2017-05-03

# Problem 100: Arranged probability

Description:

If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, `P(BB) = (15/21)x(14/20) = 1/2`.

The next such arrangement, for which there is exactly 50% chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs.

By finding the first arrangement to contain over `10^12 = 1,000,000,000,000` discs in total, determine the number of blue discs that the box would contain.

Solution:
10v>012p >01g11g:01g3*\4*+01p3*\2*+11p02g12g:02g3*\4*-0v
##1^p110<|!`*:**"@}}"/4-g20-*g212-g10+*2g112p21-*2\*3p2<
##>p102p^>402g2*+12g2*+01g2*+11g2*-8/.@
Start
??
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Output:
Stack:   (0)

Explanation:

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. 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.

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.

 Interpreter steps: 1 649 Execution time (BefunExec): 0ms (? MHz) Program size: 56 x 3 (fully conform befunge-93) Solution: 756872327473 Solved at: 2017-05-03

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