I studied a definition for time complexity: Let $M$ be a deterministic Turing Machine. The running time of $M$ is said to be:

for a function $t: \mathbb{N} \to \mathbb{N}$ ($\mathbb{N}$ is natural number) and for all $x$ belonging to $\Sigma^*$, $M$ halts in at most $O(t(|x|)$ steps.

My question is that $x$ belonging to $\Sigma^*$ means $x$ could very well be $\varepsilon$ (empty string) for which $|x| = 0$. How is $t(0)$ computed as 0 is not a natural number.

  • 2
    $\begingroup$ en.wikipedia.org/wiki/Natural_number $\endgroup$
    – kaya3
    Commented May 22 at 17:50
  • $\begingroup$ Wikipedia/Is 0 a natural number? That's a fun discussion!! Thanks for the laughs $\endgroup$
    – Stef
    Commented May 22 at 18:36
  • $\begingroup$ Is that word for word the definition in the book? It's weird. "For all x, M halts in at most O(t(|x|)) steps"? If x is already quantified by "for all", then t(|x|) is a constant and it doesn't make sense to talk about big-O... $\endgroup$
    – Stef
    Commented May 22 at 18:39
  • $\begingroup$ @Stef yes, and the same has been mentioned for s(|x|) where 's' is the space complexity function. I think O(t|x|) point is irrelevant in discussion as the entire point of using big-O is to ignore certain stuff like multiplication by constants and lower order terms. What's really bothering me is that I am not able to understand what part of this definition feels weird, is it 0 being a natural number (or not) or is it x belonging to Σ*. $\endgroup$ Commented May 23 at 6:26
  • $\begingroup$ @OM_anand Whether or not |x|=0 is allowed by the definition really couldn't matter less. The only thing important in the definition is what happens in the limit. You could add any arbitrary constraint like |x| > 0 or |x| > 1 or |x| > 42 or |x| > 137 in that definition and the resulting definitions would all be equivalent. What is important in that definition is what happens when |x| grows, not when |x| is small. But the definition is a bit sloppily written. $\endgroup$
    – Stef
    Commented May 23 at 13:22

1 Answer 1


$\mathbb{N}$ can be defined as the set of non-negative integers 0, 1, 2, ...

You should check the definition in the book you're reading.

However, there is no contradiction regardless of how $\mathbb{N}$ is defined. It might just be a tiny mistake that I'm sure you're able to fix.

  • $\begingroup$ Thanks for the quick reply. It was mentioned that ℕ here is a "natural number". So I had begun to think that instead of x belonging to Σ* (U Σ^i; i>=0), they would have probably meant x belonging to Σ+(U Σ^i; i>0). Not sure if that fix makes sense. $\endgroup$ Commented May 22 at 9:25

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