Comp Sci Why is NTM Time Complexity Guessing Polynomial?

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The discussion clarifies that the time complexity of a non-deterministic Turing machine (NTM) is polynomial due to its ability to explore multiple configurations simultaneously without overhead. In the context of the HAMPATH problem, although there are O(m^m) configurations to check, the NTM can effectively "choose" the correct configuration that leads to acceptance in polynomial time. This is because NTMs operate on the principle of non-determinism, allowing them to branch into all possible configurations at once. Thus, the perceived exponential growth in configurations does not translate to exponential time complexity for NTMs. The key takeaway is that NTMs can leverage their non-deterministic nature to achieve polynomial time complexity in certain decision problems.
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Homework Statement
Not homework but something I don't understand: Why is the time-complexity of ## N_1 ## below not exponential due to the overhead of going over all possible configurations?
Relevant Equations
NTM = nondeterministic Turing machine

Running time of a nondeterministic Turing machine that is a decider is the depth of its configuration tree ( the length of the longest branch in the tree ).
1688116665509.png

( Source: Introduction to the theory of computation, Michael Sipser, 3rd edition )

I know the computation of a non-deterministic Turing machine ( NTM ) which is a decider is defined to be the depth of its configuration tree.

In the above example of showing that ## HAMPATH \in NP ##, where it's highlighted in red - I understand that choosing a specific list of nodes costs polynomial time which is the depth of the configuration tree. However, we seem to be choosing every possible option for a list of ## m ## vertices, thus, we have at least ## O(m^m) ## configurations to check, so why is the running time of ## N_1 ## polynomial? shouldn't it be exponential?
 
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CGandC said:
Why is the time-complexity of ## N_1 ## below not[corrected] exponential due to the overhead of going over all possible configurations?
Because it is an NTM: there is no overhead in branching to any finite number of configurations. Alternatively we can imagine that it can "luckily" choose any configuration that leads to acceptance.
 
I understand now, thanks.
 
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