbours they may share with seems
well-suited to applications. In Netflix Games sharing bestows a benefit on neigh-bours, but this need not be the case. Gutin et al. [8] add constrained sharing to the Susceptible-Infected-Removed (SIR) model of disease transmission of Kermack and McKendrick [10]. Gutin et al. interpret constrained sharing as ‘social distancing’ restrictions imposed on a population and document how the reach of an epidemic is curtailed when such measures are in place.
4 One example from Gerke et al. was of a group of individuals who each want to attend
a unique equilibrium are as rare as they are useful. Uniqueness is rare due to the mathematical structure of economic models (formally, the best-response map of Nash [12, 13] rarely admits only one fixed point). Uniqueness is useful as (i) it saves the analyst from an ‘equilibrium selection’ headache - justifying why one equilibrium is more likely to emerge than another, and (ii) allows those who studygame-design to be confident in generating a particular outcome (since only one outcome is stable). It is for this reason that models with unique equilibria are so highly coveted (see for example the model of currency attacks in [11]), and why we believe the study of conditions under which uniqueDP-Nash subgraphs andD-sets exist will be of great interest to the economics community.
3
Preliminaries
In the rest of the paper, we will often writeGinstead of (G, κ) when the weight functionκis clear from the context. We will often omit the subscriptGinNG(x) anddG(x) when the graphGunder consideration is clear from the context. We will often shorten the termDP-Nash subgraph to Nash subgraph.
In the rest of this section, we provide two simple assumptions for the rest of the paper which will allow us to simplify some of our proofs. In both assumptions, (G, κ) is a weighted graph.
Assumption 1:For allu∈V we haveκ(u)≤d(u).
Assumption 1 does not change the set ofDP-Nash subgraphs of any weighted graph as ifκ(u)> d(u) we may letκ(u) =d(u) without changing min{κ(u), d(u)}. Due to this assumption, we can simplify the definition of a DP-Nash subgraph of a weighted graph (G, κ). A spanning subgraph H of Gis called a DP-Nash
subgraph if H is bipartite with partite sets D, P called the D-set and P-set
of H, respectively, such that no vertex of P is isolated and for every x ∈ D, dH(x) =κ(x).Note that Assumption 1 may not hold for a subgraph of (G, κ) if the subgraph uses the same weight function κrestricted to its vertices.
Assumption 2:If uv is an edge inG, thenκ(u)>0 orκ(v)>0.
This assumption does not change our problem due to the following:
Proposition 1. LetG∗be obtained fromGby deleting all edgesuvwithκ(u) =
κ(v) = 0. Then(D, P;E0)is a Nash subgraph of (G, κ)if and only if(D, P;E0)
is a Nash subgraph of(G∗, κ).
Proof. Let uv be any edge in Gwith κ(u) =κ(v) = 0 and let (D, P;E0) be a
Nash subgraph of (G, κ). Note that uv 6∈ E0 as if u∈ D then E0 contains no edge incident withuand ifv∈D thenE0 contains no edge incident withv and if u, v ∈ P then E0 does not contain the edge uv. This implies (D, P;E0) is a Nash subgraph of (G∗, κ).
Conversely if (D, P;E0) is a Nash subgraph of (G∗, κ) then (D, P;E0) is a Nash subgraph of (G, κ) as both graphs have the same weight function andG∗
4
Characterisation of Weighted Graphs with Unique
D-set
We begin this section by introducing some definitions and additional notation. For a set F of edges of a graph H and a vertex x of H, NF(x) = {y ∈ V(H)|xy∈F}anddF(x) =|NF(x)|.For a vertex setQof a graphH,NH(Q) =
S
x∈QNH(x).Define X(G, κ),Y(G, κ) andZ(G, κ) as follows. If (G, κ) is clear from the context these sets will be denoted by X,Y andZ, respectively.
X =X(G, κ) :={x|κ(x) =d(x)}
Y =Y(G, κ) :=N(X)\X Z=Z(G, κ) :=V(G)\(X∪Y)
Lemma 1. Letu∈V(G)and letX=X(G, κ). If|NG(u)∩X| ≤κ(u)then there
exists a Nash subgraph (D, P;E) of (G, κ) where u∈D and NG(u)∩X ⊆P.
Furthermore, if|NG(u)∩X|< κ(u)andw∈NG(u)\X then there exists a Nash
subgraph(D, P;E)of(G, κ)whereu∈D and{w} ∪(NG(u)∩X)⊆P.
Proof. Letu∈V(G) such that|NG(u)∩X| ≤κ(u). Recall that by Assumption 1 we haveκ(v)≤dG(v) for allv∈V(G). Let E∗ denote an arbitrary set ofκ(u) edges incident with u, such that NG(u)∩X ⊆ NE∗(u). Let T0 = NE∗(u),
G0 = G\({u} ∪T0) and (P0, D0;E0) a Nash subgraph of G0, which exists by Theorem 1. LetP=P0∪T0 and letD=D0∪ {u}.
Initially let ˆE=E0∪E∗. Clearly every vertex inP has at least one edge into D. Now letv∈D be arbitrary. IfdEˆ(v)6=κ(v) (recall thatκ(v)≤dG(v)) then we observe thatv∈D0 and
dEˆ(v) =dE0(v) = min{dG0(v), κ(v)}< κ(v).
Since v either has no edge to u or does not lie inX, observe that we can add κ(v)−dG0(v) edges to ˆE between v and T0 resulting in dˆ
E(v) = κ(v). After doing the above for everyx∈Dwe obtain a Nash subgraph (D, P; ˆE) ofGwith the desired properties. This completes our proof of the case|NG(u)∩X| ≤κ(u). The same proof can be used for the case |NG(u)∩X|< κ(u) if we choose E∗
such thatw∈T0. ut
Lemma 2. If X∪Z is not an independent set in(G, κ), then there exist DP
-Nash subgraphs ofGwith different D-sets. IfX∪Z is an independent set in G
then there exists a DP-Nash subgraph (D, P;E0) of (G, κ) where D = X ∪Z
andP =Y.
Proof. First assume thatX∪Z is not independent inGand thatuvis an edge
We now consider the case when u, v ∈ Z. As uv is an edge in G we may without loss of generality assume thatκ(v)≥1 (by Assumption 2). As|NG(v)∩ X|= 0< κ(v), Lemma 1 implies that there is a Nash subgraph (D0, P0;E0) in Gwherev∈D0 andu∈P0. As|NG(u)∩X|= 0≤κ(u), Lemma 1 implies that there is a Nash subgraph (D00, P00;E00) inGwhereu∈D00. As there exist Nash subgraphs whereu∈P0 and whereu∈D00, we are done in this case.
Now letX∪Z be independent inGand letP =Y and D=X∪Z. LetE0 contain all edges between X and Y as well as anyκ(z) edges from z to P for allz∈Z. AsY =N(X)\X we conclude that (D, P;E0) is a Nash subgraph of
(G, κ). ut
To state our characterisation result for weighted graphs possessing a unique D-set, we need some additional definitions and two properties.
Given a weightκon a graphG, for any subsetU ⊆V(G) letUκdenote a set of vertices obtained fromU by replacing each vertex,u∈U, by itsκ(u) copies. Note that ifκ(u) = 0 then the vertex uis not inUκ and|Uκ|=P
u∈Uκ(u). Given a weighted graph (G, κ), let Gaux be a bipartite graph with partite setsR0=X∪Z andY0=Yκ. For a vertexy∈Y,there is an edge from a copy ofy tor∈R0 in Gauxif and only if there is an edge fromy torinG.
Let Yκ>0 ⊆ Y consist of all vertices y ∈ Y with κ(y) > 0. For every set
∅ 6=W ⊆Yκ>0 let
L(W) ={x∈X∪Z | |N(x)∩W|> d(x)−κ(x)}.
We now define the propertiesM∗(G, κ) andO∗(G, κ):
M∗(G, κ) holds if for every set∅ 6=W ⊆Yκ>0 there is no matching fromL(W) toWκ of size|L(W)|in Gaux.
O∗(G, κ) holds if for every set∅ 6=W ⊆Yκ>0 we have|L(W)|>|Wκ|.
Theorem 2. If X ∪Z is not independent in G then (G, κ) has at least two
differentD-sets. IfX∪Z is independent inGthen the following three statements are equivalent:
(a) Ghas a unique D-set;
(b) M∗(G, κ) holds; (c) O∗(G, κ)holds.
Proof. The case of X ∪Z being not independent follows from Lemma 2. We
will therefore assume that X∪Z is independent in Gand prove the rest of the theorem by showing that (a) ⇒ (b) ⇒ (c) ⇒ (a). The following three claims complete the proof.
Claim A: (a)⇒ (b).
Proof of Claim A:Suppose that (a) holds but (b) does not. As (b) is false,
LetD1 =W, P1 = L(W) andG2 = G−(P1∪D1). Let (D2, P2;E20) be a DP-Nash subgraph of G2, which exists by Theorem 1. We will now prove the following six subclaims.
Subclaim A.1:For every y∈Y,we have |N(y)∩X| ≥κ(y) + 1.
Proof of Subclaim A.1:Assume that Subclaim A.1 is false and there exists
a vertexy ∈Y such that|N(y)∩X| ≤κ(y). By Lemma 1, there exists a Nash subgraph (D0, P0;E0) in G where y ∈ D0. By Lemma 2, there exists a Nash subgraph (D00, P00;E00) inGwherey ∈P00 (asP00=Y). Therefore (a) is false, a contradiction.
Sublaim A.2:Ifu∈D1=W thenN(u)∩X ⊆P1. Furthermore,|N(u)∩
P1| ≥κ(u) + 1.
Proof of Subclaim A.2:Letu∈D1 (and thereforeu∈W) be arbitrary and
let r ∈ N(u)∩X be arbitrary. We will show that r ∈ P1, which will prove the first part of the claim. As r ∈X we have dG(r) = κ(r). This implies that
|N(r)∩W| ≥1>0 =dG(r)−κ(r).Hence,r∈L(W) =P1 as desired.
We now prove the second part of Subclaim A.2. Sinceu∈Y,Subclaim A.1 implies that|N(u)∩X| ≥κ(u) + 1. As every vertex inN(u)∩X also belongs toP1,we have|N(u)∩P1| ≥κ(u) + 1.
Subclaim A.3:There exists a Nash subgraph (D1, P1;E01) ofG[D1∪P1].
Proof of Subclaim A.3:P1 andD1 were defined earlier so we will now define
E10. Let all edges of the matching M belong to E10. That is if u0v ∈ M and u0∈V(Gaux) is a copy ofu∈V(G), then add the edgeuvtoE0
1. We note that every vertex inP1is incident to exactly one of the edges added so far and every vertex u∈D1 is incident to at mostκ(u) such edges. By Subclaim A.2 we can add further edges betweenP1 andD1 such that every vertexu∈D1 is incident with exactlyκ(u) edges fromE10.
Subclaim A.4:Everyu∈(X∪Z)\L(W) has at leastκ(u) neighbours in
Y \W.
Proof of Subclaim A.4:Asu6∈L(W) we have that|NG(u)∩W| ≤d(u)−κ(u). This implies that|NG(u)\W| ≥κ(u). AsX∪Z is independent this implies that uhas at leastκ(u) neighbours inY \W,as desired.
Recall that (D2, P2;E20) is a Nash subgraph of G2 and by Subclaim A.3, (D1, P1;E10) is a DP-Nash subgraph ofG[D1∪P1].
Subclaim A.5:There exists a Nash subgraph (P1∪P2, D1∪D2, E10∪E20∪E∗)
ofGfor someE∗.
Proof of Subclaim A.5:Let P =P1∪P2, D =D1∪D2 and E0 =E10 ∪E20.
Clearly every vertex inP is incident with an edge inE0.
First consider a vertex u ∈ D1. By Subclaim A.2, u has at least κ(u) + 1 neighbours in P1 in G. Therefore, by Subclaim A.3, uis incident with exactly κ(u) edges ofE10 and so also withκ(u) edges ofE0.
Now consideru∈D2. Note thatuis incident with min{κ(u), dG2(u)} edges
therefore assume that u 6∈ X ∪Z \L(W), which implies that u ∈ Y \ W. By Subclaim A.1, u has at least κ(u) + 1 neighbours in X in G. Therefore, dG2(u) +|N(u)∩P1| ≥κ(u) + 1 (as every edge from uto X is counted in the
sum on the left hand side of the inequality). Thus, if min{κ(u), dG2(u)}< κ(u), then we can add edges fromutoP1 toE0 untiluis incident with exactlyκ(u) edges ofE0. Continuing the above process for alluand lettingE∗be the added edges, we obtain the claimed result.
Subclaim A.6:Claim A holds.
Proof of Subclaim A.6: By Lemma 2 there exists a DP-Nash subgraph,
(D, P, E0), with D = X ∪Z and P = Y. By Subclaim A.5, there exists a DP-Nash subgraph of Gwhere some vertices of Y belong to D, contradicting the fact that (a) holds. This completes the proof of Subclaim A.6, and therefore also of Claim A.
Claim B: (b)⇒ (c).
Proof of Claim B:Suppose that (b) holds but (c) does not. As (c) does not
hold there exists a∅ 6=W ⊆Yκ>0 such that|L(W)| ≤ |W|. Assume thatW is chosen such that|W|is minimum possible with this property. As (b) holds there is no matching betweenWκandL(W) inGauxsaturating every vertex ofL(W). By Hall’s Theorem, this implies that there exists a set S ⊆ L(W) such that
|NGaux(S)| <|S|. Note that NG(S) ⊆W such that NGaux(S) contains exactly
the copies ofNG(S). Note that|NG(S)| ≤ |NGaux(S)|<|S| asW ⊆Yκ>0. Let
W0 =W \NG(S). By definition we have
L(W0) ={x∈X∪Z| |NG(x)∩W0|> d(x)−κ(x)}.
We will now prove the following subclaim.
Subclaim B.1:L(W0)⊆L(W)\S.
Proof of Subclaim B.1:Letu∈L(W0) be arbitrary and note that|NG(u)∩
W| ≥ |NG(u)∩W0|> d(u)−κ(u).This implies thatu∈L(W).
We will now show that u 6∈ S. If u ∈ S, then N(u) ⊆ N(S), sou has no neighbours inW0 =W\N(S). Therefore,|NG(u)∩W0|= 0, and as we assumed that d(v)≥κ(v) for allv∈V(G), the following holds
|NG(u)∩W0|= 0≤d(u)−κ(u).
Therefore,u6∈L(W0), a contradiction. This implies thatu6∈Sand therefore L(W0)⊆L(W)\S.
By Subclaim B.1, we have that|L(W0)| ≤ |L(W)|−|S|<|L(W)|−|NG(S)|=
|W0|. This contradicts the minimality of|W|, and therefore completes the proof of Claim B.
Claim C: (c) ⇒(a).
Proof of Claim C:Suppose that (c) holds but (a) does not. By Lemma 2 and
IfYκ>0⊆P, then no vertex of X∪Z can belong toP as it would have no edge toD (asX ∪Z is independent). Therefore,X∪Z ⊆D in this case. Due to the definition of X (andY) and the fact thatX ⊆D, we have thatY ⊆P, which implies that D = X ∪Z and Y = P, which is a contradiction to our assumption that D6=X∪Z.
So we may assume that Yκ>0 6⊆ P. This implies that Yκ>0∩D 6=∅. Let W =Yκ>0∩D. We now prove the following subclaim.
Subclaim C.1:L(W)⊆P.
Proof of Subclaim C.1:Let w∈L(W) be arbitrary. Hence,|NG(w)∩W|> d(w)−κ(w). Ifw∈D, then whas at least κ(w) neighbours in P inG. By the above it has at least d(w)−κ(w) + 1 neighbours in W ⊆D, contradicting the fact that w has d(w) neighbours. This implies that w 6∈ D. Therefore, w∈ P and as w∈L(W) is arbitrary, we must haveL(W)⊆P.
We now return to the proof of Claim C. Recall thatX ∪Z is independent andL(W)⊆X∪W and every vertex inP has at least one edge toDinE0. By Subclaim C.1, L(W) ⊆P, which implies that there are at least |L(W)| edges from L(W) to W, as W = Yκ>0∩D. As there are at most θ = P
w∈W κ(w) edges fromW toL(W) we must have|L(W)| ≤θ=P
w∈Wκ(w) =|W κ|.
The above is a contradiction to (c). This completes the proof of Claim C and
therefore also of the theorem. ut
We immediately have the following:
Corollary 1. All Nash subgraphs of (G, κ) have the same D-set if and only if
X∪Z is independent inGandO∗(G, κ)holds.
Note that ifκ(x) = 0 for all x∈GthenX =V(G) andY =∅. In this case O∗(G, κ) vacuously holds and there is a unique Nash subgraph of (G, κ) with D-setV and emptyP-set.
5
Complexity of Uniqueness of Nash Subgraph
Theorem 3. DP-Nash Subgraph Uniquenessis in P.
Proof. Let (G, κ) be a weighted graph, and letX =X(G, κ), Y =Y(G, κ) and
Z =Z(G, κ) be as defined in the previous section. IfX∪Z is not independent then there exist distinct Nash subgraphs in (G, κ) by Lemma 2. So we may assume that X∪Z is independent. By Lemma 2 there exists a Nash subgraph (D, P;E0) inGwhereD=X∪Z andP =Y.
IfZ6=∅, then letz∈Z be arbitrary. InE0 we may pick anyκ(z) edges out ofz, as every vertex inY has an edge toX inE0. As d(z)> κ(z) we note that by picking different edges incident withz we get distinct Nash subgraphs ofG. We may therefore assume thatZ=∅.
We will now prove the following two claims which complete the proof of the theorem since the existence of a matching inGaux−xsaturating its partite set Y0 can be decided in polynomial time for everyx∈X.
Claim A:If for every x∈X there exists a matching inGaux−xsaturating
Y0 then there is only one Nash subgraph inG.
Proof of Claim A:We will first show that if the statement of Claim A holds
then O∗(G, κ) holds. Suppose that O∗(G, κ) does not hold. This implies that there is a set∅ 6=W ⊆Yκ>0such that|L(W)| ≤ |Wκ|. Note that, asZ =∅, we haveL(W) =NG(W)∩X. AsW 6=∅andW ⊆Y, we have thatNG(W)∩X6=∅. Letx∈NG(W)∩X be arbitrary. Now the following holds.
|(NG(W)∩X)\ {x}|=|L(W)| −1≤ |Wκ| −1<|Wκ|
This implies that there cannot be a matching in Gaux−xsaturatingY0, a contradiction. Thus,O∗(G, κ) must hold. By Corollary 1 we have that all Nash subgraphs must therefore have the same D-set. By Lemma 2 we have that all Nash subgraphs (D, P;E0) must therefore have D = X and P = Y. By the definition ofX we note thatE0 must contain exactly the edges betweenX and Y, and therefore there is a unique Nash subgraph in G.
Claim B: If for some x∈ X there is no matching in Gaux−x saturating
Y0, then there are at least two distinct Nash subgraphs inG.
Proof of Claim B:Letx∈X be defined as in the statement of Claim B. By
Hall’s Theorem there exists a set S0 ⊆ Y0 such that |NGaux(S0)\ {x}| <|S0|.
Let S ⊆ Y be the set of vertices for which there is a copy in S0. Note that (NG(S)∩X)\{x}=NGaux(S0)\{x}and|S0| ≤ |Sκ|, which implies the following.
|NG(S)∩X| ≤ |(NG(S)∩X)\ {x}|+ 1 =|NGaux(S0)\ {x}|+ 1
<|S0|+ 1≤ |Sκ|+ 1.
As all terms above are integers, this implies that|NG(S)∩X| ≤ |Sκ|. AsL(S) = NG(S)∩X by the definition ofL(S), we note that |L(S)| ≤ |Sκ| and therefore O∗(G, κ) does not hold, which by Corollary 1 implies that there are distinct Nash subgraphs in G (even with distinctD-sets). This completes the proof of Claim B and therefore also of the theorem. ut
6
Complexity of Uniqueness of
D-set
IfZ =∅thenGhas a uniqueD-set if and only ifDhas a unique Nash subgraph (this follows from the proof of Theorem 3). Thus, if Z =∅ then by Theorem 3 it is polynomial to decide whether Ghas a uniqueD-set.
k= 1. However, Theorem 5 shows that fork≥2,D-set Uniquenessisco-NP -complete.
Theorem 4. Let (G, κ)be a weighted graph and letκ(x) = 1 for allx∈V(G).
Let X = {x | dG(x) = 1}, Y = N(X) and Z = V(G)\(X ∪Y). Then all
DP-Nash subgraphs have the same D-set if and only if X ∪Z is independent
and |NG(y)∩X| ≥2 for all y ∈Y. In particular,D-set Uniqueness is inP
in this case.
Proof. If X∪Z is not independent then we are done by Lemma 2, so assume
that X ∪Z is independent. By Lemma 2, there exists a DP-Nash subgraph, (D, P;E0), such that Y = P. If |NG(y)∩X| < 2 for some y ∈ Y, then by Lemma 1 there exists aDP-Nash subgraph, (D0, P0;E00), ofG,where y ∈ D0. This implies that there existsDP-Nash subgraphs whereybelongs to itsD-set and wherey belongs to itsP-set, as desired.
We now assume thatX∪Zis independent and|NG(y)∩X| ≥2 for ally∈Y. We will prove that all DP-Nash subgraphs have the sameD-set in (G, κ) and we will do this by proving thatO∗(G, κ) holds, which by Corollary 1 implies the desired result.
Recall thatO∗(G, κ) holds if for every set∅ 6=W ⊆Y we have|L(W)|>|W|
(as Yκ>0 =Y and Wκ =W). Let W be arbitrary such that ∅ 6=W ⊆Y. By the definition of L(W), we have that |L(W)| ≥ |N(W)∩X|. As no vertex in X has edges to more than one vertex in Y (as dG(x) = 1) we have that
|N(W)∩X|=P
w∈W|N(w)∩X| ≥2|W|. Therefore, we have
|L(W)| ≥ |N(W)∩X| ≥2|W|>|W|.
implying thatO∗(G, κ) holds, as desired. ut
The following result is proved by reductions from 3-SAT. This reduction is direct for the case ofk = 2,where for an instance I of 3-SAT formula, we can construct a weighted graph (G, κ) such that κ(x) = 2 for every vertex xof G and (G, κ) at least twoD-sets if and only ifI is satisfiable. In the case ofk≥3, we first trivially reduce from 3-SAT to k-out-of-(k+ 2)-SAT, where a CNF formulaF hask+ 2 literals in every clause andF is satisfied if and only if there is a truth assignment which satisfies at leastkliterals in every clause. Then we reduce fromk-out-of-(k+ 2)-SATto the complement ofD-set Uniqueness. While the main proof structure is similar in both cases, the constructions of (G, κ) are different. The full proof can be found in Appendix B.
Theorem 5. Let k ≥2 be an integer. D-set Uniqueness is co-NP-complete
for weighted graphs (G, κ)withκ(x) =kfor all x∈V(G).
7
Conclusions
polynomials in n. Indeed, we can consider every non-empty subsetS of V(G) in turn and check whether S is the D-set of a Nash subgraph of (G, κ) using network flows. Conditional on the Strong Exponential Time Hypothesis holding, one can show that there exists aδ >0 such thatUniquenessD-setcannot be solved in time-O∗(2nδ). A natural open question is to compute a maximum such valueδ.
Consider a weighted graph (K3n, κ), where n ≥ 1 and κ(v) = 2 for every v ∈V(K3n).Observe that every p-size subset of V(K3n) for n≤p≤3n−2 is aD-set. Thus, a weighted graph can have an exponential number ofD-sets and hence of Nash subgraphs. This leads to the following open questions: (a) What is the complexity of counting all Nash subgraphs of a weighted graph? (b) Is there anO∗(dp(G, κ))-time algorithm to generate all Nash subgraphs of (G, κ), where dp(G, κ) is the number of Nash subgraphs in (G, κ)?
References
1. Bramoull´e, Y., Kranton, R.: Public goods in networks. Journal of Economic Theory 135(1), 478–494 (2007)
2. Corneil, D., Perl, Y.: Clustering and domination in perfect graphs. Discrete Appl. Math.9, 27–39 (1984)
3. Cygan, M., Pilipczuk, M., Wojtaszczyk, J.O.: Capacitated domination faster than O(2n). Inf. Process. Lett.111(23-24), 1099–1103 (2011)
4. Gale, D., Shapley, L.S.: College admissions and the stability of marriage. The American Mathematical Monthly69(1), 9–15 (1962)
5. Galeotti, A., Goyal, S., Jackson, M.O., Vega-Redondo, F., Yariv, L.: Network games. The Review of Economic Studies77(1), 218–244 (2010)
6. Gerke, S., Gutin, G., Hwang, S.H., Neary, P.R.: Netflix games: Local public goods with capacity constraints. arXiv:1905.01693 (2019)
7. Goddard, W., Henning, M.A.: Independent domination in graphs: A survey and recent results. Discrete Mathematics313(7), 839–854 (2013)
8. Gutin, G., Hirano, T., Hwang, S.H., Neary, P.R., Toda, A.A.: The effect of social distancing on the reach of an epidemic in social networks. arXiv:2005.03067 (2020) 9. Kao, M., Chen, H., Lee, D.: Capacitated domination: Problem complexity and
approximation algorithms. Algorithmica72(1), 1–43 (2015)
10. Kermack, W.O., McKendrick, A.G.: A contribution to the mathematical theory of epidemics. Proceedings of the Royal Society of London. Series A, Contain-ing Papers of a Mathematical and Physical Character115(772), 700–721 (1927), http://www.jstor.org/stable/94815
11. Morris, S., Shin, H.S.: Unique equilibrium in a model of self-fulfilling currency attacks. The American Economic Review88(3), pp. 587–597 (1998)
12. Nash, J.: Non-cooperative games. The Annals of Mathematics 54(2), 286–295 (September 1951)
13. Nash, J.F.: Equilibrium points in n-person games. Proceedings of the National Academy of Sciences of the United States of America36(1), 48–49 (1950) 14. Roth, A.E.: Deferred acceptance algorithms: history, theory, practice, and open
Appendix A: Proof of Theorem 1
The proof is a slight modification of the proof in [6].
Recall that by Assumption 1, for allv ∈V we haveκ(v)≤d(v). The proof proceeds by induction on the numbern of vertices of G. Ifn= 1, then G is a Nash subgraph withD=V(G) andP =∅. Now assume the claim is true for all graphs with fewer thann≥2 vertices, and let Gbe a graph onnvertices.
Case 1: There is a vertex u of degree equal κ(u).LetB be the star with
center {u} and leavesNG(u) and letG0=G−B. If G0 has no vertices then B is clearly a Nash subgraph of G with D-set {u} and P-set NG(u). Otherwise, by induction hypothesis,G0 has a Nash subgraph H0 = (D0, P0;E). Construct a subgraph H of Gfrom the disjoint union of H0 and B as follows. For every v∈D0 withdG0(v)< κ(v),add exactlyκ(v)−dG0(v) edges ofGbetweenv and
NG(u).SetD=D0∪ {u}andP =P0∪N(u).To see thatH is a Nash subgraph of G, observe that (a) H is a spanning bipartite subgraph of G as H0 and B are bipartite and the added edges are between D and P only, (b) every vertex x∈D has degree in H equal toκ(x),and (c) every vertex y∈P is of positive degree (since it is so in bothB andH0).
Case 2: For every vertex z ∈ V(G), dG(z)> κ(z). Let u be an arbitrary
vertex. Delete any dG(u)−κ(u) edges incident to u and denote the resulting graph by L. Observe that every Nash subgraph of L is a Nash subgraph ofG since no vertex inLhas degree less thanκ(u).This reduces Case 2 to Case 1. ut
Appendix B: Proof of Theorem 5
Case 1: k = 2. We will reduce from 3-SAT. Let I = C1∧C2∧ · · · ∧Cm be
an instance of 3-SAT. Let v1, v2, . . . , vn be the variables in I. Assume without loss of generality that n is even (otherwise, we add to I a new clause which contains three new variables). We will now build a graphGand letκ(x) = 2 for every vertexxofG, such that there existDP-Nash subgraphs ofGwith distinct D-sets if and only ifI is satisfiable.
First define G1 as follows. Let Wi = {wi,wi¯ } for all i = 1,2, . . . , n. Let V(G1) = W1 ∪W2∪ · · · ∪Wn∪ {r1, r2, . . . , rn}. Let E(G1) contain all edges betweenW2i−1 andW2i and the edger2ir2i+1 fori= 1,2, . . . ,n2 (wherern+1= r1) and edges betweenriandWifori= 1,2, . . . , n. The graphG1is illustrated below when n= 6.
w1
¯
w1
w2
¯
w2
r2 r3
w3
¯
w3
w4
¯
w4
r4 r5
w5
¯
w5
w6
¯
w6
r6
r1
Let Gs
1 be the graph obtained from G1 after subdividing every edge once. Let u(e) denote the new vertex used to subdivide the edge e∈ E(G1) and let U ={u(e)|e∈E(G1)}.The graph Gs
w1 ¯ w1 w2 ¯ w2
r2 r3
w3 ¯ w3 w4 ¯ w4
r4 r5
w5 ¯ w5 w6 ¯ w6 r6 r1
LetQ={q1, q2}, letX∗ ={x∗1, x∗2, x∗3, x∗4, x∗5}, and for everyi= 1,2, . . . , n, letZi ={z1
i, z2i, z3i, zi4, zi5}. LetG2 be obtained fromGs1 by adding the vertices
Q∪X∗∪Z1∪Z2∪ · · · ∪Zn. Furthermore, add all edges fromWi toZi, all edges
from Zi toq1 for i= 1,2, . . . , n, and all edges from Qto X∗. The graph G2 is illustrated below whenn= 6.
w1 ¯ w1 w2 ¯ w2
r2 r3
w3 ¯ w3 w4 ¯ w4
r4 r5
w5 ¯ w5 w6 ¯ w6 r6 r1 q1 q2
Z1 Z2 Z3 Z4 Z5 Z6
Q X∗
We now constructGfromG2by adding the vertices{c1, c2, . . . , cm}and the following edge for each j = 1,2, . . . , m. If the clauseCj contains the literal vi then add an edge fromcj towiand ifCjcontains the literal ¯vithen add an edge from cj to ¯wi. Finally, add an edge fromcj to q1 for eachj = 1,2, . . . , m. This completes the construction of G. IfI = (¯v2∨v3∨v5)∧ · · · andI contains six variables thenGis illustrated below.
w1 ¯ w1 w2 ¯ w2
r2 r3
w3 ¯ w3 w4 ¯ w4
r4 r5
w5 ¯ w5 w6 ¯ w6 r6 r1 q1 q2
c1 cm
Z1 Z2 Z3 Z4 Z5 Z6
Q X∗
We will now show that there existDP-Nash subgraphs of G with distinct D-sets if and only if I is satisfiable. We prove this using the following three claims.
Claim A:There exists aDP-Nash subgraph(D, P;E0)of Gsuch thatP =
V(G1)∪QandD=V(G)\P.
Proof of Claim A: This claim follows from Lemma 2. Indeed, X(G, κ) =
X∗∪U, Y(G, κ) = Q∪V(G1) and Z(G, κ) = V(G)\(X(G, κ)∪Y(G, κ)). Observe thatX(G, κ)∪Z(G, κ) is an independent set. Thus,Y is aP-set inG.
Claim B:IfIis satisfiable then there exists aDP-Nash subgraph,(D, P;E0),
such that P 6=V(G1)∪Q.
Proof of Claim B:Assume thatIis satisfiable and letτbe a truth assignment to the variables inI which satisfies I.Construct P andD as follows. We start by constructing P \U and D\U. Since D = V(G)\P, we will describe only P \U. For everyi = 1,2, . . . , n, ifvi is true inτ then add the vertex wi to P otherwise ¯wi to P. AddQtoP.This completes construction ofP\U.
We will now distribute the vertices u(e) in P and D for each e ∈ E(G1) and constructE0 such that (D, P;E0) is aDP-Nash subgraph of Gwith P 6= V(G1)∪Q. LetE0contain all edges betweenX∗andQ. For eachz∈Zi add the edge zq1 to E0 and add the edge between z and the vertex in Wj that belongs to P to E0 fori= 1,2, . . . , n. For each j = 1,2, . . . , madd the edge cjq1 to E0 and the edge fromcj to the vertexwi ifviis a true literal inIand to the vertex
¯
wi if ¯vj is a true literal inI(just pick one true literal inCj).
Initially add allu(e) toP fore∈V(G1). We will move some of these vertices toD below. For eachi= 1,2, . . . ,n2 proceed as follows.
– If w2i−1, w2i ∈D: Move the vertex u( ¯w2i−1w2i) to¯ D. Now add the edges shown in the below picture toE0.
r2i−1
w2i−1
¯
w2i−1
w2i
¯
w2i
r2i r2i+1
– If w2i−1,w¯2i ∈D: Move the vertex u( ¯w2i−1w2i) to D. Now add the edges shown in the below picture toE0.
r2i−1
w2i−1
¯
w2i−1
w2i
¯
w2i
r2i r2i+1
r2i−1
w2i−1
¯
w2i−1
w2i
¯
w2i
r2i r2i+1
– If ¯w2i−1,w2i¯ ∈D: Move the vertex u(w2i−1w2i) to D. Now add the edges shown in the below picture toE0.
r2i−1
w2i−1
¯
w2i−1
w2i
¯
w2i
r2i r2i+1
After the above process we obtain the desiredDP-Nash subgraph (D, P;E0).
Claim C: If there exists a DP-Nash subgraph (D, P;E0) such that P 6=
V(G1)∪QthenI is satisfiable.
Proof of Claim C:Assume that there exists aDP-Nash subgraph (D, P;E0)
of G such that P 6= V(G1)∪Q. We now show the following subclaims which complete the proof of Claim C and therefore of Case 1.
Subclaim C.1:X∗⊆D andQ⊆P.
Proof of Subclaim C.1:Suppose thatQ6⊆P. This implies that each vertex
in X∗ has at most one neighbour in P and therefore X∗ ⊆ P. However, then each vertex in X∗ has an edge to D inE0, but these five edges are all incident with {q1, q2}, a contradiction to κ(q1) = κ(q2) = 2. Therefore, Q ⊆ P. This immediately implies that X∗ ⊆ D, as if some x ∈ X∗ belonged to P then it would have no edge toD inE0.
Subclaim C.2:|D∩Wi| ≤1for all i= 1,2,3, . . . , n.
Proof of Subclaim C.2:Suppose that|D∩Wi|>1 for somei∈ {1,2, . . . , n}. This implies that {wi,wi¯ } ⊆ D. In this case we must have Zi ⊆ P as every z ∈Zi has exactly one neighbour in P (namelyq1). By Subclaim C.1 we note that q1 ∈ P, which implies that every vertex in Zi has an edge into D in E0 and all these five edges must be incident with Wi. This is a contradiction to κ(wi) =κ( ¯wi) = 2 and thus to|D∩Wi|>1.
Subclaim C.3:|D∩Wi| = 1 for all i = 1,2,3, . . . , n and cj ∈ D for all
j= 1,2, . . . , m
Proof of Subclaim C.3:Let D0 =D∩V(G1). IfD0 =∅, then we note that
u(e) ∈ D for every e ∈E(G1), cj ∈ D for every j = 1,2, . . . , m, and Zi ⊆D for every i= 1,2, . . . , n, by Subclaim C.1. By Subclaim C.1 we note that P = V(G1)∪Q, a contradiction to our assumption in the begining of the proof of Claim C. ThereforeD06=∅.
LetG0be the subgraph ofG1induced by the vertices inD0. By Subclaim C.2 we note that the maximum degree ∆(G0) ofG0 is at most 2, as G1 is 3-regular and every vertex inG1is adjacent to both vertices inWjfor somej. This implies
Suppose that|E(G0)|<|D0|+|CP|. LetE00 denote all edges inG1that are incident with at least one vertex fromD0. As G1 is 3-regular we note that the following holds.
|E00|= 3|D0| − |E(G0)|>3|D0| −(|D0|+|CP|) = 2|D0| − |CP|
However, we note that u(e) ∈ P for every e ∈ E00 (as it has degree two inG and a neighbour inD). These|E00|edges as well as the|CP| edges fromCP to D in E0 are all incident with D0 in E0, a contradiction to|E00|+|CP|>2|D0| (as κ(s) = 2 for all vertices s ∈ V(G)). Therefore |E(G0)| ≥ |D0|+|CP|. As
|E(G0)| ≤ |D0|this implies that|E(G0)|=|D0|and|CP|= 0.
As∆(G0)≤2 this implies thatG0is a collection of cycles (that is, 2-regular). By Subclaim C.2 we note that the only possible cycle in G0 is the cycle of length 2ncontaining allr1, r2, . . . , rn and exactly one vertex from each set Wi. Therefore D0 contains exactly one vertex from each Wi for i = 1,2, . . . , n, as desired. As|CP|= 0 we also note thatcj∈D for allj = 1,2, . . . , m.
Subclaim C.4:I is satisfiable if we letvi be true whenwi∈P and let vi be
false when wi ∈D for alli= 1,2, . . . , n.
Proof of Subclaim C.4:For eachj = 1,2, . . . , m proceed as follows. By Sub-claim C.3 we note thatcj ∈D and therefore has two edges toP inE0. At least one of these edges most go to a vertex inV(G1) ascj is only incident with one edge that is not incident with V(G1) (namely cjq1). Ifcjwi ∈E0 thenwi ∈P andviis true andviis a literal ofCj, soCj is satisfied. Alternatively ifcjwi¯ ∈E0 then ¯wi∈P (aswi∈Dand Subclaim C.3) andviis false and ¯viis a literal ofCj, so againCj is satisfied. This implies thatCj is satisfied for allj = 1,2, . . . , m, which completes the proof of Case 1.
Case 2: k ≥ 3. We will reduce from k-out-of-(k+ 2)-SAT. That is, every
X∗(k2+ 1)
Y∗(k−1)
y0
Z1 (2k+ 1)
w1 w¯1
W1
X1
k(k−1) 2
Y1
(k−2)
X10
k(k−1) 2
Z2 (2k+ 1)
w2 w¯2
W2
X2
k(k−1) 2
Y2
(k−2)
X20
k(k−1) 2
Z3 (2k+ 1)
w3 w¯3
W3
X3
k(k−1) 2
Y3
(k−2)
X30
k(k−1) 2
c1 c2
LetI=C1∧C2∧ · · · ∧Cmbe an instance ofk-out-of-(k+ 2)-SAT where
each clause Ci contains k+ 2 literals. Let v1, v2, . . . , vn be the variables in I. We will now build a graph G, where κ(x) = k for every vertex x of G. For each i = 1,2,3, . . . , n, let Wi = {wi,wi¯ }, let Xi and Xi0 be vertex sets of size
k(k−1)
2 ,letYi be a vertex set of sizek−2, and letZi a vertex set of size 2k+ 1. Furthermore, let X∗ be a vertex set of size k2+ 1, let Y∗ be a vertex set of sizek−1,andC={c1, c2, . . . , cm}.Lety0 be a vertex and define a graphGas follows (see the illustration above).
V(G) ={y0} ∪Y∗∪X∗∪C∪(∪n
i=1(Wi∪Xi∪X 0
i∪Yi∪Zi))
We first add the following edges toGfor alli= 1,2, . . . , n(where subscriptn+ 1 is equivalent to 1): All edges fromY∗ toZi, fromZitoWi, fromWitoXi, from XitoYi, fromYitoXi0, and fromXi0 toWi+1. Furthermore, add all edges from y0 toX∗ and all edges fromX∗toY∗. We then for allj= 1,2, . . . , madd edges fromcj to the vertices corresponding to the literals in the clause. (In the figure above,C1 contains literals ¯v1,v2, v3, . . .¯ andC2contains literals ¯v2,¯v3, . . ..)
Recall thatκ(v) =kfor allv∈V(G). Define X,Y andZ as follows.
LetX =X∗∪(∪n
i=1(Xi∪Xi0)). LetY =N(X)−X ={y0} ∪Y∗∪(∪n
i=1(Wi∪Yi)). LetZ =V(G)\(X∪Y) =C∪(∪n
We will now show that there exist Nash subgraphs ofG with at least two distinct D-sets if and only ifI is satisfiable. We prove this using the following claims.
Claim A: There exists a Nash subgraph (D, P;E0) of G such that P =Y
andD=X∪Z.
Proof of Claim A:This claim follows from Lemma 2. Indeed,X(G, κ) =X,
Y(G, κ) =Y andZ(G, κ) =Z.Observe thatX∪Zis an independent set. Thus, Y is a P-set inG.
Claim B:IfIis satisfiable then there exists a Nash subgraph(D, P;E0)such
that P 6=Y.
Proof of Claim B:Assume thatIis satisfiable and letτbe a truth assignment to the variables in Iwhich satisfiesI.ConstructP and Das follows.
– Ifτ(vi) =truethen letwi ∈P and ¯wi∈D.
– Ifτ(vi) =falsethen let ¯wi∈P andwi∈D.
– LetXi∪Xi0 belong toP for alli= 1,2, . . . , n.
– LetZi∪Yi belong toD for alli= 1,2, . . . , n.
Let{y0} ∪Y∗belong toP andC∪X∗toD. This definesP andD. We now defineE0. LetE0 contain all edges betweenX∗ and{y0} ∪Y∗. For every vertex inZi(i= 1,2, . . . , n) add an edge to the vertex inWithat belongs toP and add all edges toY∗. For each vertex inWi (i= 1,2, . . . , n) that belongs toD addk edges toXi. For eachi= 1,2, . . . , naddk(k−1)/2−kedges fromYi toXi and k(k−1)/2 edges fromYitoXi0in such a way that each of thek−2 vertices inYi is incident with kedges and each vertex in Xi and Xi0 is incident with exactly one edge (fromWi∪Yi). Finally, for eachj = 1,2, . . . , m pickktrue literals in Cjand add an edge fromcj towiifviis one of the true literals and add an edge from cj to ¯wi if ¯vi is a true literal. This completes the construction ofE0. We note that (D, P;E0) is a Nash subgraph withP 6=Y.
Claim C:If there exists a Nash subgraph (D, P;E0) such thatP 6=Y then
I is satisfiable.
Proof of Claim C: Assume that there exists a Nash subgraph (D, P;E0) of
Gsuch that P 6=Y. We now show the following subclaims which complete the proof of Claim C and therefore of the theorem.
Subclaim C.1:X∗⊆D and{y0} ∪Y∗⊆P.
Proof of Subclaim C.1:Suppose that{y0} ∪Y∗6⊆P. This implies that each vertex inX∗has at mostk−1 neighbours inP and thereforeX∗⊆P. However, then each vertex in X∗ has an edge to D in E0, but these k2+ 1 edges are all incident with the k vertices {y0} ∪Y∗, a contradiction to κ(x) = k for every x∈X∗. Therefore,{y0} ∪Y∗⊆P. This immediately implies thatX∗⊆D, as if some x∈X∗ belonged toP then it would have no edge toD in E0.
Subclaim C.2:|D∩Wi| ≤1for all i= 1,2,3, . . . , n.
has at mostk−1 neighbours inP (namely the vertices inY∗). By Subclaim C.1 we note thatY∗ ⊆P, which implies that every vertex inZi has an edge intoD inE0and all these 2k+ 1 edges must be incident withWi. This is a contradiction toκ(wi) =κ( ¯wi) =kand thus to |D∩Wi|>1.
Subclaim C.3: |D ∩Wi| = 1 for all i = 1,2, . . . , n and cj ∈ D for all
j= 1,2, . . . , m
Proof of Subclaim C.3:By Subclaim C.2 we note that|D∩Wi| ≤1. Suppose
that|D∩Wi|= 0 for somei∈ {1,2, . . . , n}. We consider the following two cases.
Case C.3.1: |D∩Wi|= 0 for all i= 1,2, . . . , n.If D∩Yi 6=∅ for somei,
then Xi∪Xi0 ⊆P, as every vertex,x, in this set has dG(x) =κ(x) =kand at least one neighbour is inD. However thek(k−1) vertices inXi∪Xi0will all have edges into D∩Yi (as|D∩Wi|= 0 for alli= 1,2, . . . , n) in E0, a contradiction toD∩Yi being incident with at mostk|Yi|=k(k−2) edges. Therefore we may assume thatD∩Yi=∅ for alli= 1,2, . . . , n.
By Subclaim C.1 we note thatY ={y0} ∪Y∗∪(∪n
i=1(Wi∪Yi))⊆P. In this case we must have Y =P, asG−Y is independent. This is a contradiction to P 6=Y. This completes Case C.3.1.
Case C.3.2: |D∩Wi| 6= 0 for some i ∈ {1,2, . . . , n}. Without loss of
generality assume that |D∩W1| 6= 0. This implies thatX1∪Xn ⊆P, as every vertex, x∈X1∪Xn, hasdG(x) =κ(x) =kand at least one neighbour is in D. Without loss of generality assume that the vertex inD∩W1has at least as many edges toXn as toX1 inE0. This implies that there are at mostk/2 edges from D∩W1toX1inE0. Ask/2< k(k−1)/2 we note thatY1∩D6=∅. This implies that X10 ⊆P, as every vertex,x∈X10, has dG(x) =κ(x) = k and at least one neighbour is inD. AsX1∪X10 ⊆P,X1∪X10 has at least|X1|+|X10|=k(k−1) edges intoD∩(W1∪Y1∪W2). This implies that W2∩D 6=∅, as there can be at mostk/2 +k(k−2) edges fromD∩(W1∪Y1) toX1∪X10 inE0. Furthermore there are at least k/2 edges fromD∩W2 to X10 in E0. Continueeing the above process we note that W3∩D 6=∅, W4∩D 6= ∅, etc. This contradicts the fact that |D∩Wi|= 0 for somei. This completes Case C.3.2.
By the above two cases we note that |D∩Wi| = 1 for all i = 1,2, . . . , n. This implies thatXi∪Xi0 ⊆P for alli= 1,2, . . . , n. Therefore there are at least nk(k−1) edges from∪n
i=1(Xi∪Xi0) toD∩(∪i=1n (Wi∪Yi)). As|D∩Wi|= 1 for alli= 1,2, . . . , nthere can be at most nk+nk(k−2) such edges. This implies that there are exactlynk(k−1) edges from∪n
i=1(Xi∪Xi0) toD∩(∪ni=1(Wi∪Yi)) and there are no edges fromD∩(∪n
i=1(Wi∪Yi)) to any vertex inP that does not lie in ∪n
i=1(Xi∪Xi0). This implies that cj ∈D for all j = 1,2, . . . , m, as if some cj ∈P then it would not have any edge toD.
Subclaim C.4:I is satisfiable if we letvi be true whenwi∈P and let vi be
false when wi ∈D for alli= 1,2, . . . , n.
