Larger coverage of hashes

[DonaldTsang]

	224	256	384	512
BMW	- [ ]	- [ ] Faster than MD5	- [ ]	- [ ] Faster than SHA1
ECHO	- [ ]	- [ ] Faster than SHA1	- [ ]	- [ ] Slower than SHA256
ECHO single pipe	- [ ]	- [ ] Faster than SHA1	- [ ]	- [ ] Faster than SHA256
BLAKE2	- [ ] (2s/224)	- [ ] Faster than MD5 (BLAKE2s)	- [ ] (2b/384)	- [ ] Faster than SHA1 (BLAKE2b)
SHAVite3	- [ ]	- [ ] Faster than MD5	- [ ]	- [ ] Faster than SHA1
CubeHash	- [ ] (8/32)	- [ ] (8/32 but for 256-bit)	- [ ] (8/32)	- [ ] Faster than MD5 (CubeHash8/32)
SHABAL	- [ ]	- [ ] Faster than SHA256	- [ ]	- [ ] Faster than SHA512
Skien512	- [ ]	- [ ] Faster than SHA256	- [ ]	- [ ] Faster than SHA256
BLAKE	- [ ]	- [ ] Faster than SHA256	- [ ]	- [ ] Faster than SHA256
BBLAKE	- [ ]	- [ ] Faster than SHA256	- [ ]	- [ ] Slower than SHA256
SKEIN1024	- [ ] (384~448bit)	- [ ] Slower than SHA256 (512bit)	- [ ] (768bit)	- [ ] Faster than SHA256 (1024bit)
SHA3	- [ ]	- [ ] Faster than SHA256	- [ ]	- [ ] Slower than SHA256
Keccak Treed2	- [ ]	- [ ] Faster than MD5	- [ ]	- [ ] Faster than SHA256
SHA2	- [ ]	- [ ]	- [ ]	- [ ]

[zebra-lucky]

Yep, this code is very draft and ported from JS version. Need to refactor and speedup.

[DonaldTsang]

@zebra-lucky Cubehash and keccak should be easy Here are the notes I made for Cubehash

def s(x):
  for i in range(16): x[16+i]=(x[16+i]+x[i])%(1<<32) # step 1
  for i in range(16): x[i]=((x[i]<<7)+(x[i]>>25))%(1<<32)# step 2
  for i in range(8): x[i],x[8+i]=x[8+i],x[i] # step 3
  for i in range(16): x[i]^=x[16+i] # step 4
  for i,j in range(0,16,4), [0,1]: x[16+i+j],x[18+i+j]=x[18+i+j],x[16+i+j] # step 5
  for i in range(16): x[16+i]=(x[16+i]+x[i])%(1<<32) # step 6
  for i in range(16): ((x[i]<<11)+(x[i]>>21))%(1<<32)# step 7
  for i,j in [0,8], range(4): x[i+j],x[4+i+j]=x[4+i+j],x[i+j] # step 8
  for i in range(16): x[i]^=x[16+i] # step 9
  for i in range(0,16,2): x[i],x[16+i]=x[16+i],x[i] # step 10
  return x

def cubehash(msg,i,r,b,f,h):
  assert isinstance(msg,bytes)
  ctx=[h/8,b,r]+[0 for _ in range(29)]
  for _ in range(i): ctx = s(ctx) # initialization
  msg=[msg[n:n+b] for n in range(0,len(msg),b)]
  msg[-1] += (b'\x80'+b'\x00'*(b-len(msg[-1])-1)) if len(msg[-1])!=b else b''
  for m in msg:
     int.from_bytes(???, byteorder='big') # convert and then XOR into ctx
     for _ in range(r): ctx = s(ctx)
  ctx[31]^=1
  for _ in range(f): ctx = s(ctx)
  # return the first h/8 bytes of the ctx

[zebra-lucky]

Thank you for comments!

Sorry, at the moment I have no time to continue enhance code of this repo. This was a dirty try to port JS version without redoing code structure. There was a think to limit expectation of length of input data to all hashes to prev output lenght, and refactor all code structure, but no time left currently.

[DonaldTsang]

@zebra-lucky Does it get the same hash with the reference implementations in C? Just wondering.

[DonaldTsang]

@zebra-lucky here is a short version for cubehash https://github.com/tearsofphoenix/cubehash/blob/master/index.js and one for keccak https://github.com/XKCP/XKCP/blob/master/Standalone/CompactFIPS202/Python/CompactFIPS202.py https://github.com/XKCP/XKCP/blob/master/Standalone/KangarooTwelve/Python/K12.py

[DonaldTsang]

Cubehash and keccak should be easy Here are the notes I made for Cubehash

def s(x):
  for i in range(16): x[16+i]=(x[16+i]+x[i])%(1<<32) # step 1
  for i in range(16): x[i]=((x[i]<<7)+(x[i]>>25))%(1<<32)# step 2
  for i in range(8): x[i],x[8+i]=x[8+i],x[i] # step 3
  for i in range(16): x[i]^=x[16+i] # step 4
  for i,j in range(0,16,4), [0,1]: x[16+i+j],x[18+i+j]=x[18+i+j],x[16+i+j] # step 5
  for i in range(16): x[16+i]=(x[16+i]+x[i])%(1<<32) # step 6
  for i in range(16): ((x[i]<<11)+(x[i]>>21))%(1<<32)# step 7
  for i,j in [0,8], range(4): x[i+j],x[4+i+j]=x[4+i+j],x[i+j] # step 8
  for i in range(16): x[i]^=x[16+i] # step 9
  for i in range(0,16,2): x[i],x[16+i]=x[16+i],x[i] # step 10
  return x

def cubehash(msg,i,r,b,f,h):
  assert isinstance(msg,bytes)
  ctx=[h/8,b,r]+[0 for _ in range(29)]
  for _ in range(i): ctx = s(ctx) # initialization
  msg=[msg[n:n+b] for n in range(0,len(msg),b)]
  msg[-1] += (b'\x80'+b'\x00'*(b-len(msg[-1])-1)) if len(msg[-1])!=b else b''
  for m in msg:
     int.from_bytes(???, byteorder='big') # convert and then XOR into ctx
     for _ in range(r): ctx = s(ctx)
  ctx[31]^=1
  for _ in range(f): ctx = s(ctx)
  # return the first h/8 bytes of the ctx