I really don't see the point of expressing it in combinations when I already know that the sum is ((n(n+1))/2)^2; are you asking for a combinatorics analogy/visualization of the series?

Well, the point is to prove. So it doesn't really matter even if you already know what the sum is. I'm not sure what you mean by analogy or visualization.

As in, I don't actually know what you're asking me to do with problem #2 . This is more confusing than USAMO.

Usually for a problem, there are two big parts. The problem, and the solution. And the solution divides into two parts, the steps, and the sum. I've given you the problem, and you know the sum. I'm asking you to give me the steps, using combination/permutation.

So. Did I get #3?

Maybe show a picture... The choice of numbers (19 railguns) makes the solution fairly obvious (hexagon-shaped lattice)



and then nukes strategically placed to cover as much of the "safe zone" as possible.

About the nukes, if it just grazes the enemy and doesn't completely engulf it, does it still count as a kill? or do they have to hit the center of the enemy?

The touch of anything would equate to a kill. However, you want it to have a 100% kill. If the slightest chance of it surviving comes true, then you die. So you want it to have the most possible killing ratio. And in this case, saturation is possible.
Well. Strictly speaking, Vrael, you never got the #3. You multiplied radius by pi, but not by 2pi which is the actual length of the circle's circumference. There is still a chance that the target will still be alive even after your volley. Well with your solution, even when you use all 6 nukes, it won't work.

Oops, you're right, forgot the 2 in the circumference calculation. Space the railguns evenly on a circle with radius 878m centered on the target origin 2.5 seconds ago and the solution is fixed. Also wrong.

Well, the hexagon shapped lattice is dead wrong, a single nuke will cover more of the space available to the target. To give you some idea, in a cross section of the 19 railguns in hexagon form the furthest distance covered is 1339m (take the line along any of the three groups of five in a row), versus the 1500 diameter of a single nuke. I do have a picture of my solution, but you have to copy/paste in the url cause SEN is being noob. Fixed.

Post has been edited 3 time(s), last time on Jun 15 2010, 8:48 am by Vrael.

Quote from BeDazed

The touch of anything would equate to a kill. However, you want it to have a 100% kill.

There's a percentage of kill between 0 and 100%, exclusive? I thought that any kill was 100%

Oh. Probability of the kill.

Messing with when your railguns will be is cheating <_<. The problem strictly states that your decide your actions at the moment, and the total time will never be more or less than 5 seconds. In that sense, given the time frame, the sphere that is on the furthest reach still has some chance to live. But, I've noticed the #3 problem is pretty easy to solve. Try #2.

I'll draw a new picture with the actual solution.

Number 2:
sum of cubes from 1 to n = ((n(n+1))/2)^2

Then we notice that n(n+1) looks like a factorial, especially since you're asking about combinations.
so we have (n+1)! / (2!(n-1)!) = n(n+1)/2
So if we call n+1 = A, then A - 2 = (n - 1)
Then if we take a set with A objects and we want to choose K of them,

A choose K = A! / ( K!(A-K)! )
which is exactly what we have above, (notice that 2! = 2)
n+1 = A
2 = K
(n+1)! / 2!( n+1 - 2)!
= (n+1)! / 2!(n-1)!
If we square that expression, we get exactly the formula for the sum of cubes from 1 to n, so the sum of cubes from 1 to n is (the number of ways to choose 2 objects out of n)^2.

Post has been edited 1 time(s), last time on Jun 15 2010, 4:12 am by Vrael.

Actually, the solution to #3 is more likely less structured than one might expect. With a lot of effort calculus could probably be used to discover the most area... Actually I think it can... *boots up mathematica*

k guys, brb in an hour.

EDIT:

Okay, so you probably messed up here, you either want to improve its acceleration (try km not m), or make the object smaller, because I can end it with a single shot as of now. All it takes is one shot to the center of the sphere, the object cannot move fast enough to evade.



The orange object is the target, the blue transparent sphere + wire frame is its potential location. Yeah, pretty obvious.

Post has been edited 1 time(s), last time on Jun 15 2010, 5:23 am by Sacrieur.

In physics, the distance traveled with given time and acceleration is S = v't + 1/2at^2. So if you took 0 for v', and 75 for a, 5 for t- then you will have automatically noticed your calculations are off. 1/2 x 75 x 25 is 937.5m. The random acceleration toward any direction from 0 to 75 means that the object could be within any location drawn from the 937.5m radius sphere. The object itself is 0.15 km, which is 150 meters. Read. The. Thread.

@Vrael
Don't think that's the right proof for #2. Although I do think it is heading in the right direction. There's more than one way to prove it so... Anyways, the ending doesn't 'prove' that the sum of cubed from 1 to n is (n(n+1)/2)^2. It's just a coincidence that the sum of cubed is the sum of one's squared. Strictly speaking, if we apply units to n(n+1)/2, then we immediately know that squared sum of one's cannot be the same. And that the numbers that match are only coincidence. I haven't tried that proof, so I don't know if it'll end up correctly or not. However, we do know that proof currently isn't correct. I forgot to mention that you can use any other functions as long as your result ends up with combination/permutation, of course with units that match.

1m + 2m + 3m + ..... + n (m) = n(n+1)/2 (m)
so if you square this, it is (n(n+1)/2)^2 (m)^2
while
(1m)^3 + ...... + (n(m))^3 = is (n(n+1)/2)^2 (m)^3

Post has been edited 3 time(s), last time on Jun 15 2010, 7:31 am by BeDazed.

Actually I just did a conversion from m to km wrong >.>, lol, that's embarrassing.

Here's a completely new solution, pretty sure its actually right this time:


Just used the 6 nukes, no railguns needed I think. The blue circle represents 2 nukes, one into the page and one out of the page.


To be perfectly honest, I have no idea what the question is asking. Did you just want us to prove that the sum of cubes is equal to (n(n+1)/2)^2? What I would do is just prove it by induction then say, "oh look, its related to n choose 2 in this way."

Also, Sacrieur, where did you get mathmatica from? Did you get it legally? I want it, but I don't want to pay for it, its like 130 bucks or something

Wait, to me you said you want a legit one, but now you say you don't want to pay? O.o
How do you expect that to work?

Maybe I should do the entire proof, and just poke holes in it.

I had to get it to work on a calc class I took over the internet. I was given it. I wouldn't ever pirate something like that

Vrael, I actually did a great deal of thinking about this problem, I'll keep posting my progress here.

The more I thought about the problem the more my scope of it changed. For one, the target has a total time of 5s (5.025 to be exact) to move at max acceleration in any direction for the rail gun to hit it because we're observing it 2.5 seconds ago, so it could have moved to any of the position within the blue sphere in that 2.5 seconds. So, it could of moved more because it has 2.5 more (2.525) seconds to travel anywhere, considering we fire immediately to get there. Nukes have a slightly smaller range than the rail gun, but for our purposes we can use the same amount for both.



The blue sphere is the target's range of motion in 2.5 seconds and the green sphere is the target's range of motion in 5 seconds. The orange sphere is o/c the target. For both nukes and rail guns anywhere in the green sphere goes.

What I did figure out is that rail guns are not points of impact, they take the form of lines. When you think of it this way the problem becomes more clear. There's more to it than that, because one shot kills them all, the total death range of the rail gun takes the form of a cylinder.

Now, what you're trying to find, mathematically, is the maximum area you can take up in that green sphere using 19 cylinders and 6 spheres. But of course, it isn't that easy. While the spheres have no restrictions, the cylinders originate from a point that is 2.5 light years away. This means I need to come up with an algorithm that will find the total volume taken up by the position of each cylinder and sphere as a whole. This is some crazy shit.

Why do I feel like we're doing your math homework?

I do want a legit one, but I don't want to pay. I was hoping he might have gotten it through some student method for cheaper. I actually might just fork over the cash and buy it though.

Still don't know what you're talking about lol

Where is this extra .025 seconds coming from? Relativistic equations?

2.5 light-seconds, not light years. If you look at one of my previous posts you'll notice that BeDazed mentioned that the railgun fire will form a cone, but I showed that the angle is so small you can approximate it with a cylinder. Point being, you don't need to really calculate the area, you can assume that each railgun fire will destroy an entire horizontal cylinder through the target space, so you don't need to worry about that space. I'm pretty sure I solved this already with the 6 nukes though, you might want to take a look at my picture, up a few posts. It's a good technical problem, but I'd stop short of calling it crazy shit.

Post has been edited 1 time(s), last time on Jun 15 2010, 8:38 pm by Vrael.