Posts for NxCy


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I think when it comes to ROM hacks, there's usually some expectation that the run will involve gameplay that is substantively different to the original game. I haven't watched the other TAS of this game, but going by this run alone, I didn't see anything that hasn't been done before. It looked like basically just holding right + jumping through a couple of levels that look more or less identical to ones from the original game. Of course I don't blame HappyLee for that... I find the original SMB TASes interesting for historical / nostalgia reasons, but I've never really found SMB to be a very exciting game for speedrunning. Therefore, although I'm sure the run is optimised, I decided to vote no. As for the whole debate about passwords/the rules etc. (which admittedly I haven't fully read), with respect to HappyLee, who is clearly a great TASer, I don't really find the run interesting enough for the discussion to even be very meaningful. I don't really know how the site works, but I would hope that the rules are as they are and continue to evolve so as to accommodate the most entertaining content, and to weed out runs that are obviously at odds with the site's mission. If a new run came along that was met with high praise, but was in conflict with some detail of one of the rules, I'm sure there would be majority support for amending the rules to accommodate the new run. I don't think this run is anything like that, and even if you could win the argument that this is the `true any%', having it obsolete the full game TAS would seem like a hollow victory to me.
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I have a very different method for solving the problem. It’s by no means as simple or elegant as FractalFusion’s approach, but I think it’s pretty cool and seems to get the right answer. I don’t know anything about martingales or Markov chains, but I do know a bit about scattering theory in physics, and I was able to translate the problem into a scattering problem. Apologies in advance as this will be a bit long and probably not very straight-forward to follow. Imagine a network of nodes and links that looks something like this – 1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – The numbers (nodes) represent how many coins you have, and the dashes, which connect different nodes are going to allow you to move left or right within the network. The physics problem I’m imagining might be something like a series of optical waveguides (lines) with connections (numbers) at which light can scatter (either forwards or backwards). The idea is that we’re going to inject light into the network from the left side and find the probability (or intensity) of the light that makes it to the end. Three things can happen at each turn in the game and each one corresponds to a different type of scattering in the network: lose a coin (p = 5/9) = scatter left in the network gain a coin (p = 1/3) = scatter right one node gain two coins (p = 1/9) = scatter right two nodes (more on how this works shortly) To model the light in the network I’m going to suppose there are 4 different channels in each connection, a, b, c and d. Channel a is for all light that propagates to the left. Channels b, c and d will all be for right-propagating light but have different functions with respect to the game. Channel b will account for the case of when you win 1 coin (move right one node), channel c for when you win 2 coins (move right two nodes), and d will let me ‘skip ahead’ so that the light I initially inject into to the system can straight to go node 5 (since the game starts with 5 coins). The light in each link can thus be described by a vector of four components and scattering at the nodes can be described by setting up 4x4 scattering matrices ( https://en.wikipedia.org/wiki/S-matrix ) at the nodes. I need three different scattering matrices: one for before the entry point (node 5), one for at the entry point, and one for afterwards. I'm going to use plain variables for channels on the left side of a node, and a prime for channels on the right side of the node. For nodes 1, 2, 3, 4 we have the matrix equation (forgive the formatting) a 5/9 0 0 5/9 b b' = 1/3 1 0 1/3 c c' 1/9 0 0 1/9 d d' 0 0 1 0 a' The way this works is that light injected into the skip channel (d) stays in there and won't be scattered into any other channel. This channel acts independently of all the others. Light that comes into the node from channel c on the left all goes into channel b on the right. This ensures that all the light that scatters into channel c from the left effectively skips ahead one node (so it corresponds to winning 2 coins in the game). At the entry node (node 5) we have a 5/9 0 5/9 5/9 b b' = 1/3 1 1/3 1/3 c c' 1/9 0 1/3 1/9 d d' 0 0 0 0 a' which is more or less the same matrix, but now light coming from the skip channel gets evenly distributed among all the other options, which means light is now entering into the network at this point. For nodes to the right of the entry point the 'skip' channel becomes useless and we just have a 5/9 0 0 5/9 b b' = 1/3 1 0 1/3 c c' 1/9 0 0 1/9 d d' 0 0 0 0 a' There may well be a simpler way to set everything up, but this approach seemed fairly intuitive to me. Now we need to get the scattering matrix for the entire network so we can relate the channels at the far left to the far right. Then we can just inject light with intensity 1 into the skip channel at the left end and see how much comes out at the right end. One way to do this is to convert each 4x4 scattering matrix into a 4x4 transfer matrix (see for example http://assets.press.princeton.edu/chapters/s8695.pdf ). You can work these out by essentially finding the 'partial inverses' of the scattering matrices ( https://en.wikipedia.org/wiki/Partial_inverse_of_a_matrix ), although you need to be a bit careful with the matrix block structure. Once you have the transfer matrices you can just multiply them together to get the transfer matrix for the whole system, which can then be converted back into a scattering matrix. If I use M to denote transfer matrices, the transfer matrix for the whole system is just M_right ^ 3 * M_entry * M_left ^ 4 Convert this back into a scattering matrix S and finally compute S * [0, 0, 1, 0]. My little Python program outputs [a, b', c', d'] = [0.45451636, 0.45483641, 0.09064723, 0] Finally, since the network stopped at node 8, we need to remember that I win if I'm in either channel b or c. b is light that entered 9 directly from node 8 (you won 1 coin when you had 8) and c is light heading towards an imaginary node 10 from node 8 (you won 2 coins from node 8). So we just need to add 0.45483641 + 0.09064723 = 0.5454836405947694 which seems to be at least close to the right answer.
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I’m not sure if any of this really answers the question, but here are some random thoughts. I’ve always found questions like ‘what is the value of 8 / 2 * 4?’ a bit artificial. I think the problem is that, after a certain point, these kinds of expressions don’t really occur in the mathematics we usually do in school or beyond. When I (and I assume this is true for most people) actually sit down and do numerical or algebraic calculations by hand, I never use symbols like / or ÷. Instead, I tend to write everything using fractions and, if necessary, parentheses. Since a fraction is written vertically, there’s no ‘horizontal ambiguity’ when reading an expression from left to right; it’s clear that the top and bottom of the fraction are on equal terms and I never need to think about the order of operations (which, to be honest, I don’t remember anyway). When I see a question like ‘what is 8 / 2 * 4?’, my first reaction is something like ‘tell me what language you’re speaking first and then I’ll do the calculation for you’. I may even ask for clarification: 'do you mean (8/2) * 4 or 8/(2*4)?'. Like with Masterjun’s English sentence example, I don’t think it’s the responsibility of the listener to have to resolve ambiguities in the message they receive. It should be the responsibility of the speaker to produce a clear, unambiguous message in the first place. I suppose that someone might respond by saying 'the expression is already unambiguous if you follow the standard order of operations'. The only scenario I can think of where an expression like 8 / 2 * 4 would occur in my life is in computer code (or writing a message on a forum I suppose). In this case, given a particular programming language, we can appeal to the rules of the language to parse the expression unambiguously. I don’t have a lot of experience with different languages, but I assume these rules are consistent across most mainstream languages and also align with the order of operations we learn in school. As for why the rules should be as they are, I think for me the main reason is that the alternative seems confusing. If a division sign were to incorporate more than the numbers or expressions immediately before or after it, then things like 1 / 3 + 4 / 2 would quickly become very confusing for me. In practice, I get a bit paranoid about fractions as they are and I’m not afraid of redundant parentheses. If I need to write a fraction with something other than just numbers, I tend to write things like ()/() before filling in the contents of the parentheses, safe knowing that anything else I write in my line of code isn’t going to be affected by this particular use of the / sign. To summarise, I guess I don't really care about the question in the context of mathematics, as it's one I haven't had to worry about in over 20 years. I'd be happy to go with either result depending on what the person asking the question prefers, but I may heavily caution them against interpreting 8 / 2 * 4 as 8 /(2 * 4), as I think they are likely to run into problems doing more complicated calculations. In the context of computer programming, well just stick it into Python and see what comes out.
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Dacicus wrote:
NxCy wrote:
Hence a(n,p) = (p+1)*F(n-p-1) for p < n-1.
I believe that there are errors in the derivation of this formula. If you take n = 5 and p = 1, for example, it yields a(5, 1) = 2 * F(3) = 4. But that Wikipedia link shows that a(5, 1) should be 5.
Yeah, you're right. I forgot to include cases where the 1s are mixed between the numbers in the composition without the 1s (in your example my argument misses 2+1+2). I'll have to think about it some more.
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This turned out to be harder than I thought. I wouldn't be surprised if there's an easier way to do it, but here's what I found. If we call the duration of an eighth note 1, then a bar has duration 8. Because of ties, it's possible to construct notes of any duration from 1 to 8. Suppose first there are no rests, then a rhythm will be an ordered sequence of notes whose durations sum to 8. For example, four quarter notes would be 2 + 2 + 2 + 2 = 8. These are the compositions of an integer https://en.wikipedia.org/wiki/Composition_(combinatorics). The number of compositions of n is given by 2^(n-1), which is the number of rhythms that don't include any rests. To deal with rests, note that rests can't be tied (as there is no sound). If we look through all the compositions of n, we can replace every instance of a 1 (which corresponded to an eighth note) with an rest of duration 1 to get a new rhythm. We only need to do this for 1s though: replacing a 2 with a rest of duration 2 would give the same rhythm as the same composition with the 2 replaced with 1+1 and both 1s replaced with rests, which we have already considered. If there are p 1s in a certain composition, all choices of notes/rests for the 1s gives a total of 2^p rhythms. The total number of rhythms of duration n is hence given by sum from p=0 to p=n of 2^p * a(n,p) where a(n,p) is the number of compositions of n containing p 1s. Now we need to find a(n,p). It's easy to see that a(n,n) = 1 and a(n,n-1) = 0. For p < n-1, a composition of n containing p 1s contains precisely p 1s and a composition of n-p that does NOT contain any 1s. Since order matters, there are (p+1) choices for the locations of the 1s. The number of compositions of n-p not containing any 1s turns out to be equal to F(n-p-1) where F(n) is the n'th Fibonacci number (F(0) = 0, F(1) = 1, F(n) = F(n-1) + F(n-2)). Admittedly, I got this part using a recurrence relation I found by google searching. Hence a(n,p) = (p+1)*F(n-p-1) for p < n-1. Finally then, the number of rhythms of duration n is given by 2^n + sum from p=0 to p=n-2 of 2^p * (p+1) * F(n-p-1) When n =8 I get 1257. This does include the rhythm containing only rests (i.e. silence). If you don't count that I get 1256.
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Dacicus wrote:
FractalFusion wrote:
Just for clarification: Is a tie between two eighth notes equivalent to a quarter note, or are they distinct? (similarly for other ties of two or more notes)
They are equivalent.
I also have a question. Are two rhythms distinct if they're related by a shift/delay in time? For example, if I have quarter notes on beats 1 and 3 (rests on 2 and 4), is that distinct from having quarter notes on beats 2 and 4 (rests on 1 and 3)?
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You beat me to it :P I did it more or less the same way. Let p be the golden ration. Starting from 5.12.3 here https://dlmf.nist.gov/5.12, I made the substitution t = z^p. Then setting a = 1/p, b = p - 1/p = 1, you get 1/p * B(1/p, 1) = I, where I is the desired integral. Since B(1/p, 1) = p, it follows that I = 1.
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This was quite fun indeed, although finding a geometric solution probably took me longer than it would have taken to just do the differentiation :P The function f(x) can be thought of as the sum of the magnitudes of the vectors (x+2, 3) and (x-4, 5). If you draw a picture and think about it, you'll see that minimizing f(x) is equivalent to the following problem: what point C on the central horizontal axis in the figure is such that AC + CB is minimal? This is solved by reflecting the triangle through the axis and constructing the straight line from A to B' (purple line in the diagram). In this case, the angles to A and B from C are equal, which means (4-x)/5 = (2+x)/3. Solving this gives x = 1/4, which is the same as what Dacicus found. edit: just realized my solid/dash styling is inconsistent, but I can't be bothered changing it now. Sorry!
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I found by experimenting that the final two digits of powers of 7 cycle through four different values: 01, 07, 49 and 43. Therefore we need to reduce the exponent of the bottom 7 modulo 4. This can be done using an application of Euler's theorem. Basically in reducing numbers mod 4 you can reduce powers mod phi(4) = 2. The exponent of the bottom 7 is itself a tower of 7s. Writing it as 7^(7^...^7), clearly the thing in brackets is odd, so 7^...^7 = 1 mod 2. Therefore, modulo 4 we have 7^(7^...^7) = 7^1 = 3. This was the power of the bottom 7 of the original number and 3 corresponds to an ending of 43.
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I also don't think this category makes a lot of sense. A natural comparison can be made with the recently published minimum A presses run, as both runs restrict the use of the same button. For the most part this run looks like a normal any% run, but Mario has to keep doing wall jumps to climb up pipes and walls. For me this doesn't add anything interesting. The game ins't traversed in a new, innovative way; it's just that an extra step has been added to otherwise regular movement. I felt like the minimum A presses run required more of a rethink of how to get around different obstacles and pushed the game more to its limits with the strategies involved. I don't think this category is better than minimum A presses, and I don't see that it is different enough from other runs to warrant publication as a new category. Therefore I voted no.
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chillex_drsado wrote:
I appreciate the compliments and the thanks! As for that level, there are many gaps and enemies that Diddy can't really scale himself in a way that's faster. While most of them should be minor, the biggest timeloss as Diddy, as well as the dealbreaker for me was the enemy at 13:25. To progress past that as Diddy, I had to both swap at the beginning, which itself takes up 1.25 seconds, and wait for the enemy to be in a position where I could jump over it. Dixie's hover ability also allows me to take full advantage of her roll speed and skip over krockhead platforms, whereas I couldn't do it with Diddy. By the time I had progressed far enough into the level, it became less worth to do the rest as Diddy because I could goal swap at the end. I do plan to write more tidbits in my level-by-level breakdown tomorrow. I have some errands I need to run, so it won't be today.
Thanks for the response. Just out of curiosity I tried playing around with the level I mentioned a bit with Diddy and I think you're indeed correct. I also didn't realize when watching that those levels have ice physics and so some of your movement that I thought looked odd actually makes sense now :P
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This was a great TAS. Although I really enjoyed the first stage, I was wondering if there would be much more to the run or if each subsequent stage would just be more of the same. Thankfully there was a lot of variety overall and I think the length was just about perfect: long enough so that you got to show off a lot of cool ideas with each stage feeling different, and short enough so that it didn't have a chance to feel repetitive. My favourites were probably stages 7 and 10. Easy yes vote. Thanks for the run!
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Acumenium wrote:
I'm not sure why this is so hard to understand. Right now, it's not a challenge, because it's a TAS, there's no such thing as difficulty. At the same time, the goal is bleakly arbitrary. This isn't relatable. The last sentence is also hypocritical. If I have no right to an opinion or right to express it, neither do you. You're accusing me of a bigger podium than I have by doing the same thing you are accusing me of.
It's hard to understand because you say things like "But this isn't even a TAS" and then "because it's a TAS". In any case, I don't really see hypocrisy in what I said. Whether or not something is a TAS, insofar as it can be considered publishable on TASvideos, is decided by the community (or at least by the judges). I wouldn't claim that a movie shouldn't be published because it isn't a TAS unless I were confident that my saying so is consistent with the community's standard of what a TAS is. I think I've already explained why I'm not impressed by the word 'arbitrary'. As for relatability, I personally think most of the ACE movies are extremely unrelatable and uninteresting, but I think they deserve to be published.
Acumenium wrote:
tl;dr: In a vacuum, this run doesn't have an issue. It isn't very interesting to me but it doesn't have to be. It being accepted represents serious flaws in the publishing system as a whole. It should've been immediately rejected given what many other runs and branches have to put up with, but it wasn't. It was rather accepted for every reason those types of runs are rejected.
Perhaps this what really mattered to you all along. I only initially responded because I felt comments about how this movie isn't a TAS or is arbitrary/unrelatable were unfair when discussing this run in isolation. I won't comment on broader points about the site's publication system because I don't really have an opinion on it (I'm more of a casual viewer these days), but you should certainly bring it up in the appropriate place if you have an problem with it.
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Warp wrote:
NxCy wrote:
Yes, and I think we can use the same example. f(x) = exp(-1/x^2) g(x) = x^2 Then lim(x->0) [ f(x)^g(x) ] = lim(x->0) [ exp(-1) ] = 1/e but [ lim(x->0) f(x) ] ^ [ lim(x->0) g(x) ] = 0^0 is undefined
I don't think that's a valid argument because you could use the exact same argument to argue that the rule lim(a/b) = lim(a) / lim(b) is invalid (even though it's one of the basic laws of limits).
It's been a while since I've studied analysis, but that rule you mentioned has some assumptions behind it. I believe the simplest version goes: if lim(a) = A and lim(b) = B where B is not equal to 0 then lim(a/b) = A/B. It can probably be extended to cases where lim(a) or lim(b) = infinity, but you have to be careful with indeterminate forms. As for one function to the power of another, we have lim(x->c) f(x)^g(x) = lim(x->c) exp[ g(x)*ln(f(x)) ] = exp[ lim(x->c) g(x)*ln(f(x)) ] where I've probably assumed f is positive and used the fact that exp is a continuous function. If the limits of g(x) and ln(f(x)) are nice then we can use the limit product rule to get the limit to be exp[ g(c)*ln(f(c)) ], which then unravels to get f(c)^g(c) as you wanted, but I think it's clear that there are several things that can go wrong, for example the limit of f at c may be 0, in which case you would end up with ln(0) and (at least the simplest version of) the limit product rule no longer applies. So I think the answer is 'sometimes yes, but not in general'. If anyone has a more clear-cut answer that'd be nice.
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Warp wrote:
Btw, a somewhat related question: I saw someone in another forum claim that this equality does not hold, and can't be assumed: lim(x->c)(f(x)g(x)) = lim(x->c)(f(x))lim(x->c)(g(x)) Is that the case?
Yes, and I think we can use the same example. f(x) = exp(-1/x^2) g(x) = x^2 Then lim(x->0) [ f(x)^g(x) ] = lim(x->0) [ exp(-1) ] = 1/e but [ lim(x->0) f(x) ] ^ [ lim(x->0) g(x) ] = 0^0 is undefined
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Acumenium wrote:
Ordinarily no, TASes are not supposed to relate to RTAs. But this isn't even a TAS. It's an arbitary player-enforced goal. I also can't see how this differs very much from a normal non-ABC run of Super Mario Bros. other than it being painfully slow. - The same levels as the warps run are used. - No new screens are visited. - New techniques are shown that immediately proceed minutes of waiting in the floor. - It is a TAS of a challenge that is supposed to be player relatable and yet none of the new tricks used to get stuck in the floors repeatedly and so on are actually possible in real time. The last one is particularly relevant, I think, as that removes a major entertainment factor of the ABC in the first place. It didn't lead to an interesting movie at all. Play the normal TAS at 0.5x speed, it'll still go faster and be more interesting than watching an enemy waddle across the ground for minutes at a time. If a low-level challenge of an RPG had you able to beat the first boss at level 3 at the earliest, or level 1 with the use of TAS tools like save/load states and so on, a TAS of it isn't very interesting for the purpose of watching someone try to beat it.
It's funny because I think of the four points you listed the first three are reasonable and the last is quite baffling. If you didn't enjoy the movie or didn't find it interesting because it was slow and wasn't different enough to you, I think that's completely reasonable feedback and I respect your opinion. I can't understand at all your final point and it looks like I'm not the only one. At the risk of repeating what has already been said: how is the movie any less relatable than any other TAS? You can try and beat the game yourself in as few A presses as possible. I guess you could do it in a few hundred. The TAS will inevitably do it in fewer presses and you really have no chance of getting close. Similarly you could try to beat a game as fast as possible. The TAS will be faster and you have no realistic chance of competing unless it's a very simple game. If you think the lack of comparability to human runs is a reason for it not being entertaining to you, then fair enough, but understand that for a lot of people this isn't really relevant, especially on TASvideos. I think a lot of people actually enjoy TASes precisely because they're so different to RTAs. Lastly, I don't think you get to define what is or isn't a TAS. There are lots of TASes (and speed runs) that put time as a secondary goal.
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I'm not really a fan of using words like arbitrary in these discussions. My quick google search yields the following definition: "based on random choice or personal whim, rather than any reason or system". Minimizing A presses (and then time) is a clearly defined goal and was presumably chosen because it's a technically interesting challenge with notably different gameplay to the traditional speed-first categories (how interesting you find it is obviously subjective). I don't consider that to be random choice or mere personal whim. But even if you think it is arbitrary because the goal choice is still in some sense subjective or random, so what? Why is it a problem if it leads to an interesting movie with notably different gameplay to already existing categories? If you weren't entertained then fair enough, but I don't understand comments like "no because it's arbitrary". I think whenever these offbeat categories come up the movie itself is the argument for why the seemingly arbitrary goal choice is a good one: if the movie is good then the category is good. I don't think a TAS being comparable to RTA runs is really relevant in relation to potential publication on TASvideos.
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To parrot most people in this thread, this movie is an excellent technical achievement. I think optimizing button presses is a fairly well-known goal and in this particular game it leads to interesting gameplay that doesn't appear in other categories. The movie is more entertaining and interesting if you think about the goal while watching it, but even at face value it shows off some pretty crazy movement, some of which at least is obviously conducive to the goal (e.g. walking through the floor to avoid jumping over pipes). I think overall it was moderately entertaining; certainly good enough for the site. Most importantly, as a casual viewer, I think the movie is far better than the walkathon published on the site, which, at least from memory, felt like a couple of cool tricks amidst 20 minutes of otherwise uninteresting gameplay. If that movie was deemed suitable for publication then I don't see why this shouldn't be. Thanks to the team for the movie! Yes vote.
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This is my first time seeing this hack, which looks very interesting and well made. The movement required throughout is much more complicated than in the original game, so I imagine you had a lot of fun figuring out the fastest roll/jump patterns for each level. The movement looks quite well optimized overall. The only real thing that stuck out to me is that Dixie gets used a lot, for example in some of the swamp levels in world 3. Was it really faster to use her there (e.g. the level starting at 13:22)? It was quite hard to find instances where she was faster than Diddy in the original game, especially after the first two worlds. In any case I enjoyed watching this. Thanks for making it!
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Nice! That's exactly the intended solution. Apologies for my framing of the problem. I'll have to avoid words like "trick" and "magic" in the future :P If anyone's interested I saw it at around 26:25 in this video https://youtu.be/TCZ3YwbcDaw
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OmnipotentEntity wrote:
You can add the final bit by handing the cards either face up or face down, which is very magic-trick-like, in that there's a trick. And that gives you the exact number of bits required to specify 1 of 48 unique cards.
There is a way to always correctly determine the final card solely from the four cards that were passed and their order. There's no additional trickery involved (handing the cards face up vs face down or providing additional information by other means). I guess there are probably multiple ways to do it, but the solution I saw is quite elegant.
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Maybe not exactly a maths problem, but here's a cool little trick I saw in a Youtube video recently. Two people work together to perform a card trick. The first person chooses five cards randomly from a standard deck. After analysing the cards, he chooses one of the five cards to keep for himself, arranges the remaining four cards in an order of his choice and passes them to the second person. The second person, after observing the four cards they received, correctly announces the identity of the fifth card that was kept by the first person. How is the trick performed?
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Consider the nine angles x = pi/18, 3pi/18, 5pi/18, 7pi/18, 9pi/18, 11pi/18, 13pi/18, 15pi/18 and 17pi/18 These all satisfy the equation cot(9x) = 0. The idea is to find a polynomial whose roots are the tangents of these angles (excluding 9pi/18 = pi/2 for which the tangent is infinite). Let t = tanx. Iterating tan(3x) = ( 3t - t^3 )/( 1 - 3t^2 ), we can find an expression for tan(9x) in terms of t. The denominator of tan(9x) (or the numerator of cot(9x)) comes out as 9t^8 - 84t^6 + 126t^4 - 36t^2 + 1 The roots of this polynomial are the values of tan(x). Only even powers of t are present because the values of tan(x) come in opposite sign pairs (e.g. tan(pi/18) = -tan(17pi/18)). Let y = t^2, then 9y^4 - 84y^3 + 126y^2 - 36y + 1 = 0 has roots y1 = tan^2(pi/18), y2 = tan^2(3pi/18), y3 = tan^2(5pi/18), y4 = tan^2(7pi/18). Note that tan^4(3pi/18) = (1/sqrt3)^4 = 1/9. Therefore, using some properties of roots of polynomials, we end with tan^4(pi/18) + tan^4(5pi/18) + tan^4(7pi/18) = tan^4(pi/18) + tan^4(3pi/18) + tan^4(5pi/18) + tan^4(7pi/18) - 1/9 = y1^2 + y2^2 + y3^2 + y4^2 - 1/9 = (y1 + y2 + y3 + y4)^2 - 2(y1y2 + y1y3 + ...) - 1/9 = (-84/9)^2 - 2(126/9) - 1/9 = 59
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Joined: 11/19/2007
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Some personal thoughts on the topic: It seems to me that the crux of the paradox lies in the conflict between 'infinities don't exist in the real world' and that the real world infinity in the Zeno example emerges intuitively and doesn't doesn't seem to violate any physical principles. I would first say that I understand 'infinities don't exist in the real world' to mean that infinity can never be the result of a measurement. This almost seems true to me by definition, as any numerical measurement will necessarily yield a numerical answer and infinity is not a number (it's certainly not a rational number at any rate, which, as far as I know, is what all numerical measurements end up being). At best, you may measure something to be extremely large realtive to other related things and say that is is 'practically infinite', but this of course is not the same as actually 'measuring an infinity'. This doesn't logically exclude the possibility that there may exist an infinite number of things in the universe, but just says that we would never be able to conclude this to be the case experimentally, i.e. by actually counting stuff. The universe could therefore, at least in principle until demonstrated otherwise, be infinite in size and infinitely divisible without contradicting the idea that infinities don't exist in the real world in the way I've described here. Having said that, I don't really think there is a paradox, because there is no measured infinity. When I move from A to B, who's to say how many 'events' have occurred? I could reasonably say one event has occured, namley the event of moving from A to B. Of course I can increase the number of events indefinitely by taking ever more intermeidate points, operating under the definition that an event is the motion between two points that I've measured, but even discounting various physical limitations that prevent us from being able to measure arbitrarily small distances or times in practice, we are still never going to measure an infinity, because whatever we ultimately end up measuring is going to be some number. In mathematics, like you said, the paradox doesn't exist because we can deal with the relevant infinite sums. So even if spacetime is ultimately infinitely divisible, I don't think there's any problem. By the way, as far as I know the Planck length is just the legnth scale at which gravity becomes unignorable for elementary particles. Since a good theory of quantum gravity doesn't exist yet, it's also the legnth scale beyond which we don't really know what's going on. It's not that the universe is some kind of lattice where the Planck scale is the smallest unit of length (feel free to correct me if there do exist theoreis like this, as I'm definitely not very well read on this topic).
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Warp wrote:
What is the area of the shaded part, compared to the entire circle?
Am I misunderstanding something or is there not enough information to solve this? Seems like there are too many degrees of freedom. For example, you can make the upper blue segment tiny and shift the bottom blue segment up so that it effectively occupies half of the outer circle (imagine the bottom blue segment is a semi circle with the same diameter as the outer circle). Alternatively, you could make the blue area almost zero by moving the two horizontal chords very close to each other (imagine the blue circles are huge and you're taking tiny segments of them).
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