重构 Luhn 算法的 Ruby 代码

Question

帮我重构实现Luhn algorithm，描述如下：

The formula verifies a number against its included check digit, which is usually appended to a partial account number to generate the full account number. This account number must pass the following test:

From the rightmost digit, which is the check digit, moving left, double the value of every second digit; if the product of this doubling operation is greater than 9 (e.g., 8 × 2 = 16), then sum the digits of the products (e.g., 16: 1 + 6 = 7, 18: 1 + 8 = 9).

Take the sum of all the digits.

If the total modulo 10 is equal to 0 (if the total ends in zero) then the number is valid according to the Luhn formula; else it is not valid.

Assume an example of an account number "7992739871" that will have a check digit added, making it of the form 7992739871x:

Account number 7 9 9 2 7 3 9 8 7 1 x

Double every other 7 18 9 4 7 6 9 16 7 2 -

Sum of digits 7 9 9 4 7 6 9 7 7 2 =67

The check digit (x) is obtained by computing the sum of digits then computing 9 times that value modulo 10 (in equation form, (67 × 9 mod 10)). In algorithm form:

Compute the sum of the digits (67).

Multiply by 9 (603).

The last digit, 3, is the check digit. Thus, x=3.

以下是我的实现，它可以工作，但我相信可以做得更好。

def credit_check(num)
verify = num.to_s.split('').map(&:to_i)

half1 = verify.reverse.select.each_with_index { |str, i| i.even? }
half1 = half1.inject(0) { |r,i| r + i }

# This implements rule 1
half2 = verify.reverse.select.each_with_index { |str, i| i.odd? }     
double = half2.map { |n| n * 2 }
double = double.map { |n| n.to_s.split('') }
double = double.flatten.map(&:to_i) 
double = double.inject(0) { |r,i| r + i }

final = double + half1   

puts final % 10 == 0 && (num.to_s.length > 12 && num.to_s.length < 17) ? "VALID" : "INVALID"
end

显然，我在所有这些方面都是菜鸟。但我感谢任何帮助，包括适当的风格！

Answer 1

如何使用嵌套注入方法

 half2  = verify.reverse.select.each_with_index { |str, i| i.odd? }
 double = half2.map { |n| n * 2 }

 double = double.inject(0){|x,y| x + y.to_s.split("").inject(0){|sum, n| sum + n.to_i}}

Answer 2

建议：

尝试将您的代码封装在 class 中并提供直观的 public API。在私有方法中隐藏算法的内部细节。
在最多5行的class中将规则分解成小方法，谨慎地打破这个规则。关注 Sandi Metz Rules.
研究问题并找到与问题相关的域名；用它来命名小方法。
注重可读性。记住这句话："Programs must be written for people to read, and only incidentally for machines to execute." 来自 SICP.
阅读Ruby style guide以改进代码格式；是的，找一个更好的编辑器。
遵循这些似乎会使代码更加冗长。但它会提高可读性并有助于维护。此外，如果您甚至在个人项目中也倾向于遵循它，那么这个过程将深深地刻在您的身上，很快就会成为第二天性。

考虑到这些，通过以下尝试解决问题：

class CreditCard
  VALID_LENGTH_RANGE = 12..17

  def initialize(number)
    @number = number.to_s
  end

  def valid?
    valid_length? && check_sum_match?
  end

  private

  def valid_length?
    VALID_LENGTH_RANGE.include? @number.length
  end

  def check_sum_match?
    check_sum.end_with? check_digit
  end

  def check_sum
    digits = check_less_number
             .reverse
             .each_char
             .each_with_index
             .map do |character, index|
      digit = character.to_i
      index.even? ? double_and_sum(digit) : digit
    end

    digits.reduce(:+).to_s
  end

  def check_less_number
    @number[0..-2]
  end

  def check_digit
    @number[-1]
  end

  def double_and_sum(digit)
    double = digit * 2
    tens = double / 10
    units = double % 10

    tens + units
  end
end

因此您可以按如下方式使用它：

CreditCard.new(222222222224).valid? # => true
CreditCard.new(222222222222).valid? # => false

Answer 3

我会像这样实现该算法：

def credit_card_valid?(num)
  digits = String(num).reverse.chars.map(&:to_i)
  digits.each_with_index.reduce(0) do |acc, (value, index)|
    acc + if index.even?
            value
          else
            double_value = value * 2
            if double_value > 9
              double_value.to_s.split('').map(&:to_i).reduce(&:+)
            else
              double_value
            end
          end
  end % 10 == 0
end

好吧，该代码适用于维基百科中的那些示例:)

这里有一些建议给你：

在你的函数中去掉 prints/puts 到标准输入，只是 return 价值。对于这个函数 boolean true/false 很好。
ruby 社区使用“？”在 return false/true
别忘了正确格式化你的代码，但也许你还没有学会如何在 Whosebug 上做它（我还没有:)
使用 2 个空格缩进代码

重构 Luhn 算法的 Ruby 代码

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