仿射密码破解的同余计算问题 #832
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HinataYiacc
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@HinataYiacc 这两个在模意义下是相等的。 |
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你好,仿射密码破解这部分有个公式我不太理解。
文中提到:
通过:
y1 = (ax1 + b) mod 26 ①
y2 = (ax2 + b) mod 26 ②
①-②可以得到公式:y1 - y2 = (a(x1-x2)) mod 26
但是对于上述推导,如果:
1 = 6 % 5
2 = 12 % 5
1 - 2 ≠ (6-12)mod 5
应该是:(y1 - y2)mod 26 = (a(x1-x2)) mod 26
是否还有其他约束条件我没注意
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