Lecture 10¶

Numerical Functions¶

Definition

A Turing machine $M$ computes $f: \mathbb{N}^K \rightarrow \mathbb{N}$ if $\forall n_1, \dots, n_k \in \mathbb{N}$,

\[ M(\text{bin}(n_1), \text{bin}(n_2), \dots, \text{bin}(n_k)) = \text{bin}(f(n_1, \dots, n_k)), \]

where $\text{bin}(x)$ is the binary encoding of $x$.

And we say $f$ is computable.

Primitive Recursive Function¶

Basic Functions¶

zero function

\[ \text{zero}(n_1, \dots, n_k) = 0. \]
identity function

\[ \text{id}_{k, j}(n_1, \dots, n_k) = n_j. \]
successor function

\[ \text{succ}(n) = n + 1. \]

Theorem

Basic functions are computable.

Composition¶

If we have a computable function $g: \mathbb{N}^k \rightarrow \mathbb{N}$ and several computable functions $h_1, \dots, h_k: \mathbb{N}^l \rightarrow \mathbb{N}$. The way to contruct the following function $f: \mathbb{N}^l \rightarrow \mathbb{N}$ is called composition.

\[ f(n_1, \dots, n_l) = g(h_1(n_1, \dots, n_l), \dots, h_k(n_1, \dots, n_l)). \]

Recursive Definition¶

If we have two computable functions $g: \mathbb{N}^k \rightarrow \mathbb{N}$ and $h: \mathbb{N}^{k + 2} \rightarrow \mathbb{N}$. The way to contruct the following function $f: \mathbb{N}^{k + 1} \rightarrow \mathbb{N}$ is called recursive definition.

$f(n_1, \dots, n_k, 0) = g(n_1, \dots, n_k)$.
$\forall t \in \mathbb{N}$, $f(n_1, \dots, n_k, t + 1) = h(n_1, \dots, n_k, t, f(n_1, \dots, n_k, t))$.

Example

A typical recursive function is factorial: $f(0) = 1$ and $f(n + 1) = (n + 1)f(n)$. In this case, we have

$k = 0$.
$g(n) = 1$.
$h(n, f(n)) = (n + 1)f(n)$.

Definition

Primitive recursive functions are the basic functions and those obtained from the basic functions by applying composition and recursive definition a finite number of times.

Theorem

Primitive recursive functions are computable.

Definition

If the output of the primitive recursive function is either 0 or 1, then it's called a predicate.

Theorem

If $g$, $h$ are primitive recursive and $p$ is a predicate, then

\[ f(n_1, \dots, n_k) = \left\{\begin{aligned} g(n_1, \dots, n_k), & \text{ if } p(n_1, \dots,n _k), \\ h(n_1, \dots, n_k), & \text{ otherwise }. \end{aligned}\right. \]

is also primitive recursive.

Here we give some examples of primitive recursive functions.

Example

$\text{plus2}(n) = n + 2$

\[ \text{succ}(\text{succ}(n)) \]
$\text{plus}(m, n) = m + n$

\[ \left\{ \begin{aligned} \text{plus}(m, 0) &= m = \text{id}_{1, 1}(m), \\ \text{plus}(m, n + 1) &= \text{plus}(m, n) + 1 = \text{succ}(\text{plus(m, n)}). \end{aligned} \right. \]

But it's not strict enough actually. The following is the a stricter version. Since it's trivial, we needn't actually expand it commonly.

\[ \left\{ \begin{aligned} \text{plus}(m, 0) &= m = \text{id}_{1, 1}(m), \\ \text{plus}(m, n + 1) &= \text{succ}(\text{plus(m, n)}) = h(m, n, \text{plus}(m, n)). \end{aligned} \right. \]

where

\[ h(m, n, \text{plus}(m, n)) = \text{succ}(\text{id}_{3,3}(m, n, \text{plus}(m, n))). \]
$\text{mult}(m, n) = m \cdot n$

\[ \left\{ \begin{aligned} \text{mult}(m, 0) &= 0 = \text{zero}(m), \\ \text{mult}(m, n + 1) &= \text{mult}(m, n) + m = \text{plus}(\text{mult(m, n)}, m). \end{aligned} \right. \]
$\text{exp}(m, n) = m^n$

\[ \left\{ \begin{aligned} \text{exp}(m, 0) &= 1 = \text{succ}(\text{zero}(m)), \\ \text{exp}(m, n + 1) &= \text{exp}(m, n) \cdot m = \text{mult}(\text{exp(m, n)}, m). \end{aligned} \right. \]
constant function: $f(n_1, \dots, n_k) = c$

\[ \underbrace{\text{succ}(\dots(\text{succ}}_{c \text{ times}}(\text{zero}(n_1, \dots, n_k)) \dots ) \]
positive / sign function: $\text{positive}(n) = \text{sgn}(n) = \left\{\begin{aligned} 1, n > 0, \\ 0, n = 0. \end{aligned}\right.$

\[ \left\{ \begin{aligned} \text{sgn}(0) &= 0, \\ \text{sgn}(n + 1) &= 1. \end{aligned} \right. \]
predecessor function: $\text{pred}(n) = \left\{\begin{aligned} &n - 1, &n > 0, \\ &0, &n = 0. \end{aligned}\right.$

\[ \left\{ \begin{aligned} \text{pred}(0) &= 0, \\ \text{pred}(n + 1) &= n = \text{id}_{2, 1}(n, \text{pred}(n)) = n. \end{aligned} \right. \]
non-negative substraction $m \sim n = \max(m - n, 0)$

\[ \left\{ \begin{aligned} m \sim 0 &= m, \\ m \sim (n + 1) &= (m \sim n) \sim 1 = \text{pred}(m \sim n). \end{aligned} \right. \]
$\text{iszero}(n) = \left\{\begin{aligned} &1, &n = 0, \\ &0, &n > 0. \end{aligned}\right.$

\[ 1 \sim \text{positive}(n) \]
$\text{geq}(m, n) = \left\{\begin{aligned} &1, &m \ge n, \\ &0, &m < n. \end{aligned}\right.$

\[ \text{iszero}(n \sim m) \]
$\text{eq}(m, n) = \left\{\begin{aligned} &1, &m = n, \\ &0, &m \neq n. \end{aligned}\right.$

\[ \text{geq}(m, n) \cdot \text{geq}(n, m) \]
$\text{rem}(m, n) = m\ \%\ n$.

\[ \left\{ \begin{aligned} \text{rem}(0, n) &= 0, \\ \text{rem}(m + 1, n) &= \left\{\begin{aligned} & 0, && \text{if $m + 1$ is divisible by $n$}, \\ & \text{rem}(m, n), && \text{otherwise}. \end{aligned}\right. \end{aligned} \right. \]

where $m + 1 \text{ is divisible by } n \Leftrightarrow \text{eq}(\text{rem}(m, n), \text{pred}(n))$.
$\text{div}(m, n) = \lfloor m / n \rfloor$. ($n \neq 0$)

\[ \left\{ \begin{aligned} \text{div}(0, n) &= 0, \\ \text{div}(m + 1, n) &= \left\{\begin{aligned} & \text{div}(m, n) + 1, && \text{if $m + 1$ is divisible by $n$}, \\ & \text{div}(m, n), && \text{otherwise}. \end{aligned}\right. \end{aligned} \right. \]
$\text{digit}(m, n, p) = a_{m - 1}$.

\[ \text{div}(\text{rem}(n, p^m), p^{m - 1}) \]
$\text{sum}_f(m, n) = \sum\limits_{k = 0}^{n} f(m, k)$, where $f$ is primitive recursive.

\[ \left\{ \begin{aligned} \text{sum}_f(m, 0) &= f(m, 0), \\ \text{sum}_f(m, n + 1) &= \text{sum}_f(m, n) + f(m, \text{succ}(n)). \end{aligned} \right. \]

Wrong Consideration

\[ \text{sum}_f(m, n) = \underbrace{f(m, 0) + f(m, 1) + \dots + f(m, n)}_{n \text{ times}} \]

Notice that the theorem that sum of two primitive recursive functions is primitive recursive can not induce that sum of $n$ primitive recursive function is primitive recursive, when $n$ is a variable and is an input of the function.
$\text{mult}_f(m, n) = \prod\limits_{k = 0}^{n} f(m, k)$, where $f$ is primitive recursive.

\[ \left\{ \begin{aligned} \text{mult}_f(m, 0) &= f(m, 0), \\ \text{mult}_f(m, n + 1) &= \text{mult}_f(m, n) \cdot f(m, \text{succ}(n)). \end{aligned} \right. \]
$g_p(n) = \left\{\begin{aligned} &1, && \text{if } \exists n' \le n, p(n') = 1 , \\ &0, &&\text{otherwise}. \end{aligned}\right.$, where $p$ is a predicate.

\[ g_p(n) = p(0) \vee p(1) \vee \dots \vee p(n) = \text{positive} \left(\sum_{n'=0}^n p(n')\right) = \text{positive}(\text{sum}_p(n)) \]

Note

If $p$ and $q$ are predicates, then

\[ \begin{aligned} p \wedge q &= p \cdot q \\ p \vee q &= 1 \sim \text{iszero}(p + q) \end{aligned} \]

Computability¶

Suppose we define

\[ \begin{aligned} \text{PR} &= \{\langle f\rangle | f \text{ is a primitive recursive numerical function}\}, \\ \text{C} &= \{\langle f\rangle | f \text{ is a computable numerical function}\}. \end{aligned} \]

Lemma

$\text{PR}$ is decidable.

Lemma

$\text{C}$ is undecidable.

Proof

Assume that $C$ is decidable. Then $C' = \{\langle f\rangle | f \text{ is a unary computable numerical function}\} \subseteq C$ is also decidable.

Then $C'$ is lexicographically Turing enumerable. Thus we can enumerate $C'$ by

\[ g_1, g_2, g_3, \dots \]

Now we construct a TM $M$ = on input $n$,

enumerate $C'$ to get $g_n$.
compute $g_n(n)$.
output $g_n(n) + 1$.

Then $g^*(n) = g_n(n) + 1$ is unary computable numerical function and thus $g^* \in C'$.

However, $\forall n \in \mathbb{N}$, $g^* \neq g_n$, which leads to a contradiction.

So we can conclude that

\[ \text{PR} \subsetneq \text{C} \]

That's not good since all computable function can't build up by that simple functions!

But now we consider add an additional operation of the primitive recursive function: minimalization.

Minimalization¶

Definition

Suppose $g: \mathbb{N}^{k + 1} \rightarrow \mathbb{N}$, and we define

\[ f(n_1, \dots, n_k) = \left\{ \begin{aligned} & \text{minimum $n_{k + 1}$ such that $g(n_1, \dots, n_k, n_{k + 1}) = 1$}, && \text{if such that $n_{k + 1}$ exists}. \\ & 0, && \text{if such that $n_{k + 1}$ does not exist}. \end{aligned} \right. \]

Then we call $f$ is a minimalization of $g$, denoted by

\[ \mu_m [g(n_1, \dots, n_k, m) = 1]. \]

Definition

A function $g$ is minimalizable if $g$ is computable and $\forall n_1, \dots, n_k$, $\exists m \ge 0$, such that $g(n_1, \dots, n_k, m) = 1$.

Example

\[ \log(m, n) = \lceil \log_{m + 2} (n + 1) \rceil = \mu_p[\text{geq}((m + 2)^p, n + 1) = 1]. \]

This function is minimalizable.

Theorem

If $g: \mathbb{N}^{k + 1} \rightarrow \mathbb{N}$ is minimalizable, then $\mu_m[g(n_1, \dots, n_k, m) = 1]$ is computable.

Theorem

The problem that

Given a function $g$, is $g$ minimalizable?

is undecidable.

最后更新: 2024.01.29 15:36:20 CST
创建日期: 2023.12.12 22:53:54 CST