How Hard is Exact Simulation of Algorithms, and a Related Operation on Complexity Classes

Question

Teaser

Since the problem is longish, here is a special case that captures its essence.

Problem: Let A be a deterministic algorithm for 3-SAT. Is the problem of completely simulating the algorithm A (on every instance of the problem). P-Space hard?

(More precisely, are there reasons to believe this task is P-Space hard, does something in this direction follow from standard CC conjectures, and is there hope to prove that this task is X-hard for some complexity class X which is presumed to be strictly above NP.)

Related questions: are-pspace-complete-problems-inherently-less-tractable-than-np-complete-problems;

EDITED UPDATE: There are various interpretation for "Completely simulating A". And there may be different interesting answers according to the interpretation. (Also Ryan Williams proposed an interpretation for simulating a non deterministic algorithm.) For a certain way to associate a decision problem to the computational task "Completely simulate A", Joe Fitzsimons found an algorithm A for which this associated decision problem is still in NP. If "completely simulate" refers to being able to output the entire register of the computer at a given step $i$ then for Joe's algorithm it seems that $P^{NP}$ is what is needed. For this version (I think, but am not sure) Ryan's answer sketches a $P^{NP}$-hardness argument. Joe remarked that if of you are required to supply the entire registers (which is not any more a decision problem) it is not surprising that you need to step up and the complexity classes are not the same.

Anyway, if we require to output the state of the registers at a prescribed step $i$ then the answers of Ruan and Joe suggests (but again, I am not sure about it) that $NP^+$ is essentially $P^{NP}$. We can speculate that by this interpretation the operation pushes up one step higher in the polynomial hierarchy, and that $PH^+ =PH$.

In any case by these interpretations the answer to my teaser question is NO.

I had a more drastic interpretation for "completely simulating an algorithm A" in mind. (But perhaps Joe's and Ryan's interpretation is more interesting.) My interpretation by "completely simulating algorithm A" is that you output the state of the registers at every step $i$. In particular, if the algorithm is not polynomial your output is also not polynomial. Under this drastic interpretation I wondered if we should believe that for every algorithm A, $C_A$ is P-SPACE hard, and what can we prove.

Motivation:

This question was motivated by a lecture by Paul Goldberg (slides, video, paper) describing a paper with Papadimitriou and Savani. They showed that it P-space complete to find any equilibria that are computed by the Lemke-Howson algorithm. The problem to find some equilibrium point is only PPAD-complete. This gap is quite amazing and similar results are described already in Papadimitriu's well-known paper: The Complexity of the Parity Argument and Other Inefficient proofs of Existence (1991). (It is known that PPAD-complete problems cannot even be NP-hard (unless terrible things happen so this is far down in the complexity world compared to P-space).

What the question is about

My question is about similar gaps for even older and more classical computational complexity problems. (Maybe this is already familiar.)

Given a computational problem we can distinguish between three issues:

a) Solve the problem algorithmically

b) Reach the same solution as a specific algorithm A

c) Simulate the entire algorithm A

Of course c) is at least as hard as b) which is at least as hard as a). The results mentioned above shows a gap between the computational difficulty of tasks a) and b) for the problem of computing equilibria. We would like to understand the situation (and mainly the gap between a) and c)) for other computational problems.

The question:

The basic form of the question with an example

We start with a computational problem, Problem X

An example can be

Problem X: Solve an instance of SAT with n variables

we also specify

A: an algorithm that performs Problem X

and we pose a new problem

Problem Y: Exactly simulate algorithm A

and we are interested in the computational difficulty of Problem Y. We wish to understand the class of such problems Y for all algorithms A that solve the original Problem X. Especially we want to know how easy can problem Y be (or how hard must it be) if we are allowed to choose the algorithm A at will.

The proposed operation on complexity classes

Start with a complexity class $C$ which is described by some computational task. Given an algorithm A to perform every instance of this computational task, consider a new complexity class $C_A$ which is described by the computational task of completely simulating $A$. Then we can (hopefully) define an "ideal" of complexity classes

$C^+ = \{C_A:$ for all algorithms A}.

If we let $P$ to describe whatever a digital computer can do in polynomial time (so I don't want to restrict attention e.g. to decision problems) then $P^+$ is the ideal spanned by $P$ itself.

Finally, My Questions

My questions are:

1) Does the definition make sense (in the wide sense of the word sense). Is it well known or the same as (or similar to) some well known thing. (My formulation was informal and in particular when we move from specific problems like SAT to a complexity class like NP we have to worry about various things that I neglected.)

The next two questions assume that the definition can make sense or salvaged to make sense.

2) Suppose we equip ourselves with all the standard conjectures regarding computational complexity. Can we say what $C^+$ is supposed to be for some familiar complexity classes. (E.g. $C=NP$, $C$=P-space,..)? EDIT: Several people pointed out that $PSPACE^+=PSPACE$. So > we can ask instead what is $(P^{NP})^+$? is $PH^+=PH$?

Can we guess what are the complexity classes $C$ so that $C^+$ is the ideal spanned by $C$?

So the question how easy can the computational task of simulating an algorithm A for 3-SAT (when we can choose the algorithm to make it as easy as possible) is an interesting special case.

3) Is there hope to actually prove something about this operation?

Of course, if you prove that all complexity classes in $NP^+$ are P-space hard this will show that $P=NP$ implies $P=PSPACE$, which (I think) would be a huge and highly unexpected result. But if you show that all complexity classes in $NP^+$ are hard to something say in the third level of the polynomial hierarchy (e.g. $\Delta_3^{\bf P}$) this would only imply things that we already know, things that follow from the fact that $P=NP$ causes PH to collapse.

Answer 1

Problem: Let A be a deterministic algorithm for 3-SAT. Is the problem of completely simulating the algorithm A (on every instance of the problem) P-Space hard?

I don't understand the statement of this problem. But I think there is a natural way to formalize your more general question which may shed a little light on it. Maybe I am completely missing your point, but hopefully you still find something interesting to read here.

1) Is the definition makes sense (in the wide sense of the word sense). Is it well known or the same as (or similar to) some well known thing. (My formulation was informal and in particular when we move from specific problems like SAT to a complexity class like NP we have to worry about various things that I neglected.)

One way to make sense of the task exactly simulate algorithm $Y$ is the following. Let's fix the model to be one-tape Turing machines for simplicity; what I will say can apply to any typical computational model.

For each algorithm $Y$ and input $x$, we can define its computation history $H_Y(x,i,j)$: Given integers $i$ and $j$ which range from $0$ to the running time of $Y$, $H_Y(x,i,j)$ equals the content of the $j$th cell of the tape of Turing machine $Y$ on input $x$ in time step $i$. (And if the tape head is reading the $j$th cell in the $i$th step, include that too along with the machine's state.) Of course, computation histories come up all the time in complexity theory.

Now, one could argue that any algorithm which can decide the language

$C_Y = \{(x,i,j,\sigma)~|~H_Y(x,i,j)=\sigma\}$

(or simulate the function $H_Y$ on all inputs) is solving the task exactly simulate algorithm $Y$, because it has the ability to print every little part of every possible computation of algorithm $Y$. Certainly, given an oracle for $C_Y$ one could do a step-by-step simulation of $Y$.

2) Suppose we equip ourselves with all the standard conjectures regarding computational complexity. Can we say what C+ is supposed to be for some familiar complexity classes. (E.g. C = NP, C = P-space,..)? Can we guess what are the complexity classes C so that C+ is the ideal spanned by C?

This is still an interesting question, under the above proposal. For deterministic time classes, nothing changes. $P^+$ is just $P$: we can decide $C_Y$ in polytime, for every polytime algorithm. Similarly $EXP^+ = EXP$. Also $PSPACE^+$ is still $PSPACE$. For nondeterministic time complexity classes, the situation becomes more interesting. In that case, an algorithm $Y$ can have multiple computation histories, so $H_Y$ is no longer well-defined. However we still want to print some computation history, so our "exact simulation" would have to isolate a specific nondeterministic computation history and then print its values. For an $NP$ algorithm $Y$, one can say that $C_Y \in P^{NP}$: we can use the $NP$ oracle to binary search for the "first" accepting computation history (in lex order), then once we have obtained it, print the relevant bits. For an $NEXP$ algorithm $Y$, one can say $C_Y \in EXP^{NP}$, for similar reasons.

Can we put $NP^+$ in a smaller class? I don't know. Notice we cannot simply redefine

$C_Y = ${$(x,i,j,\sigma)~|~$ there exists a $H_Y$ such that $H_Y(x,i,j)=\sigma$}

to try to put $C_Y$ in $NP$, because we need the history string to be the same for all quadruples involving $x$, in order to "exactly simulate algorithm $Y$".

Anyway, this issue of not being able to "print a witness" to an $NEXP$ computation without going to $EXP^{NP}$ does arise in some situations, such as circuit complexity. If $NEXP$ has polynomial size circuits, then is it also the case that $C_Y \in P/poly$ for every nondeterministic $2^{n^k}$ time $Y$? Impagliazzo, Kabanets, and Wigderson answered this question affirmatively in 2001. Their proof is extremely interesting, invoking tools from derandomization and diagonalization (why would derandomization be necessary for such a result?) and it turns out to be a very useful theorem for proving circuit lower bounds for $NEXP$ functions.

Is there hope to actually prove something about this operation?

Maybe... that depends on whether you are happy with the above formalization of your question. For a deterministic 3-SAT algorithm $Y$, I think the above exact simulation of $Y$ problem would be $P^{NP(1)}$-hard, where $P^{NP(1)}$ is polynomial time with one query to $NP$. (The annoying syntax of StackExchange requires that I write (1) instead of the alternative. Earlier I said $C_Y$ should be $P^{NP}$-hard, but I am not sure of the details: you may see how to generalize the below.)

We give a polytime many-one reduction from every $L \in P^{NP(1)}$ to $C_Y$. Given such an $L$, let $M$ be a machine recognizing $L$ that does a single $NP$ query. WLOG, that query is a SAT formula. Also WLOG, the SAT query can be "postponed" until the very last step of the computation, in the following sense: there is a polynomial time algorithm $A(x)$ which prints a formula $F$ and bit $b$, such that for all $x$,

$M(x)$ accepts iff $A(x)=(F,b)$ such that ($SAT(F)$ XOR $b$) = 1.

($M$ may reject iff $F$ is satisfiable, or it may accept; the bit $b$ captures this.)

For simplicity, let's say when $Y$ ends its computation, it moves its tape head to cell 1, writes "accept" or "reject" in that cell, and loops forever. Then, asking if $(F,T,1,accept) \in C_Y$ for sufficiently large $T$ will tell us if $F$ is accepted. (In general, we just need that it's efficient to compute the instance $y$ of $C_Y$ which will tell us.) Note this already shows that $C_Y$ is both $NP$-hard and $coNP$-hard; the latter is true because $(F,T,1,reject) \in C_Y$ iff $F$ is not satisfiable.

The final reduction from $L$ to $C_Y$ is: given $x$, run $A(x)$ getting $(F,b)$. If $b=0$ then output $(F,T,1,accept)$, else if $b=1$ output $(F,T,1,reject)$.

For every $x$ we are producing (in polynomial time) a $y$ such that $M(x)$ accepts iff $y \in C_Y$.

(Thanks to Joe for demanding that I be clearer about this part.)

But I don't see (for example) how to get $\Sigma_2 P$-hardness. To reduce $\Sigma_2$-SAT to the problem, it would appear you'd need to write a 3-CNF SAT instance $x$ which simulates your deterministic algorithm $Y$ within it (where $Y$ is solving Tautologies on various subproblems). But as $Y$ has no given time bound, that 3-CNF $x$ could be huge, so you don't necessarily get a polynomial-time reduction. Unless I am missing something.

Answer 2

I initially posted an incorrect answer, so hopefully this is an improvement.

I'm going to start out by considering the 3SAT example. In your comment on Ryan's answer you say

I am interesting how hard it is to simulate a deterministic algorithm Y for 3-SAT. And the question is if no matter what Y is this is P-space hard.

The answer to this is that it is not PSPACE-hard for at least some Y, assuming that NP$\neq$PSPACE, but that there exist other algoriths for which it is PSPACE-hard. Showing the latter is trivial: We simply construct an algorithm which interprets the bit string representing the 3SAT formula instead as a TQBF problem, which it then solves before solving the 3SAT instance. Obviously there is nothing special about TQBF in this case, so the algorithm can be made arbitrarily hard to simulate. So we should only care about lower bounds on simulation of any algorithm for a given problem, rather than a specific algorithm.

With that in mind, we construct the following algorithm to solve 3SAT:

Take a register of $n$ bits (initially all set to 0) to contain a trial solution, together with a register containing the 3SAT instance, a counter register of size $\lceil\log_2 (c + 1)\rceil$ initially set to 1 and and two flag bits (call these the fail flag and the accept flag). Here $c$ is the number of clauses. The algorithm then proceeds as follows:

• If the value of the clause counter $k$ is less than or equal to $c$, and the trial solution is not $111......1$, check whether clause $k$ is satisfied. If not set the fail bit. Increment the counter.
• If the value of the clause counter $k$ is less than or equal to $c$, and the trial solution is $111......1$, check whether clause $k$ is satisfied. If it is not, set the fail flag. Increment the counter.
• If $k$ exceeds $c$, and the trial solution is not $111......1$, check if the fail flag is set. If so then increment the trial solution, reset the counter $k$ to 1, and unset the fail flag. If the fail flag was not set, set the accept flag, set the clause counter $k$ to zero, set the trial solution to zero and halt.
• If $k$ exceeds $c$, and the trial solution is $111......1$, check if the fail flag is set. If the fail flag was not set, set the accept flag. Set the clause counter $k$ to zero, set the trial solution to zero and halt.

When the algorithm halts, the state of the accept flag tells you whether or not the 3SAT formula can be satisfied.

Now, if I want to compute the state of the algorithm at time $i$ I have an algorithm to compute this in polynomial time with a single call to an NP oracle as follows:

Note that for any $i$, assuming the accept bit has not yet been set, the state of the registers can be calculated in polynomial time, since the value of $k$ and the trial solution register $t$ are simply functions of $i$. Determining if the fail flag is set can be done in polynomial time, simply by checking whether the current state of the trial solution register satisfies all clauses less than or equal the current value of $k$. If and only if this not the case, the fail flag is set. The accept flag is set to zero.

Similarly, if the accept bit has already been set, the state of the registers can be efficiently computed, since everything is zero except the accept bit, which is set.

So the only hardness comes in determining whether the accept bit is set. This is equivalent to determining whether the given 3SAT instance has a solution less than $t$. If it does, then the accept bit must necessarily be set, and if it does not, then the accept bit must necessarily be zero. Clearly this problem is itself NP-complete.

Thus the state of the system at step $i$ can be determined by in polynomial time with a single query to an NP oracle.

Important update: Initially I had misread Ryan's formulation of exact simulation as a decission problem, and so my claim that the decission version was in NP was incorrect. Given the problem of deciding whether bit $j$ at step $i$ on input $x$ as formulated in Ryans answer, then there are 3 possibilities:

1. This bit is constant in independent of whether $F$ is satisfiable. As there are only two possible states for the system at this time (both of which can be computed in P) determining if this is the case, and if so, whether the value is equal to $\sigma$ is in P.
2. The bit is equal to $\sigma$ if $F\in SAT$, and unequal otherwise. This problem is clearly in NP, as the satisfying assignment of $F$ acts as a witness.
3. The bit is equal to $\sigma$ if $F\in UNSAT$ in which case the problem is then in coNP.

Clearly deciding which one of these three is the case can be done in polynomial time, simply by comparing the value that bit takes if $F\in SAT$ and if $F\in UNSAT$. Thus the exact simulation problem is in NP $\cup$ coNP. Thus, as Ryan's lower bound and my upper bound match, assuming both are correct, I think we have an answer: $C_Y = NP\cup coNP$.

Note that there is nothing special about 3SAT in this case, and the same could be done for any problem in NP. Further, the same trick can be used for any non-deterministic complexity class, not just NP, which seem to be the hard ones to simulate. For deterministic classes you can simply run the best algorithm and halt at step $i$. So with this in mind, full simulation of at least some deterministic algorithm for a problem is only as hard as solving the problem itself.

Answer 3

Very interesting thought! Here's an extended comment that was too long to post as such:

Regarding the definition in (1) as such, that is:

Start with a complexity class $C$ which is described by some computational task. Given an algorithm A to perform every instance of this computational task, consider a new complexity class $C_A$ which is described by the computational task of completly simulating $A$. Then we can (hopefully) define an "ideal" of complexity classes: $C^+ = \{C_A:$ for all algorithms $A\}$.

I believe you're going to quickly run into an issue with undecidability for non-trivial membership in $C^+$. In particular, given the description of two TMs $M$ and $M^{\prime}$, it's well-known that deciding whether they accept the same language (i.e. $L(M) \stackrel{?}{=} L(M^{\prime})$ is undecidable in general by reduction from the Halting Problem.

Further, given the description of a Turing Machine , there is always a trivial simulation: Just construct a new Turing Machine $M^*$ that calls $M$ as a subroutine (using its description), which accepts if $M$ accepts and rejects if $M$ rejects. This only costs a linear overhead. Specifically, if $M$ runs in $t(n)$ computational steps, then $M^*$ runs in $O(t(n))$ (whereby "run," I mean "simulates $M$ exactly").

This suggests to me that if there's any serious traction to be gained here, the precise way you go about making the definition of the ideal of a complexity class is going to be fairly important. In particular, if you intend to show that the ideal of, say, $NP$ is $PSPACE$-hard, you'll have to rule out the notion of a trivial, linear-time simulation of the $NP$ machines in question.

With respect to the homotopy methods described in the talk/paper and GPS's $PSPACE$-completeness result, I believe the "gap" you're witnessing between $PPAD$-completeness and $PSPACE$-hardness is due to the distinction between finding any NE (in the sense of END OF LINE) and finding a unique NE given a specific, initial configuration (in the sense of OTHER END OF LINE).