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= Under Construction = This page will be created/preened at a Sage Days that will take place August 7th - 10th. If you have cryptography related interactions that you are interested in adding to this page, please do. You can also contact Amy Feaver at [email protected] (insert my name in 'firstname' and 'lastname') 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! |
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This page was first created at Sage Days 103, 7-9 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 Holly Paige Chaos, Amy Feaver, Eva Goedhart, Sara Lapan 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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{{{#!sagecell print "Put your message in quotes, and your desired shift: " @interact def shift_cipher(message = input_box(default="'secrets hi'", width = 50), shift=slider(0,25,1,3)): |
You can use this interact to encrypt a message with a shift cipher. {{{#!sagecell #Last edited 8/7/19 2:45pm pretty_print(html("<h1>Shift Cipher Encryptor</h1>")) print "Put your message inside the provided quotes (with no additional quotes or apostrophes!), and select your desired shift: " @interact def shift_cipher(message = input_box(default='"secrets"', width = 50), shift=slider(0,25,1,3)): |
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print "This is your encrypted text shifted by ",shift,":" print C }}} === 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. {{{#!sagecell #Last edited 8/7/19 2:56pm pretty_print(html("<h1>Shift Cipher Decryptor</h1>")) print "Enter the encrypted text in quotes, and enter a guess for the shift amount:" @interact def shift_decrypt(text = input_box('"KL"'), shift_by = input_box(0)): S = ShiftCryptosystem(AlphabeticStrings()) ciphertext = S.encoding(text) decrypt = S.deciphering(shift_by%26,ciphertext) print "If the shift was by", shift_by,", then the original message was:" print decrypt decrypt = S.brute_force(ciphertext) print "These are the possibilities for the plaintext:" print decrypt decrypt = S.brute_force(ciphertext,ranking = "chisquare") print "These are the possibilities ranked by likelihood with the chi-squared function:" print decrypt }}} == 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. === Affine Cipher Encryption === by Sarah Arpin, Alexis Newton You can use this interact to encrypt a message with the affine cipher. Notice that the only choices for a can be numbers that are relatively prime to 26. This cipher will encipher a letter m of your message as a*m + b. {{{#!sagecell # Last edited 8/7/2019 2:01pm pretty_print(html("<h1>Affine Cipher Encryptor</h1>")) print "Put your message in between the provided quotes (with no additional quotes or apostrophes!), and select your desired a and b: " @interact 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]): A = AlphabeticStrings() S = AffineCryptosystem(A) message = S.encoding(message) C = S.enciphering(a,b, message) print "This is your encrypted text:" print C }}} === 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 these are not known, brute force can be used to get a list of possible decrypted messages. The chi-squared function ranks these results by likelihood according to letter frequency. {{{#!sagecell #Last edited 8/7/2019 3:01pm pretty_print(html("<h1>Affine Cipher Decryptor</h1>")) print "Enter the encrypted text in quotes, and enter a guess for the a and 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 }}} == Substitution Cipher == by Catalina Camacho-Navarro A substitution cipher encrypts messages by assigning each letter of the alphabet to another letter. For instance, if A is assigned to F, then all A's in the original message will be substituted with F's in the encrypted message. Brute force or frequency analysis can be used to decrypt a message encrypted with a substitution cipher. {{{#!sagecell pretty_print(html('<h1> Substitution Cipher')) print "Select your letter substitutions and enter your message in quotes." from string import ascii_uppercase left_over_letters=[0] +[let for let in ascii_uppercase] @interact def _(A =selector(left_over_letters, default=0)): if A!=0: left_over_letters.remove(A) # print left_over_letters @interact def _(B =selector(left_over_letters, default=0)): if B!=0: left_over_letters.remove(B) # print left_over_letters @interact def _(C =selector(left_over_letters, default=0)): if C!=0: left_over_letters.remove(C) @interact def _(D =selector(left_over_letters, default=0)): if D!=0: left_over_letters.remove(D) @interact def _(E =selector(left_over_letters, default=0)): if E!=0: left_over_letters.remove(E) @interact def _(F =selector(left_over_letters, default=0)): if F!=0: left_over_letters.remove(F) @interact def _(G =selector(left_over_letters, default=0)): if G!=0: left_over_letters.remove(G) @interact def _(H =selector(left_over_letters, default=0)): if H!=0: left_over_letters.remove(H) @interact def _(I =selector(left_over_letters, default=0)): if I!=0: left_over_letters.remove(I) @interact def _(J =selector(left_over_letters, default=0)): if J!=0: left_over_letters.remove(J) @interact def _(K =selector(left_over_letters, default=0)): if K!=0: left_over_letters.remove(K) @interact def _(L =selector(left_over_letters, default=0)): if L!=0: left_over_letters.remove(L) @interact def _(M =selector(left_over_letters, default=0)): if M!=0: left_over_letters.remove(M) @interact def _(N =selector(left_over_letters, default=0)): if N!=0: left_over_letters.remove(N) @interact def _(O =selector(left_over_letters, default=0)): if O!=0: left_over_letters.remove(O) @interact def _(P =selector(left_over_letters, default=0)): if P!=0: left_over_letters.remove(P) @interact def _(Q =selector(left_over_letters, default=0)): if Q!=0: left_over_letters.remove(Q) @interact def _(R =selector(left_over_letters, default=0)): if R!=0: left_over_letters.remove(R) @interact def _(S =selector(left_over_letters, default=0)): if S!=0: left_over_letters.remove(S) @interact def _(T =selector(left_over_letters, default=0)): if T!=0: left_over_letters.remove(T) @interact def _(U =selector(left_over_letters, default=0)): if U!=0: left_over_letters.remove(U) @interact def _(V =selector(left_over_letters, default=0)): if V!=0: left_over_letters.remove(V) @interact def _(W =selector(left_over_letters, default=0)): if W!=0: left_over_letters.remove(W) @interact def _(X =selector(left_over_letters, default=0)): if X!=0: left_over_letters.remove(X) @interact def _(Y =selector(left_over_letters, default=0)): if Y!=0: left_over_letters.remove(Y) @interact def _(Z =selector(left_over_letters, default=0)): if Z!=0: left_over_letters.remove(Z) @interact def _(text=input_box(default="'MESSAGE'",label="Message")): new_ordering=[A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z]; new_key=[ord(new_ordering[i])-65 for i in range(26)] alphabet=AlphabeticStrings() Es=SubstitutionCryptosystem(alphabet) Key = alphabet(new_key) e = Es(Key) TEXT0=text TEXT=alphabet.encoding(TEXT0) print "Ciphertext:", e(TEXT) }}} == Playfair Cipher == by Catalina Camacho-Navarro Based on code from Alasdair McAndrew at trac.sagemath.org/ticket/8559. A playfair cipher is 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. === Playfair Encryption === Use this interact to encrypt a message using the playfair cipher. {{{#!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 pretty_print(html("<h1>Playfair Cipher Encryptor</h1>")) 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) }}} === Playfair Decryption === ##Playfair decryption ##PLAYFAIR CIPHER ## CATALINA CAMACHO-NAVARRO ##Based on code from Alasdair McAndrew at //trac.sagemath.org/ticket/8559 ##Last edited 8/9/19 at 1:55pm {{{#!sagecell 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_decrypt(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]:#locate the first letter of di in PF i0=i j0=j if PF[i][j]==di[1]: i1=i j1=j if (i0<>i1) & (j0<>j1):## if di[0] and di[1] are not in the same column or row, switch to corners in the same row return PF[i0][j1]+PF[i1][j0] if (i0==i1) & (j0<>j1):## if di[0] and di[1] are in the same row, then switch left return PF[i0][(j0-1)%5]+PF[i1][(j1-1)%5] if (i0<>i1) & (j0==j1):## if di[0] and di[1] are in the same column, then switch up 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_decrypt(pl1,word): # decrypts a plaintext "pl" with a Playfair cipher pl=change_to_plain_text(pl1) if len(pl1)%2<>0: raise TypeError('The lenght of the ciphertext is not even') pl2=makeDG(pl) if pl2<>pl: if 'J' in pl: raise TypeError('The ciphertext contains a J') if len(pl2)<>len(pl): raise TypeError('The ciphertext contains digraphs with repeated letters') PF=makePF(word) # using a keyword "word" tmp='' for i in range(len(pl2)//2): tmp+=pf_decrypt(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 def playfair_decrypt_options(pl): ##Modifies the output of the playfair_decrypt by erasing replacing I's or deleting X pl_noI=pl.replace('I','J') if pl.endswith('X'): pl_no_last_X=pl[:-1] else: pl_no_last_X=pl pl_noX=pl for ch in pl_noX: if (ch=='X') & (pl.find(ch)<>0): if pl_noX[pl_noX.find(ch)-1]==pl_noX[pl_noX.find(ch)+1]: pl_noX=pl_noX.replace('X','') return([pl,pl_noI,pl_noX,pl_no_last_X]) pretty_print(html("<h1>Playfair Cipher Decryptor</h1>")) print 'Enter your ciphertext and a guess for the key to construct you polybius square.' print 'Warning: both the message and the key must be in quotes.' @interact def _(Ciphertext=input_box(default="'LYXAXGDA'"),Key=input_box(default="'key'", label='Guess key'),showmatrix=checkbox(True, label='Show polybius square')): print 'These are some of the possibilities for the plaintext:' print playfair_decrypt_options(playfair_decrypt(Ciphertext,Key)) if showmatrix: poly=makePF(Key) print '----------------------' for i in range(5): print(poly[i]) }}} == Frequency Analysis Tools == Frequency analysis is a technique for breaking a substitution cipher that utilizes the frequencies of letters appearing 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. === 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 of text that matched the frequencies of letters in the English language relatively well, to make this message easier to decrypt. {{{#!sagecell #Last Edited 8/8/19 at 2:36pm pretty_print(html("<h1>Letter Frequency Counter</h1>")) print "This interact prints a bar graph showing the distribution of 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 a suggested translation of the input text by matching frequencies of letters in the input to frequencies of letters in the English language. With the output you will see that some letters were substituted incorrectly, 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. {{{#!sagecell #Last edited 8/8/19 at 2:54pm pretty_print(html("<h1>Frequency Analysis Decryption Guesser</h1>")) print "Warning: the shorter the input text is, the less accurate 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 }}} == Vigenère Cipher == A Vigenère cipher is an example of a polyalphabetic cipher. Using a secret codeword as the key, the Vigenère encrypts each letter of a message by shifting it according to the corresponding letter in the key. For example, we will use the key "CAT" to encrypt our default text "secrets hi". To do this the message is first broken up into three-letter chunks, because the key is three letters long, to be "SEC RET SHI". Next each letter of the chunk is shifted by the value of the corresponding letter in the key. The standard shifts are A=0, B=1, C=2, etc. So in our example, S shifts by C=2 letters to U, E shifts by A=0 letters and remains at E, and C shifts by T=19 letters to V. Thus "SECRETSHI" becomes UEVTEMUHB when encrypted. To decrypt the message, simply use the keyword to undo the encryption process. Cryptography by Simon Rubinstein-Salzedo was used as reference for this interact. === Vigenère Cipher Encryption === by Holly Paige Chaos, Rebecca Lauren Miller, Katherine Stange Use this interact to encrypt a message using the Vigenère Cipher. {{{#!sagecell #This encrypts your message: Final 8/7/19. Written by Rebecca Lauren Miller, Holly Paige Chaos, Katherine Stange. pretty_print(html("<h1>Vigenère Cipher Encryptor</h1>")) print "Put your message and codeword inside the quotes: " @interact def vigenere_cipher(message = input_box(default ="'secrets hi'", width = 50), code_word = input_box(default="'cat'", width = 50)): A = AlphabeticStrings() message2 = A.encoding(message) code_word2 = A.encoding(code_word) system = VigenereCryptosystem(A,len(code_word2)) ciphertext = system.enciphering(code_word2,message2) |
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print C }}} |
print ciphertext }}} === Vigenère Cipher Decryption === by Holly Paige Chaos, Rebecca Lauren Miller, Katherine Stange If you used the Vigenère Cipher to encrypt a message, you can use this interact to decrypt by inputting your key and encrypted text. {{{#!sagecell #Last edited 8/7/19 at 12:00pm pretty_print(html("<h1>Vigenère Cipher Decryptor</h1>")) print "Put your encrypted message and codeword inside the quotes: " @interact def vigenere_cipher(message = input_box(default ="'UEVTEMUHB'", width = 50), code_word = input_box(default="'cat'", width = 50)): A = AlphabeticStrings() message2 = A.encoding(message) code_word2 = A.encoding(code_word) system = VigenereCryptosystem(A,len(code_word2)) ciphertext = system.deciphering(code_word2,message2) print "Deciphered message:" print ciphertext }}} == One-Time Pad == by Sarah Arpin, Alexis Newton One-time pad is an encryption method that cannot be cracked. It requires a single-use shared key (known as a one-time pad) the length of the message or longer. In this method, every letter is first converted to numbers using the standard A=0, B=1, C=2, etc. Then each character in the message is multiplied modulo 26 by the number in the corresponding position in the key. This is then converted back to letters to produce the encrypted text. {{{#!sagecell #Last edited 8/7/2019 5:12pm from random import randrange dictt = {'a':0,'b':1,'c':2,'d':3,'e':4,'f':5,'g':6,'h':7, 'i':8,'j':9,'k':10,'l':11,'m':12,'n':13,'o':14,'p':15,'q':16, 'r':17,'s':18,'t':19,'u':20,'v':21,'w':22,'x':23,'y':24,'z':25 } pretty_print(html("<h1>One-Time Pad Encryptor</h1>")) 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((dictt[message[i]] + one_time_pad[i]).mod(26)) letter_cipher_text="" for i in cipher_text: letter_cipher_text += (chr(i+97)) print "Your one-time pad is:" print one_time_pad print "" print "Your encrypted message is:" print letter_cipher_text }}} == Hill Cipher == The Hill cipher requires some basic knowledge of Linear Algebra. In this encryption method, an invertible n x n matrix of integers modulo 26 is used as the key. The message is first converted to numbers and spit into chunks size n. These chunks are then converted to n x 1 vectors and multiplied by the key modulo 26 to produce 1 x n vectors. The integers from these vectors are converted back letters to produce the encrypted text. === Hill Cipher Encryption === by Holly Paige Chaos, Alexis Newton Use this interact to encrypt a message with the Hill cipher. Be sure to use an invertible matrix so that your message can be decrypted! {{{#!sagecell #Last edited 8/8/19 at 1:47pm pretty_print(html("<h1>Hill Cipher Encryptor</h1>")) print "Please select 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 Use this interact to decrypt messages encrypted by the Hill cipher. Remember that this only works if the message was encrypted using an invertible matrix as the key! {{{#!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 }}} == Modular Arithmetic (Preliminaries for RSA, Diffie-Hellman, El Gamal) == This section gives visual representations of the modular arithmetic necessary for RSA, Diffie-Hellman, and El Gamal. === Modular Arithmetic Multiplication Table === by Rebecca Lauren Miller, Kate Stange Given a positive integer n, this prints the multiplication mod n. Each entry in this table corresponds to the product of the row number by the column number, modulo n. {{{#!sagecell #Last edited 8/9/19 at 12:30pm pretty_print(html("<h1>Multiplication Table modulo n</h1>")) print "This tool creates a multiplication table modulo 𝑛." @interact def modular_multiplication_tables(n = input_box(default = 7, width = 25)): R = IntegerModRing(n) rows = [['*']+[str(r) for r in R]]+[[i]+[i*r for r in R] for i in R] print table(rows, frame=True) }}} === Modular Exponentiation === by Rebecca Lauren Miller, Kate Stange Given a modulus n and a nonnegative exponent a this displays a graph where each integer between 0 and n-1 is mapped to its ath power, mod n. {{{#!sagecell #Last edited 8/9/19 at 2:46pm pretty_print(html("<h1>Arrow Diagram modulo n</h1>")) print "Input your modulus, 𝑛, and an integer, 𝑎. The output will be an arrow diagram picture of 𝑥↦𝑎𝑥 on the ring ℤ/𝑛ℤ, i.e. the elements modulo 𝑛." @interact def modular_multiplication_graph(n = input_box(default = 7, width = 25), a = input_box(default = 2, width = 25)): R = IntegerModRing(n) left=[' '+str(r)+' ' for r in R] right=[' '+str(r)+' ' for r in R] pre_pos=graphs.CompleteBipartiteGraph(len(left),len(right)).get_pos() G = DiGraph() pos={} for (i,v) in enumerate(left+right): G.add_vertex(v) pos[v]=pre_pos[i] for l in range(n): G.add_edge(left[l],right[lift(R(a*l))]) show(G.plot(pos=pos)) }}} === Discrete Log Problem === by Sara Lapan The discrete logarithm, log(x) with base a, is an integer b such that a^b^ = x. Computing logarithms is computationally difficult, and there are no efficient algorithms known for the worst case scenarios. However, the discrete exponentiation is comparatively simple (for instance, it can be done efficiently using squaring). This asymmetry in complexity has been exploited in constructing cryptographic systems. Typically, it is much easier to solve for x in x=a^b^ (mod m) when a,b,m are known, than it is to solve for b when x,a,m are known. Interact to find x given a,b,m: {{{#!sagecell pretty_print(html("<h1>Discrete Log Problem</h1>")) print('This will evaluate x=a^b (mod m). Choose your base (a), exponent (b), and modulus (m). These should all be positive integers.') @interact def DLP_solve(a=input_box(default=5),b=input_box(default=25),m=input_box(default=47)): if (not a in ZZ) or (not b in ZZ) or (not m in ZZ) or (a<=0) or (b<=0) or (m<=0): print "*********** ERROR: a,b,m should all be positive integers. ***********" else: a=Integer(a) b=Integer(b) m=Integer(m) print('This is the evaluation process using squares:') print('') C=b.str(base=2) L=len(C) S=[] T=[] # print "The base 2 expansion of",b,"is",C ans=str(a) ans_num=a for i in range(L): pow=L-i-1 #print C[pow],"copy(ies) of",2,"^",i,"=",2^i # Convert to integer: # Integer(C[i],base=2) S+=[(2^pow)] print "Step",i+1,":",str(a)+"^"+str(2^i),"=",ans,"=",ans_num,"mod",m #ans_num= a^(i+1) %m ans=str(ans_num)+"^"+str(2) ans_num= (ans_num)^2%m if C[pow]=="1": T+=[2^i] else: T expansion = "" STR="" STR_eval="" STR_eval_num=1 while len(T)>1: expansion += str(T[-1])+"+" STR += "("+str(a)+"^"+str(T[-1])+")" STR_eval += "("+str(a^(T[-1])%47)+")" STR_eval_num = STR_eval_num*(a^(T[-1])%47) T.remove(T[-1]) expansion+=str(T[0]) STR += "("+str(a)+"^"+str(T[0])+")" STR_eval += "("+str(a^(T[0])%47)+")" STR_eval_num = STR_eval_num*(a^(T[0])%47) STR_eval_num = STR_eval_num%47 print "Step",L+1,":",str(a)+"^"+str(b),"=",STR,"=",STR_eval,"=",STR_eval_num,"mod",m print " Since, as a sum of powers of 2,",str(25)+" is "+expansion+"." print "CONCLUSION: "+str(STR_eval_num)+" = "+str(a)+"^"+str(b)+" mod",m,". It takes ",L+1,"steps to calculate with this method." }}} == RSA == Named for the authors Rivest, Shamir, and 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. === RSA, From Alice's Perspective === by Sarah Arpin, Eva Goedhart Babette sent Alice an encrypted message. You, as Alice, will provide information so that you can read Babette's message. {{{#!sagecell #Last edited 8/9/19 at 1:53pm print "Hi, Alice! Let's set up RSA together." print "" print "1. Input two PRIVATE distinct primes, p and q, that are each greater than 10." print " The size of the primes depends on the size of Babette's message. Her message requires p,q > 10. A longer and stronger encrypted" print " message requires larger primes." print "" print "2. Input a PUBLIC integer, e, which needs to be relatively prime to the the Euler phi function of the product pq, φ(pq)." print " If e is not relativley prime to φ(pq), then we can not decrypt the message." @interact def rsa(p = input_box(default = 11,label = "p: "), q = input_box(default = 23,label = "q: "),e = input_box(default = 7,label = "e:")): p = ZZ(p) q = ZZ(q) e = ZZ(e) if p == q: print "*********** Make sure p and q are different.***********" if p < 10: print "*********** Make p larger. ***********" if q < 10: print "*********** Make q larger. ***********" if not p.is_prime(): print "*********** p needs to be prime. ***********" if not q.is_prime(): print "*********** q needs to be prime. ***********" phi = (p-1)*(q-1) if not gcd(e,phi) == 1: print "*********** e must be replatively prime to φ(pq) - see factorization below. ***********" print "" print "φ(pq) = ",phi.factor() print "" 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," where N = pq and the factorization of N is kept secret." print "" print "Alice's PRIVATE key is: p =",p,", q = ",q,", d = ",d,", where the decryption key d is the inverse of e modulo φ(N), i.e., de = 1 (mod 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 "" print "3. Babette took her plaintext message and converted into integers using ASCII. Then she encrypted it by raising each integer to the e-th power modulo N and sent the result to Alice:" print "" print " ", encrypted_ascii print "" print "4. To decrypt, we raise each integer of the lisy above to the d =",d,", modulo N =",N,":" print "" print " ",decrypted_ascii print "" decrypted_secret = "" for ascii in decrypted_ascii: decrypted_secret += chr(ascii) print "5. Going from the integers in ASCII to the plaintext in letters, we figure out the secret is: " print "" print " ",decrypted_secret print "" print "************************************************************************************************" print "REMARK: Babette encrypted her message one character at a time." print "Usual protocal dictates that the entire message is concatenated with leading zeros removed." print "This will require that N = pq is larger than the integer value of the full message." print "************************************************************************************************" }}} === RSA, From Babette's Perspective === by Sarah Arpin, Eva Goedhart {{{#!sagecell #Last edited 8/9/19 2:40pm print "Hi, Babette! Let's send a message to Alice using her PUBLIC key (N, e) with RSA." print "" print "1. Input Babette's secret message for Alice in between the quotation marks below." print " Make sure that there are no apostrophes or extra quotation marks in your message." @interact def rsa(message = input_box(default = "'Secrets for Alice'",label="Message:")): 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) ascii_secret = [] for char in message: ascii_secret.append(ord(char)) print "2. Using ASCII, we take the characters in our message and convert each of them into integers." print "" print " ",ascii_secret print "" print "Alice's PUBLIC key is given to be (N, e) = (",N,",",e,")." print "" print "4. We encode the list of numbers by raising each integer to the e-th power modulo N. Recall that e is called the encryption key. This is what get's sent to Alice:" encrypted_ascii = [] for ascii in ascii_secret: encrypted_ascii.append(power_mod(ascii,e,N)) print "" print " ",encrypted_ascii print "" print "5. To decrypt, Alice raises each integer to the d-th power modulo N, where d is Alice's PRIVATE decryption key." decrypted_ascii = [] for ascii in encrypted_ascii: decrypted_ascii.append(power_mod(ascii,d,N)) print "" print " ", decrypted_ascii print "" decrypted_secret = "" for ascii in decrypted_ascii: decrypted_secret += chr(ascii) print "6. Going from the integers to letters using ASCII, we find that Babette's message was " print "" print " ",decrypted_secret }}} === RSA With Digital Signatures === by Sarah Arpin, Eva Goedhart {{{#!sagecell }}} #== Diffe-Hellman Key Exchange == |
Sage Interactions - Cryptography
This page was first created at Sage Days 103, 7-9 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 Holly Paige Chaos, Amy Feaver, Eva Goedhart, Sara Lapan 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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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
You can use this interact to encrypt a message with a shift cipher.
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.
Affine Cipher Encryption
by Sarah Arpin, Alexis Newton
You can use this interact to encrypt a message with the affine cipher. Notice that the only choices for a can be numbers that are relatively prime to 26. This cipher will encipher a letter m of your message as a*m + b.
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 these are not known, brute force can be used to get a list of possible decrypted messages. The chi-squared function ranks these results by likelihood according to letter frequency.
Substitution Cipher
by Catalina Camacho-Navarro
A substitution cipher encrypts messages by assigning each letter of the alphabet to another letter. For instance, if A is assigned to F, then all A's in the original message will be substituted with F's in the encrypted message. Brute force or frequency analysis can be used to decrypt a message encrypted with a substitution cipher.
Playfair Cipher
by Catalina Camacho-Navarro
Based on code from Alasdair McAndrew at trac.sagemath.org/ticket/8559.
A playfair cipher is 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.
Playfair Encryption
Use this interact to encrypt a message using the playfair cipher.
Playfair Decryption
Frequency Analysis Tools
Frequency analysis is a technique for breaking a substitution cipher that utilizes the frequencies of letters appearing 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.
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 of text that matched the frequencies of letters in the English language relatively well, to make this message easier to decrypt.
Frequency Analysis Decryption Guesser
by Rebecca Lauren Miller, Katherine Stange
This interact prints a suggested translation of the input text by matching frequencies of letters in the input to frequencies of letters in the English language. With the output you will see that some letters were substituted incorrectly, 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.
Vigenère Cipher
A Vigenère cipher is an example of a polyalphabetic cipher. Using a secret codeword as the key, the Vigenère encrypts each letter of a message by shifting it according to the corresponding letter in the key. For example, we will use the key "CAT" to encrypt our default text "secrets hi". To do this the message is first broken up into three-letter chunks, because the key is three letters long, to be "SEC RET SHI". Next each letter of the chunk is shifted by the value of the corresponding letter in the key. The standard shifts are A=0, B=1, C=2, etc. So in our example, S shifts by C=2 letters to U, E shifts by A=0 letters and remains at E, and C shifts by T=19 letters to V. Thus "SECRETSHI" becomes UEVTEMUHB when encrypted. To decrypt the message, simply use the keyword to undo the encryption process. Cryptography by Simon Rubinstein-Salzedo was used as reference for this interact.
Vigenère Cipher Encryption
by Holly Paige Chaos, Rebecca Lauren Miller, Katherine Stange
Use this interact to encrypt a message using the Vigenère Cipher.
Vigenère Cipher Decryption
by Holly Paige Chaos, Rebecca Lauren Miller, Katherine Stange
If you used the Vigenère Cipher to encrypt a message, you can use this interact to decrypt by inputting your key and encrypted text.
One-Time Pad
by Sarah Arpin, Alexis Newton
One-time pad is an encryption method that cannot be cracked. It requires a single-use shared key (known as a one-time pad) the length of the message or longer. In this method, every letter is first converted to numbers using the standard A=0, B=1, C=2, etc. Then each character in the message is multiplied modulo 26 by the number in the corresponding position in the key. This is then converted back to letters to produce the encrypted text.
Hill Cipher
The Hill cipher requires some basic knowledge of Linear Algebra. In this encryption method, an invertible n x n matrix of integers modulo 26 is used as the key. The message is first converted to numbers and spit into chunks size n. These chunks are then converted to n x 1 vectors and multiplied by the key modulo 26 to produce 1 x n vectors. The integers from these vectors are converted back letters to produce the encrypted text.
Hill Cipher Encryption
by Holly Paige Chaos, Alexis Newton
Use this interact to encrypt a message with the Hill cipher. Be sure to use an invertible matrix so that your message can be decrypted!
Hill Cipher Decryption
by Holly Paige Chaos, Alexis Newton
Use this interact to decrypt messages encrypted by the Hill cipher. Remember that this only works if the message was encrypted using an invertible matrix as the key!
Modular Arithmetic (Preliminaries for RSA, Diffie-Hellman, El Gamal)
This section gives visual representations of the modular arithmetic necessary for RSA, Diffie-Hellman, and El Gamal.
Modular Arithmetic Multiplication Table
by Rebecca Lauren Miller, Kate Stange
Given a positive integer n, this prints the multiplication mod n. Each entry in this table corresponds to the product of the row number by the column number, modulo n.
Modular Exponentiation
by Rebecca Lauren Miller, Kate Stange
Given a modulus n and a nonnegative exponent a this displays a graph where each integer between 0 and n-1 is mapped to its ath power, mod n.
Discrete Log Problem
by Sara Lapan
The discrete logarithm, log(x) with base a, is an integer b such that ab = x. Computing logarithms is computationally difficult, and there are no efficient algorithms known for the worst case scenarios. However, the discrete exponentiation is comparatively simple (for instance, it can be done efficiently using squaring). This asymmetry in complexity has been exploited in constructing cryptographic systems. Typically, it is much easier to solve for x in x=ab (mod m) when a,b,m are known, than it is to solve for b when x,a,m are known. Interact to find x given a,b,m:
RSA
Named for the authors Rivest, Shamir, and 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.
RSA, From Alice's Perspective
by Sarah Arpin, Eva Goedhart
Babette 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
#== Diffe-Hellman Key Exchange ==