Difference between revisions of "Test"
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<math>\frac{1}{\displaystyle1+\frac{1}{\displaystyle 1+\sqrt{5}}}</math> | <math>\frac{1}{\displaystyle1+\frac{1}{\displaystyle 1+\sqrt{5}}}</math> | ||
+ | |||
+ | <wikitex> | ||
+ | Let $Q$ be any finite set, and $\mathcal B=2^Q$ be the collection of the subsets of | ||
+ | $Q$. Let $f:\mathcal B\rightarrow \mathbb R$ be a function assigning real numbers to | ||
+ | the subsets of $Q$ and suppose $f$ satisfies the following conditions: | ||
+ | :(i) $f(A)\ge 0$ for all $A\subseteq Q$, $f(\emptyset)=0$, | ||
+ | :(ii) $f$ is monotone, i.e. if $A\subseteq B\subseteq Q$ then $f(A)\le f(B)$, | ||
+ | :(iii) $f$ is submodular, i.e. if $A$ and $B$ are different subsets of $Q$ then | ||
+ | $$f(A)+f(B)\ge f(A\cap B) + f(A\cup B).\eqno{(2)}$$ | ||
+ | </wikitex> |
Revision as of 14:01, 18 November 2011
<math>\frac{1}{\displaystyle1+\frac{1}{\displaystyle 1+\sqrt{5}}}</math>
<wikitex> Let $Q$ be any finite set, and $\mathcal B=2^Q$ be the collection of the subsets of $Q$. Let $f:\mathcal B\rightarrow \mathbb R$ be a function assigning real numbers to the subsets of $Q$ and suppose $f$ satisfies the following conditions:
- (i) $f(A)\ge 0$ for all $A\subseteq Q$, $f(\emptyset)=0$,
- (ii) $f$ is monotone, i.e. if $A\subseteq B\subseteq Q$ then $f(A)\le f(B)$,
- (iii) $f$ is submodular, i.e. if $A$ and $B$ are different subsets of $Q$ then
$$f(A)+f(B)\ge f(A\cap B) + f(A\cup B).\eqno{(2)}$$
</wikitex>