GEOGRAPHY PLAYED ON AN N-CYCLE TIMES A 4-CYCLE

M. S. Hogan^{1}

Department of Mathematics and Computer Science, University of Prince Edward Island, Charlottetown, PE C1A 4P3, Canada

D. G. Horrocks^{2}

Department of Mathematics and Computer Science, University of Prince Edward Island, Charlottetown, PE C1A 4P3, Canada

                            dhorrocks@upei.ca

Received:12/3/02, Accepted: 4/29/03, Published: 4/30/03

Abstract

In the game of Geography, two players alternate moving a coin along the edges of a directed graph. The coin may not be moved to the same vertex more than once, and the last player who is able to move wins. We analyze this game played on the cartesian product of cycles C_{n} × C_{4}.

1. Introduction

Geography, as described in [2] and [3], is a game for two players, played on a directed graph. A coin is initially placed on a vertex and, on his turn, a player moves the coin along a directed edge to an adjacent previously unoccupied vertex. The players move alternately and, under normal play, the last player who is able to move wins.

Some particular cases of Geography have been studied. For example, Kotzig’s (or Modular) Nim is played on a directed graph with vertex set {0, 1, 2, …, n − 1}. There is a directed edge from x to y if and only if y − x (mod n) is contained in a given set of integers, called the move set. Many instances of this game have been solved; we refer the reader to [1] and [4] for the results.

We will assume that the reader is familiar with standard terminology of combinatorial game theory, as described in [1]. In particular, for an impartial game G, the outcome class of G, denoted G(G), is P if the previous player has a winning strategy, and N if the next player wins. We will refer to our two players as Alice and Bob, with Alice playing first.

1The author was partially supported by an NSERC Undergraduate Summer Research Award.

2The author was partially supported by NSERC.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #G02 2

We now define the graphs to be considered here. Let C_{n} denote the directed cycle with vertex set V (C_{n}) = {0, 1, …, n − 1} and edge set E(C_{n}) = {(0, 1), (1, 2), …, (n − 2, n − 1), (n − 1, 0)} where (x, y) denotes an edge directed from x to y. The cartesian product C_{m} × C_{n} has vertex set V (C_{m}) × V (C_{n}) and there is a directed edge from (a, b) to (c, d) if and only if a = c and (b, d) is a directed edge in C_{n}, or b = d and (a, c) is a directed edge in C_{m}.

Nowakowski and Poole [5] investigated the game of Geography played on a cartesian product of two directed cycles. In particular, the following theorem may be found in [5].

Theorem 1 If n = 2, or if both n and m are even, then G(C_{n} × C_{m}) = N.

Furthermore, the main result of [5] is the determination of the outcome class of C_{n} × C_{3} for all positive integers n.

2. Geography played on C_{n} × C_{4}

In this paper, we consider the game of Geography played on C_{n} × C_{4}. The following is an immediate consequence of Theorem 1.

Theorem 2 If n is even then G(C_{n} × C_{4}) = N.

Therefore, we need only consider the case in which n is odd and in the sequel we will assume that this is the case. The rest of this paper is devoted to proving the following result.

Theorem 3 If n is an odd positive integer then

G(C_{n} × C_{4}) = \{ \begin{matrix}P & if n ≡ 11 (mod 12), \\ N & otherwise. \end{matrix}

We will visualize the graph C_{n} ×C_{4} as a rectangular board having 4 rows and n columns. The rows are numbered 0, 1, 2, 3 from top to bottom and the columns are numbered 0, 1, 2, …, n − 1 from left to right: the ordered pair (c, r) refers to the cell at column c and row r. We will assume, without loss of generality, that the coin is initially placed at (0, 0) and will denote this on the diagrams by X. A legal move is to move right (in the first coordinate) or down (in the second coordinate), including “wrap around” moves. Every time play reaches the last column, we will say that one pass has occurred. We will denote the players’ moves by A and B for Alice and Bob, respectively. It will be convenient, on occasion, to refer to the ith move by player Y; this will be done using the notation Y_{i}. Further, a move that is forced upon player Y will be denoted Y! and parentheses will be used when it does not matter which of the two available moves is played.

The cell (c, r) is called a sink if it has not yet been occupied but its followers, (c + 1, r) and (c, r + 1), have both been played. Thus a player who is able to reach a sink wins the game. An important notion, introduced in [5], is that of a closing-off sequence. In the game C_{n} × C_{4}, such a sequence of moves has one of the following forms.

\begin{aligned}S_{1} & = & (a, i) & → & (a + 1, i) & → & (a + 1, i + 1) & → & (a + 1, i + 2) & → & (a + 1, i + 3) \\ S_{2} & = & (a, i) & → & (a, i + 1) & → & (a + 1, i + 1) & → & (a + 1, i + 2) & → & (a + 1, i + 3) \\ S_{3} & = & (a, i) & → & (a, i + 1) & → & (a, i + 2) & → & (a + 1, i + 2) & → & (a + 1, i + 3) \end{aligned}

Notice that a closing-off sequence creates a sink at (a, i+3) (assuming that it has not yet been occupied) and that the game will end no later than at this cell on the next pass. As we will see in the next section, a player who is the first to complete a closing-off sequence on the first pass wins the game.

3. Preliminary Lemmas

In this section, we establish two lemmas, namely Lemma 2 and Lemma 3, which present conditions under which one player may force a win. They will be used repeatedly in the proof of Theorem 3.

Lemma 1 If the first two moves of the game are in the first column then Alice has a winning strategy.

Proof. We have that Alice’s first move is to (0, 1) and Bob replies to (0, 2). Suppose now that Alice moves to (0, 3): Bob is then forced to respond with (1, 3).

Alice’s winning strategy now is to always play down. Note that this move is always available to her and that Bob is always the first player to reach a new column. Eventually then, Bob will be the first to play on the last column and, as the game cannot wrap around to the first column, Alice will win. □

Therefore, should Alice begin the game by moving down to (0, 1), Bob is forced to respond by moving across to (1, 1). In the sequel then, we will assume that, on the first pass, neither (0, 2) nor (0, 3) are occupied.

Lemma 2 A player who completes a closing-off sequence on the first pass, and is the first to do so, has a winning strategy.

Proof. Suppose that Alice is the first player to complete a closing-off sequence and does so on the first pass by moving to (a + 1, i). This move creates a sink at (a, i) and we show that Alice wins the game by moving to this vertex on the second pass.

First, notice that since n is odd, colouring the vertices in a checkerboard fashion shows that Alice will be the player to reach (a, i) on the second pass.

Second, it must be shown that the game will not end before Alice reaches (a, i). We observe that, since the game cannot proceed beyond column a on the second pass, it will end earlier if and only if another sink is both created and occupied on the same pass. In order for this to happen, the players must make four consecutive moves (two moves each) in the same column. We claim that this cannot happen if Alice always moves across when it is possible to do so.

Clearly, up to the second last column of the first pass, Alice can always move across. At the last column, however, Alice may be forced to move down once (if Bob has played to (n − 1, 0), for example). However, play will not “go around” in the second coordinate in the last column because in order for this to happen, Bob must move to either (n − 1, 2) or (n − 1, 3). In either case, since we assume that neither (0, 2) nor (0, 3) are occupied on the first pass, the across move will be available to Alice. Finally, once the second pass has begun, it is impossible for play to go around in any one column since each column contains at least one occupied cell. Therefore, regardless of how the players move, play must continue until Alice reaches the sink at (a, i) and wins. □

Lemma 3 If the last two moves on the first pass are to (n − 2, 1) and (n − 1, 1) then Alice wins.

Proof. Colouring the cells in a checkerboard fashion, we see that, since n is odd, Bob must be the player to move to (n − 2, 1) and it is Alice’s response to (n − 1, 1). Thus a sink is created at (n − 1, 0). Bob’s response will be to either (n − 1, 2) or (0, 1). In either event, Alice’s winning strategy is to reply to (0, 2) which, by the assumption following Lemma 1, will be unoccupied.

As in the proof of Lemma 2, once the second pass has begun, it is impossible for play to go around in any one column since each column contains at least one occupied cell. Therefore, play will eventually reach the second last column and the first player to do so will be Alice at (n − 2, 3), or Bob at (n − 2, 0) or (n − 2, 2). If Alice moves to (n − 2, 3) then Bob’s next move is to either (n − 1, 3) or (n − 2, 0). In either event, Alice may reply to (n − 1, 0) and win. If Bob moves to (n − 2, 2) then Alice replies to (n − 2, 3) and has a winning strategy as described above. Finally, should Bob move to (n − 2, 0) then Alice replies to (n − 1, 0) and wins. □

4. Two Special Move Sequences

4.1 Playing the 12

Our analysis of the game C_{n} × C_{4} will depend upon the value of n modulo 12. The underlying reason for this is that, on the first pass, a player may force his opponent into repeating a block of moves comprising 12 columns. We call this playing the 12.

Let us suppose that, on the first pass, Bob is the first player to reach column c and does so by playing to (c, 0), denoted by B_{0} on the diagram below. Alice may then dictate the play for the next 12 columns so that Bob will be forced to be the first player to reach the (c + 12)th column by playing to (c + 12, 0). The following diagram describes Alice’s strategy. Recall that ! indicates a forced move and parentheses indicate that either move may be made.

B_{0} A_{1} (B_{1})  (B_{7}) A_{8} B_{8}!
(B_{1}) A_{2} B_{2}! A_{3} (B_{3})
(B_{3}) A_{4} B_{4}! A_{5} (B_{5})
(B_{5}) A_{6} B_{6}! A_{7} (B_{7})

Bob’s first, third, fifth, and seventh moves may be either across or down. Notice, however, that Bob’s other moves are forced. If, for example, B_{2} is to (2, 2) then, by Lemma 2, Alice wins by moving to (2, 3) thereby completing a closing-off sequence.

4.2 Playing the 10

In a manner similar to playing the 12, Alice may, at the beginning of the game, force a sequence of moves that comprises 10 columns. To do this, Alice plays down to (0, 1) on her first move and, by the remark following Lemma 1, we may assume that Bob responds with (1, 1).

The complete sequence of moves is called playing the 10 and is shown in the diagram below.

X  (B_{6}) A_{7} B_{7}!
A_{1} B_{1}! A_{2} (B_{2})
(B_{2}) A_{3} B_{3}! A_{4} (B_{4})
(B_{4}) A_{5} B_{5}! A_{6} (B_{6})

Notice that, as in playing the 12, Bob’s third, fifth, and seventh moves are forced, in order to prevent a closing-off move from Alice.

5. Proof of Theorem 3

This section is devoted to the proof of our main result, Theorem 3. The argument is in six cases, depending upon the value of the odd integer n modulo 12.

5.1 The case n ≡ 1 (mod 12)

5.11 n = 1

The game C_{1} × C_{4} is trivially a first player win.

5.12 n > 1

In this case, n = 13 + 12m for some m ≥ 0. Alice’s winning strategy is to begin by playing the 10, followed by playing the 12 m times, and finally to finish with a block of 3 columns. The analysis is in two cases, according to Bob’s second move in the final block of 3 columns.

CASE 1: Bob’s second move in the final block of 3 is across

By Lemma 2, Alice wins by closing-off at (n − 1, 3).

X  (B) A B! A ⋯ A B! A B
A B! A (B)  (B) ⋯  A!
(B) A B! A (B)  ⋯  B!
(B) A B! A (B)  ⋯ (B)  A

CASE 2: Bob’s second move in the final block of 3 is down

After Bob moves to (n − 2, 1), Alice responds with (n − 1, 1) and wins by Lemma 3.

X  (B) A B! A ⋯ A B! A
A B! A (B)  (B) ⋯  B A
(B) A B! A (B)  ⋯
(B) A B! A (B)  ⋯ (B)

5.2 The case n ≡ 3 (mod 12)

5.21 n = 3

Consider first the game C_{3} ×C_{4}. Alice’s winning strategy is to begin the game by playing across. If Bob’s first move is down then Alice responds by moving across and wins by Lemma 3.

On the other hand, if Bob’s first move is across then Alice’s second move is forced to be down. If Bob’s second move is down then Alice plays the closing-off move (2, 3) and wins on her fourth move as shown below.

\begin{aligned}X & A_{1} & B_{1} \\ & & A_{2}! \\ & & B_{2} \\ B_{3}! & A_{4} & A_{3} \end{aligned}

If Bob’s second move is across then Alice wins on her fifth move as shown below.

\begin{aligned}X & A_{1} & B_{1} \\ B_{2} & A_{3} & A_{2}! \\ & B_{3}! \\ A_{5} & A_{4} & B_{4}! \end{aligned}

5.22 n > 3

We have n = 15 + 12m for some m ≥ 0. Alice’s winning strategy is to open by playing the 10, then to play the 12 m times, and finally to finish with a block of 5 columns. By moving to (n − 1, 1) Alice wins by Lemma 3.

X  ⋯  (B) A B! A ⋯ A B! A (B)
A B! A ⋯  (B) ⋯  (B) A B! A
(B) ⋯  ⋯
⋯ B! A (B)  ⋯ (B)

5.3 The case n ≡ 5 (mod 12)

Alice’s winning strategy is to play the 12 as often as required and then to close with the block of 5 columns as shown below. By moving to (n − 1, 1), Alice wins by Lemma 3.

X A ⋯ (B) A B! A ⋯ (B) A B! A (B)
(B) ⋯  (B) ⋯  (B) A B! A
⋯  ⋯
⋯ A (B)  ⋯ A (B)

5.4 The case n ≡ 7 (mod 12)

5.41 n = 7

We consider first the game C_{7} × C_{4}. Observe that Bob’s third move is forced to be down, otherwise Alice may move to (6, 1) and win by Lemma 3. As shown below, Alice may create a sink by moving to (3, 3) on the second pass.

X A_{1} (B_{1})
(B_{1}) A_{2} B_{2}! A_{3}
B_{3}! A_{4} B_{4}!
B_{5}! A_{6}! B_{6}! A_{7}  A_{5}

After Bob responds to A_{7} with either (3, 0) or (4, 3), Alice may force Bob’s every move thereafter by moving to (4, 0), (6, 0), (0, 1), (1, 2) and finally to (3, 2) at which point she wins.

5.42 n > 7

In this case, n = 19 + 12m for some m ≥ 0. The analysis is in two cases depending upon Bob’s third move.

CASE 1: Bob’s third move is down

Alice’s winning strategy is to play the 12 (m + 1) times and then to finish with a block of 7 columns as shown below. Notice that Bob’s penultimate move in the first pass must be down, otherwise Alice may move to (n − 1, 1) and win by Lemma 3. Bob is forced to begin the second pass at (0, 3) and subsequently Alice creates a sink by moving to (3, 3).

X A_{1} (B_{1})  ⋯ (B_{7}) A_{8} B! A (B)
(B_{1}) A_{2} B_{2}! A_{3}  ⋯  (B) A B! A
B_{3} A_{4} ⋯  B! A B!
B! A! B! A  ⋯ A_{7} (B_{7})  A

On the second pass, the first player to reach the last column will be Alice at (n − 1, 0) or Bob at (n − 1, 1). In either case, Alice begins the third pass at (0, 1) and, two moves later, wins the game at (3, 2).

CASE 2: Bob’s third move is across

Alice begins with a block of 14 columns. She then plays the 12 m times and concludes with a block of 5 columns as shown below. By Lemma 3, this is a winning strategy for Alice.

X A (B)  (B) A B! A ⋯
(B) A B! A B_{3} A (B)  (B) ⋯
(B) A B! A (B)  ⋯
(B) A B! A (B)  ⋯
⋯  (B) A B! A (B)
⋯  (B) A B! A

⋯

⋯ B! A (B)

5.5 The case n ≡ 9 (mod 12)

5.51 n = 9

We consider first C_{9} ×C_{4}. The analysis is in two cases, depending upon Bob’s first move.

CASE 1: Bob’s first move is down

The analysis is in two subcases, depending upon Bob’s third move.

CASE 1A: Bob’s third move is across

Notice that Bob’s fourth move must be down, otherwise Alice may play her fifth move at (8, 1) which wins by Lemma 3. Alice begins the second pass at (0, 2) which creates a sink.

X A_{1}
B_{1} A_{2} B_{2}! A_{3} B_{3} A_{4}
A_{6}  B_{4}! A_{5}! B_{5}!

On the second pass, the first player to reach the last column is Alice at (8, 0), or Bob at (8, 1) or (8, 3). If Bob plays to (8, 1) then Alice replies to (0, 1) and wins. If Alice plays to (8, 0) then Bob is forced to reply to (8, 1) after which Alice moves to (0, 1) and wins. Finally, if Bob moves to (8, 3) then Alice moves to (8, 0) and wins on her next move as described above.

CASE 1B: Bob’s third move is down

On the second pass, Alice may create a sink by moving to (3, 3), as shown below.

X A_{1}
B_{1} A_{2} B_{2}! A_{3}
B_{3} A_{4} B_{4}! A_{5} (B_{5})
B_{6}! A_{7}! B_{7}! A_{8}  (B_{5}) A_{6}

On the second pass, the first player to reach the last column is Alice at (8, 0) or Bob at (8, 1). In either case, Alice begins the third pass at (0, 1) and is able to force a win at (3, 2).

CASE 2: Bob’s first move is across

Bob’s fourth move is forced to be down lest Alice reply with (8, 1) and win by Lemma 3. On the second pass, Alice may create a sink by moving to (5, 3), as shown below.

X A_{1} B_{1} A_{2} (B_{2})
(B_{2}) A_{3} B_{3}! A_{4}
B_{4}! A_{5} B_{5}!
B_{6}! A_{7}! B_{7}! A_{8}! B_{8}! A_{9}  A_{6}

On the second pass, the first player to reach the last column is Alice at (8, 0) or Bob at (8, 1). In either case, Alice begins the third pass at (0, 1) and, three moves later, wins at (5, 2).

5.52 n > 9

We have n = 21 + 12m for some m ≥ 0. We require two cases, according to Bob’s first move.

CASE 1: Bob’s first move is down

The analysis is in two subcases, depending upon Bob’s third move.

CASE 1A: Bob’s third move is across

Alice’s winning strategy is to open with a block of 14 columns, followed by playing the 12 m times, and then to finish with a block of 7 columns, as shown below. Alice begins the second pass by moving to (0, 2), thereby creating a sink at (0, 1).

X A  (B) A B! A ⋯
B_{1} A B! A B_{3} A (B)  (B) ⋯
A (B) A B! A (B) ⋯
(B) A B! A (B)  ⋯
⋯  (B) A B! A (B)
⋯  (B) A B! A (B)
⋯  (B) A B!
⋯ B! A (B)

The first player to reach the last column on the second pass is Alice at (n − 1, 0), or Bob at (n − 1, 1) or (n − 1, 3). If Bob has moved to (n − 1, 1) then Alice wins immediately by moving to (0, 1). If Alice has moved to (n − 1, 0) then Bob is forced to (n − 1, 1) and Alice wins as above. Finally, if Bob has moved to (n − 1, 3) then Alice wins by moving to (n − 1, 0).

CASE 1B: Bob’s third move is down

Alice’s winning strategy is to play the 12 (m+1) times, followed by a block of 9 columns, as shown below. On the second pass, Alice moves to (3, 3) which creates a sink at (3, 2).

X A  (B) A B! A ⋯
B_{1} A B! A  (B) ⋯
B_{3} A B! A (B)  ⋯
B! A! B! A  (B) A B! A (B)  ⋯

\begin{matrix}⋯ & & (B) & A & B! & A & (B) \\ ⋯ & & & & & (B) & A & B! & A & (B) \\ ⋯ & & & & & & & & (B) & A & B! & A & (B) \\ ⋯ & B! & A & (B) & & & & & & & & (B) & A \end{matrix}

The first player to reach the last column on the second pass is Alice at (n − 1, 0) or Bob at (n − 1, 1). In either case, Alice may begin the third pass at (0, 1) and is able to force a win at (3, 2).

CASE 2: Bob’s first move is across

The analysis is in two subcases, depending upon Bob’s fourth move.

CASE 2A: Bob’s fourth move is across Alice’s winning strategy is to open with a block of 16 columns, followed by playing the 12 m times, and then to close with a block of 5 columns, as shown below. By Lemma 3, Alice will win the game.

X A B_{1} A (B)  (B) A B! ⋯
(B) A B! A B_{4} A (B)  ⋯
(B) A B! A (B)  ⋯
(B) A B! A (B)  ⋯
⋯  (B) A B! A (B)
⋯  (B) A B! A

⋯

⋯ B! A (B)

CASE 2B: Bob’s fourth move is down

Alice’s winning strategy is to open with a block of 14 columns, followed by playing the 12 m times, and then to close with a block of 7 columns, as shown below. On the second pass, Alice moves to (5, 3) which creates a sink at (5, 2).

X A B_{1} A (B)  (B) A B! A ⋯
(B) A B! A  (B) ⋯
A (B)  B_{4} A B! A (B)  ⋯
(B) A B! A! B! A  (B) A B! A (B)  ⋯
⋯  (B) A B! A (B)
⋯  (B) A B! A (B)
⋯  (B) A B!
⋯ B! A (B)

On the second pass, the first player to reach the last column is Alice at (n − 1, 0) or Bob at (n − 1, 1) or (n − 1, 3). If Bob has moved to (n − 1, 3) then Alice will respond with (n − 1, 0) so that, in any case, Alice will begin the third pass at (0, 1) and will win at (5, 2).

5.6 The case n ≡ 11 (mod 12)

This is the only case in which Bob wins. His winning strategy is to open with the block of 11 columns shown below, and then to play the 12 as often as required. On the first pass, after Bob plays to (n − 1, 0), Alice is forced to respond with (n − 1, 1). Bob may then play the closing-off move (n − 1, 2).

X (A_{1})  (A_{7}) B_{7} A! B ⋯
(A_{1}) B_{1} A_{2}! B_{2} (A_{3})  (A) ⋯
(A_{3}) B_{3} A_{4}! B_{4} (A_{5})  ⋯
(A_{5}) B_{5} A_{6}! B_{6} (A_{7})  ⋯

⋯  (A) B A! B ⋯  (A) B
⋯  (A) ⋯  A!
⋯  ⋯  B
⋯ A! B (A)  ⋯ A! B (A)

The proof of Theorem 3 is now complete.

References

[1] E.R. Berlekamp, J.H. Conway, and R.K. Guy, Winning Ways For Your Mathematical Plays, A K Peters (2001).

[2] A.S. Fraenkel and S. Simonson, Geography, Theoret. Comput. Sci. (Math Games) 110 (1993), 197–214.

[3] A.S. Fraenkel, E.R. Scheinerman and D. Ullman, Undirected Edge Geography, Theoret. Comput. Sci. (Math Games) 112 (1993), 371–381.

[4] A.S. Fraenkel, A. Jaffray, A. Kotzig, and G. Sabidussi, Modular Nim, Theor. Comput. Sci. (Math Games) 143 (1995), 319–333.

[5] R.J. Nowakowski and D.G. Poole, Geography Played on Products of Directed Cycles, Games of No Chance, MSRI Publications 29 (1996), 329–337.