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= Sage Interactions - Cryptography - Under Construction = This page was be created at Sage Days 103, 7-10 August 2019 by Sarah Arpin, Catalina Camacho-Navarro, Holly Paige Chaos, Amy Feaver, Eva Goedhart, Rebecca Lauren Miller, Alexis Newton, and Nandita Sahajpal. If you have cryptography related interactions that you are interested in adding to this page, please do so. You can also contact Amy Feaver at [email protected] if you have interactions that you are interested in having us add to the page during Sage Days. We will consider them and add them in if we can! |
= Sage Interactions - Cryptography = This page was first created at Sage Days 103, 7-10 August 2019 by Sarah Arpin, Catalina Camacho-Navarro, Holly Paige Chaos, Amy Feaver, Eva Goedhart, Rebecca Lauren Miller, Alexis Newton, and Nandita Sahajpal. Text edited by Amy Feaver, Eva Goedhart, and Alexis Newton. This project was led by Amy Feaver. We acknowledge Katherine Stange, who allowed us to use code from her cryptography course as a starting point for many of these interacts. Dr. Stange's original code and course page can be found at http://crypto.katestange.net/ If you have cryptography-related interactions that you are interested in adding to this page, please do so. You can also contact Amy Feaver at [email protected] |
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The shift cipher is a classical cryptosystem that takes plaintext and shifts it through the alphabet by a given number of letters. For example, a shift of 2 would replace all A's with C's, all B's with D's, etc. When the end of the alphabet is reached, the letters are shifted cyclically back to the beginning. Thus, a shift of 2 would replace Y's with A's and Z's with B's. === Shift Cipher Encryption === |
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A classical cryptosystem that takes your plaintext and shifts it through the alphabet by a given number of letters. -EG === Shift Cipher Encryption === |
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print "Put your message in between the provided quotes (with no additional quotes or apostrophes!), and select your desired shift: " | print "Put your message inside the provided quotes (with no additional quotes or apostrophes!), and select your desired shift: " |
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If you know that your encrypted message is a shift cipher, you can use the known encryption to decrypt. If this is not known, brute force can be used to get possible decrypted messages. |
by Sarah Arpin, Alexis Newton If you know that your message was encrypted using a shift cipher, you can use the known shift value to decrypt. If this is not known, brute force can be used to get 26 possible decrypted messages. The chi-squared function ranks the brute force results by likelihood according to letter frequency. |
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An affine cipher combines the idea of a shift cipher with a multiplicative cipher. In this particular example, we map consecutive letters of the alphabet to consecutive numbers, starting with A=0 (you can also do this cipher differently, and starting with A=1). The user selects two values, a and b. The value a is the multiplier and must be relatively prime to 26 in order to guarantee that each letter is encoded uniquely. The value b is the addend. Each letter's value is multiplied by a, and the product is added to b. This is then replaced with a new letter, corresponding to the result modulo 26. -AF === Affine Cipher Encryption === |
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print "Notice that the only choices for a can be numbers that are relatively prime to 26" | |
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def affine_cipher(message = input_box(default='"secrets"', width = 50), a=[1,3,4..12,14,15..25], b =[0..25]): | def affine_cipher(message = input_box(default='"secrets"', width = 50), a=[1,3,5,7,9,11,15,17,19,21,23], b =[0..25]): |
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=== Affine Cipher Decryption === by Sarah Arpin, Alexis Newton If you know that your message was encrypted using an affine cipher, you can use the known a and b values to decrypt. If this is not known, brute force can be used to get possible decrypted messages. The chi-squared function ranks the results by likelihood according to letter frequency. {{{#!sagecell #Last edited 8/7/2019 3:01pm print "Enter the encrypted text in quotes, and enter a guess for the a and the b:" @interact def shift_decrypt(text = input_box('"XNSILPCVA"'), a=[1,3,5,7,9,11,15,17,19,21,23,25], b =[0..25]): S = AffineCryptosystem(AlphabeticStrings()) ciphertext = S.encoding(text) decrypt = S.deciphering(a,b,ciphertext) print "If the a =", a, "and the b =",b, ", then the original message was:" print decrypt decrypt = S.brute_force(ciphertext,ranking="none") print "\nThese are the possibilities for the plaintext:" print decrypt decrypt = S.brute_force(ciphertext,ranking = "chisquare")[:10] print "\nThese are the top 10 possibilities ranked by likelihood with the chi-squared function:" print decrypt }}} |
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}}} === Playfair Cipher === A special type of substitution cipher in which the plaintext is broken up into two-letter digraphs with some restrictions. Those digraphs are encrypted using a Polybius square, (i.e. a 5x5 grid in which each letter of the alphabet is its own entry with the exception of ij which are placed together). The positions of the letters in the digraph determine how the digraph is encrypted. {{{#!sagecell }}} == Frequency Analysis Decryption Tool == {{{#!sagecell |
}}} == Playfair Cipher == by Catalina Camacho-Navarro Based on code from Alasdair McAndrew at trac.sagemath.org/ticket/8559 A special type of substitution cipher in which the plaintext is broken up into two-letter digraphs with some restrictions. Those digraphs are encrypted using a Polybius square, (i.e. a 5x5 grid in which each letter of the alphabet is its own entry with the exception of ij which are placed together). The positions of the letters in the digraph determine how the digraph is encrypted. -EF {{{#!sagecell ##PLAYFAIR CIPHER def change_to_plain_text(pl): plaintext='' for ch in pl: if ch.isalpha(): plaintext+=ch.upper() return plaintext def makePF(word1): #creates 5 x 5 Playfair array beginning with "word" word=change_to_plain_text(word1) alph='ABCDEFGHIKLMNOPQRSTUVWXYZ' pf='' for ch in word: if (ch<>"J") & (pf.find(ch)==-1): # ensures no letter is repeated pf+=ch for ch in alph: if pf.find(ch)==-1: #only uses unused letters from alph pf+=ch PF=[[pf[5*i+j] for j in range(5)] for i in range(5)] return PF def pf_encrypt(di,PF): # encrypts a digraph di with a Playfair array PF for i in range(5): for j in range(5): if PF[i][j]==di[0]: i0=i j0=j if PF[i][j]==di[1]: i1=i j1=j if (i0<>i1) & (j0<>j1): return PF[i0][j1]+PF[i1][j0] if (i0==i1) & (j0<>j1): return PF[i0][(j0+1)%5]+PF[i1][(j1+1)%5] if (i0<>i1) & (j0==j1): return PF[(i0+1)%5][j0]+PF[(i1+1)%5][j1] def insert(ch,str,j): # a helper function: inserts a character "ch" into tmp='' # a string "str" at position j for i in range(j): tmp+=str[i] tmp+=ch for i in range(len(str)-j): tmp+=str[i+j] return tmp def playfair(pl1,word): # encrypts a plaintext "pl" with a Playfair cipher pl=change_to_plain_text(pl1) PF=makePF(word) # using a keyword "word" pl2=makeDG(pl) tmp='' for i in range(len(pl2)//2): tmp+=pf_encrypt(pl2[2*i]+pl2[2*i+1],PF) return tmp def makeDG(str): # creates digraphs with different values from a string "str" tmp=str.replace('J','I') # replace all 'J's with 'I's c=len(tmp) i=0 while (c>0) & (2*i+1<len(tmp)): if tmp[2*i]==tmp[2*i+1]: tmp=insert("X",tmp,2*i+1) c-=1 i+=1 else: c-=2 i+=1 if len(tmp)%2==1: tmp+='X' return tmp print('Enter your message and the key to construct you polybius square. Warning: both the message and the key must be in quotes.') @interact def _(Message=input_box(default="'message'"),Key=input_box(default="'key'"),showmatrix=checkbox(True, label='Show polybius square')): if showmatrix: poly=makePF(Key) for i in range(5): print(poly[i]) print '\nCiphertext:',playfair(Message,Key) }}} == Frequency Analysis Tools == Frequency analysis is a technique for breaking a substitution cipher that is based on the frequencies that letters appear (in large chunks of text) in the English language. For example, E is the most common letter in the English language, so if a piece of encrypted text had many instances of the letter Q, you would guess that Q had been substituted in for E. The next two interacts create a couple of basic tools that could be useful in cracking a substitution cipher. -AF === Letter Frequency Counter === by Rebecca Lauren Miller, Katherine Stange This tool looks at the percentage of appearances of each letter in the input text, and plots these percentages. The encrypted input text is a bit strange, but was constructed by Amy Feaver to give a short block text that matched the frequencies of letters in English relatively well, to make this message easier to decrypt. -AF {{{#!sagecell #Last Edited 8/8/19 at 2:36pm print "This interact prints a bar graph of the distribution of the letters in the input text. Warning: the smaller the input text the less accurate the distribution will be. Letters that do not occur will not appear in the graph." # Initial text is "Greetiiiings my name is Weeegbert Deuce the True Eater of the Toupee. Hear ye, hear ye! Dee dee dee. A head of these liger cubs carrying the trippy tomahawks are coming fo' thee. Take shelters in the tombs. Tammy ran to the other townspeople and aardvarks. What is her ETA? Her ETA please! Toil, bring your food cups and oil and be swift. The women and the child Occotion CIII should pick bamboo at Atitisoting. See? Nanna Wu Shacah's inner noodle cups: not nuutty sesame notions." @interact def frequencyAnalysis(text = input_box('"Nyllappppunz tf uhtl pz Dlllnilya Klbjl aol Aybl Lhaly vm aol Avbwll. Olhy fl, olhy fl! Kll kll kll. H olhk vm aolzl spnly jbiz jhyyfpun aol aypwwf avthohdrz hyl jvtpun mv aoll. Ahrl zolsalyz pu aol avtiz. Ahttf yhu av aol vaoly avduzwlvwsl huk hhykchyrz. Doha pz oly LAH? Oly LAH wslhzl! Avps, iypun fvby mvvk jbwz huk vps huk il zdpma. Aol dvtlu huk aol jopsk Vjjvapvu JPPP zovbsk wpjr ihtivv ha Hapapzvapun. Zll.Uhuuh Db Zohjho z puuly uvvksl jbwz: uva ubbaaf zlzhtl uvapvuz."', width = 150)): alphabet = AlphabeticStrings() englishText = alphabet.encoding(text) distribution = englishText.frequency_distribution() L1 = englishText.frequency_distribution().function() L1=[x for x in L1.items()] L1.sort(key=lambda x:x[0]) labels, ys = zip(*L1) import numpy as np import math from matplotlib import pyplot as plt xs = np.arange(len(labels)) plt.bar(xs, ys, align='center') plt.xticks(xs, labels) #Replace default x-ticks with xs, then replace xs with labels plt.ylim(0,.2) plt.yticks(ys) plt.xlabel('Letters (Some may be missing)') plt.ylabel('Frequency') plt.show() }}} === Frequency Analysis Decryption Guesser === by Rebecca Lauren Miller, Katherine Stange This interact prints suggested translation of the input text, by matching frequencies of letters in the input to letter frequencies in the English language. With the output you will see that some letters were substituted in correctly, and others were not. Usually frequency analysis requires additional work and some trial and error to discover the original message, especially if the input text is not long enough. -AF {{{#!sagecell #Last edited 8/8/19 at 2:54pm print "Warning: the shorter the input text the less accuate the distribution will be." @interact # Initial text is "Greetiiiings my name is Weeegbert Deuce the True Eater of the Toupee. Hear ye, hear ye! Dee dee dee. A head of these liger cubs carrying the trippy tomahawks are coming fo' thee. Take shelters in the tombs. Tammy ran to the other townspeople and aardvarks. What is her ETA? Her ETA please! Toil, bring your food cups and oil and be swift. The women and the child Occotion CIII should pick bamboo at Atitisoting. See? Nanna Wu Shacah's inner noodle cups: not nuutty sesame notions." def frequencyAnalysis(text = input_box('"Nyllappppunz tf uhtl pz Dlllnilya Klbjl aol Aybl Lhaly vm aol Avbwll. Olhy fl, olhy fl! Kll kll kll. H olhk vm aolzl spnly jbiz jhyyfpun aol aypwwf avthohdrz hyl jvtpun mv aoll. Ahrl zolsalyz pu aol avtiz. Ahttf yhu av aol vaoly avduzwlvwsl huk hhykchyrz. Doha pz oly LAH? Oly LAH wslhzl! Avps, iypun fvby mvvk jbwz huk vps huk il zdpma. Aol dvtlu huk aol jopsk Vjjvapvu JPPP zovbsk wpjr ihtivv ha Hapapzvapun. Zll.Uhuuh Db Zohjho z puuly uvvksl jbwz: uva ubbaaf zlzhtl uvapvuz."', width = 150)): alphabet= AlphabeticStrings() englishText =alphabet.encoding(text) L1 = englishText.frequency_distribution().function() L1=[x for x in L1.items()] L1.sort(key=lambda x:x[1],reverse=True) alphafreq = ['E','T','A','O','I','N','S','H','R','D','L','U','C','M','F','W','Y','P','V','B','G','K','J','Q','X','Z'] translator={} for i in range(0, len(L1)): translator.update({str(L1[i][0]):alphafreq[i]}) answer="" print "\nThe suggested substitutions, based on letter frequency are:" print translator for char in englishText: answer+= translator[str(char)] print "\nThe suggested translation is:\n", answer |
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by Holly Paige Chaos, Rebecca Lauren Miller, and Kate Stange | |
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{{{#!sagecell #This encrypts your message: Final 8/7/19. Written by Rebecca Lauren Miller, Holly Paige Chaos, Kate Strange. |
by Holly Paige Chaos, Rebecca Lauren Miller, Katherine Stange {{{#!sagecell #This encrypts your message: Final 8/7/19. Written by Rebecca Lauren Miller, Holly Paige Chaos, Katherine Stange. |
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{{{#!sagecell #This decrypts your message: Final 8/7/19. Written by Rebecca Lauren Miller, Holly Paige Chaos, Kate Strange. |
by Holly Paige Chaos, Rebecca Lauren Miller, Katherine Stange {{{#!sagecell #Last edited 8/7/19 at 12:00pm |
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== One-Time Pad == by Sarah Arpin, Alexis Newton {{{#!sagecell #Last edited 8/7/2019 5:12pm from random import randrange dictt = {'a':1,'b':2,'c':3,'d':4,'e':5,'f':6,'g':7,'h':8, 'i':9,'j':10,'k':11,'l':12,'m':13,'n':14,'o':15,'p':16,'q':17, 'r':18,'s':19,'t':20,'u':21,'v':22,'w':23,'x':24,'y':25,'z':26 } print "Enter your message to be encrypted via one-time pad in the Plain Text box below:" @interact def one_time_pad(plain_text = input_box("'message'",label="Plain Text:")): #This code takes in a plain text, converts all of the letters to numbers, and then creates a one-time pad for encryption message = [] for char in plain_text: if char.isalpha(): message.append(char.lower()) size = len(message) one_time_pad = [] for i in range(size): one_time_pad.append(randrange(26)) cipher_text = [] for i in range(size): cipher_text.append(1+(dictt[message[i]] + one_time_pad[i]).mod(26)) letter_cipher_text="" for i in cipher_text: letter_cipher_text += (chr(i+96)) print "Your one-time pad is:" print one_time_pad print "" print "Your encrypted message is:" print letter_cipher_text }}} |
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by Holly Paige Chaos and Rebecca Lauren Miller {{{#!sagecell |
=== Hill Cipher Encryption === by Holly Paige Chaos, Alexis Newton {{{#!sagecell #Last edited 8/8/19 at 1:47pm print "Please input the size of your key:" @interact def hill_cipher(Size=['2','3','4']): if Size=='2': print "Please input your message (in quotes) and numbers for your key:" @interact def two_matrix(message=input_box(default='"Alexis smells"'), a=input_box(default=1), b=input_box(default=3), c=input_box(default=3), d=input_box(default=4)): S = AlphabeticStrings() E = HillCryptosystem(S,Size) R = IntegerModRing(26) M = MatrixSpace(R,Size,Size) A = M([[a,b],[c,d]]) print "This is your key:" print A invertible = A.is_invertible() if invertible==false: print "WARNING! You will not be able to decrypt this message because your matrix is not invertible." e = E(A) message=E.encoding(message) print "This is your encrypted message:" print e(S(message)) if Size=='3': print "Please input your message (in quotes) and the numbers in your square matrix key:" @interact def three_matrix(message=input_box(default='"Alexis smells"'), a=input_box(default=1), b=input_box(default=2), c=input_box(default=3), d=input_box(default=5), e=input_box(default=2), f=input_box(default=6), g=input_box(default=7), h=input_box(default=9), i=input_box(default=9)): S = AlphabeticStrings() E = HillCryptosystem(S,3) R = IntegerModRing(26) M = MatrixSpace(R,3,3) A = M([[a,b,c],[d,e,f],[g,h,i]]) print "This is your key:" print A invertible = A.is_invertible() if invertible==false: print "WARNING! You will not be able to decrypt this message because your matrix is not invertible." e = E(A) message=E.encoding(message) print "This is your encrypted message:" print e(S(message)) if Size=='4': print "Please input your message (in quotes) and the numbers in your square matrix key:" @interact def four_matrix(message=input_box(default='"Alexis smells"'), a=input_box(default=17), b=input_box(default=8), c=input_box(default=7), d=input_box(default=10), e=input_box(default=0), f=input_box(default=17), g=input_box(default=5), h=input_box(default=8), i=input_box(default=18), j=input_box(default=12), k=input_box(default=6), l=input_box(default=17), m=input_box(default=0), n=input_box(default=15), o=input_box(default=0), p=input_box(default=17)): S = AlphabeticStrings() E = HillCryptosystem(S,4) R = IntegerModRing(26) M = MatrixSpace(R,4,4) A = M([[a,b,c,d],[e,f,g,h],[i,j,k,l],[m,n,o,p]]) print "This is your key:" print A invertible = A.is_invertible() if invertible==false: print "WARNING! You will not be able to decrypt this message because your matrix is not invertible." e = E(A) message=E.encoding(message) print "This is your encrypted message:" print e(S(message)) }}} === Hill Cipher Decryption === by Holly Paige Chaos, Alexis Newton {{{#!sagecell #Last edited 8/8/19 at 1:47pm dictt = {'a':1,'b':2,'c':3,'d':4,'e':5,'f':6,'g':7,'h':8, 'i':9,'j':10,'k':11,'l':12,'m':13,'n':14,'o':15,'p':16,'q':17, 'r':18,'s':19,'t':20,'u':21,'v':22,'w':23,'x':24,'y':25,'z':26 } print "Please select the size of your key:" @interact def decrypt_hill(size=['2','3','4']): if size=='2': print "Please input your encrypted message and your key:" @interact def two_decrypt(coded_text=input_box(default='"HSVAKSCYLENB"'), a=input_box(default=1), b=input_box(default=3), c=input_box(default=3), d=input_box(default=4)): R = IntegerModRing(26) M = MatrixSpace(R,2,2) a = M([[a,b],[c,d]]) print "The key:" print a message = [] for char in coded_text: if char.isalpha(): message.append(char.lower()) cipher_text = [] for i in range(len(message)): cipher_text.append(dictt[message[i]]-1) new_text = [] for i in range(len(cipher_text)-1): b = matrix(Integers(26),1,2,[cipher_text[i],cipher_text[i+1]]) if i%2 ==0: x = b*a.inverse() x.column(0) for i in x[:][0]: new_text.append(i) final_text = "" for i in range(len(new_text)): new_text[i]=Integer(new_text[i]) final_text += chr(97+new_text[i]) print "The decrypted text:" print final_text if size=='3': print "Please input your encrypted message and your key:" @interact def three_decrypt(coded_text=input_box(default='"FGYHQTCSGKYB"'), a=input_box(default=1), b=input_box(default=2), c=input_box(default=3), d=input_box(default=5), e=input_box(default=2), f=input_box(default=6), g=input_box(default=7), h=input_box(default=9), i=input_box(default=9)): R = IntegerModRing(26) M = MatrixSpace(R,3,3) a = M([[a,b,c],[d,e,f],[g,h,i]]) print "The key:" print a message = [] for char in coded_text: if char.isalpha(): message.append(char.lower()) cipher_text = [] for i in range(len(message)): cipher_text.append(dictt[message[i]]-1) new_text = [] for i in range(len(cipher_text)-2): b = matrix(Integers(26),1,3,[cipher_text[i],cipher_text[i+1],cipher_text[i+2]]) if i%3 ==0: x = b*a.inverse() x.column(0) for i in x[:][0]: new_text.append(i) final_text = "" for i in range(len(new_text)): new_text[i]=Integer(new_text[i]) final_text += chr(97+new_text[i]) print "The decrypted text:" print final_text if size=='4': print "Please input your encrypted message (In quotes) and your key:" @interact def four_decrypt(coded_text=input_box(default='"UIBBSMUGGXTX"'), a=input_box(default=17), b=input_box(default=8), c=input_box(default=7), d=input_box(default=10), e=input_box(default=0), f=input_box(default=17), g=input_box(default=5), h=input_box(default=8), i=input_box(default=18), j=input_box(default=12), k=input_box(default=6), l=input_box(default=17), m=input_box(default=0), n=input_box(default=15), o=input_box(default=0), p=input_box(default=17)): R = IntegerModRing(26) M = MatrixSpace(R,4,4) a = M([[a,b,c,d],[e,f,g,h],[i,j,k,l],[m,n,o,p]]) print "The key:" print a message = [] for char in coded_text: if char.isalpha(): message.append(char.lower()) cipher_text = [] for i in range(len(message)): cipher_text.append(dictt[message[i]]-1) new_text = [] for i in range(len(cipher_text)-3): b = matrix(Integers(26),1,4,[cipher_text[i],cipher_text[i+1],cipher_text[i+2],cipher_text[i+3]]) if i%4 ==0: x = b*a.inverse() x.column(0) for i in x[:][0]: new_text.append(i) final_text = "" for i in range(len(new_text)): new_text[i]=Integer(new_text[i]) final_text += chr(97+new_text[i]) print "The decrypted text:" print final_text |
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Named for the authors Rivest, Shamir, Aldeman, RSA uses exponentiation and modular arithmetic to encrypt and decrypt messages between two parties. Each of those parties has their own secret and public key. To see how it works, following along while Alicia and Bernadette share a message. -EG {{{#!sagecell }}} |
Named for the authors Rivest, Shamir, Aldeman, RSA uses exponentiation and modular arithmetic to encrypt and decrypt messages between two parties. Each of those parties has their own secret and public key. To see how it works, following along while Alice and Babette share a message. -EG === RSA, From Alice's Perspective === by Sarah Arpin, Eva Goedhart Babette has sent Alice an encrypted message. You , as Alice, will provide information so that you can read Babette's message. {{{#!sagecell #Last edited 8/8/19 at 11:42am print "Hi, Alice! Let's set up RSA together." go = True ---- /!\ '''Edit conflict - other version:''' ---- ---- /!\ '''Edit conflict - your version:''' ---- ---- /!\ '''Edit conflict - other version:''' ---- @interact ---- /!\ '''Edit conflict - your version:''' ---- ---- /!\ '''End of edit conflict''' ---- ---- /!\ '''End of edit conflict''' ---- while go: p = ZZ(raw_input("Provide a SECRET decently large prime (>10): ")) if p.is_prime() and p>10: go = False elif p<=10: print "Larger prime, please." elif not p.is_prime(): print "Prime, please." go = True while go: q = ZZ(raw_input("Provide a SECRET different decently large prime (>10): ")) if q.is_prime() and p!=q and q>10: go = False elif p<=10: print "Larger prime, please." elif not p.is_prime(): print "Prime, please." elif p == q: print "Different prime, please." phi = (p-1)*(q-1) print "So far, you can compute:" print "N = pq =",p*q print "phi(N) = (p-1)(q-1) =",phi.factor() go = True while go: e = ZZ(raw_input("Provide a PUBLIC exponent that is relatively prime to phi(N):")) if gcd(e,phi) == 1: go = False ---- /!\ '''Edit conflict - other version:''' ---- ---- /!\ '''Edit conflict - your version:''' ---- ---- /!\ '''Edit conflict - other version:''' ---- ---- /!\ '''Edit conflict - your version:''' ---- @interact ---- /!\ '''End of edit conflict''' ---- ---- /!\ '''End of edit conflict''' ---- def rsa(): N = p*q R = IntegerModRing(phi) d = (e^(R(e).multiplicative_order()-1)).mod(phi) print "Alice's public key is: N = pq =",N,", e =",e,"." print "Alice's private key is: p =",p,", q = ",q,", d = ",d,", where the decryption key d is the inverse of e modulo phi(N)." secret_message_from_babette = "Hi Dr. Strange!" ascii_secret = [] for char in secret_message_from_babette: ascii_secret.append(ord(char)) encrypted_ascii = [] for ascii in ascii_secret: encrypted_ascii.append(power_mod(ascii,e,N)) decrypted_ascii = [] for ascii in encrypted_ascii: decrypted_ascii.append(power_mod(ascii,d,N)) print "Babette's encrypted message to you is: ", encrypted_ascii print "" print "To decrypt, we raise each one of these to the ",d,", modulo ",N,":" print decrypted_ascii print "" decrypted_secret = "" for ascii in decrypted_ascii: decrypted_secret += chr(ascii) print "Going from ascii to letters, we figure out the secret is: " print decrypted_secret }}} === RSA, From Babette's Perspective === by Sarah Arpin, Eva Goedhart {{{#!sagecell #Last edited 8/8/19 at 12:30pm print "Hi, Babette! Let's send a message to Alice using RSA." p = next_prime(100) q = next_prime(p) phi = (p-1)*(q-1) e = 13 N = p*q R = IntegerModRing(phi) d = (e^(R(e).multiplicative_order()-1)).mod(phi) print "Alice's public key is: N =",N,", e =",e,"." message = raw_input("Type a message for Alice:") ascii_secret = [] for char in message: ascii_secret.append(ord(char)) print "We turn these characters into ascii:" print ascii_secret print "" print "Then we encode them by raising each ascii number to the e-th power modulo N." encrypted_ascii = [] for ascii in ascii_secret: encrypted_ascii.append(power_mod(ascii,e,N)) print encrypted_ascii print "" @interact def rsa(): print "Alice receives our secret and uses her private key to decrypt the message." decrypted_ascii = [] for ascii in encrypted_ascii: decrypted_ascii.append(power_mod(ascii,d,N)) print decrypted_ascii print "" decrypted_secret = "" for ascii in decrypted_ascii: decrypted_secret += chr(ascii) print "Going from ascii to letters, she figures out your message was: " print decrypted_secret }}} === RSA With Digital Signatures === by Sarah Arpin, Eva Goedhart {{{#!sagecell }}} |
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== One-time Pad == |
Sage Interactions - Cryptography
This page was first created at Sage Days 103, 7-10 August 2019 by Sarah Arpin, Catalina Camacho-Navarro, Holly Paige Chaos, Amy Feaver, Eva Goedhart, Rebecca Lauren Miller, Alexis Newton, and Nandita Sahajpal. Text edited by Amy Feaver, Eva Goedhart, and Alexis Newton. This project was led by Amy Feaver.
We acknowledge Katherine Stange, who allowed us to use code from her cryptography course as a starting point for many of these interacts. Dr. Stange's original code and course page can be found at http://crypto.katestange.net/
If you have cryptography-related interactions that you are interested in adding to this page, please do so. You can also contact Amy Feaver at [email protected]
goto interact main page
Contents
Shift Cipher
The shift cipher is a classical cryptosystem that takes plaintext and shifts it through the alphabet by a given number of letters. For example, a shift of 2 would replace all A's with C's, all B's with D's, etc. When the end of the alphabet is reached, the letters are shifted cyclically back to the beginning. Thus, a shift of 2 would replace Y's with A's and Z's with B's.
Shift Cipher Encryption
by Sarah Arpin, Alexis Newton
Shift Cipher Decryption
by Sarah Arpin, Alexis Newton
If you know that your message was encrypted using a shift cipher, you can use the known shift value to decrypt. If this is not known, brute force can be used to get 26 possible decrypted messages. The chi-squared function ranks the brute force results by likelihood according to letter frequency.
Affine Cipher
An affine cipher combines the idea of a shift cipher with a multiplicative cipher. In this particular example, we map consecutive letters of the alphabet to consecutive numbers, starting with A=0 (you can also do this cipher differently, and starting with A=1). The user selects two values, a and b. The value a is the multiplier and must be relatively prime to 26 in order to guarantee that each letter is encoded uniquely. The value b is the addend. Each letter's value is multiplied by a, and the product is added to b. This is then replaced with a new letter, corresponding to the result modulo 26. -AF
Affine Cipher Encryption
by Sarah Arpin, Alexis Newton
Affine Cipher Decryption
by Sarah Arpin, Alexis Newton
If you know that your message was encrypted using an affine cipher, you can use the known a and b values to decrypt. If this is not known, brute force can be used to get possible decrypted messages. The chi-squared function ranks the results by likelihood according to letter frequency.
Substitution Cipher
by Catalina Camacho-Navarro
A simple cipher to encrypt messages in which each letter is assigned to another letter. Brute force or frequency analysis can be used to decrypt. -EG
Playfair Cipher
by Catalina Camacho-Navarro
Based on code from Alasdair McAndrew at trac.sagemath.org/ticket/8559
A special type of substitution cipher in which the plaintext is broken up into two-letter digraphs with some restrictions. Those digraphs are encrypted using a Polybius square, (i.e. a 5x5 grid in which each letter of the alphabet is its own entry with the exception of ij which are placed together). The positions of the letters in the digraph determine how the digraph is encrypted. -EF
Frequency Analysis Tools
Frequency analysis is a technique for breaking a substitution cipher that is based on the frequencies that letters appear (in large chunks of text) in the English language. For example, E is the most common letter in the English language, so if a piece of encrypted text had many instances of the letter Q, you would guess that Q had been substituted in for E. The next two interacts create a couple of basic tools that could be useful in cracking a substitution cipher. -AF
Letter Frequency Counter
by Rebecca Lauren Miller, Katherine Stange
This tool looks at the percentage of appearances of each letter in the input text, and plots these percentages. The encrypted input text is a bit strange, but was constructed by Amy Feaver to give a short block text that matched the frequencies of letters in English relatively well, to make this message easier to decrypt. -AF
Frequency Analysis Decryption Guesser
by Rebecca Lauren Miller, Katherine Stange
This interact prints suggested translation of the input text, by matching frequencies of letters in the input to letter frequencies in the English language. With the output you will see that some letters were substituted in correctly, and others were not. Usually frequency analysis requires additional work and some trial and error to discover the original message, especially if the input text is not long enough. -AF
Vigenère Cipher
Using a secret code word, encrypt each letter by shifting it the corresponding letter in the code word. -EG
Vigenère Cipher Encryption
by Holly Paige Chaos, Rebecca Lauren Miller, Katherine Stange
Vigenère Cipher Decryption
by Holly Paige Chaos, Rebecca Lauren Miller, Katherine Stange
One-Time Pad
by Sarah Arpin, Alexis Newton
Hill Cipher
Hill Cipher Encryption
by Holly Paige Chaos, Alexis Newton
Hill Cipher Decryption
by Holly Paige Chaos, Alexis Newton
RSA
Named for the authors Rivest, Shamir, Aldeman, RSA uses exponentiation and modular arithmetic to encrypt and decrypt messages between two parties. Each of those parties has their own secret and public key. To see how it works, following along while Alice and Babette share a message. -EG
RSA, From Alice's Perspective
by Sarah Arpin, Eva Goedhart
Babette has sent Alice an encrypted message. You , as Alice, will provide information so that you can read Babette's message.
RSA, From Babette's Perspective
by Sarah Arpin, Eva Goedhart
RSA With Digital Signatures
by Sarah Arpin, Eva Goedhart