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With a 64-bit known polynomial, future output of an LFSR can be trivially predicted from the last 64 bits output. Even if the 64-bit polynomial is unknown, the last 128 bits are enough, using the Berlekamp–Massey algorithm. Thus indeed, the LFSR-based PRNG in the hardware described in a section 27 of the document linked to in question, with some additional ...

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The Berlekamp-Massey algorithm is an iterative method for finding the shortest LFSR that can generate a given sequence of bits. The given sequence might or might not be generated by an LFSR: the Berlekamp-Massey algorithm does not care. It just finds the shortest LFSR that can generate the given sequence, and if the sequence has been generated by an LFSR of ...

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If there was no non-linearity, then every bit of keystream output would be a (known) linear function of the unknown key bits. Consequently, in a known-plaintext attack scenario, each bit of known keystream output would allow us to write a linear equation on the unknown key bits. If we have a 128-bit key, there are 128 boolean unknowns (variables), so once ...

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If the initial state is $b_0,b_1,\dots,b_{k-1}$ and the recurrence relation is $b_k = \sum_{i = 0}^{k-1} a_ib_i$, then in linear-algebraic terms we have $$\begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_k \end{pmatrix} = U \begin{pmatrix} b_0 \\ b_1 \\ \vdots \\ b_{k-1} \end{pmatrix},$$ and more generally $$\begin{pmatrix} b_{n} \\ b_{n+1} \\ \vdots \\ b_{n+k-1}... 10 The obvious way to do this is to generate N words, and use logical operations to combine them in a single word such that each bit of the output word is a 1 with probability approximately 0.1 (and the individual bits are uncorrelated). In the simplest case, you could generate 3 words, and just AND them together into a single one. In C, this would be: ... 9 The variable of a polynomial is traditionally noted x, not X; and, when dealing with LFSRs, the polynomial is seldom considered a function. Thus I'll rewrite the polynomial as P(x)=x^4+x^3+1, a polynomial of degree n=4. This is a polynomial with coefficients in the field \operatorname{GF}(2), also noted \mathbb Z_2 or (\{0,1\},+,\cdot) [note: ... 9 For text known to be ASCII encoded as octets with high-order bit of octets at zero, that reveals one bit of the output of the LFSR out of 8. It allows finding the original state of the LFSR from (say) the first 51 octets of the ciphertext by merely solving a system of linear boolean equations; then decipher the rest. There is no guesswork involved. For 7-... 8 Bluetooth has gone through a few transitions in protocol. Bluetooth BR/EDR uses E0/SAFER for a cipher and Bluetooth LE uses AES-CCM. Fundamentally, when you write a specification as an engineer, you are looking for a minimally viable CMOS implementation. You make choices that are not always the best from a cryptographic standpoint because you have space ... 7 The fundamental property that you will use is that the output of an A-bit LSFR (using the A-bit feedback polynomial P) xored with the output of a B-bit LSFR (using the B-bit polynomial Q) is the same as the A+B bit LSFR with the feedback polynomial P \cdot Q. And, don't forget, when you multiply the feedback polynomials, you use polynomial ... 7 Sure enough, if you use xor with the words, what you do have is the bitwise LFSR repeated 32 times in parallel, each in a different state, but each one having a 2^{16}-1 period, so the period of your construction should be 2^{16}-1 as well. If LFSR period wasn't exactly 2^{16}-1 this will be another history. 6 This is basically the problem of finding a generator g in a group G of order 2^n-1, where the factorization of 2^n-1 is unknown. Here is a pragmatic solution that should work fine in practice. Use ECM factoring to find all small factors of 2^n-1; in this way, find as many prime factors of 2^n-1 as you can. Now, pick g uniformly at random. ... 6 Well, one way to look at this is to notice that, if the feedback polynomial is prime, then the result of starting with state E, and then stepping the LFSR forward N steps is the value 2^N \cdot E, where we do the computation in GF(2^K), using the polynomial representation (with the feedback polynomial being the polynomial). In addition, I believe ... 6 There were popular once upon a time; They were very easy to implement in the hardware. Think about you only need latch and \oplus. For combining functions some \wedge operation. Their periodic properties are well studied, the minimal polynomial and characteristic polynomial etc. You can see a glimpse in the classical book of Solomon Golomb. They are ... 5 Berlekamp-Massey is designed for the situation where you have observed 2n consecutive output bits from a n-bit LFSR. It doesn't work if the observed bits are scattered randomly, at random non-contiguous offsets in the stream. Information-theoretically, a minimum of 2n bits of output are needed to reconstruct the LFSR. Intuitively, this is because ... 5 I have argued so 15 years ago, and not been proven wrong since. Basically, A5/1, with a n-bit state, offers a resistance of roughly 2n/3 bits of security. With n = 64, the resistance is very low, thus amenable to not only direct breaking, but also all kinds of trade-offs. All the attacks published so far are dances around that resistance level of ... 5 Galois LFSR In a Galois LFSR with polynomial P of degree n, the state is a vector of n bits assimilated to the binary coefficients of terms of degree n-1 to 0 of a polynomial; we note both the state and its polynomial S_j when at step j\ge0 . The next state is computed as S_{j+1}=S_j\,x\bmod P. Equivalently, the state is shifted towards the ... 5 Consider a Galois LFSR with feedback polynomial x^3 + x + 1 and initialization 1 + 0x + 0x^2, that is, the shift register shifts its initial content (1,0,0) rightwards, and the output bit (the rightmost bit) is fed back into the shift register into the 1 and the x position. The initial loading of the LFSR is the Galois field element 1, and ... 5 First of all, LFSR's are not secure when they are used alone. They have good statistical properties and one can calculate their period given the feedback polynomial. The LFSR is maximal-length if and only if the corresponding feedback polynomial is primitive. In this case, it visit all states except the all-zero state if it started with a non-zero sate. To ... 5 The number of consecutive plaintext-ciphertext pairs (x_i,y_i) of bits necessary is 256 if we know the LFSR taps (equivalently: the reduction polynomial) in advance, for the reason explained in the question. 512 is when we do not, because it creates more unknowns: the coefficients of the polynomial, in addition to the initial state. If we know the LFSR ... 4 There are no issues with negative indices, even the first (or zero-th depending on how you want to count them) iteration. The quantity d is called the discrepancy. During the N-th iteration, d is the difference between s_N, the N-th bit of the given sequence for which you are finding the LFSR, and the bit computed by the LFSR that you have ... 4 Background: An infinite sequence c_0,c_1,c_2,\dots is generated by a (linear feedback shift register (LFSR) with) polynomial f(x) = \sum_{i=0}^n f_i x^i if for any j,$$\sum_{i=0}^n c_{j+i} f_{n-i} = 0 \text.$$We can consider the sequence as a power series \sum_{i=0}^{\infty} c_i x^i. If we multiply this power series with the polynomial f(x), we ... 4 It is very easy to predict and should not be used as a main cryptographic component on its own. Firstly, many k-decimations s_{kt} of LFSR sequences (those with gcd(k,2^n-1)=1) are shifted LFSR sequences themselves so this can be used for prediction. If you can find a template T=\{0,t_1,\ldots,t_{n-1}\} (with 0<t_1<t_2<\cdots<t_{n-1} ... 4 A binary sequence of period 105 will have characteristic polynomial that is a divisor of x^{105}-1. Since x^{105}-1 has irreducible factors of degrees 12, 4, 3, and 1, it is possible to have an LFSR of length 4+3=7 generate a sequence of period 105, and that is exactly what has happened here. Note that$$x^7+x^6+x^5+x+1 = (x^3+x+1)(x^4+x^3+1)$$... 4 The polynomial factorisation of X^{2^L-1}+1 into irreducible factors gives you all the polynomials g_i(X) that can be used as LFSR polynomials in generating any sequence of period 2^L-1. Say your goal is to generate a sequence of this period with linear complexity c. Assume$$(X^{2^L-1}+1)=\prod_{i=1}^v g_i(X), there will be no repeated factors. ...

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You made a mistake in the propagation of the feedback. If you follow the LFSR very slowly this is what you get: initial state: +--------------------------------------------------------------+ | | | | | | +-----+ +-----+ | +-----+ | +-----+ | +-----+ | | | | | | v | | ...

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In this answer I will consider the Galois LFSR mentionned at this question: Sequence output by a Galois type LFSR see the image below. First we assume that the 5 positions of bits are numbered from left to right: 0 .. 4 The Galois representation is as follows: +--------------------------------------------------------------+ | | ...

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First up: a 5 bit LFSR is horribly small, so getting the initial seed which was used is a piece of cake. But indeed, it’ll basically boil down to a (rather quick) brute-force job where – unless that file contained purely random data – you’ld have good chances detecting the initial state (resulting in successful decryption) by just looking at the first ...

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The matrix representation seems preferable in your case because You indicate that you already know how to solve your problem in this form (by matrix exponentiation) You said you are not concerned about performance The matrix representation is preferable in general becase You can apply the tools of linear algebra to understand it LFSR representation (...

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TL;DR: If you have a similar question, then use the online Berlakamp-Massey algorithm to get the minimal polynomial by providing the given keystream, then get the C++ code from github and then replace your tap positions, seed, and ciphertext and then compile and run. The answer is attraction How to get it What is given: We have given the following ...

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why are LFSRs represented as polynomials? Because it gives useful insights into the behavior of the LFSR. If the feedback terms are represented by the polynomial $P$, and if the initial state of the LFSR is represented by the polynomial $Q$, then the state of the LFSR after $n$ steps will be the polynomial $x^n Q \bmod P$ (where the calculation is done on ...

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