Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In combinatorial game theory, the Norton product of a game $G$ and a positive (short) game $U$ in canonical form is defined recursively by:

$G \cdot U = \{G^L \cdot U + (U+I)\,|\, G^R \cdot U - (U+I) \}$ if $G$ is not an integer

where $I$ ranges over all $U^L - U$ and $U-U^R$ (the incentives of $U$). If $G = n$ is an integer we use the obvious definition for $n \cdot U$, i.e., the sum of $|n|$ copies of $U$ or $-U$.

It is known that this product has nice properties: $G \cdot U$ is independent of the form of $G$, the product is monotonic ($G \ge H$ implies $G \cdot U \ge H \cdot U$), and the product distributes over addition. This is all in Winning Ways (see the Extras to Volume 1, Chapter 8).

By some experimentation using CGSuite, it appears that the mean value operator behaves like a homomorphism with respect to products: $$\mu(G \cdot U) = \mu(G) \cdot \mu(U)$$ where $\mu$ denotes the mean value of a game, the product on the left is Norton product, and the product on the right is the ordinary product of rational numbers. Is this true? I have not seen this fact (if it is true) mentioned before.

(Here we assume all games are short (i.e. have only finitely many positions) so that the means are defined, and are dyadic rationals.)

Edit: Here is a simple verification when $G = \frac{1}{2} = \{0\,|\,1\}$ and $U = \frac{p}{2^q}$ ($p,q$ positive integers). Then $I = \frac{1}{2^q}$ and therefore $$G \cdot U = \left\{ \frac{p+1}{2^q}\,|\,-\frac{1}{2^q}\right\}$$ and the mean value of the right side is $\frac{p}{2^{q+1}}$ as expected.

share|cite|improve this question

This is true. First we prove it for the case where $G$ is a dyadic rational. If $G = p/2^q$, then by the linearity of the mean value and Norton product, $$2^q \mu(G \cdot U) = \mu(p \cdot U) = p \cdot \mu(U)$$ and hence $\mu(G \cdot U) = (p/2^q) \cdot \mu(U) = \mu(G)\mu(U)$ as required.

In the case where $G$ is a general short game, there exists a dyadic rational $t$ (the temperature of $G$) such that for every positive integer $n$, $$n \mu(G) - t \le n \cdot G \le n \mu(G) + t.$$ This characterization of the mean value is found in On Numbers and Games, Chapter 9, Theorem 59.

By the linearity and monotonicity of Norton product, we have $$(n \mu(G) - t) \cdot U \le n \cdot (G \cdot U) \le (n \mu(G) + t) \cdot U$$ and taking the mean value throughout (since mean value is also linear and monotonic) $$\mu((n \mu(G) - t) \cdot U) \le n \cdot \mu(G \cdot U) \le \mu((n \mu(G) + t) \cdot U).$$ The numbers $n \mu(G) \pm t$ in the first and last terms above are dyadic rationals; hence, using the already-proved dyadic rational case, $$(n \mu(G) - t) \cdot \mu(U) \le n \cdot \mu(G \cdot U) \le (n \mu(G) + t) \cdot \mu(U).$$ Now dividing throughout by $n$ and taking the limit as $n \to \infty$ gives the desired result $\mu(G \cdot U) = \mu(G) \cdot \mu(U)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.