Can every 2 player game be represented as a sum of rock-paper-scissors games and seed games? I'm currently trying to formulate the following sentence rigorously:
"Given any deterministic two player game (a game such that if two players play multiple times, the result is the same every time), the result of the game can be written as a linear combination of rock-paper-scissors  games and seed games (ones in which the higher seed always beats the lower seed)"

Given some finite set $X$, let $Y=X\times X\setminus \Delta X$ ($\Delta X$ is the diagonal of $X$).
Let $F$ be the set of functions $f:Y\to \{-1,1\}$ such that for any $(a,b)\in Y$, $f(a,b)+f(b,a)=0$.
Given any total ordering $<$ on $X$, we define $g_<:Y\to \{1,-1\}$ as the function $g_<(a,b)=\begin{cases}1&a>b\\-1&a<b\end{cases}$. We call any such function a seed function.
Given any distinct $a,b,c\in X$, let $S_{a,b,c}=\{(a,b),(b,c),(c,a)\}$. We define $h_{a,b,c}:Y\to \{-1,0,1\}$ as the function $$h_{a,b,c}(x,y)=\begin{cases}1&(x,y)\in S_{a,b,c}\\-1&(y,x)\in S_{a,b,c}\\0&\text{otherwise}\end{cases}$$We call any such function a RPS (rock-paper-scissors) function.
For any $f\in F$, is there always a linear combination of seed and RPS functions that equals $f$? If so, is there always such a combination with at most one seed function?
What if I changed $F$ to be the set of functions $f:Y\to[-1,1]$ with $f(a,b)+f(b,a)=0$. Would that change any of the answers above?
 A: Choose two different elements $a$ and $b$ in $X$ and let $x_1,\ldots, x_n$ be any fixed enumeration of the remaining ones. We will define two total orderings, $\prec_1$ and $\prec_2$ as follows:
$$a \prec_1 b \prec_1 x_1 \prec_1 \ldots \prec_1 x_n$$
$$x_n \prec_2 \ldots \prec_2 x_1 \prec_2 a \prec_2 b$$
and consider the linear combination $g=\frac{1}{2}g_{\prec_1} + \frac{1}{2}g_{\prec_2}$ of the associated seed functions. The only non-zero values of $g$ are $g(b,a)=1$ and $g(a,b)=-1$ since for any other pair of inputs, the two orderings disagree about their relative order and the corresponding seed functions cancel out. Repeating the same construction for different pairs of $(a,b)$ and adding them all up can thus be used to build any function in $F$ (whether in the $\{0,1\}$ or $[0,1]$ case); even without using any RPS functions at all.
Now, if we restrict ourselves to just one seed function, we might not be so lucky. Let $S$ be an operator which transforms any game function $f$ into function $Sf: X\to \mathbb{R}$ defined as follows:
$$Sf(a)=\sum_{x\in X}f(a,x)$$
This operator is linear and for any RPS function, we have $Sh_{a,b,c}(x)=0$ for all $x$. When it comes to seed functions, we have $$Sg_\prec(a)=\left|\{x\in X\ |\ x\prec a\}\right|-\left|\{x\in X\ |\ a\prec x\}\right|$$
The largest (under the ordering $\prec$) element is thus mapped to $|X|-1$, the second largest maps to $|X|-3$ and so on all the way down to $-|X|+1$.
Therefore, if a function $f$ can be represented as a sum of a single seed function and any number of RPS functions, $Sf$ must map $X$ bijectively to $\{|X|-1,|X|-3,\ldots,-|X|+1\}$ (or, in the $[0,1]$ case, some multiple of this set).
As it turns out, this necessary condition is also sufficient. The single seed function will be based on the ordering $\prec$ defined as $a\prec b \iff Sf(a)<Sf(b)$. If we consider the function $f_0:=f-g_\prec$, we have $Sf_0(x)=0$ for all $x\in X$. We will show that any $f_0$ satisfying this property is a linear combination of RPS functions; with the induction being done over the number of pairs $(a,b)$ for which $f_0(a,b)\not=0$.
If there is no such pair, we are done since $f_0$ is identically zero. Otherwise, we'll show that there is some sequence of elements $x_1,x_2,\ldots,x_k$ with $k\geq 3$ such that $f_0(x_i,x_{i+1})\not=0$ for $1\leq i\leq k$ (where the last index corresponds to $(x_k,x_1)$ in a wrap-around way). We can then consider the function $$c:=h_{x_1,x_2,x_3}-h_{x_2,x_3,x_4}+\ldots +(-1)^{k-1} h_{x_{k-2},x_{k-1},x_k}$$
It has the following property:
$c(a,b)=\begin{cases}
+1 & a=x_k\land b=x_{k+1}\\
-1 & a=x_{k+1}\land b=x_k\\
0 & \mathrm{otherwise} \\
\end{cases}$
Now, we can consider the function $f_0^* := f_0-f_0(x_1,x_2)c$. It agrees with $f_0$ everywhere other than on pairs $(x_i,x_{i+1})$ (and their reversed forms, of course). We also have $f_0^*(x_1,x_2)=0$, so $f_0^*$ has fewer non-zero values than $f_0$ and we can apply induction hypothesis to build it from RPS games too.
In order to see that such a sequence $\langle x_i\rangle$ exists, we can look at all sequences of distinct elements for which $f(x_i,x_{i+1})\not =0$ for $1\leq i<k$. Since $f_0$ had at least one non-zero value, $f_0(a,b)\not =0$, the sequence $\langle a,b\rangle$ is a valid one. Now, consider any such sequence which cannot be extended further. Since $f_0(x_{k-1},x_k)\not=0$ but $Sf_0(x_k)=0$, there must be at least one element $x$ distinct from $x_{k-1}$ and $x_k$ such that $f_0(x_k,x)\not =0$. However, the sequence cannot be extended further, so $x$ must be present in the sequence already. In other words, $x=x_m$ for some $1\leq m<k-1$. We are now done; the subsequence $\langle x_m,x_{m+1},\ldots,x_k\rangle$ has the desired property of $f_0(x_i,x_{i+1})\not =0$ for all $i$ (considered in the wrap-around way).

Now, this whole explanation was really unnecessarily complicated and can be simplified if we rephrased the problem in a different setting. Let $X$ be the set of nodes of an oriented graph with weighted edges. An edge from $a$ to $b$ will be present if and only if $f(a,b)>0$, with $f(a,b)$ being its weight. A seed function corresponds to a tournament (= complete oriented graph), while an RPS game is an oriented cycle on three nodes; weights of all edges being $1$ in both cases.
We can use RPS games to build oriented cycles of any length -- if we have already built oriented cycle $\langle x_1,\ldots,x_k\rangle$ (of weights $1$), we can extend it by node $x_{k+1}$ by adding game $(x_k, x_{k+1}, x_1)$. This adds edges $(x_k,x_{k+1})$, $(x_{k+1},x_1)$ and cancels out the edge $(x_{k+1},x_1)$ since the new RPS game includes its opposite, $(x_1,x_{k+1})$.
RPS games keep certain quantities intact. One such invariant for each node is the total weight of all incoming edges minus total weight of all outgoing ones. If a graph can be built from just one seed function and some RPS ones, these invariants need to match those for a seed function. On the other hand, if our graph's invariants match those of some seed function, subtracting this seed function from the original yields a graph in which all of these invariants are zero -- for each node, the sum of weights of all incoming edges is equal to the sum of weights of all outgoing ones.
This implies that every node that has an incoming edge must also have at least one outgoing one. Thus, if we start from any node and follow the outgoing edges, we will eventually reach a node we have visited before and never get stuck. Doing so forms an oriented cycle $C$ (note that the weights of the edges in the cycle might be different). We can then use RPSes to build a cycle $C^*$ of weights $1$ over the same nodes, scale it so that it matches weight of some edge from cycle $C$ and subtract it from the original graph. Doing so cancels out at least one edge and doesn't introduce any news ones; so repeating this process eventually eliminates all cycles. Since we have already observed that unless the graph is empty, it contains at least one cycle... it must end up empty.
