Talk:Nimber

Standard value?

Nice article! I didn't understand this:

(2) The nimber square of a Fermat 2-power x is equal to the standard value of 3x/2.

What does "standard value of 3x/2" mean? AxelBoldt 01:50 Jan 5, 2003 (UTC)

Thanks. The part that you asked about has been edited. It means the value of 3x/2 under the ordinary multiplication of natural numbers, so that the nimber square of 2 is 3, the nimber square of 4 is 6, the nimber square of 16 is 24, the nimber square of 256 is 384, etc. David McAnally 21:22 Jul 1, 2004 (AEST)

Nested exponents

Other than the clumsy workaround 2^{2^n} or some sort of graphic (such as appears on the Fermat number page), is there any way to get Wikipedia to deal with nested exponents? I changed to the above workaround, as the previous format looked like 2²n in my browser. MJSS 20:29, 3 Sep 2004 (UTC)

If worse comes to worst you could resort to TeX and write: $2^{2^{n}}$. Formerly, TeX looked good when "displayed" and often looked terrible when embedded in lines of text. Now it may look less than beautiful, but usually is nowhere near as bad as it used to be. Of course, if you're going to "display" it, thus

$\int _{0}^{\infty }1\,dx$

then using TeX works well. Michael Hardy 22:17, 3 Sep 2004 (UTC)

Oh -- one more thing. The paragraph I wrote above was already indented, so I need to indent the "displayed" TeX twice. Michael Hardy 22:18, 3 Sep 2004 (UTC)

There is a similar problem with omega^omrga^omega at the point where it was pointed out that the set of nimbers less than omega^omega^omega is an algebraically closed field. Like 2^2^n, it does not look right, although it once did.

David McAnally 25 Septemeber, 2004

What does the following look like to you?

$\omega ^{\omega ^{\omega }}\neq 2^{2^{n}}.$

Specifically where in the article is the problem you're referring to?

Michael Hardy 00:14, 26 Sep 2004 (UTC)

Nimber multiplication of natural numbers

We explain that nimber addition of natural numbers corresponds to the XOR operator on their binary representations. Is there an XOR-related statement for nimber multiplication of natural numbers that can be added to the article? —Quantling (talk | contribs) 20:25, 27 October 2025 (UTC)

Not as easy a statement as that. There is a recursive algorithm obtained by dividing 2^k-bit numbers into pairs of 2^(k-1)-bit numbers, and then using the associate and distributive law together with the fact that the nim-square of a nimber x with the binary representation 1+(power-of-two number of 0's) has the binary representation x|(x>>1). A brief description is on page 53 in Conway's On Numbers and Games. We also have a couple of cryptic lines that with effort can be understood as describing this algorithm towards the end of our article's multiplication section. With some care this can be implemented to run in time roughly (#bits)^{log_2 3), analogous to the Karatsuba algorithm, which I implemented long ago; my implementation can be found on https://ics.uci.edu/~eppstein/numth/ —David Eppstein (talk) 21:20, 27 October 2025 (UTC)

Thank you! It seems that getting XOR into this will be tough. But having followed a few leads and tangents spurred on by your response, maybe we could add some sort of motivation for the explicit definition of multiplication. I am not familiar enough with the topic to be able to give citations (nor to know whether something is obvious enough to be WP:CALC) but I am hoping that your expertise might turn this into something. Specifically, something like:

Intuition behind definition of nimber multiplication: Base case: We define x ⊗ 0 = 0 and 0 ⊗ y = 0. Recursion: If x′ < x then x′ ⊕ x ≠ 0. Likewise, if y′ < y then y′ ⊕ y ≠ 0. We want nimber multiplication to be a field multiplication and, in particular, we want the product of two non-zero values to be non-zero; so we want (x′ ⊕ x) ⊗ (y′ ⊕ y) ≠ 0. We want nimber multiplication to distribute over nimber addition, so this last expression becomes (x′ ⊗ y′) ⊕ (x′ ⊗ y) ⊕ (x ⊗ y′) ⊕ (x ⊗ y) ≠ 0. This can be written as x ⊗ y ≠ (x′ ⊗ y′) ⊕ (x′ ⊗ y) ⊕ (x ⊗ y′). Finally, this last expression is satisfied if we define x ⊗ y = mex{(x′ ⊗ y′) ⊕ (x′ ⊗ y) ⊕ (x ⊗ y′) | for all x′ < x, y′ < y}. It turns out that this expression also satisfies other criteria for defining a field.

I'm a bit of an amateur at this, so I apologize if this is more of a jumble than an encyclopedic entry. But if your corrections could turn it into something worth putting in the article, I would appreciate that. —Quantling (talk | contribs) 02:18, 28 October 2025 (UTC)

Here's an alternative that might be worth covering. A nimber is the Grundy value of an impartial game like Nim. Nimber addition G1 ⊕ G2 is the value of the compound game where a player may choose whether to play on board G1 or board G2. I believe that nimber multiplication G1 ⊗ G2 corresponds to some game created from G1 and G2, but what are the rules for that game? Joule36e5 (talk) 04:16, 28 October 2025 (UTC)

"The reason why multiplication is defined only for numbers, not for arbitrary games, is that there are games $G$, $G'$, $H$ with $G=G'$ but $GH\neq G'H$." —David Eppstein (talk) 04:48, 28 October 2025 (UTC)

I think that quote is about surreal multiply, not nimber multiply. Joule36e5 (talk) 09:37, 28 October 2025 (UTC)

I made an edit to the article. Please feel free to improve it, or revert it if it is beyond salvation. Please provide feedback here as well. Thank you —Quantling (talk | contribs) 02:47, 9 November 2025 (UTC)

Asterisk * (Unicode 0x2A) vs. ∗(Unicode 0x2217)

@OfficialURL et al., the article text informs us that Nimbers are often denoted using a star notation. For me (on my system in my browswer) using Unicode 0x2A to indicate that star puts the star immediately adjacent to the number, much as does {{tmath}}: ⁠$*\omega$⁠. However, the using Unicode 0x2217 instead puts a gap between the asterisk and the following symbol: ⁠$\ast ~\omega$⁠. To me 0x2A looks better. What do you think? —Quantling (talk | contribs) 17:07, 6 April 2026 (UTC)

I agree. My intention was just to center the asterisk, since *ω displays on my browser as a superscript, which is not correct. See e.g. page 179 of Siegel's Combinatorial Game Theory. viiii (talk) 17:43, 6 April 2026 (UTC)

If you're having trouble with consistent display of unicode versions of mathematics formulas, the solution is to switch to LaTeX: <math>\star\omega</math> gives $\star \omega$ (a five-point star), or <math>\ast\omega</math> gives $\ast \omega$ (a six-point star). I'd have to check ONAG/WW to see which type of star they prefer. —David Eppstein (talk) 18:45, 6 April 2026 (UTC)

Consistency doesn't seem to be the problem. So far those that are good or bad have been so consistently. I made a bold edit to use ⁎ω 0x204E LOW ASTERISK, instead of *ω 0x2A ASTERISK or ∗ω 0x2217 ASTERISK OPERATOR. It is giving me the right vertical position without the unwanted horizontal spacing. —Quantling (talk | contribs) 19:04, 6 April 2026 (UTC)

I checked ONAG and WW now that I'm in my office where my copies are. They use an asterisk (not a five-point star), set in an operator position (slightly above the baseline, like ASTERISK OPERATOR and unlike LOW ASTERISK). —David Eppstein (talk) 20:07, 6 April 2026 (UTC)

Thank you for looking at that. If it is not much trouble ... what can you tell us about the horizontal spacing used in those books? That is, how much gap is there between the asterisk and the omega (or other nimber)? —Quantling (talk | contribs) 20:42, 6 April 2026 (UTC)

Not much. Like what I see for all the examples here (including the 0x2217 example in your fourth-paragraph reply) but not the extra space that I see in your 0x2217 example in your first paragraph, where to demonstrate that space you used LaTeX with a hard space (tilde). —David Eppstein (talk) 22:03, 6 April 2026 (UTC)

I get too much horizontal gap for Unicode 0x2217 in {{math|∗1, ∗''ω'', ∗(''ω''+1)}}:

∗1, ∗ω, ∗(ω+1)

so maybe LaTeX is our best bet. What is the preference between {{tmath|\ast 1, \ast\omega, \ast(\omega+1)}}:

⁠$\ast 1,\ast \omega ,\ast (\omega +1)$⁠

and using an ordinary asterisk in {{tmath|*1, *\omega, *(\omega+1)}}:

⁠$*1,*\omega ,*(\omega +1)$⁠?

They look identical to my eyes. —Quantling (talk | contribs) 17:05, 7 April 2026 (UTC)

I don't see a difference either. For that matter, the spacing appears the same as these to me in your unicode version, which however has too little spacing around the + sign. —David Eppstein (talk) 22:01, 7 April 2026 (UTC)

I didn't find evidence for this notation, instead adopting the convention where expressions interpreted as ordinary addition and multiplication are enclosed in square brackets, and I have carried this throughout the rest of the article. Probably, the directed sum and product symbols should be replaced, too, with the usual symbols for the operators. If you can find a source, please add it back in the article. –LaundryPizza03 (dc̄) 21:42, 28 April 2026 (UTC)