# Tag Info

35

What methods would they use? Since WW2, we know the security of Enigma machines was weakened by the reflector, resulting in two problems: No difference between en- and decryption, which means that if K ↦ T, then T ↦ K. No letter can be encrypted by itself because electricity can not travel the same way back, which results in a reduction of encryption ...

18

The example is using a shorthand notation for the rotors that somewhat obscures the way they actually work. For example, the first rotor in your example, BDFHJLCPRTXVZNYEIWGAKMUSQO, actually applies the following permutation of the alphabet: ABCDEFGHIJKLMNOPQRSTUVWXYZ ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓ BDFHJLCPRTXVZNYEIWGAKMUSQO Applying this rotor in the reverse ...

18

Are all encryption algorithms with fixed-point free permutations inherently flawed? Yes - when fixed points, or the lack of them, is knowable and detectable. This is a violation of multiple modern semantic security definitions. For example, this means that plaintexts with repeating symbols are distinguishable with high probability from plaintexts that ...

17

Depends on the exact model. Wikipedia is your friend: "Combining three rotors from a set of five, the rotor settings with 26 positions, and the plugboard with ten pairs of letters connected, the military Enigma has 158,962,555,217,826,360,000 (nearly 159 quintillion) different settings." More in detail: if you consider an Enigma with 3 rotors out of 5 ...

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Are all encryption algorithms with fixed-point free permutations inherently flawed? No, they are not inherently flawed. Consider the following cipher: Let $k_0$ be a key for AES-256, and let $k_1$ be a key for a hypothetical Advanced Derangement Standard, ADS-256. To encrypt the $n^{\mathit{th}}$ message $m$, the ciphertext is $$c = m \oplus \bigl[\... 12 Since this is an historical question, I am going to digress and make some historical corrections. In science, we give credit for important inventions to the people who published. If it turns out that someone else invented it earlier and didn't publish, they don't get credit. Obviously, they should be mentioned in passing or a footnote in the interests of ... 11 No, padding would make the message much easier to crack. This is a great example of why cryptography is left to the professionals (I am not a professional cryptographer, I'm not even a very good amateur one). Amateurs tend to just make things worse. First problem is the Enigma had no way to produce a "null". It was only capable of producing letters. The ... 9 After a year I have managed to find a suitable solution to the problem. My understanding is based on the following picture that I created, based on a simplified version that I found online (at present I cannot find the reference, if I do I will edit it in). It is supposed to be a bombe and can be understood in two parts. The top contains 26 columns, one ... 8 I didn't want to delete this question, but it seems like after pondering this for a week I finally understand right when I seek help online. When the ring setting and the rotor position all increase by the same amount, they cancel and so the ring pretty much stays the same. Although the contact points of the ring to the alphabet ring is different, it makes ... 7 The answer depends on the Enigma model, on the number of rotors among which the active rotors are chosen, on the number of wires used for the reflector, and on what one accounts for as part of a setting. The discrepancy between the two numbers around is because the position of the rotors (except the left one, which notch is inactive) can be accounted for - ... 7 There is a full breakdown of key size on a website talking about the “Technical Details of the Enigma Machine”. To sum up: If all rotor combinations are included then you have a possible 3*10^{114} possible keys. However, that didn't happen (the operators would need to keep \frac{26!}{26}=1.5*10^{25} rotors at hand if they didn't allow repeats). By ... 7 Having had the privilege of using one of the original machines, I can tell you the sequence is the following: set the plugs, and set the rotors. Once you push the key, you can feel the force of the rotors turn, and the light turns on for the encrypted letter. The light turn off when you take pressure off the key. This was on a 3-wheel Enigma, and I do ... 7 Block ciphers operators from \{0,1\}^n \to \{0,1\}^n. Each key selects one permutation among all possible permutations n! and this is very small one if you compare 2^{128} to 128^{128} For block ciphers like AES; One of them is identity, which key selects, we don't know. We don't know even it is selectable by one of the key spaces 2^{128},2^{196},... 6 This example is correct. The inversed versions are the inverse permutation; that is, if the forward direction is the permutation P, then the inverse permutation P^{-1} has the property that P^{-1}(P(X)) = X for all X. That is, if X is a plaintext letter, and we run it through in the forward direction (giving us P(X)), and then run it through in ... 6 No, there was no way on the Enigma machine to change the behavior of the rotor rotation. Rotation on the Enigma was a fixed mechanical property of the rotors like the gears in a clock. However, other cryptographers noted this vulnerability in the Enigma and improved it in follow on rotor machines particularly the British Typex and the American SIGABA aka ... 5 There's two missing pieces. First, the ring setting changes the output letter, it doesn't rotate the whole exit pattern. Second, the rotors are advanced before the letter is encrypted. If your rotor (Enigma I Rotor I) is set up like this with the ring at A. abcdefghijklmnopqrstuvwxyz ekmflgdqvzntowyhxuspaibrcj Then if you advance the ring to B all the ... 5 This doesn't answer your question, but you may find clues in two wartime papers of Alan Turing which were formerly classified were declassified and published in the UK National Archives in 2015, and are now available on the arXiv: Alan M. Turing, ‘The Applications of Probability to Cryptography’, UK National Archives HW 25/37, 1942, arXiv:1505.04714. ... 5 If I understand correctly, you want a function that for each input string p assigns a permutation over an alphabet L. If the number of elements in L is small enough, the permutation set P(L) will be enumerable. More precisely, |P(L)| = |L|!. There exists a surjective function f:\{0,1\}^k \to P(L) that for each bit string s of length k ... 5 There was at least one late Enigma derivative, the Russian M-125 Fialka, that did modify the reflector to allow a letter to encrypt to itself. Curiously, rather than simply making the cipher alphabet size an odd number (which would've been natural enough for Russian, with its 33-letter Cyrillic alphabet), the designers of the Fialka instead used a rather ... 5 There are three nice answers here, each supported with well thought out arguments. It seems to me that not only is it difficult to distinguish between fixed point free and non fixed point free encryption mappings, as shown by @SqueamishOssifrage's answer, it is not the pure encryption mapping E(k,x) that is used in many situations but offsets like$$E(k,x+...

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I suspect it was a semi-deliberate feature. That is, while it probably wasn't a design goal in and of itself, it neatly solved a mechanical issue that would otherwise have required a more complicated and failure-prone solution. What was the issue? Simply, it was making the third wheel only advance one step at a time, rather than 26 steps in a row. That's ...

4

It looks like with no leakage or errors, Enigma is still secure. Quoting the Enigma@Home project website: Enigma@Home is a wrapper between BOINC and Stefan Krah's M4 Project. “The M4 Project is an effort to break 3 original Enigma messages with the help of distributed computing. The signals were intercepted in the North Atlantic in 1942 and are believed ...

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On page 5 of the document you linked in the comment it says the following in section 22: Satzzeichen Es werden ausgedrückt: Punkt durch x, Doppelpunkt durch xx, Fragezeichen durch ud, Komma durch y, Trennungsstrich, Bruchstrich, Bindestrich durch yy, Klammer durch kk. Satzzeichen sind im allgemeinen nicht entbehrlich, Schlusspunkt ist nicht ...

4

out of 325 pairs to choose 10 pairs, there are 325C10 possibilities to connect pairs with 10 wires. This accounts for the fact that we can not choose a pair that has both letters identical to an earlier pair, but misses that we can not choose a pair that has any letter common with an earlier pair. A correct line of thinking is: imagine that the 20 ends of ...

3

My amateur understanding, osmosed from everything I've avidly read, is that Alan Turing independently discovered and applied Bayes theorem to the cracking of the M4 Enigma. Bayes theorem could be described as 'inference'. Turing didn't have a direct hand in attacking Fish nor the building of Colossus but his probabilistic approach was adopted by the ...

3

I guess you already figured it out but my guess is that you forgot to move the first rotor 1 step before starting the encryption. That also impacts what letter it arrives to on next rotor and so on. The complete path is KBD-A>SB>A - A>ETW>A - B>(1)R3>K - J>(2)R2>B - B>(3)R1>D - D>UKW>H - H>(3)R1>D - D>(2)R2&...

3

Ok, here we go. Let's use Rotor I as an example. The wiring for this rotor looks like this: EKMFLGDQVZNTOWYHXUSPAIBRCJ On every rotor, there originally was a dot, for more detail, read K ROSSER's answer. We need to know the position of this dot. It's always where the ring setting letter is (So, if the ring setting is A, it's where the A is). Since all the ...

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The easiest way to understand the path of the current through the rotors is to make up six tables for the three wheels, three going forward and three going backwards plus a table for the reflector. The simulator was used with rotors III, II, I and B reflector. The simulator advances the rotor when the first key is pressed, i.e., the first entry. To ...

3

How secure is this cipher? At first glance, not very. It would appear to be vulnerable to a ciphertext-only attack, for example, the attacker can recover the plaintext given a ciphertext of about 10k (actually, he probably can deal with less), even assuming that all the attacker initially knows is that the plaintext is "ASCII English", and he has no ...

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A great page with everything Enigma is Frode Weierud's CryptoCellar: http://cryptocellar.web.cern.ch/cryptocellar/Enigma/index.html The main topic headings from the page: Enigma Publications Historical Documents Cryptanalytical Documents The Enigma Series Decoding Projects Patents and Manuals General Information Enigma Messages and Keys Enigma ...

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