57 lines
1.9 KiB
Markdown
57 lines
1.9 KiB
Markdown
|
Let's say `b` is the number of blue disks, `r` the number of red disks
|
||
|
|
and `n` is the total number.
|
||
|
|
From the problem description we can infer this:
|
||
|
|
~~~
|
||
|
|
b+r = n (1)
|
||
|
|
0 < b < n (2)
|
||
|
|
0 < r < n (3)
|
||
|
|
n > 10^12 (4)
|
||
|
|
(b/n)*((b-1)/(n-1)) = 1/2 (5)
|
||
|
|
~~~
|
||
|
|
|
||
|
|
Now we can user (1) and (2) to get a formula for `b` and `n`:
|
||
|
|
|
||
|
|
~~~
|
||
|
|
(b/n)*((b-1)/(n-1)) = 1/2
|
||
|
|
2*b*(b-1) = n * (n-1)
|
||
|
|
2*b^2 - 2*b = n^2 - n
|
||
|
|
2 * (b^2 - b) = n^2-n
|
||
|
|
0.5 * (2b)^2 - (2b) = (n)^2 - (n)
|
||
|
|
2*(b^2) - 2b = (n)^2 - (n)
|
||
|
|
|
||
|
|
b = 1/2 ( sqrt(2n^2 - 2n + 1) + 1 )
|
||
|
|
~~~
|
||
|
|
|
||
|
|
For the last formula we search for integer solutions.
|
||
|
|
We can now either solve this manually with diophantine equations,
|
||
|
|
or we ask [Wolfram|Alpha](www.wolframalpha.com/input/?i=2*b*b-2b+%3D+n*n-n).
|
||
|
|
Which gives us the following two formulas:
|
||
|
|
|
||
|
|
~~~
|
||
|
|
s = sqrt(2)
|
||
|
|
|
||
|
|
b = (1/8) * ( 2*(3-2*s)^m + s*(3-2*s)^m + 2*(3+2*s)^m - s*(3+2*s)^m + 4)
|
||
|
|
n = (1/4) * ( -(3-2*s)^m - s*(3-2*s)^m - (3+2*s)^m + s*(3+2*s)^m + 2)
|
||
|
|
|
||
|
|
m element Z, m >= 0
|
||
|
|
~~~
|
||
|
|
|
||
|
|
We can see both formulas contain the expression sqrt(2), which is not
|
||
|
|
only fractional but also irrational. Which is a problem with the strict integer
|
||
|
|
operations in befunge.
|
||
|
|
|
||
|
|
But we can sidestep this by using a special number notation `r * 1 + s * sqrt(2)`.
|
||
|
|
In every step we calculate the "real" part of the number plus a multiple of `sqrt(2)`.
|
||
|
|
This is kinda like the common notation of [imaginary numbers](https://en.wikipedia.org/wiki/Imaginary_number).
|
||
|
|
|
||
|
|
Now all we have to do is create algorithms for addition and multiplication in our new number format.
|
||
|
|
|
||
|
|
~~~
|
||
|
|
(r1 + s1 * sqrt(2)) + (r2 + s2 * sqrt(2)) = (r1+r2) + (s1+s2) * sqrt(2)
|
||
|
|
(r1 + s1 * sqrt(2)) * (r2 + s2 * sqrt(2)) = (r1*r2+2*s1*s2) + (s1*r2+r1*s2) * sqrt(2)
|
||
|
|
~~~
|
||
|
|
|
||
|
|
Now we can use the formulas from Wolfram|Alpha until we find a value for `n > 10^12`.
|
||
|
|
|
||
|
|
In the end this problem wasn't that hard to code when all the preparations were done.
|
||
|
|
Also it's pretty fast.
|