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Let the binary relation computed by a nondeterministic transducer be the relation between input strings and the possible output strings the transducer can produce (and accept) for the given input string.

It is easy to see that the nondeterministic transducers using polynomial time can compute different binary relations than the nondeterministic transducers using logarithmic space. (Trivial examples like $\{(1^n,ww^R):n\in\mathbb N, |w|\leq n\}$ suffice to see this. Edit: This assumes that the output is generated symbol by symbol, see András Salamon's comment. If random access to write only output memory is assumed, then $\{(1^n,w):n\in\mathbb N, w\in D_2, |w|\leq 2n\}$ where $D_2$ is the Dyck language of all strings of matching brackets over two pairs of brackets, say $\{ [,], (,) \}$ is an example. However, proving that this example really works is non-trivial.)

Can anything of interest be concluded from this basic observation? For example, concluding that there must exist an oracle $A$ such that $NL^A \neq NP^A$ would be of interest, but things don't work that way. Or concluding that directly programming in NP is easier that directly programming in NL, because more powerful subroutines are available. That sort of sounds true, even so subroutines generally can have more that a single input string and produce more than a single output string. (So one could try to prove that if all those possible subroutines would be equal, then there could not exist an oracle for which the relativized decision problems are different. But even this will probably fail.)

Edit:

  • WLOG, we can restrict us to NL tranducers that never repeat a configuration. These are exactly those NL transducers that only use polynomial time.
  • The random access to write only output memory is well outside NL. But it is still inside NP, so the composition of those NL transducers could be done inside NP. This should be good enough to conclude things about powerful subroutines or oracles, if this is possible at all. (Even the symbol by symbol output tape was outside of NL, it was just less obvious.)
  • The oracles $A$ such that $NL^A \neq NP^A$ are white box oracles, where the problem $A$ is (expected to be) outside of the considered complexity classes. The conclusions about subroutines from the basic observation might tell us something about black box oracles from the same or weaker complexity classes. Or they might at least provide a motivation to prove black box separations between complexity classes relative to oracles from weaker complexity classes.
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Thomas Klimpel
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We know that $NL \subseteq NP$, but that $NL \subsetneq NP$ is not known. So one cannot outright say that $NL$ and $NP$ are different.

We only know for sure $NL \subsetneq PSPACE \subsetneq EXPSPACE$ by space hierarchy theorem. And $P \subsetneq EXPTIME$,and $NP \subseteq NEXPTIME$ by time hierarchy theorem.

We only know for sure is $$L\subseteq NL \subseteq P \subseteq NP \subseteq PH \subseteq PSPACE =NPSAPCE \subseteq EXPTIME \subseteq NEXPTIME\subseteq EXPSPACE$$

Only above four inclusions are known to be strict. All the other inclusions are unknown and open problems. (Though most but not all have unproven belief that each of them is strict).

See https://en.wikipedia.org/wiki/PSPACE#Relation_among_other_classes.

Also, I think, we might be able to prove that there exists a language $B$ for which $NL^B \neq NP^B$. Probably $U_B$ and $B$ of Baker, Gill and Soloway will work here too. And we are also able to show that $NL^A = NP^A$ for $A = TQBF$, the $PSPACE$ complete problem to decide the true quantified Boolean formulas.

Sarvottamananda
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