A Short Program in MuPAD that Computes in the Limit a Function f : N → N Which Eventually Dominates Every Computable Function g : N → N

1. The Collatz Problem Leads to a Short Computer Program That Computes in the Limit a Function $\gamma :\mathbb{N}\to \{0,1\}$ of Unknown Computability

Definition 1

(cf. [10], pp. 233–235). A computation in the limit of a function $f:\mathbb{N}\to \mathbb{N}$ is a semi-algorithm which takes as input a non-negative integer n and for every $m\in \mathbb{N}$ prints a non-negative integer $\xi (n,m)$ such that $\underset{m\to \infty }{lim}\xi (n,m)=f\left(n\right)$.

By Definition 1, a function $f:\mathbb{N}\to \mathbb{N}$ is computable in the limit when there exists an infinite computation which takes as input a non-negative integer n and prints a non-negative integer on each iteration and prints $f\left(n\right)$ on each sufficiently high iteration.

It is known that there exists a limit-computable function $f:\mathbb{N}\to \mathbb{N}$ which is not computable, see Theorem 1. Every known proof of this fact does not lead to the existence of a short computer program that computes f in the limit. So far, short computer programs can only compute in the limit functions from $\mathbb{N}$ to $\mathbb{N}$ whose computability is proven or unknown.

Lemma 1.

For every $n\in \mathbb{N}$,

$${\displaystyle \frac{\mathrm{sign}(n-1)·(2n+(1-{(-1)}^n)·(5n+2))}{4}}=\left\{\begin{array}{cc}\hfill 0,& \mathrm{if}n=1\hfill \\ \hfill {\displaystyle \frac{n}{2}},& \mathrm{if}n\mathrm{is}\mathrm{even}\hfill \\ \hfill 3n+1,& \mathrm{if}n\mathrm{is}\mathrm{odd}\mathrm{and}n\ne 1\hfill \end{array}\right.$$

MuPAD is a part of the Symbolic Math Toolbox in MATLAB R2019b. By Lemma 1, the following program in MuPAD computes in the limit a function $\gamma :\mathbb{N}\to \{0,1\}$.

input("Input a non-negative integer n",n):

while TRUE do

print(sign(n)):

n:=sign(n-1)*(2*n+(1-(-1)^n)*(5*n+2))/4:

end_while:

The computability of $\gamma$ is unknown, see [1], p. 79. The Collatz conjecture implies that $\gamma \left(n\right)=0$ for every $n\in \mathbb{N}$.

2. A Limit-Computable Function $f:\mathbb{N}\to \mathbb{N}$ Which Eventually Dominates Every Computable Function $g:\mathbb{N}\to \mathbb{N}$

For $n\in \mathbb{N}$, let

$$E_n=\{1=x_k,x_i+x_j=x_k,x_i·x_j=x_k:i,j,k\in \{0,\dots ,n\}\}$$

Theorem 1.([9] p. 118). There exists a limit-computable function $f:\mathbb{N}\to \mathbb{N}$ which eventually dominates every computable function $g:\mathbb{N}\to \mathbb{N}$.

We present an alternative proof of Theorem 1. For $n\in \mathbb{N}$, $f\left(n\right)$ denotes the smallest $b\in \mathbb{N}$ such that if a system of equations $\mathcal{S}\subseteq E_n$ has a solution in ${\mathbb{N}}^{n+1}$, then $\mathcal{S}$ has a solution in ${\{0,\dots ,b\}}^{n+1}$. The function $f:\mathbb{N}\to \mathbb{N}$ is computable in the limit and eventually dominates every computable function $g:\mathbb{N}\to \mathbb{N}$, see [12]. The term "dominated" in the title of [12] means "eventually dominated". Flowchart 1 shows a semi-algorithm which computes $f\left(n\right)$ in the limit, see [12].

3. A Short Program in MuPAD That Computes f in the Limit

Flowchart 2 shows a simpler semi-algorithm which computes $f\left(n\right)$ in the limit.

Lemma 2.

For every $n,m\in \mathbb{N}$, the number printed by Flowchart 2 does not exceed the number printed by Flowchart 1.

Proof. 

For every $(a_0,\dots ,a_n)\in {\{0,\dots ,m\}}^{n+1}$,

$$E_n\supseteq \{1=x_k:(k\in \{0,\dots ,n\})\wedge (1=a_k)\}\cup$$

$$\{x_i+x_j=x_k:(i,j,k\in \{0,\dots ,n\})\wedge (a_i+a_j=a_k)\}\cup$$

$$\{x_i·x_j=x_k:(i,j,k\in \{0,\dots ,n\})\wedge (a_i·a_j=a_k)\}$$

   □

Lemma 3.

For every $n,m\in \mathbb{N}$, the number printed by Flowchart 1 does not exceed the number printed by Flowchart 2.

Proof. 

Let $n,m\in \mathbb{N}$. For every system of equations $\mathcal{S}\subseteq E_n$, if $(a_0,\dots ,a_n)\in {\{0,\dots ,m\}}^{n+1}$ and $(a_0,\dots ,a_n)$ solves $\mathcal{S}$, then $(a_0,\dots ,a_n)$ solves the following system of equations:

$$\{1=x_k:(k\in \{0,\dots ,n\})\wedge (1=a_k)\}\cup$$

$$\{x_i+x_j=x_k:(i,j,k\in \{0,\dots ,n\})\wedge (a_i+a_j=a_k)\}\cup$$

$$\{x_i·x_j=x_k:(i,j,k\in \{0,\dots ,n\})\wedge (a_i·a_j=a_k)\}$$

   □

Theorem 2.

For every $n,m\in \mathbb{N}$, Flowcharts 1 and 2 print the same number.

Proof. 

It follows from Lemmas 2 and 3.    □

Definition 2.

An approximation of a tuple $(x_0,\dots ,x_n)\in {\mathbb{N}}^{n+1}$ is a tuple $(y_0,\dots ,y_n)\in {\mathbb{N}}^{n+1}$ such that

$$(\forall k\in \{0,\dots ,n\}(1=x_k\Rightarrow 1=y_k))\wedge$$

$$(\forall i,j,k\in \{0,\dots ,n\}(x_i+x_j=x_k\Rightarrow y_i+y_j=y_k))\wedge$$

$$(\forall i,j,k\in \{0,\dots ,n\}(x_i·x_j=x_k\Rightarrow y_i·y_j=y_k))$$

Observation 1.

There exists a set $\mathcal{A}\left(n\right)\subseteq {\mathbb{N}}^{n+1}$ such that

$$\mathrm{card}\left(\mathcal{A}\left(n\right)\right)⩽2^{\mathrm{card}\left(E_n\right)}=2^{n+1+2·{(n+1)}^3}$$

and every tuple $(x_0,\dots ,x_n)\in {\mathbb{N}}^{n+1}$ possesses an approximation in $\mathcal{A}\left(n\right)$.

Observation 2.

$f\left(n\right)$ equals the smallest $b\in \mathbb{N}$ such that every tuple $(x_0,\dots ,x_n)\in {\mathbb{N}}^{n+1}$ possesses an approximation in ${\{0,\dots ,b\}}^{n+1}$.

Observation 3.

For every $n,m\in \mathbb{N}$, Flowcharts 1 and 2 print the smallest $b\in \{0,\dots ,m\}$ such that every tuple $(x_0,\dots ,x_n)\in {\{0,\dots ,m\}}^{n+1}$ possesses an approximation in ${\{0,\dots ,b\}}^{n+1}$.

The following program in MuPAD implements the semi-algorithm shown in Flowchart 2.

input("Input a non-negative integer n",n):

m:=0:

while TRUE do

X:=combinat::cartesianProduct([s $s=0..m] $t=0..n):

Y:=[max(op(X[u])) $u=1..(m+1)^(n+1)]:

for p from 1 to (m+1)^(n+1) do

for q from 1 to (m+1)^(n+1) do

v:=1:

for k from 1 to n+1 do

if 1=X[p][k] and 1<>X[q][k] then v:=0 end_if:

for i from 1 to n+1 do

for j from i to n+1 do

if X[p][i]+X[p][j]=X[p][k] and X[q][i]+X[q][j]<>X[q][k] then v:=0 end_if:

if X[p][i]*X[p][j]=X[p][k] and X[q][i]*X[q][j]<>X[q][k] then v:=0 end_if:

end_for:

end_for:

end_for:

if max(op(X[q]))<max(op(X[p])) and v=1 then Y[p]:=0 end_if:

end_for:

end_for:

print(max(op(Y))):

m:=m+1:

end_while:

For $n\in \mathbb{N}$, $h\left(n\right)$ denotes the smallest $b\in \mathbb{N}$ such that if a system of equations $\mathcal{S}\subseteq \{x_i+1=x_k,x_i·x_j=x_k:i,j,k\in \{0,\dots ,n\}\}$ has a solution in ${\mathbb{N}}^{n+1}$, then $\mathcal{S}$ has a solution in ${\{0,\dots ,b\}}^{n+1}$. From [12] and Lemma 3 in [11], it follows that the function $h:\mathbb{N}\to \mathbb{N}$ is computable in the limit and eventually dominates every computable function $g:\mathbb{N}\to \mathbb{N}$. A bit shorter program in MuPAD computes h in the limit.

4. A Limit-Computable Function $\beta :\mathbb{N}\to \mathbb{N}$ of Unknown Computability Which Eventually Dominates Every Function $\delta :\mathbb{N}\to \mathbb{N}$ with a Single-Fold Diophantine Representation

The Davis-Putnam-Robinson-Matiyasevich theorem states that every listable set $\mathcal{M}\subseteq {\mathbb{N}}^n$$(n\in \mathbb{N}\setminus \{0\left\}\right)$ has a Diophantine representation, that is

$$(a_1,\dots ,a_n)\in \mathcal{M}⟺\exists x_1,\dots ,x_m\in \mathbb{N}W(a_1,\dots ,a_n,x_1,\dots ,x_m)=0\left(\mathtt{R}\right)$$

(1)

for some polynomial W with integer coefficients, see [6]. The representation (R) is said to be single-fold, if for any $a_1,\dots ,a_n\in \mathbb{N}$ the equation $W(a_1,\dots ,a_n,x_1,\dots ,x_m)=0$ has at most one solution $(x_1,\dots ,x_m)\in {\mathbb{N}}^m$.

Hypothesis 1

([2,3,4,5] pp. 341–342, [7] p. 42, [8] p. 745). Every listable set $\mathcal{X}\subseteq {\mathbb{N}}^k$$(k\in \mathbb{N}\setminus \{0\left\}\right)$ has a single-fold Diophantine representation.

For $n\in \mathbb{N}$, $\beta \left(n\right)$ denotes the smallest $b\in \mathbb{N}$ such that if a system of equations $\mathcal{S}\subseteq E_n$ has a unique solution in ${\mathbb{N}}^{n+1}$, then this solution belongs to ${\{0,\dots ,b\}}^{n+1}$. The computability of $\beta$ is unknown.

Theorem 3.

The function $\beta :\mathbb{N}\to \mathbb{N}$ is computable in the limit and eventually dominates every function $\delta :\mathbb{N}\to \mathbb{N}$ with a single-fold Diophantine representation.

Proof. 

This is proved in [12]. Flowchart 3 shows a semi-algorithm which computes $\beta \left(n\right)$ in the limit, see [12].     □

5. A Short Program in MuPAD That Computes $\beta$ in the Limit

Flowchart 4 shows a simpler semi-algorithm which computes $\beta \left(n\right)$ in the limit.

Lemma 4.

For every $n,m\in \mathbb{N}$, the number printed by Flowchart 4 does not exceed the number printed by Flowchart 3.

Proof. 

For every $(a_0,\dots ,a_n)\in {\{0,\dots ,m\}}^{n+1}$,

$$E_n\supseteq \{1=x_k:(k\in \{0,\dots ,n\})\wedge (1=a_k)\}\cup$$

$$\{x_i+x_j=x_k:(i,j,k\in \{0,\dots ,n\})\wedge (a_i+a_j=a_k)\}\cup$$

$$\{x_i·x_j=x_k:(i,j,k\in \{0,\dots ,n\})\wedge (a_i·a_j=a_k)\}$$

   □

Lemma 5.

For every $n,m\in \mathbb{N}$, the number printed by Flowchart 3 does not exceed the number printed by Flowchart 4.

Proof. 

Let $n,m\in \mathbb{N}$. For every system of equations $\mathcal{S}\subseteq E_n$, if $(a_0,\dots ,a_n)\in {\{0,\dots ,m\}}^{n+1}$ is a unique solution of $\mathcal{S}$ in ${\{0,\dots ,m\}}^{n+1}$, then $(a_0,\dots ,a_n)$ solves the system $\widehat{\mathcal{S}}$, where

$$\widehat{\mathcal{S}}=\{1=x_k:(k\in \{0,\dots ,n\})\wedge (1=a_k)\}\cup$$

$$\{x_i+x_j=x_k:(i,j,k\in \{0,\dots ,n\})\wedge (a_i+a_j=a_k)\}\cup$$

$$\{x_i·x_j=x_k:(i,j,k\in \{0,\dots ,n\})\wedge (a_i·a_j=a_k)\}$$

By this and the inclusion $\widehat{\mathcal{S}}\supseteq \mathcal{S}$, $\widehat{\mathcal{S}}$ has exactly one solution in ${\{0,\dots ,m\}}^{n+1}$, namely $(a_0,\dots ,a_n)$.    □

Theorem 4.

For every $n,m\in \mathbb{N}$, Flowcharts 3 and 4 print the same number.

Proof. 

It follows from Lemmas 4 and 5.    □

The following program in MuPAD implements the semi-algorithm shown in Flowchart 4.

input("Input a non-negative integer n",n):

m:=0:

while TRUE do

X:=combinat::cartesianProduct([s $s=0..m] $t=0..n):

Y:=[max(op(X[u])) $u=1..(m+1)^(n+1)]:

for p from 1 to (m+1)^(n+1) do

for q from 1 to (m+1)^(n+1) do

v:=1:

for k from 1 to n+1 do

if 1=X[p][k] and 1<>X[q][k] then v:=0 end_if:

for i from 1 to n+1 do

for j from i to n+1 do

if X[p][i]+X[p][j]=X[p][k] and X[q][i]+X[q][j]<>X[q][k] then v:=0 end_if:

if X[p][i]*X[p][j]=X[p][k] and X[q][i]*X[q][j]<>X[q][k] then v:=0 end_if:

end_for:

end_for:

end_for:

if q<>p and v=1 then Y[p]:=0 end_if:

end_for:

end_for:

print(max(op(Y))):

m:=m+1:

end_while: