Pseudo-Boolean function

The submodular set functions can be viewed as a special class of pseudo-Boolean functions, which is equivalent to the condition

${\displaystyle f({\boldsymbol {x}})+f({\boldsymbol {y}})\geq f({\boldsymbol {x}}\wedge {\boldsymbol {y}})+f({\boldsymbol {x}}\vee {\boldsymbol {y}}),\;\forall {\boldsymbol {x}},{\boldsymbol {y}}\in \mathbf {B} ^{n}\,.}$

This is an important class of pseudo-boolean functions, because they can be minimized in polynomial time. Note that minimization of a submodular function is a polynomially solvable problem independent on the presentation form, for e.g. pesudo-Boolean polynomials, opposite to maximization of a submodular function which is NP-hard, Alexander Schrijver (2000).

If f is a quadratic polynomial, a concept called roof duality can be used to obtain a lower bound for its minimum value.[3] Roof duality may also provide a partial assignment of the variables, indicating some of the values of a minimizer to the polynomial. Several different methods of obtaining lower bounds were developed only to later be shown to be equivalent to what is now called roof duality.[3]

If the degree of f is greater than 2, one can always employ reductions to obtain an equivalent quadratic problem with additional variables. One possible reduction is

${\displaystyle \displaystyle -x_{1}x_{2}x_{3}=\min _{z\in \mathbf {B} }z(2-x_{1}-x_{2}-x_{3})}$

There are other possibilities, for example,

${\displaystyle \displaystyle -x_{1}x_{2}x_{3}=\min _{z\in \mathbf {B} }z(-x_{1}+x_{2}+x_{3})-x_{1}x_{2}-x_{1}x_{3}+x_{1}.}$

Different reductions lead to different results. Take for example the following cubic polynomial:[4]

${\displaystyle \displaystyle f({\boldsymbol {x}})=-2x_{1}+x_{2}-x_{3}+4x_{1}x_{2}+4x_{1}x_{3}-2x_{2}x_{3}-2x_{1}x_{2}x_{3}.}$

Using the first reduction followed by roof duality, we obtain a lower bound of -3 and no indication on how to assign the three variables. Using the second reduction, we obtain the (tight) lower bound of -2 and the optimal assignment of every variable (which is ${\displaystyle {(0,1,1)}}$).

Consider a pseudo-Boolean function ${\displaystyle f}$ as a mapping from ${\displaystyle \{-1,1\}^{n}}$ to ${\displaystyle \mathbb {R} }$. Then ${\displaystyle f(x)=\sum _{I\subseteq [n]}{\hat {f}}(I)\prod _{i\in I}x_{i}.}$ Assume that each coefficient ${\displaystyle {\hat {f}}(I)}$ is integral. Then for an integer ${\displaystyle k}$ the problem P of deciding whether ${\displaystyle f(x)}$ is more or equal to ${\displaystyle k}$ is NP-complete. It is proved in [5] that in polynomial time we can either solve P or reduce the number of variables to ${\displaystyle O(k^{2}\log k).}$ Let ${\displaystyle r}$ be the degree of the above multi-linear polynomial for ${\displaystyle f}$. Then [5] proved that in polynomial time we can either solve P or reduce the number of variables to ${\displaystyle r(k-1)}$.