Lecture 23: Computational Complexity
Why Computer Science Stack Exchange recommends it
Computer Science Stack Exchange askers cite it as the lecture they watched when they hit a confusing point about decision problems, EXP versus NP, or reductions, and ask for clarification.
Recommended by 1 people across 1 different months, 2016 to 2021.
What it is
A 51-minute MIT 6.006 lecture in which Erik Demaine introduces computational complexity. He covers decision problems, the counting argument that most problems are uncomputable, the classes P, NP, EXP and R, and the idea of reductions and NP-hardness.
What people said
Recommends Yamar69 · 2019-07 · Computer Science Stack Exchange · on “Show that there are infinitely more problems than we will ever be able to compute”
I was looking at this [this video] reading of MIT on computational complexity and on minute 15:00 Erik Demaine embarks on a demonstration to show what is stated in the title of this question. However I cannot follow his reasoning, in practice what he says is this: we can state a decision problem as string of $0$ and $1$ which in practice is the truth table of the function. He goes on to say that a decision problem is an infinite string of bits while a program is a finite string of bits and up to here no problem. What I don't understand is the continuation of the proof from this point on: Decision problems are in $\mathbb{R}$ cause you can put a decimal point before the string representing the pro
Cites rb612 · 2016-05 · Computer Science Stack Exchange · on “Is EXPTIME "solvable" or "checkable" in exponential time?”
According to this [LINK] video, EXP has problems that are exponentially difficult to check. But according to this [this video] video, EXP are problems that are exponentially difficult to solve. It would make sense to me, that if EXP contains problems that are exponentially difficult to check, that they would contain problems that are harder than NP-complete ones. However, if it's true that EXP is problems that are just solvable in exponential time, why wouldn't they be equal? Wouldn't EXP therefore be a subset of NP (and so would R), as problems that are harder (like R) are still solvable in "non-deterministic" time? Because the term "non-deterministic" extends to any finite amount of time, corre
Cites Mr. Johnny Doe · 2020-09 · Computer Science Stack Exchange · on “Why can't we say that a Neural Network is a NP problem solver?”
From this video lecture from MIT [this video] there is mention how NP complexity works with finding a "lucky" algorithm and luck can never be accounted for. The example he draws on is just seeking the next yes till you find some final output. For those reasons, I had assumed that even SGD is NP complex, and by extension Neural Networks. And my logic was that, through readings it seems like a Back Prop algorithm is always looking for a way to reach towards the global minima by taking the most negative step. However, from my reading, it seems like Neural Networks have a computational Complexity of P. And as a result of the fact that most people believe $P\neq NP$ then a NN can only take a guess/app
Cites geckos · 2021-04 · Computer Science Stack Exchange · on “Is reduction symmetric?”
I was watch this lecture [this video] , and at this point he says a lot about reductions, take a problem and reduce to another problem. From what I could understand this is a symmetric relation, if I can reduce A to B, then I can reduce B to A. This seems pretty intuitive to me but I don't have a formal proof, is there any (for beginners would be better) material on this proof? Still on this lecture, from what I could understand we have problems that are known to be NP-hard, if I have a problem that I don't know what is but I can reduce to NP-hard problem then it must be NP-hard. Another question, supposing that reduction is symmetric if any NP problem can be reduced to P problem that would prove