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社区首页 >问答首页 >AES - SBOX雪崩效应问题

AES - SBOX雪崩效应问题
EN

Cryptography用户
提问于 2022-11-26 19:22:34
回答 1查看 173关注 0票数 1

我正在制作一个程序来验证AES块密码的SBOX在一定次数的迭代中的雪崩效应。该程序在输入列表(state)的一个元素中随机更改一位,然后将subBytes操作应用于原始列表和修改列表。在此之后,它计算这两个结果列表之间的不同位数。

代码语言:javascript
运行
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sbox = [
    0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5, 0x30, 0x01,   0x67, 0x2b, 0xfe, 0xd7, 0xab, 0x76,
    0xca, 0x82, 0xc9, 0x7d, 0xfa, 0x59, 0x47, 0xf0, 0xad, 0xd4, 0xa2, 0xaf, 0x9c, 0xa4, 0x72, 0xc0,
    0xb7, 0xfd, 0x93, 0x26, 0x36, 0x3f, 0xf7, 0xcc, 0x34, 0xa5, 0xe5, 0xf1, 0x71, 0xd8, 0x31, 0x15,
    0x04, 0xc7, 0x23, 0xc3, 0x18, 0x96, 0x05, 0x9a, 0x07, 0x12, 0x80, 0xe2, 0xeb, 0x27, 0xb2, 0x75,
    0x09, 0x83, 0x2c, 0x1a, 0x1b, 0x6e, 0x5a, 0xa0, 0x52, 0x3b, 0xd6, 0xb3, 0x29, 0xe3, 0x2f, 0x84,
    0x53, 0xd1, 0x00, 0xed, 0x20, 0xfc, 0xb1, 0x5b, 0x6a, 0xcb, 0xbe, 0x39, 0x4a, 0x4c, 0x58, 0xcf,
    0xd0, 0xef, 0xaa, 0xfb, 0x43, 0x4d, 0x33, 0x85, 0x45, 0xf9, 0x02, 0x7f, 0x50, 0x3c, 0x9f, 0xa8,
    0x51, 0xa3, 0x40, 0x8f, 0x92, 0x9d, 0x38, 0xf5, 0xbc, 0xb6, 0xda, 0x21, 0x10, 0xff, 0xf3, 0xd2,
    0xcd, 0x0c, 0x13, 0xec, 0x5f, 0x97, 0x44, 0x17, 0xc4, 0xa7, 0x7e, 0x3d, 0x64, 0x5d, 0x19, 0x73,
    0x60, 0x81, 0x4f, 0xdc, 0x22, 0x2a, 0x90, 0x88, 0x46, 0xee, 0xb8, 0x14, 0xde, 0x5e, 0x0b, 0xdb,
    0xe0, 0x32, 0x3a, 0x0a, 0x49, 0x06, 0x24, 0x5c, 0xc2, 0xd3, 0xac, 0x62, 0x91, 0x95, 0xe4, 0x79,
    0xe7, 0xc8, 0x37, 0x6d, 0x8d, 0xd5, 0x4e, 0xa9, 0x6c, 0x56, 0xf4, 0xea, 0x65, 0x7a, 0xae, 0x08,
    0xba, 0x78, 0x25, 0x2e, 0x1c, 0xa6, 0xb4, 0xc6, 0xe8, 0xdd, 0x74, 0x1f, 0x4b, 0xbd, 0x8b, 0x8a,
    0x70, 0x3e, 0xb5, 0x66, 0x48, 0x03, 0xf6, 0x0e, 0x61, 0x35, 0x57, 0xb9, 0x86, 0xc1, 0x1d, 0x9e,
    0xe1, 0xf8, 0x98, 0x11, 0x69, 0xd9, 0x8e, 0x94, 0x9b, 0x1e, 0x87, 0xe9, 0xce, 0x55, 0x28, 0xdf,
    0x8c, 0xa1, 0x89, 0x0d, 0xbf, 0xe6, 0x42, 0x68, 0x41, 0x99, 0x2d, 0x0f, 0xb0, 0x54, 0xbb, 0x16
]
sboxInv = [
    0x52, 0x09, 0x6a, 0xd5, 0x30, 0x36, 0xa5, 0x38, 0xbf, 0x40, 0xa3, 0x9e, 0x81, 0xf3, 0xd7, 0xfb,
    0x7c, 0xe3, 0x39, 0x82, 0x9b, 0x2f, 0xff, 0x87, 0x34, 0x8e, 0x43, 0x44, 0xc4, 0xde, 0xe9, 0xcb,
    0x54, 0x7b, 0x94, 0x32, 0xa6, 0xc2, 0x23, 0x3d, 0xee, 0x4c, 0x95, 0x0b, 0x42, 0xfa, 0xc3, 0x4e,
    0x08, 0x2e, 0xa1, 0x66, 0x28, 0xd9, 0x24, 0xb2, 0x76, 0x5b, 0xa2, 0x49, 0x6d, 0x8b, 0xd1, 0x25,
    0x72, 0xf8, 0xf6, 0x64, 0x86, 0x68, 0x98, 0x16, 0xd4, 0xa4, 0x5c, 0xcc, 0x5d, 0x65, 0xb6, 0x92,
    0x6c, 0x70, 0x48, 0x50, 0xfd, 0xed, 0xb9, 0xda, 0x5e, 0x15, 0x46, 0x57, 0xa7, 0x8d, 0x9d, 0x84,
    0x90, 0xd8, 0xab, 0x00, 0x8c, 0xbc, 0xd3, 0x0a, 0xf7, 0xe4, 0x58, 0x05, 0xb8, 0xb3, 0x45, 0x06,
    0xd0, 0x2c, 0x1e, 0x8f, 0xca, 0x3f, 0x0f, 0x02, 0xc1, 0xaf, 0xbd, 0x03, 0x01, 0x13, 0x8a, 0x6b,
    0x3a, 0x91, 0x11, 0x41, 0x4f, 0x67, 0xdc, 0xea, 0x97, 0xf2, 0xcf, 0xce, 0xf0, 0xb4, 0xe6, 0x73,
    0x96, 0xac, 0x74, 0x22, 0xe7, 0xad, 0x35, 0x85, 0xe2, 0xf9, 0x37, 0xe8, 0x1c, 0x75, 0xdf, 0x6e,
    0x47, 0xf1, 0x1a, 0x71, 0x1d, 0x29, 0xc5, 0x89, 0x6f, 0xb7, 0x62, 0x0e, 0xaa, 0x18, 0xbe, 0x1b, 
    0xfc, 0x56, 0x3e, 0x4b, 0xc6, 0xd2, 0x79, 0x20, 0x9a, 0xdb, 0xc0, 0xfe, 0x78, 0xcd, 0x5a, 0xf4, 
    0x1f, 0xdd, 0xa8, 0x33, 0x88, 0x07, 0xc7, 0x31, 0xb1, 0x12, 0x10, 0x59, 0x27, 0x80, 0xec, 0x5f, 
    0x60, 0x51, 0x7f, 0xa9, 0x19, 0xb5, 0x4a, 0x0d, 0x2d, 0xe5, 0x7a, 0x9f, 0x93, 0xc9, 0x9c, 0xef, 
    0xa0, 0xe0, 0x3b, 0x4d, 0xae, 0x2a, 0xf5, 0xb0, 0xc8, 0xeb, 0xbb, 0x3c, 0x83, 0x53, 0x99, 0x61, 
    0x17, 0x2b, 0x04, 0x7e, 0xba, 0x77, 0xd6, 0x26, 0xe1, 0x69, 0x14, 0x63, 0x55, 0x21, 0x0c, 0x7d
]
def subBytes(state):
    result = [None] * 16
    for i in range(len(state)):
        result[i] = sbox[state[i]]
    return result

def subBytesInv(state):
    result = [None] * 16
    for i in range(len(state)):
        result[i] = sboxInv[state[i]]
    return result

#Changes a random bit in one of the elements of the state
def changeBit(state):
    result = state[:]
    result[random.randrange(0,16)] ^= 1 << random.randrange(0,8)
    return result

#Return index of first element that is different in the list
def findDiff(list1, list2):
    for i in range(len(list1)):
        if list1[i] != list2[i]:
            return i
    return -1

#Count number of different bits between two integers
def countDiffBits(a, b):
    return bin(a ^ b).count('1')


#Default array
state=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]


average = []
for i in range(10000):
    #Create new state with random bit changed
    changed_state = changeBit(state)

    #Apply bin() to each element and print
    print(state)
    sbox_state = subBytes(state) 
    print(list(map(bin,sbox_state)))

    print(changed_state)
    sbox_changed_state = subBytes(changed_state) 
    print(list(map(bin,sbox_changed_state)))

    index = findDiff(state,changed_state)
    print("Changed bits: " + str(countDiffBits(sbox_state[index],sbox_changed_state[index])))

    average.append(countDiffBits(sbox_state[index],sbox_changed_state[index]))

print("Statistics: " + str(Counter(average)))

这是输出:

代码语言:javascript
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Statistics: Counter({5: 2611, 3: 2323, 4: 2236, 2: 1682, 6: 656, 1: 340, 8: 77, 7: 75})

但是,正如您所看到的,在10000次迭代中,有340个例子在两个列表之间只改变了一个位,这意味着雪崩效应没有发生。我现在的问题是,这是代码中的某种错误,还是我误解了这个概念。

EN

回答 1

Cryptography用户

回答已采纳

发布于 2022-11-26 22:21:22

但是,正如您所看到的,在10000次迭代中,有340个例子在这两个列表之间只更改了一个位。

比预期的要高一点。对于AES SBox,有24对输入(计数02和0a作为0a和02的相同对)与单个比特输入差导致一个单比特输出差;从随机抽样的10000对中,人们会期望看到大约234个对。

更新:我看到了问题:您的代码将这对的第一组设置为0到15之间的随机值(而不是255)。结果表明,AES有三个这样的对,它们产生一个单比特差分输出(02,0a04,1408,28);这给出了一个预期的312匹配( 10,000);它和所看到的340之间的差异可以归因于随机性。

我误解了这个概念。

实际上,AES并不依赖sbox来雪崩( MixCollumns和ShiftRows的组合很好地做到了这一点);相反,它依赖sbox具有较低的概率来抵抗线性和差分特性(而少量1位-> 1位差分对的存在并不与此相冲突)。

票数 2
EN
页面原文内容由Cryptography提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://crypto.stackexchange.com/questions/102984

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