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Solving problems related to Marginal Contribution Nets

Submitted by: @import:stackexchange-cs··
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Problem

So, I encoutered this problem in examination:


Consider the following marginal contribution net:


$\{a \wedge b\} \to 5$


$\{b\} \to 2$


$\{c\} \to 4$


$\{b \wedge \neg c\} \to −2$


Let $v$ be the characteristic function defined by these rules. Give the values of the
following:


i) $v(\emptyset)$


ii) $v(\{a\})$


iii) $v(\{b\})$


iv) $v(\{a, b\})$


v) $v(\{a, b, c\})$

My answer is below, but I am not sure.


i) $v(\emptyset) = -2$


ii) $v(\{a\}) = 0 - 2$


iii) $v(\{b\}) = 2 - 2$


iv) $v(\{a, b\}) = 5 + 2 - 2$


v) $v(\{a, b, c\}) = 5 + 4 + 2 - 2$

If anybody knows how to solve this kind of problem, could you confirm?

Exact same problem is shown on page 4 of following paper:
Marginal Contribution Nets: A Compact Representation Scheme for Coalitional Games (by Samuel Ieong and Yoav Shoham)

Solution

I got the following answer:

  • $v(\emptyset)=0$



  • $v(\{a\}) = 0$



  • $v(\{b\}) = 2-2 = 0$



  • $v(\{a,b\}) = 5+2-2 = 5$



  • $v(\{a,b,c\}) = 5+2+4 = 11$



Here's a general approach to doing this calculation.
Consider a set $A\subseteq \{a,b,c\}$. Define a map $I_A:\{a,b,c\}\to\{\text{True},\text{False}\}$ as follows:
$$I_A(x) = \left\{ \begin{array}{l@{~~}l} \text{True} & \mbox{if}~x\in A \\
\text{False} & \mbox{otherwise}
\end{array}\right.$$
This is the so-called indicator function of set $A$.

Now take each element of the marginal contribution net and evaluate whether it is true or false given $I_A$. If it is true, add the contribution. If not, add no contribution. More explicitly, given element $\psi\to c$, add $c$ only if $I_A\models\psi$ holds.

Here the notation $I_A\models\psi$ simply means evaluate the truth of the formula $\psi$ given the assignment of the variables $\{a,b,c\}$ to true or false.

Consider $v(\{a,b\})$. This corresponds to the map
$I_{\{a,b\}}=\{a \mapsto\text{True},b\mapsto\text{True},x\mapsto\text{False}\}$. Given this assignment of truth values, we can conclude that:

  • $a\wedge b$ is true, so $5$ is contributed.



  • $b$ is true, so $2$ is contributed.



  • $c$ is false, so nothing is contributed.



  • $b\wedge\neg c$ is true, so $-2$ is contributed.



Therefore the total contribution is $5+2-2=5$.

Context

StackExchange Computer Science Q#1319, answer score: 6

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