The official Challenge 10 tips and hints thread
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18th December 2023 at 7:58 am #92864Adam_YafaiParticipant
As funny as it was, I did question the initial artistic decision behind allocating only 4 pixels towards the clues
18th December 2023 at 7:58 am #92835ZB123ParticipantHarry, are there 4 ways of going about this? Or less… If you know what I mean
18th December 2023 at 7:58 am #92836ZB123ParticipantFor part a that is
18th December 2023 at 7:58 am #92837ZB123Participantor for b? I just realised something if so…
18th December 2023 at 7:59 am #92866JCParticipantIs 10b the bifid cipher?
18th December 2023 at 9:00 am #92767HarryKeymasterThe fourth hint for Challenge 10A
Challenge 9A was a very simple cipher in which the letters in positions 1,3,5, 7, 9 etc were left along and the letters in positions 2,4,6,8,10 … were shifted by the letters to their left. Many of you noted that this is a special case of a Hill cipher, Why not look that up?
18th December 2023 at 9:00 am #92765HarryKeymasterThe third official hint for Challenge 10B
Since the fragments showed plaintext squares top left and bottom right of the page, we might guess this is a Four Square encryption. We can use that along with our bigram count to start filling in the square. At this stage these are only guesses and further analysis will be needed!
18th December 2023 at 11:34 pm #92878snugglyhedgehogParticipantI thought my hill-climbing code was wrong… turns out that there’s only so much you can find with randomness in one go. finally found the main bit, now onto more specific decryption. thanks harry for another great year!
18th December 2023 at 11:38 pm #92885madnessParticipantIf you’re bored with 10A and 10B, try the H³ challenge, in a separate topic.
18th December 2023 at 11:39 pm #92879Flappyasdf_2023ParticipantHey Harry,
In terms of prizes for induvial; how fast must we have solved 10B by? (I did it yesterday). Also is it simply the first 2 people to solve, or is the explanation of how it was solved also relevant in making that decision?
Thanks!
[We do take the explanations into account, and have asked some people to write to us to help with that. Best wishes, Harry]
18th December 2023 at 11:39 pm #92880JakeParticipantThanks for the hints!
I had guessed that it was one of the so-and-so square ciphers, such as Playfair and its relatives, so it’s nice to be correct there. But from my searches I couldn’t find a guide on cracking it with pencil and paper – all seem to need a certain amount of programming skill. Do you know of a good outline I can read to tackling these kinds of ciphers in a traditional slow path way?
[The hints we have lined up over the next two weeks will (we hope) guide you through it. All the best, Harry]
19th December 2023 at 9:00 am #92769HarryKeymasterThe fifth hint for Challenge 10A
If we think this is a Hill cipher and that the names JODIE and HARRY are likely to appear at the start or the finish of the text, then we might see the pairs
JO, DI
OD, IE
HA, RR
AR, RY
in those positions. Assuming that we are using the convention A->0, B->1, C-> 2 and so on this would give us a series of equations that might allow us to work out the Hill cipher matrix. Take a look at Unit 88 in Madness’s book to see how to use that information.
20th December 2023 at 9:00 am #92784HarryKeymasterThe sixth hint for Challenge 10A
If Jodie sent this message and signed it we will see the bigrams OD, IE at the end and these are represented by the vectors (14, 3) and (8, 4).
What we actually see are the letter pairs RU, MQ which correspond to the vectors (17,20) and (12,16), so the encryption works to give
14a+3b = 17
14c+3d=20
8a+4b =12
8c+4d=16
We can solve these four equations in four unknowns to get the encryption parameters. Try looking up solving systems of equations online, but be careful doing so in mod 26 arithmetic!
21st December 2023 at 9:00 am #92768HarryKeymaster21st December 2023 at 9:00 am #92788HarryKeymasterThe seventh hint for Challenge 10A
The inverse matrix is given by switching the diagonal entries and negating the off diagonal so we get 2,-1,-1,1.
Now you use that matrix to decrypt the cipher text as if you were encrypting the plain text, and out should come the message. You can do it by hand, or you can use a python script or a spreadsheet. Have fun!
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