# Constructive comparison in bidding combinatorial games

A class of discrete Bidding Combinatorial Games that generalize alternating normal play was introduced by Kant, Larsson, Rai, and Upasany (2022). The major questions concerning optimal outcomes were resolved. By generalizing standard game comparison techniques from alternating normal play, we propose an algorithmic play-solution to the problem of game comparison for a class of bidding games that include game forms that are defined numbers. We demonstrate a number of consequences of this result that, in some cases, generalize the classical results in alternating play (from Winning Ways and On Numbers and Games). We state a couple of thrilling conjectures and open problems for readers to dive into this promising path of bidding combinatorial games.

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## 1. Introduction

In a recent paper [1], we introduce Discrete Bidding Combinatorial Games that generalize alternating normal play. All essential structure problems regarding outcomes were resolved. In this paper we study the same families of games, by generalizing standard game comparison techniques from alternating normal play; we propose a constructive i.e. algorithmic solution to the problem of game comparison.

For further references and background, see [1]; let us review the basic definitions and main results of that paper.

A game is a triple , where we take a note of Left’s part of the budget, . Here indicates who holds the tie-breaking marker. If is understood, we write . The game form is . 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.111The formal birthday is the rank of the literal form game tree. Since we do not yet study reduction techniques of games, we will not distinguish between birthday and formal birthday. The game is a 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 a player does not want to win an auction at 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). An instance of bidding is an ordered pair,

, where Left bids and Right bids .

• ; Left outbids Right, i.e. .

• ; Left outbids Right, while including the marker, .

• ; Left wins a tie, i.e. .

• ; Right outbids Left, i.e. .

Note that the third item is included in the second; but from a recreational play point of view, those bids are distinguished.222If Left does not want to hold the marker in the next round, then she should of course include it to the current bid; if she prefers to keep the marker in case she wins the bid, she would act differently.. Observe that in case of a tie, the marker shifts owner. This automatic rule is to the core of our generalization of alternating play. Namely corresponds to alternating normal play rules.

###### Theorem 1 (First Fundamental Theorem, [1]).

Consider the bidding convention where the tiebreaker may be included in a bid. For any game there is a pure strategy subgame perfect equilibrium, computed by standard backward induction.

An important consequence of our auction definition (above item (ii)) is that the last move wins is equivalent with a player who cannot move loses. A player is dominating if they have strictly larger budget, or in case of holding the marker weakly larger budget.

###### Corollary 2 (Last Move Wins, [1]).

If the dominating player has an option from which the other player cannot move, then they win the game.

The pure subgame perfect equilibrium of a game is the partial outcome , where by convention the total order of the results is , i.e. ‘Left wins’ ‘Right wins’. The outcome of the game is , defined via the tuple of partial outcomes as

 o(G)=(o(G,\savestack\tmpbox\stretchto\scaleto\scalerel∗[\widthofTB]⋀0.5ex\stackon[1pt]TB\tmpbox),…,o(G,\savestack\tmpbox\stretchto\scaleto\scalerel∗[\widthof0]⋀0.5ex\stackon[1pt]0\tmpbox),o(G,TB),…,o(G,0)).

Note that the outcomes inherit a partial order from the total order of the partial outcomes.

The next two results have good use in this paper.

###### Theorem 3 (Monotonicity, [1]).

Consider a fixed , with . Then .

###### Theorem 4 (Marker Worth, [1]).

Consider . Then .

The main result of paper [1] will not be used here, but it is interesting in its own right. An outcome is feasible if it satisfies Monotonicity and Marker Worth.

###### Theorem 5 ([1]).

Consider any total budget . An outcome, say , is feasible if and only if there is a game such that .

## 2. Sums and partial orders of bidding combinatorial games

Let us recall the definitions of disjunctive sum of game forms and partial order of games. The disjunctive sum of the game forms and is defined recursively as:

 G+H={G+HL,GL+H∣G+HR,GR+H},

Where , in case , and otherwise the set is not defined and omitted. The conjugate of a game form is the game form where players have swapped positions, and is recursively defined as . A follower of a game is any subgame of that can occur in a play sequence (possibly empty) from .

The partial order of games is defined as usual. Consider games . Then if, for all games , . Game equality satisfies if and . The games ( is greater than ) if but . The games and are confused if and ; this is denoted .

A game form is a number if for all and for all , , and all options are numbers. We will later prove that all numbers are invertible, and the inverse is the conjugate form. And we will prove that numbers are closed under addition, so they are a subgroup of all bidding games. Since we observe infinitely many invertible elements, we also have infinitely many s.

The empty game satisfies: for all , . Namely, independent of play sequence, the final auction in a follower of will appear at the game (an idempotent). We already introduced the name . We argue that this is the natural name: if a game , then we may omit it in a sum of games. that is, for all games , if , then . In fact, the proof is immediate by definition (and does not depend on the specifics of the auction).

###### Theorem 6.

Consider any total budget. If , then, for any , .

###### Proof.

We must demonstrate that for all , . Since , we have that for all , . Take . ∎

## 3. A main theorem for numbers

Our first aim is a main theorem, which states that, under certain conditions, Left weakly prefers a game form before the neutral element if and only if she wins playing without the marker. We begin by proving it for the special case of games that are numbers.

###### Theorem 7 (Main Theorem, Case Study Numbers).

Consider any total budget and a number game . Then if and only if .

###### Proof.

By definition, if and only if and , if and only if and .

: Take , and . Then since , so implies .
:

1. Let’s assume that and then to prove .

1. Suppose that Left’s optimal starting bid in the game is . Since we have , so for any Right’s choice in the game we will have following observations:

1. If then such that .

2. If then such that .

3. If then we will have .

Now in the game she will bid and for Right’s choice we will have following cases:

1. If then Left will win the bid and will play in the game to the same as in the observation (i) and game will reach to . In this position by using observation (i) and applying induction we will get .

2. If then Left will win the bid and will play in the game to the same as in the observation (ii) and game will reach to . In this position by using case(ii) and applying induction we will get

3. If then Right will win the bid and game will reach either for any or for any (If then Right will play to and if then Right will play to and if as well as then obviously either this case will not arise or Left is the winner). Here for by using observation (iii) and applying induction we will get . For , since G is a number so, and thus . Hence we will have using induction and budget monotonicity.

2. Suppose that Left’s optimal starting bid in the game is . Since we have , so for Right’s choice in the game we will have following observations:

1. If then such that .

2. If then we will have .

Now in the game she will bid and for Right’s choice we will have following cases:

1. If then Left will win the bid and will play in the game to the same as in observation (i) and game will reach to . In this position by using case(ii) and applying induction we will get

2. If then this will arise the same scenario as in subsubcase (iii) of previous subcase, where the outcome is Left winning.

2. Next let’s assume that and then to prove .
In this case suppose that Left’s optimal starting bid in the game is . Since we have , so for Right’s choice in the game we will have following observations:

1. If then such that .

2. If then we will have .

3. If then we will have and .

Next since , so let’s consider the game . In this game it is possible for Right to bid , then for any we will have . Now for any choice of Left’s bid, it is again possible for Right to bid , then Left will win the bid and there will exist a Left’s move such that . Let’s call this as observation .

Now in the game she will bid and for Right’s choice we will have following subcases:

1. If then Left will win the bid and will play in the game to the same as in observation (i) and game will reach to . In this position by using observation (i) and applying induction we will get .

2. If then Right will win the bid and game will reach either for any or for any (If then Right will play to and if then Right will play to and if as well as then obviously either this case will not arise or Left is the winner). Here for by using case (i) and applying induction with observation (ii) we will get . For , we will do the analysis for one more step in which she will bid . Then for Right’s choice following cases will arise:

1. If then Left will win the bid and will play in the game to the same as in observation (i) and game will reach to . Since is a number so, . Since from previous subcase , so using budget monotonicity and hence .

2. If then Left will win the bid and will move in the game (existence of Left’s move in any is guaranteed by observation (iv)). So, the game will go to . Now using observation (iv) and induction we will have .

3. If then Right will win the bid and game will reach either or . For since is a number so, . Now by applying induction using case (i), observation (iii), and budget monoticity, we will get . Now since is a number, so is also a number. So, we will also have . Now for repeating the same analysis of this subcase, either we will get the outcome or we will have more budget for Left, which eventually will fall in different scenario where the outcome is .

3. If then Right will win the bid and game will reach either or for any or, or for any . From previous subcase we have and . Next for by using observation (iii) and applying induction we will get . For since is a number, so . Thus . Hence we will have using induction and budget monotonicity.

Thus if and then we will have . Hence if and only if . ∎

## 4. A constructive main theorem

In alternating play, it is well known and easy to prove that if Left wins both and playing second, then she wins playing second. If she wins playing second and playing first, then she wins playing first. The following lemma generalizes these results, to a particular bidding setting.

Consider a game . Left plays a 0-bid strategy if she bids at each follower of . Left has an optimal 0-bid strategy, if the 0-bid is optimal at each follower of .

Consider any total budget . Suppose, for fixed marker owner , . If Left has an optimal 0-bid strategy in and , then . Suppose . If Left has an optimal 0-bid strategy in and , then .

###### Proof.

We must prove,

1. if , then ;

2. if , then ;

3. if , then ;

where, in each case, we assume that Left has an optimal 0-bid strategy in , and . The proof is by induction.

Case (i): Let be an optimal Left bid in . If , then there exists a Left option in (since Right can pass). If then Right will win the bid (and a Left option in might not exist).

Suppose that Left bids in . If Left wins the bid, she plays to , and wins by induction.

If Right ties Left’s -bid, then he plays to

 (GR+H,\savestack\tmpbox\stretchto\scaleto\scalerel∗[\widthofp+ℓ+q]⋀0.5ex\stackon[1pt]p+ℓ+q\tmpbox)or(G+HR,\savestack\tmpbox\stretchto\scaleto\scalerel∗[\widthofp+q+ℓ]⋀0.5ex\stackon[1pt]p+q+ℓ\tmpbox)

(note that in this case ). But, by monotonicity, since Left has an optimal 0-bid in , then . And since is an optimal Left bid in , we get . Hence by induction, using (ii) and (iii),

 o(GR+H,\savestack\tmpbox\stretchto\scaleto\scalerel∗[\widthofp+ℓ+q]⋀0.5ex\stackon[1pt]p+ℓ+q\tmpbox)=o(G+HR,\savestack\tmpbox\stretchto\scaleto\scalerel∗[\widthofp+q+ℓ]⋀0.5ex\stackon[1pt]p+q+ℓ\tmpbox)=L.

If Right outbids Left by not using the marker, then he plays to or (note that in this case ). But, by monotonicity since Left has an optimal 0-bid in , . Moreover, ; namely since is Left’s optimal bid in , she has a defence strategy if Right outbids her. Hence by induction, using (i),

 o(GR+H,p+ℓ+1+q)=o(G+HR,p+q+ℓ+1)=L.

Altogether .

Case (ii). Let be an optimal Left bid in . A winning Left option exists, because she holds the marker.

She bids the same in the game . If she wins the bid and she plays to and otherwise she plays to . In the first case, she wins by induction, since . In the second case, she wins by induction, since .

If Right outbids Left, by monotonicity , he plays to or . But, since Left has an optimal 0-bid in by monotonicity, . And moreover, Left has a defence to a Right move in , so . Hence, by induction,

 o(GR+H,\savestack\tmpbox\stretchto\scaleto\scalerel∗[\widthofp+ℓ+1+q]⋀0.5ex\stackon[1pt]p+ℓ+1+q\tmpbox)=o(G+HR,\savestack\tmpbox\stretchto\scaleto\scalerel∗[\widthofp+q+ℓ+1]⋀0.5ex\stackon[1pt]p+q+ℓ+1\tmpbox)=L

Altogether .

Case (iii). Let be Left’s optimal bid in . She bids the same in and if she wins the bid, she plays to and wins by induction. Note that exists if , and otherwise she cannot win the bid in , because Right owns the marker. But in she might win the bid if and Right passes. In this case, she has a winning move in to , and so, by using (i), by induction .

Suppose Right wins the bid. Since Left owns the marker, we assume by monotonicity that he bids , and plays to

Observe that . Therefore induction together with monotonicity gives . And by induction, since by assumption .

Altogether . ∎

One of the most celebrated results in alternating play theory states that Left wins playing second if and only if . We generalize this constructive-, algorithmic-, recursive-, play-, local-comparison to bidding play.333Constructive comparison appears under many names in the literature. For a game to apply to the test, it is required to satisfy another recursive test: “Does Left have an optimal -bid strategy?”

The next result is our main theorem (second fundamental theorem) of bidding play.

###### Theorem 9 (Main Theorem).

Consider a game form and any total budget. Suppose that Left has an optimal -bid strategy in . Then if and only if .

###### Proof.

By definition, if for all and for all , implies .

: Assume . Take and . By , the implication gives .

:

Case 1: Suppose and , where Left has an optimal -bid strategy in . We must show that .

Consider a Left optimal bid in , say . Assume that she bids in .

If she wins the bid, by induction she has a move to , which she wins.

If Right ties the bid and plays in the -component, Left wins by induction, using Case 2 below. Hence assume he ties and plays to

 (1) (GR+X,\savestack\tmpbox\stretchto\scaleto\scalerel∗[\widthofp+ℓ]⋀0.5ex\stackon[1pt]p+ℓ\tmpbox).

Now, since by assumption, Left has an optimal 0-bid strategy in , Lemma 8 applies; since , by monotonicity Left wins (1).

If Right outbids Left, Marker Worth implies that his result cannot be better than in the previous paragraph.

Case 2: Suppose next that and , where Left has an optimal 0-bid strategy in . We must show that . Note that by the assumption , Left has a (winning) move in (Right can bid 0 and force Left to move).

Consider an optimal Left bid in , and assume that Left bids in . If she wins the bid by including the marker, she has a move to , and otherwise she has a move to . In one of these cases Left wins by induction, since, by assumption, or . Similarly, if Right outbids Left and plays in the -component, Left wins by induction.

Hence suppose Right outbids Left and plays to

 (2)

Left bids , and unless Right outbids Left, she plays in the -component to , such that , and wins by induction (since she has an optimal 0-bid strategy in ) and by Monotonicity. Assume that Right outbids Left and plays to

 (3)

Then we use Marker Worth together with Lemma 8. Namely implies , and Left has an optimal 0-bid strategy in . Combine this with , to see that Left wins (3) (by using also Monotonicity).

If Right outbids Left and instead plays to

 (4)

then observe that, since Left has an optimal 0-bid strategy in , then . And so, by Marker Worth, . Therefore, since Left has an optimal 0-bid strategy in , Lemma 8 together with Monotonicity implies

 o(GRR+X,\savestack\tmpbox\stretchto\scaleto\scalerel∗[\widthofp+2ℓ+2]⋀0.5ex\stackon[1pt]p+2ℓ+2\tmpbox)= L.

We summarize the result in terms of how it is used in applications.

###### Corollary 10 (Constructive Comparison Tests).

Consider any total budget.

• Suppose that Left has an optimal 0-bid strategy in . Then Left wins if Right holds the marker if and only if .

• Suppose that Right has an optimal 0-bid strategy in . Then Right wins if Left holds the marker if and only if .

• Suppose that the non-marker holder has an optimal 0-bid strategy in . Then the non-marker holder wins if and only if .

• Suppose that Left has an optimal 0-bid strategy in . Then Left wins independently of holding the marker if and only if .

• Suppose that Right has an optimal 0-bid strategy in . Then Right wins independently of holding the marker if and only if .

###### Proof.

This follows from Theorem 9. ∎

The statement of the Main Theorem (Theorem 9) raises some questions, especially with respect to the similar theorem for games that are numbers in Section 3.

###### Problem 11.

In view of the similarities of Theorems 7 and 9, some questions arise:

1. Is it true that “Left has an optimal 0-bid strategy in ” if and only if “for every Right option , a follower of , there is an answer by Left, ”?

2. Is it true that these properties hold if and only if is a number?

3. May we put the Left-bidding-0 strategy inside the equivalence in Theorem 9, i.e. is it true that “ if and only if and Left has an optimal 0-bid strategy in ”?

Item (iii) does not seem to be valid, since the first implication in the proof does not give a -bidding strategy. But the reverse direction holds.

We observe that we may not remove the proviso of a Left winning -bid strategy in the statement of Theorem 9:

###### Observation 12 (No Generic 0-optimal Strategy).

We demonstrate that, for any non-zero total budget, there is a game form such that , but Left does not optimally bid 0 at each follower.

For any total budget , there is a game such that , and . It suffices to take