In computer science, the satisfiability problem is the task of determining whether a boolean problem can be satisfied, that is, made to be true, by some combination of inputs. For example, the expression: A AND NOT B is satisfied whenever A is true and B is false, while the expression: A AND NOT A cannot be satisfied.
A naive approach to satisfiablilty would be to check every possible input and see if the expression evaluates to true.
However, the number of combinations that must be checked is 2n, where n is the number of variables in the expression. This is what is meant by exponential time.
The exponential time hypothesis conjectures that you cannot do better, that is, the number of operations needed to solve satisfiability will always be exponential in the size of the problem, no matter what algorithm is chosen.
If this hypothesis is true, it would imply that P≠NP, because satisfiability has been shown to be in NP, and in order for it to be in P there would have to exist an algorithm that solves it in polynomial time.
> the number of operations needed to solve satisfiability will always be exponential in the size of the problem, no matter what algorithm is chosen
A common complexity theory misunderstanding is that statements like this applies to all or even most instances-- and it leads to objectively bad engineering of the form "oh look this is actually 3sat, it's unsolvable".
(Great fodder for a midwit meme: idiot and genius say "my heuristics go brrr!"-- midwit says "No no no! NP problems can't be efficiently solved!")
Any specific instance may be efficiently solvable (and perhaps all instances coming from your particular problem set -- a subset of all possible 3sats).
So I would add to that paragraph "at least for some 3sat instances".
There's a set of problems that are (it is conjectured) easy to verify but hard to solve. One of them, called 3SAT, looks for values (a, b, c, ...) such that an expression like ((a or b or c) and (b or not-c or e) and ...) evaluates to true. And it turns out that all of these supposedly hard problems can be converted to 3SAT, so if you find a solution to one you find a solution to all.
The Exponential Time Hypothesis in effect says that 3SAT can only be solved by brute force--trying every possible combination until you find the right one. If your 3SAT expression has 100 variables, then this will take about 2^100 tries (exponential time).
What they found was an algorithm faster than brute force. 3SAT is still potentially one of these hard problems, but finding anything faster than brute force is a major advance.
The exponential time hypothesis is essentially that NP-complete problems' best solutions are all exponential.
NP-complete problems are essentially those where you need to make some kind of choice over a superlinearly-growing set of options where each option can be solved in polynomial time. Like, for example, finding a route between cities on a map under a certain total distance. You might need to go over all choices of routes.
The ETH essentially assumes that the best solution (in terms of complexity, i.e. how it grows) you can get for these kinds of problems is exponential. OpenAI just proved that this is false, you can have solutions that are below exponential complexity (but still above polynomial - if anyone finds a polynomial solution then we'd have proven that P=NP)
> NP-complete problems are essentially those where you need to make some kind of choice over a superlinearly-growing set of options where each option can be solved in linear time.
Seems like a weird way to describe NP, and I'm not even sure it's true?
NP is the class of decision problems where if the answer is YES and someone slips you a hint, you can solve them quickly (or quickly figure out that the hint was useless and report that). Where quickly means in polynomial time.
There's nothing about linear time in the classic definition.
Sorry, you're correct. I meant "polynomial" rather than linear.
NP = nondeterministically polynomial. i.e. if you could evaluate all choices in parallel (or just "know" the right choice), then you'd get a polynomial runtime. That's what I was trying to informally hint at, but I ended up using the wrong term. I'll edit my original post.
That's equivalent to your description - the "hint" is knowing which of the options to choose.
The reason why superlinearity comes into play is that if you had a linearly growing amount of solutions to test then you'd just have a sum of polynomials, which is itself polynomial.
the language around complexity domains is really confusing.
P means "this problem can be solved in polynomial # of steps relative to the problem size"
NP means "we don't know if the problem can be solved in polynomial # of steps relative to problem size, but we know we can check if the result is correct in a polynomial # of steps relative to the problem size"
!ETH: it was widely suspected that the flagship problem in NP, 3SAT, requires an exponential # of steps relative to problem size. We now know that 3SAT requires AT MOST something more than polynomial but strictly less than exponential to solve.
"P = NP" means "if you can check it in polynomial time, there's some way to solve it in polynomial time". "P != NP" means "there are some problems checkable in polynomial time for which there is no polynomial way to obtain the result"
Today's new proof makes "P = NP" "feel more likely", because we know that the flagship problem in NP is not exponential.
This isn't quite true, problems in NP can still require exponential time to solve.
Solving an arbitrary NP problem looks like first reducing that problem to SAT and then solving the equivalent SAT problem. We know (via the Cook–Levin theorem) that any NP problem can be reduced to SAT in polynomial time, and we now know that solving SAT is subexponential in terms of the size of the SAT problem, but it's possible for the conversion to SAT to grow the problem enough that it's exponential in terms of the original problem.
But even if every problem in NP is sub-exponential (which, as tfpgh notes, this result doesn't prove), how does that make it "more likely" that every such problem is solvable in a polynomial # of steps? I feel like it's a bit like claiming that, because Alex Honnold has shown that a human can free solo El Capitan in less than 4 hours, that makes it "more likely" that I could. To me, his success does not budge the latter probability at all.
It seems to me that the logic only goes the other way ... if ETH had been proven then it would follow that P != NP. With !ETH we really don't know any more about P ?= NP then we did before. Reasoning that P = NP is "more likely" strikes me as a form of a fallacy of affirming the consequent.
This is not a proof that NP != EXP. NP is poly time reduceable to 3SAT + 3SAT is sub-exponential does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
I like this https://www.reddit.com/r/explainlikeimfive/comments/33fmrq/e...
2n should read as 2^n
Unless OP already solved p=np...
that threw me, thanks
> the number of operations needed to solve satisfiability will always be exponential in the size of the problem, no matter what algorithm is chosen
A common complexity theory misunderstanding is that statements like this applies to all or even most instances-- and it leads to objectively bad engineering of the form "oh look this is actually 3sat, it's unsolvable".
(Great fodder for a midwit meme: idiot and genius say "my heuristics go brrr!"-- midwit says "No no no! NP problems can't be efficiently solved!")
Any specific instance may be efficiently solvable (and perhaps all instances coming from your particular problem set -- a subset of all possible 3sats).
So I would add to that paragraph "at least for some 3sat instances".
My 2nd grader once brought home a homework sheet with some simple problems involving combinations of coins and I was like "this is unsolvable!"
There's a set of problems that are (it is conjectured) easy to verify but hard to solve. One of them, called 3SAT, looks for values (a, b, c, ...) such that an expression like ((a or b or c) and (b or not-c or e) and ...) evaluates to true. And it turns out that all of these supposedly hard problems can be converted to 3SAT, so if you find a solution to one you find a solution to all.
The Exponential Time Hypothesis in effect says that 3SAT can only be solved by brute force--trying every possible combination until you find the right one. If your 3SAT expression has 100 variables, then this will take about 2^100 tries (exponential time).
What they found was an algorithm faster than brute force. 3SAT is still potentially one of these hard problems, but finding anything faster than brute force is a major advance.
The exponential time hypothesis is essentially that NP-complete problems' best solutions are all exponential.
NP-complete problems are essentially those where you need to make some kind of choice over a superlinearly-growing set of options where each option can be solved in polynomial time. Like, for example, finding a route between cities on a map under a certain total distance. You might need to go over all choices of routes.
The ETH essentially assumes that the best solution (in terms of complexity, i.e. how it grows) you can get for these kinds of problems is exponential. OpenAI just proved that this is false, you can have solutions that are below exponential complexity (but still above polynomial - if anyone finds a polynomial solution then we'd have proven that P=NP)
> NP-complete problems are essentially those where you need to make some kind of choice over a superlinearly-growing set of options where each option can be solved in linear time.
Seems like a weird way to describe NP, and I'm not even sure it's true?
NP is the class of decision problems where if the answer is YES and someone slips you a hint, you can solve them quickly (or quickly figure out that the hint was useless and report that). Where quickly means in polynomial time.
There's nothing about linear time in the classic definition.
Sorry, you're correct. I meant "polynomial" rather than linear.
NP = nondeterministically polynomial. i.e. if you could evaluate all choices in parallel (or just "know" the right choice), then you'd get a polynomial runtime. That's what I was trying to informally hint at, but I ended up using the wrong term. I'll edit my original post.
That's equivalent to your description - the "hint" is knowing which of the options to choose.
The reason why superlinearity comes into play is that if you had a linearly growing amount of solutions to test then you'd just have a sum of polynomials, which is itself polynomial.
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the language around complexity domains is really confusing.
P means "this problem can be solved in polynomial # of steps relative to the problem size"
NP means "we don't know if the problem can be solved in polynomial # of steps relative to problem size, but we know we can check if the result is correct in a polynomial # of steps relative to the problem size"
!ETH: it was widely suspected that the flagship problem in NP, 3SAT, requires an exponential # of steps relative to problem size. We now know that 3SAT requires AT MOST something more than polynomial but strictly less than exponential to solve.
"P = NP" means "if you can check it in polynomial time, there's some way to solve it in polynomial time". "P != NP" means "there are some problems checkable in polynomial time for which there is no polynomial way to obtain the result"
Today's new proof makes "P = NP" "feel more likely", because we know that the flagship problem in NP is not exponential.
This isn't quite true, problems in NP can still require exponential time to solve.
Solving an arbitrary NP problem looks like first reducing that problem to SAT and then solving the equivalent SAT problem. We know (via the Cook–Levin theorem) that any NP problem can be reduced to SAT in polynomial time, and we now know that solving SAT is subexponential in terms of the size of the SAT problem, but it's possible for the conversion to SAT to grow the problem enough that it's exponential in terms of the original problem.
thanks for the correction. Corrected.
But even if every problem in NP is sub-exponential (which, as tfpgh notes, this result doesn't prove), how does that make it "more likely" that every such problem is solvable in a polynomial # of steps? I feel like it's a bit like claiming that, because Alex Honnold has shown that a human can free solo El Capitan in less than 4 hours, that makes it "more likely" that I could. To me, his success does not budge the latter probability at all.
It seems to me that the logic only goes the other way ... if ETH had been proven then it would follow that P != NP. With !ETH we really don't know any more about P ?= NP then we did before. Reasoning that P = NP is "more likely" strikes me as a form of a fallacy of affirming the consequent.
3SAT can't be solved in sub-exponential time
Fwiw never thought it was a well-named conjecture
*Can be solved
This proves that EXP != NP
This is not a proof that NP != EXP. NP is poly time reduceable to 3SAT + 3SAT is sub-exponential does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
I'm stating the conjecture as requested
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