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