Complexity and subcomplexities, Part 2

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Index

First three properties of complexity $\mathit{rn}$
Subcomplexities
Ceiling of a subcomplexity
Maximum of two subcomplexities. A truncated subcomplexity.
Subcomplexity $m_5$

First three properties of complexity $\mathit{rn}$

Every Ł-string features character $1$ at least one time. It follows that $\mathit{rn}(n)\ge 1$ for every natural number $n=1\ 2\ \ldots$ . On the other hand, since Ł-string $''1''$ represents $1$ , it follows that $\mathit{rn}(1) \le 1$ , hence:
\[ \mathit{rn}(1) = 1\]

That’s the first of the three properties. Now let $a\ b\in\mathbb N$ be two arbitrary natural numbers; and let Ł-strings $''A''\ \ ''B''$ be such that:

$\mathit{val}(''A'') = a$ and $\mathit{val}(''B'') = b$ ;
$\mathit{rn}(a) = \mathit{RN}(''A'')$ and $\mathit{rn}(b) = \mathit{RN}(''B'')$

Then Ł-string $''AB+''$ represents (has value) $a\ b\ +$ . It follows that:
\[ \mathit{rn}(a\ b\ +)\ \ \le\ \ \mathit{RN}(”AB+”)\ \ =\ \ \mathit{RN}(”A”)\ \mathit{RN}(”B”)\ + \]

\[ =\ \ \mathit{rn}(a)\ \mathit{rn}(b)\ + \]

We see that

\[ \forall_{a\ b\ \in\ \mathbb N}\quad \mathit{rn}(a\ b\ +)\quad\le\quad \mathit{rn}(a)\ \mathit{rn}(b)\ + \]

and this is the second property of complexity $\mathit{rn}$ . A very similar proof justifies also the third property:

\[ \forall_{a\ b\ \in\ \mathbb N}\quad \mathit{rn}(a\ b\ \bullet)\quad\le\quad \mathit{rn}(a)\ \mathit{rn}(b)\ + \]

Subcomplexities

DEFINITION 0 A real positive function $\gamma : \mathbb N \rightarrow (0;\infty)$ is called a subcomplexity $\Leftarrow:\Rightarrow$ $\gamma$ satisfies the following three properties:

$\gamma(1)\le 1$
$\forall_{a\ b\ \in\ \mathbb N}\quad \gamma(a\ b\ +)\ \le\ \gamma(a)\ \gamma(b)\ +$
$\forall_{a\ b\ \in\ \mathbb N}\quad \gamma(a\ b\ \bullet)\ \le\ \gamma(a)\ \gamma(b)\ +$

We can see from the First three properties that the complexity function $\mathit{rn}$ is a subcomplexity. Furthermore, a simple induction on natural numbers shows that $\mathit{rn}$ is the greatest subcomplexity, meaning that for every subcomplexity $\gamma$ the following inequality holds:

\[ \forall_{n\in\mathbb N}\ \ \gamma(n)\le\mathit{rn}(n) \]

Ceiling of a subcomplexity

We will see that the ceiling of a subcompolexity is a subcomplexity.

Let $t\in\mathbb R$ be an arbitrary real number. Then the floor $\lfloor t\rfloor$ and the ceiling $\lceil t\rceil$ are defined as the integers which satisfy inequalities:
\[ t-1\ <\ \lfloor t\rfloor\ \le t \] \[ t\ \le\ \lceil t\rceil\ <\ t+1 \] Such integers are unique for every real $t$ . They have the following properties for every real $t$ and integer $n$ :

$t\ge n\qquad\Rightarrow\qquad \lfloor t\rfloor \ge n$
$t \le n\qquad\Rightarrow\qquad \lceil t\rceil \le n$

We will apply the latter property. Let $\gamma$ be an arbitrary subcomplexity. Let $\delta :\mathbb N\rightarrow (0;\infty)$ be given by:

\[ \forall_{n\in\mathbb N}\quad \delta(n) := \lceil \gamma(n)\rceil \]

(one could write $\delta := \lceil\gamma\rceil$ ). Then $0 < \gamma(1)\le 1$ hence $\delta(1) = 1$ . Thus $\delta$ has the first property of subcomplexities.

Now let $a\ b\in\mathbb N$ be two arbitrary natural numbers. Then

\[\gamma(a\ b\ +)\ \ \le\ \ \gamma(a) \gamma(b)\ +\ \ \le\ \ \delta(a)\ \delta(b)\ + \]

The $\delta$ -expression on the right is an integer. Thus, by the definition of $\delta$ , and by the general property of the operation ceiling (see above), we get:

\[ \delta(a\ b\ +)\ \ \le\ \ \delta(a)\ \delta(b)\ + \]

Thus $\delta$ has the second property of a su8bcomplexity. A proof of the third property:

\[ \delta(a\ b\ \bullet)\ \ \le\ \ \delta(a)\ \delta(b)\ + \]

is very similar.

Maximum of two subcomplexities. A truncated subcomplexity.

Let $\alpha\ \beta : \mathbb N\rightarrow (0;\infty)$ be two arbitrary subcomplexities. Define their maximum $\gamma : \mathbb N\rightarrow (0;\infty)$ in the usual way:
\[ \forall_{n\in\mathbb N}\quad \gamma(n)\ \ :=\ \ \max(\alpha(n)\ \ \beta(n)) \]

Then, as it is easy to prove, $\gamma$ is a subcomplexity too.

Let $\gamma:\mathbb N\rightarrow (0;\infty)$ be an arbitrary subcomplexity. Let $M\in (0;\infty)$ be an arbitrary positive real number. Define the truncated function $\gamma_M$ as follows:

\[ \forall_{n\in\mathbb N}\quad \gamma_M(n)\ \ :=\ \ \min(\gamma(n)\ \ M) \]

It is easy to prove that the truncated function $\gamma_M$ is a subcomplexity–call it the truncated subcomplexity.

POTENTIAL APPLICATION Let $\gamma$ be a subcomplexity, and let $M\in\mathbb N$ be such that $\gamma(k) = \mathit{rn}(k)$ for every $k=1\ldots M$ . Then define $\delta:\mathbb N\rightarrow (0;\infty)$ as follows:

$\forall_{k\in\mathbb N\setminus\{M\}}\quad \delta(k)\ :=\ \gamma(k)$
$\delta(M)\ :=\ \mathit{zn}(M)$

Then $\delta$ is a subcomplexity too–call it an improved subcomplexity.

Subcomplexity $m_5$

Let’s define function $m_5:\mathbb N\rightarrow (0;\infty)$ as follows:

\[ \forall_{k\in\mathbb N}\quad m_5(k)\ \ :=\ \ \min(k\ \ 5) \]

First of all $m_5(1)=1$ which means that function $m_5$ has the first subcomplexity property.

Now let $a\ b\in\mathbb N$ be two arbitrary natural numbers. Then

$\max(a\ b)\le 5\qquad\Rightarrow\qquad m_5(a\ b\ +)\ \ \le\ \ a\ b\ +\ \ =\ \ m_5(a)\ m_5(b)\ +$
$\max(a\ b)\ge 5\qquad\Rightarrow\qquad m_5(a\ b\ +)\ \ \le\ \ 5\ \ \le\ \ m_5(a)\ m_5(b)\ +$

which show that $m_5$ has the second property of a subcomplexity:

\[ \forall_{a\ b\in \mathbb N}\quad m_5(a\ b\ +)\ \ \le\ \ m_5(a)\ m_5(b)\ + \]

Furthermore, if $\min(a\ b) = 1$ then $a\ b\ \bullet\ =\ \max(a\ b)$ and $\{1\ \max(a\ b)\}\ =\ \{a\ b\}$ , hence

\[ \min(a\ b) = 1\quad\Rightarrow\quad m_5(a\ b\ \bullet)\ =\ \ m_5(\max(a\ b)) \]
\[ <\ \ \ 1\ \ m_5(\max(a\ b))\ \ +\ \ \ =\ \ \ m_5(a)\ m_5(b)\ + \] -- i.e. the third subcomplexity property holds when $\min(a\ b) = 1$ . Next, if $\min(a\ b) = \max(a\ b) = 2$ (so that $a=b=2$ ) then
\[ m_5(a\ b\ \bullet)\ =\ 4\ \ =\ \ m_5(a)\ m_5(b)\ + \]

Finally, let the remaining possibility hold: $\min(a\ b)\ \ge 2$ and $\max(a\ b) \ge 3$ . Then $m_5(\min(a\ b))\ \ge\ 2$ and $m_5(\max(a\ b))\ \ge\ 3$ hence