$\mu$-operator

$\mu$ on predicates

Let a property on non-negative integers be given. Informally, the $\mu$-operator looks for the smallest number satisfying the given property. The $\mu$-operator is also known as the (unbounded  ) minimization operator or the (unbounded) search operator. Formally,

Definition. Let $\Phi(\boldsymbol{x},y)$ be an $(m+1)$-ary predicate  (property) over $\mathbb{N}$, the set of natural numbers ($0$ included here), with $m$ a non-negative integer ($\boldsymbol{x}$ is $m$-ary). Define

 $A_{\Phi}(\boldsymbol{x}):=\{y\in\mathbb{N}\mid\Phi(\boldsymbol{x},y)\},$

The $\mu$-operator on $\Phi$ is given by

 $\mu y\Phi(\boldsymbol{x},y):=\left\{\begin{array}[]{ll}\min A_{\Phi}(% \boldsymbol{x})&\textrm{if }A_{\Phi}(\boldsymbol{x})\neq\varnothing,\\ \textrm{undefined}&\textrm{otherwise.}\end{array}\right.$

The notation $\mu y\Phi(\boldsymbol{x},y)$ reads “the smallest $y$ such that $\Phi(\boldsymbol{x},y)$ is satisfied”. Note that $y$ is used both as a variable and a number that the variable $y$ represents.

Note that $\mu y\Phi(\boldsymbol{x},y)$ is a partial function  on $\boldsymbol{x}$ ($y$ is a bounded   variable). In other words, the $\mu$-operator is a function that takes an $m+1$-ary predicate to an $m$-ary partial function. When $m=0$, $\mu$ is either an integer, or $\varnothing$.

For example, suppose $\Phi(x,y)$ is the property $(x-y)(x+y)\geq 10$. Then $\mu y\Phi(x,y)=0$ iff $x\geq 4$, and undefined otherwise.

The reason why the $\mu$-operator is called the search operator comes from recursive function  theory. The search for the smallest $y$ such that $\Phi(\boldsymbol{x},y)$ begins with testing for $\Phi(\boldsymbol{x},0)$. If the test fails ($\Phi(\boldsymbol{x},0)$ is false), then testing for $\Phi(\boldsymbol{x},1),\Phi(\boldsymbol{x},2),\Phi(\boldsymbol{x},3),\cdots$ are successively performed. The testing stops when a $y$ with $\Phi(\boldsymbol{x},y)$ is found. This $y$ is also the smallest $y$ satisfying $\Phi$. Nevertheless, the testing can conceivably go on indefinitely, hence the name unbounded. There is also a bounded version of $\mu$-operation  :

Definition. Let $\Phi(\boldsymbol{x},y)$ be given as above. Define

 $A_{\Phi}(\boldsymbol{x},y):=\{z\in\mathbb{N}\mid\Phi(\boldsymbol{x},z)\mbox{ % and }z\leq y\}.$

The bounded $\mu$-operator on $\Phi$ is given by

 $\mu z\leq y\Phi(\boldsymbol{x},z):=\left\{\begin{array}[]{ll}\min A_{\Phi}(% \boldsymbol{x},y)&\textrm{if }A_{\Phi}(\boldsymbol{x},y)\neq\varnothing,\\ y+1&\textrm{otherwise.}\end{array}\right.$

Thus the bounded $\mu$-operator takes an $(m+1)$-ary predicate (on $(\boldsymbol{x},z)$, where $z$ is a free variable   ) to an $(m+1)$-ary total function (on $(\boldsymbol{x},y)$, as $z$ is now bounded by $\mu$).

$\mu$ on total functions

The $\mu$ operator can also be defined on functions. We first discuss the case of $\mu$ on total functions.

Definition. Given a total $(m+1)$-ary function $f:\mathbb{N}^{m+1}\to\mathbb{N}$, define

 $A_{f}(\boldsymbol{x}):=\{y\in\mathbb{N}\mid f(\boldsymbol{x},y)=0\},$

The $\mu$-operator on $f$ is given by

 $\mu yf(\boldsymbol{x},y):=\left\{\begin{array}[]{ll}\min A_{f}(\boldsymbol{x})% &\textrm{if }A_{f}(\boldsymbol{x})\neq\varnothing,\\ \textrm{undefined}&\textrm{otherwise.}\end{array}\right.$

This definition is actually equivalent      to the one regarding predicates, in the following sense: given a total function $f$ with arity $m+1$, define predicate $\Phi_{f}(\boldsymbol{x},y)$ of arity $m+1$, as “$f(\boldsymbol{x},y)=0$”. Then

 $\mu yf(\boldsymbol{x},y)=\mu y\Phi_{f}(\boldsymbol{x},y).$

Conversely, given an $(m+1)$-ary predicate $\Phi(\boldsymbol{x},y)$, define an $(m+1)$-ary function $f_{\Phi}(\boldsymbol{x},y):=1\dot{-}\chi_{\Phi}(\boldsymbol{x},y)$, where $\chi_{\Phi}$ is the characteristic function   of $\Phi$. Then

 $\mu y\Phi(\boldsymbol{x},y)=\mu yf_{\Phi}(\boldsymbol{x},y).$

Note also that $\mu f$ may be partial even if $f$ is total, since it is possible $f(\boldsymbol{x},y)\neq 0$ for all $y$, and the search will go on indefinitely. For example, let $f(x,z,y)=x^{2}+z^{2}-y^{2}$ if $x^{2}+z^{2}\geq y^{2}$, and $1$ otherwise. Clearly, $f$ is a total function. It is easy to see that $\mu yf(3,4,y)=5$, while $\mu yf(1,2,y)$ is undefined.

$\mu$ on partial functions

The definition of the $\mu$-operator on total functions can be generalized to partial functions. However, in recursive functions theory, an additional condition is imposed in order to make the generalization  .

Definition. Given a partial $(m+1)$-ary function $f:\mathbb{N}^{m+1}\to\mathbb{N}$, define

 $A_{f}(\boldsymbol{x}):=\{y\in\mathbb{N}\mid f(\boldsymbol{x},y)=0\mbox{ and }(% \boldsymbol{x},z)\in\operatorname{dom}(f)\mbox{ for all }z\leq y\},$

The $\mu$-operator on $f$ is given by

 $\mu yf(\boldsymbol{x},y):=\left\{\begin{array}[]{ll}\min A_{f}(\boldsymbol{x})% &\textrm{if }A_{f}(\boldsymbol{x})\neq\varnothing,\\ \textrm{undefined}&\textrm{otherwise.}\end{array}\right.$

The extra condition that $f(\boldsymbol{x},z)$ be defined for all $z\leq y$ is needed in order to avoid situations where testing for $f(\boldsymbol{x},z)=0$ gets stuck in an infinite loop (in a Turing machine  or a URM) because $f(\boldsymbol{x},z)$ is undefined, before $y$ is ever reached for testing. If we drop this extra condition, then it is possible to find a partial recursive function $f$ such that $\mu f$ is not recursive.

Remarks.

Title $\mu$-operator muoperator 2013-03-22 19:05:47 2013-03-22 19:05:47 CWoo (3771) CWoo (3771) 12 CWoo (3771) Definition msc 03D20 minimization operator minimization-operator search operator search-operator