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thatguy wrote:
So the easiest way to visualise this is to draw a graph of f(x) = lnx/x, and then substitute in x and y. If f(x) > f(y), then xy > yx. Interestingly, this gives the relation a transitive property: If xy > yx, and yz > zy, then you can guarantee that xz > zx.
To take this just a little bit further, it can be shown with just a little calculus that lnx/x peaks at x = e and decreases thereafter. If both x and y are greater than e, then x^y > y^x if and only if x>y. Likewise, if both x and y are less than e, then x^y > y^x if and only if x<y. If x and y straddle e, f(x) must be compared directly with f(y).
On an unrelated note, here's a new math problem, which is not really a problem. I'm interested in very basic examples of algebraic geometry, specifically up to quadratic forms. I have a sort of obsession with solving systems of equations using the substitution method because when one encounters a system of equations with quadratic terms, the elimination and matrix methods offer no help. The substitution method is ironclad.
But it can also be tedious. As a simple example, consider an elastic collision in one dimension. We know kinetic energy and linear momentum are conserved, giving us two equations in two variables:
1/2*m_1*v_10^2 + 1/2*m_2*v_20^2 = 1/2*m_1*v_1f^2 + 1/2*m_2*v_2f^2
m_1*v_10 + m_2*v_20 = m_1*v_1f + m_2*v_2f
where m_1 and m_2 are the respective masses, v_10 and v_20 are the masses' initial velocities, and v_1f and v_2f are the masses' final velocities. (I know the underscores and carets are tiring to the eyes, but I don't wish to format the equations right now.) Given any four values-- typically m_1, m_2, v_10, and v_20-- that's just two equations and two variables and so all that remains is a little algebra.
Well, "a little algebra" can quickly become a nightmare. You end up with a quadratic equation in either v_1f or v_2f and a lot of coefficients flying around everywhere. It ends up being easier to just memorize the solution or...
... being clever with your physics. By recognizing that the physical process is subject to a time-reversal symmetry, you can determine that a particularly easy way of solving this problem involves entering the center of mass frame, negating the initial velocities, and then exiting the center of mass frame. Altogether, the solution looks like this:
v_com|G = (m_1*v_10|G + m_2*v_20|G)/(m_1 + m_2)
v_10|com = v_10|G - v_com|G
v_20|com = v_20|G - v_com|G
v_1f|com = - v_10|com
v_2f|com = - v_20|com
v_1f|G = v_1f|com + v_com|G
v_2f|G = v_2f|com + v_com|G
As you can see, the subscripts have become even more complicated. Here, the vertical bar stands for "relative to", G means ground (or the lab frame), and com means center of mass. On the bright side, there's not a squared term in sight! What was a quadratic equation becomes a linear system of equations. Granted, it's now seven equations and seven variables, but they're actually quite easy to solve since each equation only has a dependence on the ones that preceded it.
So that's the kind of thing I'm looking for. I can think of two similar physics problems that I haven't yet fully deciphered. First, what if an elastic collision takes place in two dimensions? Second, what about Compton scattering? In the case of Compton scattering, our three equations are
h*c/lambda + m*c^2 = h*c/lambda' + sqrt(m^2*c^4 + m^2*v^2/(1-v^2/c^2))
h/lambda = h/lambda'*cos(theta) + m*v/sqrt(1-v^2/c^2)*cos(phi)
0 = h/lambda'*sin(theta) - m*v/sqrt(1-v^2/c^2)*sin(phi)
where our variables typically are v and phi. This reduces to a simple equation:
lambda' - lambda = h/(m*c) * (1 - cos(theta))
This can be derived with the substitution method, but it takes about thirty minutes to an hour. (Of course, the elimination and matrix methods are again no help.) A textbook I referred to pointed out that the derivation can be significantly simplified by writing the conservation of momentum in vector form and taking the dot product between the first photon's momentum and the second photon's momentum, which naturally gives a single equation that does not involve phi. I think it's a nice trick, but I have a few issues with it. It's a math trick that must be memorized, not a physics trick, and it still leaves you with two nonlinear equations.
My question is whether there is a reliable method for solving quadratic equations in several variables, preferably a method that reduces the problem to a system of linear equations and if possible, one that can be given physical justification, as I did with the one-dimensional elastic collision example. Again, I know the answer lies somewhere in algebraic geometry, but I'm afraid I have no background in the subject and don't know where to start.
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I'm feeling bored and charitable, so I did a RAM search. I don't have definitive results, but here's what I've come up with so far.
Here's a summary of my search process:
Started a race.
Set RAM Search to automatically search for addresses with zero changes. This dropped my framerate to about 1 FPS...
Drove up to an item box.
Turned off automatic RAM searching.
Picked up an item (banana bunch), waited for it to register in my inventory and stop blinking.
(Saved the state and my RAM file.)
Used my bananas. I made sure no karters were behind me.
Searched for addresses that had changed more than once.
Reset the change count.
Turned around, ran into a banana.
Searched again for addresses that had changed more than once.
Edit: I started a new race and picked up a new banana bunch. The results I got were inconsistent with the list of RAM addresses I included in that post. Maybe it's correct, maybe not. In any case, I tried.
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ThunderAxe31 wrote:
This is probably useless, but i want to point out that I noticed something strange in stage 0. Look at the terrain map:
In the right up corner, there are four ceiling tiles to prevent Zook from jumping above screen. It is actually pretty easy to pass through them but still, nothing in particular happens.
I wonder if they did put these four ceiling tiles as a quick workaround to some bad glitch, then later were fixed by an actual improvement of the code: that would explain why they did not implement such ceilings in level 6 as well.
I still hope to discover some way to walk above the screen like in Super Mario Bros.
Why not hex edit those tiles out, then see what happens?
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It should be emphasized that playarounds are held to at least the same standards as "conventional" TASes. I'd argue they're held to even higher standards.
Few playarounds have been published and there are some consistent themes among them. The majority of them (26 out of 41 movies) are fighting games, probably because that genre lends itself to playarounds over strict "fastest completion" TASes. Six additional movies are in the sports genre, probably for the same reason. There are also games that are part of iconic series: Zelda, Mario, and Tetris. Those movies were probably published because of their respective series' popularity and the fact that they tend showcase glitches deep within the game. Those glitches were, in turn, discovered because of their game's popularity and the fact that their code has been picked apart.
That leaves just a few games that are (at least partly) outside of those major categories: Brain Age, Scribblenauts, Gradius, Oddworld, and Family Feud. Like the other games, these games tend to showcase major glitches and/or unexpected gameplay.
In short, it is quite possible to get a playaround published, but because of the inherent speed/entertainment tradeoff in that goal choice, the entertainment value has to be through the roof.
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ShilohPell wrote:
Yeah, I kinda figured that might be the case in the end. I mean having a bot to sift through permutations sounds awesome but the way Basic Bot appears to be doing it looks super inelegant. "Well, I think you'll need to hit the jump button about 15% of the time? I guess?" I think I'll continue learning LUA so I can write my own bot if the best this one can do is the monkey-with-a-stick approach.
Learning Lua is fun, useful, and kind of liberating when it comes to TASing. But I also say there's no need to reinvent the wheel. If the task you're looking to complete is short and simple, I suggest using my script if possible. Ultimately, it's up to you, though.
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ShilohPell wrote:
So I discovered Basic Bot and played around with it some. I understand you provide it with a memory address and let it run through permutations of inputs to try and optimize a section. But it seems to me whenever I set it up, it ends up running through hundreds of thousands of permutations of what basically amounts to junk data and even after hours and hours of letting it run, it still can't manage to figure out how to jump over a low wall. Am I setting it up incorrectly? How can I make the most of this tool?
I've never used Basic Bot, but as I understand it, it tries "random" button presses until it finds a solution. In that case, you're going to be stuck with hundreds of thousands of junk permutations.
Three years ago, I wrote this script that also does a brute-force search but greatly cuts down on its parameter space by omitting inputs where a button is pressed and released excessively. Theoretically, Basic Bot will eventually produce the optimal input for any game, given enough time. Practically, it's only good for maybe ten frames of input. My bot has the potential to miss perfect input, but can be used for several seconds of input (or much more or less, depending on how complicated you define the input).
Please be aware I haven't updated or worked with this bot in three years, so I can't do a whole lot more than point you in that direction and hope that it answers your question.
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Say, I don't know if this has been answered in the thread already, but I have a question about this pursuit problem.
It would appear to me that there is a certain scale+rotation symmetry in this problem. The condition after a certain time t matches the initial condition, just scaled down and rotated slightly. As such, it seems logical to me that the curvature of the path should not vary with where the animals are on the path. This is an obvious contradiction with the approximating figure and the known result, since constant curvature would imply circular paths. Does anyone have a relatively simple resolution to this contradiction?
Staring at the figure, I can almost see how this symmetry would imply the differential equation r' = r, characterizing the curve. Can someone put the last pieces together?
Edit: Aaaand, immediately after posting, I see that the scaling increases the curvature, just like scaling a circle down produces a circle of greater curvature. Derp.
I'll let my post stand and hang my head in shame.
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GenericMadScientist wrote:
Bobo the King wrote:
Some people here are incredibly smart mathematicians and I've never had any luck obfuscating an otherwise straightforward problem, so I imagine this challenge will not prove as difficult as I would hope.
Prove that
where n is a positive integer. (I'm actually pretty sure n can be any positive real number, if that helps.)
I hope I transcribed that correctly...
The binomial series tells us that, if |x| < 1, then we have
By multiplying each factor of the product by n and then dividing by n^k, and then substituting y = -x we get
As y → 1, the terms of the infinite sum monotonically decrease, and they all have the same sign, so we can interchange the sum and the limit and so the left hand side converges to the sum in Bobo's post, plus 1. Meanwhile if α > 0 then the RHS converges to 0. Now subtract 1 from both sides and set α to 1/n.
Absolutely correct!
And wow! I drew out a lurker! Welcome to the boards!
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Some people here are incredibly smart mathematicians and I've never had any luck obfuscating an otherwise straightforward problem, so I imagine this challenge will not prove as difficult as I would hope.
Prove that
where n is a positive integer. (I'm actually pretty sure n can be any positive real number, if that helps.)
I hope I transcribed that correctly...
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This is pretty funny, but this April Fools submission would have reached legendary status if you had included a photo of a homemade version of the final burger.
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Proposal to make the opening of the submission text or the publication text as follows:
Jennifer is a rebellious teenage orphan who is kindly adopted by the Barrows family. Upon arriving at their mansion, she ungratefully punches a hole in their wall, steals the Barrows' car, and drives through their garage door.
Clock Tower's core message is that teenage girls are the true monsters.
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Fantastic work, FractalFusion! It will take me a few read-throughs to absorb it all. What I especially love is that the probability of guessing a card correctly is substantially better than 50 percent. I can easily believe that with some (possibly sub-conscious) card counting, confirmation bias, and a little bit of luck 58 percent could turn into 70 percent in someone's mind.
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thatguy wrote:
I am pretty convinced the probability would tend to 50/50 as the pile of cards gets large.
Better than that, based on the two examples I outlined and a little reasoning, I'm pretty convinced that you can do better than 50/50 every time you play.
Suppose the piles are even (maybe at the start of the game). Your goal is to finish the game while at least breaking even. You wager the first card is red. If it is, great. If not, you lose a "point" but now wager that the next card will be red (since the red stack is necessarily shorter). This process repeats itself-- if the next card is red, great, you're back to even, and if not, you again wager that the next card will be red. The idea is that if you are at a deficit of red cards, the remaining red cards must eventually show up somewhere in the main pile. You're sure to eventually get back to 50/50 regardless of what cards you draw because even in the worst case scenario, you will be down by n (red) cards with exactly n cards remaining, bringing you back to 50/50.
thatguy wrote:
One way you could attack this analytically is a Pascal's Triangle-like formula that works out the N reds, M blacks scenario from the N-1 reds, M blacks scenario and the N reds, M-1 blacks scenario. That probability will be given by:
p(N, M) = (N/(N+M))*p(N-1, M) + (M/(N+M))*p(N, M-1)
You could then build up a triangle of results quickly, but it would still take hundreds of calculations to get to p(26, 26).
EDIT: okay that formula doesn't work because it just produces a triangle full of ones. I don't know whether the formula just needs to be tweaked or whether the idea is totally wrong.
It sounds like you're trying to prove this by induction. I had some trouble with that tactic since I don't yet know of a way to generate a list of permutations of n+2 cards from a list of permutations of n cards. As far as I can tell, the problem is nontrivial.
You have given me an idea, though. What if we ignore the first, say, 42 cards drawn and focus on the 2048 possibilities for the last ten cards to be drawn. (Note that by doing it this way, we no longer necessarily have an even number of red and black cards.) Assume that all of the player's net score comes from these last ten cards. Assign a probability to each possibility, multiply by the calculated final score, and then add them all up to find a good estimate on the overall efficiency of this guessing method. The error in this estimate should be under 0.5% because we are neglecting as negligible cases where the last eleven cards are all red or black.
I'll try that approach later today.
Thanks for the link! I wanted to figure out where I went wrong. Laurent Lessard's proof is (of course) more elegant than mine (though I think my variables were easier to keep track of). The key difference between our proofs is that he exploited a symmetry, stating that there is an equal probability that the triangular region defined by the x- and y-intercepts will lie in the first quadrant (greater than zero probability of concavity) as that it will lie in the third quadrant (zero probability of concavity). Lessard also computed the area of this triangle directly, avoiding the two inner integrals I produced.
After toying around with the bounds on my integrals, I found that the problem was in my bounds on yIV. The key error that I made was assuming yIV is only bounded by the condition that yIV be less than yII/xII*xIV. For certain values of yII, xII, and xIV, this allows yIV to exceed 1, which is not allowed. Correcting this bound would be messy, so Lessard had the right idea by exploiting symmetry.
Okay, new challenge. A few years ago I had a landlady who was a little batty. She was usually nice enough, but was really into a bunch of New Agey bullshit. For example, What the &%@#! do We Know? was one of her favorite movies. Lest you think I'm being too gossipy, she actually wrote and self-published a book about how to rent out a room in your house and included several anecdotes involving me.
Anyway, she had a long-distance boyfriend named Jack. Jack once boasted to her that he had some psychic abilities. He could shuffle a deck of cards, put it face down in front of him, and predict with roughly 70 percent accuracy whether the next card would be red or black, placing the the revealed card in a corresponding pile to the left or right of the deck, depending on its color.
Of course, my first instinct was that Jack was totally full of it. I actually still think he was, but on further reflection, I realized I might be able to approach Jack's number using just a little strategy and no psychic abilities whatsoever. To maximize the probability of choosing the next card correctly, I could keep track of which pile of revealed cards, red or black, was taller. My method would be to always assume the next card is from the shorter pile or if the two piles are equal, without loss of generality, assume the next card is red. I suspect that consciously or not, Jack was doing something similar.
So my question is What percentage of cards, on average, can you correctly guess using this strategy? I'll assume equal numbers of red and black cards in the deck for simplicity.
A few extreme cases are worth considering. One can always correctly guess the last card, no matter the size of the deck, guaranteeing that you can beat 50 percent on average. For an infinitely large deck, the probability is exactly 50 percent because each overturned card offers no information on the state of the next card. And for a deck of just two cards, you can correctly guess the color of the cards 75 percent of the time. The fact that even in the best-case scenario you do just five percent better than Jack's boast leads me to believe he was full of it.
From there, the math gets tricky. I'll cover just two more cases: a deck of four cards and a deck of six cards.
With four cards, there are six ways of arranging the deck. I'll use A and B instead of red and black for better readability. I'll also assume the player's default strategy is to pick A when the piles are equal. Here are how all six possibilities play out, correct guesses in bold:
{A,A,B,B} -- 3 correct
{A,B,A,B} -- 4 correct
{A,B,B,A} -- 3 correct
{B,A,A,B} -- 3 correct
{B,A,B,A} -- 2 correct
{B,B,A,A} -- 2 correct
This gives a total of 17 correct guesses for every 24 overturned cards, a 70.83 percent success rate.
Now I'll repeat this analysis for a deck of 6 cards, which covers 20 possible shufflings:
{A,A,A,B,B,B} -- 4 correct
{A,A,B,A,B,B} -- 4 correct
{A,A,B,B,A,B} -- 5 correct
{A,A,B,B,B,A} -- 4 correct
{A,B,A,A,B,B} -- 5 correct
{A,B,A,B,A,B} -- 6 correct
{A,B,A,B,B,A} -- 5 correct
{A,B,B,A,A,B} -- 5 correct
{A,B,B,A,B,A} -- 4 correct
{A,B,B,B,A,A} -- 4 correct
{B,A,A,A,B,B} -- 4 correct
{B,A,A,B,A,B} -- 5 correct
{B,A,A,B,B,A} -- 4 correct
{B,A,B,A,A,B} -- 4 correct
{B,A,B,A,B,A} -- 3 correct
{B,A,B,B,A,A} -- 3 correct
{B,B,A,A,A,B} -- 4 correct
{B,B,A,A,B,A} -- 3 correct
{B,B,A,B,A,A} -- 3 correct
{B,B,B,A,A,A} -- 3 correct
That's a total of 82 correct guesses for every 120 cards, a 68.33 percent success rate, already a little worse than Jack's claim.
And that's as far as I've gotten. Based on this very short trend (75 --> 70.83 --> 68.33), it appears that the success rate crashes fairly rapidly toward 50 percent as the deck size increases. In this sense, 52 would be "closer to infinity than to 2". Can anyone confirm this? What is the approximate probability of success using this strategy on a deck of 52 cards?
I suggest beginning by writing a computer program to confirm my results above and approximating the asymptotic behavior. Is it a negative exponential? An inverse power relation? Something else? Compute the first ten or so results, then extrapolate this out to 52 cards. (There are about 500 trillion different ways of arranging 52 cards distinguished only by their face color, which is at the upper limit of what can be calculated by supercomputers, so unless you can find an exact formula, it would be wise to do an extrapolation.)
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arflech wrote:
I ended up doing a numerical integration and got a figure a few standard deviations outside your Monte Carlo simulation, a bit over 92%.
Odd. I just did an integration myself and obtained a smaller figure, about 87%. I'll edit this post in a few minutes to show my work.
Edit: So I obtained a solution. It doesn't agree with my earlier Monte Carlo simulation or aflech's result, but at least we're all in the same ballpark. Here's my work.
First, define the following coordinates:
xI
yI
xII
yII
xIV
yIV
These correspond to the x- and y-coordinates of three of our points. The subscripts indicate which quadrant each resides in. NOTE: I found it much easier to work with all positive numbers. Even though xII is in the second quadrant, I am taking it to be the positive distance from the y-axis. Likewise for yIV and the x-axis. I encourage you to draw some pictures as you read my argument below.
We wish to find the probability that these three points form a concave portion of the quadrilateral. We may pick xII, yII, and xIV as we please and, through suitable choices of the remaining variables, concave portions can be admitted. Therefore, we may write
xII ∈ (0, 1)
yII ∈ (0, 1)
xIV ∈ (0, 1)
With yIV, we encounter our first nontrivial restriction. We must pick yIV such that the line adjoining points II and IV passes above the origin-- otherwise, no convex region can be created and the probability is zero. It can be shown that this condition is equivalent to
yIV ∈ (0, yII*xIV/xII)
Next, we have a restriction on xI. This must lie between 0 and the x-intercept defined by the line adjoining points II and IV. This condition can be written as
xI ∈ (0, yII*(xII+xIV)/(yII+yIV) - xII)
Only one restriction left. We demand that yI lie below the line adjoining points II and IV for the now-given value of xI. This condition is...
yI ∈ (0, yII - (yII+yIV)/(xII+xIV)*(xII+xI))
Welp, that's a monster. All this needs to be plugged into a sextuple-integral. I guess I'll write it out in monospaced text in the hopes that it will be more readable:
The integral of
the integral of
the integral of
the integral of
the integral of
the integral of
1 dy_I dx_I dy_IV dx_IV dy_II dx_II
as y_I ranges from 0 to y_II - (y_II + y_IV)/(x_II + x_IV)*(x_II + x_I)
as x_I ranges from 0 to y_II*(x_II + x_IV)/(y_II + y_IV) - x_II
as y_IV ranges from 0 to y_II*x_IV/x_II
as x_IV ranges from 0 to 1
as y_II ranges from 0 to 1
as x_II ranges from 0 to 1
I'm sure the integral isn't that hard to evaluate but after setting it up, I decided I'd spare myself that agony and used my calculator. It spit out an exact answer of (2*ln(2)-1)/12 or about 0.0322.
Now, that's the probability that just one side is concave. It can be easily seen that no more than one region of a quadrilateral that doesn't self-intersect (like this one) can be concave. As such, we simply take this probability and multiply it by four to find the probability that any of the four sides is concave. Subtract this value from 1 to obtain (4 - 2*ln(2))/3 ≈ 0.8712.
Hey, what can I say? I tried.
Edit 2: I forgot to mention that my prediction of P(one region concave) = 0.0322 does not agree with my Excel spreadsheet. That's showing a probability of something like 0.0228 + 0.0002. I have no idea if that would help anyone figure out the real solution.
The classic challenge, in which you have to determine the probability that a quadrilateral constructed from points selected at random in unit squares of each quadrant, is very difficult. I tried finding a closed-form solution but gave up when writing and evaluating the integrals proved difficult. I also had a hard time splitting up the problem into disjoint probabilities that could be added or multiplied.
So I cheated. I set up an Excel spreadsheet that picks 10,000 sets of eight numbers (our four coordinates) at random. I produced additional columns that returned booleans based on whether three adjacent points formed a concave or convex region as well as another column that checked whether all four were convex. I then refreshed the numbers about 30 times and took the mean and standard error of the output. I can now say with good confidence that the probability that the quadrilateral is convex is 0.9086 + 0.0004.
I may continue this analysis with an even larger set of random numbers, but in any case, you are more than welcome to check a computed result against my Monte Carlo method.
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PikachuMan wrote:
Whenever I use TAStudio, I notice that my input gets erased when frames get added. Plus, whenever I clone frames, it creates junk input. This was never a problem until 1.11.9. But it does do it on certain N64 games on as early as 1.11.4!
I am dependent on TAStudio, but when my input gets erased, I don't like it. It needs to be fixed.
If memory serves me, you'll want to uncheck "Recording Mode". If memory does not serve me, someone else can help you.
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OmnipotentEntity wrote:
The problem with the duck swimming in a straight line from the point where the fox's theta and the duck's theta are equivalent, is the fox can also head directly to that point and does not have to go the long way around.
The duck needs to lead the fox the long way around initially until the fox is committed.
I disagree. If the duck is outside the radius at which it can keep up with the angular speed of the fox (what I called rho above), it is surely going at a lower angular speed than the fox, meaning the fox will benefit most from continuing its path around the pond.
Suppose the fox is crafty enough to see what the duck is doing and reverses direction at some instant to go after it "the short way". Then the duck can instantly use the same strategy it had been using, just mirrored about an axis through the center of the pond and its current point. The fox surely can't keep up with it because it is now chasing it the long way around. If the fox continues to switch strategies, it will surely end up even farther away from the duck when the duck reaches land. This seems like a game theory problem to me.
By the way, I'm still a little confused about the duck's straight line strategy. In the instant before the duck reaches the shore, it would seem to me that the duck would do just a little bit better if it changed course to head straight for the shore. If it did so, it would end up a little bit ahead of the fox. Even though I understand all of the reasoning above and in my previous post, it seems very strange to me that there might be another strategy not based on a straight line that results in the duck being farther ahead of the fox.
I'll need to sit down and work this out in detail.
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I have an answer that's just a bit larger than rhebus's. I built off of his answer in which the duck follows two different strategies. As before, the duck starts by swimming such that it is moving diametrically opposite the fox at the same angular speed. I then built off of his suspicion that there is a better strategy for phase 2 than the straight-line dash to the shore. I based my strategy on two assumptions:
1) The duck should first move tangential to the circle it traced at the end of phase 1. This is because phase 1's strategy worked up to that point and I expect smoothness. Maybe this assumption is unfair.
2) The duck will end phase 2 moving directly toward the shore. This makes sense because once the duck is infinitesimally close to the shore, its best move should be to swim straight toward land, regardless of where the fox is at that time.
Based on these two assumptions as well as my demand that I be able to calculate the path length without resorting to Wolfram Alpha, I hypothesized that the duck's second strategy will be to swim in a new circle of radius (R - rho), where R is the radius of the pond and rho is radius of the duck's circle at the end of strategy 1. It was shown in solving for part 1's strategy that rho is v_1/v_2 * R, where v_1 is the duck's speed and v_2 is the fox's speed. (Sorry for failing to non-dimensionalize everything. We physicists get antsy when the units don't work out!)
By the end of its motion, relative to the center of the pond, the duck will have traced out an angle theta from the time it began strategy 2 to its final escape. Extend a line from the center of the pond through the duck's position at the start of strategy 2 all the way to the end of the pond. Using this line as well as the line from the center of the pond to the duck's escape point, you can see that this creates an isosceles triangle whose sides are R, R, and R-rho. I then used the law of cosines to obtain theta:
cos(theta) = 1 - 1/2*(1-v_1/v_2)^2
We need to know what angle was subtended by the duck's new circular path. I called this phi and from the isosceles triangle I drew above, you can easily see that
theta + 2*phi = π.
Next, I demanded that the fox barely miss the duck at the problem's conclusion. This criterion is:
(R - rho)*phi/v_1 = R*(π + theta)/v_2
After a little algebra (including substituting in for rho and phi), this reduces to
1/2*(v_2/v_1 - 1)*(π - theta) = (π + theta)
Finally, I found it easier to work with a new variable instead of v_1 and v_2. I decided to define x as so:
x = 1 - v_1/v_2
and correspondingly
v_2/v_1 = 1/(1-x).
So let me summarize these equations...
cos(theta) = 1 - 1/2*x^2 (Law of cosines on isosceles triangle defined by center of pond and duck's circular path.)
theta + 2*phi = π (Angles on this isosceles triangle must add to 180o.)
1/2*x*(π - theta) = (1-x)*(π + theta) (Fox and duck end at same point.)
v_2/v_1 = 1/(1-x) (Solve for this.)
Using the first three equations together, I was able to eliminate theta and phi, leaving a single transcendental equation in x. I then plugged this into my calculator to find a numerical solution. The solution it gave was x = 0.7698. Finally, I computed v_2/v_1, as written above, to find the ratio of the fox's speed to the duck's is 4.344. This is, of course, a little bit faster than π+1 = 4.142.
As rhebus did, I openly wonder if there's a yet faster value, based on a different strategy for phase 2.
Edit: Welp, I took a look at another message board with the solution. It turns out that there is another strategy that allows for an even faster fox. I'll offer a hint: the new strategy is much simpler than the one I just came up with (assuming I'm reading it correctly).
Edit 2: I just realized my answer above cannot be optimal for a mind-numbingly obvious reason: the curve minimizing the duck's time spent in strategy 2 will be a straight line. In other words, the duck could have swam in a straight line toward the same point I deduced above, saving some time (though even this would not be optimal).
Using this condition as well as one of the criteria I outlined above, the optimal solution should be apparent.
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thatguy wrote:
Well, to a physics graduate and amateur/recreational mathematician like myself, it's not obvious why the Riemann Hypothesis would imply that the primes are "randomly" distributed (for a suitable definition of "random"). The only link that I can understand that it has to primes at all is the Euler identity:
infinite_sum(1/n)^s = infinite_product(p^s/p^s-1)
where p are the prime numbers. It's not obvious how you go from there to saying that the zeta function has anything to do with the distribution of primes.
As a fellow physics graduate, ya got me. I only know what I see on popular websites and online discussions like this one. I suggest you do the same thing I did and Google "Riemann hypothesis distribution of primes" and peruse what pops up. I think it's fascinating stuff but I don't have much patience for it since it's difficult for me to understand, not all that practical, and being studied by people orders of magnitude smarter than I am.
In the right up corner, there are four ceiling tiles to prevent Zook from jumping above screen. It is actually pretty easy to pass through them but still, nothing in particular happens. I wonder if they did put these four ceiling tiles as a quick workaround to some bad glitch, then later were fixed by an actual improvement of the code: that would explain why they did not implement such ceilings in level 6 as well. I still hope to discover some way to walk above the screen like in Super Mario Bros.
where n is a positive integer. (I'm actually pretty sure n can be any positive real number, if that helps.) I hope I transcribed that correctly...
By multiplying each factor of the product by n and then dividing by n^k, and then substituting y = -x we get
As y → 1, the terms of the infinite sum monotonically decrease, and they all have the same sign, so we can interchange the sum and the limit and so the left hand side converges to the sum in Bobo's post, plus 1. Meanwhile if α > 0 then the RHS converges to 0. Now subtract 1 from both sides and set α to 1/n.