Python version of a quick check of the Freestart collision published here https://sites.google.com/site/itstheshappening/

sha1_freestart.py

import struct, binascii

def _left_rotate(n, b):
    return ((n << b) | (n >> (32 - b))) & 0xffffffff
    
def sha1(IV, message):
    # Initialize variables:
    (h0, h1, h2, h3, h4) = IV
    
    # Pre-processing:
    original_byte_len = len(message)
    original_bit_len = original_byte_len * 8
    # append the bit '1' to the message
    message += b'\x80'
    
    # append 0 <= k < 512 bits '0', so that the resulting message length (in bits)
    #    is congruent to 448 (mod 512)
    message += b'\x00' * ((56 - (original_byte_len + 1) % 64) % 64)
    
    # append length of message (before pre-processing), in bits, as 64-bit big-endian integer
    message += struct.pack(b'>Q', original_bit_len)
    # Process the message in successive 512-bit chunks:
    # break message into 512-bit chunks
    for i in range(0, len(message), 64):
        w = [0] * 80
        # break chunk into sixteen 32-bit big-endian words w[i]
        for j in range(16):
            w[j] = struct.unpack(b'>I', message[i + j*4:i + j*4 + 4])[0]
        # Extend the sixteen 32-bit words into eighty 32-bit words:
        for j in range(16, 80):
            w[j] = _left_rotate(w[j-3] ^ w[j-8] ^ w[j-14] ^ w[j-16], 1)
    
        # Initialize hash value for this chunk:
        a = h0
        b = h1
        c = h2
        d = h3
        e = h4
    
        for i in range(80):
            if 0 <= i <= 19:
                # Use alternative 1 for f from FIPS PB 180-1 to avoid ~
                f = d ^ (b & (c ^ d))
                k = 0x5A827999
            elif 20 <= i <= 39:
                f = b ^ c ^ d
                k = 0x6ED9EBA1
            elif 40 <= i <= 59:
                f = (b & c) | (b & d) | (c & d) 
                k = 0x8F1BBCDC
            elif 60 <= i <= 79:
                f = b ^ c ^ d
                k = 0xCA62C1D6
    
            a, b, c, d, e = ((_left_rotate(a, 5) + f + e + k + w[i]) & 0xffffffff, 
                            a, _left_rotate(b, 30), c, d)
    
        # sAdd this chunk's hash to result so far:
        h0 = (h0 + a) & 0xffffffff
        h1 = (h1 + b) & 0xffffffff 
        h2 = (h2 + c) & 0xffffffff
        h3 = (h3 + d) & 0xffffffff
        h4 = (h4 + e) & 0xffffffff
    
    # Produce the final hash value (big-endian):
    return '%08x%08x%08x%08x%08x' % (h0, h1, h2, h3, h4)
    
IV1 = [0x506b0178, 0xff6d1890, 0x202291fd, 0x3ade3871, 0xb2c665ea]
M1 = '9d443828a5ea3df086eaa0fa7783a7363324484daf702aaaa3dab679d8a69e2d543820eda7fffb52d3ff493fc3ff551efbffd97f55feeef2085af312088688a9'
hash1 = sha1(IV1, binascii.unhexlify(M1))

IV2 = [0x506b0178, 0xff6d1891, 0xa02291fd, 0x3ade3871, 0xb2c665ea];
M2 = '3f44383881ea3deca0eaa0ee5183a72c3324485dab702ab66fdab66dd4a69e2f943820fd13fffb4eefff493b7fff5504dbffd96f71feeeeee45af306048688ab'
hash2 = sha1(IV2, binascii.unhexlify(M2))

print hash1
print hash2
print hash1 == hash2

If you want it without padding which will give the hash same hash values as in the paper comment out line 14 to 21