Henning Makholm
Reputation
393/400 score
 5h comment Checking whether a Language is Regular @OveAhlman: Hmm, that does make an amount of sense. 5h comment Checking whether a Language is Regular @techno: It is so close to the argument inside Myhill-Nerode that I don't think an explicit invocation of it is necessary. But in classwork I would certainly include that line you quote -- which is what shows that you understand how the handwavy "finite memory" connects to the actual automata formalism. 6h comment Checking whether a Language is Regular @Ove: I don't understand the fixation people have on the pumping lemma for this kind of questions. The Myhill-Nerode theorem is much easier to apply and more precise too. In my opinion the only thing the pumping lemma has going for it is a memorable name. 6h answered Checking whether a Language is Regular 8h comment Birthday problem(probability) @soulless: If what you want to compute is the probability that there is exactly one group of three sharing a birthday and nobody else share birthdays even just in in pairs, then your formula looks correct. But of you allow people to share birthdays in pairs in addition to the triple, then you need stronger stuff than you have there. 9h comment Birthday problem(probability) @soulless: Do you count cases where there are 4 people sharing a birthday? Or cases where there are two groups of 3 people each that share birthdays? The wording "3 people having the same birthday" is ambiguous by itself. 9h comment Birthday problem(probability) @A.S. Not necessarily, but you'd still need to take the possibility that there are two or more such triples into account -- otherwise a naive calculation would double-count the cases where there are two triplets and produce a too high probability. 9h comment Birthday problem(probability) We can compute the expected number of triples that share a birthday easily: $\binom n3\frac{1}{365^2}$, and as long as that's small (say, $n\le 50$ which gives an expectation of $\approx 0.14$ that should be a reasonable approximation of the probability that there's at least one such group. But for highter $n$ we get clearly too high an approximation because then there's a significant chance of having two or more triples -- for example for $n=100$ the expected number of triples is 1.21. 9h comment Birthday problem(probability) @soulless: The problem is that after you have chosen those three people you need the probability that those three people have the same birthday (that's easy, $\frac{1}{365^2}$) and no other group of three people have the same birthday (which is hard). 1d comment How big are regular (hyperbolic) polygons? @nbubis: Our results are not really different -- you can convert mine into yours by the double-angle formula for hyperbolic consines: $\cosh(2x)=2\cosh^2(x)-1$, and then $\cos^2\frac\pi n = \frac12+\frac12\cos\frac{2\pi}{n}$ and $\cos^2+\sin^2=1$. 1d revised How big are regular (hyperbolic) polygons? added 283 characters in body 1d answered How big are regular (hyperbolic) polygons? 1d comment How big are regular (hyperbolic) polygons? Does it matter that the surface is embedded in $\mathbb R^3$? Your questions seem so be only intrinsic properties of the surface. 1d awarded Nice Answer 1d comment When solving a big Rubik cube (100x100x100), do you reduce the solution to like 50x50x50, and then 25x25x25, and then like 10x10x10 and then 3x3x3? Are you asking whether each person who answers would prefer such a strategy (which would be too subjective for SE, I think),or whether such a reduction strategy is necessary for solving big cubes? 1d comment Understanding Bell's inequality vs. quantum mechanics @FrankScience: Yes, the model I describe here is the one that Bell's inequaliy says cannot work. The point of the entire reasoning is that actual quantum mechanics predicts results that cannot come from any theory that follows this model -- and there fore (take-home lesson here!) it would be futile to try to search for a theory that both fits into the $A$-$B$-$H$ model and produces the same prediction as quantum mechanics does. 1d comment Understanding Bell's inequality vs. quantum mechanics (...) instead it argues that every possible theory that can be described in the "local realistic" format I've sketched will fail to derive the same probability as quantum mechanics does, no matter whether it uses complex, numbers, Hilbert spaces, or something completely different from that. 1d comment Understanding Bell's inequality vs. quantum mechanics There are two parts to this. The first is to understand how to derive the probabilities according to the quantum-mechanical formalism. That's how Hilbert spaces, state projections and so on come, and it looks like you're reasonably on top of that (at least good enough that I don't feel competent to correct you). The other one is to understand the particular class of alternative theories that cannot produce the same probabilities -- which is what my answer here is about. And this part of the problem is specifically not about Hilbert spaces; (...) 1d comment Understanding Bell's inequality vs. quantum mechanics @Frank: It is improtant here to distinguish between the same random variable and two different random variables that just happen to have the same distribution. The latter is the case if the two variables can have different values in a single run of the experiment. In my model $A$ and $B$ might well have the same distribution, but they are certainly different variables, because they represent two independent choices of which measurement to make of each particle. 1d comment Understanding Bell's inequality vs. quantum mechanics @FrankScience: How a typical Bell-inequality test is performed is that first two entangled particles are produced somehow, then they are transported to two different detectors some distance from each other (these are the "two ends" of the experiment). Then, simultaneously at each detectors a random decision about which direction to measure the particle's spin is made and this measurement is carried out quickly enough that light-speed signals about the direction the other measurement was done in cannot yet have reached the detector.