Strictly speaking it doesn’t make sense for a single . But non-strictly speaking I mean that is arbitrary and the bound one can prove for the circuit complexity is a polynomial that does not depend on .

]]>I don’t know whether it would be easier, but that’s an interesting suggestion. A related suggestion is to try to establish (1) with formula complexity replacing circuit complexity. That again would potentially lead to very interesting results that fell short of solving the P vs NP problem.

]]>*Thanks — corrected.*

Hm… In the obvious game, by setting the winning set A to contain all sequences with x_2=1, we can force the second player to choose a specific move (x_2=0 in our case), and thus allow the first player to move twice. Thus, we can assume that each player has k consecutive moves.

Now, take any P-space complete game, for example Node Kayles game (where players mark vertices in an undirected graph, and you cannot choose a marked vertex or any of its neighbours, and loses a player who cannot make a legitimate move), and let G be any hard instance of this game with m vertices. Let all vertices in G are marked as 0-1 sequences of length k. Let n=mk, and consider the following obvious game. If the first k moves of player 1 do not describe a vertex of G, then the sequence does not belong to A. Otherwise, if the next k moves of player 2 do not describe a legitimate move in a Node Kayles game, then the sequence belongs to A, and so on. As a result, the first player wins this “obvious” game if and only the first player wins the corresponding Node Kayles game. For every given sequence it is easy to say if it belongs to A or not, therefore A has low circuit complexity. However, there is no “simple” winning strategy unless either P=PSPACE or we define simplicity such that “simple” does not imply “efficiently computable”…

Does it make sense?

“My instinct is that they don’t.”

If they don’t, thet you current open task

“think further about whether there might be some notion of “simple” strategy for which it can be shown that games with payoff sets of low circuit complexity have simple winning strategies.”

may be solved only with “simple” strategy witch is not efficiently computable…

On the other hand, if they are, it would be a very interesting result indeed.

]]>I agree that the second worry doesn’t rule out an approach along these lines, but I still find it a worrying sign that some of the hardest NP functions, such as the clique function, are not hard from this point of view.

However, your first point is one that I agree with: it would be interesting to know whether functions of low circuit complexity have efficient strategies (either for the 0-set or for the 1-set). My instinct is that they don’t. If, for example, one creates a highly pseudorandom function, then it seems to me to be very unlikely that there would be a polynomial-computable function (incidentally, I think it should be polynomial in rather than because the payoff set is not part of the input) for deciding how to play optimally.

However, that is not a very good argument, since the weirdness of a computable function could be matched by the weirdness of a computable strategy. For example, it might be that one could build up a circuit for the strategy in terms of a circuit for the function. So it would be good to try to find either a proof that efficient strategies exist, which would be a very interesting result I think, or a convincing heuristic argument that they don’t.

When I get time, I think I’ll add a page suggesting this question (including your thoughts above and crediting it and them to you).

]]>Completely agree that proving the statement S: “every set A of low circuit complexity have an efficiently computable strategy” would not directly solve P vs NP problem for the reasons you described. However, it would be at least a good starting point, because, with any definition of simplicity I can imagine, a simple strategy is efficiently computable, therefore the statement T: “every set A of low circuit complexity have a simple strategy” is much stronger than S, therefore we cannot hope proving T before we even know how to prove S.

I do not completely agree with second worry: Having T, be will only need to show that ONE (not any) NP-complete function does not have a simple strategy. Ok, so we would need to choose a non-monotone example at this stage.

]]>I’ve wondered about this definition of “simple”, but it has some problems. First, the whole point of any definition of simplicity is that it should be something we can use to prove results about computational complexity. If to prove a lower bound on circuit complexity we are required to show that there is no polynomially computable strategy of some kind, then the obvious question arises of how we are going to do that, given that proving superpolynomial lower bounds is more or less the problem we started with (but perhaps harder in this case, since strategies are complicated objects to analyse).

Secondly, if you take any monotone game, then there is a very simple optimal strategy (at least in the version of the game where each player has their own set of coordinates to play): Player I always plays 1 and Player II always plays 0. Since there are plenty of NP-complete monotone functions, this is fairly worrying.

]]>Perhaps you should raise that through https://github.com/TiddlySpace/tiddlyspace, rather than here.

]]>I would call a winning strategy for player i is “simple” if there is an efficient algorithm (say, polynomial in 2^n), which, given payoff set A and a sequence of current moves, returns the next move for player i. I understand that the definition of simplicity invovling “efficient algorithm” is maybe not what you wanted, but this is exactly what I intuitively understand by “simple strategy”. For example, chess is determined, but there is no simple stragety in the sense above, and therefore the game is interesting to play. And if it would be, and I knew it, I could easily win against any opponent. So, for me this seems to be THE CORRECT definition of “simple strategy”. And, by the way, you question with this definition of simplicity seems to be interesting.

]]>I now have an example that shows that the modified version of the game where both players declare their parities at the beginning is not a Ramsey lift. The general idea behind the construction can probably be used to rule out a number of other over-simple candidates for Ramsey lifts but I haven’t looked into that yet.

]]>You are of course right. When I wrote that, I must have been misremembering a question I had thought about that didn’t have a simple answer like that. In fact, in my notes (written before I wrote the TiddlySpace document) I made the observation you made above. I then commented that it didn’t seem obvious which player won the game defined by the parity function, so that may be a better question. I’ll change the “open task” to that.

]]>I am reading your TiddlySpace description, looking to the first “open problem” I can contribute. In section “A more complicated class of games” you asked a question “Open Task: Analyse this (3_SAT) game for some very simple payoff sets, such as x_1=1”.

First, let us clarify the problem formulation. If you mean that 3-SAT clauses players can use are GIVEN AND LISTED, and there are m of them, then your statement “rules force the game to terminate: it terminates when there is just one sequence that satisfies all the clauses” would be wrong, if there are at least two sequences that satisfies all m clauses.

So, I conclude players can use ANY clauses they want subject to rules 1 and 2 in game description. Then, I claim player 1 wins your game in 4 moves:

Move 1. x_1=1 or x_2=1 or x_3=1

Move 2. x_1=1 or x_2=1 or x_3=0

Move 3. x_1=1 or x_2=0 or x_3=1

Move 4/ x_1=1 or x_2=0 or x_3=0

(no matter what player 2 does)

After move 4, all the remaining feasible sequences have x_1=1, and player 1 is guaranteed to win. ]]>

hmmm, ok, on further thought this may be related to some of your musings on xor of functions (have to go look at those again maybe). let $f_n(x)$ be a function that is meant to “approximate” a target function $g_n(x)$ where the latter is NP complete. then $f_n(x)$ Xor $g_n(x)$ is the number of “errors” in using $f_n(x)$ as an “approximation” of $g_n(x)$. the total error is the sum of all those bits. this is a common pattern in monotone circuit lower bounds proofs incl razborovs. but note this is also just the hamming distance between the output bit of the two function’s truth tables. (suspect something like this concept also shows up in natural proofs). figure it is in coding theory also somewhere.

similar to this question

cnf/dnf conversion to minimize errors

I think you’re right that what I did in the pages was different: very roughly speaking, Player I declares a strategy for the first few moves and Player II declares an outcome for those moves that is consistent with Player I’s strategy. But that’s still a “one-shot” lift. The question of whether one could get from zero to simplifying all sets in a series of gradual steps is one that I don’t know the answer to and I agree that it would be well worth answering.

]]>I see the point: basic sets are probably “too simple” so a lower bound would be good news and confirm the need to define simplicity a bit higher in the hierarchy. Even a lower bound showing that the alphabet must square would morally confirm this, since one expects an iterative construction, and squaring at each iteration is more than we can afford.

However, here is one question that arises (and I really hope it wasn’t asked in the proof-tree already and I missed it). Let’s look at the lift that provides the “trivial” doubly exponential upper bound on the alphabet size. That lift is built in one shot. Is there a way to think of that lift, or one that has the same effect of providing a doubly exponential upper bound, as being built iteratively, by an iteration step that, say, squares the alphabet at each step? Just to imagine how this could go, perhaps each iteration step would be designed to take care of one more move in the game, so the auxiliary information would not be based on a full strategy but just a strategy for the first few moves (I found in the pages something along the lines of considering a few moves only, but the goal looked different). If this iterative construction existed, the hope would be to exploit it as a skeleton on which to build other more efficient lifts by redesigning each step to fit whatever goal we are after (e.g. simplifying a set of low circuit complexity).

]]>Can you be more precise about what the payoff set is? That is, under what circumstances does the game stop and one of the players get declared as the winner?

]]>its really quite simple =)

havent worked this all out, but something like the following. the goal is to compute the NP complete function for a specific number of input bits. the score of a player is how many “errors” remain (simple calculation using the hamming difference in the truth table, partially constructed function vs actual function). the “board” consists of the circuit constructed so far. each player adds one gate to the DAG constructed so far. the question is, are there any thms in game theory that describe that certain games must take a large amt of moves for either player to succeed, that would be key? (lower bound)… am not an expert on game theory….

Can you describe more precisely how this game works?

]]>It’s not the first case of somebody trying to relate the theory of Borel sets to computational complexity. As far as I know, that was Michael Sipser. As I said elsewhere, there are worrying disanalogies between Borel sets and sets of low circuit complexity that suggest that this approach to the P versus NP problem is an extremely long shot.

]]>