1. Introduction
Normal play combinatorial games are sequential games, usually played under the alternating move convention, and where “the player who moves last wins” [BCG2001, C1976, S2013]. Or wait, should we rather say, “a player who cannot move loses”? Of course, in alternating play, the latter implies the former playing from a nonterminal game. The former is more intuitive from a recreational play perspective, but the latter is the correct one, from a recursive point of view; the neutral element, the 0game, has no move options for either player, and so the current player loses. Moreover, if is the starting position, there is no “last player”. The recursive point of view is fundamental to the theory, in computing prefect play outcomes via backward induction.
We will generalize alternating play Combinatorial Game Theory (CGT) into an infinite class of game families under the normal play convention, by combining the standard game forms with a certain bidding convention, the socalled Richman auction [LLPU1996]. As usual, the players are called Left (female) and Right (male), and games are finite, i.e. there is no cycle or infinite play sequence. If a game is played by the discrete bidding convention [DP2010, LPR2021], then there is a total budget partitioned between the players as , with , and where Left’s part of the budget is . The player who bids more moves next, and if bids are equal then a tiebreaker determines the current player (as in standard CGT we identify ‘next’ with ‘current’).
At each stage of play, exactly one of the players holds a tiebreaker, a tiebreaking marker, and this is denoted by (Left holds the marker in the game ) or (if Right holds the marker). If a player wins a bid strictly, then this player hands over the bidding amount to the other player, and plays their move. If bids are equal then the player with the marker wins the bid, they play their move, and the marker together with the bid shifts over to the other player.
The marker owner can also choose to explicitly announce the marker with the bid, in which case the marker together with a winning bid shifts owner, even if the winning bid is strict. This additional rule will be important in restoring some familiar normalplay properties, in particular a move from should be favourable to the currently stronger player (first player in alternating play). Without this rule, since no player wants the tiebreaker if they have no move, the marker holder would benefit from a tie, and the other player would benefit from a strict win or loss bidding. This situation is a normal form (matrix) game with no pure equilibrium, and removes any familiar tool from normal play theory. Examples 1 and 2 in Section 3 elaborate on this theme.
2. The basic set up
A game is a triple , where we take a note of Left’s part of the budget, . If is understood, we write . If we do not wish to specify , we may write or just . Games are recursively defined, with , where or if is terminal, i.e. the current player, Left or Right, cannot move. In case , then is terminal irrespective of move order. Games are finite and contain no cycles, i.e. each game has finitely many options, and the birthday (rank of game tree) is finite, implying that each play sequence is finite. The game is a subgame (or follower) of a game if there exists a path of moves, perhaps empty, (in any order of play) from to .
If the current player cannot move, they lose. Hence nobody wants to win the Richman auction from a terminal position.
Suppose, without loss of generality, that Left has the marker. There are four cases to establish the next game (in case there is one):

; A Left move after bidding more, i.e. .

; A Left move, after including the marker with .

; A Left move after winning a tie, i.e. .

; A Right move after bidding more, i.e. .
Observe that in case of a tie, the marker is transferred. This automatic rule is to the core of our generalization of alternating play. Namely corresponds to alternating normal play rules.
In the classical continuous Richman auctions [LLPU1996], the tiebreaker has infinitesimal value with respect to any bid. We adapt a discrete bidding convention, which is more natural from a recreational play point of view, but where the analysis of the tiebreaker becomes nontrivial. In fact, the normal play winning implication in the first paragraph of the paper, does not apriori hold, and moreover ambiguity on winning strategy may appear so traditional requirements for combinatorial games such as pure strategy subgame perfect equilibria may break. With some care of picking a sound tiebreaking convention this issue will be remedied, so that standard backward induction techniques (Section 4) will produce optimal outcomes that generalize normal play outcomes with new lattice structures (Section 8).
3. Games in pure subgame perfect equilibria
Let us review why the tiebreaking rule used in [DP2010] would require mixed strategies in equilibrium in this study. A basic observation is that (independently of specifics of rule) the game is losing for the player who holds the tiebreaking marker. Namely, the player without the marker will bid 0, in a successful attempt to lose that bid. And indeed, this resembles alternating normal play, where the set of 0games is the unique equivalence class of zugzwang positions, where no player has incentive to move.
Let us dwell a bit more on the choice of tiebreaker. Two motivating examples will guide us further up the road.
Example 1.
Consider and the game under the Richman bidding convention where the tiebreaker shifts player when used to resolve a tie, but not otherwise. Suppose player Left has budget $1 together with the marker. Both players have the terminal game as the option, and so both players prefer not to own the marker at the next position. This means that Left wants a tie, but Right wants either player to win the bid strictly. There are two possibilities for Left to reach her goal, and there are two ways for Right. Namely Left prefers the bidding pair or , whereas Right wants or . See Table 1. This situation violates the normal play assertion: “last move wins” in case Left wins the bid by the bidding pair . She gets the last move but loses the game. Moreover, this bidding rule gives too much emphasis on the marker; both players’ goal of the game would be to not hold the marker when the game ends, rather than focusing on “who moves last”. But the marker was introduced merely as a device to resolve ties. Hence, in this paper we will adapt the bidding convention in Example 2, where Left can assure a win of the game by going all in at the second last bid; last move wins.
Bids  

L  R  
R  L 
Example 2.
We alter the bidding convention in Example 1, so that the marker owner, here player Left, may include the marker in her bid, and hand it over in case of winning the bid strictly. In case of a tie, of course the marker swap is still mandatory. With total budget , Left has four bidding alternatives as displayed in Table 2. The final winner is displayed in each case, and depending on Right bidding $0 or $1. Note that row 4 dominates the other rows, so Left will bid . She gets the last move and wins the game.
[.5] Bids  

[.5]  L  R 
[.5]  R  L 
[.5]  L  R 
[.5]  L  L 
In the classical continuous version Richman auction, because bids are real numbers, the tiebreaker has infinitesimal value, and so the theory is not sensitive to variations on tiebreaker transfer.
For convenience, let us prove that bidding games generalize alternating play.
Theorem 3.
Consider . Then bidding play is identical to alternating play. The current player is the player who holds the marker.
Proof.
By the rules of bidding play, the marker shifts owner if the players tie a bid. Since , every bid is a tie. Therefore the marker alternates between the players, and by definition a tie is resolved by assigning the current player to the marker owner. ∎
4. The first fundamental theorem of bidding play
Let us mention a couple of possibilities for discrete bidding and its tiebreaking marker.

a player who wins the bid decides who is the current player;

a player who wins the bid is the current player;

the marker alternates between the players;

the marker owner may include it in the bid;

the marker stays with one of the players;

the marker shifts owner, if and only if it has been used to resolve equal bids.

if the marker owner wins the bid, the marker gets transferred.
In [DP2010], they study the variation (a). In [LPR2021] we study (b) together with (iv).
Let us begin by proving that, for every game , our variation using (b) and (ii) has pure strategy subgame perfect bidding equilibria. From now on, we will study the variation using (b) and (ii).
Theorem 4 (First Fundamental Theorem).
Consider the bidding convention where the tiebreaking marker may be included in a bid. For any game there is a pure strategy subgame perfect equilibrium, computed by standard backward induction.
Proof.
Suppose without loss of generality that Left holds the marker. A terminal game has a pure strategy equilibrium; a player that holds the marker but does not have any option loses when the other player bids . This holds also for the terminal games or , when Right and Left does not have any option, respectively.
Suppose that is nonterminal. By induction, we may assume that each option is in a pure strategy subgame perfect equilibrium.
Suppose that the winner of the bid loses the game. Then the other player has no incentive to change their bid.
Therefore, we may assume that one of the players has a winning bid, which the other player cannot or does not want to outbid, together with a winning option (which by assumption is in a PSPE). Without loss of generality, say Left wins , by bidding and Right does not win the game if he outbids Left or ties in case of owning the marker.
Case, Left holds the marker: Suppose that Right does not tie Left’s bid, i.e. Right bids . By assumption, he does not gain by deviating by outbidding Left. And, by deviating by lowering his bid, Left remains the winner of the auction. If Right ties Left’s bid, i.e. , then Left wins the bid by marker ownership. Right can deviate by lowering his bid. But this would not change the outcome, since, if it would, Left would have included the marker in a strict winning bid. (By assumption he does not gain by a potential increase of his bid.)
Case, Left does not hold the marker: Then the winning is strict, and by assumption Right does not gain by outbidding Left. And, lowering the bid does not change the winner of the bid. ∎
By this result, henceforth we will refer to perfect play with the meaning perfect pure bids and a perfect move by the player who wins the bid.
Bids are typically simultaneous, but in view of perfect play and Theorem 4, the order of bidding is irrelevant.
Consider a given stage of play. A player who has the larger part of the budget, or half the budget together with the marker, is called the (currently) dominating player.
Lemma 5 (Last Move Wins).
If the dominating player has an option from which the other player cannot move, then they win the game.
Proof.
Suppose first that the dominating player, say Left, does not hold the marker. Then she goes all in, and moves such that Right cannot move. Since Right still holds the marker, he will lose the game. Namely, at her final bid, Left bids 0, and hence Right wins this final bid.
Suppose next that the dominating player, say Left, holds the marker. She goes all in, and includes the marker in her bid. She wins the bid and moves such that Right cannot move. Again, the game ends as in the previous paragraph. ∎
By this lemma, we may use the term last move wins with the same meaning as in alternating normal play. Henceforth, we adapt the following wording convention: alternating play means the classical normal play games [BCG2001], and bidding games is our generalization of those games. All our games will be normal play; we vary the move order convention, but not the winning condition.
5. A motivating result and a generalization of the impartial game tree
Standard outcome classes in alternating combinatorial game theory are
Left wins, the Next player wins, the Previous player wins, and Right wins, respectively.
Let be an impartial game. That is, for all followers of , . For impartial games, in alternating play, only the outcomes and applies.
The following result, which was discovered at the “Chocolate Café” between sessions at the CGTC3 in Lisbon 2019, motivates this work; it encouraged us to take a closer look at the proposed class of bidding games.
Theorem 6.
Consider and an impartial game form . If , then Left wins bidding if and only if alternating normal play is an position. If , then Left wins , unless , in which case the player who holds the marker loses.
Example 7.
Figure 1 displays seven game trees, of which the naming of the first four should be familiar to the reader. The two game trees on the bottom row, and , have nonstandard names due to their special interest to this study (note ). Let us review some behavior of each of these seven games with respect to differences and similarities between alternating play and bidding play. In the player without the marker wins. In the dominating player wins. The game is a Left win (independently of who owns the marker). Right can win the game , if he holds a sufficient budget (even with ). How? The index in the game indicates how many consecutive bids a player needs to win, to win the game, if . In alternating play, of course ,^{1}^{1}1See Section 9 for a brief discussion of equivalence classes of games. and does not deserve a special name. The game is discussed just below.
Consider the following generalization of impartial games, to symmetric ending games. If one of the players has a terminal move, then so has the other player. And if one of the players cannot move then neither can the other player. That is, is a symmetric ending game if for all followers , if and only if , and if and only if .
The family of symmetric ending games is larger than the impartial games in that it admits the alternating play outcomes and . For example, the literal form has outcome . Note that the alternating play equivalent form does not have a symmetric ending.
Recall that the dicot game forms are defined recursively by: if one of the players has a move, then so does the other, and each child is a dicot. The family of dicots contributes a major step from impartial games towards the full family of partizan games. In alternating play, the symmetric ending games are equivalent to the dicots. To see this, note that each Left move to , where Right does not have a move to (and vice versa) may be replaced with a move to the game (this does not hold any longer in generic bidding play).
The point here is that the symmetric ending condition permits an similar induction proof as that for impartial bidding games (Theorem 6). In particular, note that the only terminal position is .
A player is dominating if they have the larger budget, or in case of holding the marker, the weakly larger budget. A player is strictly dominating if they have a strictly larger budget than their opponent.
Theorem 8.
Consider a symmetric ending game , and let . If is terminal, then the marker owner loses. Otherwise the dominating player wins, unless is even and the players have an equal share of the budget. In this case Left wins bidding if and only if normal play is an or position (and vice versa if Right holds the marker).
Proof.
In each case, we must present a strategy for the proposed winner. If is terminal, the nonmarker holder passes and loses the move, but wins the game.
Next, consider a nonterminal , with . The strictly dominating player, here Left, bids , until there is a move to a terminal position. By the symmetric ending condition, if one of the players can terminate the game, so can the other. Left goes all in and wins the last move, independently of whether she holds the marker.
For the remainder of the proof, we consider an even total budget and a nonterminal game . Suppose first that Left has a terminal move. Then is an position in alternating normal play, since the symmetric ending condition ensures that if the position is (or ), then it does not have a move to a terminal position. Therefore, in the bidding variation, the marker holder, here Left, bids and wins the last move.
Suppose next that is a nonterminal  or position in alternating play. We must show that Left will lose the bidding variation. Right, who does not have the marker, bids . If Left bids , then Left wins the move and, by definition of  and positions, has to move to an  or position, and Right gets the marker. That is losing for Left, by induction. If Left bids , Right becomes the dominating player, and wins (by the second paragraph in the proof), because is nonterminal. Indeed, if were terminal, then alternating play .
Suppose next that is an  or position in alternating play, for which there is no move to a terminal position. We must prove that Left, who holds the marker, wins. Left bids . If Right passes, then he gets the marker, and Left can move to an alternating play  or position, and win by induction. If Right bids , then he will win the bid but Left becomes a dominating player, and so wins by using the second paragraph of the proof, since by assumption no option of is terminal. ∎
6. Outcomes in bidding games
The alternating play outcomes generalize in a word notation, where . For example “” means that Left wins when Left starts and Right wins when Right starts. Similarly for bidding play, the winner in pure subgame perfect equilibrium, given a current player (the winner of the current bid), is either L (Left) or R (Right), with the usual total order . This induces a partial order of outcomes that generalizes alternating play.
For example, with , one might envision 16 partially ordered candidate outcome classes, in word notation, with the largest (smallest) outcome LLLL (RRRR), and so on. Here the leftmost letter symbolizes the outcome in perfect play when Left has budget 1 and the marker; the second letter corresponds to the result in perfect play when Left has the marker but no budget, and symmetrically for the last two letters. An example of an incomparable pair of outcomes would be LRLR and RLRL.
Our notion of outcome is a tuple of pure subgame perfect equilibria, representing, given any budget partition, who is the winner in perfect play.
In alternating play, each zugzwang (a game where no player has an incentive to move) belongs to the outcome class . Here, we define a zugzwang as: no player wants to win the bid. Therefore, it is not the monetary strength, but the marker ownership, that defines the zugzwang concept; obviously both players will bid , whenever they do not want to move, and the marker will decide who is the current player (i.e. in perfect play, the losing player). Therefore, with , the only zugzwang is the outcome RRLL. An outcome such as RLRL would be rare, if it appears at all, since Right wins without either money or marker, but loses if he is given a dollar Next, we prove that such outcomes are impossible; outcomes are monotone.
The pure subgame perfect equilibrium of a game is denoted . Sometimes we refer to this as a ‘partial outcome’. The outcome of the game is , defined via the tuple of partial outcomes as
Since this notation can be quite lengthy, we instead adapt word notation. For example, instead of we simply write RRLL.
Theorem 9 (Outcome Monotonicity).
Consider a fixed , with . Then .
Proof.
Let us first state the partial outcomes on options of , derived by the assumption, , where we first assume that , i.e. Right holds the marker. Left has a bid that wins the game.
Case 1: Suppose that Left wins by bidding . Then, she knows that:
A) If , there is a Left option, , such that ;
B) For all Right options, , , if ;
C) For all Right options, , if . This case is a tie, or Right includes the marker in the bid;
We must present a Left strategy that beats every Right strategy in the game . Suppose that Left bids in the game . Then, by induction on the birthday of the game tree:
A1) If , , where is the same as in A above;
B1) For all , for all , ;
C1) For all , for all , (in case of a tie or Right included the marker in the bid).
Note that in case there is no , then B, B1, C and C1 are vacuously true. This proves that , in case .
Case 2: Suppose that Left wins by bidding , . Then:
A) If , there is a Left option, , such that . That is, Right can tie this bid, in which case Left loses the marker;
B) If , there is a Left option, , such that or such that . Left can choose to include the marker or not to include the marker;
C) If , then for all Right options, , and all , , if ;
In the game , Left bids the same, and plays the same in case of winning a bid. Right now has the smaller budget . The results follow by induction on the birthday of the game tree, in details:
A1) If , then if Right ties the bid, Left may play as in A above to get ,
B1) If , then Left wins the bid, and with is as in B above, or , where the latter case includes the marker in the bid;
C1) If , then for all , .
This proves that , which together with Case 1, conclude the proof. ∎
If these monotone properties hold, the outcome is monotone.
A player is never forced to become the current player via monetary advantage, since the 0bid is always available. If one of the players did not have the 0bid available, the monotonicity may break. Here we study games where all bids are available to Left, and analogously for Right.
Observation 10.
There are 9 monotone outcome classes for . And in general, there are monotone outcomes for .
For example, with , the monotone outcomes are LLLL, LLLR, LLRR, LRLL, LRLR, LRRR, RRLL, RRLR and RRRR.
L R  LR LR  LLR LRR  LLRR LLRR  
L L  LL LL  LLL LLL  LLLL LLLL  
L L  LL LR  LLL LLR  LLLR LLLR  
R L  LR LR  LRR LLR  LLRR LLRR  
R L  LR LL  LRR LLL  LLRR LLLL  
L L  LL LL  LLL LLL  LLLL LLLL  
L R  LR LR  LLR LRR  LLRR LLRR 
Do all monotone outcomes appear? Can we find a game form for each and each monotone outcome? For all outcomes are trivially monotone, and indeed, all appear by day 1, namely , , and . For all monotone outcomes, except LLRR, appear for games born by day 2. Let us look at this outcome where a player loses with a dollar budget, but wins with the marker alone. Can the marker be worth more than a dollar? The answer is no.
Theorem 11 (Marker Worth).
Consider . Then .
Proof.
Assume that Left wins by bidding (otherwise we are done). We claim that Left wins by bidding , unless , in which case she bids . To prove this, we observe the cases for Right’s bidding amount.



.
In case of A, Right might bid or . The case of Right bidding will be treated by using instead B.
In case A, Right bids and plays to , if there exists a Right option (o.w. we are done). But, by assumption, Left wins , if , so by induction she wins .
In case B, Right wins the bid and plays to , if exists (o.w. done). By monotonicity, Left wins this game. Namely, since , then , and by assumption, Left wins .
In case C, Left wins the bid and plays to some . There is a Left option , since by assumption Left wins playing first in by bidding ; even in the case of she might win the bid (if Right bids 0), so she has a defence, by playing some . Indeed, if Right can tie , then by assumption, she wins by moving to some . If Right cannot tie, because , then bidding or wins the game by assumption. In the first case, this results in some game , which Left wins. Left bids in the game and moves to , which she wins by induction (here we used that implies ). In the second case, Left wins the game by assumption, and so by monotonicity, she wins .
Altogether, Left wins given that Left wins . ∎
Let us count the number of feasible outcomes for a given .
Let us denote the triangular numbers by .
Proposition 12.
For a given total budget , there are feasible outcomes.
Proof.
Simple counting to do. ∎
A challenge will be to demonstrate that all feasible outcomes appear for some game form. We resolve this in Theorem 17.
7. Do all feasible outcomes appear?
Now we see that, for , all feasible outcomes appear by day 2, namely take the rows 1,2,3,4,6 in Table 3 and include the conjugates of rows 3, 4 and 6. What about the outcomes for ? There are 13 feasible outcome classes. But, will games born by day 2 suffice? For example, what about the feasible outcome LLRLLL? The nearest in the table is LRRLLL. It turns out that a solution is a game of day 3; namely has the desired outcome.
Example 13.
Consider total budget . The outcome of is . Right wins if and only if he has budget and no marker. Clearly, if Right has the marker he loses. If he does not have the marker, he will pass, and so Left will play to and hand over the marker. Now, he wins if and only if he wins two consecutive bids, which is possible if and only if he has budget .
Another feasible outcome that does not appear in Table 3 is LRRLRR. A game of day 3 has this outcome.
Example 14.
We have that . That is, Left wins if and only if she has budget , with or without the marker. With budget , she starts by bidding , and wins the marker. Right has to move to down. Now Left has , and wins the next two bids to win the game. If Left starts with budget , then she wins two bids, moving to and then . Right wins any other game. For example with budget , he begins by bidding 0 and gets either the marker or another dollar. Thus he wins after Left’s move to . If he starts with , then he begins by bidding 0. If Left bids 0, he plays to . Left cannot afford to lose the first move, and so she bids . Right gets all budget and wins from . The other cases follow by monotonicity.
Together with the games in Table 3 and conjugate forms, we have now found games for all outcomes with . Below, in Table 4, we will find game forms for all budget partitions with , and in Figure 2 we display two game trees from this table. Greater total budgets offer more challenges, but in Theorem 17 we will provide a general construction.
To this pursuit, we introduce a short form, , for a feasible outcome. Here and is the smallest Left budget for which she wins, when Left has the marker and does not have the marker respectively. Hence, by Theorem 11,
(1) 
An outcomeconjugate is obtained when the Ls and Rs are reversed and swapped symmetrically. Hence, a conjugate short form of is . Note that outcomeconjugation preserves feasibility.
Three of the outcomes in Table 4 are their own conjugates. Hence, we list games, and note .
Outcome  Shortform  Game 

1  
0 
Lemma 15.
Consider a total budget and two Left budgets and such that , and where . Suppose that Left holds the marker in both sequences or neither of them. Then Right can win at least one more consecutive bid in a bidding sequence that starts from rather than .
Proof.
Fix a budget , and consider Left budgets and respectively.
Suppose first that Left holds the marker in both games. By , Right can over bid Left in the first sequence. If , then Right cannot over bid Left in the second setting. Therefore we assume that , so Right can over bid both Left’s budget in both settings.
The first Left budget sequence starts as
And the second Left budget sequence starts as
Thus the Left budgets in the first sequence will mimic the second sequence’s entries, delayed one step.
Suppose next that Right holds the marker in both games. By Marker Worth Lemma, we may assume that he uses the marker in the first bid, and henceforth over bids strictly, in both sequences. Therefore the first Left budget sequence starts as
And the second Left budget sequence starts as
Thus the Left budgets in the first sequence will be smaller than the second sequence’s entries, delayed one step.
∎
Lemma 16.
Fix the marker owner and consider a pair of bidding sequences that starts at the Left budgets and , respectively, where say . In each sequence, at each bidding stage, the player with the larger budget overbids the smaller budget. The sequence pair terminates, at stage , if the dominating player shifts in exactly one of the sequences at the entries. The sequence pair terminates.
Proof.
It suffices to prove that the condition in Lemma 15 will be satisfied at some stage .
Suppose first that Left holds the marker. The first sequence will flip dominating player at some point, say after bids. The corresponding Left budget is . If the second sequence shifted dominating player before this stage, we are done. Hence suppose that the second sequence flipped dominating player at the same stage , with Left budget . The difference between this pair of Left budgets is . At this point Right has become the dominating player in both sequences, with difference between the pair of Right budgets say , since . Thus, the absolute difference is strictly increasing also at each stage when the dominating player is shifting. Therefore the condition in Lemma 15 will eventually be satisfied.
In case Left does not hold the marker, Right can tie Left’s budget at the first stage, and then over bid. The rest of the proof is in analogy with the previous paragraph.
∎
Let us call a bid a pass, if the player without the marker bids 0.
Theorem 17 (Main Theorem).
Consider any total budget . An outcome, say , is feasible if and only if there is a game such that .
Proof.
For the other direction, let us view the short form of a feasible outcome, . For any total budget and any , with , we will build a game such that . Recall, in case of Left holding the marker, player Right wins the game if Left has budget , and Left wins with budget . That is if and only if . Similarly, if and only if (i.e. in case of Right holding the marker, player Right wins if Left has budget , and Left wins with budget ).
By monotonicity, it suffices to study two Left starting budgets,
(2) 
Our construction will satisfy: Left wins but loses . The proof when Right holds the marker is analogous; construction for Right options will be symmetric.
Namely, the construction will rely on play beginning with a pass round. Lets assume that Left holds the marker by the start of play. If Right passes, Left cannot benefit by a monetary overbidding. Either way, she must start playing, and by Theorem 11, the marker is never worth more than a dollar.
We begin by verifying that, under the assumption that our construction will hold, the passing player cannot benefit by instead overbidding the opponent. Hence, suppose that Right over bids Left’s bid. We have to prove that Left wins , assuming that she wins . To this purpose, observe that , since . The Right option will be constructed to satisfy Left passing in case Right holds the marker, and where Left wins if and only if her original budget is greater than . And indeed, in the symmetric case, when Right moves in to , Left wins because her budget is greater than .
The intuition of the construction is as follows: We may assume that a dominating player overbids the other player whenever they face a threat of an imminent loss. We will design terminal threats, i.e. options to the game , for those instances, where a currently nondominating player must win the game. Such threats will eventually flip the dominating player, and at this point we wish to let the player who must win terminate the game. However, it may happen that the dominating player flips simultaneously in both game sequences, say those initiated by the respective Left starting budgets and . In this case, we cannot yet let the currently dominating player, say Left, have a terminating option, but we will rely on Lemma 15. Namely the game will proceed, with Left, who is now dominating, facing terminal threats. She must overbid Right, until at least one of the bidding sequences flips the dominating player. If exactly one sequence flips, then we terminate the game, and otherwise, repeat. Indeed Lemma 15 assures that this process will finish in a finite number of steps. See Figure 3 for an example of such a constructed game tree.
Let us now detail the construction, given the tipping point parameter , when Left holds the marker, from some short form . Let . We will describe a pair of sequences and of monetary Leftbudgets, starting with , , as in (2), while we define a sequence of followers, , of , depending on the tipping point alone.
Case 1: The tipping point . Left starts as the dominating player in precisely one of the games, namely in . We let . Then , since Right cannot benefit by overbidding, and Left cannot force Right to start.
Case 2: The tipping point . Left starts as the dominating player in both game sequences. But we must satisfy . We create a threat . Hence Left must strictly overbid Right’s budget (Right holds the marker). This is the setting of Lemma 16. Hence, we may apply Case 1, replace with for some finite . Here the intermediate game forms , for are given by Case 2 or Case 3 depending on who is the dominating player.
Case 3: The tipping point . Right starts as the dominating player in both sequences. But we must satisfy . We create a threat . Hence Right must weakly overbid Left’s budget (Right holds the marker). This is the setting of Lemma 16. Hence, we may apply Case 1; replace for with some finite . Here the intermediate game forms , for are given by Case 2 or Case 3 depending on who is the dominating player.
∎
8. Outcome lattices
At this point, one might wonder: what exactly are the structures of the feasible outcomes? Obviously for any , we have a partially ordered lattice structure, with join outcome and meet outcome . Let us provide a recursive construction of this family of lattices. Recall the outcome diamond for normal play, i.e. : and , with . This is displayed to the left in Figure 4, by using the short form notation (the smallest Left budget for which she wins, with and without the marker, respectively).
The infinite joinsemilattice (Figure 5) has nodes of the form , with , and where . There is a directed edge from to if and only if and where and (that is the directed edges produce the natural partial order between such pairs of integers). For any , we construct a lattice, , by including the nodes for which , together with the nodes where . And inherits the directed edges from .
Theorem 18.
Far all , represents the partial order of the feasible outcomes in their short form representations. A pair of feasible outcomes and satisfies if and only if there is a directed path in from to .
Proof.
This is direct by construction and the definition of short form feasible outcome. ∎
9. Future directions
This work has produced a complete solution for the structure of the outcome classes of bidding combinatorial games that generalize alternating normal play. The most natural application of these outcomes is to play a disjunctive sum of games. Given a finite number of game components, a current player moves in precisely one of the game components. A sum of two games and is written . The outcome alone does not suffice to understand such compositions of games, but one can define a more comprehensive partial order by letting if, for all games , . In our follow up paper [KLRU2], we find a constructive solution for game comparison given a certain proviso. We refer the reader to that paper for further open problems and future directions.