What is actually happening in the Hackenbush advantage measurement?











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I'm reading Berlekamp/Conway/Guy's Winning Ways for Your Mathematical Plays. Here:



enter image description here





I am a little bit confused: What is happening here? It seems to me that we know that a game with a unique red edge is a $1-$move advantage for red. But we still can't know what is the advantage value for $(a)$, so we call the advantage of red and blue $r,b$. Then for $(a)$, we have $r,b$ advantages.



For $(b)$, we have $r+1,b-1$ advantages. Now $(c)$ is a zero position, it seems this allow us to write the following advantage equations: $2r+1=0, 2b-1=0$ and from this we can know the advantage value of a certain game for each player.



Is my interpretation correct? I am asking what is the "moral of the story", it seems that whenever we don't know the value of a game, we can try to "compose it" with some other games (such as the game with a single red or blue edge which we know it's value) until it forms a zero position, from which we can write a system of equations, solve and find the advantage value of each player in our unknown game.










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  • 1




    The answers are good, but I think part of the problem might be that the passage (and most of the book) you're reading is presented in a style as if the authors and readers are trying to discover the theory together (it's just that the authors already have many relevant case studies at hand), as opposed to a regular textbook where the theory is known and the author tries to explain it. For more traditional texts on the subject, see "Lessons in Play" by Albert, Nowakowski, and Wolfe or "An Introduction to Combinatorial Game Theory" by L. R. Haff & W. J. Garner.
    – Mark S.
    54 mins ago















up vote
3
down vote

favorite
1












I'm reading Berlekamp/Conway/Guy's Winning Ways for Your Mathematical Plays. Here:



enter image description here





I am a little bit confused: What is happening here? It seems to me that we know that a game with a unique red edge is a $1-$move advantage for red. But we still can't know what is the advantage value for $(a)$, so we call the advantage of red and blue $r,b$. Then for $(a)$, we have $r,b$ advantages.



For $(b)$, we have $r+1,b-1$ advantages. Now $(c)$ is a zero position, it seems this allow us to write the following advantage equations: $2r+1=0, 2b-1=0$ and from this we can know the advantage value of a certain game for each player.



Is my interpretation correct? I am asking what is the "moral of the story", it seems that whenever we don't know the value of a game, we can try to "compose it" with some other games (such as the game with a single red or blue edge which we know it's value) until it forms a zero position, from which we can write a system of equations, solve and find the advantage value of each player in our unknown game.










share|cite|improve this question


















  • 1




    The answers are good, but I think part of the problem might be that the passage (and most of the book) you're reading is presented in a style as if the authors and readers are trying to discover the theory together (it's just that the authors already have many relevant case studies at hand), as opposed to a regular textbook where the theory is known and the author tries to explain it. For more traditional texts on the subject, see "Lessons in Play" by Albert, Nowakowski, and Wolfe or "An Introduction to Combinatorial Game Theory" by L. R. Haff & W. J. Garner.
    – Mark S.
    54 mins ago













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





I'm reading Berlekamp/Conway/Guy's Winning Ways for Your Mathematical Plays. Here:



enter image description here





I am a little bit confused: What is happening here? It seems to me that we know that a game with a unique red edge is a $1-$move advantage for red. But we still can't know what is the advantage value for $(a)$, so we call the advantage of red and blue $r,b$. Then for $(a)$, we have $r,b$ advantages.



For $(b)$, we have $r+1,b-1$ advantages. Now $(c)$ is a zero position, it seems this allow us to write the following advantage equations: $2r+1=0, 2b-1=0$ and from this we can know the advantage value of a certain game for each player.



Is my interpretation correct? I am asking what is the "moral of the story", it seems that whenever we don't know the value of a game, we can try to "compose it" with some other games (such as the game with a single red or blue edge which we know it's value) until it forms a zero position, from which we can write a system of equations, solve and find the advantage value of each player in our unknown game.










share|cite|improve this question













I'm reading Berlekamp/Conway/Guy's Winning Ways for Your Mathematical Plays. Here:



enter image description here





I am a little bit confused: What is happening here? It seems to me that we know that a game with a unique red edge is a $1-$move advantage for red. But we still can't know what is the advantage value for $(a)$, so we call the advantage of red and blue $r,b$. Then for $(a)$, we have $r,b$ advantages.



For $(b)$, we have $r+1,b-1$ advantages. Now $(c)$ is a zero position, it seems this allow us to write the following advantage equations: $2r+1=0, 2b-1=0$ and from this we can know the advantage value of a certain game for each player.



Is my interpretation correct? I am asking what is the "moral of the story", it seems that whenever we don't know the value of a game, we can try to "compose it" with some other games (such as the game with a single red or blue edge which we know it's value) until it forms a zero position, from which we can write a system of equations, solve and find the advantage value of each player in our unknown game.







combinatorial-game-theory






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asked 4 hours ago









Billy Rubina

10.3k1458134




10.3k1458134








  • 1




    The answers are good, but I think part of the problem might be that the passage (and most of the book) you're reading is presented in a style as if the authors and readers are trying to discover the theory together (it's just that the authors already have many relevant case studies at hand), as opposed to a regular textbook where the theory is known and the author tries to explain it. For more traditional texts on the subject, see "Lessons in Play" by Albert, Nowakowski, and Wolfe or "An Introduction to Combinatorial Game Theory" by L. R. Haff & W. J. Garner.
    – Mark S.
    54 mins ago














  • 1




    The answers are good, but I think part of the problem might be that the passage (and most of the book) you're reading is presented in a style as if the authors and readers are trying to discover the theory together (it's just that the authors already have many relevant case studies at hand), as opposed to a regular textbook where the theory is known and the author tries to explain it. For more traditional texts on the subject, see "Lessons in Play" by Albert, Nowakowski, and Wolfe or "An Introduction to Combinatorial Game Theory" by L. R. Haff & W. J. Garner.
    – Mark S.
    54 mins ago








1




1




The answers are good, but I think part of the problem might be that the passage (and most of the book) you're reading is presented in a style as if the authors and readers are trying to discover the theory together (it's just that the authors already have many relevant case studies at hand), as opposed to a regular textbook where the theory is known and the author tries to explain it. For more traditional texts on the subject, see "Lessons in Play" by Albert, Nowakowski, and Wolfe or "An Introduction to Combinatorial Game Theory" by L. R. Haff & W. J. Garner.
– Mark S.
54 mins ago




The answers are good, but I think part of the problem might be that the passage (and most of the book) you're reading is presented in a style as if the authors and readers are trying to discover the theory together (it's just that the authors already have many relevant case studies at hand), as opposed to a regular textbook where the theory is known and the author tries to explain it. For more traditional texts on the subject, see "Lessons in Play" by Albert, Nowakowski, and Wolfe or "An Introduction to Combinatorial Game Theory" by L. R. Haff & W. J. Garner.
– Mark S.
54 mins ago










2 Answers
2






active

oldest

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up vote
3
down vote













A Hackenbush game has a value that is a number, so you just need one number for the value of the position, not $r,b$ separately. Yes, one way to value a position is to compose it with known positions and find a combination that has $0$ value, then use algebra to determine the value of the unknown position. If we let the value of position $a$ be $v$, the value of position $c$ is $2v-1$. Once we prove that is $0$ we can find $v=frac 12$ by algebra.



Another way is to look at the options in a position. The red above blue position is ${0|1}$ because blue can move to $0$ and red can move to $1$. There is a theorem coming that the value of ${a|b}$ is the simplest number that fits between $a$ and $b$. For ${0|1}$ that is $frac 12$






share|cite|improve this answer




























    up vote
    3
    down vote













    Your idea is correct. From a broader perspective, the set of (red-blue) Hackenbush positions (up to equivalence) form a totally ordered abelian group (called the surreal numbers): they have operations of addition and subtraction and a relation $leq$ which satisfy all the usual properties. Now, it's a theorem that any totally ordered abelian group $G$ satisfying a certain extra "finiteness" condition (the Archimedean axiom) is isomorphic to a subgroup of the real numbers. Namely, if you fix some element of $G$ to call "$1$", for every other element $gin G$ you can consider the set of rational numbers $frac{m}{n}$ such that $mcdot gleq ncdot 1$. This set forms a Dedekind cut in the rational numbers and so determines a real number. It can then be shown that mapping $g$ to this real number is an isomorphism of ordered abelian groups from $G$ to a subgroup of $mathbb{R}$.



    Now, in the case of Hackenbush, the set of finite Hackenbush positions (up to equivalence) satisfies the Archimedean axiom, and so this theorem applies. That means that when we identify some element to be $1$, there is a canonical way to identify such positions as real numbers. We choose to let "$1$" be a position with $1$-move advantage for Left, so that we can loosely think of the number associate to a position as "the number of moves that Left is ahead by". It turns out then that the subgroup of the real numbers corresponding to finite Hackenbush positions is the group of dyadic rationals.






    share|cite|improve this answer





















    • Yes. But the construction of these numbers is recursive, right? In the book, it seems that given $1$, I can construct $1/2$ and I need it to construct $3/2$.
      – Billy Rubina
      42 mins ago










    • I'm not sure what you mean by that. There is recursion involved but I'm not sure what you're talking about exactly or what you see as a problem.
      – Eric Wofsey
      39 mins ago











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    2 Answers
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    up vote
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    A Hackenbush game has a value that is a number, so you just need one number for the value of the position, not $r,b$ separately. Yes, one way to value a position is to compose it with known positions and find a combination that has $0$ value, then use algebra to determine the value of the unknown position. If we let the value of position $a$ be $v$, the value of position $c$ is $2v-1$. Once we prove that is $0$ we can find $v=frac 12$ by algebra.



    Another way is to look at the options in a position. The red above blue position is ${0|1}$ because blue can move to $0$ and red can move to $1$. There is a theorem coming that the value of ${a|b}$ is the simplest number that fits between $a$ and $b$. For ${0|1}$ that is $frac 12$






    share|cite|improve this answer

























      up vote
      3
      down vote













      A Hackenbush game has a value that is a number, so you just need one number for the value of the position, not $r,b$ separately. Yes, one way to value a position is to compose it with known positions and find a combination that has $0$ value, then use algebra to determine the value of the unknown position. If we let the value of position $a$ be $v$, the value of position $c$ is $2v-1$. Once we prove that is $0$ we can find $v=frac 12$ by algebra.



      Another way is to look at the options in a position. The red above blue position is ${0|1}$ because blue can move to $0$ and red can move to $1$. There is a theorem coming that the value of ${a|b}$ is the simplest number that fits between $a$ and $b$. For ${0|1}$ that is $frac 12$






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        A Hackenbush game has a value that is a number, so you just need one number for the value of the position, not $r,b$ separately. Yes, one way to value a position is to compose it with known positions and find a combination that has $0$ value, then use algebra to determine the value of the unknown position. If we let the value of position $a$ be $v$, the value of position $c$ is $2v-1$. Once we prove that is $0$ we can find $v=frac 12$ by algebra.



        Another way is to look at the options in a position. The red above blue position is ${0|1}$ because blue can move to $0$ and red can move to $1$. There is a theorem coming that the value of ${a|b}$ is the simplest number that fits between $a$ and $b$. For ${0|1}$ that is $frac 12$






        share|cite|improve this answer












        A Hackenbush game has a value that is a number, so you just need one number for the value of the position, not $r,b$ separately. Yes, one way to value a position is to compose it with known positions and find a combination that has $0$ value, then use algebra to determine the value of the unknown position. If we let the value of position $a$ be $v$, the value of position $c$ is $2v-1$. Once we prove that is $0$ we can find $v=frac 12$ by algebra.



        Another way is to look at the options in a position. The red above blue position is ${0|1}$ because blue can move to $0$ and red can move to $1$. There is a theorem coming that the value of ${a|b}$ is the simplest number that fits between $a$ and $b$. For ${0|1}$ that is $frac 12$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        Ross Millikan

        290k23195368




        290k23195368






















            up vote
            3
            down vote













            Your idea is correct. From a broader perspective, the set of (red-blue) Hackenbush positions (up to equivalence) form a totally ordered abelian group (called the surreal numbers): they have operations of addition and subtraction and a relation $leq$ which satisfy all the usual properties. Now, it's a theorem that any totally ordered abelian group $G$ satisfying a certain extra "finiteness" condition (the Archimedean axiom) is isomorphic to a subgroup of the real numbers. Namely, if you fix some element of $G$ to call "$1$", for every other element $gin G$ you can consider the set of rational numbers $frac{m}{n}$ such that $mcdot gleq ncdot 1$. This set forms a Dedekind cut in the rational numbers and so determines a real number. It can then be shown that mapping $g$ to this real number is an isomorphism of ordered abelian groups from $G$ to a subgroup of $mathbb{R}$.



            Now, in the case of Hackenbush, the set of finite Hackenbush positions (up to equivalence) satisfies the Archimedean axiom, and so this theorem applies. That means that when we identify some element to be $1$, there is a canonical way to identify such positions as real numbers. We choose to let "$1$" be a position with $1$-move advantage for Left, so that we can loosely think of the number associate to a position as "the number of moves that Left is ahead by". It turns out then that the subgroup of the real numbers corresponding to finite Hackenbush positions is the group of dyadic rationals.






            share|cite|improve this answer





















            • Yes. But the construction of these numbers is recursive, right? In the book, it seems that given $1$, I can construct $1/2$ and I need it to construct $3/2$.
              – Billy Rubina
              42 mins ago










            • I'm not sure what you mean by that. There is recursion involved but I'm not sure what you're talking about exactly or what you see as a problem.
              – Eric Wofsey
              39 mins ago















            up vote
            3
            down vote













            Your idea is correct. From a broader perspective, the set of (red-blue) Hackenbush positions (up to equivalence) form a totally ordered abelian group (called the surreal numbers): they have operations of addition and subtraction and a relation $leq$ which satisfy all the usual properties. Now, it's a theorem that any totally ordered abelian group $G$ satisfying a certain extra "finiteness" condition (the Archimedean axiom) is isomorphic to a subgroup of the real numbers. Namely, if you fix some element of $G$ to call "$1$", for every other element $gin G$ you can consider the set of rational numbers $frac{m}{n}$ such that $mcdot gleq ncdot 1$. This set forms a Dedekind cut in the rational numbers and so determines a real number. It can then be shown that mapping $g$ to this real number is an isomorphism of ordered abelian groups from $G$ to a subgroup of $mathbb{R}$.



            Now, in the case of Hackenbush, the set of finite Hackenbush positions (up to equivalence) satisfies the Archimedean axiom, and so this theorem applies. That means that when we identify some element to be $1$, there is a canonical way to identify such positions as real numbers. We choose to let "$1$" be a position with $1$-move advantage for Left, so that we can loosely think of the number associate to a position as "the number of moves that Left is ahead by". It turns out then that the subgroup of the real numbers corresponding to finite Hackenbush positions is the group of dyadic rationals.






            share|cite|improve this answer





















            • Yes. But the construction of these numbers is recursive, right? In the book, it seems that given $1$, I can construct $1/2$ and I need it to construct $3/2$.
              – Billy Rubina
              42 mins ago










            • I'm not sure what you mean by that. There is recursion involved but I'm not sure what you're talking about exactly or what you see as a problem.
              – Eric Wofsey
              39 mins ago













            up vote
            3
            down vote










            up vote
            3
            down vote









            Your idea is correct. From a broader perspective, the set of (red-blue) Hackenbush positions (up to equivalence) form a totally ordered abelian group (called the surreal numbers): they have operations of addition and subtraction and a relation $leq$ which satisfy all the usual properties. Now, it's a theorem that any totally ordered abelian group $G$ satisfying a certain extra "finiteness" condition (the Archimedean axiom) is isomorphic to a subgroup of the real numbers. Namely, if you fix some element of $G$ to call "$1$", for every other element $gin G$ you can consider the set of rational numbers $frac{m}{n}$ such that $mcdot gleq ncdot 1$. This set forms a Dedekind cut in the rational numbers and so determines a real number. It can then be shown that mapping $g$ to this real number is an isomorphism of ordered abelian groups from $G$ to a subgroup of $mathbb{R}$.



            Now, in the case of Hackenbush, the set of finite Hackenbush positions (up to equivalence) satisfies the Archimedean axiom, and so this theorem applies. That means that when we identify some element to be $1$, there is a canonical way to identify such positions as real numbers. We choose to let "$1$" be a position with $1$-move advantage for Left, so that we can loosely think of the number associate to a position as "the number of moves that Left is ahead by". It turns out then that the subgroup of the real numbers corresponding to finite Hackenbush positions is the group of dyadic rationals.






            share|cite|improve this answer












            Your idea is correct. From a broader perspective, the set of (red-blue) Hackenbush positions (up to equivalence) form a totally ordered abelian group (called the surreal numbers): they have operations of addition and subtraction and a relation $leq$ which satisfy all the usual properties. Now, it's a theorem that any totally ordered abelian group $G$ satisfying a certain extra "finiteness" condition (the Archimedean axiom) is isomorphic to a subgroup of the real numbers. Namely, if you fix some element of $G$ to call "$1$", for every other element $gin G$ you can consider the set of rational numbers $frac{m}{n}$ such that $mcdot gleq ncdot 1$. This set forms a Dedekind cut in the rational numbers and so determines a real number. It can then be shown that mapping $g$ to this real number is an isomorphism of ordered abelian groups from $G$ to a subgroup of $mathbb{R}$.



            Now, in the case of Hackenbush, the set of finite Hackenbush positions (up to equivalence) satisfies the Archimedean axiom, and so this theorem applies. That means that when we identify some element to be $1$, there is a canonical way to identify such positions as real numbers. We choose to let "$1$" be a position with $1$-move advantage for Left, so that we can loosely think of the number associate to a position as "the number of moves that Left is ahead by". It turns out then that the subgroup of the real numbers corresponding to finite Hackenbush positions is the group of dyadic rationals.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            Eric Wofsey

            177k12202328




            177k12202328












            • Yes. But the construction of these numbers is recursive, right? In the book, it seems that given $1$, I can construct $1/2$ and I need it to construct $3/2$.
              – Billy Rubina
              42 mins ago










            • I'm not sure what you mean by that. There is recursion involved but I'm not sure what you're talking about exactly or what you see as a problem.
              – Eric Wofsey
              39 mins ago


















            • Yes. But the construction of these numbers is recursive, right? In the book, it seems that given $1$, I can construct $1/2$ and I need it to construct $3/2$.
              – Billy Rubina
              42 mins ago










            • I'm not sure what you mean by that. There is recursion involved but I'm not sure what you're talking about exactly or what you see as a problem.
              – Eric Wofsey
              39 mins ago
















            Yes. But the construction of these numbers is recursive, right? In the book, it seems that given $1$, I can construct $1/2$ and I need it to construct $3/2$.
            – Billy Rubina
            42 mins ago




            Yes. But the construction of these numbers is recursive, right? In the book, it seems that given $1$, I can construct $1/2$ and I need it to construct $3/2$.
            – Billy Rubina
            42 mins ago












            I'm not sure what you mean by that. There is recursion involved but I'm not sure what you're talking about exactly or what you see as a problem.
            – Eric Wofsey
            39 mins ago




            I'm not sure what you mean by that. There is recursion involved but I'm not sure what you're talking about exactly or what you see as a problem.
            – Eric Wofsey
            39 mins ago


















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