文件名称:New-folder-(4)
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We begin with choosing two random large distinct primes p and q. We
also pick e, a random integer that is relatively prime to (p-1)*(q-1). The
random integer e is the encryption exponent. Let n = p*q. Using Euclid s
greatest common divisor algorithm, one can compute d, the decryption
exponent, such that:
e*d = 1 (mod (p-1)*(q-1))
Both plaintext m and ciphertext c should be in the set of nonnegative
integers. Furthermore, before encrypting a plaintext message m, we need
to make sure that 0 <= m < n. If m is greater than the modulus n, the result
c of the encryption will not be a unique one-to-one mapping m to c.
From one of the theorems of Euler s, we know that for all integers m,
med = m (mod n) -We begin with choosing two random large distinct primes p and q. We
also pick e, a random integer that is relatively prime to (p-1)*(q-1). The
random integer e is the encryption exponent. Let n = p*q. Using Euclid s
greatest common divisor algorithm, one can compute d, the decryption
exponent, such that:
e*d = 1 (mod (p-1)*(q-1))
Both plaintext m and ciphertext c should be in the set of nonnegative
integers. Furthermore, before encrypting a plaintext message m, we need
to make sure that 0 <= m < n. If m is greater than the modulus n, the result
c of the encryption will not be a unique one-to-one mapping m to c.
From one of the theorems of Euler s, we know that for all integers m,
med = m (mod n)
also pick e, a random integer that is relatively prime to (p-1)*(q-1). The
random integer e is the encryption exponent. Let n = p*q. Using Euclid s
greatest common divisor algorithm, one can compute d, the decryption
exponent, such that:
e*d = 1 (mod (p-1)*(q-1))
Both plaintext m and ciphertext c should be in the set of nonnegative
integers. Furthermore, before encrypting a plaintext message m, we need
to make sure that 0 <= m < n. If m is greater than the modulus n, the result
c of the encryption will not be a unique one-to-one mapping m to c.
From one of the theorems of Euler s, we know that for all integers m,
med = m (mod n) -We begin with choosing two random large distinct primes p and q. We
also pick e, a random integer that is relatively prime to (p-1)*(q-1). The
random integer e is the encryption exponent. Let n = p*q. Using Euclid s
greatest common divisor algorithm, one can compute d, the decryption
exponent, such that:
e*d = 1 (mod (p-1)*(q-1))
Both plaintext m and ciphertext c should be in the set of nonnegative
integers. Furthermore, before encrypting a plaintext message m, we need
to make sure that 0 <= m < n. If m is greater than the modulus n, the result
c of the encryption will not be a unique one-to-one mapping m to c.
From one of the theorems of Euler s, we know that for all integers m,
med = m (mod n)
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下载文件列表
RSA.frm
frmView.frm
frmView.frx
Project.vbp
Project.vbw
RSA.bas
frmView.frm
frmView.frx
Project.vbp
Project.vbw
RSA.bas
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