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Is that an incentive to review matrix multiplications before tomorrow?
🙂It wasn’t (it would be a little early in the competition to be introducing Hill ciphers!) but it never hurts to be prepared! (Mind you, I might mean being prepared for CC20026 or CC2027 … ) Harry
If I were you (and I am pretty sure that I am not), then I would study past seasons to see what could be expected. I would notice that matrix transposition and Vigenère come up often. And if I were still being you, then I would work to make sure I knew how to solve such ciphers quickly.
Speaking of commutation, and speaking of ciphers, have you noticed that if you encrypt a document with a Vigenère cipher and then encrypt the result with another Vigenère cipher, then the overall result is the same as if you switch the order of the two encryptions?
Now try it with two Beaufort encryptions. Here the order matters. Why?
The way I like to think about it is by putting the 26 letters evenly spaced on a circle. A Vigenère cipher is like a set of rotations of that circle, but Beaufort is like a set of reflections. Each key letter is a different angle of rotation for the V, and a different axis of reflection for the B.
Now here is something to really blow your mind: That text that was encrypted with two Beauforts can be decrypted with one Vigenère. Figure out what that means.
For the mathematicians in the audience, think dihedral group.
CC20026 is a while away… I really hope we’ll be prepared by then…
So if Beaufort was encrypted with two keys, say k1 and k2, then
C_i= (k2_i – (k1_i – [text_i]))%26 = ((k2_i – k1_i) + [text_i])%26 which is by definition using the Vigenère encryption
So the decryption would just be the single Vigenère key (k2 – k1)%26 ?
I haven’t studied dihedral groups yet, but I do look forward to doing some reading on them!
Hello this is my first year competing is there any mathematical equations or any equation really that you would recommend I learn in advance.
If possible could you also give the cipher they go with I already know how Index Of coincidence works at a basic level btw.I recommend reading our codebreaking guide which you can download on the resources page, and then working through Madness’s book which you can get from the BOSS library page. Beyond that I am sure others will have advice – so I will open this discussion to them! Harry
@Gen_ruikt
The BOSS Training Division page/Resources section has pretty much everything I can recommend as well!
Other than that, you might find it useful to write up some python (or your preferred programming language) scripts to decipher the most frequently appeared challenges by looking at past years archive. Always keep an eye out for modifications/variations to known ciphers in the later challenges. But most importantly, enjoy the process of codebreaking and the story! Hope that helps!
The index of coincidence (IoC) is the probability that a random pair of letters from a ciphertext are the same. Or at least, that’s one use for that statistic: there’s no reason you can’t use it on blocks of 2, 3, 4, 5, … letters. Then it becomes even more useful!
Perhaps you know already that most English plaintexts of reasonable length will have an IoC around 0.067.
The IoC is unchanged by substitution (with a single alphabet) or transposition. This means you can fiddle around with a ciphertext until you see the magic value of around 0.067 (3B was ~0.0648), then you know that that section is a simple substitution (and/or transposition). For example:
A Baconian (5-bit binary) substitution will exhibit an English-like IoC when you take blocks of 5 letters.
A Vigenère cipher (each column of the plaintext encrypted with a separate Caesar shift) exhibits an English-like IoC when you take the letters in each column; when you get the right number of columns the average of the IoC values of each column will be around 0.067 too. You can use this fact to work out the number of columns used from a Vigenère ciphertext.
It’s also useful for the later ciphers, which might include multiple stages of obfuscation: sometimes you know when you’ve got past the first stage because the IoC of the resulting letters (or pairs, triples, etc.) is around 0.067. For instance, 9B from 2019 gives you an IoC of ~0.0662 once you figure out the initial obfuscation.
So IoC is an excellent statistic to get to grips with. Other than that, I’d recommend doing some programming in a spreadsheet or Python, and looking at the library resources on the 2×2 hill cipher, which comes up quite frequently (2021, 2022, 2023 and others). Hill ciphers are probably the most complicated maths that will be used as a cipher in the challenge.
Sorry if this is too much information at once: Madness’ book & the other resources provides a more gradual introduction.
Thanks for the tips I will give a look at the Library and other resources
I was having a look at the hill cipher because of an earlier post and I kind of understand matrix multiplication, does it ever go past 2×2 like 3×2 or something like that for example?
Way beyond any math I have ever done though lolOh yes, it can go way beyond 3×3! Muahaha! (Sorry, it’s passed Halloween now, I need to switch the sound effects to something less spooky, Harry)
Also this is really random I’m making the assumption that most people doing this like puzzles can anybody solve a rubiks cube or any variation id love to hear about some of the harder puzzles that people have solve ( so far its the pentaminx or axis cube for me) and if anybody want I know there are plenty of tutorials online I can try put one in here see if I’m able to explain it(probably not lol)
… shouldn’t be too bad right…right?
also for sound effects November sucks doesn’t it because its to late for spooky sfx but to early for the Christmas onesHow about the gentle rustle of Autumn leaves as they fall to earth? Or are the Elves the only ones who can hear that? Harry
@Gen_ruikt
I’m a big Rubik’s Cube fan! Here’s a cute bit of maths related to it. Every now and then, you see videos claiming that you can solve a Rubik’s Cube from any position by just repeating a certain sequence of moves over and over again. (For example, they might claim that if you move the right side up and then the upper side left and keep repeating this, you’ll eventually solve the cube.) Now, anyone who has solved a Rubik’s Cube knows intuitively that this is nonsense, but how can we prove it?
We note that the transformations of the Rubik’s Cube form a group (if you aren’t familiar with this concept, then I’m sure everyone will be happy to recommend their favourite algebra texts). In the language of group theory, the nonsense claim is simply that the Rubik’s Cube group is cyclic. So we just need to figure out if our group is cyclic.
One important property of cyclic groups is that they are abelian (the order in which you perform transformations doesn’t matter), but for the Rubik’s Cube, order does matter. To see this, compare moving the right side, then the top side, with moving the top side, then the right side. Thus, the Rubik’s cube group can’t be cyclic. If someone tries to tell you otherwise, you can chuckle at the confused person who believes in non-abelian cyclic groups.
You might be thinking that the above argument only works if we consider the sequence of moves as an indivisible unit that we somehow apply simultaneously (and you would be correct). If we allow the person performing the repeated action to stop mid-sequence, then such an “always solving sequence” can be found. Can you come up with one such sequence? [Hint: the easiest one to find is really trivial, but you won’t be able to write it down explicitly.]
In terms of puzzles, I mostly stick to the 3Ă—3Ă—3, but I’m a big fan of blind-solving. I find the techniques used for blind solving are much more mathematically interesting than the speed-solving methods. If you don’t yet know how to solve a Rubik’s Cube blindfolded, it may be a fun project to try to come up with your own method for doing so (some familiarity with algebra may help). However this is a hard problem and I certainly wasn’t mathematically mature enough to solve it when I first learnt blind-solving.
What is cc20026?
Harry in the first post of this forum what do you mean by being prepared for cc20026 and cc 2027 what are those is it cipher challenge 2026 and 27
Yes! There is of course a typo in cc20026! Harry
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