Digraph Placement Games

Let $G=(V,E)$ a directed graph and $\phi\colon V\to\{\text{blue},\text{red}\}$ a $2$-coloring of the vertices of $G$. Note that $\phi$ need not be a proper coloring. The Digraph placement game played on $G$, denoted simply by $G$ when the fact that we are playing a Digraph placement game is clear from context, is described as follows: Suppose there are two players, Left (blue) and Right (red). On her turn Left chooses a blue vertex $v$ of $G$ that has not been deleted, and deletes all vertices in $N^{+}[v]$. On his turn Right chooses a red vertex $u$ of $G$ that has not been deleted, and deletes all vertices in $N^{+}[u]$. If at the beginning of her turn, all blue vertices have been deleted, then Left loses. Similarly, if at the beginning of his turn, all red vertices have been deleted, then Right loses. When necessary we denote the Digraph placement game played on digraph $G=(V,E)$ with $2$-coloring $\phi$ by $G=(V,E,\phi)$. The set of all Digraph placement games is the ruleset Digraph placement.

Given a Digraph placement game $G$ and a sequence of vertices $u_{1},\dots,u_{t}$ such that $u_{i}$ is not an out-neighbour of any vertex $u_{j}$ where $j<i$, we denote the game resulting from deleting vertices $u_{1},\dots,u_{t}$ and their out-neighbourhoods in $G$ by $G/[u_{1},\dots,u_{t}]$. To make for easier to read expressions, we denote $G/[u]$ simply as $G/u$ when only a single vertex and its out-neighbours are removed. We make no assumptions about the colours of removed vertices to allow for non-alternating play.

For readers familiar with the combinatorial game Node Kayles, observe that Digraph placement can be seen as a directed version of partisan Node Kayles. For readers unfamiliar, partisan Node Kayles is played on a $2$-colored simple graph, where players take turns choosing a vertex of their color, call it $u$, then deleting the closed neighbourhood of $u$, $N[u]$, from the graph.

Any Digraph placement game is a two-player game with perfect information, no chance, and where a player loses if they cannot move on their turn. As a result, every Digraph placement game is an example of a normal play combinatorial game. The players in a Digraph placement game necessarily have different options on their move, such games are called partisan, which simply implies that players need not have all of the same moves. See Figure 1 for two examples of Digraph placement games. The first example is disconnected and equal to a sum of simple combinatorial games, the second is connected and is a equal to the same sum of games.

This paper is intended for a general mathematical audience. We assume familiarity with basic notation from graph theory, such as the in or out neighbourhood of a vertex, and induced subgraphs. We do not assume the audience is familiar with combinatorial game theory. We follow many conventions from combinatorial game theory, see [1, 3, 30].

A combinatorial game is a two-player game which involves no randomness or hidden information. Normally the players in a combinatorial game are denoted Left and Right. For technical reasons, a specific instance of a game is called a game, whereas the set of all games played with the same rules is called a ruleset. Examples of combinatorial game rulesets include Chess, Checkers, and Go. A combinatorial game is short if there exists an integer $k$, such that after $k$ turns the game must end, no matter the choices by either player, and at any point during the game, both players have at at most a finite number of moves. This prevents infinite sequences of play or infinite numbers of followers. Games which allow infinite sequences of play or allow a player to have infinitely many moves on their turn are called long. A combinatorial game is played under the normal play convention if a player loses if and only if they cannot move on their turn. In this paper all the combinatorial games we consider are normal play and short. Hence, if we say that $X$ is a game, suppose that $X$ is short.

Given a game $X$ we let $L(X)$ be the set of games Left can move to if they move first in $X$, and we let $R(X)$ be the set of games Right can move to if they move first. We say a member of $L(X)$ is a Left option of $X$ and a member of $R(X)$ is a Right option of $X$. Often a game $X$ will be written as $\{Y_{1},\dots,Y_{k}\mathrel{|}Z_{1},\dots,Z_{t}\}$ where $L(X)=\{Y_{1},\dots,Y_{k}\}$ and $R(X)=\{Z_{1},\dots,Z_{t}\}$.

Given a game $X$ written as $\{Y_{1},\dots,Y_{k}\mathrel{|}Z_{1},\dots,Z_{t}\}$ and a game $X^{\prime}$ written as
$\{Y^{\prime}_{1},\dots,Y^{\prime}_{k}\mathrel{|}Z^{\prime}_{1},\dots,Z^{\prime}_{t}\}$ we say that $X\cong X^{\prime}$ if and only if $Y_{i}\cong Y^{\prime}_{i}$ and $Z_{j}\cong Z^{\prime}_{j}$ for all $i$ and $j$. The game with no moves from either player, $\{\mathrel{|}\}$, is denoted by $0$. So $0\cong\{\mathrel{|}\}$. If $X\cong X^{\prime}$ then we say that $X$ and $X^{\prime}$ have the same literal form. Notice that literal forms are useful as they allow one to say that two games appearing in different rulesets are literally the same, even when the rulesets are dissimilar

Games having the same literal form is too strong a notion of equivalence to be useful in many contexts. To that end we define another standard notion of equivalence in games, called equality. Given games $X$ and $Y$, the game $X+Y$ is the game with $L(X+Y)=\{X^{L}+Y:X^{L}\in L(X)\}\cup\{X+Y^{L}:Y^{L}\in L(Y)\}$ and $R(X+Y)=\{X^{R}+Y:X^{R}\in L(X)\}\cup\{X+Y^{R}:Y^{R}\in R(Y)\}$. This is often less formally written as

We say that $X+Y$ is the disjunctive sum of $X$ and $Y$. Although it may not be immediately clear, the game $X+Y$ is the game played by each player choosing either $X$ or $Y$, but not both, to make a move in on their turn. As players lose when they have no move this new game played on two smaller games is well defined. Given a game $X$ we say the outcome of $X$ is Left win if Left will win moving first or second, is Right win if Right will win moving first or second, is next player win if whoever moves next will win, or is previous player win if whoever moves first will lose. Now, we say games $Y$ and $Z$ are equal, written as $Y=Z$, if and only if for all games $X$, the outcome of $X+Y$ is the same as the outcome of $X+Z$.

Observe that if $G$ is a Digraph placement game, whose underlying graph is disconnected, then $G$ is a disjunctive sum of its components. Also, notice that even if $G$ is a connected Digraph placement game, then $G$ can easily become disconnected through players deleting vertices. Hence, understanding disjunctive sums, and therefore values, is critical to understanding Digraph placement games.

We say the class of games equal to $X$, is the value of $X$. For each game $X$, the games with value $X$ have a unique simplest representative, which is called the canonical form of $X$. For a proof of this see Theorem 2.7 and Theorem 2.9 from Chapter II of [30].

Modulo equality, the set of all games under this addition operation forms an abelian group. This group, and other algebraic objects like it, are the central object of study in combinatorial game theory. For more on the foundations of combinatorial game theory we recommend [1, 3, 30].

Placement games, also called medium placement games, are a class of rulesets introduced in [7], satisfying the following properties,

1. 1.

   the game begins with an empty game board, and

2. 2.

   a move is to place pieces on one, or more, parts of the board subject to the rules of the game, and

3. 3.

   the rules must imply that if a piece can be legally placed in a certain position on the board, then it was legal to place this piece in that position at any time earlier in the game, and

4. 4.

   once played, a piece remains on the board, never being moved or removed.

Note here that deleting a vertex $v$ in Digraph placement can be viewed as placing a piece onto $v$, with the added rule that if an in-neighbour of a vertex $u$, or $u$ itself, has a piece placed on it, then no further piece may be placed onto $u$. In this way, the set of vertices with pieces placed on them at the end of the game will form a minimal directed dominating set. So Digraph placement is a placement game.

Placement games that satisfy the additional constraint, that if a position can be reached by a sequence of legal moves, then any sequence of moves which reach this position is legal, are called strong placement games. This class of games was introduced by Faridi, Huntemann, and Nowakowski in [15], while the theory of strong placement games has been expanded upon in [16, 23, 24, 26]. Notice that Digraph placement is not a strong placement ruleset.

Placement games include many well studied rulesets such as Nim [5], Hackenbush [3, 13], Col [13, 18], Snort [3, 27, 25, 29], NoGo [7, 10], Poset Games [11, 18, 17, 19, 31], Node Kayles [4, 7, 20], Arc Kayles [8, 22, 29], Domineering [2, 6, 26, 35], and Teetering Towers [12]. Note that throughout the paper, when discussing rulesets such as Hackenbush and Poset Games, we mean blue, green, red Hackenbush and Poset Games. Domineering is an example of a strong placement game while Hackenbush is an example of a placement game which is not a strong placement game.

These rulesets are diverse in notable ways. For example, one can calculate the winner of a game of Nim or Teetering Towers in polynomial time [5, 12], whereas in other rulesets such as Snort, Col, and Poset Games calculating the winner of a given position is PSPACE-complete [18, 19, 29]. As another example of the diversity of possible rulesets that are presentable as placement games, observe that rulesets such as Nim both players always have the same option, while in other rulesets such as Domineering, players never the same options.

Given a ruleset $R$, one of the natural questions to ask is, for which games $X$ does there exist a game $G$ in $R$ where $G=X$. For example, it was shown by Conway [13] that every game of Col is equal to $x$ or $x+\mathord{\ast}$ for a number $x$. Alternatively, Huntemann [23] showed that many other kinds of values such a nimbers and switches appear in strong placement games. For many rulesets $R$, there exists a game $X$ such that for all games $G$ in $R$, $G\neq X$. For example, if $R$ is an impartial ruleset, then no game in $R$ is equal to $1\cong\{0\mathrel{|}\}$. If for all $X$ there exists a game $G$ in $R$ such that $G=X$, then we say that the ruleset $R$ is universal.

Surprisingly, no ruleset was known to be universal until 2019. Even more interestingly, when a ruleset was discovered to be universal by Carvalho and Pereira dos Santos in [9], this ruleset, called Generalised Konane, was well established in the literature [14, 21, 28] and did not have a mathematical origin. Deciding a winner in Konane, the game which inspires Generalised Konane, is known to be PSPACE-complete [21] and the game originates from the ancestral peoples of Hawaii.

Since 2022, several more universal rulesets have been discovered. The first of these was given by Suetsugu [32] who showed another ruleset, called Turning Tiles, was universal. Unlike Konane, Turning Tiles is not an established ruleset, having been first defined in [32]. Determining the winner in Turning Tiles is also PSPACE-complete [37]. The next year, in 2023, Suetsugu [34] defined two more rulesets, called Go on Lattice and Beyond the Door, which they showed were also universal. By reducing Go on Lattice and Beyond The Door to Turning Tiles it was also shown that determining the winner of a game of Go on Lattice or Beyond the Door is also PSPACE-complete [37].

The primary contribution of this paper is to show that Digraph placement is a universal ruleset. This is significant because Digraph placement is the first placement ruleset which has been shown to be universal. This is particularly notable because Digraph placement, and placement games in general, provide an opportunity to build a new bridge between combinatorial game theory and other areas of mathematics, in particular graph theory. This is not only because Digraph placement is played on a graph, as other rulesets, for example Go on Lattice, can be generalised to be played on general graphs. Rather, placement games are much closer to traditional problems in graph theory, such as graph coloring and domination, compared to Generalised Konane or the rulesets introduced by Suetsugu. For example Col is a game where players take turns proper coloring vertices of a graph until no vertices remain that can be properly colored. As another example, in Node Kayles players take turns picking vertices, so that the sets of all chosen vertices form an independent set.

Like other placement games played on graphs such as Col, Snort, and Node Kayles, many tools from graph theory can be applied to study Digraph placement games. Given Digraph placement is universal, this may allow future research to study the space of all short games in new ways. Furthermore, there is also the possibility of tools from combinatorial game theory being helpful in graph theory.

The rest of the paper is structured as follows. Section 2 is devoted to introducing definitions and more necessary considerations. This includes standard notation from combinatorial game theory. Next, in Section 3 we prove that many well known rulesets are special cases of Digraph placement. Section 4 extends this discussion by proving that Digraph placement is a universal ruleset. This should be viewed as the main contribution of our paper. In Section 5 we consider the related extremal question of given a game $X$, what is the smallest integer $n$ such that there is a Digraph placement game $G$ on $n$ vertices, where $G=X$? We conclude with a discussion of future work.

We must introduce some standard notation and ideas from the combinatorial game theory literature. We require these tools for proving various statements throughout the paper. What follows will be familiar to combinatorial game theorists. Such readers can proceed directly to Section 3.

As noted in the introduction, the set of all normal play games modulo equality forms an abelian group under the disjunctive sum operation. However, we have yet to address what an additive inverse of a game $X$ is. We clarify this now. Given a game $X=\{Y_{1},\dots,Y_{k}\mathrel{|}Z_{1},\dots,Z_{t}\}$, we define the negative of $X$, denoted $-X$, as follows,

Notice that $-X$ is essentially $X$, except the roles of each player are reversed. Hence, if $G$ is a Digraph placement game, then $-G$ is the Digraph placement game played on the same digraph, where blue vertices in $G$ are red in $-G$, and red vertices in $G$ are blue in $-G$.

Also, note that $0\cong\{\mathrel{|}\}$ is the additive identity, due to the fact that $X+0\cong X$, and any games with the same literal forms are equal. Hence, we note that $X+(-X)$, which can be written as $X-X$ is equal to $0$. Notice that whoever moves first in $X-X$ and in $0$ loses. This is no coincidence as $X-X+0\cong X-X$ must have the same outcome as $0+0\cong 0$ by the definition of disjunctive sum. We further note that $X-Y$ is previous player win if and only if $X-Y=0$ if and only if $X=Y$.

Using this observation we extend our equivalence relation, equality, into a partial order. Given games $X$ and $Y$, we say $X\leq Y$ if Right wins moving second in $X-Y$. Equivalently, $X<Y$ if and only if Right wins moving first or second. Symmetrically, we $X\geq Y$ if Left wins $X-Y$ moving second and $X>Y$ if and only if Left wins $X-Y$ moving first or second. We note that if $X\leq Y$ and $X\geq Y$, then this implies $X=Y$. We use $X\mathrel{\fltrivb}Y$ to denote $X\not\geq Y$. This implies, that $X$ is Left win if and only if $X>0$, that $X$ is right win if and only if $X<0$, that $X$ is previous player win if and only if $X=0$, and $X$ is next player win if and only if $X\not\geq 0$ and $X\not\leq 0$.

Finally, we observe that there are many pairs of games $X$ and $Y$ such that $X\not\geq Y$ and $X\not\leq Y$. For example the game $\{1\mathrel{|}-1\}\cong\{\{0\mathrel{|}\}\mathrel{|}\{\mathrel{|}0\}\}$ satisfies $\{1\mathrel{|}-1\}\not\geq 0$ and $\{1\mathrel{|}-1\}\not\leq 0$. We also note that $\{1\mathrel{|}-1\}\not\geq 1\cong\{0\mathrel{|}\}$ and $\{1\mathrel{|}-1\}\not\leq 1\cong\{0\mathrel{|}\}$. Notice that $X\not\geq Y$ and $X\not\leq Y$ if and only if $X-Y$ is next player win.

These observations provide a very useful way of proving how the value of two games $X$ and $Y$ are related. That is, if we want to check if $X\leq Y$, then we simply play $X-Y$ and determine if Left loses moving first. Such a proof, is called a playing proof. Moreover, these inequalities provide a means to study when a game $X$ is better for one player, than a game $Y$. If $X$ is at least as good for Left than $Y$, then $X\geq Y$.

These inequalities are important in defining dominated options and numbers. Given a game $X=\{Y_{1},Y_{2},\dots,Y_{k}\mathrel{|}Z_{1},Z_{2}\dots,Z_{t}\}$, we say a Left option $Y_{1}$ is dominated if there exists another Left option, call it $Y_{2}$, such that $Y_{1}\leq Y_{2}$. Here we can think of $Y_{2}$ being a better move for Left than $Y_{1}$. Because $Y_{2}$ is a better move, we can ignore $Y_{1}$, since Left will never move to $Y_{1}$ when using an optimal strategy. Thus, when $Y_{1}$ is a dominated Left option, we can remove $Y_{1}$ without changing the value of $X$. That is,

Similarly, if $Z_{1}\geq Z_{2}$, then we say that $Z_{1}$ is a dominated Right option. As before, we can remove a dominated option without changing the value of $X$,

It is important to note that canonical forms have no dominated options for either player.

The second way we will use these inequalities outside of playing proofs, is by considering a notable class of games called numbers. We say that a game $X$ is a number if and only if all $X^{L}\in L(X)$ and all $X^{R}\in R(X)$ satisfy $X^{L}<X^{R}$. We point out that it was shown in [13] that every number $x$ is equal to a Hackenbush stalk, that is a blue, red path in Hackenbush with a single ground edge. We also note that this class of games in called numbers, because if we allow for long games, then the field of real numbers injects into the set of number games, which form a field using the disjunctive sum as addition and a multiplication operation we do not define here. For more on this see [13].The set of short numbers is equal to the image of the dyadic rationals under this map. Given a number $X$, we say we often write $X=x$, if the image of this map sends the real number $x$ to the value of $X$, and we use $x$ to denote the canonical form of $X$. For example, $1\cong\{0\mathrel{|}\}$ and $\frac{1}{2}\cong\{0\mathrel{|}1\}$. We note that integers, that is games $X=n$ where $n$ is an integer, are the unique games in which one or both players may have no option.

The final definition we must give in this preliminaries section is a game’s birthday. Given a game $X$, we let the birthday of $X$ be the greatest integer $b$, such that there exists a sequence of moves beginning in $X$ of length $b$. If $X$ is a short game, then $b$ is finite, if $X$ is a long game then $b$ is an ordinal. When we say that the canonical form of a game $X$ is the unique simplest game equal to $X$, we mean that the canonical form of $X$, call it $C$, has a birthday strictly less than any other game equal to $X$.

Whereas Digraph placement is first defined in this paper, we show that many placement games from well studied rulesets are instances of Digraph placement. For a formal statement of this see Theorem 3.1. This result is a natural analogue to Theorem 4.4 from [15] which shows that every strong placement game is an instance of a position in a given ruleset played on a simplicial complex.

Let $R$ be a placement ruleset and $X$ a game in $R$. We can record all moves in $X$ as triples $(x,T,p)$ where $T$ is the type of piece to be placed, $x$ is the location of the piece on the game board, and $p$ is the player who can legally make this move if it is their turn. If Left and Right can both place piece $T$ at location $x$ on their turn, then we record two triples: $(x,T,\text{Left})$ and $(x,T,\text{Right})$, respectively. For a placement game $X$, we let $M(X)$ denote the moves of $X$ expressed as triples. For any follower $Y$ of $X$, $M(Y)\subseteq M(X)$ by the definition of a placement game. The options of $Y$ are not necessarily a subset of the options of $X$.

Now we define a class of placement games which we show characterize the literal forms of Digraph placement. A placement ruleset $R$ is a conflict placement ruleset if for all games $X$ in $R$, and all moves $(x,T,p)$ in $X$ there exists a set of moves in $X$, $A_{x,T,p}$, such that $(x,T,p)$ is a legal move in a follower $Y$ of $X$ if and only if $Y$ is obtained from $X$ by a sequence of legal moves $(x_{1},T_{1},p_{1}),\dots,(x_{n},T_{n},p_{n})$ and

If $X$ is a game in a conflict placement ruleset, then we say $X$ is a conflict placement game. In a conflict placement game $X$ and a move $(x,T,p)$ in $X$, we call the set $A_{x,T,p}$ the conflict set of $(x,T,p)$.

Notice that if $X$ is a short conflict placement game, then for all $(x,T,p)\in M(X)$, $(x,T,p)$ is in its own conflict set. Otherwise, the player $p$ can make the move $(x,T,p)$ an arbitrary number of times, which contradicts our assumption that $X$ is short. Hence, if a short conflict placement ruleset $R$ allows for a player to place a piece in the same position as an already placed piece, then the two pieces must be different types, otherwise this would allow an infinite sequence of play. Recall that all games we consider are short.

Observe that the rulesets Nim, Col, Snort, Poset Games, Node Kayles, Arc Kayles, Domineering, and Teetering Towers are all conflict placement rulesets. For example, in Domineering a move $(x,T,p)$ has conflict set $\{(x^{\prime},T^{\prime},p^{\prime}):x\cap x^{\prime}\neq\emptyset\}$, and in Poset Games if we record moves as elements of the poset, a move $x$ has conflict set $\{y:y\leq x\}$. In particular, Digraph placement is a conflict placement ruleset. Given a Digraph placement game $G$, and a vertex $v$ in $G$, the conflict set of the move which deletes $v$, is the set of moves given by either player deleting a vertex in $N^{-}[v]$. We will prove that every conflict placement game has the same literal form as a Digraph placement game.

Familiar placement rulesets Hackenbush and NoGo are not conflict placement rulesets. For each ruleset there exist games where a move $(x,T,p)$ can be illegal after some set of moves $S$, where $\lvert S\rvert\geq 2$, are played but if a subset of the moves in $S$ are played then $x$ remains legal. For the reader who is familiar with the ruleset Hackenbush see Figure 2 for an example of a game in Hackenbush which is not a conflict placement game.

There are immediate consequences of this concerning the computational complexity of deciding a winner in a Digraph placement game.

Despite there being literal forms such as the example in Figure 2 which do not appear as Digraph placement games, in this section we show that Digraph placement is a universal ruleset. See Theorem 4.5 for this result. Our proof is constructive and requires some preliminaries from the literature. Before proving Theorem 4.5 we consider four relevant lemmas, the first two of which are familiar to those familiar with combinatorial game theory.

The first lemma we must prove is that every short number is equal to some Digraph placement game. For completeness we state it formally as follows.

The next step in the proof of the main theorem requires a complicated construction (gadget). Our construction is recursive. Let’s first explore a relatively simple case. Given Digraph placement games $G$ and $H$ we aim to construct a Digraph placement game equal to $\{G\mathrel{|}H\}$. Hence, we take a copy of $G$, a copy $H$, and some additional vertices and edges. To get a Left option to $G$ we add a blue vertex which has an edge to every vertex not in our copy of $G$ and to get an option to $H$ we add a red vertex with an edge to every vertex not in our copy of $H$. To ensure that the moves in either the copy of $G$ or the copy $H$ do not affect the value we include additional red and blue vertices that ensure dominated incentives for these moves. The inclusion of these extra vertices (and some edges) results in a game with value $\{-1,G\mathrel{|}1,H\}$. We will argue in the proof of Theorem 4.5 that the existence of such a Digraph placement game is sufficient to imply the existence of a Digraph placement game equal to $\{G\mathrel{|}H\}$. Our formal construction is given for games with multiple options for each player. See Lemma 4.4 for the value of what we construct.

For fixed but arbitrary positive integers $k$ and $t$, let $G_{1},\dots,G_{k}$ and $H_{1},\dots,H_{t}$ be Digraph placement games. Let $n$ be a positive integer. We define a new Digraph placement game denoted by $n\langle G_{1},\dots,G_{k}\mathrel{|}H_{1},\dots,H_{t}\rangle$; this game is played on the directed graph $G$. The vertex set of $G$, $V(G)$, is the disjoint union of the vertex sets of the digraphs $G_{1},\dots,G_{k}$ and $H_{1},\dots,H_{t}$ and vertex sets $\{b_{1},\dots,b_{k}\},\{r_{1},\dots,r_{t}\}$, and $\{x_{1},\dots,x_{n}\},\{y_{1},\dots,y_{n}\}$. The $2$-coloring of $G$, $\phi$, is as follows;

• •

  if $v\in V(G_{i})$, $\phi(v)=\psi_{i}(v)$ where $\psi_{i}$ is the $2$-coloring of $G_{i}$, and

• •

  if $v\in V(H_{j})$, $\phi(v)=\psi_{k+j}(v)$ where $\psi_{k+j}$ is the $2$-coloring of $H_{j}$, and

• •

  for all $1\leq i\leq k$, $\phi(b_{i})=\text{blue}$, and

• •

  for all $1\leq j\leq t$, $\phi(r_{j})=\text{red}$, and

• •

  for all $1\leq i\leq n$, $\phi(x_{i})=\text{blue}$, and

• •

  for all $1\leq j\leq n$, $\phi(y_{j})=\text{red}$.

We define the edge set of $G$ as follows. Let $u,v\in V(G)$,

1. 1.

   if $u,v\in V(G_{i})$, then $(u,v)\in E(G)$ if and only if $(u,v)\in E(G_{i})$, and

2. 2.

   if $u,v\in V(H_{j})$, then $(u,v)\in E(G)$ if and only if $(u,v)\in E(H_{j})$, and

3. 3.

   if $u\in V(G_{i})$ and $v\in V(G_{j})$ where $i\neq j$, then $(u,v),(v,u)\in E(G)$, and

4. 4.

   if $u\in V(H_{i})$ and $v\in V(H_{j})$ where $i\neq j$, then $(u,v),(v,u)\in E(G)$, and

5. 5.

   if $u\in V(G_{i})$ and $v\in V(H_{j})$, then $(u,v),(v,u)\in E(G)$, and

6. 6.

   if $u,v\in\{b_{1},\dots,b_{k}\}\cup\{r_{1},\dots,r_{t}\}$, then $(u,v),(v,u)\in E(G)$, and

7. 7.

   if $u\in\{b_{1},\dots,b_{k}\}\cup\{r_{1},\dots,r_{t}\}$ and $v\in\{x_{1},\dots,x_{n}\}\cup\{y_{1},\dots,y_{n}\}$, then
   $(u,v),(v,u)\in E(G)$, and

8. 8.

   if $v=b_{i}$ and $u\in V(G_{i})$, then $(u,v)\in E(G)$ and $(v,u)\not\in E(G)$, and

9. 9.

   if $v=r_{j}$ and $u\in V(H_{j})$ , then $(u,v)\in E(G)$ and $(v,u)\not\in E(G)$, and

10. 10.

    if $v=b_{j}$, and $u\in V(G_{i})$ for $i\neq j$, or $u\in V(H_{l})$ for any $l$, then $(u,v),(v,u)\in E(G)$, and

11. 11.

    if $v=r_{i}$, and $u\in V(H_{j})$ for $j\neq i$, or $u\in V(G_{l})$ for any $l$, then $(u,v),(v,u)\in E(G)$, and

12. 12.

    if $u,v\in\{x_{1},\dots,x_{n}\}\cup\{y_{1},\dots,y_{n}\}$, then $(u,v),(v,u)\not\in E(G)$ , and

13. 13.

    if $u\in V(G_{i})$ or $u\in V(H_{j})$ for any $i$ or $j$, and $v=x_{l}$, then $(v,u)\in E(G)$, while $(u,v)\in E(G)$ if and only if $\phi(u)=\text{blue}$.

14. 14.

    if $u\in V(G_{i})$ or $u\in V(H_{j})$ for any $i$ or $j$, and $v=y_{l}$, then $(v,u)\in E(G)$, while $(u,v)\in E(G)$ if and only if $\phi(u)=\text{red}$.

This describes the Digraph placement game $n\langle G_{1},\dots,G_{k}\mathrel{|}H_{1},\dots,H_{t}\rangle$. Given sets of Digraph placement games $L=\{G_{1},\dots,G_{k}\}$ and $R=\{H_{1},\dots,H_{t}\}$, we let $n\langle L\mathrel{|}R\rangle=n\langle G_{1},\dots,G_{k}\mathrel{|}H_{1},\dots,H_{t}\rangle$.

We now consider some examples. Let $0$ denote the empty graph. Let

such that $\phi(u)=\text{blue}$ and $\phi(v)=\text{red}$; notice that $H=\mathord{\ast}$. See Figure 3 for a figure depicting $1\langle 0,H\mathrel{|}0,H\rangle$. In Figure 1, the component of the (left) game equal to $\mathord{\downarrow}=\{\mathord{\ast}\mathrel{|}0\}$ is $1\langle H\mathrel{|}0\rangle$.

We now proceed to Lemma 4.4.