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I'm trying to write a very general equation to calculate some $Value$ which relies on a context-dependent function $g$.

How do I concisely communicate that $g$ maps to different functions under different contexts?

So far I have:

\begin{gather} &\text{Value (a, b, c)} = {g(a,b)} + 10\\ \ \text{where:}&\\ g \mapsto\\ & A(i,j) = \begin{cases} 1 ,& \text{if some condition}\\ 0 ,& \text{otherwise} \end{cases}\\ \nonumber\ & B(i,j) = \begin{cases} 1 ,& \text{if some condition}\\ 0 ,& \text{otherwise} \end{cases}\\ \nonumber & C(i,j) = \begin{cases} 1 ,& \text{if some condition}\\ 0 ,& \text{otherwise} \end{cases}\\ \nonumber \end{gather}

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The Iverson brackets could be useful in this case. Let $P$ be a proposition. We write \begin{align*} [[P]]= \begin{cases} 1&\qquad \text{if $P$ is true}\\ 0&\qquad \text{if $P$ is false} \end{cases} \end{align*}

This way we can write \begin{align*} g(a,b)=A(a,b)[[X(a,b)]]+B(a,b)[[Y(a,b)]]+C(a,b)[[Z(a,b)]] \end{align*}

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  • $\begingroup$ Interesting, never seen that before -- but I'm not sure I can use it in the way that I want. I'm trying to emphasize that A B C map directly to g with full mutual exclusivity $\endgroup$ – tetris11 Mar 4 '17 at 21:35
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    $\begingroup$ @tetris11: This aspect is not part of your question. So, you mean $\max_{a,b}\left\{[[X(a,b)]]+[[Y(a,b)]]+[[Z(a,b)]]\right\}=1$? $\endgroup$ – Markus Scheuer Mar 4 '17 at 21:47
  • $\begingroup$ Ah, sorry. No not exactly, X Y and Z are just random statements - I would ignore them entirely (I just put them in order to give context to the $g$ parameters. I've modified my original question. $\endgroup$ – tetris11 Mar 4 '17 at 22:09
  • $\begingroup$ what I'm really trying to say is "$g$ maps directly to one of A, or B, or C depending on some parameter (say $c$)". I hope that's a little clearer $\endgroup$ – tetris11 Mar 4 '17 at 22:11

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