It seems that another encrypted message has been intercepted. The encryptor seems to have learned their lesson though and now there isn't any punctuation! Can you still crack the cipher?
This time around, we don't even have spaces or full stops!
We can use a similar approach to last time, but this time we may have to use frequency analysis (as suggested by the last flag!) and even bi- and trigram analysis. This means finding common letters, or common groupings of 2/3 letters, and comparing it to what would typically occur in a regular english text.
However, first off, there is very clearly a flag at the end:
qcuhUIE{K6F4G_4K41R515_15_73A10B5_702E03EU}
And we can use the same initial approach as last time and update the alphabet accordingly.
so there is a word P*TITIO*. According to a crossword solver, that's either petition or petitios. But ahead of it, there's even more:
COgPnTITIOk
This looks like COMPETITION! Let's throw that in.
After spotting lots more words like cybersecurity etc, I get the alphabet:
alphabet = 'mjuanedsc-oxgkhqvftibzwpr-'
And we use the decrypt again:
from string import ascii_uppercase, ascii_lowercase
alphabet = 'mjuanedsc-oxgkhqvftibzwpr-'
text = 'isnfnnpctitnznfmxhisnfwnxxntimjxctsnascdstushhxuhgqbinftnubfciruhgqnicichktckuxbackdurjnfqmifchimkabturjnfusmxxnkdnisntnuhgqnicichktehubtqfcgmfcxrhktrtingtmagckctifmichkebkamgnkimxtwscusmfnznfrbtnebxmkagmfonimjxntocxxtshwnznfwnjnxcnznisnqfhqnfqbfqhtnhemscdstushhxuhgqbinftnubfciruhgqnicichkctkhihkxrihinmuszmxbmjxntocxxtjbimxthihdnitibankitckinfntinackmkanpucinamjhbiuhgqbinftucnkunanenktcznuhgqnicichktmfnheinkxmjhfchbtmeemcftmkauhgnahwkihfbkkckdusnuoxctitmkanpnubickduhkecdtufcqitheenktnhkisnhisnfsmkactsnmzcxrehubtnahknpqxhfmichkmkacgqfhzctmichkmkaheinksmtnxngnkitheqxmrwnjnxcnznmuhgqnicichkihbusckdhkisnheenktcznnxngnkitheuhgqbinftnubfcirctisnfnehfnmjniinfznscuxnehfinusnzmkdnxctgihtibankitckmgnfcumkscdstushhxtebfisnfwnjnxcnznismimkbkanftimkackdheheenktczninuskcvbntctnttnkicmxehfghbkickdmkneenuicznanenktnmkaismiisnihhxtmkauhkecdbfmichkehubtnkuhbkinfnackanenktcznuhgqnicichktahntkhixnmatibankitihokhwisncfnkngrmtneenuicznxrmtinmusckdisngihmuicznxrisckoxconmkmiimuonfqcuhuiectmkheenktcznxrhfcnkinascdstushhxuhgqbinftnubfciruhgqnicichkismitnnotihdnknfminckinfntickuhgqbinftucnkunmghkdscdstushhxnftinmusckdisngnkhbdsmjhbiuhgqbinftnubfcirihqcvbnisncfubfchtcirghiczmickdisngihnpqxhfnhkisncfhwkmkankmjxckdisngihjniinfanenkaisncfgmusckntisnexmdctqcuhUIE{K6F4G_4K41R515_15_73A10B5_702E03EU}'.lower()
dec = ''
for c in text:
if c in ascii_uppercase:
dec += ascii_uppercase[alphabet.index(c.lower())]
elif c in ascii_lowercase:
dec += ascii_lowercase[alphabet.index(c)]
else:
dec += c
print(dec)
# picoctf{n6r4m_4n41y515_15_73d10u5_702f03fc}
As we can see from the flag (and also the hint), ngram analysis was the way to go.
