The game of Flipping Coins
aa r X i v : . [ m a t h . C O ] F e b THE GAME OF FLIPPING COINS
Anthony Bonato Ryerson University, Toronto, Ontario, Canada [email protected]
Melissa A. Huggan Ryerson University, Toronto, Ontario, Canada [email protected]
Richard J. Nowakowski Dalhousie University, Halifax, Nova Scotia, Canada [email protected]
Abstract
We consider flipping coins , a partizan version of the im-partial game turning turtles , played on lines of coins. Weshow the values of this game are numbers, and these are found byfirst applying a reduction, then decomposing the position into aniterated ordinal sum. This is unusual since moves in the middleof the line do not eliminate the rest of the line. Moreover, when G is decomposed into lines H and K , then G = ( H : K R ). Thisis in contrast to hackenbush strings where G = ( H : K ). Keywords : Combinatorial Game Theory, ordinal sum, flipping coins
1. Introduction
In Winning Ways Volume 3 [2], Berlekamp, Conway, and Guy intro-duced turning turtles and considered many variants. Each gameinvolves a finite row of turtles, either on feet or backs, and a move is to Supported by an NSERC Discovery grant. Supported by an NSERC Postdoctoral fellowship. Supported by an NSERC Discovery grant. moebius (flip up to 5 coins) playedwith 18 coins, involves
M¨obius transformations ; mogul (flip up to7 coins) on 24 coins, involves the miracle octad generator developedby R. Curtis in his work on the Mathieu group M and the Leechlattice, [5, 6]; ternups (flip three equally spaced coins) requires ternaryexpansions; and turning corners , a two-dimensional version wherethe corners of a rectangle are flipped, needs nim-multiplication.We consider a simple partizan version of turning turtles , alsoplayed with coins. We give a complete solution and show that it involvesordinal sums. This is somewhat surprising since moves in the middleof the line do not eliminate moves at the end. Compare this with hackenbush strings [2], and domino shave [4].We will denote heads by 0 and tails by 1. Our partizan version will beplayed with a line of coins, represented by a 0-1 sequence, d d . . . d n ,where d i ∈ { , } . To this position, we associate the binary number P ni =1 d i i − . Left moves by choosing some pair of coins d i , d j , i < j ,where d i = d j = 1 and flips them over so that both coins are 0s. Rightalso chooses a pair d k , d ℓ , k < ℓ , with d k = 0 and d ℓ = 1, and flips themover. If j is the greatest index such that d j = 1, then d k , k > j , will bedeleted. For example,1011 = { , , | , } . The game eventually ends since the associated binary number decreaseswith every move. We call this game flipping coins .The game is biased to Left. If there are a non-zero even number of1s in a position, then Left always has a move; that is, she will win. Leftalso wins any non-trivial position starting with 1. However, there arepositions that Right wins. The two-part method to find the outcomesand values of the remaining positions can be applied to all positions.First, apply a modification to the position (unless it is all 1s), which2educes the number of consecutive 1s to at most three. After this reduc-tion, build an iterated ordinal sum, by successively deleting everythingafter the third last 1, this deleted position determines the value of thenext term in the ordinal sum. As a consequence, the original position isa Right win, if the position remaining at the end is of the form 0 . . . G is found as follows. Let G L be the position after the Left move that removes the rightmost 1s. Let H be the string G \ G L ; that is, the substring eliminated by Left’s move.Let H R be the result of Right’s best move in H . Now, we have that G = G L : H R . In contrast, the ordinal sums for hackenbush strings and domino shave [4], involve the value of H not H R .The proof of Theorem 3.3 is given in Section 4.1. The final sectionincludes a brief discussion of open problems.Finally we pose a question for the reader, which we answer at the endof Section 4.1: Who wins 0101011111 + 1101100111 + 0110110110111and how?
2. Numbers
All the values in this paper are numbers and this section contains allthe necessary background to make the paper self-contained. For furtherdetails, consult [1, 7]. Positions are written in terms of their options,that is G = { G L | G R } . Definition 2.1. [1, 2, 7]
Let G be a number whose options are numbersand let G L , G R be the Left and Right options of the canonical form of .1. If there is an integer k , G L < k < G R , or if either G L or G R does not exist, then G is the integer, say n , closest to zero thatsatisfies G L < n < G R .2. If both G L and G R exist and the previous case does not apply,then G = p q , where q is the least non-negative integer such thatthere is an odd integer p satisfying G L < p q < G R . The properties of numbers required for this paper are contained inthe next two theorems.
Theorem 2.2. [1, 2, 7]
Let G be a number whose options are numbersand let G L , G R be the Left and Right options of the canonical form of G . If G ′ and G ′′ are any Left and Right options respectively, then G ′ G L < G < G R G ′′ . Theorem 2.2 shows that if we know that the property holds, we needto only consider the best move for both players in a number.We include the following examples.(a) 0 = { | } = {− | } = {− | } ;(b) − { | − } = {− | − } ;(c) 1 = { | } = { | } ;(d) = { | } = { | } .For games G and H , to show that G ≥ H , we need to show that G − H ≥
0. Meaning, we need to show that G − H is a Left winmoving second. For more information, see Sections 5.1, 5.8, and 6.3 ofthe second edition of [1].Let G and H be games. The ordinal sum of G , the base , and H , the exponent , is G : H = { G L , G : H L | G R , G : H R } . Intuitively, playing in G eliminates H but playing in H does not affect G . For ease of reading, if an ordinal sum is a term in an expression,then we enclose it in brackets. 4ote that x : 0 = x = 0 : x since neither player has a move in 0.We demonstrate how to calculate the values of other positions with thefollowing examples.(a) 1 : 1 = { | } = 2;(b) 1 : − { | } = ;(c) 1 : = { , (1 : 0) | (1 : 1) } = { , | { | }} = { | } = ;(d) : 1 = { , ( : 0) | } = { , | } = { | } = ;(e) (1 : −
1) : = ( : ) = { , ( : 0) | , ( : 1) } = { , | , } = { | } = .Note that in all cases, players prefer to play in the exponent ratherthan the base. This is true in all cases, but in this paper all the expo-nents will be positive. Theorem 2.3.
Let G be a number all of whose options are numbers,and let H > be a number.1. G L < ( G : H L ) < ( G : H ) < ( G : H R ) < G R .2. If H = 0 , then G : H = G . If H > , then ( G : H ) > G .Proof. For item (1), by definition, G L is a Left option of G : H L andboth are Left options of G . Thus G L < ( G : H L ) < G by Theorem 2.2.The proof is similar for the Right options.For item (2), if H = 0, then G : 0 = { G L | G R } and this is G .Suppose H >
0. Let Right move first in ( G : H ) − G . If Right moves to G R − G or ( G : H ) − G L , then Left responds to G R − G R and G L − G L respectively and wins. If Right moves to ( G : H R ) − G , then since H R >
0, ( G : H R ) − G > hereditary closed set of positionsof a ruleset if it is closed under taking options. This game satisfiesruleset properties introduced in [3]. In particular, the properties arecalled the
F1 property and the
F2 property and are defined formally asfollows. 5 efinition 2.4. [3]
Let S be a hereditary closed ruleset. Given a posi-tion G ∈ S , the pair ( G L , G R ) ∈ G L × G R satisfies the F1 property ifthere is a G RL ∈ G R L such that G RL > G L or there is a G LR ∈ G L R such that G LR G R . Definition 2.5. [3]
Let S be a hereditary closed ruleset. Given a posi-tion G ∈ S , the pair ( G L , G R ) ∈ G L × G R satisfies the F2 property ifthere is a G RL ∈ G R L and there is a G LR ∈ G L R such that G LR G RL . As proven in [3], if all positions of the ruleset satisfy one of theseproperties, the value of the position is a number.
Theorem 2.6. [3]
Let S be a hereditary closed ruleset. Every positionin S satisfies either the F1 or the F2 property if and only if everyposition is a number.
3. Main results
Before considering the values and associated strategies, we considerthe outcomes; that is, we partially answer the question: “Who winsthe game?” The full answer requires an analogous analysis to findingthe values.
Theorem 3.1.
Let G = d d . . . d n . If d d . . . d n contains an evennumber of s, or if d = 1 and there are least two s, then Left wins G .Proof. A Right move does not decrease the number of 1s in the position.Thus, if in G , Left has a move, then she still has a move after any Rightmove in G . Consequently, regardless of d , if there are an even numberof 1s in G , it will be Left who reduces the game to all 0s. Similarly, if d = 1 and there are an odd number of 1s, Left will eventually reduce G to a position with a single 1; that is, to d = 1 and d i = 0 for i > d = 0 and an odd number of 1s, is more in-volved. The analysis of this case is the subject of the remainder of thepaper. We first prove the following. Theorem 3.2.
All flipping coins positions are numbers. roof. Let G be a flipping coins position. If only one player has amove, then the game is an integer. Otherwise, let L be the Left moveto change ( d i , d j ) from (1 ,
1) to (0 , R be the Right move tochange ( d k , d ℓ ) from (0 ,
1) to (1 , L and R can be played in eitherorder. In this case G LR = G RL . Thus, the F2 property holds. If thereare only three distinct indices, then two of the bits are ones. If Leftmoves first, then d i = d j = d k = 0. If Right moves first, then there arestill two ones remaining after his move. After Left moves, the positionbecomes G (0 , ,
0) and hence, G L = G RL . The F1 property holds.There are no more cases since there must be at least three distinctindices. Since every position satisfies either the F1 or F2 property then,by Theorem 2.6, every position is a number.Given a position G , the following algorithm returns a value. Algorithm:
Let G be a flipping coins position. Let G = G .1. Set i = 0.2. Reductions: Let α and β be strings of coins, and either can beempty.(a) If G = α j β , j ≥
1, then set G = α j β .(b) If G = α β , and β contains an even number of 1s, thenset G = α β .(c) Repeat until neither case applies, then go to Step 3.3. If G i is 0 r r ≥ a p i q i a ≥ p i + q i ≥
0; then go toStep 5.Otherwise, G i = α a p i q i p i + q i ≥ a > α . Set Q i = 0 p i q i ,G i +1 = α a . Go to Step 4.4. Set i = i + 1. Go to Step 3. 7. If G i = 0 r
1, then set v i = − r . If G i = 1 a p i q i
1, then set v i = ⌊ a ⌋ + pi + qi . Go to Step 6.6. For j from i − v j = v j +1 : pj + qj − .7. Return the number v .The algorithm implicitly returns two different results:1. For Step 3, the substrings, Q , Q , . . . , Q i − , G i , partition the re-duced version of G ;2. The value v .First we illustrate the algorithm with the following example. Let G =10011110110110111011110011. The reductions, applied to the under-lined digits, give that:10011110110110111011110011 = 10011110110110111011110011= 100111101101101111010011= 1001111011011101010011= 10011110111001010011= 100111110001010011= 1010110001010011 . Step 3 partitions the last expression into 101(011)(000101)(0011) sothat the ordinal sum is given by v = (cid:18)(cid:18)
12 : 12 (cid:19) : 164 (cid:19) : 18= 1025716348 . Now let H = 01001110110111011101. The reductions give that:01001110110111011101 = 01001110110111011101= 010011101110011101= 0100111100011101= 01010100011101 . v = (cid:18)(cid:18) − (cid:19) : 132 (cid:19) : 1= − . The next theorem is the main result of the paper.
Theorem 3.3 (Value Theorem) . Let G be a flipping coins position.If v is the value obtained by the algorithm applied to G , then G = v . In the next section, we derive several results that will be used to proveTheorem 3.3. The proof of Theorem 3.3 will appear in Section 4.1.
4. Best moves and Reductions
The proofs in this section use induction on the options. An alter-nate but equivalent approach is to regard the techniques as induc-tion on the associated binary number of the positions. The proofsrequire detailed examination of the positions and we will use notationsuitable to the case being considered. Often, a typical position willbe written as a combination of generic strings and the substring un-der consideration. For example, 111011000110101 might be parsed as(11101)(100011)(0101), and written α β or more compactly as α β .We require several results before being able to prove Theorem 3.3. Webegin by proving a simplifying reduction, followed by the best moves foreach player, and then the remaining reductions used in the algorithm.As an immediate consequence of Theorems 3.2 and 2.2 we have thefollowing. Corollary 4.1.
Let α , β , and γ be arbitrary substrings of coins. Wethen have that α β γ > α β γ . Moreover, for an integer r ≥ wehave that β r > β . roof. Recall that by Theorem 3.2 all flipping coins positions arenumbers. Thus, Theorem 2.2 applies.A Right option of α β γ is α β γ and so we have that α β γ >α β γ . Similarly, a Left option of β r β we have that β r >β . Next we prove the best moves for each player. Right wants to playthe zero furthest to the right and the 1 adjacent to it. Left wants toplay the two ones furthest to the right. Theorem 4.2.
Let G be a flipping coins position, where in G , k and n are the greatest indices such that the associated coin is . Let i be the greatest index such that d i = 0 . Left’s best move is to play ( d k , d n ) , and Right’s best move is to play ( d i , d i +1 ) .Proof. We prove this theorem by induction on the options. Note thatwe use the equivalent binary representation of the game position. Ifthere are three or fewer coins, then by exhaustive analysis the theoremis true.Let G be d d . . . d n . We begin by proving Left’s best moves. Let k and n be the two largest indices, where d k = d n = 1, and let i and j , i 1) to G (1 , , , G (1 , , , 1) to G (0 , , , G (1 , , , − G (0 , , , > 0. For the movesto be different, at least three of i, j, k, n are distinct. Further, by thechoice of the indices, we now know that i < k .We first assume the four indices are distinct. In this case, we havethat i < j < k < n . By applying Corollary 4.1 twice, we have that G (1 , , , > G (1 , , , > G (0 , , , . We may assume then, without loss of generality, that j = k or j = n .Now consider G ( d i , d k , d n ) = G (1 , , j = k , G (1 , , > G (0 , , j = n , G (1 , , > G (0 , , k be the largest index such that d k = 0 and d k +1 = 1 and also let10 , j , i < j be indices with d i = 0 and d j = 1. The claimed best move is d k , d k +1 and this must be compared to the arbitrary Right move d i , d j .The original position is G ( d i , d j , d k , d k +1 ) = G (0 , , , D = G (1 , , , − G (0 , , , > 0. For themoves to be different, there must be at least three distinct indices.Suppose Right plays in the first summand of D . Note that, by in-duction, the best moves of Left and Right are known.(1) First, suppose j < k . By induction, Right’s best move in the firstsummand of D , is to D ′ = G (1 , , , − G (0 , , , i < j , then G (1 , , , 0) is a Right option of G (0 , , , 0) and thus, D ′ is non-negativeby Corollary 4.1.(2) If j = k + 1, then there are only three distinct indices. The origi-nal game is G ( d i , d k , d k +1 ) = G (0 , , 1) and D = G (1 , , − G (0 , , G (1 , , 0) is a Right option of G (0 , , D is non-negativeby Corollary 4.1.(3) If j > k + 1, then the original game, with the indices in increas-ing order, is G ( d i , d k , d k +1 , d j ) = G (0 , , , 1) and D = G (1 , , , − G (0 , , , G (1 , , , − G (0 , , , 1) and Leftresponds to G (1 , , , − G (1 , , , 0) = 0, and Left wins.In all cases, if Right moves in the first summand of D , then Leftwins. Next, we consider Right moving in the second summand of D = G (1 , , , − G (0 , , , d ℓ = 1 if n > ℓ > k + 1.(1) If n > k + 2, then Right’s best move in the second summand is tochange d n − , d n from (1 , 1) to (0 , n = k + 2. 112i) If j < k + 1, then G ( d i , d j , d k , d k +1 , d k +2 ) = G (0 , , , , 1) and D = G (1 , , , , − G (0 , , , , G (1 , , , , − G (0 , , , , G (1 , , , , − G (0 , , , , G (0 , , , − G (0 , , , 0) = 0, and she wins. For these cases, we onlylist the original position. The strategy for both cases is as just de-scribed.(2ii) If j = k + 1, then G ( d i , d k , d k +1 , d k +2 ) = G (0 , , , 1) and D = G (1 , , , − G (0 , , , j = k + 2, then G ( d i , d k , d k +1 , d k +2 ) = G (0 , , , 1) and D = G (1 , , , − G (0 , , , n = k + 1.(3i) If j < k + 1, then let ℓ < k + 1 be the largest index such that d ℓ = 1.If j < ℓ , then we have G ( d i , d j , d ℓ , d k , d k +1 ) = G (0 , , , , 1) and D = G (1 , , , , − G (0 , , , , G (1 , , , , − G (0 , , , , G (1 , , , , − G (0 , , , , 0) which isnon-negative since G (1 , , , , 0) is a Right option of G (0 , , , , j = ℓ , then G ( d i , d j , d k , d k +1 ) = G (0 , , , 1) and D = G (1 , , , − G (0 , , , G (1 , , , − G (0 , , , G (0 , , , − G (0 , , , 0) = 0, and Left wins.(3ii) If j = k + 1, then G ( d i , d k , d k +1 ) = G (0 , , 1) and D = G (1 , , − G (0 , , D , proving the result.Suppose in a position that the coins of the best Right move are12ifferent from those of the best Left move. The next lemma essentiallysays that the position before and after one move by each player areequal. It is phrased in a way that is useful for reducing the length ofthe position. For the Algorithm, it suffices to prove the result for β being empty. However, it is useful, certainly for a human, to reducethe length of the position as much as possible. Lemma 4.3. If α ≥ , then we have that α a = α a .Proof. Let H = α a − α a . We need to show H = 0. We haveseveral cases to consider.(1) If a ≥ 2, then playing the same move in the other summand aregood responses. After two such moves we have either α a − − α a − = 0 , by induction,or α a − α a − = α a − − α a − = 0 , by induction . (2) If a = 1, then H = α − α α − α , then Right movesto α − α 101 = 0.ii. Right plays in the second summand to α − α, then Leftmoves to α − α . Since ( α L = α , then α > α .iii. Right plays in the first summand to α − α α − α 101 = 0.iv. Left plays in the second summand to α − α 11, then Rightmoves to α − α 11 = α − α 11 = 0, by induction.(3) If a = 0, then H = α − α 1. There are several cases to consider.i. If Left or Right play in the first summand, then the response isin the first summand giving α − α α = β b , b ≥ 0. We have that β b − β b β b − β b . Here, Right responds to β b − − β b , which by induction is equal to β b − − β b = 0.iii. Right plays in the second summand. For a Right move to exist,then α = β a , a ≥ 0. Thus, H = β a − β a 1, and Rightmoves to β a − β . Left responds by moving to β a − β .We then have that ( β a L = β ; thus, β a > β . Hence,we find that β a − β > H thereby proving the result.There are reductions that can be applied to the middle of the posi-tion, but extra conditions are needed. Lemma 4.4. Let α and β be arbitrary binary strings where either (a) β starts with a , or (b) β starts with and has an even number of s.We then have that α β = α β. Proof. Let H = α β − α β . We need to show that H = 0. Wehave several cases to consider.1. If β is empty or β = 1 a , then H = 0 by Lemma 4.3. Therefore,we may assume that β has at least one 1 and one 0.2. If β = 1 γ β must end in a 1), then the best moves, in bothsummands, are pairs of coins in β and − β . If each player copiesthe opponent’s move in the other summand, then this leads to α β − α β → α β ′ − α β ′ and the latter expression is equal to 0, by induction.3. If β = 1 γ 1, then β = 0 γ γ β and − β and are the best responses to each other.We then derive that α β − α β → α β ′ − α β ′ = 0 , by induction.14n all cases H = 0, and this concludes the proof.In Lemma 4.4, the conditions are necessary. An example is:3 / = 1001 = 1 / . Here, β starts with a 0 and has an odd number of 1s.These reduction lemmas are important in evaluating a position. Thereduced positions will end in 011 or 01. By considering the exact endof the string, specifically, if there are at least two 0s (in one specialcase three 0s), then we can find an ordinal sum decomposition. Thedecomposition is determined by where the third topmost 1 is situated.The next result is the start of the ordinal sum decomposition of aposition. The exponent is the value of the Right option of the substringbeing removed. Lemma 4.5. If a ≥ and p and q are non-negative integers such that p + q ≥ , then α a p q α a : 12 p + q − . Proof. We prove that α a p q − (cid:18) α a : 12 p + q − (cid:19) = 0 . Note that in Theorem 2.3 we have that playing in the base of α a : p + q − is worse than playing in the exponent. We have two cases toconsider. Case 1: Left plays in the first summand and Right in the second. Righthas a move in the exponent (moves to 0) since 2 p + q − ≥ α a − ( α a : 0) = α a − α a = 0 . Case 2: Right plays in the first summand and Left plays in the secondsummand.Consider 15 a p q − (cid:18) α a : 12 p + q − (cid:19) . We have two sub-cases.(a) Assume 2 p + q − = 0. After the two moves we have the position α a r s − (cid:18) α a : 12 p + q − (cid:19) , where 2 r + s = 2 p + q − . By induction, we have that α a r s α a : 12 r + s − = α a : 12 p + q − . Thus, α a r s − (cid:0) α a : p + q − (cid:1) = 0.(b) Assume 2 p + q − q = 1, p = 0. The original positionis α a − ( α a : 1) . After the two moves we have the position α a − α a − (note thatLeft has no move in the exponent). By Lemma 4.3, α a 11 = α a − .Hence, we have that α a − α a − = 0 and the result follows.The values of the positions not covered by Lemma 4.5 are given next. Lemma 4.6. Let a , p , and q be non-negative integers. We then havethat p − p, and a p q j a k + 12 p + q . Proof. Let G = 0 p 1. Left has no moves and Right has p . Note that in1 a , Left has ⌊ a ⌋ moves and Right has none.Now, let G = 1 a p q 1. We proceed by induction on p + q . In allcases, Left’s move is to 1 a , that is, to ⌊ a ⌋ . If p = 0 and q = 0 then G = 1 a 11, which has value ⌊ a ⌋ + = ⌊ a ⌋ + 1. Assume that p + q = k ,16 > 0. If q > 0, then G = {⌊ a ⌋ | a p q − } . By induction, we havethat G = (cid:26)j a k (cid:12)(cid:12)(cid:12) j a k + 12 p + q − (cid:27) = j a k + 12 p + q . If q = 0, then G = {⌊ a ⌋ | a p − } . By induction, we have that G = (cid:26)j a k (cid:12)(cid:12)(cid:12) j a k + 12 p − (cid:27) = j a k + 12 p , and the result follows. We now have all of the tools to prove Theorem 3.3. Proof of Theorem 3.3. Let G be a flipping coins position. Step 2reduces the string of coins. The reductions in Step 2(a) are those ofLemma 4.3 and Lemma 4.4 part(a). The reductions in Step 2(b) arethose of Lemma 4.4 part(b). In all cases, these lemmas show that eachnew reduced position is equal to G .In Step 3, we claim G i = β for any β . This is true for i = 0 byLemma 4.3. If i > 0, then at each iteration of Step 3, the last two1s are removed from G i − . Now, the original reduced position wouldbe G = β γ , where γ has an even number of 1s. Lemma 4.4 part(b)would apply eliminating the three consecutive 1s. Now either G i is oneof 0 r r ≥ a p i q i a ≥ p i + q i ≥ 0, or G i = α a p i q i p i + q i ≥ a > 0. In the latter case, the index is incremented and thealgorithm goes back to Step 3.Step 5 applies when Step 3 no longer applies, i.e., G i is one of 0 r r ≥ a p i q i a ≥ p i + q i ≥ 0. Now, v i is the value of G i ,as given in Lemma 4.6.Lemma 4.5 shows that for each j < i , G j = G j +1 : pj + qj , theevaluation in Step 6. Thus, the value of G is v , and the theoremfollows.The question: “Who wins 0101011111+1101100111+0110110110111and how?” from Section 1 can now be answered.17irst, we have that0101011111 = 01011011 = (cid:18) (cid:19) = (cid:18)(cid:18) 01 : 12 (cid:19) : 12 (cid:19) = (cid:18)(cid:18) − (cid:19) : 12 (cid:19) = − (cid:18) 12 : 1 (cid:19) = 340110110110111 = 0110110111 = 0110111 = 0111 = 0 . Thus, we have that0101011111 + 1101100111 + 0110110110111 = − . Left’s only winning move is to01010111 + 1101100111 + 0110110110111 = − 34 + 34 + 0 = 0 . Her best moves in the second position gives a sum of − + +0 = − ,and in the third yields − + − = − . Left loses both times. 5. Open Problems Natural variants of flipping coins involve increasing the number ofcoins that can be flipped from two to three or more. A brief computersearch suggests that the only version where the values are numbers isthe game in which Left flips a subsequence of all 1s and Right a sub-sequence of 0s ended by a 1. We conjecture that a similar ordinal sumstructure will arise in these variants. Other variants have values thatinclude switches, tinies, minies, and other three-stop games. However,some variants, when the reduced canonical values are considered, onlyseem to consist of numbers and switches. A more thorough investiga-tion should shed light on their structures.Instead of playing on a string, we can play on a directed acyclic graph(or DAG). We know that the impartial game twins , played on a DAG,18educes to a simple multi-heap game. Of course, instead of coins, otherobjects could be used. In a line of dice, for example, the end dice wouldbe turned to a lower number. References [1] M. H. Albert, R. J. Nowakowski and D. Wolfe. Lessons in Play:An Introduction to Combinatorial Game Theory . 1st edition, 2007;2nd edition, 2019. MA: A K Peters.[2] E. R. Berlekamp, J. H. Conway and R. K. Guy. Winning ways foryour mathematical plays . , (2nd ed.), MA: A K Peters, Ltd.2001.[3] A. Carvalho, M. A. Huggan, R. J. Nowakowski, C. P. dos Santos. A Note on Numbers , Preprint 2021.[4] A. Carvalho, M. A. Huggan, R. J. Nowakowski, C. P. dos Santos. Ordinal Sums, clockwise hackenbush , and domino shave ,Preprint 2021.[5] J. H. Conway. The Golay Codes and The Mathieu Groups. In:Sphere Packings, Lattices and Groups. Grundlehren der mathe-matischen Wissenschaften (A Series of Comprehensive Studies inMathematics), vol 290. (1983) Springer, New York, NY.[6] R. T. Curtis. Eight octads suffice.