Official help for 3b
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22 Oct 17 at 1:11 am #31621
Anonymous
InactiveSo, by now hopefully you have taken the standard advise and solved part a. This will give you a big hint as to what type of cipher to do more research on. You should also have noticed that there are just 5 characters used in the cipher text. How can you encode the whole alphabet with just 5 characters? More hints to follow…
22 Oct 17 at 1:36 pm #31633InactiveI know the cipher and what I’m supposed to do, but how am I supposed to figure out the order?
22 Oct 17 at 1:39 pm #31642InactiveIf you know the cipher type, and just need to figure out the order, then think about how you would do that for other (simpler) ciphers.
23 Oct 17 at 12:27 am #31644Inactivethank you so much! I just realised I completely skipped a thing I do when I decrypt stuff
23 Oct 17 at 12:28 am #31645InactiveNevermind I tried with the method I use but it just comes out even more messed up…
23 Oct 17 at 12:29 am #31649InactiveI also know what the cipher is but don’t know the order.
23 Oct 17 at 10:29 am #31669InactiveGod knows I completely forgot part B was on a deadline xD Need to do it asap but too tired now
23 Oct 17 at 1:45 pm #31674Inactivewhen will the next hint be published?
23 Oct 17 at 3:11 pm #31683InactiveOFFICIAL CLUE: If you don’t know the type of cipher, solve part a. Check out the forum page for hints on that (I just revealed the type of cipher for part a). If you have part a solved, then you know the type of cipher. Now to think, what might be a common theme with the 5 letters which are used in the cipher text? Remember it was written back in Roman times.
23 Oct 17 at 7:08 pm #31687InactiveI understand how this cipher might work now, but decrypting it never makes sense, and there is little info out there about the cipher, so I have no idea how to decrypt it
24 Oct 17 at 1:26 am #31692InactiveOFFICIAL CLUE: As you have probably found out from part a, this is a polybius cipher. This means the alphabet has been put in a 5×5 grid (2 letters combined to make 25 letters total). Then each column and row is given a heading (the 5 characters which appear in the cipher text). You read off the row and column to give you 2 characters in the cipher text for each 1 character in the plain text. Now you need to figure out the square grid, and the row/column headings. More hints will follow.
24 Oct 17 at 6:18 pm #31704InactiveDo I need to change the cipher text into numbers?
24 Oct 17 at 7:19 pm #31708Inactive@Koala , no you don’t, each 2 letters stands for 1 letter, instead of numbers in the columns or rows in the table, you replace them with the letters in the text, but you need to figure out the order they are in (which I myself have yet to do)
24 Oct 17 at 7:24 pm #31712InactiveEXAMPLE: An example of the Polybius cipher is given below. In this example I am assuming there are no letter ‘z’ in the message, therefore I can just ignore this letter in the grid below:
– 1 2 3 4 5
1 a b c d e
2 f g h i j
3 k l m n o
4 p q r s t
5 u v w x yTo encode the message: “hello” you would read off the row & column for each letter, so for h you would get 23. The message hello would be:
23 15 32 32 35.Hope this helps – and no, the above is not the exact polybius square that was used to encode 3b.
25 Oct 17 at 4:00 pm #31719InactiveSpent the day on 3b, finally given up, worrying that I am beaten on challenge 3. Oh well.
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