PTAS vs. FPTAS input

Question

I am trying to understand what is the PTAS, FPTAS and what is the difference between them.

I found this analysis:

PTAS definition vs. FPTAS

but I cannot understand what do we mean by saying:

"....time complexity is polynomial in the input size and also polynomial in 1/ϵ" ?? Ok n has to do with the linear O(n) worst case that the algorithm runs. But O(1/ε) how does it work?

Answer 1

For example, the running time $O(2^{1/\varepsilon} \cdot n^2)$ corresponds to a PTAS, since the running time depends polynomialy on $n$, but not on $1/\varepsilon$ since it is in the exponent. The running time $O(1/\varepsilon \cdot n \log{n})$ depends polynomialy on $n$ and $1/\varepsilon$, so it corresponds to a FPTAS.

NB, you also need to show that your algorithm has an approximation ratio of $(1 + \varepsilon)$ (or -) to show the algorithm is in fact a (F)PTAS.