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

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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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    $\begingroup$ 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. $\endgroup$ – Mark S. Dec 16 '18 at 20:29
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    $\begingroup$ Could you define what you mean by “$ r,b $ advantages” or refer to a definition? As the answer by Ross Millikan indicates, you only need a single “value” to specify the value of a game, and for Hackenbush that value turns out to be a number, which is just a dyadic fraction (denominator is $ 2 ^n $) if the game is finite. You also say “For (b), we have $ r+1,b−1 $ advantages”, which (like the sequel) suggests you are applying some rules of how to combine “advantages”; could you perhaps also make them explicit? $\endgroup$ – PJTraill Dec 17 '18 at 0:27
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    $\begingroup$ Perhaps a source of confusion is the unstated assumption (though perhaps it is fairer to regard it as a hypothesis to be taken for a test-drive) in the second paragraph on page 5 that one can add games together by combining them, and that (if they have numerical values) the value of that sum of games is the sum of the values. On page 12 under Do our Methods work? they address doubts the reader may have, and allude to the mathematical discussion coming up in Chapter 2. As @MarkS. indicates, in this book you have to think like a collaborating researcher and intuit where they are going. $\endgroup$ – PJTraill Dec 17 '18 at 0:46
  • $\begingroup$ @MarkS. Thanks for the recommendations! I'll look at them. $\endgroup$ – Billy Rubina Dec 17 '18 at 2:38
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    $\begingroup$ @PJTraill I can't because I'm trying to guess what is happening first. I guess that is the maximum infomation I can provide. $\endgroup$ – Billy Rubina Dec 17 '18 at 2:53

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 $g\in G$ you can consider the set of rational numbers $\frac{m}{n}$ such that $m\cdot g\leq n\cdot 1$. 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.

  • $\begingroup$ 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 $3/2$. $\endgroup$ – Billy Rubina Dec 16 '18 at 20:41
  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey Dec 16 '18 at 20:44
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    $\begingroup$ @BillyRubina: See page 7 for all the numbers you need to construct $ \frac 1 2 $ and $ \frac 3 2 $: $ 0 $ is constructed as $ \{ | \} $, from that you construct $ -1 = \{ | 0 \} $ and $ +1 = \{ 0 | \} $. Then you get $ \frac 1 2 = \{ 0 | 1 \} $, $ 2 = \{ 1 | \} $ and $ \frac 3 2 = \{ 1 | 2 \} $. $\endgroup$ – PJTraill Dec 17 '18 at 0:35
  • $\begingroup$ @EricWofsey It was something like what PJTraill said. $\endgroup$ – Billy Rubina Dec 17 '18 at 2:35

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 $\frac 12$


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