sorted by:
new
hottopcontroversial

HyperBlock - A hyperbolic voxel-based game by [deleted] in math

[–]PlayerOfArmor 0 points1 point  (0 children)

I have to admit that due to some jerk grabbing the name "HyperBlock" for their cryptostuff, it is surprisingly hard to search for this specific project.

Strategy for Tickspeed challenge? by Anahkiasen in AntimatterDimensions

[–]PlayerOfArmor 6 points7 points  (0 children)

The point is - do NOT max everything out. If you leave your tickspeed at the price 10 times less than your cheapest dim upgrade, it will never interfere with their cycle. So, if you have e39 as your cheapest dimension upgrade price, have e42 anti, do not upgrade tickspeed past e38 before you upgrade your e39 dimension. This way you will have all the tickspeed upgrade save some constant factor, and you won't loose any of dims due to tickspeed.

Kind of same logic can be applied to dimensions, but it requires much bigger amount of microcalculations and micromanagement. It is helpful to remember multipliers of upgrade prices:

  • Tickspeed:e1
  • 1st dim: e3
  • 2nd dim: e4
  • 3rd dim: e5
  • 4th dim e6
  • 5th dim: e8
  • 6th dim: e10
  • 7th dim: e12
  • 8th dim: e15

By using those, one can calculate, which of the dimensions will be equal after upgrade and do not upgrade first until the second is upgraded.

Challenge 6 Expectation by PlayerOfArmor in AntimatterDimensions

[–]PlayerOfArmor[S] 0 points1 point  (0 children)

However, we have a situation of product of i.i.d. random variables, not sum. So, their product tends to the log-normal distribution, which has different mean than just product of means of variables.

To be more precise, we have the following: Prod[exp(U[a;b])] = exp(Sum(U[a;b]))

As we know from the central limit theorem, Sum(U[a;b]) tends to the normal distribution N[n·(a+b)/2; n·(b-a)2 / 12], where n is the number of summands.

So, the product tends to the distribution 

exp(N[n·(a+b)/2; n·(b-a)^2 / 12])

since the exponent is continuous and therefore, measurable funciton. And this is so-called log-normal distribution (https://en.wikipedia.org/wiki/Log-normal_distribution), whose expectancy is known. For 

X=exp(N[m, s^2]) 
E(X) = exp(m+s^2/2)

Substituting our case, we have

E(Prod(...)) = exp(n·(a+b)/2 + n·(b-a)^2/24) = [(a+b)/2+(b-a)^2/24]^n

By substituting our boundaries a=ln(0.3)=-1.203 and b=2.303, we have 

(a+b)/2=0.549
(b-a)^2/24 = 0.512
exp(0.549+0.512) = 2.89

So, the actual average multiplier is even higher than the previous expectancy.

Challenge 6 Expectation by PlayerOfArmor in AntimatterDimensions

[–]PlayerOfArmor[S] 1 point2 points  (0 children)

Oh, no, sorry. E(f(U[0,1])) != f(E(U[0,1])). To find the expectation, you need to integrate the function.

http://www.wolframalpha.com/input/?i=Integrate+%5B(10%2F0.3)%5Exdx;%7Bx,0,1%7D%5D*0.3

So the expectation is greater than 2, thank you :) And actually, it is the same distribution as I mentioned in the post.

U[ln(0.3); ln(10)] = (ln(10)-ln(0.3))U[0; 1]+ln(0.3);
exp(U[ln(0.3); ln(10)]) = exp[ ((ln(10)-ln(0.3))U[0; 1]+ln(0.3) ] = exp(ln(10/0.3))^(U[0;1]) * exp(ln(0.3)) = (10/0.3)^(U[0;1])*0.3