107

He can't. The commonly-accepted answer was provided by Marius, and looks good on the surface. But consider that Kleptomaniacs are smart. Klep Kleppington III -a particularly wily kleptomaniac- could intercept the package and place his own padlock on it and send it back to Mike under the pretense that it is Jane's padlock. Mike, not foreseeing this kink in ...


95

Solution:


42

Scroll to the end if you just want the answer. I give a detailed explanation of how I reached it. First, I hypothesized that the permutation has two cycles: one of length 9, and one length 17. The length-17 one contains T and R, and the length-9 one contains E. How did I get there? Look at the string JR. at the start of the second paragraph. That's a ...


35

Let's say S1 has the list {a,b} and say S2 has {a,c}. S1: {a,b}, {c,d}, {e,f}, {g,h} - one of these is my list. S2: {a,c}, {b,d}, {e,g}, {f,h} - one of these is mine. Now both S1,S2 know their sets are inside {a,b,c,d}, whereas because of symmetry, it could be {e,f,g,h} or {a,b,c,d} as far as eavesdroppers are concerned. So S1 and S2 play a game ...


34

I think it's pretty obvious. ;) Alternatively,


34

This is just a visual representation of @Marius answer ;)


23

Why, her name was Explanation: First, what everyone has deduced so far: Then, How do you like them apples?


22

This procedure is called the Burrows-Wheeler transform. It is useful for data compression, because if the starting text contains many identical substrings ("the", "and", ...), then the Burrows-Wheeler transformation will contain many runs of identical characters. Applying the inverse algorithm on that Wikipedia page, with a suitable guess of the location of ...


21

The title means We can


21

The paper says: How the deciphering works: Now my best explanation of the joke is Here is the paper together with the plaintext:


20

We don't need to brute-force $256^n$ possibilities – since the overly large number is written in base $1253$ (that's what the polynomial is for) and ASCII characters are at most $128$, we can just express the large number in base $1253$ and extract the digits exactly as is. Wolfram Alpha gives a decomposition of 112:48:108:121:110:111:109:49:64:108:...


19

First, I copy-pasted the numbers into Mathematica: str = "181 ... 29"; Next I convert the string to a list of numbers: numbers = FromDigits /@ StringSplit[str]; Now, is this a decimal ASCII string? FromCharacterCode[Take[numbers, 10]] (* "µæ\.1dµæ\.1dµæ\.1dµ" *) Evidently not. Next I look at the letter frequencies: Histogram[numbers, {1}] It looks ...


19

It was a pain to solve this, but fun. You can test this with the last diary picture: Now for the real puzzle: Applying the decode algorithm with the correct key gives you:


18

The answer is: Which is a reference to one of my favourite movies: The first step was to decrypt the text at the bottom of the note using This yields: Next find all of the clues in order. The cat clue: "envyf tb mvtmnt" decrypts to "rails go zigzag". The other three clues are bin: ksadeplpzd, gel: ieassruoke, and fur: pqyqfnssaf. Every other clue in ...


18

The decoded message ... First, let's transcribe the card layout, so that we can tackle it more easily: Frequency analysis tells us ... Let's make some assumptions: Let's rearrange the input according to these assumptions: Let's try to find some structure. But ... So, here goes:


17

Solution Key Countries of the participants Reverse-Engineered Ciphering Process Now let's apply this to the original messages. Reverse-Engineered Deciphering Process Reasoning One Last Thing To Do: Name of the Cipher Notes


17

Let $+$ and $-$ denote the "standard" arithmetic operations modulo 26 on the English alphabet, done letterwise to strings. Here, encrypting text $T$ with a given one-time pad $P$ is simply calculating $T+P$ to get the ciphertext $X$, and decrypting reverses the process: $T=X-P$. Note that, since $+$ is "made up of" regular mod-26 addition, it is ...


15

(No spoiler tags because they're messing with the line breaks...) First we do a trivial reshaping of the image to un-distort it: img = Import["http://i.stack.imgur.com/fIAv4.png"] rectified = Image[ArrayReshape[ImageData[img, Automatic], {400, 399, 3}], "Byte"] After noticing that the black pixels are not all exactly black, I extracted the lowest bit from ...


15

The "P.S." hints that To use that hint, we must


15

Thanks to the others here, I managed to figure it out. Sorry about the terrible formatting, I'm pretty new at this. Solution: Explanation: Credits:


15

I am happy and I know it. The message reads: How is it encoded? The whole story:


15

Any commutative hash function will do.  Using RSA makes this relatively easy, I think. So Alice and Beth both establish their secret primes, and, in a twist, keep everything secret.  $ % Make EA, EB, DA and DB look like functions; i.e., *not* italic: \def\EA{\operatorname{EA}} \def\EB{\operatorname{EB}} \def\DA{\operatorname{DA}} \def\DB{\operatorname{DB}} $...


13

It's going to be a long phone call. I haven't taken computer security in a while so I hope I got the process right, I was just kinda going off the top of my head.


13

The final answer is: The clue "listen" pertains to audio files embedded in the images. The legend for the blue layer is four cards. The file is an OGG Opus file. The audio is someone saying "If you haven't found any clues yet, here's what you should be looking for" followed by what sounds like a dial up modem (yes I'm that old). The blue layer ...


13

I think the checkmark should actually go to Lord of dark for actually finding the answer to the original question, but here's the missing bit of the explanation: (Which makes the "no-computers" tag a bit ironic.)


12

Updated after the question's correction I think the next valid password is Reasoning: Here is a possible recovery of most of the missing and mistyped characters: I also added if the password seem to pass or fail (based on the difference between the current and the next timestamp: an attempt is considered fail if there was another attempt right after it) ...


12



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I believe the word you're looking for is OK, so how did I figure that out? Well, let's start with the obvious outer layer first: Now, since this is a crypto puzzle, and since your little verse happens to mention Of course, we need to know the key to have any hope of decoding that, but This still isn't exactly readable, but it's also definitely not the ...


12

The answer is: How the transformation function works: Quite challenging, but very nice puzzle! Previous answer below: Alright, nothing close to a solution yet, but I'll just post some of my observations here and maybe it helps someone else solve it:


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