I'm trying to create a text file with every possible distribution of 100% into n containers? So that for 4 containers, it would look something like this:
0.97, 0.01, 0.01, 0.01
0.96, 0.01, 0.01, 0.02
0.96, 0.01, 0.02, 0.01
0.96, 0.02, 0.01, 0.01
...
Any ideas on a good way to accomplish this?
Based on your response in the comments above, here's a recursive solution in Ruby:
$resolution = 100
$res_f = $resolution.to_f
def allocate(remainder, bin_number, all_bin_values, number_of_bins)
if bin_number >= number_of_bins
all_bin_values << remainder / $res_f
puts all_bin_values.join(", ")
all_bin_values.pop
else
remainder.downto(1) do |i|
if remainder - i >= number_of_bins - bin_number
all_bin_values << i / $res_f
allocate(remainder - i, bin_number + 1, all_bin_values, number_of_bins)
all_bin_values.pop
end
end
end
end
num_bins = (ARGV.shift || 4).to_i
allocate($resolution, 1, [], num_bins)
The number of containers defaults to 4, but you can override that at run time by providing a command-line argument.
ADDENDUM
I was surprised by your comment that a looped version "was way too slow". All else being equal, looping should be faster than recursion and that was the case when I timed the iterative version given here:
resolution = 100
res_f = resolution.to_f
resolution.downto(1) do |first|
remainder1 = resolution - first
remainder1.downto(1) do |second|
remainder2 = remainder1 - second
remainder2.downto(1) do |third|
fourth = remainder2 - third
printf "%g, %g, %g, %g\n", first / res_f,
second / res_f, third / res_f, fourth / res_f if fourth > 0
end
end
end
Although this is faster, the downside is that if you wanted a different number of containers the code would have to be modified accordingly by adding additional nested loops.
