Baskerhound chuckled. "You wouldn't hold that small escapade against an old friend after all these years, would you? Anyway, I have a personal vendetta against this kidnapper — they call him Sam the Snatcher. He has the conceit of being the world's greatest gamester. What nerve."

"It seems that he wants to play a game he calls "Ultimate Tic-Tac-Toe" against the FBI. If we win, he will charge only 10 million dollars per hostage. Rates have gone up since you were my guest."

Ecco smiled again. I suspected he still missed Baskerhound's hideouts and his companions.

"So, what's the game, Dr. Baskerhound?" 13-year-old Liane asked.

"I'm glad someone is taking me seriously around here," Baskerhound replied and then directed his explanation to Liane.

"Imagine a 4 X 4 grid (not 3 X 3). We will play a tic-tac-toe variant on this. Players alternate play as in tic-tac-toe with X moving first and then O moving second. A threesome for X is a set of three cells that form a line vertically, horizontally, or diagonally that all contain Xs. Similarly for a threesome for O. Since the grid is 44, there are 16 cells, so eight alternating turns of the two players will fill the board.

"Now for the rules: If a player gets a threesome before his last possible turn, then he loses. If he gets a threesome on his last turn, he wins. The game is of the 'sudden death' variety. As soon as there is a loser or a winner, the game stops. So if both players could get a threesome on the last turn, then the first player wins anyway. If either player loses on some turn (by getting a threesome too early), then it doesn't matter if the other player would be forced to lose on his next move. The player who loses first still loses. If neither player gets a threesome by the end of the game, the game is a draw. The big question is: Does either player have a winning strategy (that is, can they force a win)? If not, can either player force a draw?

"Fortunately, we don't have to answer the big question. Sam has given us two game puzzles. Train your intuition on the first one. Can X force a win from this configuration? (In Figure 1(a), it's X's move.)"

"I think so," Liane replied.

"First move as in Figure 1(b). By case analysis, nothing O can do will prevent the X player from being able to place X in a square that will give him two opportunities to win in the last turn. X has two ways to win on the next turn."

"Nicely done, young mathematician," Baskerhound said. "Now for Sam's second puzzle in Figure 2. The asterices represent moves by X and O. All you know is that neither has a threesome and that there are six Xs and six Os among the perimeter. Can X force a win in this case?"

Liane was able to answer the question. Baskerhound left to talk to his "handlers."

Still, his original question weighed heavily in the air. Was a winning strategy possible for X in general or could O always force a draw? I never heard the answer.

Another variant of the game is the Come-Back variant. In that variant, O wins if he obtains a threesome on the eighth turn regardless of whether X has already obtained a threesome on his eighth turn. On the other hand, early threesomes still face a sudden death rule. In that case, does O have a winning strategy?

Last Month's Solution

Placing parks at the points: (3,1), (4,1), (5,1), (7,1), and (8,1) giving an initial configuration of

000000000000

000000000000

000000000000

0P0110000000

0P0000000000

0P0000000000

000000000000

0P0000000000

0P0000000100

000000000100

000000000000

000000000000

retards full development of all nonpark land by 15 years.

Reader Notes

Several readers found solutions to the "Who Rules?" puzzle (DDJ, September 2001) including: Michael Birken, Wyatt Matthews, Chad Harrington, Pieter Gosselink, Geoffrey Probert, Bob Byard, Bernhard Kuehl, Laszlo Vecsey, Jim Woodgate, Andrew Calafato, Jim Love, Scott Romanowski, Tomas G. Rokicki, Grisha Golberg, Richard Ford, Pearl Pauling, Rodney Meyer, Hans Knorr, Andrew Palfreyman, Yves Piguet, Ketil Malde, Wim Nuij, and Robin Golden.

Birken and Vecsey were the first to observe that units 1 and 5 might be interchanged in the solution to the first question, yielding

3

|

+ — 7

| |

| + — 1

| | |

| | + — 2

| | |

| | + — 6

| | |

| | + — 8

| |

| + — 5

|

+ — 12

|

+ — 4

|

+ — 9

|

+ — 10

|

+ — 11

and

3

|

+ — 7

| |

| + — 1

| |

| + — 5

| |

| + — 2

| |

| + — 6

| |

| + — 8

|

+ — 12

|

+ — 4

|

+ — 9

|

+ — 10

|

+ — 11

Several readers — including Birken, Matthews, Kuehl, and Goldberg — found multiple solutions to the second problem, while identifying 6 and 8 as Silky and Slit.

DDJ

Figure 1: (a) X moves next. Can X guarantee a win? (b) X moves in the spot X.1. No matter where O moves, X can next move in the center in such a way that X has two ways to move in its last move to get a threesome.