Mixing
Take away, subtraction or nim?
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During most of my life my math skills were an absolute joke mostly due to bad teatchers (since 1st grade till 11th grade) but at some point, though late, of my life fortunatelly I was blessed with a bit of understanding of math, and what a beautiful world it is. I'm saying this in advance in order to humbly ask to whoever can to please help me in the most simple way possible in order for me to understand because I know that I'm no Cauchy.
Here's the problem: In a bar a guy comes to me with a game. The rules are as follow:
1- There was a matrix 3*5 of coins on the tabble.
2- Two players required to play.
3- Sets of 1,2 or 3 coins could be drawed at time.
4- The player who wipes the tabble clean wins.
Now here's the twist, that fellow droped out of school very early so his mathematical skills are very limited let's say, so I'm guessing that there must be a simple way to understand how to win. Never the less he aced this game with such a confidence and he really won every single time no matter the opponent. I searched for an answer but so far only got confused.
Of my refection I figyred out that:
1- If it's my turn and there are 5, 6 or 7 coins it's a win to me.
2- The only way to assure a win is to let the other player with 8 coins on the tabble thus leadding to 100% of chance for me to win.
The question is what's the strattegy to win and what happens when both players kwon it?
thank you.
discrete-mathematics
asked Jul 19 at 5:07
mxajd
163
add a comment |Â
up vote
0
down vote
favorite
During most of my life my math skills were an absolute joke mostly due to bad teatchers (since 1st grade till 11th grade) but at some point, though late, of my life fortunatelly I was blessed with a bit of understanding of math, and what a beautiful world it is. I'm saying this in advance in order to humbly ask to whoever can to please help me in the most simple way possible in order for me to understand because I know that I'm no Cauchy.
Here's the problem: In a bar a guy comes to me with a game. The rules are as follow:
1- There was a matrix 3*5 of coins on the tabble.
2- Two players required to play.
3- Sets of 1,2 or 3 coins could be drawed at time.
4- The player who wipes the tabble clean wins.
Now here's the twist, that fellow droped out of school very early so his mathematical skills are very limited let's say, so I'm guessing that there must be a simple way to understand how to win. Never the less he aced this game with such a confidence and he really won every single time no matter the opponent. I searched for an answer but so far only got confused.
Of my refection I figyred out that:
1- If it's my turn and there are 5, 6 or 7 coins it's a win to me.
2- The only way to assure a win is to let the other player with 8 coins on the tabble thus leadding to 100% of chance for me to win.
The question is what's the strattegy to win and what happens when both players kwon it?
thank you.
discrete-mathematics
asked Jul 19 at 5:07
mxajd
163
leave a multiple of $4$.
– Lord Shark the Unknown
Jul 19 at 5:16
Thank you for the quick repply. But here's the thing if he goes first and takes 3 out I have got no chace to win, assuming of course that he's proficient at the game. And this is what has been bothering me. Like I said, my math skills are not very strong but I'd say that if I start and take 3 out (of 15) I'll always win no matter what because doesn't matter if he takes 1, 2 or 3, I can then leave him with the magical multiple of 4, the 8. Same goes the other way arround.
– mxajd
Jul 19 at 5:26
If there are 15 coins, and your opponent goes first, then the most important thing for you to do is to not wager any money on the game.
– Lord Shark the Unknown
Jul 19 at 5:39
@LordSharktheUnknown Wise words, thank you.
– mxajd
Jul 19 at 5:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
During most of my life my math skills were an absolute joke mostly due to bad teatchers (since 1st grade till 11th grade) but at some point, though late, of my life fortunatelly I was blessed with a bit of understanding of math, and what a beautiful world it is. I'm saying this in advance in order to humbly ask to whoever can to please help me in the most simple way possible in order for me to understand because I know that I'm no Cauchy.
Here's the problem: In a bar a guy comes to me with a game. The rules are as follow:
1- There was a matrix 3*5 of coins on the tabble.
2- Two players required to play.
3- Sets of 1,2 or 3 coins could be drawed at time.
4- The player who wipes the tabble clean wins.
Now here's the twist, that fellow droped out of school very early so his mathematical skills are very limited let's say, so I'm guessing that there must be a simple way to understand how to win. Never the less he aced this game with such a confidence and he really won every single time no matter the opponent. I searched for an answer but so far only got confused.
Of my refection I figyred out that:
1- If it's my turn and there are 5, 6 or 7 coins it's a win to me.
2- The only way to assure a win is to let the other player with 8 coins on the tabble thus leadding to 100% of chance for me to win.
The question is what's the strattegy to win and what happens when both players kwon it?
thank you.
discrete-mathematics
asked Jul 19 at 5:07
mxajd
163
During most of my life my math skills were an absolute joke mostly due to bad teatchers (since 1st grade till 11th grade) but at some point, though late, of my life fortunatelly I was blessed with a bit of understanding of math, and what a beautiful world it is. I'm saying this in advance in order to humbly ask to whoever can to please help me in the most simple way possible in order for me to understand because I know that I'm no Cauchy.
Here's the problem: In a bar a guy comes to me with a game. The rules are as follow:
1- There was a matrix 3*5 of coins on the tabble.
2- Two players required to play.
3- Sets of 1,2 or 3 coins could be drawed at time.
4- The player who wipes the tabble clean wins.
Now here's the twist, that fellow droped out of school very early so his mathematical skills are very limited let's say, so I'm guessing that there must be a simple way to understand how to win. Never the less he aced this game with such a confidence and he really won every single time no matter the opponent. I searched for an answer but so far only got confused.
Of my refection I figyred out that:
1- If it's my turn and there are 5, 6 or 7 coins it's a win to me.
2- The only way to assure a win is to let the other player with 8 coins on the tabble thus leadding to 100% of chance for me to win.
The question is what's the strattegy to win and what happens when both players kwon it?
thank you.
discrete-mathematics
asked Jul 19 at 5:07
mxajd
163
asked Jul 19 at 5:07
mxajd
163
asked Jul 19 at 5:07
mxajd
163
asked Jul 19 at 5:07
mxajd
163
163
leave a multiple of $4$.
– Lord Shark the Unknown
Jul 19 at 5:16
Thank you for the quick repply. But here's the thing if he goes first and takes 3 out I have got no chace to win, assuming of course that he's proficient at the game. And this is what has been bothering me. Like I said, my math skills are not very strong but I'd say that if I start and take 3 out (of 15) I'll always win no matter what because doesn't matter if he takes 1, 2 or 3, I can then leave him with the magical multiple of 4, the 8. Same goes the other way arround.
– mxajd
Jul 19 at 5:26
If there are 15 coins, and your opponent goes first, then the most important thing for you to do is to not wager any money on the game.
– Lord Shark the Unknown
Jul 19 at 5:39
@LordSharktheUnknown Wise words, thank you.
– mxajd
Jul 19 at 5:49
add a comment |Â
leave a multiple of $4$.
– Lord Shark the Unknown
Jul 19 at 5:16
Thank you for the quick repply. But here's the thing if he goes first and takes 3 out I have got no chace to win, assuming of course that he's proficient at the game. And this is what has been bothering me. Like I said, my math skills are not very strong but I'd say that if I start and take 3 out (of 15) I'll always win no matter what because doesn't matter if he takes 1, 2 or 3, I can then leave him with the magical multiple of 4, the 8. Same goes the other way arround.
– mxajd
Jul 19 at 5:26
If there are 15 coins, and your opponent goes first, then the most important thing for you to do is to not wager any money on the game.
– Lord Shark the Unknown
Jul 19 at 5:39
@LordSharktheUnknown Wise words, thank you.
– mxajd
Jul 19 at 5:49
leave a multiple of $4$.
– Lord Shark the Unknown
Jul 19 at 5:16
leave a multiple of $4$.
– Lord Shark the Unknown
Jul 19 at 5:16
Thank you for the quick repply. But here's the thing if he goes first and takes 3 out I have got no chace to win, assuming of course that he's proficient at the game. And this is what has been bothering me. Like I said, my math skills are not very strong but I'd say that if I start and take 3 out (of 15) I'll always win no matter what because doesn't matter if he takes 1, 2 or 3, I can then leave him with the magical multiple of 4, the 8. Same goes the other way arround.
– mxajd
Jul 19 at 5:26
Thank you for the quick repply. But here's the thing if he goes first and takes 3 out I have got no chace to win, assuming of course that he's proficient at the game. And this is what has been bothering me. Like I said, my math skills are not very strong but I'd say that if I start and take 3 out (of 15) I'll always win no matter what because doesn't matter if he takes 1, 2 or 3, I can then leave him with the magical multiple of 4, the 8. Same goes the other way arround.
– mxajd
Jul 19 at 5:26
If there are 15 coins, and your opponent goes first, then the most important thing for you to do is to not wager any money on the game.
– Lord Shark the Unknown
Jul 19 at 5:39
If there are 15 coins, and your opponent goes first, then the most important thing for you to do is to not wager any money on the game.
– Lord Shark the Unknown
Jul 19 at 5:39
@LordSharktheUnknown Wise words, thank you.
– mxajd
Jul 19 at 5:49
@LordSharktheUnknown Wise words, thank you.
– mxajd
Jul 19 at 5:49
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
-1
down vote
You have to know a bit about graph theory. The main idea is to obtain a directed graph that represents the game and the player who is in the kernel wins.
What is a kernel y directed graph theory? it is an independent set of vertices $N$ such that for every vertex $vnotin N$ there exists an arrow from $v$ to a vertex in $N$.
Now, how would this directed graph look like? well every vertex must represent an amount of coins there are on the board. There is an arrow from $v$ to $w$
if $v-w leq 3$.
So the set of vertices is $0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$ and the
arrows are as I described above. Since this is an acyclic digraph it has a kernel. In this case the kernel is the set $N= 0,4,8,12$ which are the multiples of 4. So if you want to win, you must make sure to always leave an amount of coins that is a multiple of four.
Sorry I dont have a picture of the graph, I also don't know how to make a graph with latex.
answered Jul 19 at 5:28
allizdog
592115
Thank you for the reply. True, the k*4 is the answer. But In this scenario if both players are aware of this information it's just a matter of whoever goes first to take 3 out. So the answer to my question would be this conclusion right?
– mxajd
Jul 19 at 5:31
yes, if both players know this, the first one to step into the kernel wins.
– allizdog
Jul 19 at 5:42
What a bummer. But hey thank you so much @allizdog , at least I have my answer.
– mxajd
Jul 19 at 5:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
You have to know a bit about graph theory. The main idea is to obtain a directed graph that represents the game and the player who is in the kernel wins.
What is a kernel y directed graph theory? it is an independent set of vertices $N$ such that for every vertex $vnotin N$ there exists an arrow from $v$ to a vertex in $N$.
Now, how would this directed graph look like? well every vertex must represent an amount of coins there are on the board. There is an arrow from $v$ to $w$
if $v-w leq 3$.
So the set of vertices is $0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$ and the
arrows are as I described above. Since this is an acyclic digraph it has a kernel. In this case the kernel is the set $N= 0,4,8,12$ which are the multiples of 4. So if you want to win, you must make sure to always leave an amount of coins that is a multiple of four.
Sorry I dont have a picture of the graph, I also don't know how to make a graph with latex.
answered Jul 19 at 5:28
allizdog
592115
Thank you for the reply. True, the k*4 is the answer. But In this scenario if both players are aware of this information it's just a matter of whoever goes first to take 3 out. So the answer to my question would be this conclusion right?
– mxajd
Jul 19 at 5:31
yes, if both players know this, the first one to step into the kernel wins.
– allizdog
Jul 19 at 5:42
What a bummer. But hey thank you so much @allizdog , at least I have my answer.
– mxajd
Jul 19 at 5:48
add a comment |Â
up vote
-1
down vote
You have to know a bit about graph theory. The main idea is to obtain a directed graph that represents the game and the player who is in the kernel wins.
What is a kernel y directed graph theory? it is an independent set of vertices $N$ such that for every vertex $vnotin N$ there exists an arrow from $v$ to a vertex in $N$.
Now, how would this directed graph look like? well every vertex must represent an amount of coins there are on the board. There is an arrow from $v$ to $w$
if $v-w leq 3$.
So the set of vertices is $0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$ and the
arrows are as I described above. Since this is an acyclic digraph it has a kernel. In this case the kernel is the set $N= 0,4,8,12$ which are the multiples of 4. So if you want to win, you must make sure to always leave an amount of coins that is a multiple of four.
Sorry I dont have a picture of the graph, I also don't know how to make a graph with latex.
answered Jul 19 at 5:28
allizdog
592115
Thank you for the reply. True, the k*4 is the answer. But In this scenario if both players are aware of this information it's just a matter of whoever goes first to take 3 out. So the answer to my question would be this conclusion right?
– mxajd
Jul 19 at 5:31
yes, if both players know this, the first one to step into the kernel wins.
– allizdog
Jul 19 at 5:42
What a bummer. But hey thank you so much @allizdog , at least I have my answer.
– mxajd
Jul 19 at 5:48
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
You have to know a bit about graph theory. The main idea is to obtain a directed graph that represents the game and the player who is in the kernel wins.
What is a kernel y directed graph theory? it is an independent set of vertices $N$ such that for every vertex $vnotin N$ there exists an arrow from $v$ to a vertex in $N$.
Now, how would this directed graph look like? well every vertex must represent an amount of coins there are on the board. There is an arrow from $v$ to $w$
if $v-w leq 3$.
So the set of vertices is $0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$ and the
arrows are as I described above. Since this is an acyclic digraph it has a kernel. In this case the kernel is the set $N= 0,4,8,12$ which are the multiples of 4. So if you want to win, you must make sure to always leave an amount of coins that is a multiple of four.
Sorry I dont have a picture of the graph, I also don't know how to make a graph with latex.
answered Jul 19 at 5:28
allizdog
592115
You have to know a bit about graph theory. The main idea is to obtain a directed graph that represents the game and the player who is in the kernel wins.
What is a kernel y directed graph theory? it is an independent set of vertices $N$ such that for every vertex $vnotin N$ there exists an arrow from $v$ to a vertex in $N$.
Now, how would this directed graph look like? well every vertex must represent an amount of coins there are on the board. There is an arrow from $v$ to $w$
if $v-w leq 3$.
So the set of vertices is $0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$ and the
arrows are as I described above. Since this is an acyclic digraph it has a kernel. In this case the kernel is the set $N= 0,4,8,12$ which are the multiples of 4. So if you want to win, you must make sure to always leave an amount of coins that is a multiple of four.
Sorry I dont have a picture of the graph, I also don't know how to make a graph with latex.
answered Jul 19 at 5:28
allizdog
592115
answered Jul 19 at 5:28
allizdog
592115
answered Jul 19 at 5:28
allizdog
592115
answered Jul 19 at 5:28
allizdog
592115
592115
Thank you for the reply. True, the k*4 is the answer. But In this scenario if both players are aware of this information it's just a matter of whoever goes first to take 3 out. So the answer to my question would be this conclusion right?
– mxajd
Jul 19 at 5:31
yes, if both players know this, the first one to step into the kernel wins.
– allizdog
Jul 19 at 5:42
What a bummer. But hey thank you so much @allizdog , at least I have my answer.
– mxajd
Jul 19 at 5:48
add a comment |Â
Thank you for the reply. True, the k*4 is the answer. But In this scenario if both players are aware of this information it's just a matter of whoever goes first to take 3 out. So the answer to my question would be this conclusion right?
– mxajd
Jul 19 at 5:31
yes, if both players know this, the first one to step into the kernel wins.
– allizdog
Jul 19 at 5:42
What a bummer. But hey thank you so much @allizdog , at least I have my answer.
– mxajd
Jul 19 at 5:48
Thank you for the reply. True, the k*4 is the answer. But In this scenario if both players are aware of this information it's just a matter of whoever goes first to take 3 out. So the answer to my question would be this conclusion right?
– mxajd
Jul 19 at 5:31
Thank you for the reply. True, the k*4 is the answer. But In this scenario if both players are aware of this information it's just a matter of whoever goes first to take 3 out. So the answer to my question would be this conclusion right?
– mxajd
Jul 19 at 5:31
yes, if both players know this, the first one to step into the kernel wins.
– allizdog
Jul 19 at 5:42
yes, if both players know this, the first one to step into the kernel wins.
– allizdog
Jul 19 at 5:42
What a bummer. But hey thank you so much @allizdog , at least I have my answer.
– mxajd
Jul 19 at 5:48
What a bummer. But hey thank you so much @allizdog , at least I have my answer.
– mxajd
Jul 19 at 5:48
add a comment |Â
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leave a multiple of $4$.
– Lord Shark the Unknown
Jul 19 at 5:16
Thank you for the quick repply. But here's the thing if he goes first and takes 3 out I have got no chace to win, assuming of course that he's proficient at the game. And this is what has been bothering me. Like I said, my math skills are not very strong but I'd say that if I start and take 3 out (of 15) I'll always win no matter what because doesn't matter if he takes 1, 2 or 3, I can then leave him with the magical multiple of 4, the 8. Same goes the other way arround.
– mxajd
Jul 19 at 5:26
If there are 15 coins, and your opponent goes first, then the most important thing for you to do is to not wager any money on the game.
– Lord Shark the Unknown
Jul 19 at 5:39
@LordSharktheUnknown Wise words, thank you.
– mxajd
Jul 19 at 5:49