# Proof that “$\uparrow$ is the unique solution of $tiny(G) = G$”

Tiny & miny games can be defined as: $$tiny(G) = \{0||0|-G\}$$ $$miny(G) = -tiny(G) = \{G|0||0\}$$

From the Wikipedia page for tiny and miny:

Similarly curious, mathematician John Horton Conway noted, calling it "amusing," that "$$\uparrow$$ is the unique solution of $$tiny(G) = G$$"

$$\uparrow$$ is generally defined as $$\{0|*\}$$. Interestingly, $$tiny(0)=\{0||0|0\}=\{0|*\}=\uparrow$$ & $$miny(0)=\{0|0||0\}=\{*|0\}=\downarrow$$.

How is $$\uparrow$$ the unique solution to $$tiny(G) = G$$?

Additionally, if not apparent, what makes this amusing?

I will use $$\pmb{+}_G$$ for $$\text{tiny-}G$$ throughout, but the unicode symbol ⧾ is arguably more correct for that plus sign.

## Why Amusing?

This is a typo in the wikipedia page at the time of writing. On page 215 of On Numbers and Games, it actually says:

...it is amusing to verify that for any game $$G$$, we have $$\pmb{+}_{\pmb{+}_{\pmb{+}_G}}=\uparrow$$, so that in particular, $$\uparrow$$ is the unique solution of $$G=\pmb{+}_G$$.

Without asking, I can't be certain why John H. Conway found it amusing, but I personally find it amusing to work through why something complicated and with an arbitrary game parameter like $$\pmb{+}_{\pmb{+}_{\pmb{+}_G}}\cong\{ 0\Vert0|-\{ 0\Vert0|-\{ 0\Vert0|-G\} \} \}$$ simplifies to something as simple as $$\uparrow\cong\{ 0\Vert0|0\}$$.

## Why $$\uparrow$$?

Verifying the claims in the quote above is a problem in Chapter 5 of Lessons in Play: An Introduction to Combinatorial Game Theory (tinies and minies are introduced in section 5.4). It's an amusing exercise, so I don't want to spoil the whole thing, but I can clarify the second part a bit.

Once we have $$\pmb{+}_{\pmb{+}_{\pmb{+}_G}}=\uparrow$$, then proving that $$\uparrow$$ is the unique solution (up to equality) to $$\pmb{+}_G=G$$ does not require any game theory. It's a general fact that if we have a function $$f:X\to X$$ and a particular $$y\in X$$ such that $$\forall x\in X,f(f(f(x)))=y$$, then $$y$$ is the only solution to $$f(x)=x$$. Can you see why?

• Thank you! This is incredibly helpful. I will take a crack at proving $\pmb{+}_{\pmb{+}_{\pmb{+}_G}}=\uparrow$ for myself. Also, I am not sure what exactly is meant by $*\cong\{ 0\Vert0|0\}$ It seems this would imply $*\cong\{0|*\}$ and thus $*\cong\uparrow$. What does $\cong$ mean in this case? Is confused with? – meowzz Nov 10 '18 at 16:46
• That was a typo of mine. I meant $\uparrow$ there, not $*$. $\cong$ can mean "isomorphic to"/"identical to" since in CGT $=$ is merely an equivalence relation. – Mark S. Nov 10 '18 at 17:11
• Got it! What symbol in CGT is best for fuzzy /“is confused with” ($\cong$, $=$, ?)? Additionally, does fuzzy (equal?) isomorphic? Or? – meowzz Nov 10 '18 at 17:28
• @meowzz That's a little off topic, but I've just added the standard symbols for "confused with" to my answer distinguishing "confused with" from "fuzzy". "Isomorphic" is a general/common term across mathematics. In the context of combinatorial games, Lessons in Play says "two games are isomorphic if they have identical game trees". This has nothing to do with the relation of being confused; two isomorphic games are necessarily equal, so they can't possibly be confused with each other. – Mark S. Nov 10 '18 at 17:46
• @meowzz, Do you already know theorems about (in)equality of combinatorial games? You'll need at least a couple to do this exercise from the middle of a textbook. If you don't have the background already, I'd really recommend tracking down a copy of Lessons in Play. And if you do, but are stuck in the calculations, you could post a new question and show all your work/where you're getting stuck. – Mark S. Nov 11 '18 at 3:24