MA111: Contemporary mathematics 15 3 4 5 6 7 8 9 10 11 12 13 14 16 1 17 18 19 20 21 Schedule: HW 1 is due Tuesday, Oct 6th, 2015 Mini-Exam 2 is in-class on Thursday, Oct 8th, 2015 HW 2 is due Tuesday, Oct 13th, 2015 HW 3 is due Thursday, Oct 15th, 2015 HW 4 is due Tuesday, Oct 20th, 2015 Exam 2 is in-class on Thursday, Oct 22nd, 2015 Entrance Slip (due 5 min past the hour): 2 Today we use numbers to make using the codes easier. G A E I O U 1 2 3 4 5 X F H V K L M N W J P Q R S T Use a shift of 5 (so that d =3 becomes B C D Y Z K =8) to encrypt the message: “ this quiz is too easy ”
While we are passing out the worksheet... 11 1 2 3 4 5 6 7 8 9 10 12 X 13 14 15 16 17 18 19 20 21 What is 16 + 5? Is there a simpler way of describing the vowel shift? Please turn in your entrance slips. What about a shift of 10? What about 11? W F A E I O U 1 2 3 4 V 5 G P T S R H Q N M L K J Use a shift of 5 (so that d =3 becomes B C D Y Z L =8) to encrypt the message: “ this quiz is too easy ” Where does t =16 go? Z =21 What about 20 + 5? Where does y =20 go?
Old words General words encryption (how to convert plaintext to ciphertext) decryption (the reverse, cipher to plain) cipher (both encryption and decryption methods) key (a small secret that lets you change the cipher) Shift cipher Encrypt: shift vowels and consonants right by an amount according to the key Decrypt: shift vowels and consonants left by an amount according to the key plaintext (plain message, “ can you keep a secret ”) ciphertext (hidden version, “ DEP ZUA LIIQ E TIDSIV ”)
New words: shift cipher with numbers 18 6 7 8 9 10 11 12 13 14 15 16 17 19 A 20 21 To encrypt with shift cipher, add the key to the number, using wrap-around if too big (subtract 5 if a vowel, or subtract 21 if a consonant) To decrypt with shift cipher, subtract the key from the number, using wrap-around if too small, (add 5 if a vowel, or add 21 if a consonant) For example if the shift key is 7, then And to decrypt, 5 4 3 J E I O U 1 2 3 4 5 F G 2 H K S 1 X W V L T R Q P N M B C D Y Z g = 5 → P =12, since 5 + 7 = 12 and w =18 → F =4, since 18 + 7 = 25 and 25 − 21 = 4 . P =12 → g = 5, since 12 − 7 = 5 and F = 4 → w =18, since 4 − 7 = − 3 and − 3 + 21 = 18 .
New words: double-it cipher The double-it cipher has no key (we’ll fjx that next week). To encrypt, double the number using wrap-around. To decrypt, …fjll in the decoder wheel? (we’ll fjnd a faster way next week)
Exit quiz 9 W X 1 2 3 4 5 6 7 8 10 T 11 12 13 14 15 16 17 18 19 20 V S Decode this message knowing that it is encoded using a shift R A E I O U 1 2 3 4 5 21 F G H J K L M N P Q cipher that takes b to P “ Kvifi ror hvi ebozeyg tu? ” B C D Y Z
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