Math 1350 (FALL 2008) Prelim 1 (10/6/2008) 1. Solution. For the Atbash cipher for the Roman alphabet we have

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1 Math 1350 (FALL 2008) Prelim 1 (10/6/2008) 1 1. (8 pts) Decrypt the following message, which was enciphered using the Atbash cipher. GSVYZ WMVDH RHGRN VUORV HGSVT LLWMV DHRHB LFIVG SVKRO LG Solution. For the Atbash cipher for the Roman alphabet we have The plaintext is First 13 letters A B C D E F G H I J K L M Last 13 letters Z Y X W V U T S R Q P O N THEBA DNEWS ISTIM EFLIE STHEG OODNE WSISY OURET HEPIL OT, that is, THE BAD NEWS IS TIME FLIES, THE GOOD NEWS IS YOU RE THE PILOT. 2. (8 pts) Use the Vigenère cipher to decipher ZWAHECOO using the keyword TIME. Solution. Let y be the numerical equivalent of the ciphertext, let x be the numerical equivalent of the plaintext and let k be the numerical equivalent of the keyword. Since the enciphering formula of the Vigenère cipher is y = (x+k) MOD 26, solving for x gives us x = (y k) MOD 26. ciphertext Z W A H E C O O y keyword T I M E T I M E k y k x = (y k) MOD plaintext G O O D L U C K Therefore, the answer is GOOD LUCK.

2 Math 1350 (FALL 2008) Prelim 1 (10/6/2008) 2 3. (8 pts) Decipher the following, which was enciphered using a keyword columnar transposition with the keyword BAILOUT: HMACE HISTK HINTF EEAST OEVSS RPEUL EITKR DTARO CTDIF TCSOE ENOSE N Decipher the message. Solution. The keyword is 7 characters long. Since the ciphertext is 56 characters long, and 56 = 7 8, we arrange the ciphertext in 7 columns and 8 rows, in a way determined by the order of our keyword s letters in the alphabet, as follows: B A I L O U T T H E S T O C K M A R K E T H A S P R E D I C T E D N I N E O U T O F T H E L A S T F I V E R E C E S S I O N S Now the original message can be read off row by row: THE STOCK MARKET HAS PREDICTED NINE OUT OF THE LAST FIVE RECESSIONS (Paul A. Samuelson)

3 Math 1350 (FALL 2008) Prelim 1 (10/6/2008) 3 4. (10 points) Decrypt the following English sentence, which was encrypted using a shift cipher. PUTHA OLTHA PJZFV BKVUA BUKLY ZAHUK AOPUN ZFVBQ BZANL ABZLK AVAOL T Show your work. Hint: Note that the most frequent ciphertext letter is A. Solution. Our counting reveals that A occurs nine times in the ciphertext. Since E is the most common occurring letter in English, we make the tentative assumption that ciphertext A corresponds to plaintext E, i.e. C A = P E. Since the numerical equivalents of A and E are 0 and 4, respectively, the deciphering shift must be k = 22. For this shift, PUTHA decrypts as LQPDW, which does not seem to be the beginning of an English word or phrase. Since T is the second most common occurring letter in English, let us assume that ciphertext A corresponds to plaintext T, i.e. C A = P T. Since the numerical equivalents of A and T are 0 and 19, respectively, the deciphering shift must be k = 19. For this shift, PUTHA decrypts as INMAT, which seems promising. Let us decrypt the second five-letter group of the ciphertext: OLTHA decrypts as HEMAT. Shifting the English alphabet using k = 7, we obtain the table(s) cipher A B C D E F G H I J K L M plain T U V W X Y Z A B C D E F cipher N O P Q R S T U V W X Y Z plain G H I J K L M N O P Q R S Thus, the given English sentence decrypts as: PUTHA OLTHA PJZFV BKVUA BUKLY ZAHUK AOPUN ZFVBQ BZANL ABZLK AVAOL T becomes INMAT HEMAT ICSYO UDONT UNDER STAND THING SYOUJ USTGE that is, TUSED TOTHE M, IN MATHEMATICS YOU DON T UNDERSTAND THINGS, YOU JUST GET USED TO THEM. (Johann von Neumann)

4 Math 1350 (FALL 2008) Prelim 1 (10/6/2008) 4 5. (8 pts) The ciphertext TUFBC was enciphered using an affine cipher. Given that the ciphertext letters T and U correspond to the plaintext letters C and R, respectively, find (a) the encipherment formula. Solution. Let a and b be the enciphering multiplier and the enciphering shift of the given affine cipher. Since we are given that C T = P C and C U = P R, we have 19 2a + b (mod 26) and 20 17a + b (mod 26) Subtracting the first congruence from the second gives us 1 15a (mod 26) or, equivalently, 15a 1 (mod 26) Since the inverse of 15 modulo 26 is 7, multiplying the above congruence by 7 gives us a (mod 26), The only (and obvious) acceptable solution of this is a=7. With a = 7, the first equation above gives us b 19 2a = = 5 (mod 26). The only (and obvious) acceptable solution of this is b = 5. Therefore, the encipherment formula is (b) the plaintext. Solution. Since the decipherment formula is y = (7x + 5) MOD 26. a 1 (y b) = 7 1 (y 5) 15(y 5) (mod 26). x = 15(y 5) MOD 26 or x = 15y + 3 MOD 26 We need only decipher the letters F, B and C. We have ciphertext T U F B C y y x = (15y + 3) MOD plaintext C R A S H Therefore, the plaintext is CRASH.

5 Math 1350 (FALL 2008) Prelim 1 (10/6/2008) 5 6. (8 pts) The following text was encrypted using a mixed alphabet with a keyword. The first two words of the plaintext are BEWARE OF. Decrypt the text and recover the keyword. Explain your method. OEWHP ELABP EEGQO EHPDK BBDAT Q Solution. Using the given cribs, we start to construct an encryption alphabet: Ciphertext H O E A ciphertext L P W Between A and L there are 8 spaces for eight missing letters. So continuing in this way we easily get Ciphertext H O E A B C D F G I J ciphertext K L M N P W X Y Z It seems that the keyword is a five-letter word H O E, which contains exactly two of Q, R, S, T, U, V. We may form two English words by these combinations: HOUSE and HORSE. Using HOUSE as the keyword we easily see that not all words of the ciphertext are deciphered properly (check that!). Using HORSE, we have the following encryption alphabet: Ciphertext H O R S E A B C D F G I J ciphertext K L M N P Q T U V W X Y Z The given text decrypts as OEWHP ELABP EEGQO EHPDK BBDAT Q BEWARE OF GREEKS BEARING GIFTS (Laocoon)