0
$\begingroup$

per this post

$t = x^2$ means the problem is solvable in "Polynomial" time.

per this post

in the form

$$a_{n}x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{2}x^{2}+a_{1}x+a_{0} { > \boldsymbol{=0}}$$

plug in $n=100000000000$ with $a_n=1$ and $a_k=0$ for all $k \neq n$

then we see that $x^{100000000000}$ is a polynomial

it doesn't matter if its 2, 3, or a billion, as long as the n is a finite number

in this context, does $t = x^{100000000000}$ still mean the problem is solvable in "Polynomial" time?

$\endgroup$
  • 2
    $\begingroup$ I suggest you read up on what a polynomial is. $\endgroup$ – dkaeae Jun 7 '19 at 7:00
  • $\begingroup$ I'm voting to close this question as it is not about CS but rather about basic math terminology. $\endgroup$ – dkaeae Jun 7 '19 at 7:01
2
$\begingroup$

Yes, $x^{100000000000}$ is a polynomial. Your first quote gives an example; your second is the definition.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.