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Synthetically, the advantages of the Montgomery ladder are that it is simple and fast. If you look at X25519, the Diffie-Hellman algorithm applied to Curve25519 and described in RFC 7748, you will see that for an n-bit Montgomery curve, multiplying a point with an n-bit scalar, you will need to compute about 10n multiplications of field elements. In more ...

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In 1985, Montgomery introduced a new clever way to represent the numbers $\mathbb{Z}/n \mathbb{Z}$ such that arithmetic, especially the modular multiplications become easier. Peter L. Montgomery; Modular multiplication without trial division ,1985 We need the modulus $n$ we are working and an integer $r$ such that $\gcd(r,n) =1$ and $r>n$ Definition: ...

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You have 5 limbs because it is based on DJB's papers and as the Ed25519 paper mentions, it's using a $2^{51}$ radix representation for performance reasons. It does so in order to avoid carries when performing field multiplication. Because otherwise the carry bits need to be handled by subsequent additions, which slows down even modern CPUs. So according to ...

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Montgomery multiplication Theorem (Montgomery, 1985). For any odd integer $N$ and any integer $0 \le T < N2^k$, one has: $$T 2^{-k} \equiv \frac{T + UN}{2^k} \pmod N$$ where $U = T N' \bmod 2^k$ and $N' = -N^{-1} \bmod 2^k$. Further, one has: $$\begin{cases} T+UN \propto 2^k\\ 0\le \frac{T + UN}{2^k} < 2N\end{cases}$$ The last 2 properties show that $... 4 What a coincidence, I implemented this attack yesterday! I'm executing it right now and I can tell you that the difference at each step is around 1 reduction (as the paper suggests). See for example my logs for bits 6 to 8 of a 96-bit key: bit 6: avg M2 (#111)= 1.2072072072072073, avg M1 (#99889)= 0.1684369650311846 [1, 1, 1, 1, 1, 1, 1] bit 7: avg M2 (#... 3 The usual motivation for using Montgomery multiplication is that it significantly reduces the cost of modular reduction by changing the representation of elements. In the Montgomery ring the polynomial$A(x)$is instead represented by$a(x)=A(x)r(x)\pmod{n(x)}$where$r(x)$is a fixed element of the same degree as$n(x)$, but for which division is very ... 3 I believe that the meaning is "compute the$2n$-bit value$a[j]*b + C$, and then assign the top$n$bits to$C$and the bottom$n$bits into$S$2 Theorem 2 in [Dussé et al. 1991] states that, if we skip the final subtraction, then, for$N < R / 4$and$0 \leq A, B < 2 N$, we have$0 \leq C = \text{MonMul}(A B) < 2 N$, while keeping$C \equiv A B R^{-1} \pmod N$. I think the condition$N < R / 4$inherently holds in your case e.g. you are using larger$R$value acutually. 2 Brett just asked & answered this question : Confused about final subtraction of modulus in Montgomery Multiplication, during modular exponentiation You should increase$R$exponent by$2$. If you use$n=1024$, increase it to be $$n=1024 + 2 = 1026.$$ Recalculate the pre-compute$R'$, based on the new exponent. $$R' = 2^{(2\cdot 1026)} \bmod(M).$$ 2 Write$(x_i, y_i) = (X_i : Y_i : Z_i)$, so that$x_i = X_i/Z_i$and$y_i = X_i/Z_i$, where$Z_i \ne 0$is arbitrary. (If you are not familiar with projective coordinates or you like visuals, see an illustration of projective coordinates on a real elliptic curve.) Let's take the doubling formula for example, where$(x_3, y_3) = (x_1, y_1) + (x_1, y_1) = (...

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Where to apply Montgomery Multiplication in $GF(2^n)$? This answer really depends on how you constructed the binary extension field $GF(2^n)$. If the irreducible polynomial is trinomial or pentanomial then the reduction is already efficient! Modular Multiplication with trinomials Let $GF(2^n)$ is constructed with the trinomial $P(x)=x^m+x^n+1$ where $\alpha$...

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The issue is, the message length is now longer than the modulus $p$ and $q$ That's not true. In the $p$ track, you are raising $M \bmod p$ to the power $d \bmod p-1$; we have $M \bmod p < p$; that is, the value we are exponentiating is less that $p$, as results by Montomery multiplication. In the same way, the $q$ track also satisfies the requirements ...

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The confusion comes from the choice of representation. I'd a quick look to the referenced paper, where the autors use a 2-radix representation. Then you shoud initialise $e=\frac{m+15}{w}$ instead of $e=\frac{m+1}{w}$ as you use a 16 bit adder! The best is to read again the seminal paper of P.L. Montgomery:http://web.itu.edu.tr/~orssi/dersler/cryptography/...

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Elliptic curves can be represented in different form. The most basic equation, in which every elliptic curve can be represented is the Weierstraß equation: $$y^2 = x^3 + ax + b$$ For a Montgomery curve it must be able to be represented in the following form: $$(b)y^2 = x^3 + ax^2 + x$$ For every Montgomery curve it is possible to transform its equation to ...

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The key to Montgomery Multiplication is the prescaling of one or two of the inputs by $r^k$ as stated in this answer by kelalaka to How Does Montgomery Reduction Work. In the case of polynomial evaluation of $$p = a_0 + a_1b^1 + a_2b^2 + ... + a_mb^m \mod n$$ by Horner's scheme: $$p = (((a_m * B) + a_{m-1}) * B + ...) * B + a_0$$ $$= (((a_m b ) + a_{m-1}) b ... 1 Maybe one day someone finds this Post again and has the same questions. To you: I hope you're having a great day! Questions: The projective Arithmetic is faster, because there are only multiplications, squarings, additions to do. The affine Artihmetic is slower, because it takes a lot of time to compute the division. Especially for the big numbers used in ... 1 Actually, the answer is none. Everything works, and produces the correct results. Potential problems arise when converting back to Twisted Edwards space, and that's a different question. 1 I figured this out, the best approach is to use either Montgomery or Barrets algorithm for modulo reduction, Barrets requires a slightly higher bit multiplication but not pre-transformation, while Montgomery has a pre-transformation but multiplications are slightly faster. And if you use Montgomery you can do the entire point double/add in Montgomery form ... 1 The additive identity is 0, as usual. The multiplicative identity is the Montgomery representative for 1, namely 1\cdot R \bmod N = R \bmod N, just like the Montgomery representative for any other element x is x\cdot R \bmod N. 1 First of all, The Montgomery's Reduction algorithm requires that \operatorname{GCD}(n,R)=1 . This requirement is satisfied iff n is odd. The R is chosen as 2^l where  2^{l-1} \leq n < 2^{l}. For x < n the n-residue with respect to R is defined as;$$ x' = x \cdot r \bmod n$$than the set$$\{i \cdot R \bmod n\;|\; 0 \leq i \leq n-...

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The problem was that there was one additional thing I left out....if I'm increasing the loop iteration count and the number of bits from 2048 to 2050, then I had to choose a new R to satisfy the preconditions of Walter's optimization. $R=2^{bitwidth}$ so I needed to have $R=2^{2050}$ instead of $R=2^{2048}$. I'd now like to modify this to use multiple ...

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The above mentioned work is focused on a hardware implementation (I have this work as a PDF). I'd suggest you to search for: Colin D. Walter. Montgomery Exponentiation Needs no Final Subtractions. Electronics Letters, 35(21):1831{1832, October 1999. Colin D. Walter. Montgomery's Multiplication Technique: How to Make It Smaller and Faster. In C etin K. Koc ...

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Looks like misunderstanding to me: you are asking about Montgomery reduction but your example is about Montgomery multiplication. Montgomery reduction inputs a number in range $[0..NR-1]$ and outputs a number in range $[0..N-1]$, where $R$ - auxiliary Montgomery modulus. Montgomery multiplication inputs two numbers $A$ and $B$ in range $[0..N-1]$ and ...

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Consult the transcript from the class, there is an example he works through which is very similar to this problem. Fundamentally you're trying to solve the problem c = (T + T(-N^-1) (mod R)N)/R (mod N). I suggest creating an equation in Excel (or other tool of your choice). Test your equation using the inputs from the example problem (EXAMPLE INPUTS: T=69, N=...

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I've already replied to the question posted some days before. Montgomery multiplication is another way to perform modular multiplication in the residue system representation. The operation induced is in fact a group morphism. In $GF(p)$, $p$ prime, or in the multiplicative group $\mathbb{Z}/n\mathbb{Z}$, the transformation allows to perform modular ...

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@haster8558 I know this is late, but I am struggling with the same problem. I do have a few answers for you though... I dont understand why I the number of the word are linked to the width of the adders They aren't. I believe Robert NACIRI is wrong on this one. The number of words is only linked to the number of DIGITS (bits if you are in base 2, decimal ...

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