A simple game with $n$ points in 3D space - red triangle wins
$begingroup$
(Once again a son is torturing his father...)
Alice and Bob play a fairly simple game with $n$ predefined points in 3D space. No four points are complanar (which also implies that no three points are collinear).
They play in turns by connecting points that are not yet connected with straight segments. On her turn Alice connects a single pair of unconnected points with a single segment of red color. After that Bob connects $k$ pairs of unconnected points with $k$ segments of blue color.
Alice wins if she is able to create a triangle with red sides. If all points are connected and Alice was not able to create a red sided triangle before that, Bob wins.
Who has a winning strategy?
My thoughts: obviously, the outcome of the game heavily depends on releation between $n$ and $k$.
Suppose that $k=1$ with big enough value for $n$. Alice will simply draw segment $AB$, Bob draws his segment at will, Alice then draws $AC$, Bob blocks triangle $ABC$ by drawing segment BC, Alice draws segment $AD$ and Bob is lost. He cannot block triangles $ABD$ and $ACD$ by drawgin a single segment.
On the other side, with $k$ big enough, Bob can easily block creation of all red triangles by drawing a lot of lines.
So for $kle k_1$, Alice wins. For $kge k_2ge k_1$, Bob wins. I need some estimate for $k_1$ and $k_2$ (better than mine). Also note that the game always has a winner. Does it mean that $k_1=k_2$? I guess it does, but if we are just trying to estimate the lower and the upper bound, we can come up with two different values.
combinatorics combinatorial-game-theory
asked Jan 5 at 20:38
OldboyOldboy
8,62711036
$endgroup$
add a comment |
$begingroup$
(Once again a son is torturing his father...)
Alice and Bob play a fairly simple game with $n$ predefined points in 3D space. No four points are complanar (which also implies that no three points are collinear).
They play in turns by connecting points that are not yet connected with straight segments. On her turn Alice connects a single pair of unconnected points with a single segment of red color. After that Bob connects $k$ pairs of unconnected points with $k$ segments of blue color.
Alice wins if she is able to create a triangle with red sides. If all points are connected and Alice was not able to create a red sided triangle before that, Bob wins.
Who has a winning strategy?
My thoughts: obviously, the outcome of the game heavily depends on releation between $n$ and $k$.
Suppose that $k=1$ with big enough value for $n$. Alice will simply draw segment $AB$, Bob draws his segment at will, Alice then draws $AC$, Bob blocks triangle $ABC$ by drawing segment BC, Alice draws segment $AD$ and Bob is lost. He cannot block triangles $ABD$ and $ACD$ by drawgin a single segment.
On the other side, with $k$ big enough, Bob can easily block creation of all red triangles by drawing a lot of lines.
So for $kle k_1$, Alice wins. For $kge k_2ge k_1$, Bob wins. I need some estimate for $k_1$ and $k_2$ (better than mine). Also note that the game always has a winner. Does it mean that $k_1=k_2$? I guess it does, but if we are just trying to estimate the lower and the upper bound, we can come up with two different values.
combinatorics combinatorial-game-theory
asked Jan 5 at 20:38
OldboyOldboy
8,62711036
$endgroup$
1
$begingroup$
"I need some estimate for $k_1$ and $k_2$ (better than mine)." What are yours?
$endgroup$
– Peter Taylor
Jan 10 at 15:51
add a comment |
$begingroup$
(Once again a son is torturing his father...)
Alice and Bob play a fairly simple game with $n$ predefined points in 3D space. No four points are complanar (which also implies that no three points are collinear).
They play in turns by connecting points that are not yet connected with straight segments. On her turn Alice connects a single pair of unconnected points with a single segment of red color. After that Bob connects $k$ pairs of unconnected points with $k$ segments of blue color.
Alice wins if she is able to create a triangle with red sides. If all points are connected and Alice was not able to create a red sided triangle before that, Bob wins.
Who has a winning strategy?
My thoughts: obviously, the outcome of the game heavily depends on releation between $n$ and $k$.
Suppose that $k=1$ with big enough value for $n$. Alice will simply draw segment $AB$, Bob draws his segment at will, Alice then draws $AC$, Bob blocks triangle $ABC$ by drawing segment BC, Alice draws segment $AD$ and Bob is lost. He cannot block triangles $ABD$ and $ACD$ by drawgin a single segment.
On the other side, with $k$ big enough, Bob can easily block creation of all red triangles by drawing a lot of lines.
So for $kle k_1$, Alice wins. For $kge k_2ge k_1$, Bob wins. I need some estimate for $k_1$ and $k_2$ (better than mine). Also note that the game always has a winner. Does it mean that $k_1=k_2$? I guess it does, but if we are just trying to estimate the lower and the upper bound, we can come up with two different values.
combinatorics combinatorial-game-theory
asked Jan 5 at 20:38
OldboyOldboy
8,62711036
$endgroup$
(Once again a son is torturing his father...)
Alice and Bob play a fairly simple game with $n$ predefined points in 3D space. No four points are complanar (which also implies that no three points are collinear).
They play in turns by connecting points that are not yet connected with straight segments. On her turn Alice connects a single pair of unconnected points with a single segment of red color. After that Bob connects $k$ pairs of unconnected points with $k$ segments of blue color.
Alice wins if she is able to create a triangle with red sides. If all points are connected and Alice was not able to create a red sided triangle before that, Bob wins.
Who has a winning strategy?
My thoughts: obviously, the outcome of the game heavily depends on releation between $n$ and $k$.
Suppose that $k=1$ with big enough value for $n$. Alice will simply draw segment $AB$, Bob draws his segment at will, Alice then draws $AC$, Bob blocks triangle $ABC$ by drawing segment BC, Alice draws segment $AD$ and Bob is lost. He cannot block triangles $ABD$ and $ACD$ by drawgin a single segment.
On the other side, with $k$ big enough, Bob can easily block creation of all red triangles by drawing a lot of lines.
So for $kle k_1$, Alice wins. For $kge k_2ge k_1$, Bob wins. I need some estimate for $k_1$ and $k_2$ (better than mine). Also note that the game always has a winner. Does it mean that $k_1=k_2$? I guess it does, but if we are just trying to estimate the lower and the upper bound, we can come up with two different values.
combinatorics combinatorial-game-theory
combinatorics combinatorial-game-theory
asked Jan 5 at 20:38
OldboyOldboy
8,62711036
asked Jan 5 at 20:38
OldboyOldboy
8,62711036
asked Jan 5 at 20:38
OldboyOldboy
8,62711036
asked Jan 5 at 20:38
OldboyOldboy
8,62711036
asked Jan 5 at 20:38
OldboyOldboy
8,62711036
8,62711036
1
$begingroup$
"I need some estimate for $k_1$ and $k_2$ (better than mine)." What are yours?
$endgroup$
– Peter Taylor
Jan 10 at 15:51
add a comment |
1
$begingroup$
"I need some estimate for $k_1$ and $k_2$ (better than mine)." What are yours?
$endgroup$
– Peter Taylor
Jan 10 at 15:51
1
1
$begingroup$
"I need some estimate for $k_1$ and $k_2$ (better than mine)." What are yours?
$endgroup$
– Peter Taylor
Jan 10 at 15:51
$begingroup$
"I need some estimate for $k_1$ and $k_2$ (better than mine)." What are yours?
$endgroup$
– Peter Taylor
Jan 10 at 15:51
add a comment |
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1
$begingroup$
"I need some estimate for $k_1$ and $k_2$ (better than mine)." What are yours?
$endgroup$
– Peter Taylor
Jan 10 at 15:51