A Note on Numbers
Alda Carvalho, Melissa A. Huggan, Richard J. Nowakowski, Carlos Pereira dos Santos
AA Note on Numbers
January 26, 2021
Alda Carvalho , Melissa A. Huggan , Richard J. Nowakowski , Carlos Pereira dos Santos ISEL–IPL & CEMAPRE–University of Lisbon, [email protected] Ryerson University, [email protected] Dalhousie University, [email protected] Center for Functional Analysis, Linear Structures and Applications,University of Lisbon & ISEL–IPL, [email protected]
Abstract
When are all positions of a game numbers? We show that two propertiesare necessary and sufficient. These properties are consequences of that,in a number, it is not an advantage to be the first player. One of theseproperties implies the other. However, checking for one or the other,rather than just one, can often be accomplished by only looking at thepositions on the ‘board’. If the stronger property holds for all positions,then the values are integers.
Keywords : Combinatorial Game Theory, numbers, blue-red-hackenbush , domino shave , shove , push , lenres , polychromatic chomp , partizanturning turtles , divisors , blue-red-cherries , cutcake , erosion . When analyzing games, an early question is: is it possible that all the positionsare numbers? If that is true, then it is easy to determine the outcome of adisjunctive sum of positions, just add up the numbers. It is also easy to find thebest move, just play the summand with the largest denominator. The problemis how to recognize when all the positions are numbers.Siegel [8], page 81, exercise 3.15, states “If every incentive of G is negativethen G is a number”. This does not provide much insight or intuition. In fact,in most non-all-small-games, there are non-zero positions, some of which arenumbers and others not. Let S be a set of positions of a ruleset. It is called a hereditary closed set of positions of a ruleset (HCR) if it is closed under takingoptions. These HCR sets are the natural objects to consider.1 a r X i v : . [ m a t h . C O ] J a n here are two properties either of which, if satisfied for all followers of aposition, tells us that the position is a number. Both are aspects of the first-move-disadvantage in numbers. The first is a comparison with one move againsttwo moves. Definition 1.1 (F1 Property) . Let S be a HCR. Given G ∈ S , the pair( G L , G R ) ∈ G L × G R satisfies the F1 property if there is G RL ∈ G R L such that G RL (cid:62) G L or there is G LR ∈ G L R such that G LR (cid:54) G R .The second property involves moves by both players. Definition 1.2 (F2 Property) . Let S be a HCR. Given G ∈ S , ( G L , G R ) ∈ G L × G R satisfies the F2 property if there are G LR ∈ G L R and G RL ∈ G R L such that G RL (cid:62) G LR .In many games, the literal form of positions will tell us whether they satisfythe F1 property or the F2 property with equality. See Section 2 for examples.The two results a player should remember are: Lemma 3.2
Let S be a HCR. If, for any position G ∈ S , all pairs( G L , G R ) ∈ G L × G R satisfy the F1 property or the F2 property, then allpositions G ∈ S are numbers;and Theorem 3.5
Let S be a HCR. If, for any position G ∈ S , all pairs( G L , G R ) ∈ G L × G R satisfy the F2 property, then all positions G ∈ S areintegers.Theorem 3.1 is the central theoretical result: All the positions in a HCR setare numbers, if and only if there is no position and no number such that the sumis an N -position. This is all that is required to prove Lemma 3.2. Lemma 3.3shows that the F2 property implies the F1 property, with strict inequality. Now,it may seem that the F2 property is useless and it shouldn’t be a hypothesis ofLemma 3.2. However, in practice, it is easier to recognize that all pairs satisfyeither the F1 property or the F2 property rather than trying to prove that allpairs satisfy just the F1 property. In fact, Lemma 3.2 may be written as anecessary and sufficient condition. This is Theorem 3.4 which only uses the F1property.The F2 property is a stronger constraint than the F1 property.Theorem 3.5 shows that the F2 property implies that the numbers will beintegers.We recall the results about numbers needed for this paper. Theorem 1.3. [1, 2, 8]
Let G be a number whose options are numbers. . After removing dominated options, the form of G has at most one Leftoption and at most one Right option.2. For the options that exist, G L < G < G R .3. If there is an integer k , G L < k < G R , or if either G L or G R does notexist, then G is an integer.4. If both G L and G R exist and the previous case does not apply, then G isthe simplest number between G L and G R . The most important point to remember is item 2, that is, when a playerplays in a number the situation gets worse for them. This has an importantconsequence when games are being analyzed.
Theorem 1.4 (Number Avoidance Theorem) . [1, 2, 8] Suppose that G is anumber and H is not. If Left can win moving first on G + H , then Left can doso with a move on H . In many cases, when checking the properties, the Left and Right options willrefer to two specific moves on the ‘game board’, one by Left and one by Right.If this happens, then the actual positions will automatically give the strongerconditions, G LR ∼ = G RL or G RL ∼ = G L . Moreover, no calculations are required.Examples are given in Section 2. In these examples, we illustrate that, sometimes, only two specific moves on the‘game board’, one for each player, are sufficient. We will refer to the specificmoves by lower case letters, (cid:96) for Left and r for Right.We first sketch a proof to show that the values of polychromatic chomp (see Appendix for rules), blue-red-hackenbush strings [2, 9] are numbers,and that cutcake [2] positions are integers. We then give the properties thatthe following games satisfy: domino shave [4], shove [1], push [1], lenres [7], divisors , and partizan turning turtles [3] (see Appendix forlast two rulesets). The two games, partizan euclid [5] and partizansubtraction [6] are examples where many positions satisfy one or bothproperties. However, since there are positions which satisfy neither, then onlya few positions are numbers. Example 2.1.
Let G be a polychromatic chomp position. Let (cid:96) and r beblack and gray squares respectively. If neither G (cid:96) nor G r eliminates the other,then playing both moves, in either order, results in the same position Q , i.e., G (cid:96)r ∼ = G r(cid:96) . Suppose G (cid:96) eliminates G r , as illustrated in Figure 1. In this case,Left can play her move before or after Right’s move, i.e., G (cid:96) ∼ = G r(cid:96) .Hence, all ( G L , G R ) satisfy one or both of the properties. Therefore, byLemma 3.2, all polychromatic chomp positions are numbers.3 = G ℓ = G r = Figure 1: F1 argument in polychromatic chomp . Example 2.2.
Let G be a blue-red-hackenbush string and let (cid:96) and r bethe edges played by Left and Right, respectively. If r is higher up the stringthan (cid:96) then playing (cid:96) eliminates r . Thus, G r(cid:96) and G (cid:96) are identical. Otherwise,playing r eliminates (cid:96) , and G (cid:96)r ∼ = G r .Hence, by Theorem 3.4, all blue-red-hackenbush strings are numbers. Example 2.3.
Consider a m × n cutcake position. The moves are notindependent but almost so. For given (cid:96) and r , consider the pair of options, G L = m × ( n − (cid:96) ) + m × (cid:96) and G R = ( m − r ) × n + r × n , and their options G LR = m × ( n − (cid:96) ) + ( m − r ) × (cid:96) + r × (cid:96),G RL = ( m − r ) × n + r × ( n − (cid:96) ) + r × (cid:96). The moves cannot be interchanged and get the same board position.However, we know that if i > j then k × i (cid:62) k × j (intuitively, there are moremoves for Left in k × i than in k × j ) and similarly i × k (cid:54) j × k . The termsof G LR and G RL pair off: r × (cid:96) is in both; ( m − r ) × (cid:96) (cid:54) ( m − r ) × n ; m × ( n − (cid:96) ) (cid:54) r × ( n − (cid:96) ). Therefore G LR (cid:54) G RL and, so, ( G L , G R ) satisfiesthe F2 property. Therefore, by Theorem 3.5, G is an integer. Example 2.4.
1. Given a shove position G , if any token is pushed off theend of the strip then ( G (cid:96) , G r ) satisfies the F1 property; if not, ( G (cid:96) , G r )satisfies the F2 property.2. Given a push position G , if any token pushes the other then ( G (cid:96) , G r )satisfies the F1 property; if not, ( G (cid:96) , G r ) satisfies the F2 property.4. Given a lenres position G , if any digit in the move replaces the otherthen ( G (cid:96) , G r ) satisfies the F1 property; if not, ( G (cid:96) , G r ) satisfies the F2property.4. Given a domino shave position G , ( G L , G R ) ∈ G L × G R satisfies the F1property.5. Given a divisors position G = ( l, r ) and a pair of options, ( G L , G R ) =(( (cid:96) (cid:48) , r ) , ( (cid:96), r (cid:48) )), if (cid:96) (cid:48) = r or (cid:96) = r (cid:48) then ( G L , G R ) satisfies the F1 property;if not, ( G L , G R ) satisfies the F2 property.6. Given a partizan turning turtles position G , if G L and G R conflictthen ( G L , G R ) satisfies the F1 property; if G L and G R don’t conflict then( G L , G R ) satisfies the F2 property.Even more can be said about blue-red-cherries [1] and erosion [1].These game all satisfy the F2 property and thus, by Theorem 3.5, they areintegers. Example 2.5.
1. Given a blue-red-cherries position G , all ( G L , G R ) ∈ G L × G R . If Left removes a cherry from one end ( (cid:96) ) and Right removes acherry from the other end ( r ) then G (cid:96)r ∼ = G r(cid:96) .2. Given an erosion position G , all ( G L , G R ) ∈ G L × G R satisfy the F2property. This is vacuously true since, by the rules, it is impossible forboth players to have options at the same time.For the values all to be numbers, the properties must always be true. It isnot sufficient for most of the positions to satisfy them. Two games, that have N -positions but where many of the positions naturally satisfy one or the otherproperty, are:1. F1 : In partizan euclid [5] with G = ( p, q ) and p > q then G LR = G R or G RL = G L .2. F2 : In the partizan subtraction subset of splittles [6], let a be thelargest that can be taken. Suppose the heap size is n , and n (cid:62) a thenLeft taking (cid:96) and Right taking r results in a heap of size n − (cid:96) − r regardlessof the order. Thus G (cid:96)r ∼ = G r(cid:96) . Theorem 3.1. (Outcomes and numbers) Let S be a HCR . All positions G ∈ S are numbers if and only if there is no G ∈ S and a number x such that G + x ∈ N .Proof. ( ⇒ ) If all positions G ∈ S are numbers then, regardless of what the numbers x are, all G + x are numbers. Hence, there is no G ∈ S and a number x such5hat G + x ∈ N .( ⇐ ) Let G ∈ S . If G L = ∅ or G R = ∅ then G is an integer. Suppose that G L (cid:54) = ∅ and G R (cid:54) = ∅ . By induction, since S is hereditary closed, all G L ∈ G L and G R ∈ G R are numbers. Hence, after removing dominated options, thereare 3 possible cases:1) G = { a | a } = a + ∗ , where a is a number;2) G = { a | b } = a + b ± a − b , where a and b are numbers and a > b ;3) G = { a | b } , where a and b are numbers and a < b .If 1) or 2) then, G − a ∈ N or G − a + b ∈ N , contradicting theassumptions. Therefore, we must have 3). Now G is the simplest numberstrictly between a and b .In all cases G is a number and the theorem follows.Now, a natural question arises: Is it easy, in practice, to know if a HCRdoes not have positions such that G + x ∈ N ? In other words, is Theorem 3.1useful? Lemma 3.2 answers that question. Lemma 3.2.
Let S be a HCR . If, for any position G ∈ S , all pairs ( G L , G R ) ∈ G L × G R satisfy the F1 property or the F2 property, then all positions G ∈ S are numbers.Proof. Item 1: By Theorem 3.1, it is enough to prove that if all pairs( G L , G R ) ∈ G L × G R satisfy the F1 property or the F2 property then there isno G ∈ S and a number x such that G + x ∈ N .For the contrapositive, suppose that there is a position G ∈ S and anumber x such that G + x ∈ N . Assume that the birthday of G in suchconditions is the smallest possible.Since G + x ∈ N , there are G L + x (cid:62) G R + x (cid:54) G L , G R ) satisfies the F1 property or the F2property. If the pair satisfies the F1 property, there is G RL such that G RL (cid:62) G L or there is G LR such that G LR (cid:54) G R . If the first happens, then G RL (cid:62) G L implies G RL + x (cid:62) G L + x (cid:62)
0. That is incompatible with G R + x (cid:54)
0. If the second happens, then G LR (cid:54) G R implies G LR + x (cid:54) G R + x (cid:54)
0. That is incompatible with G L + x (cid:62)
0. In either casewe have a contradiction; the pair ( G L , G R ) cannot satisfy the F1 property.Hence, the pair ( G L , G R ) satisfies the F2 property, and there are G LR ∈ S and G RL ∈ S such that G LR (cid:54) G RL . Since G L + x (cid:62)
0, we have G LR + x (cid:54) (cid:54) G R + x (cid:54)
0, we have G RL + x (cid:54) (cid:62)
0. The second inequality allows toconclude that G LR + x (cid:54) (cid:62) G LR (cid:54) G RL . However, G LR + x (cid:54) (cid:54) G LR + x (cid:54) (cid:62)
0, implies that G LR + x ∈ N , contradicting the smallest rankassumption. Therefore, the pair ( G L , G R ) cannot satisfy the F2 property.The pair ( G L , G R ) doesn’t satisfy the F1 property or the F2 property, andthat contradicts the hypothesis. There is no G ∈ S and number x such that G + x ∈ N . Therefore, all positions G ∈ S are numbers.It is possible to have a pair of options that satisfies the F2 propertywithout satisfying the F1 property; an example of that is a pair like( G L , G R ) = ( { , ∗ | ∗} , {∗ | , ∗} ). However, the options of ∗ do not satisfy theF2 property. On the other hand, if all followers also satisfy the F2 property,then the next lemma shows that all pairs satisfy the F1 property. Lemma 3.3.
Let S be a HCR such that, given any position G ∈ S , all pairs ( G L , G R ) ∈ G L × G R satisfy the F2 property then all pairs satisfy the F1property.Proof. It is enough to prove that if a pair ( G L , G R ) ∈ G L × G R satisfies theF2 property, it also satisfies the F1 property.Suppose that a pair ( G L , G R ) satisfies the F2 property. If so, by definition,there are G LR and G RL such that G LR (cid:54) G RL . Since G LR is a right option of G L , we have G LR (cid:54) (cid:54) G L . On the other hand, by Lemma 3.2, all positions of S are numbers, so G LR cannot be incomparable with G L . Therefore, we musthave G LR > G L and consequently G RL (cid:62) G LR > G L . Thus ( G L , G R ) satisfiesthe F1 property. Theorem 3.4.
Let S be a HCR . All positions G ∈ S are numbers if and only if,for any position G ∈ S , all pairs ( G L , G R ) ∈ G L × G R satisfy the F1 property.Proof. ( ⇐ ) Consequence of Lemma 3.2.( ⇒ ) Let G ∈ S and ( G L , G R ) ∈ G L × G R . Since G is a number, G L < G R and thus G L − G R <
0. Since Right, playing first, wins G L − G R , either thereis a G LR with G LR − G R (cid:54) G RL with G L − G RL (cid:54)
0. Hence,there is G LR (cid:54) G R or G L (cid:54) G RL , and, by definition, ( G L , G R ) satisfies the F1property.By Lemma 3.3, if all pairs ( G L , G R ) satisfy the F1 property or the F2property, then all pairs satisfy the F1 property. That means that a pairsatisfying the F2 property also satisfies the F1 property. But, observe that theopposite is not true: it is possible to have a pair satisfying the F1 propertywithout satisfying the F2 property. For example, if G = = { | } (canonicalform), then the pair (0 ,
1) satisfies the F1 property and does not satisfy the F2property because G L R = ∅ . The F2 property is a stronger condition and has asurprising consequence. 7 heorem 3.5. Let S be a HCR . If, for any position G ∈ S , all pairs ( G L , G R ) ∈ G L × G R satisfy the F2 property, then all positions G ∈ S are integers.Proof. Let G ∈ S . If G L = ∅ or G R = ∅ , then G is an integer and the theoremholds. Suppose that G L (cid:54) = ∅ and G R (cid:54) = ∅ . By Lemma 3.2, all positions G ∈ S are numbers and the canonical form is G = { G L | G R } . By induction, G L and G R are integers. Since ( G L , G R ) satisfies the F2 property, there is G LR (cid:54) G RL and, by induction, both are integers. Let G RL = k . Therefore, we have that G L < k and G R > k , and, thus G is an integer. Observation . Theorem 3.5 exhibits a sufficient but not necessary condition.Consider S , a HCR whose game forms are {− | } , − −
1, and 0 (the lastthree, canonical forms). Of course, all game values of S are integers. However,regarding G = {− | } , the pair ( G L , G R ) = ( − ,
0) does not satisfy the F2property.
Acknowledgments
Alda Carvalho was partially supported by the ProjectCEMAPRE/REM-UIDB/05069/2020, financed by FCT/MCTES throughnational funds.Melissa A. Huggan was supported by the Natural Sciences and EngineeringResearch Council of Canada (funding reference number PDF-532564-2019).Richard J. Nowakowski was supported by the Natural Sciences andEngineering Research Council of Canada (funding reference number4139-2014).Carlos Santos is a CEAFEL member and has the support ofUID/MAT/04721/2019 strategic project.
References [1] M. Albert, R. J. Nowakowski, and D. Wolfe,
Lessons in Play: AnIntroduction to Combinatorial Game Theory , A. K. Peters, 2007.[2] E. R. Berlekamp, J. H. Conway, and R. K. Guy,
Winning Ways for YourMathematical Plays , Academic Press, London, 1982.[3] A. Bonato, M. A. Huggan, and R. J. Nowakowski, Partizan Turning Turtles,preprint.[4] A. Carvalho, M. A. Huggan, R. J. Nowakowski, and C. P. dosSantos, Ordinal Sums, Clockwise Hackenbush, and Domino Shave, arXiv:2011.11518. 85] N. A. McKay and R. J. Nowakowski, Outcomes of partizan Euclid.
Integers (2012/13), Proceedings of the Integers Conference 2011, Paper No. A9,15 pp.[6] G. A. Mesdal, Partizan Splittles, in
Games of No Chance 3 , CambridgeUniv. Press, 2009, pp. 447-461.[7] A. A. Siegel,
On the Structure of Games and their Posets,
Ph.D. thesis,Dalhousie University, 2011.[8] A. N. Siegel,
Combinatorial Game Theory , American Math. Soc., 2013.[9] T. van Roode,
Partizan Forms of Hackenbush Combinatorial Games , M.Sc.Thesis, University of Calgary, 2002.
Appendix 1: Rulesets divisors
Position: An ordered pair of positive integers ( l, r ).Moves: Left is allowed to replace ( l, r ) by ( l (cid:48) , r ) where l (cid:48) < l is a divisor of r .Right is allowed to replace ( l, r ) by ( l, r (cid:48) ) where r (cid:48) < r is a divisor of l .(5 , L → (2 , R → (2 , L → (1 , partizan turning turtles Position: A line of turtles. A turtle may be on its feet or on its back.Moves: Left is allowed to choose two upside-down turtles and turn them onto itsfeet. Right is also allowed to choose a pair of turtles, provided that the leftmostis on its feet and the other is on its back; his move is turning over both turtles. L → R → polychromatic chomp Position: A grid with one poison square in the lower left corner. Besides thepoison square, each square is either black or gray.9oves: On her turn, Left chooses a black square and removes it and all othersquares above or to the right of it. On his turn, Right moves analogously, buthe has to choose a gray square. ××
Position: An ordered pair of positive integers ( l, r ).Moves: Left is allowed to replace ( l, r ) by ( l (cid:48) , r ) where l (cid:48) < l is a divisor of r .Right is allowed to replace ( l, r ) by ( l, r (cid:48) ) where r (cid:48) < r is a divisor of l .(5 , L → (2 , R → (2 , L → (1 , partizan turning turtles Position: A line of turtles. A turtle may be on its feet or on its back.Moves: Left is allowed to choose two upside-down turtles and turn them onto itsfeet. Right is also allowed to choose a pair of turtles, provided that the leftmostis on its feet and the other is on its back; his move is turning over both turtles. L → R → polychromatic chomp Position: A grid with one poison square in the lower left corner. Besides thepoison square, each square is either black or gray.9oves: On her turn, Left chooses a black square and removes it and all othersquares above or to the right of it. On his turn, Right moves analogously, buthe has to choose a gray square. ×× L → ××
Position: An ordered pair of positive integers ( l, r ).Moves: Left is allowed to replace ( l, r ) by ( l (cid:48) , r ) where l (cid:48) < l is a divisor of r .Right is allowed to replace ( l, r ) by ( l, r (cid:48) ) where r (cid:48) < r is a divisor of l .(5 , L → (2 , R → (2 , L → (1 , partizan turning turtles Position: A line of turtles. A turtle may be on its feet or on its back.Moves: Left is allowed to choose two upside-down turtles and turn them onto itsfeet. Right is also allowed to choose a pair of turtles, provided that the leftmostis on its feet and the other is on its back; his move is turning over both turtles. L → R → polychromatic chomp Position: A grid with one poison square in the lower left corner. Besides thepoison square, each square is either black or gray.9oves: On her turn, Left chooses a black square and removes it and all othersquares above or to the right of it. On his turn, Right moves analogously, buthe has to choose a gray square. ×× L → ×× R → ××
Position: An ordered pair of positive integers ( l, r ).Moves: Left is allowed to replace ( l, r ) by ( l (cid:48) , r ) where l (cid:48) < l is a divisor of r .Right is allowed to replace ( l, r ) by ( l, r (cid:48) ) where r (cid:48) < r is a divisor of l .(5 , L → (2 , R → (2 , L → (1 , partizan turning turtles Position: A line of turtles. A turtle may be on its feet or on its back.Moves: Left is allowed to choose two upside-down turtles and turn them onto itsfeet. Right is also allowed to choose a pair of turtles, provided that the leftmostis on its feet and the other is on its back; his move is turning over both turtles. L → R → polychromatic chomp Position: A grid with one poison square in the lower left corner. Besides thepoison square, each square is either black or gray.9oves: On her turn, Left chooses a black square and removes it and all othersquares above or to the right of it. On his turn, Right moves analogously, buthe has to choose a gray square. ×× L → ×× R → ×× L → ××
Position: An ordered pair of positive integers ( l, r ).Moves: Left is allowed to replace ( l, r ) by ( l (cid:48) , r ) where l (cid:48) < l is a divisor of r .Right is allowed to replace ( l, r ) by ( l, r (cid:48) ) where r (cid:48) < r is a divisor of l .(5 , L → (2 , R → (2 , L → (1 , partizan turning turtles Position: A line of turtles. A turtle may be on its feet or on its back.Moves: Left is allowed to choose two upside-down turtles and turn them onto itsfeet. Right is also allowed to choose a pair of turtles, provided that the leftmostis on its feet and the other is on its back; his move is turning over both turtles. L → R → polychromatic chomp Position: A grid with one poison square in the lower left corner. Besides thepoison square, each square is either black or gray.9oves: On her turn, Left chooses a black square and removes it and all othersquares above or to the right of it. On his turn, Right moves analogously, buthe has to choose a gray square. ×× L → ×× R → ×× L → ×× R → ××