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How to use compute_hamming_distance method in avocado

Best Python code snippet using avocado_python

vinegere.py

Source:vinegere.py

...72        b3 = bytes_str[i*2:i*3+1]73        b4 = bytes_str[i*3:i*4+1]74        # Average the distances from the 4 blocks of bytes75        distances.append(76            (compute_hamming_distance(b1, b2) / i) +77            (compute_hamming_distance(b1, b3) / i) +78            (compute_hamming_distance(b1, b4) / i) +79            (compute_hamming_distance(b2, b3) / i) +80            (compute_hamming_distance(b2, b4) / i) +81            (compute_hamming_distance(b3, b4) / i) / 682        )83        key_sizes.append(i)84    return key_sizes[distances.index(min(distances))]85def compute_hamming_distance(b1, b2):86    distance_result_str = bytes_to_long(b1) ^ bytes_to_long(b2)87    binary_rep = bin(distance_result_str)[2:]88    distance = 089    for char in binary_rep:90        if char == '1':91            distance += 192    return distance93def detect_single_byte_xor(cipher_strings):94    results_list = list()95    score_list = list()96    for cipher_str in cipher_strings:97        result, key, score = single_byte_xor_solver(cipher_str)98        results_list.append(result)99        score_list.append(score)...

c16.py

Source:c16.py

1import base642import math3import c124import c135def compute_hamming_distance(b1: bytes, b2: bytes) -> int:6    """Return the number of differing bits"""7    assert len(b1) == len(b2)8    dist = 09    for i, j in zip(b1, b2):10        x = i ^ j11        dist += format(x, 'b').count('1')12    return dist13def break_repeating_key_xor(cipher: bytes) -> tuple:14    # Guess the key size in bytes15    ham_avg = []16    max_key_size = 4017    for ks in range(2, max_key_size + 1):18        # Try with 4 KEYSIZE blocks give us better avg Hamming distance then 2 KEYSIZE blocks19        dist = compute_hamming_distance(cipher[:2*ks], cipher[2*ks:4 * ks])20        ham_avg.append({'ks': ks, 'ham': dist / (2*ks)})21    ham_avg.sort(key=lambda h1: h1['ham'])22    # Try 3 keysize with smallest avg Hamming distance23    max_try = 424    plain_bytes = None25    key_bytes = None26    min_score = math.inf27    for d in ham_avg[:max_try]:28        print(f'Try key size: {d["ks"]}, Avg Hamming Distance: {d["ham"]}')29        keysize = d['ks']30        num_repeat = len(cipher) // keysize31        num_remain = len(cipher) % keysize32        # Split cipher into list of separated single-byte XOR33        single_byte_xor = []34        for i in range(0, keysize):35            cipher_i = [cipher[i + j * keysize] for j in range(0, num_repeat)]36            # Check out of index37            if num_remain and i < num_remain:38                cipher_i.append(cipher[i + num_repeat * keysize])39            single_byte_xor.append(bytes(cipher_i))40            # print(len(cipher_i), single_byte_xor[i].hex())41        single_xor_msg = []42        for i in range(0, keysize):43            decrypt_msg, score = c13.break_single_byte_xor(single_byte_xor[i])44            if decrypt_msg:45                single_xor_msg.append(decrypt_msg)46        if len(single_xor_msg) == keysize:47            # Merge keysize strings into one string48            print('Found a message')49            msg = ''50            for m in zip(*single_xor_msg):51                msg += ''.join(m)52            # Handle the remain characters53            remain = ''.join(m[len(m) - 1] for m in single_xor_msg if len(m) > num_repeat)54            msg += remain55            guess_key = c12.xor_bytes(cipher[:keysize], msg[:keysize].encode())56            if c13.compute_score(msg) < min_score:57                print('Update the new message as best guess message')58                plain_bytes = msg.encode()59                key_bytes = guess_key60    return plain_bytes, key_bytes61if __name__ == '__main__':62    s1 = 'this is a test'63    s2 = 'wokka wokka!!!'64    result = compute_hamming_distance(s1.encode(), s2.encode())65    assert result == 3766    test = compute_hamming_distance(b'B', b'A')67    assert test == 268    with open('6.txt', 'r') as f:69        data = f.read().replace('\n', '')70        cipher = base64.b64decode(data)71        plain, key = break_repeating_key_xor(cipher)72        print(f'Decrypted Message:\n{plain.decode()}')...

SkewArray.py

Source:SkewArray.py

...19    return positions20#We say that position i in k-mers p and q is a mismatch if the symbols at position i21# of the two strings are not the same. The total number of mismatches between strings p and q22# is called the Hamming distance between these strings. 23def compute_hamming_distance(p,q):24    num_of_mismatches = 025    for i in range(len(p)):26        if p[i] != q[i]:27            num_of_mismatches += 128    return num_of_mismatches29#We say that a k-mer Pattern appears as a substring of Text with at most d mismatches 30# f there is some k-mer substring Pattern' of Text having d or fewer mismatches with Pattern; 31# that is, HammingDistance(Pattern, Pattern') â¤ d. Our observation that a DnaA box may appear 32# with slight variations leads to the following generalization of the Pattern Matching Problem.33def compute_approximate_pattern_matching(Text, Pattern, d):34    positions = [] #Initializing list of positions35    for i in range(len(Text) - len(Pattern) + 1):36        if compute_hamming_distance(Text[i:i+len(Pattern)], Pattern) <= d:37            positions.append(i)38    return positions39#This function computes the number of occurrences of Pattern in Text with at most d mismatches.40def compute_approximate_pattern_count(Text, Pattern, d):41    count = 042    for i in range(len(Text) - len(Pattern) + 1):43        if compute_hamming_distance(Text[i:i+len(Pattern)], Pattern) <= d:44            count += 145    return count46text = 'AGCGTGCCGAAATATGCCGCCAGACCTGCTGCGGTGGCCTCGCCGACTTCACGGATGCCAAGTGCATAGAGGAAGCGAGCAAAGGTGGTTTCTTTCGCTTTATCCAGCGCGTTAACCACGTTCTGTGCCGACTTT'...

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