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Absolutely Mind Blasting
Discussion on Absolutely Mind Blasting within the CO2 Private Server forum part of the Conquer Online 2 category.
10/28/2010, 06:59
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#1
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Absolutely Mind Blasting
Code:
static double f(double x, int klimit)
{
double result = Double.NaN;
double numerator, denominator;
int k;
if (Math.Abs(x) < 1)
{
result = 0;
for (k = 0; k <= klimit; k++)
{
numerator = Math.Pow(-1, k) * Math.Pow(x, 1 + (2 * k));
denominator = 1 + (2 * k);
result += (numerator / denominator);
}
}
else if (Math.Abs(x) > 1)
{
result = 0;
for (k = 0; k <= klimit; k++)
{
numerator = Math.Pow(-1, k) * Math.Pow(x, (-1) - (2 * k));
denominator = 1 + (2 * k);
result += (numerator / denominator);
}
numerator = Math.PI * Math.Sqrt(x * x);
denominator = x * 2;
result = (numerator / denominator) - result;
}
else // x == 1
{
return 0.25 * Math.PI;
}
return result;
}
Code:
static double finv(double x, int klimit)
{
Func<int, int> factorial = (n) =>
{
int num = 1;
for (int j = n; j > 0; j--)
num *= j;
return num;
};
Func<double, int, double> g = (gx, gklimit) =>
{
double gres = 0;
double gnum, gdenom;
for (int gk = 0; gk <= gklimit; gk++)
{
gnum = Math.Pow(-1, gk) * Math.Pow(gx, (2 * gk));
gdenom = factorial(2 * gk);
gres += (gnum / gdenom);
}
return gres;
};
Func<double, int, double> h = (hx, hklimit) =>
{
double hres = 0;
double hnum, hdenom;
for (int hk = 0; hk <= hklimit; hk++)
{
hnum = Math.Pow(-1, hk) * Math.Pow(hx, (2 * hk) + 1);
hdenom = factorial((2 * hk) + 1);
hres += (hnum / hdenom);
}
return hres;
};
return (h(2 * x, klimit) / (g(2 * x, klimit) + 1));
}
Code:
finv( f ( x , k ) , k )
Will give you the same result as the x parameter
Well, not the exact result, but as k (klimit) increases, accuracy increases.
How's this relevant to CO?
Well, if we build on the f() function a bit we can do this...
Code:
static double f2(double y, double x, int klimit)
{
if (x > 0)
return f(y / x, klimit);
else if (y >= 0 && x < 0)
return Math.PI + f(y / x, klimit);
else if (y < 0 && x < 0)
return (-Math.PI) + f(y / x, klimit);
else if (y > 0 && x == 0)
return Math.PI / 2;
else if (y < 0 && x == 0)
return (-Math.PI) / 2;
return Double.NaN;
}
Code:
static int GetAngle(int X1, int Y1, int X2, int Y2)
{
return (int)Math.Round(f2(Y2 - Y1, X2 - X1, 5) * 180 / Math.PI);
}
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10/28/2010, 07:06
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#2
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Easy geometry homework ftw.
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10/28/2010, 07:07
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#3
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Umm wtf does this do lol?
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10/28/2010, 07:09
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#4
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Nothing -- well, that's not true but it's not worth explaining. I just found it highly fascinating.
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10/28/2010, 07:29
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#5
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Thank you for sharing that. I found it interesting because I didn't know much about the math class until now.
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10/28/2010, 10:00
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#6
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Absoultely fantastic on how that works(Never done anything with maths(Well, not much)) May come in handy. #Bookmarked
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10/28/2010, 10:21
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#7
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nice and pretty fascinating
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10/28/2010, 13:29
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#8
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WHAT IS THIS I DONT EVEN ---
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10/28/2010, 22:13
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#9
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I warned you that bricks would be shat.
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10/29/2010, 05:11
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#10
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Quote:
Originally Posted by copz1337
"Oh Absolutely" nerds these days.
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Lol, srry that the math confuses you.
Noobs these days
EDIT: xP i'm sorry, that was a bit harsh. Try and understand it. The result is interesting.
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10/29/2010, 13:25
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#11
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10 dollars says nobody that has posted in this thread understands it
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10/29/2010, 13:32
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#12
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Quote:
Originally Posted by _tao4229_
10 dollars says nobody that has posted in this thread understands it
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Agree.
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10/29/2010, 13:42
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#13
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I don't understand it.
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10/29/2010, 13:53
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#14
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I got lost, sigh. Should get more into coding.
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10/29/2010, 14:01
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#15
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This doesnt have very much to do with programming, you need to be good at maths to understand this.
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