Bounds on Zeckendorf Games
Anna Cusenza, Aiden Dunkelberg, Kate Huffman, Dianhui Ke, Micah McClatchey, Steven J. Miller, Clayton Mizgerd, Vashisth Tiwari, Jingkai Ye, Xiaoyan Zheng
aa r X i v : . [ m a t h . N T ] S e p BOUNDS ON ZECKENDORF GAMES
ANNA CUSENZA, AIDEN DUNKELBERG, KATE HUFFMAN, DIANHUI KE,MICAH MCCLATCHEY, STEVEN J. MILLER, CLAYTON MIZGERD, VASHISTH TIWARI,JINGKAI YE, AND XIAOYAN ZHENG
Abstract.
Zeckendorf proved that every positive integer n can be written uniquely as thesum of non-adjacent Fibonacci numbers. We use this decomposition to construct a two-player game. Given a fixed integer n and an initial decomposition of n = nF , the two playersalternate by using moves related to the recurrence relation F n +1 = F n + F n − , and whoevermoves last wins. The game always terminates in the Zeckendorf decomposition; depending onthe choice of moves the length of the game and the winner can vary, though for n ≥ thereis a non-constructive proof that Player 2 has a winning strategy.Initially the lower bound of the length of a game was order n (and known to be sharp)while the upper bound was of size n log n . Recent work decreased the upper bound to ofsize n , but with a larger constant than was conjectured. We improve the upper bound andobtain the sharp bound of √ n − IZ ( n ) − √ Z ( n ) , which is of order n as Z ( n ) is thenumber of terms in the Zeckendorf decomposition of n and IZ ( n ) is the sum of indices in theZeckendorf decomposition of n (which are at most of sizes log n and log n respectively). Wealso introduce a greedy algorithm that realizes the upper bound, and show that the longestgame on any n is achieved by applying splitting moves whenever possible. Contents
1. Introduction 12. Strategy to achieve the longest game 33. Upper Bound on the Game Length 84. Future Work 14References 141.
Introduction
The Fibonacci numbers are one the most interesting and famous sequences. Among their fas-cinating properties, the Fibonacci numbers lend themselves to a beautiful theorem by EdouardZeckendorf [Ze] which states that each positive integer n can be written uniquely as the sum ofdistinct, non-consecutive Fibonacci numbers. This sum is called the Zeckendorf decomposition of n and requires that we define the Fibonacci numbers by F = 1 , F = 2 , F = 3 , F = 5 ... instead of the usual , , , , ... to create uniqueness. Baird-Smith, Epstein, Flint and Miller[BEFM1, BEFM2] create a game based on the Zeckendorf decomposition. We quote from[BEFM2] to describe the game.We introduce some notation. By { n } or { F n } we mean n copies of , the first Fibonaccinumber. If we have 3 copies of F , 2 copies of F , and 7 copies of F , we write either { F ∧ F ∧ F } or { ∧ ∧ } . Date : September 22, 2020.
TBD 1HE FIBONACCI QUARTERLY
Definition 1.1 (The Two Player Zeckendorf Game) . At the beginning of the game, there is anunordered list of n F = 1 , F = 2 , and F i +1 = F i + F i − ; therefore the initial list is { F n } . On each turn, a player can do one of the following moves. (1) If the list contains two consecutive Fibonacci numbers, F i − , F i , then a player canchange these to F i +1 . We denote this move { F i − ∧ F i → F i +1 } . (2) If the list has two of the same Fibonacci number, F i , F i , then (a) if i = 1 , a player can change F , F to F , denoted by { F ∧ F → F } , (b) if i = 2 , a player can change F , F to F , F , denoted by { F ∧ F → F ∧ F } ,and (c) if i ≥ , a player can change F i , F i to F i − , F i +1 , denoted by { F i ∧ F i → F i − ∧ F i +1 } .The players alternate moving. The game ends when one player moves to create the Zeckendorfdecomposition. The moves of the game are derived from the Fibonacci recurrence, either combining termsto make the next in the sequence or splitting terms with multiple copies. A proof that thisgame is well defined and ends at the Zeckendorf decomposition can be found in [BEFM2].We introduce some further notation and state some simple results. • Let i max ( n ) be the largest index of terms in the Zeckendorf decomposition of n . Theorder of i max ( n ) is at most log n ; this follows immediately from the exponential growthof the Fibonacci numbers, as we can never use a summand larger than the originalnumber n . • Let δ i denote the number of F i ’s in the Zeckendorf decomposition of n . Then n = P i max ( n ) i =1 δ i F i . • Let Z ( n ) denote the number of terms in the Zeckendorf decomposition of n , and Z ( n ) = P i max ( n ) i =1 δ i . The order of Z ( n ) is at most log n since Z ( n ) ≤ i max ( n ) . • Let IZ ( n ) denote the sum of indices in the Zeckendorf decomposition of n , and IZ ( n ) = P i max ( n ) i =1 i δ i . The order of IZ ( n ) is at most log n ; this follows trivially from summingthe indices and recalling the largest index used is of order log n . • The original upper bound for the game was of order n log n , and the lower bound wasfound to be sharp at n − Z ( n ) in [BEFM2]. The upper bound on the number of moveswas improved to n − Z ( n ) − IZ ( n ) + 1 in [LLMMSXZ]. Since the order of Z ( n ) and IZ ( n ) are both less than n , we observe that the upper and lower bounds are both oforder n . • Finally, several deterministic games have been introduced in [LLMMSXZ]. These aredefined in terms of the priority of moves; that is, each move in a strategy will followwhichever move is available and comes first in the ordering of moves. – Combine Largest: adding consecutive indices from largest to smallest, adding 1’s,splitting from largest to smallest. – Split Largest: splitting from largest to smallest, adding consecutive indices fromlargest to smallest, adding 1’s.2 VOLUME, NUMBEROUNDS ON ZECKENDORF GAMES – Split Smallest: splitting from smallest to largest, adding 1’s, adding consecutiveindices from smallest to largest.It was shown in the same paper that the Combine Largest and Split Largest gamesboth realize the shortest game.Since the lower bound of the game has been shown to be sharp, we focus on the upper boundof the game. One of our main result is a proof of a conjecture from [BEFM2] that the longestgame on any n is achieved by applying splitting moves whenever possible. Theorem 1.2.
The longest game on any n is achieved by applying split moves or combine 1’s(in any order) whenever possible, and, if there is no split or combine 1 move available, combineconsecutive indices from smallest to largest. This algorithm is not deterministic. Thus, there are many game paths that follow thisalgorithm. For instance, it can be easily shown that the Split Smallest game described in[LLMMSXZ] is a deterministic example of this algorithm, and therefore realizes the longestgame.Now that we have an algorithm that achieves the longest game, we are interested in theupper bound of the game length. The previous upper bound was already very close to theknown lower bound (both of order n ). Nevertheless, we are able to further close the gap. Theorem 1.3.
Let a i = F i +2 − i − i > , i ∈ N ) . The upper bound of the game is given by P i max ( n ) i =1 a i δ i which is at most √ n − IZ ( n ) − √ Z ( n ) . It was originally conjectured in [BEFM2] that the number of moves in the Split Smallestgame grows linearly with n , with a constant of the golden mean squared, which is equivalentto √ . We observe that this conjecture has been shown here, since the order of Z ( n ) and IZ ( n ) are less than n .Though this bound is very close to the actual longest game, it is not a strict upper boundfor most n . In fact, we observed during the construction of this upper bound that this boundis sharp if and only if the game on n can be played with only split and combine 1 moves, andidentified all such n . Theorem 1.4.
A game can be played with only splitting and combine 1 moves, if and only if n = F k − k ≥ . In Section 2 we prove Theorem 1.2. Then in Section 3 we prove Theorems 1.3 and 1.4.Finally in Section 4 we give some possible directions for future research.2.
Strategy to achieve the longest game
We start by introducing some notation that we use in our proofs.Following the notation introduced in [LLMMSXZ], we let
M C i denote the number of com-bining moves at the index i with i ≥ , with M C the number of combine 1 moves. Similarlythe number of splitting moves at i is denoted M S i for i ≥ . We refer to combining moves at i by C i , and splitting moves at i by S i .The move of adding 1’s is usually considered a combining move as the case in [BEFM1,BEFM2, LLMMSXZ], but for the sake of our proofs, we consider combine 1 ( C ) to be asplitting move, and also refer to it as S in this section.We begin with the proof of Theorem 1.2, which is a greedy algorithm that achieves thelongest game path, starting with two lemmas.TBD 3HE FIBONACCI QUARTERLYAs a reminder, the algorithm requires moves to be in the following order: choose any splitor combine move whenever possible, then combine consecutive terms with smallest indices. Lemma 2.1.
If the aforementioned strategy gives us choice of moves at some game state G ,then starting from this game state, no matter which move we choose at this step, there exists apath that follows our strategy and has the same length as our initial path.Proof. In paths following our strategy, the only game states that allow choice of moves aregame states with at least two splitting (including S ) moves available.Let P be a game path starting from any game state that follows our strategy, and let G beany game state that P visits that allows a choice. Let P ′ be another path that starts from thesame game state as P and follows our strategy such that P ′ follows the same moves as P untilthey differ in choice of move for the first time at G .Let P choose S i and P ′ choose S j at G . We want to construct the steps for P ′ such that ithas same length as P .Since P ′ could have chosen S i , we must have at least two F i at G . Since S j does not decreasethe number of F i , we still have at least two F i after this step. Thus for the next step, P ′ canstill choose S i . By similar reasons, P ′ can always imitate the moves that P takes after S i until P takes a S j . We know that P must take S j at some step because we had at least two F j at G ,and the only moves that decrease the number of F j ’s are S j , C j , and C j +1 . Since our strategyprioritizes split moves over combine moves, we must take S j at some point in P .Thus the moves in P ′ after G are as follows: perform S j and S i in the first two steps,and then imitate the moves of P after S i until P takes S j . Since P follows our strategy ofprioritizing split moves, P ′ does also. In this way, P and P ′ take the same set of steps but indifferent order, so they reach the same game state with the same number of steps. After that, P and P ′ follow exactly same steps until game terminates.Thus, we prove that no matter which move we choose at some game state G , there exists apath that follows our strategy and has same length as our initial path. (cid:3) Lemma 2.2.
Starting from any game state, all paths that follow our strategy have the samelength.Proof.
We show that an arbitrary game path P that follows our strategy is no longer or shorterthan any other path that follow our strategy, given that they start from same game state.Suppose for the sake of contradiction that P and P ′ both follow our strategy but differ inlength. Then P and P ′ must differ by at least move. Let G be the first game state where P and P ′ differ in choice of moves. By Lemma 2.1, there exists a path P that chooses the samemove as P ′ at G , has the same length as P , and follows our strategy.Since P and P ′ differ in length, they must differ by at least one move. Thus there existsa game state G after G where P and P ′ differ in their choice for the first time. Again, byLemma 2.1, there exists a path P that chooses the same move as P ′ at G , has the samelength as P (which is equal to length of P ), and follows our strategy.Since the number of steps in any game path is finite, we can repeat this process until wefind a P k with the same length as P , but there no longer exists G k where P ′ and P k can differin choice of moves. Thus, the rest of steps are deterministic, which means P k and P ′ must beexactly the same path, and therefore both have the same length as P , a contradiction.In conclusion, all paths that start from same game state and follow our strategy have samelength. (cid:3) Lemma 2.3.
Starting from any game state, if a path does not follow our strategy, then thisgame path is either not the longest path or there exists a path that has the same length as thispath and follows our strategy.Proof.
Suppose a game path P contains at least one step that is not chosen by our strategy.We want to either find a path P ∗ that is longer than P , or construct a P ∗ that follows ourstrategy and is as long as P .We look at the last step in P that is not chosen by our strategy. Suppose the step is takenat game state G . Since we consider the last step that does not follow our strategy, all movesafter this step must follow our strategy.There are two situations when a step is not chosen by our strategy: either the combining movetaken is not the one with smallest index when no splitting move is available, or a combiningmove is taken when there is splitting move available. Note that in splitting move we includecombine ’s moves, and in combining move we exclude combine ’s. We look at these two cases.In both cases we want to find a path P ′ that is either longer than P or follows our strategy atand after G and have same length as P .First, suppose P takes a combining move that is not the smallest when no splitting move isavailable at G . In this case, there is at most one F k for any k . Let C i ( i ≥ ) be the smallestcombining move at G and let C j ( j > i ) be the combining move chosen by P at G . We studythe following sub-cases based on whether G contains a F i − term. • Case 1.1. At G , there is at least one F j − .In this case, we find a path P ′ that is longer than P . Let P ′ take the same moves as P before reaching G . At G , path P takes C j at G and reaches G ′ . Compared to G , G ′ contains one less F j − , one less F j , and one more F j +1 . Let path P ′ take C j − and S j .Then the game state it reaches also contains one less F j − , one less F j , and one more F j +1 compared to G . Thus P ′ also reaches G ′ with one more step than P . After that, P ′ can imitate the moves P takes until game terminates. In the end, P ′ is one movelonger than P . • Case 1.2. At G , there are no F j − .Again, let P ′ follow same steps as P and reach the game state G . At G , let P take C j and P ′ take C i . After that, P follows our strategy.First, we look at the steps in P . C j increases number of F j +1 by one, so if there isone F j +1 at G , we apply S j +1 . This is the only possible splitting move at this gamestate since we assumed there are no F j − . Similarly, S j +1 increases F j +2 by one, so ifthere is one F j +2 at G , we apply S j +2 . Since we have used up all F j − ’s with C j , thereare no F j − ’s before taking S j +1 , so we cannot apply S j − . Thus S j +2 should be theonly possible splitting move at this game state.We repeat the process of taking the only splitting move until we have to do a com-bining move. Note that the number of such splitting moves that can be taken by P after taking C j and before taking another combining move depends on the number ofconsecutive F k ’s ( k > j ) starting from F j +1 . Let α ≥ be the number of such moves.The move taken in P after these α steps is C i since there is no splitting move availableand C i is the smallest combining move.Now P has taken α + 1 steps and P ′ has taken 1 step. The current game statefor P and P ′ has the same number of F k ’s for all k ≤ j − since S j and the α splittingmoves in P do not affect the number of F k ’s ( k ≤ j − ), and P and P ′ both take a C i move. Thus, they can follow the same steps until either we get a F j − before we haveTBD 5HE FIBONACCI QUARTERLYto take C j in P ′ or we never get a F j − and the next step in P ′ is C j . Suppose we took β steps before we stop. We look at these two cases. – Case 1.2.1.
We get a F j − before we have to take C j in P ′ .This case is similar to Case 1.1 where we find a path longer than P . In P ′ we take C j − , S j , and then follow the α steps described in P . After that, P and P ′ reachthe same game state. Notice that after G , path P took α + 1 + β steps, and P ′ took β + 2 + α steps. Thus, by the end of the game, P ′ is one move longerthan P . – Case 1.2.2.
We never get a F j − before the next step in P ′ is C j .In this case P ′ takes the C j move and then follows the α steps described in P .Then P and P ′ reach the same game state, and share the same steps after that.Notice that the sub-paths of P and P ′ after G follow our strategy, and P ′ has thesame length as P .Second, suppose P chooses a combining move when there is a splitting move available. Let C i ( i ≥ ) be the combining move that P chooses. Then there must be at least one F i − andone F i at G . We consider the following cases based on the number of F i − ’s and F i ’s. • Case 2.1.
There are more than one F i − ’s at G (i.e., P ′ can take the move S i − ). – Case 2.1.1. i = 2 (i.e., path P takes C ).At G , path P takes C , and path P ′ takes S and S to reach the same game state.After that P ′ imitates the steps P takes. In the end, P ′ is one move longer than P . – Case 2.1.2. i = 3 (i.e., path P takes C ).At G , path P takes C , and path P ′ takes S , S , S to reach the same game state.After that P ′ imitates the steps P takes. In the end, P ′ is two moves longer than P . – Case 2.1.3. i > .At G , path P takes C i , and path P ′ takes S i − , S i , C i − to reach the same gamestate. After that P ′ imitates the steps P takes. In the end, P ′ is two moves longerthan P . • Case 2.2.
There is exactly one F i − and more than one F i at G (i.e., P ′ can take S i ). – Case 2.2.1. i = 2 (i.e., path P takes C ).At G , path P takes C , and path P ′ takes S and S to reach the same game state.After that P ′ imitates the steps P takes. In the end, P ′ is one move longer than P . – Case 2.2.2. i > .At G , path P takes C i , and path P ′ takes S i and C i − to reach the same gamestate. After that P ′ imitates the steps P takes. In the end, P ′ is one move longerthan P . • Case 2.3.
There is exactly one F i − and one F i at G .Let G ′ be the game state P reaches after taking C i at G . Since G is the last gamestate where our strategy is violated, all steps in P after G follows our strategy. Weproved in Lemma 2.2 that starting from any game state, any game path that followsour strategy has same length. Thus, there exists a path that starts from G ′ , performs S i +1 only when no other splitting move is available, and has the same length as the6 VOLUME, NUMBEROUNDS ON ZECKENDORF GAMESsub-path of P starting from G ′ . We extend this path so that it starts from the samegame state as P and follows the same steps as P until G ′ . Call the extended path P ′′ .Note that the only differences between G and G ′ are that G ′ has one more F i +1 term, one less F i term and one less F i − term. If we let P ′ follow the same moves after G that P does after G ′ , then the game states in P ′ and P always differ in only thesethree terms. Thus, P ′ may be unable to imitate P if P performs S i +1 .We avoid this problem to the maximum extent by letting P ′ follow the same stepsin P ′′ (instead of P ) until either P ′′ is forced to take S i +1 and P ′ cannot choose thesame step, or P ′ performs all splitting moves before P ′′ takes a combining move. Inboth situations, let G ′′ be the game state reached by P ′′ here and consider the nextstep in P ′ . The next step is either a splitting move or a combining move (there mustexist a next step since we can always take C i ). Notice that if there is a splitting movepossible, it is either S i − or S i since F i − and F i are the only two terms which gamestates in P ′ have more of than game states in P ′′ . – Case 2.3.1.
The next step in P ′ can be S i − or S i .Suppose the next step taken in P ′ is C i . If the step in P ′′ before reaching G ′′ is S i +1 and P ′ failed to follow this move, P ′ takes S i +1 after the C i move. Then P ′ reaches G ′′ with same number of steps as P ′′ . However, according to Case 2.1 and
Case 2.2 , we can find a path that is longer than P ′ . Thus there exist a pathlonger than P . – Case 2.3.2.
The next step in P can only be a combining move.Again, suppose the next step taken in P ′ is C i (and S i +1 if P ′ failed to follow the S i +1 move in P ′′ ). Then, it reaches G ′′ with same number of steps as P ′′ . Let P ′ follow the same moves in P ′′ after G ′′ .If C i was not the smallest possible combining move P ′ could have taken, then byour discussion about the case where the smallest combining move is not chosen,we know that there exists either a path longer than P ′ or a path that has samelength as P ′ and its sub-path (after G ) follows our strategy.If C i was the smallest possible combining move, then this step followed our strategy.Thus P ′ is a path that has the same length as P and whose sub-path (after G )follows our strategy.In both cases, we can either find a path that is longer than P or a path whose sub-pathafter G follows our strategy and have same length as P .Now we can start to construct a path P ∗ that is either longer than P or has same length as P and follows our strategy.Since the game takes finitely many steps, there are finitely many game states in P that donot follow our strategy in choosing moves. We denote all such game states in P in reversechronological order as G , G , . . . , G k , where G is the last game state where a move is notchosen in accordance with our strategy and G k is the first.We start by looking at the path after G . Since our strategy is not followed at G , it mustfall into either of the two cases discussed above. In both cases, we can either find a path thatis longer than P or find a path whose sub-path after G always follow our strategy and is aslong as P .If we find a path that is longer than P , then this is the P ∗ that we are looking for. Otherwise,we denote the path whose sub-path after G follows our strategy as P . Notice that P and P follow same moves before reaching G , and P follows our strategy at G , so the last gamestate in P where our strategy is violated is G .TBD 7HE FIBONACCI QUARTERLYBy the same argument, we can either find a path that is longer than P or find a path whosesub-path after G always follow our strategy and is as long as P . If we find a path that islonger than P , then this is the P ∗ that we are looking for. Otherwise, we find a P whosesub-path after G follows our strategy and has same length as P .We repeat this process until either we find a P ∗ that is longer than P , or we find a P k whosesub-path after G k follows our strategy and has the same length as P . Since G k is the firstgame state in P where our strategy is not followed, P k is a path that follows our strategy fromthe starting state. Thus P k is the P ∗ we want.In conclusion, starting from any game state, if there is a path P that does not follow ourstrategy, then it is either not the longest game or there exists a path P ′ that is as long as P and follows our strategy. (cid:3) Proof of Theorem 1.2.
By Lemma 2.3, we proved that a path that does not follow our strategyis either not the longest game, or there exists a path that follows our strategy and is as longas the original path. By Lemma 2.2, we know that all paths that start from same game stateand follow our strategy have the same length. Thus our strategy gives the longest game. (cid:3) Upper Bound on the Game Length
In this section, we construct and analyze the order of an upper bound on the game length.
Proof of Theorem 1.3.
The first step is to construct an upper bound on the game length. To dothis, we first look at changes in the amount of F . We start the game with no F , and end with δ of F . Every time we combine two F ’s, we get a ( F ∧ F → F ) and increase the numberof F ’s by one. Every time we split two F ’s, we get a ( F ∧ F → F ∧ F ) and increasethe number of F ’s by one. These are the only moves that increase the number of F ’s. Eachcombining move of F and F ( F ∧ F → F ) and combining move of F and F ( F ∧ F → F )decreases the number of F ’s by one. Finally splitting two F ’s ( F ∧ F → F ∧ F ) decreasesthe number of F ’s by two. These are the only moves that decrease the number of F ’s.Thus we can construct the following equation: M C − M S + M S − M C − M C = δ . (3.1)Similarly, for each ≤ i ≤ i max ( n ) , we start the game with no F i , and end with δ i of the F i . Every time we combine F i − and F i − , we increase the number of F i ’s by one. Every timewe split two F i − ’s, we increase the number of F i ’s by one. Every time we split two F i +2 ’s,we increase the number of F i ’s by one. These are the only moves that increase the number of F i ’s. Each combining move of F i − and F i and combining move of F i and F i +1 decreases thenumber of F i ’s by one. Finally splitting two F i ’s decreases the number of F i ’s by two. Theseare the only moves that decrease the number of F i ’s.Thus we have for ≤ i ≤ i max ( n ) M S i − − M S i + M S i +2 + M C i − − M C i − M C i +1 = δ i . (3.2)Since i max ( n ) is the largest index in the final decomposition, we know that for all i ≥ i max ( n ) , M S i = M C i = 0 . Thus for i ≥ i max ( n ) − , we can get rid of a few terms in the equationabove: M S i max ( n ) − − M S i max ( n ) − + M C i max ( n ) − − M C i max ( n ) − − M C i max ( n ) − = δ i max ( n ) − , (3.3) M S i max ( n ) − − M S i max ( n ) − + M C i max ( n ) − − M C i max ( n ) − = δ i max ( n ) − , (3.4) M S i max ( n ) − + M C i max ( n ) − = δ i max ( n ) . (3.5)8 VOLUME, NUMBEROUNDS ON ZECKENDORF GAMESNow we have i max ( n ) − linear equations with · i max ( n ) − variables, we write the systemof equations in matrix form. · · · · · · · · · · · · · · · · · · · · · ... ... · · · ...0 0 0 0 · · · · · · · · · · · · M C M S M S ... M S i max ( n ) − M C ... M C i max ( n ) − = δ δ δ ... δ i max ( n ) − δ i max ( n ) (3.6)Let M be the ( i max ( n ) − × (2 · i max ( n ) − matrix shown in (3.6). We express each entry m i,j in M explicitly: m i,j = if j = i − if j = i + 1 and j ≤ i max ( n ) − if j = i + 3 and j ≤ i max ( n ) − if j = i + i max ( n ) − and j ≥ − if j = i + i max ( n ) − and j ≤ · i max ( n ) − − if j = i + i max ( n ) and j ≤ · i max ( n ) − otherwise. (3.7)Let A be the left ( i max ( n ) − × ( i max ( n ) − sub-matrix of M . Notice that A is uppertriangular and has 1’s in the diagonal, so it is invertible.To solve the equation in (3.6), we write M in reduced row-echelon form: · · · · · · · · · · · · · · · · · · · · · ... ... · · · ...0 0 0 0 · · · · · · · · · · · · . (3.8)Thus the solutions for the system of equations are in the form: M C M S M S ... M S i max ( n ) − M C M C ... M C i max ( n ) − = A − δ δ δ ... δ i max ( n ) + M C -1-10...010...0 + M C -1-1-1...001...0 + · · · + M C i max ( n ) − -1-1-1...-100...1 . (3.9)TBD 9HE FIBONACCI QUARTERLYThe length of game path is the sum of all M C and
M S terms, thus can be written as h · · · · · · i M C M S M S ... M S i max ( n ) − M C M C ... M C i max ( n ) − (3.10)which is equal to (cid:2) · · · (cid:3) A − δ δ δ ... δ i max ( n ) − M C − M C − · · · − ( i max ( n ) − M C i max ( n ) − . (3.11)We calculate A − with Gauss-Jordan elimination, and let a i,j be the i, j -th entry of A − : A − = · · · a ,i max ( n ) − a ,i max ( n ) − · · · a ,i max ( n ) − a ,i max ( n ) − · · · a ,i max ( n ) − a ,i max ( n ) − ... · · · ... · · · · · · (3.12)where a i,j = 2 a i,j − − a i,j − for all ≤ i ≤ i max ( n ) − , ≤ j ≤ i max ( n ) − . Also observethat a i +1 ,j +1 = a i,j .We claim that a ,j = F j +1 − , and prove this with induction. First, we see that a , = F − , a , = F − , a , = F − − , so this claim holds for j = 1 , , . Thensuppose this claim holds for all j ′ < j , consider a ,j . a ,j = 2 a ,j − − a ,j − = 2 F j − F j − − F j + F j − + F j − − F j − − F j +1 − . (3.13)By induction, our claim holds for all j .Then we calculate (cid:2) · · · (cid:3) A − : (cid:2) · · · (cid:3) A − = (cid:2) · · · (cid:3) F − F − · · · F i max ( n ) − F i max ( n ) F − · · · F i max ( n ) − F i max ( n ) − ... · · · ... · · · F − = " · · · i max ( n ) − X i =2 ( F i − i max ( n ) X i =2 ( F i − . (3.14)10 VOLUME, NUMBEROUNDS ON ZECKENDORF GAMESWe add an extra 0 before the sequence of , , , . . . and express it explicitly as a = 0 , a j = j X i =2 ( F i − for j ≥ . Then we find a formula for j ≥ : a j = j X i =2 ( F i −
1) = j X i =1 F i − j = ( F j −
1) + ( F j +1 − − j = F j +2 − j − . (3.15)Since a = 0 = F − − is consistent with this formula, we conclude that a j = F j +2 − j − for all j .Now the game length given in (3.11) becomes i max ( n ) X j =1 a j δ j − M C − M C − · · · − ( i max ( n ) − M C i max ( n ) − . (3.16)Since all M C terms are non-negative, we ignore them for the upper bound. Thus the upperbound on game length is i max ( n ) X j =1 a j δ j = a δ + a δ + a δ + · · · + a i max ( n ) δ i max ( n ) . (3.17)This is the bound we claimed in Theorem 1.3.We now analyze the order of this bound.We show that a j ≤ √ F j − j − √ using Binet’s formula F j = 1 √ √ ! j +1 − − √ ! j +1 (3.18)and the formula a j = F j +2 − j − . Thus, a j = F j +2 F j F j − j − √ (cid:18)(cid:16) √ (cid:17) j +3 − (cid:16) −√ (cid:17) j +3 (cid:19) √ (cid:18)(cid:16) √ (cid:17) j +1 − (cid:16) −√ (cid:17) j +1 (cid:19) F j − j −
2= 3 + √ F j + − √ ! j +1 − j − . (3.19)Since − < −√ < , we have (cid:16) −√ (cid:17) j +1 ≤ (cid:16) −√ (cid:17) = −√ , so a j ≤ √ F j + 3 − √ − j − √ F j − j − √ . (3.20)TBD 11HE FIBONACCI QUARTERLYTherefore we have i max ( n ) X j =1 a j δ j ≤ i max ( n ) X j =1 √ F j − j − √ ! δ j = √ i max ( n ) X j =1 F j δ j − i max ( n ) X j =1 j δ j − √ i max ( n ) X j =1 δ j = √ n − IZ ( n ) − √ Z ( n ) . (3.21)In conclusion, the game length is at most √ n − IZ ( n ) − √ Z ( n ) . (cid:3) In this proof, equation (3.16) is the exact game length and (3.17) is the bound we gave inTheorem 1.3. From these two equations, we observe that our upper bound is strict if and onlyif the game on n can be played with only splitting and combine 1 moves. Remark 3.1.
The same method can be used to calculate game length even if the game doesnot start from all ’s by replacing δ i ’s with the difference in number of F i ’s between startingstate and final state at each position. Also if we take the M S i terms as free variables insteadof pivot variables, we can use this method to calculate the lower bound of the game. We move on to the proof of Theorem 1.4. The reason we are interested in games that canbe played with only splitting moves is to identify for which n is our upper bound in Theorem1.3 strict. Lemma 3.2. If n = F k − k ≥ , then we can play the game with only split and combine1 moves (starting from any game state).Proof. To prove this, we first prove that any game state (except the final one) in the game on n = F k − k ≥ has at least two F i ’s for some i .Suppose for the sake of contradiction that there is a game state that has at most one ofany F i and is not the final game state. Since this game state is not the final state, there aremoves that we can apply. Since this game state has at most one of any F i , we cannot applyany splitting moves. Thus, there must be some combining moves available. We apply thecombining move with the largest index, say F i − ∧ F i → F i +1 . Note that we cannot have F i +1 in this game state, or we would have chosen to combine F i and F i +1 . Thus after this move,we still have at most one of any F i . We repeat this process until we reach the final state, andeach game state we visit has at most one of any F i .Now we consider the Zeckendorf Decomposition of n = F k − . It has to be in the form of F + F + F + · · · or F + F + F + · · · because if we add to these decompositions, we canget a Fibonacci number. By assumption, we reach the final state with a combining move. Let F j − ∧ F j → F j +1 be the last step we took. Since there is a F j +1 in the final decomposition of n ,there must also be a F j − . Thus we had two F j − ’s before the last step we took. However, wejust showed that each game state we visit has at most one of any F i , which is a contradiction.Now that we have proved that any game state (except the final one) in the game on n = F k − k ≥ has at least two F i ’s for some i , we know that we can apply a splitting orcombine move at any game state until the game terminates. Thus, the game can be playedwith only splitting and combine moves. (cid:3) Lemma 3.3.
If we can play the game with only splitting and combining moves, then n canonly be in the form F k − k ≥ .
12 VOLUME, NUMBEROUNDS ON ZECKENDORF GAMES
Proof.
We consider all possible n that are not in the form of F k − k ≥ and divide theminto three cases based on their final decomposition. In each case, we prove that the game on n cannot be played with only splitting moves. Case 1 : The smallest term in the final decomposition of n is at least F .Suppose that the game on n can be played with only splitting and combine moves. Sincethe smallest term in n ’s final decomposition is at least F , we have to generate at least one F at some point of the game. Thus there must exist split moves in the game. Let t be the lastsplit move in the game.At step t , one F is generated. Since there are no F in the final decomposition, we have todo a combine move to decrease the number of F ’s, producing a F with this move. Sincethere is no F in the final decomposition, we have to do a split move, which contradicts ourassumption that step t is the last split move. Case 2 : The first two terms in the final decomposition of n are F and F .Suppose that the game on n can be played with only splitting and combine moves. Sincewe have F in the final decomposition, we have to generate a F at some point of the game.Thus there exist split moves in the game. To do the split move, we need a F , so there mustexist a split move. Let step v be the last split move in the game.At step v , one F and one F is generated. Since there is no F in the final decomposition,there exists a split move after step v , and we denote it step t . Then in step t , there is one F generated.Since both step v and t generates F , there exists a game state after step v that contains F .Therefore, there exists a combine move after step v . With this move, one F is generated.Since there is no F in the final decomposition, we have to do a split move to get rid of this F which contradicts our assumption that v is the last split move. Case 3 : The smallest term of n ’s decomposition is F or F , and the first terms in thedecomposition are not F and F .For any such n , suppose the game can be played without any combining moves exceptcombine .In the following proof, we define the gap between two terms as the difference between theirindices. Notice that for any n not in the form of F k − k ≥ , its decomposition either hasa smallest term of F or larger ( Case 1 ), or contains two consecutive terms with a gap of atleast . We make the following claim. Claim 1.
If a game is played with only split or combine moves, then no game statecontains two consecutive terms with a gap larger than . Proof of Claim 1.
Since the first step of the game is always F ∧ F → F , which onlygenerates a gap of , the claim is true for the first step. Suppose the claim is true for m th step.For the ( m + 1 )th step, if we split ’s, it generates a gap of at most , and the gap between F and other terms larger than F shrinks.If we combine ’s, we generate a gap of , and the gap between F and other terms largerthan F shrinks.Every time we split F i ∧ F i → F i +1 ∧ F i − ( i > ), we generate a gap of , which is the gapbetween F n +1 and F i − , but the gaps between F i − and any terms smaller than F n − shrink,and the gaps between F i +1 with any terms larger than F i +1 shrink. In other words, every timewe split, we generate a gap of but also make other gaps smaller.Therefore, for all the situations above, the claim is true for ( m + 1 )th step, and by inductionClaim is true.TBD 13HE FIBONACCI QUARTERLYWe now use this claim to finish our proof. Note that for any n in Case , there will be atleast two terms with a gap of at least and no terms in between them. From Claim , wealso proved that when we only use combine and splitting moves, we can never generate a gapthat is more than . Therefore, if n ’s decomposition has a gap of more than , then it is outof our consideration. For other n , we can find an i such that F i − and F i +1 are two terms inthe final decomposition, and there is neither F i − nor F i in the final decomposition.Since F i +1 is in the final decomposition, there must exist a step where S i is performed. Thisis because the first time F i +1 is generated in the game must come from the S i . So, let step t be the last S i .In order for the S i happen, there must be at least two F i ’s generated prior. Thus, there mustexist a step of splitting F i − moves before step t (this is because the first time F i is generatedin the game must come from the splitting F i − move). So, let step v be the last S i − in thegame.Since F i − is in the final decomposition, F i − is not in the final decomposition (because thefinal decomposition does not contain two consecutive Fibonacci numbers). Note that step v has generated one F i − , so there must be a S i − after step v to get rid of this F i − (here, thecombine move is also considered as a splitting move), and we can call it step r .Since step v has generated one F i and F i is not in the final decomposition, there must be a S i after step v in order to get rid of this F i , and we can call it step s .Since both step r and step s are after step v and each of them has generated one F i − , theremust exist F i − at some point after step v . As a result, there must be a S i − after step v ,and we can call it step w .Since step w has generated one F i − and F i − is not in the final decomposition, there mustbe a S i − after step w in order to get rid of this F i − . In other words, there is a S i − after step v , which contradicts with our assumption that step v is the last occurrence of S i − .Therefore, Lemma 3.3 is proved. (cid:3) The proof for Theorem 1.4 follows directly from Lemma 3.2 and 3.3.4.
Future Work
It is worth noting that while our upper bound is sharp for some n , there is still plenty ofroom for it to be tightened. It was alluded to previously in the proof of 1.3 that this couldbe done by quantifying the number of each combining moves, but such work is beyond thescope of this paper. In recent work, Cusenza et. al. [CDHKKMMTYZ] investigated winningstrategies for alliances of players in a multi-person generalization of the Zeckendorf game; onecan similarly investigate the number of moves of various strategies in these settings.Additionally, there are many ways the Zeckendorf game can be generalized (see for example[BEFM1]). With that in mind, we ask the following questions in regards to how our workrelates to other similar games. • Can our methods for analysis of game bounds be performed on generalized games? • Can we generalize the strategies suggested in this paper to achieve the longest orshortest game length in generalized games?
References [BEFM1] P. Baird-Smith, A. Epstein, K. Flynt and S. J. Miller,
The Generalized Zeckendorf Game ,the Fibonacci Quarterly (Proceedings of the 18th Conference) (2019), no. 55, 1–15. https://arxiv.org/pdf/1809.04883 .
14 VOLUME, NUMBEROUNDS ON ZECKENDORF GAMES [BEFM2] P. Baird-Smith, A. Epstein, K. Flynt and S. J. Miller,
The Zeckendorf Game , Combinatorialand Additive Number Theory III, CANT, New York, USA, 2017 and 2018, Springer Proceed-ings in Mathematics & Statistics (2020), 25–38. https://arxiv.org/pdf/1809.04881 .[CDHKKMMTYZ] A. Cusenza, A. Dunkelberg, K. Huffman, D. Ke, D. Kleber, S. J. Miller, C. Mizgerd,V. Tiwari, J. Ye and X. Zheng,
Winning Strategy for the Multiplayer and MultiallianceZeckendorf Games , preprint. https://arxiv.org/pdf/2009.03708 .[Ho] V. E. Hoggatt,
Generalized Zeckendorf theorem , Fibonacci Quarterly (1972), no. 1 (spe-cial issue on representations), pages 89–93.[Ke] T. J. Keller, Generalizations of Zeckendorf ’s theorem , Fibonacci Quarterly (1972), no.1 (special issue on representations), pages 95–102.[LLMMSXZ] R. Li, X. Li, S. J. Miller, C. Mizgerd, C. Sun, D. Xia, Z. Zhou (2020) Deterministic Zeckendorf Games, Part I , to appear in the Fibonacci Quarterly. https://arxiv.org/pdf/2006.16457 .[MN] S. J. Miller and A. Newlon,
The Fibonacci Quilt Game , Fibonacci Quarterly (2020), no.2, 157–168. https://arxiv.org/pdf/1909.01938 .[MW1] S. Miller, Y. Wang, From Fibonacci Numbers to Central Limit Type Theorems , Journal ofCombinatorial Theory, Series A (2012), no. 7, 1398–1413.[MW2] S. Miller, Y. Wang,
Gaussian Behavior in Generalized Zeckendorf Decompositions , Com-binatorial and Additive Number Theory, CANT 2011 and 2012 (Melvyn B. Nathanson,editor), Springer Proceedings in Mathematics & Statistics (2014), 159–173.[Ze] E. Zeckendorf, Représentation des nombres naturels par une somme des nombres de Fi-bonacci ou de nombres de Lucas, Bulletin de la Société Royale des Sciences de Liège (1972), pages 179–182. University of California, Los Angeles, Los Angeles, CA 90095
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