1 2
80 81 82
132 133
Posted: 1/22/2017 1:54 AM
Quote
Denial140
Active player (259)
Joined: 8/18/2013
Posts: 145
Location: location, location!
Arcade TAS of 2023
Warp wrote:
So, two questions: 1) When one is integrating something like sin(x)cos(x) dx, how and why does one come up with one of the above answers, and not the others? (It has been so long since I did any integration in high school that I don't even remember how you integrate sin(x)cos(x) dx.) 2) Small challenge: For which values of C1, C2 and C3 does it hold that sin2(x) / 2 + C1 = -cos2(x) / 2 + C2 = -cos(2x) / 4 + C3
1) For a, let u = sin(x), then du = cos(x)dx, and so the integral of sin(x)cos(x)dx = the integral of u du, which is u2/2+C, or sin2(x)/2+C. For b, let u = cos(x), then du = -sin(x)dx, and so the integral of sin(x)cos(x)dx = the integral of -u du, which is -u2/2+C, or -cos2(x)/2+C. For c, sin(2x) = 2sin(x)cos(x), so the integral of sin(x)cos(x)dx is the integral of sin(2x)/2 dx. Then let u = 2x, so du = 2dx, meaning the integral of sin(x)cos(x)dx is 1/4 of the integral of sin(u)du, so the integral of sin(x)cos(x)dx is -cos(u)/4+C, or cos(2x)/4+C. 2) It is shown in part 1 that there are constants C1, C2 and C3 such that sin2(x) / 2 + C1 = -cos2(x) / 2 + C2 = -cos(2x) / 4 + C3, as they are all antiderivatives of the same function. Therefore, substituting in any value for x will work. I choose x=0 because I feel like it. This gives C1 = -1/2+C2 = -1/4+C3. Therefore, sin2(x) / 2 + C1 = -cos2(x) / 2 + C2 = -cos(2x) / 4 + C3 for any C1, C2, C3 such that C2 = C1+1/2 and C3 = C1+1/4.
Current TAS: [SNES] Jelly Boy [NES] Street Fighter 2010
Posted: 2/1/2017 6:35 PM
Quote
Warp
Banned User, Former player
Joined: 3/10/2004
Posts: 7698
Location: Finland
One method for tesselating a sphere into triangles that are all the same shape and at least approximately equilateral, is to take a platonic solid, like the icosahedron, and subdivide each side into equal triangles. For example, each side of the icosahdedron could be divided into four triangles like this: Then the newly created vertices are moved away from the center of the sphere to its surface, to form a new 80-sided polyhedron. This 80-sided polyhedron is not a platonic solid, though, and it's not necessarily trivial to immediately see why. So why is it not a platonic solid? My hypothesis is that the resulting triangles (the ones that use the previously-existing vertices as one of their vertices) are not equilateral anymore. So, the problem: A side of an icosahedron is divided into four equal triangles, as depicted above. The three new vertices are moved away from the center of the circumscribed sphere, to its surface. What are the lengths of sides of these four new triangles?
Posted: 2/1/2017 7:26 PM
Quote
thatguy
Editor
Joined: 11/3/2013
Posts: 506
(Just sayin', I saw your drawing and immediately the Zelda theme started playing in my head. Dun dun, da da-da-da-da-da...) It's pretty trivial that this is not a platonic solid because not all the vertices are the same. There are five triangles meeting at each of the twenty "original" vertices, and six triangles meeting at each of the sixty "new" vertices. Clearly, since the shape is convex (ie angles round a vertex are less than 360) some of the triangles are pointier than equilateral triangles at the "new" vertices.
Posted: 2/1/2017 7:56 PM
(Edited: 2/2/2017 4:26 AM)
Quote
OmnipotentEntity
He/Him
Player (36)
Joined: 9/11/2004
Posts: 2623
Warp wrote:
So why is it not a platonic solid? My hypothesis is that the resulting triangles (the ones that use the previously-existing vertices as one of their vertices) are not equilateral anymore.
This is correct.
Warp wrote:
So, the problem: A side of an icosahedron is divided into four equal triangles, as depicted above. The three new vertices are moved away from the center of the circumscribed sphere, to its surface. What are the lengths of sides of these four new triangles?
If the radius of the circle is 1 then the edge length (A) of an inscribed icosohedron is csc 2pi/5 or 1.05146... The new vertexes are equidistance from the old ones, so we can use a few triangles to find the length of b, which is the length of one of the outer edges. The angle of the triangle formed by the radius, and the perpendicular bisector of A can be found to be 0.55357. B is the chord across this angle, which is also the length of one of the outer edges. B = 2*r sin (theta/2) = 0.54653... The internal triangle is a bit more tricky. The internal triangle bordered by the internal edges (C) is an equilateral triangle, because the A to B transformation is symmetric on all three corners. If we visualize looking directly into the middle of a face head-on the path of our eye to the center of the icosahedron and call that leg D We can find the length of D without the triforce extruded using the length from the middle of the face to the outer edge (0.6071...). From this we find D = 0.794654... Now we can calculate D' the length from the center to the middle of the extruded triforce face. When looking down at the extruded face we can see that the middle triangle has grown slightly, and a projection of it face on very slightly overhangs the original triangle. By how much? Well, we can get the distance from the edge A to the corner of the interior triangle with another Pythagorean argument (A/2)^2 + E^2 = B^2 => E = 0.149349..., but this isn't the shortest distance between the faces. We need to know the angle between the center of a face and the center of an edge. We can find this through the dihedral angle, which is arccos(sqrt(5)/3) = 2.4119... radians, which we can chop in half and subtract from pi/2 to get the angle we're looking for: 0.36486... Finally, we can use this angle to get the distance between the old and new faces (E'). It's 0.139518... Now we're armed with enough information to find C. D' = D + E' = 0.934172... So the distance from the center of the extruded face to a corner is 0.356822... which makes C = 0.618034... B = 0.54653... which isn't C, so it's not equilateral. EDIT: Performing a sanity check C*2 should be > A, but it's not, I'll review my work and see if I can find the error. EDIT2: Located a calculator error, fixed. Now passes sanity checks. EDIT3: Apparently, I cannot Pythagoras, fixed more calculation errors.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 2/1/2017 8:35 PM
Quote
thatguy
Editor
Joined: 11/3/2013
Posts: 506
Onto actually solving the problem: The neat thing about this is that it can actually be reduced to a two-dimensional problem, which is much easier than trying to think in three dimensions. Consider the pentagon formed by the free edges of the five faces surrounding a single vertex. They have a circumscribed circle which is just a planar slice of the sphere circumscribing the icosahedron. What is the radius of this circle? Well, if the side of the pentagon is 1 then the radius of the circle r is 1/(2*sin(pi/5)) (difficult to explain without a diagram, and a worthwhile exercise). When you turn the icosahedron's triangles into those triforces, the pentagon in the 2-d slice becomes a decagon inscribed in the same circle. By similar argument to above, this decagon has a side-length 2*sin(pi/10)*r = sin(pi/10) / sin(pi/5). That is then the side-length of the outer edges of the triforce - those triangle edges which are not part of the central triangle. The side-length for those central triangles - well that's rather harder as I suspect you do have to do some calculations in 3-d for that. I'll leave it to a more able geometer.
Posted: 2/2/2017 4:23 AM
Quote
OmnipotentEntity
He/Him
Player (36)
Joined: 9/11/2004
Posts: 2623
An easier argument I came up with while discussing this problem with a few other people. We can construct a sphere that intersects a face of the icosahedron at all of the points where the triforce self-intersects. Then we can simply use an argument that as the triforce is extruded it scales linearly. It's a much shorter argument and gives the same answer as my other approach. So that's good! b = 0.546533... c = 0.618034... Raw Mathematica Dump: Chord[r_, theta_] := 2*r*Sin[theta/2] a = Simplify[Csc[2*Pi/5]] theta1 = Simplify[ArcSin[a/2]] b = Simplify[Chord[1, theta1]] c = Simplify[(a/2)/(1 - Sqrt[b^2 - (a/2)^2])] a = 2 Sqrt[2/(5 + Sqrt[5])] theta1 = ArcSin[Sqrt[2/(5 + Sqrt[5])]] b = 2 Sin[1/2 ArcSin[Sqrt[2/(5 + Sqrt[5])]]] c = Sqrt[2/(5 + Sqrt[5])]/(1 - Sqrt[-(2/(5 + Sqrt[5])) + 4 Sin[1/2 ArcSin[Sqrt[2/(5 + Sqrt[5])]]]^2])
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 2/3/2017 12:21 AM
(Edited: 2/13/2017 8:53 AM)
Quote
Patashu
He/Him
Joined: 10/2/2005
Posts: 4014
Was bored and came up with a math challenge for myself: To roll a y sided die is to randomly select a value from 1 to y inclusive. As a probability distribution, has an expected value of (0.5+y/2), a variance of (y^2 - 1)/12 and a skewness of 0. Normally dice have a positive, integer number of sides, but I would like to extend this function to all real values of y >= 1, and the expected value for non-integer values of y should still be (0.5+y/2). For example, if I roll a 2.5 sided die, I should be able to roll either 1, 2 or 3 with non-equal weightings, and the expected value should be (0.5+2.5/2). Additionally, this function should be indistinguishable from normal die rolling whenever y is an integer. 1) What is the algorithm for rolling a real sided die that satisfies these criteria? With what probability does it roll each side? 1.5) Extend also to negative integer y, then to negative real y. 2) Does this algorithm still have a variance of (y^2 - 1)/12 for all values of y? If not, can you come up with a second algorithm that does NOT have expected value of (0.5+y/2) all the time but DOES have variance of (y^2 - 1)/12 all the time? How about skew? EDIT: Fixed variance formula. Also, here's my answer for 1: xDy, where x is integer and y real: Values from 1 to floor(y) have 1.0 weighting, value floor(y)+1 has floor(y)*frac(y)/(floor(y)-frac(y)+1) weighting. select randomly via weighting for each roll. Examples: 1.5 (should have EV 1.25) is a 1.0 weighting for 1, 1/3 weighting for 2. e.g. 3/4 chance to roll 1 1/4 chance to roll 2. EV is 1.25. 1.25 (should have EV 1.125) is a 1.0 weighting for 1, 1/7 weighting for 2, e.g. 7/8 chance to roll 1 1/8 chance to roll 2. EV is 1.125. 2.5 (should have EV 1.75) is a 1.0 weighting for 1, 1.0 weighting for 2, 2/5 weighting for 3, e.g. 5/12 chance to roll 1 5/12 chance to roll 2 2/12 chance to roll 3. EV is 1.75. 2.25 (should have EV 1.625) is a 1.0 weighting for 1, 1.0 weighting for 2, 2/11 weighting for 3 e.g. 11/24 chance to roll 1 11/24 chance to roll 2 2/24 chance to roll 3. EV is 1.625. 3.5 (should have EV 2.25) is a 1.0 weighting for 1/2/3, 3/7 weighting for 4 e.g. 7/24 chance to roll 1/2/3 3/24 chance to roll 4. EV is 2.25.
My Chiptune music, made in Famitracker: http://soundcloud.com/patashu My twitch. I stream mostly shmups & rhythm games http://twitch.tv/patashu My youtube, again shmups and rhythm games and misc stuff: http://youtube.com/user/patashu
Posted: 2/3/2017 5:48 AM
Quote
OmnipotentEntity
He/Him
Player (36)
Joined: 9/11/2004
Posts: 2623
You can't use just a uniform distribution. (ie, Ceil(U((0, y]))) If we check on the given 2.5 sided die this will give E = 1.8 vs 1.75 ideal. This feels qualitatively like it's going to require something similar to a Fourier transform, which means that likely you're going to be unable to limit the rolls to just 1, 2, and 3. But in fact any integer value will be able to be rolled, just some much less than others. If you really must have the E and also the Variance and skew. I'm not 100% sure it's even possible. But looking into similarities between this and Fourier transforms would be my first port of call.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 2/12/2017 6:23 AM
Quote
Bobthefloater
He/Him
Joined: 11/20/2015
Posts: 31
OmnipotentEntity wrote:
You can't use just a uniform distribution. (ie, Ceil(U((0, y]))) If we check on the given 2.5 sided die this will give E = 1.8 vs 1.75 ideal.
At least, you can't use a uniform distribution from 0 to y. Also, what is that variance formula? It gives a variance for a dice with 1 side equal to a quarter, when there is only 1 output value (and hence zero variance).
Posted: 2/13/2017 5:54 AM
Quote
FractalFusion
Editor, Skilled player (1938)
Joined: 6/15/2005
Posts: 3246
Glitchy TAS of 2013 Gameboy TAS of 2013 PSX TASer of 2010
Something interesting that happened with one of the Riddler "problems" last week (well, more like participatory games). (from https://fivethirtyeight.com/features/can-you-rule-riddler-nation/ )
A classic participatory game-theory problem: Submit a whole number between 1 and 1,000,000,000. I’ll then take all those numbers and find the average submission. Whoever submits the number closest to ⅔ of the mean of all of the submitted numbers wins. Submit your number
If everyone plays by the optimal solution, then the solution is obvious. Everyone chooses the number 1 (because it's the only integer x between 1 and 1,000,000,000 satisfying: x is closer to ⅔x than any other integer between 1 and 1,000,000,000) and everyone ties for the win. As you probably expected, the fact that this is participatory means that it's not so much a game-theory problem as it is a social experiment. And guess what happened. (from https://fivethirtyeight.com/features/can-you-save-the-drowning-swimmer/ )
The mean was 195,921,656*, or nearly 20 percent of the upper limit of the range. Jeffrey, our winner and resident Nostradamus, chose 196,352,731.
*(maybe the Riddler meant "⅔ of the mean was 195,921,656".)
Following that logic all the way down, the unique equilibrium is that every player submits the lowest possible number — in this case, the number 1. No one can get any closer to ⅔ of the average, and everyone ties for the win. Suffice it to say, we did not arrive at that equilibrium. In this case, the trolls prevailed. Eighty-one “solvers” submitted an answer of 1,000,000,000 and an additional 25 submitted 999,999,999 — answers that can literally never win*. Their submitted justifications ranged from “to throw off everyone else’s logical calculation” to “game theory can’t handle anarchy” to “I just want to watch the world burn.” And burn it did. Game theory is not well-equipped to deal with modern trolldom.
*(assuming that at least one other person chooses a number between 333,333,336 and 999,999,998.) You can see that deliberately uncooperative behavior is by no means restricted to political movements, internet communities, and Twitch Plays Pokemon. Even math games are not immune. (The other participatory game about deploying troops to claim castles was a little more grounded in strategy. It helps that a single choice doesn't instantly troll everyone.)
Posted: 2/13/2017 8:50 AM
Quote
Patashu
He/Him
Joined: 10/2/2005
Posts: 4014
Bobthefloater wrote:
Also, what is that variance formula? It gives a variance for a dice with 1 side equal to a quarter, when there is only 1 output value (and hence zero variance).
Ah, you're right. It should probably be (y^2 - 1)/12 instead. I'll edit the post.
My Chiptune music, made in Famitracker: http://soundcloud.com/patashu My twitch. I stream mostly shmups & rhythm games http://twitch.tv/patashu My youtube, again shmups and rhythm games and misc stuff: http://youtube.com/user/patashu
Posted: 2/14/2017 5:59 PM
Quote
Warp
Banned User, Former player
Joined: 3/10/2004
Posts: 7698
Location: Finland
In this Numberphile video Matt Parker proves that the ratio between consecutive Fibonacci numbers approaches the golden ratio. He does this by noting that the definition of the Fibonacci numbers is that each number is the sum of the two previous numbers, and then assuming that as we go along the sequence, the ratio between two consecutive pairs of numbers approaches the same value, ie, that Xn/Xn-1 = Xn-1/Xn-2, and then solves what that ratio must be so that the equality holds (which, of course, is the golden ratio). But isn't he just assuming that the ratios of consecutive Fibonacci numbers converges to a particular value? He is also assuming that the ratios of two consecutive pairs approach each other as we advance in the sequence. On which basis can these assumptions be made? Couldn't it be perfectly possible that the ratios diverge, or oscillate, as we go along the series? Can the convergence simply be assumed? Shouldn't it be proved?
Posted: 2/14/2017 7:25 PM
Quote
Nickolas
Joined: 7/13/2014
Posts: 36
Warp wrote:
In this Numberphile video Matt Parker proves that the ratio between consecutive Fibonacci numbers approaches the golden ratio. He does this by noting that the definition of the Fibonacci numbers is that each number is the sum of the two previous numbers, and then assuming that as we go along the sequence, the ratio between two consecutive pairs of numbers approaches the same value, ie, that Xn/Xn-1 = Xn-1/Xn-2, and then solves what that ratio must be so that the equality holds (which, of course, is the golden ratio). But isn't he just assuming that the ratios of consecutive Fibonacci numbers converges to a particular value? He is also assuming that the ratios of two consecutive pairs approach each other as we advance in the sequence. On which basis can these assumptions be made? Couldn't it be perfectly possible that the ratios diverge, or oscillate, as we go along the series? Can the convergence simply be assumed? Shouldn't it be proved?
No, not really. You may as well ask, "Shouldn't he start with the Peano postulates and go from there? Isn't he just assuming you can add and multiply numbers? Shouldn't it be proved?" As in everything, context is important. The context of the video is one in which you only need algebra to derive interesting and useful math, such as the relationship between Fibonacci-like sequences and the golden ratio. Formally defining the idea of series convergence and then formally proving that the ratio of successive terms converges is beyond the scope of the video. (And while it is technically possible to use nothing but algebra to show that the square of the difference of successive ratios is always decreasing -- which algebraically hints at the idea that the ratio of successive terms is approaching a fixed value -- why bother? It's a 5-minute algebra video, not a textbook.) Furthermore, the way a lot of real math is done by real mathematicians is to just sort of assume something that seems like it might be interesting or true, and to see where it leads. Afterwards you can always go back and build a framework that formally justifies your assumptions. Or not, maybe you were wrong. Furtherfurthermore, one could also argue that the point of videos like this is to inspire curiosity. If the video makes you asks questions, then perhaps you should try to answer them. (Mission accomplished!)
Posted: 2/14/2017 8:07 PM
Quote
thatguy
Editor
Joined: 11/3/2013
Posts: 506
You raise an interesting point, actually. The equation he solves has two solutions, one (the correct answer) is the golden ratio (1+sqrt(5))/2 but he has conveniently ignored (1-sqrt(5))/2. To prove convergence, you could work out the "error" (difference to the golden ratio) of Xn+1/Xn relative to Xn/Xn-1, and then show that that error is smaller by a large enough margin that as n goes to infinity error does indeed go to zero.
Posted: 2/14/2017 8:14 PM
Quote
Warp
Banned User, Former player
Joined: 3/10/2004
Posts: 7698
Location: Finland
Nickolas wrote:
You may as well ask, "Shouldn't he start with the Peano postulates and go from there? Isn't he just assuming you can add and multiply numbers? Shouldn't it be proved?"
The convergence of the ratio between two consecutive Fibonacci numbers is most certainly not an axiom. I don't think the argument "this is a convergent series, and I don't need to prove it because I don't need to prove the Peano postulates either" would fly. On what basis can you simply assume that the series converges to a specific value?
Posted: 2/14/2017 9:04 PM
Quote
FractalFusion
Editor, Skilled player (1938)
Joined: 6/15/2005
Posts: 3246
Glitchy TAS of 2013 Gameboy TAS of 2013 PSX TASer of 2010
thatguy wrote:
To prove convergence, you could work out the "error" (difference to the golden ratio) of Xn+1/Xn relative to Xn/Xn-1, and then show that that error is smaller by a large enough margin that as n goes to infinity error does indeed go to zero.
Indeed the error between Xn+1/Xn and Xn/Xn-1 converges to 0. In fact, Xn+1/Xn - Xn/Xn-1 = (-1)n/(XnXn+1). Proof: We prove Xn+1Xn-1 - (Xn)2 = (-1)n (*) by induction. First, X1=1, X2=1, X3=2, so X1X3 - X22 = 1 = (-1)2. Assume that Xn+1Xn-1 - (Xn)2 = (-1)n is true. Then Xn+2Xn - (Xn+1)2 (Xn+1 + Xn)Xn - (Xn+1)2 = Xn+1Xn + (Xn)2 - (Xn+1)2 = Xn+1(Xn - Xn+1) + (Xn)2 = -Xn+1Xn-1 + (Xn)2 = -(-1)n = (-1)n+1. So we have proved statement (*). Then Xn+1/Xn - Xn/Xn-1 = (Xn+1Xn-1 - (Xn)2)/(XnXn+1) = (-1)n/(XnXn+1), Q.E.D. Note that Xn+1/Xn = 1 + sum{2≤i≤n}((-1)i/(XiXi+1)); because the summation is alternating and the terms converge to 0, the sum converges as n→∞.
Posted: 2/15/2017 9:06 PM
Quote
Warp
Banned User, Former player
Joined: 3/10/2004
Posts: 7698
Location: Finland
It takes Alex 3 hours to walk from town A to town B, and it takes Beatrice 5 hours to walk from town B to town A. If they both start walking at the same time (and they both walk at a constant speed), how long does it take for them to meet?
Posted: 2/15/2017 9:26 PM
Quote
ShadowWraith
Joined: 8/7/2006
Posts: 344
Distance between towns = D Alex's speed = D/3 Beatrice's speed = D/5 Distance travelled by Alex = D*t/3 Distance travelled by Beatrice = D*t/5 D = D*t/3 + D*t/5 1 = t/3 + t/5 = 8t/15 t = 15/8 = 1.875 hours.
Posted: 2/17/2017 7:04 PM
Quote
thatguy
Editor
Joined: 11/3/2013
Posts: 506
This puzzle, and a solution/explanation, appears in the film Little Big League: https://www.youtube.com/watch?v=o7jVL_JSWgI&index=45&list=PLmNp3NTX4KXIqtjt2AKOr4DUIxn64xj8G
Posted: 2/17/2017 7:24 PM
Quote
thatguy
Editor
Joined: 11/3/2013
Posts: 506
A fiendish problem I came across recently (it's an old riddler problem: after seeing the Fractalfusion link I went on an archive binge.) A duck is swimming in a circular pond, with a fox standing on the bank. The fox cannot swim, and the duck can fly but cannot take off from the water - it has to get to dry land first. Once ashore it can immediately take flight and escape permanently. The fox is hungry and has the duck in its sights. It will run around the circumference of the pool to try to keep the duck trapped. The duck wants to escape, but in order to do so has to get to the shore in such a way that the fox won't be waiting for it when it arrives there. How much faster must the fox be able to move than the duck, in order to ensure that the duck cannot escape? The naive answer is pi times - assume that the duck starts from the centre of the pond and swims directly away from the fox, then the ratio of speeds is the ratio of a circle's radius to the length of a semicircular arc. However, with a cleverer strategy the duck can escape a rather faster fox.
Posted: 2/18/2017 12:23 AM
(Edited: 2/21/2017 4:03 PM)
Quote
OmnipotentEntity
He/Him
Player (36)
Joined: 9/11/2004
Posts: 2623
I do not know the best strategy for the duck, but here's one that comes to mind. The duck will always flee the fox. And the fox will always seek to be in the closest location on the circle to the duck's position. So essentially, we can model the location of the fox and the duck as a system of differential equations in polar coordinates with the following variables: Fox's speed v_f, duck's speed v_d, and radius of the pond R. Assuming we clamp the θ of the duck and the fox to (-pi, pi] rather than allowing it to take any value, the fox's position can be given as: dθ_f = v_f/R * Re{(-1)^(H(dθ_f - dθ_d))} dr_f = 0 θ(t = 0) = 0 r(t = 0) = R (Where H(x) is the Heaviside step function, and Re{z} denotes the real part of z.) The duck's position is a bit more complicated: First, we need to find the direction in which the duck is heading. For this polar coordinates are a bit of a hindrance, so we have to pop back into Cartesian for a bit. If the duck's position is a vector given by P_d = (r_d cosθ_d * i + r_d sinθ_d * j) and the fox's is P_f = (R cosθ_f * i + R sinθ_f * j) then the angle between the two is given by: arccos(P_d·(P_f-P_d)/(||P_d||||(P_f-P_dr)||)) which simplifies down to: arccos((r_d - R cos(θ_d - θ_f))/sqrt(R^2 + r_d^2 - 2*R*r_d*cos(θ_d - θ_f))) So dr_d = v_d*(r_d - R cos(θ_d - θ_f))/sqrt(R^2 + r_d^2 - 2*R*r_d*cos(θ_d - θ_f)) dθ_d = v_d*sqrt((R^2*sin^2(θ_d - θ_f))/(R^2+r_d^2-2*R*r_d*cos(θ_d - θ_f)))/r_d r_d(t = 0) = 0 θ_d(t = 0) = 0 Now we can solve this system of equations. But I doubt they have an explicit solution. I have mathematica chewing on it right now. pd = {rd[t] Cos[thetad[t]], rd[t] Sin[thetad[t]]} pf = {r Cos[thetaf[t]], r Sin[thetaf[t]]} thetadir = FullSimplify[ArcCos[pd.(pf - pd)/(Norm[pd]*Norm[pf - pd])]] DSolve[{ rd'[t] == vd*Sin[thetadir], thetad'[t] == vd*Cos[thetadir], thetaf'[t] == vf/r, rd[0] == 10^-5, thetad[0] == Pi, thetaf[0] == 0 }, {rd, thetad, thetaf}, t, Assumptions -> { t \[Element] Reals, t >= 0, r \[Element] Reals, r > 1, vd \[Element] Reals, vd > 0, vf \[Element] Reals, vf > vd, rd[t] \[Element] Reals, thetaf[t] \[Element] Reals, thetad[t] \[Element] Reals }] EDIT: Screwed up in calculating the angle of the duck direction. Revised. EDIT2: Fixed mathematica stuff.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 2/18/2017 1:50 PM
Quote
rhebus
Joined: 2/19/2010
Posts: 248
I think the answer is π+1. Essentially, there are two phases to the duck's movement: in the first phase, the duck tries to stay on the radius directly opposite the fox, no matter what the fox's movements are. The duck continues like this until she can't move fast enough to keep the fox directly opposite. At this point, she enters a second phase, where she travels directly to the water's edge, along the shortest straight line path. The first phase lasts as long as the duck can move faster around the pond (in the φ-direction, in polar coordinates) than the fox can. Call the duck's speed d and the fox's f, measured in units defined such that the radius of the pond is 1. If the duck is r from the centre, then the duck can move around the pond faster than the fox (and thus stay directly opposite him) as long as r<d/f. Having sustained this movement, responding to whatever movements the fox makes by staying directly opposite him, the duck can now start the second phase in the best possible position: she is at a point d/f along a radius toward the edge, with the fox at the end of the opposite radius. Since she can no longer outmanoeuvre him, she might as well make a run for it. She has to move a distance of (1 - d/f) at a speed d, while he has to move a distance of π at a speed f. At the point where the fox just manages to catch the duck, they will cover these distances in the same time, so (1 - d/f)/d = π/f Simplifying: (1/d - 1/f) = π/f f/d - 1 = π f/d = π+1 I suspect it might be possible to have an even better strategy in phase 2 than the straight-line dash, but I can't see it myself.
Posted: 2/18/2017 4:19 PM
Quote
Bobo_the_King
Player (79)
Joined: 8/5/2007
Posts: 865
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.
Posted: 2/19/2017 4:46 AM
Quote
OmnipotentEntity
He/Him
Player (36)
Joined: 9/11/2004
Posts: 2623
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.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 2/20/2017 8:42 AM
Quote
Bobo_the_King
Player (79)
Joined: 8/5/2007
Posts: 865
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.
1 2
80 81 82
132 133