I originally tried to ask a variation of this question on math.stack; however 1 commenter pointed out that math.stack is not a puzzle site, which made me think maybe the fine folks of puzzling could help me out! If this isn't an appropriate question for this site, I apologize. Also, I tried to post links to relevant information, however I don't have enough reputation to post more than 8 items in a post. See the variation post for relevant links.


The following is a variation of Nim-triplet addition mnemonic that is shown in On Numbers & Games by John Conway. The nim-triplets are also listed: (1,2,3), (1,4,5), (1,6,7), (2,4,6), (2,5,7), (3,4,7), (3,5,6). Any 2 numbers from a triplet sum to the 3rd (eg. 3+5=6, 3+6=5 & 5+6=3).

enter image description here

Note: the nim sum is equivalent to bitwise xor.

Inspired by the octonion multiplication mnemonic I wanted to see if I could create a multiplication mnemonic for nimbers. Much to my surprise I was able to create 1!

enter image description here

How it works Find a line (for example from 6-2), the midpoint (11) is the product. It works all along the outer edge (eg. the line from 11-9 has 2 a midpoint). For lines thru the origin, the product is 1 (eg. the line from 7-11 has ★≡∗≡1 as a midpoint).

I was trying to reconfigure my layout & stumbled upon a 'block' style mnemonic:

enter image description here

This 1 works similar to the previous except it doesn't work around the corners 🤷‍♀️

I found a few other 'simple' mnemonics & a pattern to create block mnemonics from them. Then I was able to piece together a larger mnemonic:

enter image description here

There's a lot going on (& there are still lines that can be added 😅)..

TLDR Different line types emerging from the midpoints point ↦ towards the endpoints (eg. the squiggly lines from the midpoint 2 point at ↦ the endpoints 5 & 4).

I eventually cleaned things up / moved things around a bit to come up with the following mnemonic:

enter image description here

This layout was inspired by a graph of $G_2$, specifically "the A2 Coxeter plane projection of the 12 vertices of the cuboctahedron" (full disclaimer: I don't know anything about $G_2$ or A2 Coxeter plane projections).

The following is an "alternative representation of the Fano plane" for order 2 & order 3. I think a mnemonic could be developed for nimber multiplication based on the order 3 (or something similar), possibly with 13 & 14 omitted

enter image description here

I'm not sure what the optimal configuration is for nimber multiplication up thru 15x15 (forms a closed field with no result larger than 15). Note: including 0, the 2nd degree fano plane has 8 digits & can be seen to correspond with octonions. I'm curious if this 16 digit field (including 0) might relate to sedonions somehow (super theoretical).

Which brings me to my questions:

  • Primary What is the optimum configuration for a nimonic (nimber nmemonic) which contains digits 1-15? (I suspect it may be 3d but have not yet been able to explore that avenue)
  • Secondary Given a larger nimonic, how can we determine how many smaller sub-nimonics (graph minors?) are embeded within it?
  • Bonus Does this structure/math relate to something else besides nimbers?
  • $\begingroup$ What do you mean by multiplication in the second example? E.g., can you explain more the arithmetic system used in which 6 x 2 = 11? $\endgroup$
    – justhalf
    Nov 26, 2023 at 14:17
  • $\begingroup$ @justhalf I'm still trying to wrap my head around nimber math 😅 For multiplication it uses mex (minimum excluded value) en.wikipedia.org/wiki/Nimber#Multiplication There is also addition & multiplication charts beneath for reference $\endgroup$
    – stargirl
    Nov 26, 2023 at 15:24

1 Answer 1


Multiplication of nimbers between $1$ and $15$ (or between $0$ and $2^{2^n}-1$ for any $n$) has a primitive root: a number whose powers generate all the nimbers we want. (In fact, I believe that $2^{2^{n-1}}$ will always be one possible primitive root, but I'm not sure.) In the case of nimbers $1$ through $15$, $4$ is a primitive root. So all we really need in a mnemonic is a table like the one below:

\begin{array}{c|cccccccccccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline 4^x & 1 & 4 & 6 & 14 & 5 & 2 & 8 & 11 & 7 & 10 & 3 & 12 & 13 & 9 & 15 \end{array}

(Here, exponentiation is repeated nimber multiplication; if there's another notion of nim exponentiation out there, I don't know it and I'm not using it.)

To find the nim-product of $5$ and $7$, for instance, first look them up in the bottom row, and we discover that $5 = 4^4$ and $7 = 4^8$. Because nim multiplication is associative, $4^4 \cdot 4^8 = 4^{12}$: we take the ordinary sum of the exponents, and we take it mod $15$ in case it is bigger than $15$. Then we look up $4^{12}$ in the table and get $13$.

In case using a table is boring, we can start coming up with artistic drawings, but we must be clear that we are really combining the log table above with a mnemonic for addition modulo $15$.

For example, we can use the fact that $15 = 3 \cdot 5$ to lay out the nimbers $1$ through $15$ on a repeating $3 \times 5$ grid: really, on a torus. In the table below, the white circle at a coordinate $(x,y)$ contains the nimber $4^{3x+5y}$:

toroidal "nimonic"

This is very big, but the pattern repeats: the white circles inside the red rectangle are repeated in all four directions.

Here is how to use this nimonic. If you want to multiply two numbers, find them as white circles in the grid, draw a line between them, and find the midpoint of that line.

  • If the midpoint is a black circle, that midpoint is their product.
  • If the midpoint is a left-right arrow, move any one of your endpoints $5$ spaces to the left or right (to another white circle with the same number in it) and try again.
  • If the midpoint is an up-down arrow, move any endpoint $3$ spaces up or down.
  • If the midpoint is a white circle, do both the up-down and the left-right movement.

For example, we want to multiply $5$ by $7$ again. We first find them in the center rectangle (bounded in red), and see that the midpoint of the line between them is a left-right arrow. So we replace our $7$ by the $7$ five spaces to its right, and draw another line. Its midpoint is a black circle with $13$ in it, so $5\cdot 7 = 13$.

  • $\begingroup$ Amazing! I'm going to have to read this a few times & play around with it a bit. Very impressed!! $\endgroup$
    – stargirl
    Nov 27, 2023 at 1:16
  • $\begingroup$ Looking at it as a torus is really insightful! I noticed that if u expand the red boundary box down & to the left to include 11 (& the other black numbers that would be included in the expanded box) we can just count "around" to the other side. I haven't exactly worked out the rules yet tho.. $\endgroup$
    – stargirl
    Nov 27, 2023 at 2:19
  • $\begingroup$ It seems like expanding the box down & to the right might be better? 🤔 Either way there is interesting symmetry over the black 1,2,3 line of black & white sets of numbers. $\endgroup$
    – stargirl
    Nov 27, 2023 at 3:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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