12 lines
318 B
Ruby
12 lines
318 B
Ruby
# hsfreq.rb
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require 'hailstone'
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h = Hash.new(0)
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last = 99_999
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(1..last).each {|n| h[Hailstone.hailstone(n).length] += 1}
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length, count = h.max_by {|length, count| count}
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puts "Given the hailstone sequences from 1 to #{last},"
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puts "the most common sequence length is #{length},"
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puts "with #{count} such sequences."
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