WINNER OF THE CHALLENGE
teja156 May 19th, 2019 500 Never
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- WINNER OF THE CHALLENGE ANNOUNCED!
- The given cipher text can be decrypted based on the Encryption algorithm.
- The first step is to find out the value of 'n'. Given, the second character of the original String is 'R', so we need to find a number 'n' such that n%82==73
- There are 44993 possible values of 'n' that satisfy the above criteria.
- So, this means that there are LOTS of possible decrypted texts that can be obtained from the given cipher text.
- For this reason, I'm announcing the winner as the first person who submitted any of these possible strings.
- WINNER : Subham Sahu
- WINNER's SUBMISSION STRING : RR>/B)RB/Bd/
- WINNER'S SUBMISSION TIME : 16 May, 11:49 AM
- THIS IS THE ENCRYPTION ALGORITHM CODE : https://github.com/teja156/CrackThisCipher/blob/master/encryption.py
- THIS IS THE CODE TO CHECK IF A STRING IS A POSSIBLE DECRYPTED FORM OF THE GIVEN CIPHER (ANSWERS ARE EVALUATED WITH THIS CODE, SO CHECK IT OUT TO SEE IF YOUR ANSWER WAS RIGHT) : https://github.com/teja156/CrackThisCipher/blob/master/check_possibility_match.py
- The original string in my mind when I made this challenge is 'cR709Ac20KeD'
- But no one has got this string! I understand, it's because there are literally LOTS of strings and it's difficult to extract this string from all these strings (but using regex, you could try to extract some strings that have some meaning like 'crack', 'cracked', 'youwon', etc, and by doing so, the string 'cR709Ac20KeD' could've been retrieved.)
- Here are some honorable mentions, who solved the challenge but just not the first to solve it :
- 1. Vukasin Kubet
- He submitted a whopping 2618 possible strings (https://pastebin.com/QCWv5SV1), which is really impressive. Great work :)
- 2. Tamil Selvan
- 3. Panneerselvam D
- 4. Sanjay R
- 5. Buncy Shaddai
- 6. Akash Kumar
- 7. Tanay Gupta
- 8. Shibaayan Maity
- 9. Sarathy V
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