Two players placing coins on a table- Extension
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The origin of my question comes from a common job interview question where two players take turns placing coins on a round table. The coins cannot overlap and can't be moved once they've been placed. The player which first has no available space on the table to place a coin, loses.
The intuitive strategy for the first player in this game is to place their first coin in the centre of the table, and then place their ensuing coins collinear to the central coin and his opponent's previously placed coin, as well as equidistant from the centre as his oppponent's coin.
I then question what would happen if the table was an equilateral triangle. The strategy as described above falls apart, and unfortunately I have not yet come up with a well defined strategy for the first player to win (if there exists one) without some pretty restricting assumptions. I am looking for some help with this.
discrete-mathematics recreational-mathematics game-theory algorithmic-game-theory
asked Nov 29 '18 at 18:08
Manuel BarrosManuel Barros
264
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add a comment |
$begingroup$
The origin of my question comes from a common job interview question where two players take turns placing coins on a round table. The coins cannot overlap and can't be moved once they've been placed. The player which first has no available space on the table to place a coin, loses.
The intuitive strategy for the first player in this game is to place their first coin in the centre of the table, and then place their ensuing coins collinear to the central coin and his opponent's previously placed coin, as well as equidistant from the centre as his oppponent's coin.
I then question what would happen if the table was an equilateral triangle. The strategy as described above falls apart, and unfortunately I have not yet come up with a well defined strategy for the first player to win (if there exists one) without some pretty restricting assumptions. I am looking for some help with this.
discrete-mathematics recreational-mathematics game-theory algorithmic-game-theory
asked Nov 29 '18 at 18:08
Manuel BarrosManuel Barros
264
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1
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I doubt there is one because there is no bilateral symmetry. But you can start with some discrete examples (a coin takes up k cells on a table with finite cells) and then increase the number of cells to see what happens.
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– fleablood
Nov 29 '18 at 18:25
2
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There is a winning strategy, but the winner depends on how many coins you can fit from the center of the triangle to the perimeter.
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– mlc
Nov 29 '18 at 18:47
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@fleablood An equilateral triangle does have bilateral symmetry. I think the key observation here is that it does not have 180-degree rotational symmetry, which would allow the aforementioned strategy to work.
$endgroup$
– platty
Nov 30 '18 at 1:02
add a comment |
$begingroup$
The origin of my question comes from a common job interview question where two players take turns placing coins on a round table. The coins cannot overlap and can't be moved once they've been placed. The player which first has no available space on the table to place a coin, loses.
The intuitive strategy for the first player in this game is to place their first coin in the centre of the table, and then place their ensuing coins collinear to the central coin and his opponent's previously placed coin, as well as equidistant from the centre as his oppponent's coin.
I then question what would happen if the table was an equilateral triangle. The strategy as described above falls apart, and unfortunately I have not yet come up with a well defined strategy for the first player to win (if there exists one) without some pretty restricting assumptions. I am looking for some help with this.
discrete-mathematics recreational-mathematics game-theory algorithmic-game-theory
asked Nov 29 '18 at 18:08
Manuel BarrosManuel Barros
264
$endgroup$
The origin of my question comes from a common job interview question where two players take turns placing coins on a round table. The coins cannot overlap and can't be moved once they've been placed. The player which first has no available space on the table to place a coin, loses.
The intuitive strategy for the first player in this game is to place their first coin in the centre of the table, and then place their ensuing coins collinear to the central coin and his opponent's previously placed coin, as well as equidistant from the centre as his oppponent's coin.
I then question what would happen if the table was an equilateral triangle. The strategy as described above falls apart, and unfortunately I have not yet come up with a well defined strategy for the first player to win (if there exists one) without some pretty restricting assumptions. I am looking for some help with this.
discrete-mathematics recreational-mathematics game-theory algorithmic-game-theory
discrete-mathematics recreational-mathematics game-theory algorithmic-game-theory
asked Nov 29 '18 at 18:08
Manuel BarrosManuel Barros
264
asked Nov 29 '18 at 18:08
Manuel BarrosManuel Barros
264
asked Nov 29 '18 at 18:08
Manuel BarrosManuel Barros
264
asked Nov 29 '18 at 18:08
Manuel BarrosManuel Barros
264
asked Nov 29 '18 at 18:08
Manuel BarrosManuel Barros
264
264
1
$begingroup$
I doubt there is one because there is no bilateral symmetry. But you can start with some discrete examples (a coin takes up k cells on a table with finite cells) and then increase the number of cells to see what happens.
$endgroup$
– fleablood
Nov 29 '18 at 18:25
2
$begingroup$
There is a winning strategy, but the winner depends on how many coins you can fit from the center of the triangle to the perimeter.
$endgroup$
– mlc
Nov 29 '18 at 18:47
$begingroup$
@fleablood An equilateral triangle does have bilateral symmetry. I think the key observation here is that it does not have 180-degree rotational symmetry, which would allow the aforementioned strategy to work.
$endgroup$
– platty
Nov 30 '18 at 1:02
add a comment |
1
$begingroup$
I doubt there is one because there is no bilateral symmetry. But you can start with some discrete examples (a coin takes up k cells on a table with finite cells) and then increase the number of cells to see what happens.
$endgroup$
– fleablood
Nov 29 '18 at 18:25
2
$begingroup$
There is a winning strategy, but the winner depends on how many coins you can fit from the center of the triangle to the perimeter.
$endgroup$
– mlc
Nov 29 '18 at 18:47
$begingroup$
@fleablood An equilateral triangle does have bilateral symmetry. I think the key observation here is that it does not have 180-degree rotational symmetry, which would allow the aforementioned strategy to work.
$endgroup$
– platty
Nov 30 '18 at 1:02
1
1
$begingroup$
I doubt there is one because there is no bilateral symmetry. But you can start with some discrete examples (a coin takes up k cells on a table with finite cells) and then increase the number of cells to see what happens.
$endgroup$
– fleablood
Nov 29 '18 at 18:25
$begingroup$
I doubt there is one because there is no bilateral symmetry. But you can start with some discrete examples (a coin takes up k cells on a table with finite cells) and then increase the number of cells to see what happens.
$endgroup$
– fleablood
Nov 29 '18 at 18:25
2
2
$begingroup$
There is a winning strategy, but the winner depends on how many coins you can fit from the center of the triangle to the perimeter.
$endgroup$
– mlc
Nov 29 '18 at 18:47
$begingroup$
There is a winning strategy, but the winner depends on how many coins you can fit from the center of the triangle to the perimeter.
$endgroup$
– mlc
Nov 29 '18 at 18:47
$begingroup$
@fleablood An equilateral triangle does have bilateral symmetry. I think the key observation here is that it does not have 180-degree rotational symmetry, which would allow the aforementioned strategy to work.
$endgroup$
– platty
Nov 30 '18 at 1:02
$begingroup$
@fleablood An equilateral triangle does have bilateral symmetry. I think the key observation here is that it does not have 180-degree rotational symmetry, which would allow the aforementioned strategy to work.
$endgroup$
– platty
Nov 30 '18 at 1:02
add a comment |
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I doubt there is one because there is no bilateral symmetry. But you can start with some discrete examples (a coin takes up k cells on a table with finite cells) and then increase the number of cells to see what happens.
$endgroup$
– fleablood
Nov 29 '18 at 18:25
2
$begingroup$
There is a winning strategy, but the winner depends on how many coins you can fit from the center of the triangle to the perimeter.
$endgroup$
– mlc
Nov 29 '18 at 18:47
$begingroup$
@fleablood An equilateral triangle does have bilateral symmetry. I think the key observation here is that it does not have 180-degree rotational symmetry, which would allow the aforementioned strategy to work.
$endgroup$
– platty
Nov 30 '18 at 1:02