Often, given a submodular set function that describes the values of various sets, we need to compute the values of
fractional sets. For example: we know that the value of receiving house A and house B is V, and we want to know the value of receiving 40% of house A and 60% of house B. To this end, we need a
continuous extension of the submodular set function. Formally, a set function f:2^{\Omega}\rightarrow \mathbb{R} with |\Omega|=n can be represented as a function on \{0, 1\}^{n}, by associating each S\subseteq \Omega with a binary vector x^{S}\in \{0, 1\}^{n} such that x_{i}^{S}=1 when i\in S, and x_{i}^{S}=0 otherwise. A
continuous extension of f is a continuous function F:[0, 1]^{n}\rightarrow \mathbb{R}, that matches the value of f on x\in \{0, 1\}^{n}, i.e. F(x^{S})=f(S). Several kinds of continuous extensions of submodular functions are commonly used, which are described below.
Lovász extension This extension is named after mathematician
László Lovász.F(\mathbf{x})=\sum_{S\subseteq \Omega} f(S) \prod_{i\in S} x_i \prod_{i\notin S} (1-x_i). Intuitively,
xi represents the probability that item
i is chosen for the set. For every set
S, the two inner products represent the probability that the chosen set is exactly
S. Therefore, the sum represents the expected value of
f for the set formed by choosing each item
i at random with probability xi, independently of the other items.
Convex closure Consider any vector \mathbf{x}=\{x_1,x_2,\dots,x_n\} such that each 0\leq x_i\leq 1. Then the convex closure is defined as f^-(\mathbf{x})=\min\left(\sum_S \alpha_S f(S):\sum_S \alpha_S 1_S=\mathbf{x},\sum_S \alpha_S=1,\alpha_S\geq 0\right). The convex closure of any set function is convex over [0,1]^n.
Concave closure Consider any vector \mathbf{x}=\{x_1,x_2,\dots,x_n\} such that each 0\leq x_i\leq 1. Then the concave closure is defined as f^+(\mathbf{x})=\max\left(\sum_S \alpha_S f(S):\sum_S \alpha_S 1_S=\mathbf{x},\sum_S \alpha_S=1,\alpha_S\geq 0\right).
Relations between continuous extensions For the extensions discussed above, it can be shown that f^{+}(\mathbf{x}) \geq F(\mathbf{x}) \geq f^{-}(\mathbf{x})=f^L(\mathbf{x}) when f is submodular. == Properties ==