SHA 256 pseuedocode?
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【中文标题】SHA 256 pseuedocode?【英文标题】： 【发布时间】：2012-08-09 20:28:07 【问题描述】：我一直在尝试弄清楚 SHA-256 的工作原理。我一直在为其他算法做的一件事是我已经为该算法制定了一种逐步的伪代码函数。
我尝试对 SHA256 做同样的事情，但到目前为止我遇到了很多麻烦。
我试图弄清楚***图表是如何工作的，但除了解释功能的文本部分之外，我不确定我是否做对了。
这是我目前所拥有的：
Input is an array 8 items long where each item is 32 bits.
Output is an array 8 items long where each item is 32 bits.
Calculate all the function boxes and store those values.
|I'll refer to them by function name
Store input, right shifted by 32 bits, into output.
| At this point, in the out array, E is the wrong value and A is empty
Store the function boxes.
| now we need to calculate out E and out A.
| note: I've replaced the modulo commands with a bitwise AND 2^(32-1)
| I can't figure out how the modulus adding lines up, but I think it is like this
Store (Input H + Ch + ( (Wt+Kt) AND 2^31 ) ) AND 2^31 As mod1
Store (sum1 + mod1) AND 2^31 as mod2
Store (d + mod2) AND 2^31 into output E
|now output E is correct and all we need is output A
Store (MA + mod2) AND 2^31 as mod3
Store (sum0 + mod3) AND 2^31 into output A
|output now contains the correct hash of input.
|Do we return now or does this need to be run repeatedly?
我得到所有这些加法模数了吗？还有什么是 Wt 和 Kt？ 这是否会运行一次然后你就完成了，或者它是否需要运行一定次数，输出被重新用作输入。
顺便说一下，这是链接。 http://en.wikipedia.org/wiki/SHA-2#Hash_function
非常感谢， 布赖恩
【问题讨论】：
对于初学者来说，SHA256 的输出是 32 字节。想想看：256/8=32 啊好吧...应该意识到这一点。所以图中的每个输入/输出框都是...... 32位？我想你的意思是比特。我将编辑我的起始代码以反映这一点。 【参考方案1】：看看描述算法的官方标准，变量描述在这里：http://csrc.nist.gov/publications/fips/fips180-4/fips-180-4.pdf
（哦，现在我发现我的答案迟了将近一年，啊，没关系......）
【讨论】：
以上链接是存档版本，仅供历史参考。 csrc.nist.gov/publications/fips/fips180-4/fips180-4.pdf 截止到 2015 年 8 月 5 日。【参考方案2】：initial_hash_values=[
'6a09e667','bb67ae85','3c6ef372','a54ff53a',
'510e527f','9b05688c','1f83d9ab','5be0cd19'
]
sha_256_constants=[
'428a2f98','71374491','b5c0fbcf','e9b5dba5',
'3956c25b','59f111f1','923f82a4','ab1c5ed5',
'd807aa98','12835b01','243185be','550c7dc3',
'72be5d74','80deb1fe','9bdc06a7','c19bf174',
'e49b69c1','efbe4786','0fc19dc6','240ca1cc',
'2de92c6f','4a7484aa','5cb0a9dc','76f988da',
'983e5152','a831c66d','b00327c8','bf597fc7',
'c6e00bf3','d5a79147','06ca6351','14292967',
'27b70a85','2e1b2138','4d2c6dfc','53380d13',
'650a7354','766a0abb','81c2c92e','92722c85',
'a2bfe8a1','a81a664b','c24b8b70','c76c51a3',
'd192e819','d6990624','f40e3585','106aa070',
'19a4c116','1e376c08','2748774c','34b0bcb5',
'391c0cb3','4ed8aa4a','5b9cca4f','682e6ff3',
'748f82ee','78a5636f','84c87814','8cc70208',
'90befffa','a4506ceb','bef9a3f7','c67178f2'
]
def bin_return(dec):
return(str(format(dec,'b')))
def bin_8bit(dec):
return(str(format(dec,'08b')))
def bin_32bit(dec):
return(str(format(dec,'032b')))
def bin_64bit(dec):
return(str(format(dec,'064b')))
def hex_return(dec):
return(str(format(dec,'x')))
def dec_return_bin(bin_string):
return(int(bin_string,2))
def dec_return_hex(hex_string):
return(int(hex_string,16))
def L_P(SET,n):
to_return=[]
j=0
k=n
while k<len(SET)+1:
to_return.append(SET[j:k])
j=k
k+=n
return(to_return)
def s_l(bit_string):
bit_list=[]
for i in range(len(bit_string)):
bit_list.append(bit_string[i])
return(bit_list)
def l_s(bit_list):
bit_string=''
for i in range(len(bit_list)):
bit_string+=bit_list[i]
return(bit_string)
def rotate_right(bit_string,n):
bit_list = s_l(bit_string)
count=0
while count <= n-1:
list_main=list(bit_list)
var_0=list_main.pop(-1)
list_main=list([var_0]+list_main)
bit_list=list(list_main)
count+=1
return(l_s(list_main))
def shift_right(bit_string,n):
bit_list=s_l(bit_string)
count=0
while count <= n-1:
bit_list.pop(-1)
count+=1
front_append=['0']*n
return(l_s(front_append+bit_list))
def mod_32_addition(input_set):
value=0
for i in range(len(input_set)):
value+=input_set[i]
mod_32 = 4294967296
return(value%mod_32)
def xor_2str(bit_string_1,bit_string_2):
xor_list=[]
for i in range(len(bit_string_1)):
if bit_string_1[i]=='0' and bit_string_2[i]=='0':
xor_list.append('0')
if bit_string_1[i]=='1' and bit_string_2[i]=='1':
xor_list.append('0')
if bit_string_1[i]=='0' and bit_string_2[i]=='1':
xor_list.append('1')
if bit_string_1[i]=='1' and bit_string_2[i]=='0':
xor_list.append('1')
return(l_s(xor_list))
def and_2str(bit_string_1,bit_string_2):
and_list=[]
for i in range(len(bit_string_1)):
if bit_string_1[i]=='1' and bit_string_2[i]=='1':
and_list.append('1')
else:
and_list.append('0')
return(l_s(and_list))
def or_2str(bit_string_1,bit_string_2):
or_list=[]
for i in range(len(bit_string_1)):
if bit_string_1[i]=='0' and bit_string_2[i]=='0':
or_list.append('0')
else:
or_list.append('1')
return(l_s(or_list))
def not_str(bit_string):
not_list=[]
for i in range(len(bit_string)):
if bit_string[i]=='0':
not_list.append('1')
else:
not_list.append('0')
return(l_s(not_list))
'''
SHA-256 Specific Functions:
'''
def Ch(x,y,z):
return(xor_2str(and_2str(x,y),and_2str(not_str(x),z)))
def Maj(x,y,z):
return(xor_2str(xor_2str(and_2str(x,y),and_2str(x,z)),and_2str(y,z)))
def e_0(x):
return(xor_2str(xor_2str(rotate_right(x,2),rotate_right(x,13)),rotate_right(x,22)))
def e_1(x):
return(xor_2str(xor_2str(rotate_right(x,6),rotate_right(x,11)),rotate_right(x,25)))
def s_0(x):
return(xor_2str(xor_2str(rotate_right(x,7),rotate_right(x,18)),shift_right(x,3)))
def s_1(x):
return(xor_2str(xor_2str(rotate_right(x,17),rotate_right(x,19)),shift_right(x,10)))
def message_pad(bit_list):
pad_one = bit_list + '1'
pad_len = len(pad_one)
k=0
while ((pad_len+k)-448)%512 != 0:
k+=1
back_append_0 = '0'*k
back_append_1 = bin_64bit(len(bit_list))
return(pad_one+back_append_0+back_append_1)
def message_bit_return(string_input):
bit_list=[]
for i in range(len(string_input)):
bit_list.append(bin_8bit(ord(string_input[i])))
return(l_s(bit_list))
def message_pre_pro(input_string):
bit_main = message_bit_return(input_string)
return(message_pad(bit_main))
def message_parsing(input_string):
return(L_P(message_pre_pro(input_string),32))
def message_schedule(index,w_t):
new_word = bin_32bit(mod_32_addition([int(s_1(w_t[index-2]),2),int(w_t[index-7],2),int(s_0(w_t[index-15]),2),int(w_t[index-16],2)]))
return(new_word)
'''
This example of SHA_256 works for an input string >56 characters.
'''
def sha_256(input_string):
w_t=message_parsing(input_string)
a=bin_32bit(dec_return_hex(initial_hash_values[0]))
b=bin_32bit(dec_return_hex(initial_hash_values[1]))
c=bin_32bit(dec_return_hex(initial_hash_values[2]))
d=bin_32bit(dec_return_hex(initial_hash_values[3]))
e=bin_32bit(dec_return_hex(initial_hash_values[4]))
f=bin_32bit(dec_return_hex(initial_hash_values[5]))
g=bin_32bit(dec_return_hex(initial_hash_values[6]))
h=bin_32bit(dec_return_hex(initial_hash_values[7]))
for i in range(0,64):
if i <= 15:
t_1=mod_32_addition([int(h,2),int(e_1(e),2),int(Ch(e,f,g),2),int(sha_256_constants[i],16),int(w_t[i],2)])
t_2=mod_32_addition([int(e_0(a),2),int(Maj(a,b,c),2)])
h=g
g=f
f=e
e=mod_32_addition([int(d,2),t_1])
d=c
c=b
b=a
a=mod_32_addition([t_1,t_2])
a=bin_32bit(a)
e=bin_32bit(e)
if i > 15:
w_t.append(message_schedule(i,w_t))
t_1=mod_32_addition([int(h,2),int(e_1(e),2),int(Ch(e,f,g),2),int(sha_256_constants[i],16),int(w_t[i],2)])
t_2=mod_32_addition([int(e_0(a),2),int(Maj(a,b,c),2)])
h=g
g=f
f=e
e=mod_32_addition([int(d,2),t_1])
d=c
c=b
b=a
a=mod_32_addition([t_1,t_2])
a=bin_32bit(a)
e=bin_32bit(e)
hash_0 = mod_32_addition([dec_return_hex(initial_hash_values[0]),int(a,2)])
hash_1 = mod_32_addition([dec_return_hex(initial_hash_values[1]),int(b,2)])
hash_2 = mod_32_addition([dec_return_hex(initial_hash_values[2]),int(c,2)])
hash_3 = mod_32_addition([dec_return_hex(initial_hash_values[3]),int(d,2)])
hash_4 = mod_32_addition([dec_return_hex(initial_hash_values[4]),int(e,2)])
hash_5 = mod_32_addition([dec_return_hex(initial_hash_values[5]),int(f,2)])
hash_6 = mod_32_addition([dec_return_hex(initial_hash_values[6]),int(g,2)])
hash_7 = mod_32_addition([dec_return_hex(initial_hash_values[7]),int(h,2)])
final_hash = (hex_return(hash_0),
hex_return(hash_1),
hex_return(hash_2),
hex_return(hash_3),
hex_return(hash_4),
hex_return(hash_5),
hex_return(hash_6),
hex_return(hash_7))
return(final_hash)
【讨论】：
@VladislavsDovgalecs 你应该更担心人们正在使用 SAT 求解器找到 fixed point collisions。 @gkcn 我也听说过常量选择推理，但是您应该尝试找到的是他们使用的编码，使用了多少位精度，big-endian 或 little-edian，是编码过程中去掉的小数点，是用于将 base-10 转换为 base-2 的 4 位块，最后他们决定如何处理舍入问题。 @nickandross 你写了一个更简洁的实现！你现在应该为 SHA-3 写一个！【参考方案3】：W_t 来自当前正在处理的块，而 K_t 是由迭代次数确定的固定常数。 SHA256 中每个块的压缩函数重复 64 次。每次迭代 0
我使用 Python 3.6 提供了自己的 SHA256 实现。元组 K 包含 K_t 的 64 个常数值。 Sha256 函数显示了如何在列表 W 中计算 W_t 的值。实现的重点是代码清晰而不是高性能。
W = 32 #Number of bits in word
M = 1 << W
FF = M - 1 #0xFFFFFFFF (for performing addition mod 2**32)
#Constants from SHA256 definition
K = (0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5,
0x3956c25b, 0x59f111f1, 0x923f82a4, 0xab1c5ed5,
0xd807aa98, 0x12835b01, 0x243185be, 0x550c7dc3,
0x72be5d74, 0x80deb1fe, 0x9bdc06a7, 0xc19bf174,
0xe49b69c1, 0xefbe4786, 0x0fc19dc6, 0x240ca1cc,
0x2de92c6f, 0x4a7484aa, 0x5cb0a9dc, 0x76f988da,
0x983e5152, 0xa831c66d, 0xb00327c8, 0xbf597fc7,
0xc6e00bf3, 0xd5a79147, 0x06ca6351, 0x14292967,
0x27b70a85, 0x2e1b2138, 0x4d2c6dfc, 0x53380d13,
0x650a7354, 0x766a0abb, 0x81c2c92e, 0x92722c85,
0xa2bfe8a1, 0xa81a664b, 0xc24b8b70, 0xc76c51a3,
0xd192e819, 0xd6990624, 0xf40e3585, 0x106aa070,
0x19a4c116, 0x1e376c08, 0x2748774c, 0x34b0bcb5,
0x391c0cb3, 0x4ed8aa4a, 0x5b9cca4f, 0x682e6ff3,
0x748f82ee, 0x78a5636f, 0x84c87814, 0x8cc70208,
0x90befffa, 0xa4506ceb, 0xbef9a3f7, 0xc67178f2)
#Initial values for compression function
I = (0x6a09e667, 0xbb67ae85, 0x3c6ef372, 0xa54ff53a,
0x510e527f, 0x9b05688c, 0x1f83d9ab, 0x5be0cd19)
def RR(x, b):
'''
32-bit bitwise rotate right
'''
return ((x >> b) | (x << (W - b))) & FF
def Pad(W):
'''
Pads a message and converts to byte array
'''
mdi = len(W) % 64
L = (len(W) << 3).to_bytes(8, 'big') #Binary of len(W) in bits
npad = 55 - mdi if mdi < 56 else 119 - mdi #Pad so 64 | len; add 1 block if needed
return bytes(W, 'ascii') + b'\x80' + (b'\x00' * npad) + L #64 | 1 + npad + 8 + len(W)
def Sha256CF(Wt, Kt, A, B, C, D, E, F, G, H):
'''
SHA256 Compression Function
'''
Ch = (E & F) ^ (~E & G)
Ma = (A & B) ^ (A & C) ^ (B & C) #Major
S0 = RR(A, 2) ^ RR(A, 13) ^ RR(A, 22) #Sigma_0
S1 = RR(E, 6) ^ RR(E, 11) ^ RR(E, 25) #Sigma_1
T1 = H + S1 + Ch + Wt + Kt
return (T1 + S0 + Ma) & FF, A, B, C, (D + T1) & FF, E, F, G
def Sha256(M):
'''
Performs SHA256 on an input string
M: The string to process
return: A 32 byte array of the binary digest
'''
M = Pad(M) #Pad message so that length is divisible by 64
DG = list(I) #Digest as 8 32-bit words (A-H)
for j in range(0, len(M), 64): #Iterate over message in chunks of 64
S = M[j:j + 64] #Current chunk
W = [0] * 64
W[0:16] = [int.from_bytes(S[i:i + 4], 'big') for i in range(0, 64, 4)]
for i in range(16, 64):
s0 = RR(W[i - 15], 7) ^ RR(W[i - 15], 18) ^ (W[i - 15] >> 3)
s1 = RR(W[i - 2], 17) ^ RR(W[i - 2], 19) ^ (W[i - 2] >> 10)
W[i] = (W[i - 16] + s0 + W[i-7] + s1) & FF
A, B, C, D, E, F, G, H = DG #State of the compression function
for i in range(64):
A, B, C, D, E, F, G, H = Sha256CF(W[i], K[i], A, B, C, D, E, F, G, H)
DG = [(X + Y) & FF for X, Y in zip(DG, (A, B, C, D, E, F, G, H))]
return b''.join(Di.to_bytes(4, 'big') for Di in DG) #Convert to byte array
if __name__ == "__main__":
bd = Sha256('Hello World')
print(''.join(':02x'.format(i) for i in bd))
【讨论】：
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