Artificial intelligent assistant

Understanding this description of NP completeness in TAOCP, Fascicle 6, Satisfiability. ### Context I am reading TAOCP, Vol 4, Fascicle 6, Satisfiability. ![enter image description here]( I am trying to understand what Donald Knuth is saying here. # Question Does $N^{O(1)}$ mean that for a formula of size $N$ it will take $N^{m}$ steps where $m$ is some constant like $17$ or $\pi$? For instance if $N=10$ and $m=17$, then an "efficient" algorithm will take $10^{17}$ steps.

Yes, but note that as usual this formula is for asymptotic only, as $N$ go to infinity.

$N^{O(1)}$ technically mean it's bounded above by a function of the form $N^{f(N)}$ such that $f(N)$ is eventually bounded above by a constant multiple of $1$. Parsing all that, it does mean it's eventually bounded by $N^{m}$ for some constant $m$.

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