Lua和RSA加密

2023年7月3日 1,016次阅读

我一直在Lua中使用RSA加密脚本,在BigNumbers(
http://oss.digirati.com.br/luabignum/bn/index.htm)的帮助下,我几乎有一个工作代码.但是,我被卡住了,因为在一小部分情况下,加密的原始邮件没有被解密为原始邮件,我无法弄清楚原因.请注意,这将处理非常大的数字(例如,1.08e107).我写的完整代码如下,但这里是它应该做的细分.

print(rsa_getkey())

p: 83
q: 23
n: 1909
e: 19
d: 1899
phi: 1804

以上设置键值,其中公钥由[n,e]表示,私钥由[n,d]表示.这是通过以下代码完成的：

function rsa_getkey()
  rsa_e = 0
  local primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 57, 71, 73, 79, 83, 89, 97}

  math.randomseed = os.time()

  rsa_p = primes[math.random(5,#primes)]
  rsa_q = rsa_p

  while rsa_q == rsa_p do
      math.randomseed = os.time()
      rsa_q = primes[math.random(5,#primes)]
  end

  rsa_n = rsa_p*rsa_q
  rsa_phi = (rsa_p-1)*(rsa_q-1)

  while rsa_e == 0 do
      local prime = primes[math.random(1,10)]
      if rsa_phi%prime > 0 then
          rsa_e = prime
      end
  end

  for i = 2, rsa_phi/2 do
      if ((i*rsa_phi)+1)%rsa_e == 0 then
          rsa_d = ((i*rsa_phi)+1)/rsa_e
          break
      end
  end
  return "p: ",rsa_p,"\nq: ",rsa_q,"\nn: ",rsa_n,"\ne: ",rsa_e,"\nd: ",rsa_d,"\nphi: ",rsa_phi,"\n"
end

确定密钥后,您可以加密消息.为了将纯文本(“Hello world”)转换为数字系统,我创建了一个不是100％完整的函数,但是以最基本的形式工作：

print(rsa_plaintext("Hello_world"))

1740474750625850534739

以下函数是确定消息的方式：

function rsa_plaintext(x)
  local alphanum = {A=10, B=11, C=12, D=13, E=14, F=15, G=16, H=17, I=18, J=19, K=20, L=21, M=22, N=23, O=24, P=25, Q=26, R=27, S=28, T=29, U=30, V=31, W=32, X=33, Y=34, Z=35, a=36, b=37, c=38, d=39, e=40, f=41, g=42, h=43, i=44, j=45, k=46, l=47, m=48, n=49, o=50, p=51, q=52, r=53, s=54, t=55, u=56, v=57, w=58, x=59, y=60, z=61, _=62}
      rsa_cipher = ""
  for i = 1, #x do
      local s = x:sub(i,i)
      rsa_cipher = rsa_cipher .. alphanum[s]
  end
  return rsa_cipher
end

最后,为了使这个更易于管理,我必须将其细分为几个部分.为了节省时间和代码,我将实际加密与从明文到数字格式到加密的转换相结合,尽管我已经为调试目的添加了解密.该代码还考虑将0添加到消息中以确保每个分组中的4个数字.这是我的问题所在;消息和解密应该是相同的.

print(rsa_group("Hello world"))

Msg: 1740
Encrypted: 1560
Decrypted: 1740

Msg: 4747
Encrypted: 795
Decrypted: 929

Msg: 5062
Encrypted: 1659
Decrypted: 1244

Msg: 5850
Encrypted: 441
Decrypted: 123

Msg: 5347
Encrypted: 429
Decrypted: 1529

Msg: 3900
Encrypted: 1244
Decrypted: 82

这是通过以下两个功能完成的：

function rsa_group(str)
  local cipher = {}
  local str = rsa_plaintext(str:gsub(" ","_"))
  local len = #str
  local fillin = ""
  if len%4 ~= 0 then
      fillin = string.rep(0,(4-len%4))
  end
  str = str..fillin
  for i = 1, #str, 4 do
      local s,e = i, i+3
      local part = str:sub(s,e)
      print(rsa_encrypt(part))
  end
end

function rsa_encrypt(msg)
  bnrsa_e = BigNum.new(rsa_e)
  bnrsa_n = BigNum.new(rsa_n)
  bnmsg = BigNum.new(msg)
  result = 0
  quo = BigNum.new()
  rsa_c = BigNum.new()
  result = BigNum.pow(bnmsg, bnrsa_e)
  BigNum.div(result, bnrsa_n, quo, rsa_c)

  bnrsa_c = BigNum.new(rsa_c)
  bnrsa_d = BigNum.new(rsa_d)
  result = 0
  quo = BigNum.new()
  rsa_C = BigNum.new()
  result = BigNum.pow(bnrsa_c, bnrsa_d)
  BigNum.div(result, bnrsa_n, quo, rsa_C)

  return "Msg:",msg,"\nEncrypted:",rsa_c,"\nDecrypted:",rsa_C,"\n"
end

现在,我知道这是一个很长的问题,问题本身有很多组成部分.我不知道如何找出问题所在.有什么我想念的吗？一双新鲜的眼睛可能是我的解决方案.

最佳答案 经过仔细检查,看起来消息M必须是你的两个素数的
less than the product n.在上面的测试用例中,除了第一个消息之外的所有消息都未能正确解密,因为它们大于n = 1909.

例如,考虑M刚超过n = 1909的位置：

Msg: 1910
Encrypted: 1
Decrypted: 1

Msg: 1911
Encrypted: 1222
Decrypted: 2

Msg: 1912
Encrypted: 1179
Decrypted: 3

在现实世界的例子中,n当然要大得多,因此这个问题不太可能出现.

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