跨界棋盘格算法的改进 [英] Straddling checkboard algorithm improvement

问题描述

古老的密码是我的一项爱好感谢大卫 - 卡恩的书的codebreakers ，我想在Ruby中的类实现来处理旧的密码，如虚无主义密码和< A HREF =htt​​p://en.wikipedia.org/wiki/ADFGVX相对=nofollow> ADFGVX 。对于这些，人们有用的产品的跨界棋盘。我有以下的Ruby实现，并欢迎任何改善。

ancient cryptography being one of my hobbies thanks to David Kahn's book The Codebreakers, I'm trying implement in Ruby classes to handle old ciphers such as the Nihilist cipher and ADFGVX. For these, one useful item is the Straddling checkboard. I have the following implementation in Ruby and would welcome any improvement.

类主要的基类（如果你想有一个虚拟的类）。 浓缩项＃是从给定的字删除重复字母的方法。

class Key is the base class (a virtual class if you want). Key#condensed is a method that remove duplicated letters from a given word.

class SKey < Key
  attr_reader :full_key
  attr_reader :alpha, :ralpha

  def initialize(key)
    super(key)
    @alpha = Hash.new
    @ralpha = Hash.new
    @full_key = checkboard()
    gen_rings()
  end

  # === checkboard
  #
  # Shuffle the alphabet a bit to avoid sequential allocation of the
  # code numbers
  #
  # Regular rectangle
  # -----------------
  # Key is ARABESQUE condensed into ARBESQU (len = 7) (height = 4)
  # Let word be ARBESQUCDFGHIJKLMNOPTVWXYZ/-
  #
  # First passes will generate
  #
  # A  RBESQUCDFGHIJKLMNOPTVWXYZ/-   c=0  0 x 6
  # AC  RBESQUDFGHIJKLMNOPTVWXYZ/-   c=6  1 x 6
  # ACK  RBESQUDFGHIJLMNOPTVWXYZ/-   c=12 2 x 6
  # ACKV  RBESQUDFGHIJLMNOPTWXYZ/-   c=18 3 x 6
  # ACKVR  BESQUDFGHIJLMNOPTWXYZ/-   c=0  0 x 5
  # ACKVRD  BESQUFGHIJLMNOPTWXYZ/-   c=5  1 x 5
  # ...
  # ACKVRDLWBFMXEGNYSHOZQIP/UJT-
  #
  # Irregular rectangle
  # -------------------
  # Key is SUBWAY condensed info SUBWAY (len = 6) (height = 5)
  #
  # S  UBWAYCDEFGHIJKLMNOPQRTVXZ/-   c=0  0 x 5
  # SC  UBWAYDEFGHIJKLMNOPQRTVXZ/-   c=5  1 x 5
  # SCI  UBWAYDEFGHJKLMNOPQRTVXZ/-   c=10 2 x 5
  # SCIO  UBWAYDEFGHJKLMNPQRTVXZ/-   c=15 3 x 5
  # SCIOX  UBWAYDEFGHJKLMNPQRTVZ/-   c=20 4 x 5
  # SCIOXU  BWAYDEFGHJKLMNPQRTVZ/-   c=0  0 x 4
  # ...
  # SCIOXUDJPZBEKQ/WFLR-AG  YHMNTV   c=1  1 x 1
  # SCIOXUDJPZBEKQ/WFLR-AGM  YHNTV   c=2  2 x 1
  # SCIOXUDJPZBEKQ/WFLR-AGMT  YHNV   c=3  3 x 1
  # SCIOXUDJPZBEKQ/WFLR-AGMTYHNV
  #
  def checkboard
    word = (@key + BASE).condensed.dup
    len = @key.condensed.length
    height = BASE.length / len


    # Odd rectangle
    #
    if (BASE.length % len) != 0
      height = height + 1
    end

    print "\ncheckboard size is #{len} x #{height}\n"
    res = ""
    (len - 1).downto(0) do |i|
      0.upto(height - 1) do |j|
        if word.length <= (height - 1) then
          return res + word
        else
          c = word.slice!(i * j)
          if not c.nil? then
            res = res + c.chr
          end
        end
      end
    end
    return res
  end 

  # == gen_rings
  #
  # Assign a code number for each letter. Each code number is
  # sequentially allocated from two pools, one with 0..7 and
  # the other with 80..99.
  #
  # Allocation is made on the following criterias
  # - if letter is one of ESANTIRU assign a single code number
  # - else assign of of the two letters ones
  #
  # Generate both the encoding and decoding rings.
  #
  # XXX FIXME Use of 80-99 is hardcoded
  #
  def gen_rings
    ind_u = 0
    ind_d = 80

    word = @full_key.dup
    word.scan(/./) do |c|
      if c =~ /[ESANTIRU]/
        @alpha[c] = ind_u
        @ralpha[ind_u] = c
        ind_u = ind_u + 1
      else
        @alpha[c] = ind_d
        @ralpha[ind_d] = c
        ind_d = ind_d + 1
      end
    end
  end # -- gen_rings

红宝石/ Python的/ Perl或伪code是为我好。感谢您的任何想法。

Ruby/Python/Perl or pseudo-code is fine for me. Thanks for any idea.

推荐答案

也许你应该尝试重构：我=> code 呢？有很多有用的人在那里。

Perhaps you should try Refactor :my => code instead? There are a lot of helpful people on there.

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