ok, well at the very least (having read the clarification by Warp) that it is not possible for 3 faces, because there is no polyhedron with fewer than 4 faces
this turns out to be true because it is possible to construct the dual of any polyhedron, in which each face is mapped to a vertex, each vertex to a face, and each edge to an edge, and the dual is also a polyhedron...but the dual of a 3-sided polyhedron would be a triangle, because three vertices determine at most one polygonal face
also I wasn't sure to what degree physical considerations would be needed, so for most cases I just was like "if you randomly rotated it, there would be an equal probability of any face being the one landed on if the polyhedron were placed gently on a horizontal surface at that rotation"
Finally, in that pic, the manufacturer of the 5-sided die made those customizations, but it is possible to make a triangular prism into a fair die.
Geometry problem:
Is it possible to construct a die with an odd number of faces (and at least 3 of them) so that the probability of each face (when the die is thrown normally) is the same?
It is, for all odd numbers n>3, by constructing a right prism based on a regular polygon with n-2 sides of such relative dimensions that the spherical areas of the projections of each face onto the smallest sphere containing the prism are equal.
For all even numbers n>2, if 4|n the fairest way is to attach two right pyramids based on a regular polygon with n/2 sides with such relative dimensions that all vertices lie on a sphere (forming a dipyramid), otherwise the fairest way is to construct two conical collections of n/2 kites with one common vertex and two opposite vertices each shared with a neighbor and one unshared vertex with such relative dimensions that the unshared edges lie on a regular prism based on an n-sided polygon and all vertices lie on a sphere, and attach them (this is known as a trapezohedron).
The only exceptions occur when a regular or semi-regular polyhedron with congruent faces is possible: n=4, tetrahedron; n=12, dodecahedron,; n=20, icosahedron; n=24, deltoidal icositetrahedron; 30, rhombic triacontahedron. The cube and octahedron are not exceptions, because the former is in fact a triangular trapezohedron and the latter is a square dipyramid.
For the degenerate case n=3, use an elongated right triangular prism capped with tetrahedra, so that no normal from a triangular face intersects the center of mass; for n=2, use the attachment of two tetrahedra with a smaller tetrahedron removed from one vertex, at the larger tetrahedral side, with such dimensions that no normal from a quadrilateral face intersects the center of mass; for n=1, use a unistable polyhedron; and finally for n=0, use the empty set.
Interestingly, the limiting case differs based on the parity of n: The limiting case for 2n is a cone, while the limiting case for 2n+1 is a pin; it would normally be expected for the limit to be a sphere but it doesn't seem like it...
Also I read about this more than 3 years ago, and the problem had been solved for quite a while back then; I also actually bought the 5- and 7-sided dice mentioned in the article: http://www.maa.org/editorial/mathgames/mathgames_05_16_05.html
No, I compared the stated run times: 2:24:32.63 for the TAS and 2:40:05 for SDA
but I just noticed that the description page said the in-game timer read 2:00:28 in the TAS and that when I watched the SDA run, the length (single-segment, so no funny SDA demerits involved here) of the video was over 192 minutes, while the in-game timer was just over 160 minutes...
sorry, I guess the route actually can't be improved that much, I mean the SDA runner's missteps may have been able to account for 16 minutes but not 48
I just saw the new SDA run and despite numerous mistakes and backtracking, that route ends up less than a minute longer than the currently published TAS.
Maybe not a "glitch" per se, but more of a design oversight, but in FF5 you can land the airship directly on top of a cave entrance, after which the game becomes unwinnable because you can't enter the airship anymore. If you try to enter the airship, you end up entering the cave, and when you exit the cave you appear at the tile one below the cave entrance/airship, so you are stuck.
It reminds me of the Road to Bowser's Castle in Mario & Luigi: Bowser's Inside Story (DS); there is a path you can take that will lead Bowser to an area that he cannot escape, because he can't jump.
Also I remember in 1991 I was playing The Chessmaster (SNES) and for some reason one time the brown King made 8 copies of itself on all adjacent spaces.
I think the notion of turning your head sideways is a good one; I've been playing the game and it seems from the placement of meta-information on the DeSmuME screen that internally there is no difference between holding the DS horizontally or vertically.
Another problem however may be blowing into the microphone, but perhaps that's why the raw sample option is available...
What formats are allowed? What is the format of your image?
IMO you should try a PNG, because it is an open lossless compression standard, instead of JPG (not lossless) or BMP (not compressed and not open)
lol did someone mess with the data you had given?
Anyway it is sad that most of the userbase here will need to slog through the German or Spanish versions of the site just to register an account, but once that's done I prefer to use the original Japanese version just fyi
The height of the stairs doesn't matter, because you can still have unbroken horizontal speed going up. What matters is that the stairs are closer in SMB Deluxe.
The height of the stairs dictates the length of the stairs though.
However, the end of the staircase is closer to the flagpole and that's what matters.
Yeah the encoding it pretty nice ! I really like it. Maybe every NDS games should be encoded like this when there's no action on the second screen (Where the action does not happen on both screen).
I don't know if it would be hard to make the active screen switch from the first one to the second one. Like in New Super mario Bros (when you enter a pipe, the action is at the bottom of the screen)
With careful viewing and a well-written AviSynth script it can be done
you are an NDS encoding ninja
more encoders should do what you do, possibly switching up that left pane to show the bottom screen when that is the more important one
There are more lossless solutions than H.264 lossless.
Still, *that* one was like 235MB in the original output, it's just that the step where I made the file more amenable to subtitle insertion by mencoder was the one that used a codec that yielded a 2GB file...
IMO OmnipotentEntity got an uncompressed capture, which I would use if I had that much space on my disk lol
I thought you were supposed to use H.264 lossless for initial capture; when I encoded the Metroid 100% run I never had a file above 2GB, even though it was a 40min run; I admit, it was an NES run, but is the detail so much greater that it takes about 6 times as much data per minute?
Bisqwit had a "making of" video of his semi-TAS of Chrono Cross, where he showed how he grinded a short segment over and over, perfecting it little by little, and then the video shows the final result.
Microsoft pulled VS2005 Express links from its site, although somehow it is possible to find them...but it is better to use VS2008 Express (or if you can wait until next year it is likely that VS2010 will also be made available in Express Editions)
Also you can still use the whole Visual Studio toolchain from Code::Blocks; it should even be able to automatically set it up.
The area of an equilateral triangle is s^2 * sqrt(3)/4
Should be sqrt(3/4), but yeah... you are right. It's impossible.
no, it actually is sqrt(3)/4
the height of an equilateral triangle of base s is s*sqrt(3)/2, so the area is (1/2)s*s*sqrt(3)/2=s^2*sqrt(3)/4
One way to think about this issue is to remember the Cartesian and coordinate-free formulas for the dot product of vectors: u.v=u1v1+u2v2=|u||v|cos(q), where u=(u1,u2), v=(v1,v2), and q is the angle between them. If q=60 degrees, then cos(q)=1/2, and |u|=|v|=sqrt(u1^2+u2^2), so the equation on the right simplifies to u1v1+u2v2=(u1^2+u2^2)/2.
Now let u1=n, u2=m, v1=p, and v2=q be integers, with 0<m<n and with n and m, and p and q, relatively prime; then it can first be noted that both n and m must be odd, because they cannot both be even (else they would not be relatively prime), and if only one were even then (n^2+m^2)/2 would not be an integer even though (n^2+m^2)/2=np+mq, which is an integer. Also, because n^2+m^2=p^2+q^2 (because |u|=|v|), the same holds for p and q.
To break the symmetry I will attempt to eliminate p between these two equations and then show that q cannot be an integer.
From the first equation, p=(n^2+m^2-2mq)/(2n), so p^2=(n^4+m^4+4m^2*q^2+2n^2*m^2-4mqn^2-4qm^3)/(4n^2), and substitution into that second equation yields
n^2+m^2=(n^4+m^4+4m^2*q^2+2n^2*m^2-4mqn^2-4qm^3+4n^2*q^2)/(4n^2)
which simplifies to
4n^4+4n^2*m^2=n^4+m^4+4m^2*q^2+2n^2*m^2-4mqn^2-4qm^3+4n^2*q^2
from which
4(n^2+m^2)q^2-4m(n^2+m^2)q-(3n^4+2n^2*m^2-m^4)=0
which has a discriminant (with respect to q) of
16m^2(n^2+m^2)^2+16(n^2+m^2)(3n^4+2n^2*m^2-m^4)
which simplifies to 3*16n^2(n^2+m^2)^2, with square root 4n(n^2+m^2)sqrt(3), so there is no way for q, which in the end turns out to be
(n±m*sqrt(3))/2, to be an integer after all, a contradiction.
Pick's Theorem is more elegant though.
