A Problem in Facial Digit Recasting of Positive Integers to Show P Does Not Equal NP

1. Introduction

More on terminologies, symbols and notations used in this article can be found in [3,6,8]. Our discussions also draw on the following topics 1 through 9 that can be found in some detail in the suggested references, some of which can be downloaded free of cost.

1.

Algorithms [3,4,5,10].

2.

Problem instance and size of an instance [1,4,5].

3.

A step in an algorithm [3,4,5].

4.

Polynomial-time algorithm [2,3,4,5,10].

5.

Class P and class NP of problems [2,3,4,5,9,11].

6.

Feasible solutions and proposed solutions [3,4,5,7].

7.

Certificates and certificate candidates; check and verification [4,5].

8.

Atomic sub-outputs of solutions [4,5].

9.

Bijective maps (or, bijections) [6,8].

1.1. Facial Digit Recasting

Throughout this article, $\mathbb{N}$ will denote the set of positive integers, $\mathbb{W}$ the set of non-negative integers (i.e., $\mathbb{W}=\mathbb{N}\cup \left\{0\right\}$), $\mathbb{Z}$ the set of integers and $\mathbb{R}$ the set of real numbers. This article adopts the number system with base 10. The set ${\mathbb{D}}_{10}$ defined by ${\mathbb{D}}_{10}=\left\{0,1,2,3,4,5,6,7,8,9\right\}$ is the set of all the digits base 10.

An real number y is said to be in explicit exponent form (e-e form, for brevity) if y is expressed in the form $x^k$ where (i) x and k are in $\mathbb{R}$ and (ii) the value of k is presented explicitly. Note that the value of x is, of necessity, given explicitly. For instance, $5^3$ (where $x=5$ and $k=3$) and ${125}^1$ (where $x=125$ and $k=1$) are in explicit exponent forms whereas 125 is not (since the index k is not explicit, albeit $k=1$ is understood), though $125=5^3={125}^1$.

Let $n\in \mathbb{W}$. A facial digit of n is an element of ${\mathbb{D}}_{10}$ that appears in the integral representation of n. The set $\mathcal{F}\left(n\right)$ will denote the set of all the facial digits of n. If $d\in {\mathbb{D}}_{10}$ then d is a facial digit of n if and only if $d\in \mathcal{F}\left(n\right)$.

For instance: (i) 30706 has four facial digits in all, and they are $0,3,6,$ and 7, and so $\mathcal{F}\left(30706\right)=\left\{0,3,6,7\right\}$.

(ii) If $n={25}^{35}$ then $\mathcal{F}\left(30706\right)=\left\{2,5,3\right\}$.

A facial digit will be henceforth referred to as a fit.

Let A and B be nonempty sets. A function from A to B is an assignment that assigns a unique element of B to each element of A. The notation $f:A\to B$ means f is a function from A to B. In this case, A is the domain of f and B is the codomain of f.

Let A be a nonempty set. The set of all ordered pairs [8]$\left(x,y\right)$ where $x\in A$ and $y\in A$ is denoted by $A^2$ or $A\mathsf{x}A$ and is called the Cartesian square of A. A binary operation in A is a function $f:A^2\to A$. Addition, subtraction and multiplication of real numbers are examples of binary operations in $\mathbb{R}$. Division of real numbers is a conditional binary operation in $\mathbb{R}$ in the sense that it is a function from $\mathbb{R}\mathsf{x}\left(\mathbb{R}-\left\{0\right\}\right)$ to $\mathbb{R}$; i.e., the value of this function at $\left(x,y\right)\in \mathbb{R}\mathsf{x}\mathbb{R}$ is ${\displaystyle \frac{x}{y}}$ providing $y\ne 0$. Addition, subtraction and multiplication of integers are binary operations in $\mathbb{Z}$.

If * is a binary operation in A and if x and y are elements of A then * assigns $\left(x,y\right)$ (of $A^2$) to the element $x*y$ (of A). In the operation (or, assignment) symbolized by $*:\left(x,y\right)\to x*y$, each of x and y is called an operand (of the operation *).

The binary operations addition, subtraction, multipication and division in $\mathbb{R}$ are the four fundamental binary operations. The set $\mathbb{O}$ will denote the set consisting of these four operations.

Let A be a nonempty subset of $\mathbb{R}$. An arithmetic expression over A is an expression $\mathcal{X}$ comprising finitely many binary operations and finitely many operands and having the following properties (AE1) and (AE2):

(AE1) Any binary operation appearing in $\mathcal{X}$ is a binary operation in $\mathbb{R}$.

(AE2) Every operand appearing in $\mathcal{X}$ is an element of A - i.e., no element that is not in A is an operand in $\mathcal{X}$.

Let $n\in \mathbb{N}$ and $n\ge 11$. A facial digit recasting or fit recasting of n is an equation $n={\mathcal{S}}_n\left(a_1,\dots ,a_k\right)$ (where $k\in \mathbb{N}$) - written $n={\mathcal{S}}_n$ for brevity - subject to the following requirements / provisions from (F1) through (F5):

(F1) ${\mathcal{S}}_n$ is an arithmetic expression over $\mathbb{W}-\left\{n\right\}$ in which $a_1,\dots ,a_k$ are distinct elements of $\mathbb{W}-\left\{n\right\}$. (This necessiates: for each $j=1,\dots ,k$, $a_j$ is an operand in ${\mathcal{S}}_n$ and $a_j$ may or may not be given in explicit exponential form.)

(F2) Any of $a_1$ through $a_k$ may appear more than once in the expression ${\mathcal{S}}_n$.

(F3) For each $j=1,\dots ,k$, $\mathcal{F}\left(a_j\right)\subset \mathcal{F}\left(n\right)$ and $\mathcal{F}\left(a_1\right)\cup \dots \cup \mathcal{F}\left(a_k\right)=\mathcal{F}\left(n\right)$ - i.e., each element of $\mathcal{F}\left(n\right)$ must appear at least once in ${\mathcal{S}}_n$, and no element outside $\mathcal{F}\left(n\right)$ appears anywhere in ${\mathcal{S}}_n$.

(F4) Other than $a_1$ through $a_k$ and the binary operation symbols that legitimise (F1), the only symbols allowed in ${\mathcal{S}}_n$ are the three types of pairs of nesting symbols, namely, $\left(★\right)$, $\left[★\right]$ and $\left\{★\right\}$ (that are, respectively, parentheses, square brackets and braces), where ★ denotes the appropriate mathematically consistent content enclosed by the nesting symbols.

(F5) The result of mathematically executing the expression ${\mathcal{S}}_n$ is n (thereby ensuring consistency with $n={\mathcal{S}}_n$, where the = sign means that if the expression ${\mathcal{S}}_n$ is computed mathematically then the resulting integer is n).

For brevity, a facial digit recasting of n will be called a fit-rec of n.

2. Some Definitions Connected to Fit Recasting

For $m\in \mathbb{N}$, the set ${\mathbb{N}}_{\ge m}$ is defined as ${\mathbb{N}}_{\ge m}=\left\{a\in \mathbb{N}:a\ge m\right\}$. For the remainder of this section, assume $n\in {\mathbb{N}}_{\ge 11}$.

Let $\mathcal{F}\left(n\right)=\left\{x_1,\dots ,x_q\right\}$ and $n={\mathcal{S}}_n\left(a_1,\dots ,a_k\right)$ be a fit-rec of n. For each digit $x\in \mathcal{F}\left(n\right)$, the freqency of x in ${\mathcal{S}}_n$ is denoted by $\mathsf{f}\left(x,{\mathcal{S}}_n\right)$ (or, simply, $\mathsf{f}\left(x\right)$, if ${\mathcal{S}}_n$ is clear from the context) and is defined to be the number of times x appears in the expression ${\mathcal{S}}_n$. The total frequency of the elements of $\mathcal{F}\left(n\right)$ in ${\mathcal{S}}_n$ is denoted by $\mathsf{F}\left({\mathcal{S}}_n\right)$ and is defined as $\mathsf{F}\left({\mathcal{S}}_n\right)={\Sigma }_{j=1}^q\mathsf{f}\left(x_j,{\mathcal{S}}_n\right)$.

Each $a_j$ ($j=1,\dots ,k$ ) appearing in ${\mathcal{S}}_n$ is a cog of ${\mathcal{S}}_n$. The set $\left\{a_1,\dots ,a_k\right\}$ is the set of cogs of ${\mathcal{S}}_n$. The cardinality [8] of the set $\left\{a_1,\dots ,a_k\right\}$ is the base size of ${\mathcal{S}}_n$, denoted (for obvious reasons) by $\left|{\mathcal{S}}_n\right|$.

It is important to note that $\left\{a_1,\dots ,a_k\right\}$ is a set, and so the elements $a_1$ through $a_k$ are required to be distinct as elements of $\mathbb{W}-\left\{n\right\}$ (see F1 of Section 1). This necessiates that no two of these k cogs should be equal as elements of $\mathbb{W}-\left\{n\right\}$. For instance, if a cog $a_j={11}^2$ then none of the remaining cogs should have a value 121, regardless of the form any cog is presented in.

3. Examples for ${\mathcal{S}}_N$ and $\mathsf{F}\left({\mathcal{S}}_N\right)$

1.

Let $n=112$. Then $\mathcal{F}\left(112\right)=\left\{1,2\right\}$. The equation $112={11}^2-11+2$ is a fit-rec of n. We write $n={\mathcal{S}}_n$ where ${\mathcal{S}}_n$ (or ${\mathcal{S}}_{112}$) is the expression ${11}^2-11+2$. The cogs of ${\mathcal{S}}_{112}$ are $a_1={11}^2$, $a_2=11$ and $a_3=2$. Of these, only $a_1$ is in e-e form. The set $\left\{a_1,a_2,a_3\right\}=\left\{{11}^2,11,2\right\}$ is the set of cogs of ${\mathcal{S}}_{112}$, from which the base size of ${\mathcal{S}}_{112}$ is $\left|{\mathcal{S}}_{112}\right|=3$ ($=\left|\left\{a_1,a_2,a_3\right\}\right|$). The frequencies of the elements of $\mathcal{F}\left(112\right)$ are given by: $\mathsf{f}\left(1,{\mathcal{S}}_{112}\right)=4$ (since the fit 1 appears exactly four times in ${\mathcal{S}}_{112}$) and $\mathsf{f}\left(2,{\mathcal{S}}_{112}\right)=2$ (since the fit 2 appears exactly two times in ${\mathcal{S}}_{112}$). So $\mathsf{F}\left({\mathcal{S}}_{112}\right)=\mathsf{f}\left(1\right)+\mathsf{f}\left(2\right)=6$.

2.

$n=34407$. $\mathcal{F}\left(n\right)=\left\{0,3,4,7\right\}$. A fit-rec of n is $n={\mathcal{S}}_n$ where ${\mathcal{S}}_n$ denotes the expression $300\left(70+30\right)+400\left(7+3\right)+70\mathtt{x}7-77-3-3$. Here the cogs of ${\mathcal{S}}_n$ are: $a_1=300$, $a_2=70$, $a_3=30$, $a_4=400,a_5=7,a_6=3$ and $a_7=77$, so that ${\mathcal{S}}_n$ is ${\mathcal{S}}_n\left(a_1,a_2,a_3,a_4,a_5,a_6,a_7\right)$. The set of cogs of ${\mathcal{S}}_n$ is $\left\{300,70,30,400,7,3,77\right\}$. The base size of ${\mathcal{S}}_n$ is 7, which is the cardinality of $\left\{300,70,30,400,7,3,77\right\}$. More elaborately, ${\mathcal{S}}_n$ is $a_1\left(a_2+a_3\right)+a_4\left(a_5+a_6\right)+a_2\mathsf{x}{a}_5-a_7-a_6-a_6$.

In ${\mathcal{S}}_n$, the frequencies of the elements of $\mathcal{F}\left(34407\right)$ are given by: $\mathsf{f}\left(0\right)=7$, $\mathsf{f}\left(3\right)=5$, $\mathsf{f}\left(4\right)=1$ and $\mathsf{f}\left(7\right)=6$, from which $\mathsf{F}\left({\mathcal{S}}_n\right)=\mathsf{f}\left(0\right)+\mathsf{f}\left(3\right)+\mathsf{f}\left(4\right)+\mathsf{f}\left(7\right)=19$.

3.

For $n=112$, the equation $112=11\mathsf{x}12-11\mathsf{x}2+2$ is another fit-rec of n, written $n={\mathcal{T}}_n\left(b_1,b_2,b_3\right)$ where ${\mathcal{T}}_n\left(b_1,b_2,b_3\right)$ denotes the expression $b_1{b}_2-b_1{b}_3+b_3$ with $b_1=11,b_2=12$ and $b_3=2$. Then $\mathsf{f}\left(1,{\mathcal{T}}_{112}\right)=5$, $\mathsf{f}\left(2,{\mathcal{T}}_{112}\right)=3$ and $\mathsf{F}\left({\mathcal{T}}_{112}\right)=8$.

4.

On the other hand, $112={\mathcal{Y}}_{112}$, where ${\mathcal{Y}}_{112}$ is the expression $11\mathsf{x}11-11+1+1$, although mathematically correct, is NOT a fit-rec of 112 since the element $2\in \mathcal{F}\left(112\right)$ does not appear in the expression ${\mathcal{Y}}_{112}$ (see F3 of Section 1).

5.

$n=1089$. $\mathcal{F}\left(n\right)=\left\{0,1,8,9\right\}$. A fit-rec of n is $n={\mathcal{S}}_n\left(a_1,a_2,a_3,a_4,a_5\right)$ where ${\mathcal{S}}_n\left(a_1,a_2,a_3,a_4,a_5\right)$ is the expression $80\mathtt{x}10+100\left(1+1\right)+89$ with $a_1=80$, $a_2=10$, $a_3=100$, $a_4=1$ and $a_5=89$. In ${\mathcal{S}}_{1089}$, $\mathsf{f}\left(0\right)=4$, $\mathsf{f}\left(1\right)=4$, $\mathsf{f}\left(8\right)=2$ and $\mathsf{f}\left(9\right)=1$, from which $\mathsf{F}\left({\mathcal{S}}_{1089}\right)=11$.

6.

$n=112$. $\mathcal{F}\left(n\right)=\left\{1,2\right\}$. A fit-rec of n is $n={\mathcal{S}}_n$ where ${\mathcal{S}}_n\left(a_1,a_2,a_3,a_4\right)$ is the expression $12\left(11-1\right)-11+2+1$ with $a_1=12$, $a_2=11$, $a_3=1$ and $a_4=2$. In ${\mathcal{S}}_{112}$, $\mathsf{f}\left(1\right)=7$ and $\mathsf{f}\left(2\right)=2$, from which $\mathsf{F}\left({\mathcal{S}}_{112}\right)=9$.

NOTE 3.1. By now it is clear that a given $n\in {\mathbb{N}}_{\ge 11}$ can have more than one fit-recs.

4. Some Observations About ${\mathcal{S}}_N$ and $\mathsf{F}\left({\mathcal{S}}_N\right)$

Proposition 4.1.Let $n\in {\mathbb{N}}_{\ge 11}$. Then n has a fit-rec.

Proof. Let n be a k-digit element of ${\mathbb{N}}_{\ge 11}$. Write $n=d_{k-1}\dots d_0$ where, for $j=0,\dots ,k-1$, $d_j$ is the digit (i.e., element of ${\mathbb{D}}_{10}$) in the ${10}^j$th place in the integral representation of n. Clearly $d_{k-1}\ge 1$.

Case 1: $d_0=0$. Let s denote the positive integer ${\displaystyle \frac{n}{10}}$. Let $EX{P}_1$ denote the expression ${\displaystyle \frac{d_{k-1}}{d_{k-1}}}+{\displaystyle \frac{d_{k-1}}{d_{k-1}}}+{\displaystyle \frac{d_{k-1}}{d_{k-1}}}+{\displaystyle \frac{d_{k-1}}{d_{k-1}}}+{\displaystyle \frac{d_{k-1}}{d_{k-1}}}+{\displaystyle \frac{d_{k-1}}{d_{k-1}}}+{\displaystyle \frac{d_{k-1}}{d_{k-1}}}+{\displaystyle \frac{d_{k-1}}{d_{k-1}}}+{\displaystyle \frac{d_{k-1}}{d_{k-1}}}+{\displaystyle \frac{d_{k-1}}{d_{k-1}}}+0$, subject to the following condition: if $1\notin \mathcal{F}\left(n\right)$ then the ratio ${\displaystyle \frac{n_{k-1}}{n_{k-1}}}$ is not replaced by its value 1, and if $1\in \mathcal{F}\left(n\right)$ then the ratio ${\displaystyle \frac{n_{k-1}}{n_{k-1}}}$ may be replaced by 1.

Then the digits $d_1$ through $d_{k-1}$ appear in s, and $d_0$ (which is 0) appears in $EX{P}_1$, from which $n=\left(EX{P}_1\right)s$ is a fit-rec of n.

NOTE 4.1.${\displaystyle \frac{d_{k-1}}{d_{k-1}}}$ is presented explicitly ten times as a quotient in $EX{P}_1$. Obviously the consequence of executing $EX{P}_1$ is 10. In effect, then, s is multiplied by 10, yielding n.

[For instance, if $n=2330$ then $s=233$, $d_{k-1}=d_3=2$, so that ${\displaystyle \frac{d_{k-1}}{d_{k-1}}}={\displaystyle \frac{2}{2}}$, from which we have the fit-rec $2330=233\left({\displaystyle \frac{2}{2}}+{\displaystyle \frac{2}{2}}+{\displaystyle \frac{2}{2}}+{\displaystyle \frac{2}{2}}+{\displaystyle \frac{2}{2}}+{\displaystyle \frac{2}{2}}+{\displaystyle \frac{2}{2}}+{\displaystyle \frac{2}{2}}+{\displaystyle \frac{2}{2}}+{\displaystyle \frac{2}{2}}+0\right)$ of n. Note also that ${\displaystyle \frac{2}{2}}$ is never replaced by 1 in $EX{P}_1$ since $1\notin \mathcal{F}\left(n\right)$. ]

Case 2: $d_0\ne 0$ but some other facial digit of n is 0. In this case n has at least three digits since $d_{k-1}\ne 0$. Let $s=n-d_0$. Then $n=s+d_0$ is a fit-rec of n.

Case 3: $d_0\ne 0$ and 0 is not a facial digit of n. Let $s=n-d_0$ and $t={\displaystyle \frac{s}{10}}$.

Let $EX{P}_2$ denote the expression $\left({\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}\right)$.

Then $n=t\left({\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}+{\displaystyle \frac{d_0}{d_0}}\right)+d_0$, giving the desired conclusion . Note that the quotient ${\displaystyle \frac{d_0}{d_0}}$ is presented explicitly exactly ten times in $EX{P}_2$ but this ratio may be replaced by 1 only if $1\in \mathcal{F}\left(n\right)$, a situation similar to that seen in Case 1. •

Proposition 4.2.Let $n\in {\mathbb{N}}_{\ge 11}$ and $k\in \mathbb{N}$. Then there exists a fit-rec ${\mathcal{S}}_n$ of n such that $\mathsf{F}\left({\mathcal{S}}_n\right)\ge k$.

Proof.Case 1: There is no fit-rec $n={\mathcal{S}}_n$ of n such that $\mathsf{F}\left({\mathcal{S}}_n\right)<k$. In this case, $\mathsf{F}\left({\mathcal{S}}_n\right)\ge k$ for any fit-rec $n={\mathcal{S}}_n$ of n.

Case 2: There exists a fit-rec $n={\mathcal{L}}_n$ of n such that $\mathsf{F}\left({\mathcal{L}}_n\right)<k$. Let $\mathsf{F}\left({\mathcal{L}}_n\right)=d<k$. Let q be the quotient and r the remainder upon dividing k by d - written $k=dq+r$ whre $0\le r<d$.

Sub-case 2.1. q is even. Write $q=2s$. Consider the expression $\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)+\dots +\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)$ (s times), without ever replacing $\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)$ by 0 - meaning the expression $\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)$ appears exactly s times (which also means that ${\mathcal{L}}_n$ appears exactly $2s$ times) in this repeated addition of $\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)$. Let this expression of repeated addition be denoted by $\mathcal{T}$. Clearly $\mathsf{F}\left(\mathcal{T}\right)=2sd$.

Note that $\mathcal{T}$ is an expression - we insist that $\mathcal{T}$ is not the value that results from executing the expression $\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)+\dots +\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)$ (s times).

Clearly, the result of executing $\mathcal{T}$ is 0, from which $n=\mathcal{T}+{\mathcal{L}}_n$ is a fit-rec of n. Let ${\mathcal{S}}_n$ denote the expression $\mathcal{T}+{\mathcal{L}}_n$. Then n is (obviously) the result of executing ${\mathcal{S}}_n$. Further, $\mathsf{F}\left({\mathcal{S}}_n\right)=2sd+d>d\left(2s\right)+r=dq+r=k$, completeing this case.

Sub-case 2.2. q is odd. Write $q=2s+1$. Consider the expression $\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)+\dots +\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)$ ($s+1$ times), without ever replacing $\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)$ by 0 - meaning the expression $\left({\mathcal{L}}_n-{\mathcal{L}}_n\right)$ appears exactly $s+1$ times (which also means that ${\mathcal{L}}_n$ appears exactly $2s+2$ times) in this repeated addition. Let this expression (of repeated addition) be denoted by $\mathcal{T}$.

Then $n=\mathcal{T}+{\mathcal{L}}_n$ is a fit-rec of n. Let ${\mathcal{S}}_n$ denote the expression $\mathcal{T}+{\mathcal{L}}_n$. Then $\mathsf{F}\left(\mathcal{T}\right)=d\left(2s+2\right)$, from which $\mathsf{F}\left({\mathcal{S}}_n\right)=d\left(2s+2\right)+d=d\left(2s+1\right)+2d>d\left(2s+1\right)+r=dq+r=k$, completeing this case and the proof. •

5. A Computation Problem in Fit Recasting and Its Decision Variant

A variant of a problem $\mathcal{Q}$ is a formulation of $\mathcal{Q}$ that seeks a desired type of answer to a question accompanying a given instance of $\mathcal{Q}$. Types of variants that are widely studied and used are: optimization, computation and decision.

An optimization variant of $\mathcal{Q}$ is a formulation of $\mathcal{Q}$ that asks for a solution of an optimum measure (which is either the maximum or the minimum of the concerned measure) to each instance of $\mathcal{Q}$ [1].

A computation variant of $\mathcal{Q}$ is a formulation that seeks to compute a stipulated type of solution to each instance of $\mathcal{Q}$. The stipulations must comprise finitely many conditions.

A decision variant of $\mathcal{Q}$ is a formulation of $\mathcal{Q}$ in which each instance is accompanied by a decision question (about the instance) that asks for either a “yes” or a “no” response. The basic ingredients [1] of a decision variant are: the set of instances, the set of attempted solutions (i.e., certificate candidates [4,5]) and the predicate that decides whether an input certificate candidate yields a solution.

We shall be concerned only with computation and decision variants of the problem we discuss in Section 5 through Section 7. It is common to formulate a computation problem $\mathcal{Q}$ as a decision problem to find out if $\mathcal{Q}$ is in NP (see [2,3,4,5,9,11]).

5.1. The Computation Problem FITREC$\left(n\right)$

The following is a computation problem (in facial digit recasting) we call FITREC$\left(n\right)$.

Let $n\in {\mathbb{N}}_{\ge 11}$ be the input instance. Find a fit-rec $n={\mathcal{S}}_n$ of n such that $\mathsf{F}\left({\mathcal{S}}_n\right)\ge n$.

NOTE $5.1$. If n is the input instance and r is the input size in the problem FITREC$\left(n\right)$, then $r=\lceil log\left(n\right)\rceil$. Here log denotes base 2 logarithm. Then n is exponential in r (with base 2).

5.2. A Decision Variant of FITREC$\left(n\right)$

The following is a decision variant of the computational problem seen in Subsection 5.1.

Inputs: (i) $n\in {\mathbb{N}}_{\ge 11}$ and (ii) $k\in \mathbb{N}$.

Decision question: Does there exist a fit-rec $n={\mathcal{S}}_n$ of n such that $\mathsf{F}\left({\mathcal{S}}_n\right)\ge k$?

Output: YES (meaning, such a fit-rec of n exists) or NO (meaning, no such fit-rec of n exists), as appropriate.

NOTE $5.2$. There is no need of a certificate candidate since $k\in \mathbb{N}$ ensures the output is YES by dint of Proposition 4.2. As for the inputs, it is mandatory that they be free from any error, as also that they be logically consistent with one another.

6. An Algorithm to Show FITREC$\left(n\right)$ Is in NP

A “step” in an algorithm is a primitive operation executed by the algorithm. In dealing with any algorithm, there is a presupposition that every primitive operation to be executed by the algorithm is unambiguously defined. Since we furnish algorithms only in lines of pseudocodes, we adopt the following point of view for the notion of number of steps (as in [3]): we assume that each execution of the $j^{th}$ line of the pseudocode takes $t_j$ steps (meaning, $t_j$ primitive operations).

Reading a character, updating the value stored in the address of a variable, elementary mathematical and logical operations are some examples of primitive operations in algorithms. Additional information on primitive operations are in [3].

The following algorithm is shown to verify that FITREC$\left(n\right)$ is in NP.

Algorithm FITREC-in-NP

BEGIN

1. Read n

2. if $n\in {\mathbb{N}}_{\ge 11}$

3.

then print “YES, exists a fit-rec $n={\mathcal{S}}_n$ of n such that $\mathsf{F}\left({\mathcal{S}}_n\right)\ge k$, providing $k\in \mathbb{N}$.” and STOP

4.

else print “ERROR. The problem FITREC$\left(n\right)$ requires n to be in ${\mathbb{N}}_{\ge 11}$.” and STOP

5. endif

STOP

Proposition 6.1.The algorithm FITREC-in-NPis feasible and correct.

Proof. Line 1 of the algorithm reads the value of n. This clearly terminates in a finite number of steps.

Line 2 checks if $n\in {\mathbb{N}}_{\ge 11}$. This step also terminates in a finite number of steps.

As for each of line 3 and line 4, printing the respective decisions clearly terminates in a finite number of steps. So does line 5. Note that the decision in line 3 is justified by Proposition 4.2. Consequently, FITREC-in-NP is feasible.

The correctness of the algorithm is a direct consequence of Proposition 4.2. •

Proposition 6.2.Given an input $n\in {\mathbb{N}}_{\ge 11}$, FITREC-in-NPruns in polynomial time in $\lceil log\left(n\right)\rceil$.

Proof. The total number (say, T) of steps executed by the algorithm FITREC-in-NP is the sum of the numbers of steps for all the lines executed. Suppose that one execution of the line j requires $t_j$ steps and that this line is executed exactly $r_j$ times when the algorithm is executed once. Then $t_j{r}_j$ is the number of steps consumed by the line j in one execution of the algorithm.

In one execution of the algorithm, each line is executed once. Hence, for $j=1,\dots ,5$, $t_j{r}_j=t_j$.

The input is $n\in {\mathbb{N}}_{\ge 11}$ and the input size is $\lceil log\left(n\right)\rceil$ (Note 5.1 in Subsection 5.1). The number of steps required for each of line 1 through line 4 is at most $\lceil log\left(n\right)\rceil$. The endif line requires only a constant number of steps in any algorithm [3]. Hence $T<5\lceil log\left(n\right)\rceil$. •

Proposition 6.3.To each instance n of FITREC$\left(n\right)$ there is a certificate that is verified by the algorithm FITREC-in-NPin polynomial time in $\lceil log\left(n\right)\rceil$.

Proof. By dint of Proposition 4.2, Subsection 5.2 and Proposition 6.2, the conclusion holds without the need for any certificate. •

7. FITREC$\left(n\right)$ Is in NP but Not in P

Proposition 7.1.$2^r$(and consequently $2^{r-1}$) cannot be bounded by $a{r}^b$ for any positive real constants a and b as r increases over $\mathbb{R}$.

Proof. Straightforward. •

Proposition 7.2.The problem FITREC$\left(n\right)$ is inNP .

Proof. Consequence of Proposition 6.2 and Proposition 6.3. •

Proposition 7.3.The problem FITREC$\left(n\right)$ is not inP .

Proof. Let $n\in {\mathbb{N}}_{\ge 11}$ be the input instance for the computation problem FITREC$\left(n\right)$ (Subsection 5.1) and let $r=\lceil log\left(n\right)\rceil$ be the input size. It is a straiogtforward exercise that $n\ge 2^{r-1}$.

Let ALGFIT$\left(n\right)$ be a given feasible algorithm that outputs a fit-rec $n={\mathcal{S}}_n$ of n such that $\mathsf{F}\left({\mathcal{S}}_n\right)=k\ge n$. Let $\mathcal{F}\left(n\right)=\left\{x_1,\dots ,x_q\right\}$ be the set of the facial digits of n.

To compute $\mathsf{F}\left({\mathcal{S}}_n\right)$, the algorithm ALGFIT$\left(n\right)$ has to read each facial digit $x_j$ (of n) is $\mathsf{f}\left(x_j,{\mathcal{S}}_n\right)$ times, where $\mathsf{f}\left(x_j,{\mathcal{S}}_n\right)$ is the frequency of the digit $x_j$ in ${\mathcal{S}}_n$. Then each such reading is a step executed by the algorithm. Note that all these (finitely many) readings of the frequencies have to be done by the algorithm in addition to computing the desired fit-rec $n={\mathcal{S}}_n$.

Clearly, each of these readings (of the facial digits) takes at least one step, as a result of which ALGFIT$\left(n\right)$ takes at least ${\Sigma }_{j=1}^q\mathsf{f}\left(x_j,{\mathcal{S}}_n\right)$ steps to run once fully.

But then ${\Sigma }_{j=1}^q\mathsf{f}\left(x_j,{\mathcal{S}}_n\right)=\mathsf{F}\left({\mathcal{S}}_n\right)=k\ge n\ge 2^{r-1}$. By Proposition 7.1, it follows that ALGFIT$\left(n\right)$ cannot run in polynomial time (in r). •

8. Conclusions

P ≠ NP, follows from Proposition 7.2 and Proposition 7.3.