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By Brent R.P., Zimmermann P.

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Example text

3. 8). 2 Algorithm REDC is correct. 3 1 2 3 4 5 6 7 8 57 ☛ ✞ ✟ ☎ ✝ ✡ ✆ ✠ Algorithm REDC . Input : 0 ≤ C < β 2n , N < β n , µ ← −N −1 mod β , (β, N ) = 1 Output : 0 ≤ R < β n such t h a t R = Cβ −n mod N fo r i from 0 to n − 1 do qi ← µci mod β C ← C + qi N β i R ← Cβ −n i f R ≥ β n then r e t u r n R − N e l s e r e t u r n R . 2: Montgomery multiplication (quadratic non-interleaved version). , the ci are defined by C = 2n−1 ci β i . 0 Proof. We first prove that R ≡ Cβ −n mod N : C is only modified in line 6, which does not change C mod N , thus at line 7 we have R ≡ Cβ −n mod N , which remains true in the last line.

Input : A = 0n−1 ai β i , 0 ≤ c < β . Output : Q = 0n−1 qi β i and 0 ≤ b < c such t h a t A + bβ n = cQ d ← 1/c mod β b←0 fo r i from 0 to n − 1 do i f b ≤ ai then (x, b′ ) ← (ai − b, 0) e l s e (x, b′ ) ← (ai − b + β, 1) qi ← dx mod β b′′ ← qi c−x β b ← b′ + b′′ n−1 Return qi β i , b . 0 now that Ai−1 + bβ i = cQi−1 holds for 1 ≤ i < n. We have ai − b + b′ β = x, then x + b′′ β = cqi , thus Ai + (b′ + b′′ )β i+1 = Ai−1 + β i (ai + b′ β + b′′ β) = cQi−1 − bβ i + β i (x + b − b′ β + b′ β + b′′ β) = cQi−1 + β i (x + b′′ β) = cQi .

Then step 5 gets c = ℓ and step 6 computes d = ℓ − 2h mod (2n + 1). Now γ = ℓ/N mod (2n − 1), thus γN = ℓ + (2n − 1)τ for τ ≥ 0, which gives τ ≡ −ℓ/(2n − 1) mod N . This yields δ = ℓ − 2τ , and finally (δ − d)/2 = h− τ . The cost of the algorithm is mainly that of the four convolutions AB mod n (2 ±1), cµ mod (2n −1) and γN mod (2n +1), which cost 24 M (n) altogether. However in cµ mod (2n − 1) and γN mod (2n + 1), the operands µ and N are invariant, therefore their Fourier transform can be precomputed, which saves 31 M (n).

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