Question

Turing machines, closure properties, decidability Let P(z) = coz" + ca"-! + . . . +4-12+ cn be a polynomial with a root at r = zo. Let Cmax be the largest absolute value of a . Show that 1. (a) rol S col (b) What can you conclude from (a) about the problem of deciding whether a given poly- (c) What can you conclude from (a) about the problem of deciding whether a given poly- (d) Apply the knowledge you gained in the case of the polynomial P(r- nomial with integer coefficients admits an integer root? nomial with integer coefficients admits a negative integer root? 001 18r-23. Give the resulting algorithm

Answer 1

(a) If $x_0=0$ then the inequality is immediate. Thus, we may assume that $x_0\neq 0$ ; since $x_0$ is an integer, this implies $|x_0|\geq 1$ .

Since $P(x_0)=0$ we get

$\begin{align*}0&=c_0x_0^n+c_1x_0^{n-1}+\cdots+c_{n-1}x_0+c_n\\ \Rightarrow\hspace{1.5cm}c_0x_0^n&=-c_1x_0^{n-1}-\cdots-c_{n-1}x_0-c_n\\ \Rightarrow\hspace{1.55cm}c_0x_0&=-c_1-{\frac{c_2}{x_0}}\cdots-{\frac{c_{n-1}}{x_0^{n-2}}}-{\frac{c_n}{x_0^{n-1}}}\end{align*}$

Therefore,

$\begin{align*}|c_0x_0|&=\begin{vmatrix}-c_1-{\frac{c_2}{x_0}}\cdots-{\frac{c_{n-1}}{x_0^{n-2}}}-{\frac{c_n}{x_0^{n-1}}}\end{vmatrix}\\ &\leq \begin{vmatrix}-c_1\end{vmatrix}+\begin{vmatrix}-{\frac{c_2}{x_0}}\end{vmatrix}+\cdots+\begin{vmatrix}-{\frac{c_{n-1}}{x_0^{n-2}}}\end{vmatrix}+\begin{vmatrix}-{\frac{c_n}{x_0^{n-1}}}\end{vmatrix}\\ &=\begin{vmatrix}c_1\end{vmatrix}+\begin{vmatrix}{\frac{c_2}{x_0}}\end{vmatrix}+\cdots+\begin{vmatrix}{\frac{c_{n-1}}{x_0^{n-2}}}\end{vmatrix}+\begin{vmatrix}{\frac{c_n}{x_0^{n-1}}}\end{vmatrix}\\ &\leq \begin{vmatrix}c_1\end{vmatrix}+\begin{vmatrix}c_2\end{vmatrix}+\cdots+\begin{vmatrix}c_{n-1}\end{vmatrix}+\begin{vmatrix}c_n\end{vmatrix}\\ &\leq n|c_{\max}|\end{align*}$

which proves

$\begin{align*}|x_0|&\leq {\frac{n|c_{\max}|}{|c_0|}}\end{align*}$

b) By part a), to decide whether an integer polynomial $\begin{align*}P(x)\end{align*}$ of degree $\begin{align*}n\end{align*}$ admits an integer root we only need to evaluate the polynomial at at most $\begin{align*}2n|c_{\max}|+1\end{align*}$ integers in $\begin{align*}-n|c_{\max}|,\cdots,n|c_{\max}|\end{align*}$ . Evaluating an integer polynomial at an integer point can be done in polynomial (in coefficients and degree) time. Therefore, the problem is polynomial time decidable.

c) By part a), to decide whether an integer polynomial $\begin{align*}P(x)\end{align*}$ of degree $\begin{align*}n\end{align*}$ admits a negative integer root we only need to evaluate the polynomial at at most $\begin{align*}n|c_{\max}|\end{align*}$ integers in $\begin{align*}-n|c_{\max}|,\cdots,-1\end{align*}$ . Evaluating an integer polynomial at an integer point can be done in polynomial (in coefficients and degree) time. Therefore, the problem is polynomial time decidable.

d) For this polynomial, $\begin{align*}|c_{\max}|=500\end{align*}$ and $\begin{align*}n=65\end{align*}$ , and $\begin{align*}c_0=7\end{align*}$ . Thus, by part a), if this polynomial has an integer root then it has an integer root $\begin{align*}x_0\end{align*}$ satisfying

$\begin{align*}|x_0|\leq {\frac {65\cdot 500}{7}}\leq 4642.9\end{align*}$

Therefore, the following algorithm decides whether the given polynomial admits an integer root.

Initialize $\begin{align*}x_0^+=0,~x_0^-=0\end{align*}$ .

If $\begin{align*}x_0^+,-x_0^-<4642\end{align*}$

---- evaluate $\begin{align*}P(x_0^+),~P(x_0^-)\end{align*}$

   ---- if $\begin{align*}P(x_0^+)=0\end{align*}$ then stop and return polynomial has +ve integer root $\begin{align*}x_0^+\end{align*}$

   ---- else,

   ---- if $\begin{align*}P(x_0^-)=0\end{align*}$ then stop and return polynomial has -ve integer root $\begin{align*}x_0^-\end{align*}$

---- else, $\begin{align*}x_0^+\leftarrow 1+x_0^+\end{align*}$ and $\begin{align*}x_0^-\leftarrow -1+x_0^+\end{align*}$ .

Return: polynomial has no integer root.