What is the winning strategy for this problem?
$begingroup$
Before the game starts, there are a few "points" on the desktop, and then the two players take turns to do the following operations until the operation can not be completed:
Starting from a "point" to draw a line continuously and uninterruptedly, and it does not intersect with any "point" and "line" on the way, and finally ends the curve on a "point". Later, we call the drawn line "line" and choose any point from the line to call it "point" (in fact, which point we choose as "point" will not have any impact).
However, every "point" is required to be connected with at most three "lines". A "line" passes through a "point" means that the "point" is connected with two "lines".
Meanwhile, you can also draw a "line" from a "point" to itself if satisfies the terms above.
My research shows that when the numbers of "points" in the very beginning are $ 1,2,5,6$, there are winning strategies for the second hand. However, when the numbers of "points" are $3,4,7$, there are winning strategies for the first hand. I guess that if the numbers of "points" are in the form of $4n+2, 4n+1$, the second hand has winning strategies.
My question is, is my result right? How to prove it? (And disprove it if it wrong)
discrete-mathematics game-theory combinatorial-game-theory
asked Dec 22 '18 at 10:09
Sky of warSky of war
84
$endgroup$
add a comment |
$begingroup$
Before the game starts, there are a few "points" on the desktop, and then the two players take turns to do the following operations until the operation can not be completed:
Starting from a "point" to draw a line continuously and uninterruptedly, and it does not intersect with any "point" and "line" on the way, and finally ends the curve on a "point". Later, we call the drawn line "line" and choose any point from the line to call it "point" (in fact, which point we choose as "point" will not have any impact).
However, every "point" is required to be connected with at most three "lines". A "line" passes through a "point" means that the "point" is connected with two "lines".
Meanwhile, you can also draw a "line" from a "point" to itself if satisfies the terms above.
My research shows that when the numbers of "points" in the very beginning are $ 1,2,5,6$, there are winning strategies for the second hand. However, when the numbers of "points" are $3,4,7$, there are winning strategies for the first hand. I guess that if the numbers of "points" are in the form of $4n+2, 4n+1$, the second hand has winning strategies.
My question is, is my result right? How to prove it? (And disprove it if it wrong)
discrete-mathematics game-theory combinatorial-game-theory
asked Dec 22 '18 at 10:09
Sky of warSky of war
84
$endgroup$
$begingroup$
Doesn't 1 point lead to second player winning? First player draws a loop and puts a point on the loop. Second player connects the point on the loop with the original point.
$endgroup$
– Todor Markov
Dec 22 '18 at 10:16
$begingroup$
oh, sorry, I typed that wrongly
$endgroup$
– Sky of war
Dec 22 '18 at 10:20
add a comment |
$begingroup$
Before the game starts, there are a few "points" on the desktop, and then the two players take turns to do the following operations until the operation can not be completed:
Starting from a "point" to draw a line continuously and uninterruptedly, and it does not intersect with any "point" and "line" on the way, and finally ends the curve on a "point". Later, we call the drawn line "line" and choose any point from the line to call it "point" (in fact, which point we choose as "point" will not have any impact).
However, every "point" is required to be connected with at most three "lines". A "line" passes through a "point" means that the "point" is connected with two "lines".
Meanwhile, you can also draw a "line" from a "point" to itself if satisfies the terms above.
My research shows that when the numbers of "points" in the very beginning are $ 1,2,5,6$, there are winning strategies for the second hand. However, when the numbers of "points" are $3,4,7$, there are winning strategies for the first hand. I guess that if the numbers of "points" are in the form of $4n+2, 4n+1$, the second hand has winning strategies.
My question is, is my result right? How to prove it? (And disprove it if it wrong)
discrete-mathematics game-theory combinatorial-game-theory
asked Dec 22 '18 at 10:09
Sky of warSky of war
84
$endgroup$
Before the game starts, there are a few "points" on the desktop, and then the two players take turns to do the following operations until the operation can not be completed:
Starting from a "point" to draw a line continuously and uninterruptedly, and it does not intersect with any "point" and "line" on the way, and finally ends the curve on a "point". Later, we call the drawn line "line" and choose any point from the line to call it "point" (in fact, which point we choose as "point" will not have any impact).
However, every "point" is required to be connected with at most three "lines". A "line" passes through a "point" means that the "point" is connected with two "lines".
Meanwhile, you can also draw a "line" from a "point" to itself if satisfies the terms above.
My research shows that when the numbers of "points" in the very beginning are $ 1,2,5,6$, there are winning strategies for the second hand. However, when the numbers of "points" are $3,4,7$, there are winning strategies for the first hand. I guess that if the numbers of "points" are in the form of $4n+2, 4n+1$, the second hand has winning strategies.
My question is, is my result right? How to prove it? (And disprove it if it wrong)
discrete-mathematics game-theory combinatorial-game-theory
discrete-mathematics game-theory combinatorial-game-theory
asked Dec 22 '18 at 10:09
Sky of warSky of war
84
asked Dec 22 '18 at 10:09
Sky of warSky of war
84
edited Dec 22 '18 at 10:21
Sky of war
asked Dec 22 '18 at 10:09
Sky of warSky of war
84
asked Dec 22 '18 at 10:09
Sky of warSky of war
84
asked Dec 22 '18 at 10:09
Sky of warSky of war
84
84
$begingroup$
Doesn't 1 point lead to second player winning? First player draws a loop and puts a point on the loop. Second player connects the point on the loop with the original point.
$endgroup$
– Todor Markov
Dec 22 '18 at 10:16
$begingroup$
oh, sorry, I typed that wrongly
$endgroup$
– Sky of war
Dec 22 '18 at 10:20
add a comment |
$begingroup$
Doesn't 1 point lead to second player winning? First player draws a loop and puts a point on the loop. Second player connects the point on the loop with the original point.
$endgroup$
– Todor Markov
Dec 22 '18 at 10:16
$begingroup$
oh, sorry, I typed that wrongly
$endgroup$
– Sky of war
Dec 22 '18 at 10:20
$begingroup$
Doesn't 1 point lead to second player winning? First player draws a loop and puts a point on the loop. Second player connects the point on the loop with the original point.
$endgroup$
– Todor Markov
Dec 22 '18 at 10:16
$begingroup$
Doesn't 1 point lead to second player winning? First player draws a loop and puts a point on the loop. Second player connects the point on the loop with the original point.
$endgroup$
– Todor Markov
Dec 22 '18 at 10:16
$begingroup$
oh, sorry, I typed that wrongly
$endgroup$
– Sky of war
Dec 22 '18 at 10:20
$begingroup$
oh, sorry, I typed that wrongly
$endgroup$
– Sky of war
Dec 22 '18 at 10:20
add a comment |
1 Answer
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This game is well-known and studied. It's called Sprouts.
It seems $5$ is a win for the first player (contrary to your assertion) and $6$ for
the second (which was a record for some time).
The as yet unproven conjecture is that it's a win for the first player when the number of starting points is $3,4,5$ modulo $6$. (Yours is a conjecture modulo $4$)
This has been confirmed up to $44$ points.
answered Dec 22 '18 at 10:48
Henno BrandsmaHenno Brandsma
116k349127
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add a comment |
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$begingroup$
This game is well-known and studied. It's called Sprouts.
It seems $5$ is a win for the first player (contrary to your assertion) and $6$ for
the second (which was a record for some time).
The as yet unproven conjecture is that it's a win for the first player when the number of starting points is $3,4,5$ modulo $6$. (Yours is a conjecture modulo $4$)
This has been confirmed up to $44$ points.
answered Dec 22 '18 at 10:48
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
This game is well-known and studied. It's called Sprouts.
It seems $5$ is a win for the first player (contrary to your assertion) and $6$ for
the second (which was a record for some time).
The as yet unproven conjecture is that it's a win for the first player when the number of starting points is $3,4,5$ modulo $6$. (Yours is a conjecture modulo $4$)
This has been confirmed up to $44$ points.
answered Dec 22 '18 at 10:48
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
This game is well-known and studied. It's called Sprouts.
It seems $5$ is a win for the first player (contrary to your assertion) and $6$ for
the second (which was a record for some time).
The as yet unproven conjecture is that it's a win for the first player when the number of starting points is $3,4,5$ modulo $6$. (Yours is a conjecture modulo $4$)
This has been confirmed up to $44$ points.
answered Dec 22 '18 at 10:48
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
This game is well-known and studied. It's called Sprouts.
It seems $5$ is a win for the first player (contrary to your assertion) and $6$ for
the second (which was a record for some time).
The as yet unproven conjecture is that it's a win for the first player when the number of starting points is $3,4,5$ modulo $6$. (Yours is a conjecture modulo $4$)
This has been confirmed up to $44$ points.
answered Dec 22 '18 at 10:48
Henno BrandsmaHenno Brandsma
116k349127
edited Dec 22 '18 at 19:47
answered Dec 22 '18 at 10:48
Henno BrandsmaHenno Brandsma
116k349127
answered Dec 22 '18 at 10:48
Henno BrandsmaHenno Brandsma
116k349127
answered Dec 22 '18 at 10:48
Henno BrandsmaHenno Brandsma
116k349127
116k349127
add a comment |
add a comment |
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$begingroup$
Doesn't 1 point lead to second player winning? First player draws a loop and puts a point on the loop. Second player connects the point on the loop with the original point.
$endgroup$
– Todor Markov
Dec 22 '18 at 10:16
$begingroup$
oh, sorry, I typed that wrongly
$endgroup$
– Sky of war
Dec 22 '18 at 10:20