For a simple cybersecurity project to practice the key concepts of encryption and decryption, one can create a simple program (using Python) to achieve this. For this project, we will need to utilize the Pandas library and a character list (which will be provided as a .csv).
Download: encryption/decryption key
Setup
For this, you can feed the input value or you can setup the code where it can collect a user’s input instead. For this project, we will just feed it the message to encrypt.
# import Pandas
import pandas as pd
# import encryption key via Pandas
# 'r' = read
encryptionkey = pd.read_csv(r"<insert .csv path>",
sep=',', names=['Character', 'Byte'], header=None, skiprows=[0])
# create a dataframe
df = pd.DataFrame(data=encryptionkey)
# take the columns and convert values to string
df['Character'] = df['Character'].astype(str)
df['Byte'] = df['Byte'].astype(str)
# split original message
# for example: ['g', 'r', 'e', 'e', 't', 'i', 'n', 'g', 's']
def split(message):
return [char for char in message]
# message for encryption
message = "The real test is not whether you avoid this failure, because you won't." \
" It's whether you let it harden or shame you into inaction, or whether you learn from it; whether you choose to persevere."
split_message = split(message)
print(split_message)Encryption
# declare function called 'code_message()'
def code_message():
coded_message = ""
# create for loop
# loop will be used to iterate a given number of times a block of code
# iterations are dependent on value provided
for x in range(len(split_message)):
y = split_message[x]
# establish error handling via try-except
try:
char_coded = encryptionkey.loc[encryptionkey['Character'] == y, 'Byte'].iloc[0]
print(type(char_coded))
# To handle if character is not in our decryption list
except:
print('unrecognized character')
# this is a placeholder value, hence @@@
char_coded = '@@@'
# concatenate coded_message + char_coded
coded_message = coded_message + char_coded
return coded_message
coded_message = code_message()
# display encrypted message
print('Your coded message:', code_message(), '\n')
Decryption
####### Decryption #######
def decode_message(message):
new_word = ''
message_decode = []
# similar to encryption setup, except we will use +2
# increment will be +2 instead of +1
for x in range(0, len(message), 2):
# instantiate var as slice of message
y = message[x:x + 2]
index_nb = df[df.eq(y).any(1)]
df2 = index_nb['Character'].tolist()
# transform data type to string
a = [str(a) for a in df2]
# concatenate message_decode + a
message_decode = message_decode + a
new_word = ''.join(message_decode)
return new_word
coded_message_str = str(coded_message)
print('Your decoded message:', decode_message(coded_message_str))Full Code
# import Pandas
import pandas as pd
# import encryption key via Pandas
# 'r' = read
encryptionkey = pd.read_csv(r"<insert .csv path>",
sep=',', names=['Character', 'Byte'], header=None, skiprows=[0])
# create a dataframe
df = pd.DataFrame(data=encryptionkey)
# take the columns and convert values to string
df['Character'] = df['Character'].astype(str)
df['Byte'] = df['Byte'].astype(str)
# split original message
# for example: ['g', 'r', 'e', 'e', 't', 'i', 'n', 'g', 's']
def split(message):
return [char for char in message]
# message for encryption
message = "The real test is not whether you avoid this failure, because you won't." \
" It's whether you let it harden or shame you into inaction, or whether you learn from it; whether you choose to persevere."
split_message = split(message)
print(split_message)
####### Encryption #######
# declare function called 'code_message()'
def code_message():
coded_message = ""
# create for loop
# loop will be used to iterate a given number of times a block of code
# iterations are dependent on value provided
for x in range(len(split_message)):
y = split_message[x]
# establish error handling via try-except
try:
char_coded = encryptionkey.loc[encryptionkey['Character'] == y, 'Byte'].iloc[0]
print(type(char_coded))
# To handle if character is not in our decryption list
except:
print('unrecognized character')
# this is a placeholder value, hence @@@
char_coded = '@@@'
# concatenate coded_message + char_coded
coded_message = coded_message + char_coded
return coded_message
coded_message = code_message()
# display encrypted message
print('Your coded message:', code_message(), '\n')
####### Decryption #######
def decode_message(message):
new_word = ''
message_decode = []
# similar to encryption setup, except we will use +2
# increment will be +2 instead of +1
for x in range(0, len(message), 2):
# instantiate var as slice of message
y = message[x:x + 2]
index_nb = df[df.eq(y).any(1)]
df2 = index_nb['Character'].tolist()
# transform data type to string
a = [str(a) for a in df2]
# concatenate message_decode + a
message_decode = message_decode + a
new_word = ''.join(message_decode)
return new_word
coded_message_str = str(coded_message)
print('Your decoded message:', decode_message(coded_message_str))Output
['T', 'h', 'e', ' ', 'r', 'e', 'a', 'l', ' ', 't', 'e', 's', 't', ' ', 'i', 's', ' ', 'n', 'o', 't', ' ', 'w', 'h', 'e', 't', 'h', 'e', 'r', ' ', 'y', 'o',
'u', ' ', 'a', 'v', 'o', 'i', 'd', ' ', 't', 'h', 'i', 's', ' ', 'f', 'a', 'i', 'l', 'u', 'r', 'e', ',', ' ', 'b', 'e', 'c', 'a', 'u', 's', 'e', ' ', 'y', 'o', 'u', ' ', 'w', 'o', 'n', "'", 't', '.', ' ', ' ', 'I', 't', "'", 's', ' ', 'w', 'h', 'e', 't', 'h', 'e', 'r', ' ', 'y', 'o', 'u', ' ', 'l', 'e', 't', '
', 'i', 't', ' ', 'h', 'a', 'r', 'd', 'e', 'n', ' ', 'o', 'r', ' ', 's', 'h', 'a', 'm', 'e', ' ', 'y', 'o', 'u', ' ', 'i', 'n', 't', 'o', ' ', 'i', 'n', 'a', 'c', 't', 'i', 'o', 'n', ',', ' ', 'o', 'r', ' ', 'w', 'h', 'e', 't', 'h', 'e', 'r', ' ', 'y', 'o', 'u', ' ', 'l', 'e', 'a', 'r', 'n', ' ', 'f', 'r', 'o', 'm', ' ', 'i', 't', ';', ' ', 'w', 'h', 'e', 't', 'h', 'e', 'r', ' ', 'y', 'o', 'u', ' ', 'c', 'h', 'o', 'o', 's', 'e', ' ', 't', 'o', ' ', 'p', 'e', 'r',
's', 'e', 'v', 'e', 'r', 'e', '.']
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74207768657468657220796f752061766f69642074686973206661696c7572652c206265636175736520796f7520776f6e27742e202049742773207768657468657220796f75206c65742069742068617264656e206f72207368616d6520796f7520696e746f20696e616374696f6e2c206f72207768657468657220796f75206c6561726e2066726f6d2069743b207768657468657220796f752063686f6f736520746f207065727365766572652e
Your decoded message: The real test is not whether you avoid this failure, because you won't. It's whether you let it harden or shame you into inaction, or whether you learn from it; whether you choose to persevere.
Michael is an Information Technology consultant, with a focus on cybersecurity. Every day, Michael strives to learn something new, with an aim to share it with you all!
