Skip to main content
Title
Link
David Richerby
  • 82.2k
  • 26
  • 144
  • 238

Bounds in NP Completeness proof-completeness proofs

Fixed math formatting and explicit poly-time statement.
Source Link

According to the polynomial- reductiontime-reduction definition

"If problem Y$Y$ can be reduced to problem X$X$ in polynomial time, we denote this by Y ≤(P)X $Y \leq_p X$."

If X$X$ is one of already known NP-complete problem then we can say that Y$Y$ is NP-complete.

From my understanding, P$p$ should be always polynomial and this cannot be an exponential function such as 2^n$2^n$ or 3^n$3^n$.

However, my question is if Y$Y$ or X$X$ already has a lower bound that is not polynomial, which is 2^n$2^n$ or 3^n$3^n$ then can I still say that Y$Y$ is NP-complete?

In other words, does lower bounds for X$X$ and Y$Y$ matter?

According to the polynomial- reduction definition

"If problem Y can be reduced to problem X, we denote this by Y ≤(P)X."

If X is one of already known NP-complete problem then we can say that Y is NP-complete.

From my understanding, P should be always polynomial and this cannot be an exponential function such as 2^n or 3^n.

However, my question is if Y or X already has a lower bound that is not polynomial, which is 2^n or 3^n then can I still say that Y is NP-complete?

In other words, does lower bounds for X and Y matter?

According to the polynomial-time-reduction definition

"If problem $Y$ can be reduced to problem $X$ in polynomial time, we denote this by $Y \leq_p X$."

If $X$ is one of already known NP-complete problem then we can say that $Y$ is NP-complete.

From my understanding, $p$ should be always polynomial and this cannot be an exponential function such as $2^n$ or $3^n$.

However, my question is if $Y$ or $X$ already has a lower bound that is not polynomial, which is $2^n$ or $3^n$ then can I still say that $Y$ is NP-complete?

In other words, does lower bounds for $X$ and $Y$ matter?

Post Migrated Here from stackoverflow.com (revisions)
Source Link
NoSleep
NoSleep

NP Completeness proof

According to the polynomial- reduction definition

"If problem Y can be reduced to problem X, we denote this by Y ≤(P)X."

If X is one of already known NP-complete problem then we can say that Y is NP-complete.

From my understanding, P should be always polynomial and this cannot be an exponential function such as 2^n or 3^n.

However, my question is if Y or X already has a lower bound that is not polynomial, which is 2^n or 3^n then can I still say that Y is NP-complete?

In other words, does lower bounds for X and Y matter?