Omnipotent entity: I meant the surface area of the sphere, not volume. Volume doesn't even make sense. Anyways, whatever number you calculate is going to be greater than the 0% chance it has of occuring in real life, so it's hard to have any intuition about it. (What I am saying is, "So what if it's 1% or 4% or 5%.")
I'm not going to do the calculations, but assuming uniform density, it seems likely to me that the following method is reasonable for a large family of objects including cylinders, conical frustums, and cones:
Circumscribe a sphere around the object. Then the edges of the object will split the interior of the sphere into disjoint regions. The probability that the object will land on a particular face is equal to the volume of the region associated with that particular face divided by the total volume of the sphere.
P.S. Reeds ... ?
I've never played this game, but my guess is that hydro pump is a water spell and ice beam is an ice spell, and he still needed both elements. As I recall ice beam was "super effective" against different monsters than the clearly water based techniques were.
If Genisto didn't use Acmlm's damage planning utility, they should get together, use Genisto's new path and Acmlm's utility, and iron out the problems people have brought up.
Also what would be really cool is if you could get together with Bisqwit to make BisqBot plan out walking sequences throughout the whole game. I'm sure where and when you choose to stop moving to avoid battles is not optimal, just because it would take you too long to test all the possibilities.
If you are going to attempt to be clever in such a boring and obvious fashion, you may as well read over de Brange's proof of it to see if there is any meat in his argument. Then again, you couldn't get past the first sentence, so don't.
Well, that is precisely the method I used. The name of the rule you were looking for is "similar triangles," just so you know. Obviously I made a calculation error. (I am prone to them.) If you are curious, Mathematica gives the exact representation of the solution as:
which is too complicated for me to believe there is a simpler way of arriving at this value.
Ummm, your w is greater than 4. It's very unlinkely that a 2 meter plank is standing diagonally in such a well.
The planks "intersect" at their very ends. That's why I commented it was a trivial solution ;) . An interesting exercise is to find the other trivial solution, where the longer board spans the well, but the shorter board just meets it.
Solving numerically with a different method (than you) at the moment I get w = 0.926456. We should think on it some more and then perhaps compare.
dont matter how you look at it, the user will have to know the relevant ram addresses
Probably wrong. I've never done this sort of low level debugging, but I suspect people find such information by comparing snapshots of memory before and after events occur, and noticing what changes happen and when those changes occur. This process could obviously be somewhat automated by a program.
(e.g. You feed the program a movie and tell it you dropped an item on frame 50, and you picked it up on frame 75. The program says, "Hey, here's a memory value that remained constant until frame 50, then changed once, and then remained constant until frame 75, when it changed back to what it was before frame 50.")
Objective*
Any thoughts of making a program that can work with multiple emulators, and can search memory and help you find the positions and values you are looking for?
jxq and Bomf:
Chill, my jabs at Phil are all in good fun. The (perhaps imagined) higher maturity level should exist not because we refuse to poke fun at each other, but because we don't take insult when we do. I would argue it is your posts that "accomplish the oppoosite" by making a mountain out of a molehill, but then I'd just be doing the same thing, perhaps.
Bisqwit:
Failureous, lol. you should petition Webster to get that one added :) .
You know there is no way Activision is going to make those games freeware when they are still selling like hotcakes. They are so popular that every video game store I called didn't have any in stock!
Let me take the part of a slobbering fanboy for the moment and tell you how impressed I am with your computational solution for getting refills quickly. The run is 18 times better than before now that you don't stand around trying to luck refills into dropping.
I thought my wording was pretty explicit there, the answer to your question is right there in the text you quoted. But I will explain it in another way. Consider this question: "Will the king call in prisoner n at least k number of times?" The answer to this question for any (meaningful) n and k is, "Yes, eventually."
As exciting as random homework problems from your high school calculus class are ... here is an interesting problem that might interest some of you. The formulation of the problem is very simple to understand, though I will be quite verbose to ensure that you all don't imagine there are "tricky" ways of solving the problem:
There is a king and there are his ten prisoners. The king has a dungeon in his castle that is shaped like a circle, and has ten cell doors around the perimeter, each leading to a separate, utterly sound proof room. When within the cells, the prisoners have absolutely no means of communicating with each other.
The king sits in his central room and the ten prisoners are all locked in their sound proof cells. In the king's central chamber is a table with a single chalice sitting atop it. Now, the king opens up a door to one of the prisoners’ rooms and lets him into the room ... always only one prisoner at a time! So he lets in just one of the prisoners, any one he chooses, and then asks him a question, "Since I first locked you and the other nine prisoners into your rooms, have all of you been in this room yet?" The prisoner only has two possible answers. "Yes," or, "I'm not sure." If any prisoner answers “yes” but is wrong, they all will be beheaded. If a prisoner answers “yes,” however, and is correct, all ten prisoners are granted full pardons and freed. After being asked that question and answering, the prisoner is then given an opportunity to turn the chalice upside down or right side up. If when he enters the room it is right side up, he can choose to leave it right side up or to turn it upside down, it's his choice. The same thing goes for if it is upside down when he enters the room. He can either choose to turn it upright or to leave it upside down. After the prisoner manipulates the chalice (or not, by his choice), he is sent back to his own cell and securely locked in.
The king will call the prisoners in any order he pleases, and he can call each prisoner as many times as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose.
Here’s one last monkey wrench to toss in the gears, though. The king is allowed to manipulate the cup himself, ten times, out of the view of any of the prisoners. That means the king may turn an upright cup upside down or vice versa up to ten times, as he chooses, without the prisoners knowing about it.
Assume that both the king and the ten prisoners have a complete understanding of the game as I have just explained it to you, and that the prisoners were given time beforehand to come up with a strategy they can all use.
Now you must figure out not only how to keep the prisoners alive, but how to also ensure their eventual freedom. When can any one of them be certain they've all been in the central chamber of the dungeon at least once? And how? Don't try to imagine any trickery like scratching messages in the soft gold of the chalice. The problem is as simple as it sounds. The prisoners have absolutely no way of communicating with each other except through the two orientations of the chalice. If any of them attempts any trickery at all they will all be beheaded. All the prisoners can do is turn the chalice upside down or right side up, as they choose, whenever they are called into the chamber. That is all.
Post the answer to that in binary code, and I'll set you up for a Nobel prize.
No Nobel prize in math.
Gorash wrote:
I was just about to post: "50 pages? You have to be a freaking genius, it took Wiles about 200 to make his point, and even then he lost most of his readers on page 3"
But he was long winded and his methods unnecessarily obfuscated. As usually happens his ideas have since been generalized and his proof shortened considerably.
Knux can use the scroll glitch in GH2 without dying, no fair ><. Ouzo, did you ever try to get that to work with Sonic? I was unsuccessful but that doesn't mean it's impossible ...
which is too complicated for me to believe there is a simpler way of arriving at this value.