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If the fox simply seeks to the point on the circle that the duck is closest to, rather than trying to follow the direction the duck is going, then the contradiction goes away.
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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.
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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.
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In Dust Man's stage, with the room where all of the blocks fly in slowly, I was under the possibly mistaken impression that it was both possible and faster to make the blocks fly into and block off the extreme right side of the screen.
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evknucklehead wrote:
Invariel wrote:
Though, I must admit a certain level of frustration with Giles. Not looking forward to leading Thor around by the nose either.
At least Thor doesn't need to be led very long. Just from his spot in Voloh to the exit, then from where he stops on the field back to Voloh's entrance. It's the main reason for getting the tablet outside Phillips Castle before entering Voloh, otherwise you need to lead Thor to the castle grounds as well.
Thor paths by himself for the first bit as well.
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This might be potentially related to this? Chrono Trigger SRAM Glitch
But this uses midframe resets, and I don't know if hitting a single instruction in a reset is... plausible.
I also don't know if the bridge is a valid destination from his last save point and this one.
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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.
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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])
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Wobmiar wrote:
If I understand well, then, if PURELY THEORICALLY I find someone's TAS and find a 1 frame improvement, this frame being a good frame save (when you talk about improvement in gameplay, let's assume this frame is an improvement), but everything else is the same... It's ok? Not that I want to do that or that I accuse any work here to be that.
Yes. If it contains an improvement that turns a 10000 frame movie into a 9999 frame movie, provided the quality of the gameplay is similar, then that improvement should be accepted.
For instance, SMB1.
This movie contains 17840 frames.
This movie contains 17839 frames.
If the input is mostly the same, list the other user as a co-author. If you want to be a good guy, just let him/her know about the improvement. He/She might already know about it and might be saving it for a bigger improvement.
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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.
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I was given an unfair percentage of the credit for work that was done by a large team of talented members.
Honestly, that comes with the territory of being the "face" of an operation. If you're not going out of your way to take credit then I don't see a foul.
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Warp wrote:
Warp wrote:
Speaking of the Riemann zeta function, I have had for some time now an interest in understanding the Riemann hypothesis. Just out of curiosity, as a little "hobby project". I was wondering if someone could help me with it.
In order to understand the hypothesis, I first need to understand the Riemann zeta function. In order to understand said function, I need to understand analytic continuations. And the topic already becomes really complicated (not to talk about how complicated the zeta function itself is.)
While that is indeed a good starting point for understanding qualitatively what the concept of an analytic continuation is, it's not sufficient. What you really need is to take a course in complex analysis.
I don't know your mathematical background, but you only need up until Multivariable Calculus and the basics of Complex Numbers (polar form, roots of unity, etc) before you can tackle Complex Analysis. I've been eyeing taking the Complex Analysis course at my university, so I've watched a lecture series on Complex Analysis on Youtube. Specifically, the one starting from this video: https://www.youtube.com/watch?v=exTsBIQoxcI
You should probably start with this one though, it's more indepth and has what you're interested in (analytic continuations): https://www.youtube.com/watch?v=BruPj2mUGMo&list=PLun8-Z_lTkC5wjZ-8TH99y3htILDwlji5
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The exact value of the probability for the cards is:
4610507544750288132457667562311567997623087869 /
284025438982318025793544200005777916187500000000
to 30 decimal places:
0.0162327274671946367481295565133
I solved it by using the OEIS, apparently this problem is related to the Dinner-Diner matching problem.
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Because 2100 *isn't* a leap year, we can have a 40 year gap in the otherwise regular 28 year cycle of the calendar. Which means the only possible years are 2072, 2076, 2080, 2084, 2088
Of those 5, we can check each and eliminate the ones that cycled in 1900, 1800, etc in the same way.
2080 is the same as 1872 which cycled 40 years through 1900.
2084 is the same as 1876 which cycled 40 years through 1900.
2088 is the same as 1880 which cycled 40 years through 1900.
2076 is the same as 1772 which cycled 40 years through 1800.
2072 is unique in that it couldn't cycle through 1900, 1800. It normally would have done the 40 year cycle in 1700, but because England adopted the Gregorian calendar in 1752 I guess it doesn't count.
This is what 2072's calendar looks like:
https://www.timeanddate.com/calendar/?year=2072&country=1
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Xander wrote:
I originally planned to do a 120 star TAS except I just didnt want to have to deal with estimating starbit collection. There are 2 points in an rta run where you have to stop and grind to 999 starbits except it is possible that in a TAS you might not have to. That's only a maybe and if I ran the risk of not grinding and i get to the end and it turns out I did need to grind, I would have to redo most of the TAS.
There's an easy way to determine if it's needed. Go through an RTA run and count the number of star bits available for pickup (or easily obtainable) outside of the normal grinding segments.
This will take only a few hours, and as mentioned it will save you a hell of a lot of time.
With any luck, there will be a clear surplus, and you don't have to worry about routing star bits too much. There are clearly a massive number of star bits available, and this is the most likely option (you collected over 100 without stopping on the first star.) But this way you can confirm it.
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feos wrote:
OmnipotentEntity wrote:
When do you consider the situation in Ukraine to have started? With the annexation? Or with the election/attempted Russian assassination of Yushchenko in 2004?
Some other time?
With the events that resulted in the civil war. Which actions of the Russia before then were examples of dividing the West? Or you can put it differently. Which actions of Russia not related to Ukraine were examples of dividing the West? You can answer both questions separately if you wish.
Not related to Ukraine is different from occurring before Euromaidan. For instance, covert support for Trump and Brexit is a big one. It's hard to say either of these are directly related to Ukraine.
Especially if you consider the Foundations of Geopolitics, which is apparently used as a textbook in Russian General Staff Academy. (Link is to an ancient revision of the wikipedia article, prior to the Ukrainian "situation.")
Primarily, the Russian goal for the UK is to separate it from the EU, and for the US is to stir up racism and extremism.
Additionally, Finland should be absorbed into Russia. Just letting the Finns in the audience know the endgame here. Putin isn't going to be placated.
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feos wrote:
How many examples of that can I get, before the "situation" in Ukraine has started?
When do you consider the situation in Ukraine to have started? With the annexation? Or with the election/attempted Russian assassination of Yushchenko in 2004?
Some other time?
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I'm skeptical Putin is going to be "placated." The Russian game plan is to gain power by dividing the West. And it's succeeded so far.
So now we get to motivations. Why does Putin want the West weak? I doubt the answer to that is terribly comforting to anyone. And I really doubt that Putin is going to just sit on his hands once the EU falls apart.
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Both of those answers depend on the nature of the end of the universe.
If the universe ends in a Big Rip, the answer very strongly depends on the nature of dark energy, and whether or not the ratio of its pressure and its energy density is less than, greater than or equal to -1. The ship would not survive the Big Rip.
If the universe ends in Heat Death, this could take as much as 10^2100 years to finalize, but the universe will be exceedingly uninteresting for us after about 10 to 100 quadrillion years. The ship would survive this. It would take between 71 and 76 years to complete the journey to the uninteresting bit.
Universe ends via vacuum decay. No telling how long this might take. So no good estimate.
Big Crunch will almost certainly not happen thanks to dark energy.
But overall it just might be easier to link you this: http://nathangeffen.webfactional.com/spacetravel/spacetravel.php
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