# NC = P consequences?

The Complexity Zoo points out in the entry on EXP that if L = P then PSPACE = EXP. Since NPSPACE = PSPACE by Savitch, as far as I can tell the underlying padding argument extends to show that $$(\text{NL} = \text{P}) \Rightarrow (\text{PSPACE} = \text{EXP}).$$ We also know that L $\subseteq$ NL $\subseteq$ NC $\subseteq$ P via Ruzzo's resource-bounded alternating hierarchy.

If NC = P, does it follow that PSPACE = EXP?

A different interpretation of the question, in the spirit of Richard Lipton: is it more likely that some problems in P cannot be parallelized, than that no exponential-time procedure requires more than polynomial space?

I would also be interested in other "surprising" consequences of NC = P (the more unlikely the better).

Edit: Ryan's answer leads to a further question: what is the weakest hypothesis that is known to guarantee PSPACE = EXP?

• W. Savitch. Relationships between nondeterministic and deterministic tape complexities, Journal of Computer and System Sciences 4(2):177-192, 1970.
• W. L. Ruzzo. On uniform circuit complexity, Journal of Computer and System Sciences 22(3):365-383, 1971.

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As I'm sure I'm not the only one to not know what NC is, here's a link: en.wikipedia.org/wiki/NC_%28complexity%29 – Emil Sep 24 '10 at 22:24

Yes. $NC$ can be seen as the class of languages recognized by alternating Turing machines that use $O(\log n)$ space and $(\log n)^{O(1)}$ time. (This was first proved by Ruzzo.) $P$ is the class where alternating Turing machines use $O(\log n)$ space but can take up to $n^{O(1)}$ time. For brevity let's call these classes $ATISP[(\log n)^{O(1)},\log n] = NC$ and $ASPACE[O(\log n)] = P$.

Suppose the two classes are equal. Replacing the $n$ with $2^n$ in the above (i.e., applying standard translation lemmas), one obtains

$TIME[2^{O(n)}] = ASPACE[O(n)] = ATISP[n^{O(1)}, n] \subseteq ATIME[n^{O(1)}] = PSPACE$.

If $TIME[2^{O(n)}] \subseteq PSPACE$ then $EXP = PSPACE$ as well, since there are $EXP$-complete languages in $TIME[2^{O(n)}]$.

Edit: Although the above answer is perhaps more educational, here's a simpler argument: $EXP = PSPACE$ already follows from "$P$ is contained in polylog space" and standard translation. Note "$P$ is contained in polylog space" is a much weaker hypothesis than $NC = P$.

More details: Since $NC$ circuit families have depth $(\log n)^c$ for some constant, every such circuit family can be evaluated in $O((\log n)^c)$ space. Hence $NC \subseteq \bigcup_{c > 0} SPACE[(\log n)^c]$. So $P = NC$ implies $P \subseteq \bigcup_{c > 0} SPACE[(\log n)^c]$. Applying translation (replacing $n$ with $2^n$) implies $TIME[2^{O(n)}] \subseteq PSPACE$. The existence of an $EXP$-complete language in $TIME[2^{O(n)}]$ finishes the argument.

Update: Addressing Andreas' additional question, I believe it should be possible to prove something like: $EXP=PSPACE$ iff for all $c$, every polynomially sparse language in $n^{O(\log^c n)}$ time is solvable in polylog space. (Being polynomially sparse means that there are at most $poly(n)$ strings of length $n$ in the language, for all $n$.) If true, the proof would probably go along the lines of Hartmanis, Immerman, and Sewelson's proof that $NE = E$ iff every polynomially sparse language in $NP$ is contained in $P$. (Note, $n^{O(\log^c n)}$ time in polylog space is still enough to imply $PSPACE=EXP$.)

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Thanks for the nice answer. Dexter Kozen's Theory of Computation has a nice "uniform" notation for Ruzzo's classes on page 69: $STA(f,g,h)$ where $f$ bounds space, $g$ bounds time, and $h$ bounds alternations. Then $\text{NC} = STA(\log n, {*}, (\log n)^{O(1)})$ while $\text{P} = STA(\log n, {*}, {*})$ which really highlights the construction. – András Salamon Sep 24 '10 at 18:24
Note that I am saying $NC = STA(\log n, (\log n)^{O(1)}, *)$ in the above. However I think these are the same. A machine which takes polynomial time and $O(\log n)$ space but makes only $(\log n)^{O(1)}$ alternations can be turned into another alternating machine which takes only $(\log n)^{O(1)}$ time and $O(\log n)$ space. (The other direction is obvious.) The idea is to insert more alternations so that each polynomial time existential phase and universal phase is "sped up" to run in only $(\log n)^{O(1)}$ time and $O(\log n)$ space, along the lines of Savitch's theorem. – Ryan Williams Sep 24 '10 at 19:37
what we need is some kind of greasemonkey script that automatically links something like "\NP" to the entry in the zoo. – Suresh Venkat Sep 25 '10 at 18:01

(I've seen Ryan's answer, but I just wanted to provide another perspective, which was too long to fit into a comment.)

In the $L = P \Rightarrow PSPACE = EXP$ proof, all that you need to know about L, informally, is that when blown up by an exponential, L becomes PSPACE. The same proof goes through for NL, because NL blown up by an exponential also becomes PSPACE.

Similarly, when NC is blown up by an exponential, you do get PSPACE. I like to see this in terms of circuits: NC is the class of polynomial size circuits with polylog depth. When blown up, this becomes exponential size circuits with polynomial depth. One can show that this is exactly PSPACE, once the appropriate uniformity conditions are added in. I guess if NC is defined with L-uniformity, then this will get PSPACE-uniformity.

The proof should be easy. In one direction, take a PSPACE-complete problem like TQBF and express the quantifiers using AND and OR gates of exponential size. In the other direction, try traversing the polynomial depth circuit recursively. The stack size will be polynomial, so this can be done in PSPACE.

Finally, I came up with this argument when I saw the question (and before reading Ryan's answer), so there might be bugs. Please point them out.

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One correction: NC has circuits of polynomial size and polylog depth, but this is still only polynomial depth after translation. – Ryan Williams Sep 24 '10 at 19:50
@Ryan: You're right. I'll fix that. – Robin Kothari Sep 24 '10 at 21:10

Here's a little more detail from the perspective of simulating time-space bounded Alternating Turing machine.

Suppose that $P = NC$.

Since $NC = ATISP((\log(n))^{O(1)}, O(\log(n)))$, we get $$P = ATISP((\log(n))^{O(1)}, O(\log(n))).$$

Now, consider the linear time universal simulation problem $LinU$ where we are given an encoding on a Turing machine $M$ and an input string $x$ of length $n$ and we want to know if $M$ accepts $x$ in at most $n$ steps.

We know that $LinU \in P$. Therefore, there exists a constant $c$ (sufficiently large) such that $$(*) \; LinU \in ATISP(\log^c(n), c\log(n)).$$

As a result of a padding argument (a little tricky see comments), we have $$(1) \; DTIME(n) \subseteq ATISP(\log^c(n), c\log(n)).$$

Extending the padding argument, we get $$(2) \; DTIME(n^k) \subseteq ATISP(k^c\log^c(n), kc\log(n)).$$ $$(3) \; DTIME(2^{n^k}) \subseteq ATISP(k^cn^{kc}, kcn^{k}).$$

Further, there are known results about the simulation of Alternating time-space bounded Turing machines. In particular, we know that $$ATISP(\log^c(n), c\log(n)) \subseteq DSPACE(O(\log^{c+1}(n))).$$

Therefore, we (essentially) have the following for all natural numbers $k$:

$$(2^{*}) \; DTIME(n^k) \subseteq DSPACE(k^{c+1}\log^{c+1}(n))$$ $$(3^{*}) \; DTIME(2^{n^k}) \subseteq DSPACE(n^{k(c+1)}).$$

From $(3^{*})$, we would get that $EXP = PSPACE$.

====================After Thought===================

It is important to notice that $P = NC$ implies $$ATISP((\log(n))^{O(1)}, O(\log(n))) = ATISP(\log^c(n), O(\log(n)))$$ for some constant $c$.

Any comments or corrections are welcomed. :)

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The mistake is in the step that you move from LinU to LinTime: you ignore the overhead caused by reductions to LinU from LinTime machines. Just because a universal machine for the class C is in class C' doesn't mean C is a subset of C'. – Kaveh Nov 19 '14 at 5:05
Hi Kaveh, thank you for the response. First let me describe the issue, then I'll try to clear it up. Now, say we are given a $2n$ time bounded TM $M$. Let's reduce $L(M)$ to $LinU$. Let $x$ be given. We can't just check if $(M,x)$ is in $LinU$. We have to modify $M$ slightly to obtain $M'$ and pad $x$ with $n$ characters to obtain $x'$ which has $2n$ characters in total. Then, we check if $(M',x')$ is in $LinU$. For the reduction, we map $x$ to $(M',x')$. We should be able to do this in linear time, but wait we need it to be done in $\log^c(n)$ time. – Michael Wehar Nov 19 '14 at 6:42
Now, I'll propose a way to clear it up, but honestly, your input is appreciated because there is still more for me to learn on models of alternating machines with random access tapes. – Michael Wehar Nov 19 '14 at 6:47
Some alternating machine $A$ can solve $LinU$ in $\log^c(n)$ time. Rather than computing a many one reduction like I described, let's just build a machine $A'$ that simulates $A$ on input $(M',x')$. Since $M'$ is fixed, it can be hard coded into $A'$ with no problems. The input for $A'$ will be $x$, but we need quick access to $x'$ which is $x$ with a length $n$ padding of say 0's. – Michael Wehar Nov 19 '14 at 6:55
So let's use random access to compute $n$ in binary using a form of binary search. This should take $\log^2(n)$ time. Now, $A'$ runs similar to $A$, but anytime that $A$ would use random access, $A'$ first compares the look-up address to $n$ (which was written in binary) to decide whether the character at that address is in $x$, the padding, or a blank symbol. This should blow-up the runtime by a $c\log(n)$ factor. Generalizing this construction appears to make (2) still work out (at least with just an extra log factor and larger constants). I'll review it again soon to make sure. – Michael Wehar Nov 19 '14 at 6:56