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. An open source book on the subject, mainly written by Nike Dattani, contains dozens of different quadratization methods.[4]

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:[5]

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

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 [6] 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 [6] proved that in polynomial time we can either solve P or reduce the number of variables to ${\displaystyle r(k-1)}$.