Research Article | Open Access

Yi-Chou Chen, Hang-Chin Lai, "A Non-NP-Complete Algorithm for a Quasi-Fixed Polynomial Problem", *Abstract and Applied Analysis*, vol. 2013, Article ID 893045, 10 pages, 2013. https://doi.org/10.1155/2013/893045

# A Non-NP-Complete Algorithm for a Quasi-Fixed Polynomial Problem

**Academic Editor:**Jen-Chih Yao

#### Abstract

Let be a real-valued polynomial function of the form , with degree of in An irreducible real-valued polynomial function and a nonnegative integer are given to find a polynomial function satisfying the following expression: for some constant . The constant is dependent on the solution , namely, a quasi-fixed (polynomial) solution of the polynomial-like equation . In this paper, we will provide a non-NP-complete algorithm to solve all quasi-fixed solutions if the equation has only a finite number of quasi-fixed solutions.

#### 1. Introduction and Preliminaries

Lenstra [1] found that a polynomial function (where is an algebraic number) can be solved to a polynomial function such that . It can thus be derived to find a polynomial such that there exists an as a fixed point of the polynomial function ; that is, has a polynomial solution .

Furthermore, Tung [2, 3] extended (1) to solve for the following equation: where is a real constant depending on the polynomial solution and the given positive integer .

Recently, Lai and Chen [4] extended the expression (2) to solve to satisfy the polynomial equation as the following form: where is an irreducible polynomial in and the polynomial functions are written by

*Remark 1. *Recall that a polynomial function satisfying (3) is called a quasi-fixed (polynomial) solution corresponding to the real value . This number is called a quasi-fixed (polynomial) value corresponding to the polynomial solutions .

Note that the equation may not be solvable; if it is solvable, the number of all solutions may be infinitely many, or finitely many quasi-fixed solutions If this equation has infinitely many quasi-fixed solutions, by [4, Theorem 3.5], we have(1) must be for some , ,(2)those solutions must assume a fixed form like

Moreover, if this equation has only finitely many quasi-fixed solutions, by [4, Corollary 3.3], the bound of all solutions is at most .

In this paper, we deal with the case of finitely many solutions and provide a non-NP-complete algorithm to obtain all representations of quasi-fixed solutions.

*Remark 2. *An algorithm is called “NP-complete” if the algorithm is computed in an exponential time of the input size “”; that is, the computing time in the algorithm does not exceed . Otherwise, the algorithm is called “non-NP-complete”; that is, the algorithm can be computed in a polynomial time not exceeding for some fixed real number .

The remainder of the paper is organized as follows: Section 2 introduces an example for the practical algorithm. Section 3 describes the main algorithm and its processes. Section 4 proves that the main algorithm is indeed non-NP-complete. We have given two example problems in Section 5.

#### 2. Relative Algorithm and Some Examples

Let and an irreducible polynomial function . To complete our algorithm, we need an existing algorithm. In [2], Tung has established the following algorithm.

*Algorithm 3 (Tung [5]). *
Consider the following technique.*Input.* Given is a polynomial .*Output*. All solutions satisfy
in polynomial time of , where is the computer memory of all coefficients of and . Moreover, the number of all solutions is at most .

We provide a simple example to explain our aim for solving the quasi-fixed polynomial solutions related to Algorithm 5 (in Section 3) as follows.

*Example 4. *Let
and . Then . Solve all quasi-fixed solutions of

By (7) and (8), we obtain
divides both sides by , then it becomes

It follows that
and we have
with an indeterminate . To determine , we substitute this (in (12)) into , then (8) becomes

This yields

Both sides of (14) divide by to become

According to Algorithm 3, we obtain and , so the solutions (12) of problem are

In Section 3, we provide a non-NP-complete algorithm to satisfy (3).

#### 3. Computing Procedure

In this paper, we may assume that the number of all solutions of (3) is finite and then constitute an algorithm of the approximate solutions for the quasi-fixed polynomial equation (3) as follows.

*Algorithm 5. *
Consider the following technique.*Input*. Given a polynomial , irreducible and .*Output*. All solutions satisfy
by a non-NP-complete algorithm.

The following definitions are given by [6, page 421].

*Definition 6. *(i) The content of in (4), denoted by , is defined by the greatest common divisor (g.c.d):

(ii) The primitive part of is , and is primitive if .

From (i) and (ii), we have

To complete the following algorithm, we need the following elementary property.

Lemma 7. *Let and be an irreducible polynomial in . Consider the module equation
**
The number of all solutions is thus at most . *

*Assumption*. Throughout this algorithm, for any , one can solve all solutions of the equation

The procedure of our main algorithm is described below.

*Procedures of Main Algorithm (Algorithm 5)*

*Step 0.* If and , then let and move to (Step , ).

*Step 1.* For convenience, we let
where is a primitive polynomial.

*Step 1.1.* If , we would have the problem
to deduce that . But , then . Consequently, (23) becomes

We can then solve all solutions for to get a solution set

*Step 1.2.* If and , then with . In this case, (23) becomes

*Step 1.2.1.* In case , then (26) becomes which can be solved by Algorithm 3 to obtain all solutions for the equation to get and obtain a set

*Step 1.2.2.* If , then and we can divide both sides of (26) by ; consequently,

According to Lemma 7, the solution number does not exceed , thus we may assume that is a solution of (28) with ; please note that the choice of may be larger than and we may define a solution set which collects such by setting as the following form:

Consequently, the expression (23) can go to Step .

*Step 2.* For each in (Step , ()), we can replace by in (23), and then (23) becomes

We let . Thus it follows from Definition 6(ii) that there exists which is a primitive polynomial function so that

Equation (30) then becomes

Next, we will prove that .

*Step 2.1.* If , (32) and (33) can be extrapolated as to . Since , then . Consequently, (33) becomes

We can then solve all solutions for to get a solution set

*Step 2.2.* If , then with . In this case, (33) becomes

*Step 2.2.1.* In case , (36) becomes
which can be solved by Algorithm 3 to get and obtain a set

*Step 2.2.2.* If , then and we can divide both sides of (36) by ; consequently,

By the same reasoning used in [Step ], we can solve with and obtain the set

We let

Consequently, expression (33) can go to Step .

*Step 3.* For each , we can replace by in (23). It then becomes

Moreover, we let and for a primitive polynomial , so that (42) becomes

Next, we will prove that .

*Step 3.1.* If , and (43) and (44) are extrapolated, we have

Since , we have .

Thus (44) becomes

Hence, we solve all solutions such that to obtain a solution set

*Step 3.2.* If , then for some positive integer and (44) becomes

*Step 3.2.1.* If , then (48) becomes
and, by Algorithm 3, we compute and collect
in for all . Hence, we output

*Step 3.2.2.* If , we use to divide both sides of (48) to obtain

By the same reasoning used in (Step ), we can solve with and obtain the set

We let thus expression (44) can go to Step .

Continuing this process, we get the -th step and a sequence and go to the next-th step.

*Step **.* For each , we can replace by in (23), and then (23) becomes

We let and for a primitive polynomial , then (55) becomes

Note that in Section 4, Lemma 9, we will prove that .

*Step k.1.* If , by (56) and (57), we have
since , we have .

Thus (57) becomes

Hence, we solve all solutions such that to obtain a solution set

*Step k.2.* If , then for some positive integer and (57) becomes

*Step k.2.1.* If , then (61) becomes

By Algorithm 3, we compute and obtain in , for all . Hence, we output

*Step k.2.2.* If , we use to divide both sides of (61) to obtain

By the same reasoning used in (Step ), we can solve with and obtain the set

We let thus expression (57) can go to the next step.

As explained in Remark 8, this algorithm is not an infinite loop.

*Remark 8. *(i) If , then and if , then for . This means that if , then the work finishes at step for .

(ii) This algorithm is not an infinite loop. Since , and by Corollary 13, we have for , thus this algorithm will terminate at the-th Step. We can thus say that this algorithm is a finite loop algorithm.

(iii) We may assume that the cardinal number , for ; however, if , then Algorithm 5 has an infinite number of solutions, which contradicts our assumption.

Remark 8(ii) shows that the maximum number of steps in this algorithm is . Moreover, if the-th Step happens, then we obtain by Corollary 11 and let

Thus, the possible solutions of are those solutions satisfying

Moreover, by Algorithm 3, we can get a solution set such that the bound of all solutions is “.” So the solution set of (3) is contained in

Finally, we check each element to determine whether or not

The arithmetic computing time to solve each element in only requires non-NP-complete computing time and to check that each satisfies

Non-NP-complete computing time is also needed given the cardinal number of : where , , and denote the cardinal number of , , and . The remaining work “Checks the bound of , for each .” In Section 4, Theorem 16, we prove that for .

By (74) and (75), we have the cardinal number

That is, we have to check at most “” solutions to satisfy Checking each solution only takes non-NP-complete time, thus this algorithm needs non-NP-complete time. That is, Algorithm 5 is indeed a non-NP-complete time algorithm. Note that, for each-th Step, whether belongs to or , the upper bound of all elements in may not be as large; in fact,

In the next Section 4, we will complete all necessary properties for this algorithm.

#### 4. Main Theorems

For convenience, we describe some interesting properties of the above algorithm.

Lemma 9. *Let , , , and be a parameter, if , then . *

* Proof. *Let . Then

Thus, .

The definitions of , , , and for in Algorithm 5 are for a primitive polynomial and for .

Next, we prove some properties about for .

Lemma 10. *Consider , for . *

* Proof. *From (80), we have

By definition of , we have . From Lemma 9, we obtain . That is,
and, by (81), we obtain .

By Lemma 9, , and the definition of , it follows that . Moreover, by Lemma 10, we can easily obtain the following corollaries.

Corollary 11. *Consider, for . *

To complete the main theorem, we give the following definitions.

*Definition 12. *(i) The function is said to have -power , denoted by , if and .

(ii) We say that has -power , if has -power .

The following corollary is an immediate result of Lemma 10 and Definition 12.

Corollary 13. *Consider , for . *

*Proof . *Since and by Lemma 10, we have

It follows that
so ; hence, .

Next, we obtain some properties of -power as follows.

Lemma 14. *Let , , and is defined in (4). Then *(i)*, *(ii)*, and *(iii)*, *(iv)*If and , then . *

For each , we rewrite defined in the procedures of Algorithm 5 as the following definitions.

*Definition 15. *Let , an irreducible polynomial , and be a parameter. We denote the sets , defined in Algorithm 5 as follows: (1) with , (2) satisfies and where for , and the and properties are given by the following:(*D1*): and ,(*D2*):.

From the above definitions, we obtain a bound of the cardinal number .

Theorem 16. *For each , then ( means the cardinal number). *

* Proof . *This theorem will be proven by induction on .

(i) When , claim that for any .

For this purpose, we consider

If , and let and for some coprime polynomials , .

The only possible solution with and is

That is, .

Next, we assume that , thus we deduce that .

Let
and for some , . Consider
and since by assumption, is uniquely decided, as is . The remaining task is to prove that is also uniquely determined. As with , satisfies
such that
then
it follows that must be

Consequently, .

By (1) and (2), we get that for any .

(ii) We will prove this theorem by mathematical induction. Assume that
for any with , we want to show
for any with .

Let and choose with

Indeed,

By the division rule, let be divided by , giving us
for some and .

We want to show that