I'm designing a boardgame system for which I have some properties I would like to hold, and after some trial-and-error I have found no advances on whether there is a solution to the system or not, but some interesting non-commutative diagrams of functions appeared.

To contextualize, I have four resources, call them $R_1$, $R_2$, $R_3$ and $R_4$. As an action you can use a card to exchange a resource for another. Let this action of using a card and exchanging $a \ R_1$ for $b \ R_2$ be a linear function $f_1: \mathbb{Q} \to \mathbb{Q}$ with a rational coefficient $m_1 = b/a$. That is, $f_1(a) = m_1 a$, for all $a \in \mathbb{Q}$.

This is the full non-commutative diagram of functions $f_1, \ldots, f_8$ and how they relate to the resources $R_1, \ldots, R_4$:

Image of the non-commutative diagram.

Let an $n$-action path starting at $R_j$ and ending at $R_k$ be a sequence of $n$ exchanges, starting at $R_j$ and ending at $R_k$. So, for example, there is a $2$-action path from $R_1$ onto itself; namely, $R_1 \longrightarrow_5 m_5 R_3 \longrightarrow_6 m_6 m_5 R_1$.

Finally, let an $n$-action path from $R_j$ onto itself net $x$ if, and only if, there exists an $n$-action path from $R_i$ onto itself such that $m_{i_n} m_{i_{n-1}} \ldots m_{i_1} = x$

Now onto the question itself. Are there rational coefficients $m_1, \ldots, m_8$ assigned, respectively, to the functions $f_1, \ldots, f_8$ such that, for any resource $R_j$, the following conditions are all true:

$\bullet$ Any $2$-action path from $R_j$ onto itself nets $4$;

$\bullet$ Any $3$-action path from $R_j$ onto itself nets $6$;

$\bullet$ Any $4$-action path from $R_j$ onto itself nets $16$.

Or, more generally, instead of netting $4$, $6$ and $9$, respectively, netting $a$, $b$ and $c$, with $a < b < c \in \mathbb{Q}$.

  • $\begingroup$ If I have understood the rule (not clear) then $m_1 m_2=4$ and $m_1m_2m_3=6$, yes? If so then of course $m_3=\frac 64=\frac 32$. And then $m_2m_3=4\implies m_2=\frac 83\implies m_1=\frac {12}8=\frac 32$ . Continuing in this way either yields a contradiction or lets you solve for all the $m_i$. Or have I misunderstood? $\endgroup$ – lulu Jan 29 '18 at 23:37
  • $\begingroup$ Note: if I have understood the rules correctly then it is clearly impossible as we'd have $9=m_1m_2m_3m_4=4^2$. $\endgroup$ – lulu Jan 29 '18 at 23:38
  • $\begingroup$ As you have pointed out, for $9$ and $4$ it may not be possible. What if it was $16$ and $4$, for example? It would solve this problem in particular, but it's not clear to me if it would yield any other contradictions down the line. $\endgroup$ – Gabriel Singh Jan 29 '18 at 23:40
  • $\begingroup$ Well, I may have it wrong. Are you only looking at loops that start and stop at the same place? If so then I erred in saying that $m_1m_2=4$. $\endgroup$ – lulu Jan 29 '18 at 23:41
  • $\begingroup$ You got it wrong on the indexes of $m_i$. On the picture, it would mean that $m_5 m_6 = 4 = m_6 m_5$, and also $m_1 m_2 m_6 = 6$, not $m_1 m_2 m_3$ per say. $\endgroup$ – Gabriel Singh Jan 29 '18 at 23:42

I have found, together with @lulu, a family of answers $\{a,b,c\}$, with $a,b,c \in \mathbb{Q}$. As noted in the comment section of the main post, the set $\{4,6,9\}$ is not a solution, for $m_5 m_6 = 4$ and $m_5 m_6 m_5 m_6 =16 \neq 9$.

If we no longer ask that any $4$-action path nets $c$, and instead that only a $4$-action path composed of $f_1$, $f_2$, $f_3$ and $f_4$ net $c$ (thus enabling the path $f_5 \longrightarrow f_6 \longrightarrow f_5 \longrightarrow f_6$ to not necessarily net $c$), then there is another family of solutions.

Note that every $3$-action path goes once, and only once, through either $f_5$, $f_6$, $f_7$ or $f_8$, and thus goes twice through either $f_1$, $f_2$, $f_3$ or $f_4$.

Thus if $m_i = \lambda_1$ for $i \in \{1,2,3,4\}$ and $m_j = \lambda_2$ for $j \in \{5,6,7,8\}$, then

$\bullet$ Any $2$-action path nets $\lambda_2 = a$;

$\bullet$ Any $3$-action path nets $\lambda_2 \lambda_1^2 = b$;

$\bullet$ Any $4$-action path nets $\lambda_1^4 = c$.

And so the set of solutions would be of the form $\{ \lambda_2^2, \lambda_2 \lambda_1^2, \lambda_1^4 \}$, for any $\lambda_1, \lambda_2 > 1 \in \mathbb{Q}$ such that $\lambda_2 < \lambda_1^2$.


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