如何通过单元测试产生高效的代码?

我参加了一个TDD编码Dojo,在那里我们尝试在简单的问题上练习纯TDD。然而,我突然想到,从单元测试中产生的代码并不是最有效的。现在这在大多数情况下都是好的,但是如果代码使用量增加了,那么效率就成了问题。

下面是ruby中的一个小例子:素因子分解。我遵循纯TDD方法,使测试一个接一个地通过,从而验证了我最初的验收测试(在底部评论)。 generic prime factorization algorithms 出现?为了减少问题域,假设我想得到一个 quadratic sieve

require 'shoulda'
require 'lib/prime'

class MathTest < Test::Unit::TestCase
  context "The math module" do
    should "have a method to get primes" do 
      assert Math.respond_to? 'primes'
    end
  end
  context "The primes method of Math" do
    should "return [] for 0" do
      assert_equal [], Math.primes(0)
    end
    should "return [1] for 1 " do
      assert_equal [1], Math.primes(1)
    end
    should "return [1,2] for 2" do 
      assert_equal [1,2], Math.primes(2)
    end
    should "return [1,3] for 3" do 
      assert_equal [1,3], Math.primes(3)
    end
    should "return [1,2] for 4" do 
      assert_equal [1,2,2], Math.primes(4)
    end 
    should "return [1,5] for 5" do 
      assert_equal [1,5], Math.primes(5)
    end   
    should "return [1,2,3] for 6" do 
      assert_equal [1,2,3], Math.primes(6)
    end       
    should "return [1,3] for 9" do 
      assert_equal [1,3,3], Math.primes(9)
    end        
    should "return [1,2,5] for 10" do 
      assert_equal [1,2,5], Math.primes(10)
    end                  
  end
#  context "Functionnal Acceptance test 1" do
#    context "the prime factors of 14101980 are 1,2,2,3,5,61,3853"do      
#      should "return  [1,2,3,5,61,3853] for ${14101980*14101980}" do
#        assert_equal [1,2,2,3,5,61,3853], Math.primes(14101980*14101980)
#      end
#    end
#  end
end

我用这种方法创建的简单算法

module Math
  def self.primes(n)
    if n==0
      return []
    else
      primes=[1]  
      for i in 2..n do
        if n%i==0          
          while(n%i==0)
            primes<<i
            n=n/i
          end
        end
      end      
      primes
    end
  end
end

编辑1 不 我的单元测试的标准部分,它是 新的验收测试 为回答问题而写 具体要求 从客户那里。

编辑2 浮现 换句话说:如何将迁移从琐碎的代码分解为最优的代码? 有人提到这是一种特定于问题的方法:我提供了一个示例问题,我不知道如何继续。