Encryption Lua上的Lite加密

我通过http发送短视频数据，我需要一个针对二进制数据（视频）（带或不带密钥）的快速加密/解密算法，而不改变数据的大小。我用一个键做了一些排列算法，但我在原始数据的末尾添加了任何数据。这不是一个好主意，因为数据的大小发生了变化。

编码器和解码器都必须包含此PRNG实现。
您可以更改

secret\u key.*

常量，但编码器和解码器上的常量必须相同

local set_seed, get_next_pseudo_random_byte

do
   -- pseudo-random number generator
   -- Good statistical properties of PRN sequence:
   --    uniformity,
   --    long period of 255 * 2^45 (approximately 2^53),
   --    unpredictability
   -- Compatible with Lua 5.1, 5.2, 5.3, LuaJIT

   -- all parameters in PRNG formula are derived from these 57 secret bits:
   local secret_key_6  = 58            -- 6-bit  arbitrary integer (0..63)
   local secret_key_7  = 110           -- 7-bit  arbitrary integer (0..127)
   local secret_key_44 = 3580861008710 -- 44-bit arbitrary integer (0..17592186044415)

   local floor = math.floor

   local function primitive_root_257(idx)
      -- returns primitive root modulo 257 (one of 128 existing roots, idx = 0..127)
      local g, m, d = 1, 128, 2 * idx + 1
      repeat
         g, m, d = g * g * (d >= m and 3 or 1) % 257, m / 2, d % m
      until m < 1
      return g
   end

   local param_mul_8 = primitive_root_257(secret_key_7)
   local param_mul_45 = secret_key_6 * 4 + 1
   local param_add_45 = secret_key_44 * 2 + 1

   -- state of PRNG (53 bits in total)
   local state_45 = 0 -- from 0 to (2^45-1)
   local state_8 = 2  -- from 2 to 256

   function set_seed(seed_53)
      -- set 53-bit integer as current seed 
      -- (seed is initially set to 0 when program starts)
      state_45 = seed_53 % 35184372088832
      state_8 = seed_53 % 255 + 2
   end

   local function get_random_32()
      -- returns pseudorandom 32-bit integer (0..4294967295)

      -- A linear congruential generator having full period of 2^45
      state_45 = (state_45 * param_mul_45 + param_add_45) % 35184372088832

      -- Lehmer RNG having period of 256
      repeat
         state_8 = state_8 * param_mul_8 % 257
      until state_8 ~= 1  -- skip one value to reduce period from 256 to 255 
      -- (we need it to be coprime with 2^45)

      -- Idea taken from PCG: 
      -- shift and rotate "state_45" by varying number of bits to get 32-bit result
      local r = state_8 % 32
      local n = floor(state_45 / 2^(13 - (state_8 - r) / 32)) % 2^32 / 2^r
      return floor(n % 1 * 2^32) + floor(n)
   end

   local prev_values = {}

   function get_next_pseudo_random_byte()
      if #prev_values == 0 then
         local rnd = get_random_32() -- value 0..4294967295
         local low_16 = rnd % 65536
         local high_16 = (rnd - low_16) / 65536
         local b1 = low_16 % 256
         local b2 = (low_16 - b1) / 256
         local b3 = high_16 % 256
         local b4 = (high_16 - b3) / 256
         prev_values = {b1, b2, b3, b4}
      end
      return table.remove(prev_values)
   end

end

在解码器端-几乎相同的代码，但使用减法而不是加法

-- prepare for decryption
set_seed(file_id)

-- The main loop - process file byte-by-byte

-- read next byte of encrypted file
local next_encoded_byte = ...  -- 0..255

-- decode byte
local decoded_byte = (next_encoded_byte - get_next_pseudo_random_byte()) % 256

-- write decoded byte to decrypted file
...

---

该算法应该是加密强大的吗？@EgorSkriptunoff不。它不应该是强大的。你忘了问问题，你的文章太宽泛，离题了。使用websearch查找加密算法。@Pigget我正在查找算法，但没有找到请参见

-- prepare for decryption
set_seed(file_id)

-- The main loop - process file byte-by-byte

-- read next byte of encrypted file
local next_encoded_byte = ...  -- 0..255

-- decode byte
local decoded_byte = (next_encoded_byte - get_next_pseudo_random_byte()) % 256

-- write decoded byte to decrypted file
...