E-320: Teaching Math with a Historical Perspective Oliver Knill, 2011Lecture 11: CryptologyCesar CypherIn this worksheet we crack the Caesar cypher using statistical analysis.LetterPercentageE11.16A8.50R7.58I7.54O7.16T6.95N6.65S5.74L5.49C4.54U3.63D3.38P3.17Letter PercentageM3.01H3.00G2.47B2.07F1.81Y1.78W1.29K1.10V1.01X0.29Z0.27J0.20Q0.20The frequency of letters is relevant for designing keyboards. The Kwerty keyboard for examplehas ESER and OI in prominent places.The ’top twelve’ letters help with about 80 p ercent of the text. You can remember the first 8 withthe memnonic”A SIN TO ERR”.An other thing to look for: The top pairs which appear areTH HE AN RE ER IN ON AT ND ST ES EN OF TE ED OR TI HI AS TOThe most frequent double letters are”LL EE SS OO TT FF RR NN PP CC”ExampleWe aim to decrypt the following text:xf uif qfpqmf pg uif vojufe tubuft,jo psefs up gpsn b npsf qfsgfdu vojpo,ftubcmjti kvtujdf, jotvsf epnftujd usborvjmjuz,qspwjef gps uif dpnnpo efgfodf,qspnpuf uif hfofsbm xfmgbsf,boe tfdvsf uif cmfttjoht pg mjcfsuzup pvstfmwft boe pvs qptufsjuz,ep psebjo boe ftubcmjti uijt dpotujuvujpo gps uifvojufe tubuft pg bnfsjdbDecoding:1) We count the number of letters which occur. Since we have not much time, the 8 most frequentletters are listed in this text. Can you figure out the text?f appears 39 timesu appears 29 timesp appears 25 timest appears 20 timess appears 20 timesj appears 20 timeso appears 17 timesb appears 14 timesThe Vigen`ere CipherWe learn how to encrypt messages using the Vigen`ere Cipher. This encryption was used for along time and should be seen as an important marker in the development of substitution ciphers:Julius Caesar -70Ahmad al-Qalqashandi 1400Leon Battista Alberti 1467Johannes Trithemius 1508Blaise de Vigin`ere 1586Charles Babbage 1854Friedrich Ka siski 1863Arthur Scherbius 1920Blaise de Vigen`ereAssume we have a secret key like ”ENIGMA”. Given a text like ”HARVARD IS COOL”, weencrypt it using the following table: for the first letter, we use the line starting with E, for thesecond letter, we use the line starting with N etc.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 ZA 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 ZB 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 AC C D E F G H I J K L M N O P Q R S T U V W X Y Z A BD D E F G H I J K L M N O P Q R S T U V W X Y Z A B CE E F G H I J K L M N O P Q R S T U V W X Y Z A B C DF F G H I J K L M N O P Q R S T U V W X Y Z A B C D EG G H I J K L M N O P Q R S T U V W X Y Z A B C D E FH H I J K L M N O P Q R S T U V W X Y Z A B C D E F GI I J K L M N O P Q R S T U V W X Y Z A B C D E F G HJ J K L M N O P Q R S T U V W X Y Z A B C D E F G H IK K L M N O P Q R S T U V W X Y Z A B C D E F G H I JL L M N O P Q R S T U V W X Y Z A B C D E F G H I J KM M N O P Q R S T U V W X Y Z A B C D E F G H I J K LN N O P Q R S T U V W X Y Z A B C D E F G H I J K L MO O P Q R S T U V W X Y Z A B C D E F G H I J K L M NP P Q R S T U V W X Y Z A B C D E F G H I J K L M N OQ Q R S T U V W X Y Z A B C D E F G H I J K L M N O PR R S T U V W X Y Z A B C D E F G H I J K L M N O P QS S T U V W X Y Z A B C D E F G H I J K L M N O P Q RT T U V W X Y Z A B C D E F G H I J K L M N O P Q R SU U V W X Y Z A B C D E F G H I J K L M N O P Q R S TV V W X Y Z A B C D E F G H I J K L M N O P Q R S T UW W X Y Z A B C D E F G H I J K L M N O P Q R S T U VX X Y Z A B C D E F G H I J K L M N O P Q R S T U V WY Y Z A B C D E F G H I J K L M N O P Q R S T U V W XZ Z A B C D E F G H I J K L M N O P Q R S T U V W X YNow it is your turnENIGMAE NI GMAEHARVARD IS COOLRSA EncryptionWe want to understand the basic mechanism for RSA encryption.Ron Rivest, Adi Shamir and Len Adleman.An RSA public key is a pair (n, a) where n is an integer with secret factorization n = pq andwhere a < (p − 1)(q − 1) is such that there exists b with ab = 1 mod (p − 1)(q − 1). Ana publishesthis pair. If Bob wants to send a secrete message to Ana, he transmits to Ana the messagey = xamod n .Ana can read the email by computingybmod n .Why do es it work? We use the Fermat’s little theorem which tells that xp−1− 1 is is divisibleby p and xq−1− 1 is divisible by q. But t his assumes p, q to be prime. For n = pq, we havex(p−1)(q−1)− 1 divisible by pq.Problem 1) Take p = 3 and q = 5 Verify that 2(p−1)(q−1)− 1 is divisible by n = pq.Because yb= xab= x(p−1)(q−1)= x modn, Ana gets the message, Bob has sent. But Eve has nochance to read it, because the only thing Eve can see is y and (n, …