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[Solved]: Understanding Intel's algorithm for reducing a polynomial modulo an irreducible polynomial

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I'm reading this Intel white paper on carry-less multiplication. It describes multiplication of polynomials in $\text{GF}(2^n)$. On a high level, this is performed in two steps: (1) multiplication of polynomials over $\text{GF}(2)$, and (2) reducing the result modulo an irreducible polynomial. We use the "standard" bitstring representation of polynomials, i.e. $x^3+x+1 = [1011]$.

The paper gives an algorithm for calculation of the remainder polynomial on page 16 in Algorithm 3. However, I'm having trouble understanding the reduction algorithm on pages 16-17 (Algorithm 4). Essentially, I think we need Algorithm 4 for larger fields when our or partial results don't fit 128 bits anymore. They give an example for multiplication of two polynomials in $\text{GF}(2^{128})$.

Where do the "magic constants" 63, 62, and 57 for right shifts, and the "magic constants" 1, 2, and 7 for left shifts come from?

For example, how does one generalize the algorithm for smaller fields, say $\text{GF}(2^{32})$? Would the corresponding shift values then be 15, 14, 9 and 1, 2, 7?

In the final step 4, the algorithm tells you to "XOR $[E_1:E_0]$, $[F_1:F_0]$, and $[G_1:G_0]$ with each other and $[X_3:D]$".

Why do we do this? As far as I can see, the result of this XOR operation is neither stored anywhere nor used anywhere. Is it somehow used for computing $[H_1 : H_0]$?

Asked By : Gideon

Answered By : Yuval Filmus

The Galois field $GF(2^{128})$ has many different "concrete" representations. One popular representation is using polynomials in $GF(2)[x]$ (i.e. with coefficients in $GF(2)$) modulo some irreducible polynomial of degree $128$, say $x^{128}+x^7+x^2+x^1+x^0$. The magic constants $7,2,1,0$ come from this particular irreducible polynomial. You don't see $0$ since there is no need to shift by $0$. Similarly, the magic constants $57,62,63,64$ are the complements of the constants above with respect to $64$. Again, you don't need to shift by $64$ since the registers are $64$ bits wide. If we had used some other irreducible polynomial, the constants would have been different.

Regarding step 4, $[H_1:H_0]$ results from XORing $[E_1:E_0],[F_1:F_0],[G_1:G_0],[X_3:D]$. This implements the operation of MODing by $x^{128} + x^7 + x^2 + x^1 + x^0$. The idea is that modulo this polynomial, $x^{128} = x^7+x^2+x^1+^0$; addition here is XOR. The different summands corresponds to these different monomials – see if you can find the correspondence.

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Question Source : http://cs.stackexchange.com/questions/23549

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