What is actually happening in the Hackenbush advantage measurement?











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






















  • 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.
    2 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






















  • 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.
    2 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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









Billy Rubina

10.3k1458134




10.3k1458134












  • 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.
    2 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.
    2 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.
2 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.
2 mins ago










2 Answers
2






active

oldest

votes

















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





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042817%2fwhat-is-actually-happening-in-the-hackenbush-advantage-measurement%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      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













        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 3 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

























                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























                  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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042817%2fwhat-is-actually-happening-in-the-hackenbush-advantage-measurement%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror

                      Mangá

                      Eduardo VII do Reino Unido