Does anyone have the solution for this challenge?

Question

Each challenge is designed to help introduce you to a new piece of cryptography. Solving a challenge will require you to find a "flag".

These flags will usually be in the format crypto{y0ur_f1rst_fl4g}. The flag format helps you verify that you found the correct solution.

Answer 1

Could you give more details about the challenge?

Answer 2

Youri Wesker here is an idea using join() and a formatted string. Suppose your decoded words are collected in a list. Then you can use join to combine the words and insert the underscore character between words. Use an f-string to format the flag.

#decoded holds a list of decrypted words
decoded = ["DIFFER", "BEGIN", "LOOP", "VISIT"]
print(f"crypto{{{'_'.join(decoded)}}}")

Output:
crypto{DIFFER_BEGIN_LOOP_VISIT}

Notice that in the formatted string the braces are doubled in order to let them pass as single literal braces. The third (innermost) braces are interpreted as single braces so their enclosed contents become a field that gets formatted.

If your decrypted text is all in one string, then you might prefer to use replace() to replace the spaces with underscore. Again, you could use the above formatted string to build the flag.

Answer 3

The final result of the decryption of the text With a lag of 23 AFCCBO YBDFK ILLM SFPFQ is DIFFER BEGIN LOOP VISIT. If you want to format it as a flag in the CTF style, it would be: ??

Answer 4

I'm sorry I can't provide more information than that

Answer 5

Verifying your answer with the given information:

"lag of 23" suggests the use of Caesar cipher, with a shift value of 23(backwards). With 26 letters in the English alphabet, we can also shift the letters by 26 - 23 = 3 characters ahead. If we do that, it does infact become "DIFFER BEGIN LOOP VISIT".

Writing it in your flag format might be:

crypto{d1ff3r_b3g1n_l00p_v1s1t}