Interesting post but not sure why the author calls it underrated. It seems to get a fare share of attention and love amongst those that care about such things.
Edit: presenting the problem with inverting functions is an unusual approach tho.
https://stackoverflow.com/questions/1857244/what-are-the-dif...
If you can solve the decision problem, you can solve the optimization problem by doing a binary search on the x.
That means that tsp and tso-opt are in the same ballpark of complexity. The reason tsp-opt isn't called np-complete isn't related to its complexity, but to the fact that only decision problems get to be called that.
As I said, there are problems where the decision problem and the optimization problem are completely unrelated in terms of complexity, but tsp isn't one of those.
Assuming you have an oracle for any question that you know how to verify, this is very easy to reduce:
Is there a solution in which vertices {1, 4, 6} are all positive?
First, let's establish that the question is legal: presented with a working solution, we can calculate the defining constraint of the problem, and we can check the polarity of vertices 1, 4, and 6. If this problem is in NP, verifying the constraint will take at most polynomial time. If not, not, but our question can be verified in however much time it takes to calculate the constraint.
The complexity of determining a solution by getting answers to questions of this form is interesting.
Case 1: The answer is "no" for all single-vertex sets. All vertices must be negative. This cannot actually happen, because all vertices being negative is the same thing as all vertices being positive.
Case 2: All vertices may be simultaneously positive. In this case, the answer to every question will be "yes", and we'll ask one question for every vertex in the input graph, solving the problem in sublinear time.
Case 3: We need some positive and some negative vertices. By asking about single-vertex sets, we can quickly identify a vertex that can be positive, vertex A.
We will include that vertex in every subsequent question. By the time we get there, we will have already asked about zero or more other vertices and been answered "no". We need never ask about those vertices again; they have to be negative and if we include them in any set, we'll get another "no".
So, let's ask about a never-before-examined vertex, vertex B. Can {A, B} be simultaneously positive?
If not, we can toss B into the "must be negative" pile, since we know A can be positive.
If so, we can include it in our developing solution; future questions will include vertices A and B. We still never need to reexamine any vertex that has ever been included in any of our questions.
So, unless I'm missing something, in this case we'll also produce a solution using one question per vertex in the graph.
> but for this one I don’t see the point making the distinction.
The distinction between NP-complete and NP-hard has nothing to do with the distinction between yes/no questions and open-ended questions. Those are unrelated concepts.
An NP-hard problem is at least as hard as any NP problem. It might be much harder.
An NP-complete problem is NP-hard, but it's also guaranteed to be an NP problem. That's the distinction.
My point being that transforming tsp-opt to tsp only adds a polynomial factor. So in the case of tsp-opt, it's called NP-hard rather than NP-complete because it's not a decision problem, not because it's not equivalent in terms of time complexity as a problem in NP.
People simply call it NP-hard because that term is better known than NPO or NP-equivalent.
Both NP and NP-hard are defined for the class of decision problems.
You may review the following to clarify the distinction between the various NP complexity classes:
https://en.wikipedia.org/wiki/NP-hardness#NP-naming_conventi...
>A decision problem H is NP-hard when for every problem L in NP, there is a polynomial-time many-one reduction from L to H
Under the assumption that the <= question can be answered in polynomial time, this entire algorithm will also complete in polynomial time.
What problem formulation do you have in mind?
If you want to represent the problem as a bunch of points in a Euclidean plane with free travel, that's a different (and easier) problem.
And even then, while it might take an infinite number of steps to specify the answer at infinite resolution, it will only take a finite number of steps to specify the answer at any level you're capable of writing down.
That’s basically the good old “everything is O(1) because int64/float64 has 2^64 possibilities” misconception. It’s not how complexity theory works. For any N, 2^N is still finite, we’re obviously not talking about undecidable problems.
Edit: I see that you may be responding to the “finitely enumerate” part of my comment. Sure, it wasn’t phrased well. Replace with “enumerate in P”.
What are you fixing? Suppose you want to know the answer to within one part in 10¹⁰⁰. That will take you 333 questions.
Suppose you don't actually need 100 decimal places of the answer, or more likely that even if you had them you'd be unable to use them, and you can only represent the answer to 20 decimal places. That will take you 67 questions.
You can easily enumerate this answer in P. The problem in your argument isn't that float64 only has 2^64 values. Use as many bits to hold the answer as you want. No matter how many that is, it will be a finite number, and you'll be able to specify them all in a polynomial amount of time. Each question takes polynomial time to answer and fills one bit of the solution.
I'm having some trouble with this. As far as I can see, finding a path of a given maximum length must be in NP, because it's very easy to deterministically verify that a given path has length no more than the maximum.
If determining the optimal path length is in NP, and identifying a path of that length is also in NP, how can determining an optimal path fail to be in NP?
> Hodge conjecture
I for one have no idea about topology in general. I had to look up what a nice shape was... I thought it was some mathematical term.
The only thing I remember about is topology is someone telling me how it would be easier for me to untangle cables if I knew topology but alas I never learned.
My own pet theory is that it's an evolution of speech towards "hardened" claims that you can't easily disagree with. You cannot disagree with "over/under-rated" because there's no official "rating" method. You cannot disagree with "vibes" because the source of a vibration is impossible to determine. They both allow a speaker to make a claim without evidence.
It was ever thus.
See, for example, https://blogs.illinois.edu/view/25/96439
Though I guess for a lot of young people it feels like that, with none of their friends talking about such bands.
There's another term, i'm sure, because politicians use it all the time; "I never said X" when it was very heavily implied, and the listener was led down the garden path to the conclusion, only to find out the conclusion is bitter and unpleasant. The speaker can say "oh, that's your own biases/misconceptions/dogma, i never said <something extremely specific>"
I'm in the weeds here, but a simple analogy would be: "I don't like sunny days without clouds or fog." and i say "why don't you like blue skies?" and they say "i never said that" when the salient points of a sunny day with no clouds or fog are 1) there's a sun, and 2) the sky is blue.
contradictory reply: "not at all, they were Grammy-nominated and had N number-one hits",
weasel reply: "that's not what I meant... lately they haven't been paid enough attention. way to miss the point, $NAME_CALLING".
It's a way to say "I like them" without having to refute anything other than supporting comments, aka a weasel-word.
(A) That's not going to stop anyone; (B) the claim can easily be obviously false. For example, Taylor Swift is underrated.
We have here an example of problem (B); P versus NP is possibly the single most famous problem in computer science. It isn't underrated.
(B) anything can be "obviously false" under contrived/extreme situations. try again with "Sting is underrated" and suddenly you can easily argue it either way. Try using an argument that doesn't rely on p-values being at the 3+ sigma levels, but of course you won't because you know that will invalidate your position.
"I think that the framing with inverting a function f is quite underrated."
Of course, using the title "What P vs NP is actually about" is a bit of a stretch -- "Interesting take on P vs NP" would be more honest.
I think it's just that the first time you use underrated, it seems to be referring to the problem itself, not to your specific take. With the video as context, or even just reading a few sentences in, the meaning is clearer.