Question

**DO IT AS PYTHON PLEASE**

The Trifid Cipher General Problem Description The Trifid cipher (not to be confused with the creatures from the classic science-fiction film "The Day of the Triffids") is an algorithm that enciphers a plaintext message by encoding each letter as a three-digit number and then breaking up and rearranging the digits from each letter's encoded form. For this assignment, you will create a set of Python functions that can encode messages using this cipher (these functions are described in detail following the general explanation of how the Trifid cipher works; you can skip ahead to those function specifications if you want to, but they may make more sense if you read the general explanation first) The Trifid cipher works as follows 1. Use an encryption key to construct a lookup table that will translate each letter into a 3-digit number. The lookup table (technically, there are three separate lookup tables that work together) requires 27 characters in all; in addition to the 26 letters of the alphabet, we will use the '!' character as the 27th value 2. Process the plaintext message one letter at a time, obtaining its equivalent number from the lookup table and writing the digits of that number along each of three rows (one digit per line/row) 3. After the plaintext message has been converted to digits, break each row of digits into a series of 5-digit groups Recombine the groups into a single string of digits in the following order: the first groups of each row, followed by the second groups of each row, etc 4. Each group of three digits from this combined string is then translated into a new letter via the lookup table 5. Finally, we break up the ciphertext into 5-letter groups. If the final group has fewer than 5 letters, we add Xs as necessary to pad it to a length of 5 Example Suppose that we want to use the Trifid cipher to encrypt the message "Sherlock Holmes and Doctor Watson are investigating" using the secret key "deduction". After capitalizing every letter in the key, we write out its unique letters, in the order they appear, followed by the remainder of the alphabet (plus a placeholder character to get exactly 27 characters) DEUCTIONABFGHJKLMPQRSVWXYZ! These letters will be arranged into three 3x3 grids as follows Example Suppose that we want to use the Trifid cipher to encrypt the message "Sherlock Holmes and Doctor Watson are investigating using the secret key "deduction". After capitalizing every letter in the key, we write out its unique letters, in the order they appear, followed by the remainder of the alphabet (plus a placeholder character to get exactly 27 characters) DEUCTIONABFGHJKLMPQRSVWXYZ! These letters will be arranged into three 3x3 grids as follows: Table 1 Table 2 Table 3 12 3 We can use these tables to obtain the 3-digit trigram for each letter in the plaintext message. For example, the letter "S translates to 313 (table 3, row 1, column 3). Each trigram is written vertically below the original letter (the message has beein broken into 5-letter groups) SHERL OCKHO LMESA NDDOC TORWA TSONA REINV ESTIG ATING 12113 32223 33113 31132 23123 21333 11232 11221 32231 31221 11311 12233 21111 21223 23123 22321 23233 32323 To encrypt the message, we take each "block" (in this case, all three rows of a 5-character group) and concatenate the rows into a single string of (15) digits. Every three-digit group in this string, reading left to right, is then translated back into a new letter using the same letter-trigram equivalences as before: 321321211331221 = 321 321 211 331 221 = VV B Y H 112213222311311 = 112 213 222 311 311 = E G J Q Q 221313311312233-221 313 311 312 233 -HSQR P 113312312321223 113 312 312 321 223 U RRV 131112133323123 = 131 112 133 323 123 = O E AX ! 311131123222321 311 131 123 222 321 -Q 0 1 J ν 131121122123233 = 131 121 122 123 233 = О С Т 1 P 111123223132323 = 111 123 223 132 323 = D I K N X This gives us a final enciphered message of VVBYH EGJQQ HSQRP DUUHD URRVK OEAXI QOIJV OCTIP DIKNX. Part I: Constructing the Lookup Table (15 points) (Place your answer to this problem in the "codes.ру" file) Complete the codesO function, which takes a string argument (the encryption key) and returns a dictionary that maps (uppercase) letters to three-digit trigrams. Use the pseudocode algorithm below as a guide: 1. General Preparation Convert the key to uppercase and use the string method replace ) to remove any spaces. The replace () method takes two arguments: the character sequence to be replaced, and the replacement value. In this case, use "" (a single space) as the search value and "" (the empty string) as its replacement, e.g.,myKey.replace (" Construct a list of available (not yet used) letters of the alphabet, followed by the placeholder character"! Create another list to hold your three initial lookup tables. This list should contain four elements: an arbitrary integer (or some other value) to serve as a placeholder, followed by three empty lists (the placeholder value is there so that the indices of the three lookup table lists will begin with 1 rather than 0). . Create a variable to track the index of the current lookup table (this will initially be 1) 2. Create the Lookup Tables (a) For each letter in the encryption key, check to see if it is in the list of available/unused letters. If it is, use the remove ) method to remove it from that list and append it to the current lookup table list. If the length of the current lookup table list is now 9, increment the "current lookup table" variable to point to the next list. Note that if the encryption key contains multiple copies of a given letter, that letter will only be added to the lookup table(s) one time Once you have processed the entire encryption key, add the remaining values from the "unused letters" li the current lookup table, moving to the next lookup table each time the current table's list reaches a length of 9 elements. 3. Create the Dictionary of Trigrams (a) Create a new, empty dictionary to store your letter-trigram mappings. (b) For each of the three lookup tables/lists in order (indices 1, 2, and 3 in your list from Step 1), calculate the letter trigrams as follows » Your new dictionary key will be the letter at that index The trigram for that letter is calculated as follows o first Digit the index of the current table in the list of tables (e.g., 1, 2, or 3) the index of the current letter in the current table o letter Index o secondDigit = (1etter Index 113) + 1 thirdDigit-(let terindex % 3) plus 1 o trigram- firstDigit 100+ secondDigit *10+thirdDigit (c) Return your newly-filled dictionary Examples: Note: the examples below do not show the entire lookup table that will be generated from the encryption key. Instead, they show what the resulting lookup table will return for a given plaintext character Function Call Plaintext LetterCorresponding Trigram codes (" DRAGON") 112 codes (" DRAGON") 213 codes (" DRAGON") 332 codes ("NEPTUNE") 131 II RIn codes (" NEPTUNE") I TIT 321 codes ("NEPTUNE") codes ("CHALLENGER" ) 122 nFn codes ("CHALLENGER" ) 233 codes ("CHALLENGER") 312 ITI

Answer 1

Task : Constructing The Lookup Table (complete the codes() function ; )

Python CODE ::

codes. py 1 def codes (key) key key . upper(). replace(" ",") # removing spaces Not used = list( "ABCDEFGHlJKLMNOPQRSTUVWXYZ !") Tables [0,[],[],[] ] ; Ind : 1 # Creating empty tables for letter in key: # Taking the letters from key if letter in Not used: 4 6 Tables [Ind].append (letter) Not used.remove (letter) if len (Tables [Ind]) == 9 : Ind += 1 # If current list is filled; 8 9 for letter in Not-used : # Then taking letters from remaining Tables [Ind].append (letter) if len (Tables [ Ind]) == 9 : Ind-1 Trigram- {} # Empty Trigram Dictionary for i in range (3): first = i+1 15 terInd in range (9): second(letterInd// 3)+1 third- (letter!nd % 3)+1 letter Tables[first] [letterInd] trigram first .10θ + second-10 third Trigram [letter] -trigram # FIlling Trigram 17 20 21 return Trigram # Returning Filled Trigram Dictionary 23 24 print (codes ("Deduction") 25

Execution (OUTPUT)::..

aj@d1r3wolf:/tmp/Q File Edit View Search Terminal Help aj@d1r3wolf:/tmp/Qs py codes.py 'D 111, 'E 112, 'U': 113, 'C': 121, 'T: 122, 'I 123, '0131, 'N': 132, 'A': 133, 'B': 211, 'F': 212, 'G": 213, 'H': 221, '222 : ינ, 'K': 223, 'L': 231, M: 232, "P': 233, "0': 311, 'R": 312, 'S': 313, 'V': 321, : 322, "'X': 323, Y: 331, 'Z: 332, ":333 ajedlr3wolf:/tmp/Q$

The key i have used is : "Deduction"

RAW PYTHON CODE ::

_____________________________ codes.py ____________________________________________________

def codes(key):
   key = key.upper().replace(" ",'') # removing spaces
   Not_used = list("ABCDEFGHIJKLMNOPQRSTUVWXYZ!")
   Tables = [0,[],[],[]] ; Ind = 1 # Creating empty tables
   for letter in key:   # Taking the letters from key
       if letter in Not_used:
           Tables[Ind].append(letter)
           Not_used.remove(letter)
           if len(Tables[Ind]) == 9 : Ind += 1 # If current list is filled;
   for letter in Not_used: # Then taking letters from remaining
       Tables[Ind].append(letter)
       if len(Tables[Ind]) == 9 : Ind += 1
   Trigram = {} # Empty Trigram Dictionary
   for i in range(3):
       first = i+1
       for letterInd in range(9):
           second = (letterInd // 3)+1
           third = (letterInd % 3)+1
           letter = Tables[first][letterInd]
           trigram = first*100 + second*10 + third
           Trigram[letter] = trigram # FIlling Trigrams
   return Trigram # Returning Filled Trigram Dictionary

print(codes("Deduction"))
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