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Suppose I have forgotten how to calculate $g(a,b)=a+b$.

But I have a black-box function $f(a,b,c)=a+b+c$, which I can calculate.

Presumably, $f(a,b,c)=g(a,g(b,c))=g(g(a,b),c)$ uniquely specifies $g$ (does it?).

Naturally, $g(a,b)\equiv f(a,b/2,b/2)\equiv f(a,b,0)$, but I only know this because I understand the operation "+". What if I didn't?

Question Given a function f(a,b,c)

How can we determine if this function can be written as $f(a,b,c)=a\circ b\circ c$, where $\circ$ is a binary associative operation?

I am in particular interested in knowing if $f(a,b,c)=\frac{ab+bc}{2}$ can be decomposed in this way (where the multiplications $ab$ and $bc$ are not necessarily commutative but division by two behaves as expected, i.e. $a,b,c$ are real matrices).

Edit: partial answer: the function $f:\mathbb{R}^3\rightarrow \mathbb{R}: f(a,b,c)=\frac{ab+bc}{2}$ cannot be decomposed like this, at least not assuming $g$ is differentiable. $$g(g(a,b),c)=\frac{ab+bc}{2}$$ The derivative w.r.t. $a$ is $$\frac{d}{da}g(g(a,b),c)=\frac{a}{2}=g_1(g(a,b),c)g_1(a,b)$$ The derivative w.r.t. $b$ is $$\frac{d}{db}g(g(a,b),c)=\frac{a+c}{2}=g_1(g(a,b),c)g_2(a,b)$$ ($g_1,g_2$ are the derivative of $g$ with respect to the first resp. the second parameter). So $$\frac{g_1(g(a,b),c)g_1(a,b)}{g_1(g(a,b),c)g_2(a,b)}=\frac{a}{a+c}$$ $$\frac{g_1(a,b)}{g_2(a,b)}=\frac{a}{a+c}$$ The rhs. depends on $c$ but the lhs. does not, so this $g$ does not exist.

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Wouter
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    Regarding $g(a,b)=f(a,b/2,b/2)$, you're implicitly assuming that $g(b/2,b/2) = (b/2)\circ (b/2) = b$, or that multiplication (by $2^{-1}$) works "as normally expected" (including distributivity). But that doesn't necessarily have to be the case? – Shirish Kulhari Apr 05 '18 at 09:08
  • In that example, $a,b,c$ are intended to be elements of some sensible number system like $\mathbb{R}$. We can also write $g(a,b)\equiv f(a,b,0)$, making use of the existence of a neutral element for addition. – Wouter Apr 05 '18 at 09:17
  • To answer the question on your 3rd line: Let $S$ be a set containing at least three elements $0,1,2.$ For $x,y\in S$ define $x\circ y=0;$ define $xy=1$ if $x=y=2$ and $xy=0$ in all other cases. Then, for all $x,y,z\in S,$ we have $$(x\circ y)\circ z=x\circ(y\circ z)=(xy)z=x(yz)=0.$$ – bof Apr 05 '18 at 09:34
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    For real multiplication, we have $abc=g(a,g(b,c))=g(g(a,b),c)$ where $g(x,y)=-xy.$ – bof Apr 05 '18 at 09:51

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