matlab-(含教程)基于RSA-Hash算法的文字加密系统,将文字解密到图像中并通过解密提取文字信息

def euclidAlgo(larger, smaller): if(smaller > larger): temp = larger larger = smaller smaller = temp rem = larger % smaller if(rem == 0): return smaller return euclidAlgo(smaller, rem) def extEuclidAlgo(a, b): a, b = abs(a), abs(b) x0, x1, y0, y1 = 1, 0, 0, 1 while b != 0: q, a, b = a // b, b, a % b x0, x1 = x1, x0 - q * x1 y0, y1 = y1, y0 - q * y1 return a, x0, y0 def generatePublic(p,q): n = p*q phi = (p-1)*(q-1) e = 2 while(euclidAlgo(e,phi) != 1): e = e+1 if(e >= phi): return( "not possible since e < phi(n), phi(n)={0} and e = {1}" .format(phi,e)) return e def generatePrivate(e,n): phi = eulerPhi(n) gcd = euclidAlgo(e,phi) if(gcd == 1): d = extEuclidAlgo(e,phi)[1] else: return ("not possible since the gcd of {0} and {1} is {2} and not 1" .format(e,phi,gcd)) return d def primeFactors(number): factors = [] for i in range(2, number + 1): if(number % i == 0): prime = True for j in range(2, (i//2 + 1)): if(i % j == 0): prime = False break if (prime): factors.append(i) return factors def eulerPhi(n): factors = primeFactors(n) phi = n for i in range (0, len(factors)): temp = 1 - (1/factors[i]) phi *= temp return int(phi) def encryptRSA(x,e,n): return pow(x,e,n) def decryptRSA(y,d,n): return pow(y,d,n) def paarExample(): print("Paar Example, p=3, q=11") p = 3 q = 11 e = generatePublic(p,q) d = generatePrivate(e,p*q) print("phi: {0}" .format(eulerPhi(p*q))) print("encryption key: {0}" .format(e)) print("decryption key: {0}" .format(d)) def e3p11a(): print("\nExercise 3, Problem 11a") e = 13 n = 2881 d = generatePrivate(e,n) print("decryption key: {0}" .format(d)) def e3p11b(): print("\nExercise 3, Problem 11b") e = 7 n = 2881 d = generatePrivate(e,n) print("decryption key: {0}" .format(d)) def e3p12(): print("\nExercise 3, Problem 12") e = 13 n = 35209 d = generatePrivate(e,n) print("decryption key: {0}" .format(d)) def e3p14(): print("\nExercise 3, Problem 14") n = [1997,1999,2001,2002,2003,2004,2005] phi = [0,0,0,0,0,0,0] for i in range (0,len(n)): phi[i] = eulerPhi(n[i]) print("phi({0}) = {1}" .format(n[i],phi[i])) def e3pEC(): print("\nExercise 3, Extra Credit") e = 13 n = 2881 d = generatePrivate(e,n) text = "To be, or not to be- that is the question: Whether 'tis nobler in the mind to suffer The slings and arrows of outrageous fortune Or to take arms against a sea of troubles, And by opposing end them. To die- to sleep- No more; and by a sleep to say we end The heartache, and the thousand natural shocks That flesh is heir to. 'Tis a consummation Devoutly to be wish'd. To die- to sleep. To sleep- perchance to dream: ay, there's the rub!" textEncoded = list(text.encode()) textEncrypted = [] for i in range(0,len(textEncoded)): textEncrypted.append(encryptRSA(textEncoded[i],e,n)) textDecrypted = [] for i in range(0,len(textEncrypted)): textDecrypted.append(decryptRSA(textEncrypted[i],d,n)) textDecryptedPT = bytes(textDecrypted).decode() print("Original Text: {0}" .format(text)) print("Encoded Text: {0}" .format(textEncoded)) print("Encrypted Encoded Text: {0}" .format(textEncrypted)) print("Decrypted Encoded Text: {0}" .format(textDecrypted)) print("Decrypted Decoded Text: {0}" .format(textDecryptedPT)) part = "start" while(part != "exit"): part = input("Which question would you like to display? [11a, 11b, 12, 14, or EC]: ") if(part == "11a"): e3p11a() elif(part == "11b"): e3p11b() elif(part == "12"): e3p12() elif(part == "14"): e3p14() elif(part == "EC"): e3pEC() elif(part == "all"): e3p11a() e3p11b() e3p12() e3p14() e3pEC()