Created
May 24, 2018 08:13
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Calculates Ideal Miojo Cooking time with two hourglasses
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# Solves the "Miojo with 2 hourglasses" problem | |
# cooking time and both hourglass times are passed as arguments, as in the following sample: | |
# ruby miojoproblem.rb 3 5 7 | |
# returns the minimum total time to cook the miojo accurately, or a fail message if not possible | |
# Solution is ad-hoc, can probably be optimized a lot | |
raise "Must specify 3 inputs: Miojo cooking time, hourglass 1 time, hourglass 2 time." if ARGV.length != 3 | |
cook_time = ARGV[0].to_i | |
hg_1 = ARGV[1].to_i | |
hg_2 = ARGV[2].to_i | |
raise "Cooking time must be smaller than both hourglasses' times." if cook_time >= hg_1 or cook_time >= hg_2 | |
if hg_1 % 2 == 0 and hg_2 % 2 == 0 and cook_time % 2 != 0 | |
puts "Not possible to calculate odd cooking time with even time hourglasses!" | |
return | |
end | |
i = cook_time | |
j = cook_time | |
maximum = [cook_time, hg_1, hg_2].max * 100 | |
for i in (hg_1..maximum).step(hg_1) do | |
for j in (hg_2..maximum).step(hg_2) do | |
if (i - j).abs == cook_time | |
puts "Minimum cooking time: #{[i, j].max} minutes." | |
return | |
end | |
end | |
end | |
puts "Cooking time not possible to calculate with the given hourglasses"; |
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