q2.rb

require 'benchmark/ips'

module Cond
  OPS = ['*',''].freeze
  RANGE = (1000..10000).freeze
end

class Proc0
  def self.run(num, ops, ret)
    # @note Synonymous with the following implementation, but maybe slow
    #       `return if ops.join.length.zero?`
    return if ops[0].empty? && ops[1].empty? && ops[2].empty?

    n = num.to_s
    # @note Synonymous with the following implementation, but maybe slow
    #       `exp = n.chars.reverse.zip(ops).join`
    exp = "#{n[3]}#{ops[0]}#{n[2]}#{ops[1]}#{n[1]}#{ops[2]}#{n[0]}"
    return if exp =~ /0\d/

    res = eval(exp)
    ret << n.reverse if num == res
  end
end

class Case1
  def self.run
    Array(Cond::RANGE).product(*Array.new(3, Cond::OPS)).each.with_object([]) do |(num, *ops), ret|
      Proc0.run(num, ops, ret)
    end
  end
end

class Case2
  def self.run
    ops_combi = Cond::OPS.repeated_combination(3).to_a
    Cond::RANGE.each.with_object([]) do |num, ret|
      ops_combi.each do |ops|
        Proc0.run(num, ops, ret)
      end
    end
  end
end

# run debug
if ARGV.first
  r1 = Case1.run
  r2 = Case2.run
  puts "Case1: #{r1.inspect}"
  puts "Case2: #{r2.inspect}"
  if r1 == r2
    puts 'Valid'
  else
    raise 'Invalid'
  end
end

puts 'Run benchmark'

Benchmark.ips do |x|
  x.config(time: 60)
  x.report('Case1')  { Case1.run }
  x.report('Case2')  { Case2.run }

  x.compare!
end

問題をちゃんと読んでからやってみると、どうも四則演算は全てやる必要がなさそうなことに気づく。
まぁ答えを知ってしまっているからかもしれないが。

ってことで、revision2の結果がこれ。

 ruby q2.rb
Run benchmark
Warming up --------------------------------------
               Case1     1.000  i/100ms
               Case2     1.000  i/100ms
Calculating -------------------------------------
               Case1      2.266  (± 0.0%) i/s -    136.000  in  60.152929s
               Case2      5.720  (± 0.0%) i/s -    342.000  in  60.039457s

Comparison:
               Case2:        5.7 i/s
               Case1:        2.3 i/s - 2.52x  slower

まぁそれでも2.5倍ぐらい差があるのか。