As in the previous paragraph, Lemmas 2.3
and 2.4 show that this is the case when ( n − 1) − a + i ≡ 0 mod 3. Moreover, if Algois does not close off but moves from ( a, i + 1) to ( a + 1 , i + 1), then again from Lemma 2.3 and Lemma 2.4 Berol will lose by closing-off on her next move.

This leads to a sequence of forcing moves that are actually three repetitions of a
sequence going from ( a, i) to ( a + 7 , i + 1), one for each i = 0 , 1 , 2. See Figure 1.
B 0 ∗
A 1
[ B 0] A 2 B 1 A 0 B 2 A 1 ∗ B 0 ∗
B 2
[ A 1] B 0 · · ·
A 1 ∗
B 2
[ A 1] B 0 ∗ A 2 B 1 A 0 B 2 A 1 ∗
A 1
B 0 A 2 B 1 A 0 [ B 2] A 1 · · ·
Figure 1. Forcing sequences for Algois (top) and Berol (bottom). The entry
labels A and B indicate who moved to that vertex; the subscripts are n − a + i mod 3. The brackets indicate the beginning and end of a sequence. In each
case, the other player can determine part of the sequence, but an ∗ indicates the
only time a choice is available.
Since neither can be forced into a disadvantageous closing-off move, it remains
to analyze the situations when the players have to move to the last column before
a closing-off move. In addition, we have to consider which player drops first, that
is, makes a move in the second coordinate.
334
RICHARD J. NOWAKOWSKI AND DAVID G. POOLE
In what follows, the
and
positions are calculated using Lemma 2.3 as
N
P
soon as the status of one position can be identified. All the information will be
summarized in Table 1.
Case 1. There is no closing-off, and there is a move to [ n − 1 , 0].
The following cases are independent of who dropped first:
B A B
0 A
B A 0 A
A B
0 A
A
B
A
P
P
P
N
P
B
A
B
P
N
P
N
subcase 1: B wins
subcase 2: A wins
subcase 3: A wins
The 0 indicates the initial vertex. In each case, the
position to the left of the
P
vertical line is formed by the closing-off move. The other
and
positions are
P
N
obtained from Lemma 2.3. No more need be said about these cases.
In the next two cases it matters who drops first.
A B A 0 A . . . B
A B A 0 A . . . A
B
A
B
B
subcase 4: B wins
subcase 5: A wins
The situation presented in subcase 4 is a losing one for Algois. He has played to
[ n − 1 , 0], so Berol is forced to move to ( n − 1 , 1). However, Berol does not have to move off the line i = 1: she can force Algois to have that doubtful privilege.
Whenever he does, he either moves to ( n − 1 , 2) and Lemma 2.3 shows that
(0 , 2) = ; or he moves to (2 k + 1 , 2) and thus (2 j, 2) =
for j = 0 , 1 , . . . k.
G
P
G
P
But in this case ( n− 1 , 2) =
and, again by Lemma 2.3, (2 a, 2) =
= (2 b, 2)
G
P
G
P
G
for b = j + 1 , j + 2 , . . . , a − 1, and thus Berol moves to a position on her next
P
move after she has forced Algois to drop.
In subcase 5, Algois moves to (0 , 1) and forces Berol to drop first. The analysis
follows as in subcase 4, except that now Algois wins.
Case 2. No closing-off and a move to [ n − 1 , 1]: Berol wins in all cases.
Since ( n − 1 , 0) has not been visited, it follows that ( n − 1 , 0) =
and also
G
P
that ( n − 1 , 2) =
. If Algois is forced to move to [ n − 1 , 1], Lemma 2.3 implies
G
N
that
(0 , 1) = , and therefore Berol wins by moving to (0 , 1). If Berol moves
G
P
to [ n − 1 , 1], Lemma 2.3 implies that (0 , 2) =
and
(0 , 1) =
. Thus Algois
G
P
G
N
has no good move, and Berol wins again.
Case 3. No closing-off, and a move to [ n − 1 , 2].
These cases are independent of who dropped first:
A 0 A
B
0 A
P
B
B A
P
P
N
P
B A B
B A
P
P
subcase 6: A wins
subcase 7: B wins
GEOGRAPHYLAYED ON PRODUCTS OF DIRECTED CYCLES
335
Again, in each case, the
position to the left of the line is formed by the closing-
P
off move. The other
and
positions are obtained from Lemma 2.3. No more
P
N
need be said about these cases.
Here are the cases where it matters who drops first:
0 A . . . A B
0 A . . . B A
A B
A
A B
B
P
P
N
P
P
A B A B . . . B
B A B
B
P
P
subcase 8: A wins
subcase 9: B wins
B
0 A
0 A . . . A B
0 A . . . B A
P A
A
B
N
P
P
N
P
P
A B A
A B A B A
A
A B A B A
B
N
N
subcase 10a: A wins
subcase 10b: A wins
subcase 10c: B wins
In subcase 8, Algois moves to (0 , 2); then the players must remain on the bottom
row, with Berol moving to (2 a+1 , 2). Thus (2 a+1 , 1) = , and so ( n− 1 , 1) =
G
P
G
. From Lemma 2.3 we see that
(2 a + 2 , 2) = , and thus Berol loses.
P
G
P
The analysis in subcase 9 is similar, but leads to the conclusion that Berol
wins.
In subcase 10a, if Berol moves to ( n− 1 , 0) Algois is forced to move to ( n− 1 , 1).
Now, ( n− 2 , 1) has no followers and so is a
position. By Lemma 2.3, (0 , 2) =
P
G
P
and so now Berol has no winning move. This is regardless of which player
dropped first.
Suppose that Algois drops first, from (2 a, 0) to (2 a, 1). If Berol moves to (0 , 2) instead of ( n − 1 , 0), the moves (0 , 2) 7→ (1 , 2) 7→ (2 , 2) . . . are forced until Algois moves to (2 a − 1 , 2). Thus (2 a − 1 , 1) is a
position, since it has no followers.
P
Working back we see that ( n − 1 , 2) is a
position. Thus by Lemma 2.3, we
P
conclude that (2 a, 2) is a
position and Berol loses.
N
In a similar way, we see that, if Berol drops first, from (2 a+1 , 0) to (2 a+1 , 1), and Berol moves to (0 , 2), Berol wins by moving to (0 , 2).
The preceding information is summarized in Table 1. We combine the infor-
mation from these cases and the forcing sequence argument in the next tables.
There are only three possible starting points for the sequences. Also, the drop-
ping moves can only occur at those values of n − a + i mod 3 given in Lemma 2.6.