On Tue, 2021-11-30 at 20:17 +0100, Niels Möller wrote:
> Amitay Isaacs <[email protected]> writes:
> 
> > +       .file "ecc-secp192r1-modp.asm"
> 
> Thanks, I'm looking at this file first (being the simplest, even
> though
> the security level of this curve is a bit low for current usage, so
> performance not of so great importance).

Yes. The main focus was on curves p256 and higher.  For completeness
sake I added the assembly for p192 and p224 curves.

> 
> I'm quite new to powerpc, so I'm refering to the instruction
> reference,
> and trying to learn as we go along. It seems addc is addition with
> carry
> output (but no carry input), adde is addition with carry input and
> output, and addze is addition of zero with carry input and output.
> 
> > +define(`RP', `r4')
> > +define(`XP', `r5')
> > +
> > +define(`T0', `r6')
> > +define(`T1', `r7')
> > +define(`T2', `r8')
> > +define(`T3', `r9')
> > +define(`C1', `r10')
> > +define(`C2', `r11')
> 
> As I understand it, we could also use register r3 (unused input
> argument), but we don't need to, since we have enough free scratch
> registers.

Generally I avoided using r3 as it is the first input argument, but
also serves as the return value of the function.  So unless we are
running short of registers r3 is left untouched.  (Even though there
are 32 registers, only some of them can be used without saving them
first.)  For example, r3 is used in p224 modp implementation.

> 
> > +       C void ecc_secp192r1_modp (const struct ecc_modulo *m,
> > mp_limb_t *rp)
> > +       .text
> > +define(`FUNC_ALIGN', `5')
> > +PROLOGUE(_nettle_ecc_secp192r1_modp)
> > +       ld      T0, 0(XP)
> > +       ld      T1, 8(XP)
> > +       ld      T2, 16(XP)
> > +
> > +       li      C1, 0
> > +       li      C2, 0
> > +
> > +       ld      T3, 24(XP)
> > +       addc    T0, T3, T0
> > +       adde    T1, T3, T1
> > +       addze   T2, T2
> > +       addze   C1, C1
> > +
> > +       ld      T3, 32(XP)
> > +       addc    T1, T3, T1
> > +       adde    T2, T3, T2
> > +       addze   C1, C1
> > +
> > +       ld      T3, 40(XP)
> > +       addc    T0, T3, T0
> > +       adde    T1, T3, T1
> > +       adde    T2, T3, T2
> > +       addze   C1, C1
> 
> To analyze what we are doing, I'm using the Nettle and GMP convention
> that B = 2^64 (bignum base), then p = B^3 - B - 1, or B^3 = B + 1
> (mod
> p). Denote the six input words as
> 
>   <a_5,a_4,a_3,a_2,a_1,a_0>
> 
> representing the number 
> 
>   B^5 a_5 + B^4 a_4 + B^3 a_3 + B^2 a_2 + B a_1 + a_0
> 
> The accumulation above, as I understand it, computes
> 
>   <c_1,t_2,t_1,t_0> = <a_2,a_1,a_0> + a_3 (B+1) + a_4 (B^2 + B) 
>                     + a_5 (B^2 + B + 1>
> 
> or more graphically,
> 
>       a_2 a_2 a_1
>           a_3 a_3
>       a_4 a_4
>     + a_5 a_5 a_5
>   ---------------
>   c_1 t_2 t_1 t_0
> 
> This number is < 3 B^3, which means that c_1 is 0, 1 or 2 (each of
> the
> addze instructions can increment it).
> 
> This looks nice, and I think it is pretty efficient too. It looks a
> bit
> different from what the x86_64 code is doing; maybe the latter could
> be
> improved.
> 
> > +       addc    T0, C1, T0
> > +       adde    T1, C1, T1
> > +       addze   T2, T2
> > +       addze   C2, C2
> 
> Above, c_1 is folded in at the right places, 
> 
>   <c_2,t_2,t_1,t_0>  <--  <t_2, t_1, t_0> + c_1 (B + 1)
> 
> This number is < B^3 + 3 (B+1). This implies that in the (quite
> unlikely) case we get carry out, i.e., c_2 = 1, then the value of the
> low three words is < 3 (B+1). That means that there can be no new
> carry
> out when folding c_2.
> 
> > +       li      C1, 0
> > +       addc    T0, C2, T0
> > +       adde    T1, C2, T1
> > +       addze   T2, T2
> > +       addze   C1, C1
> > +
> > +       addc    T0, C1, T0
> > +       adde    T1, C1, T1
> > +       addze   T2, T2
> 
> So I think this final folding could be reduced to just
> 
>         addc    T0, C2, T0
>         adde    T1, C2, T1
>         addze   T2, T2
> 
> There's no carry out, from this, because either C2 was zero, or T2
> was
> small, <= 3. Does that make sense?

This was the first code I wrote using the exact calculation you have
outlined above.  However, I was not sure about the sizes of the carry
(C1 and C2).  I did notice the x86 code short-cutting the C2 folding,
but the reasoning was not apparent.

Thank you for explaining the bounds calculation.  That does help me
understand why C2 folding can be simplified as you have suggested.

> 
> > +       std     T0, 0(RP)
> > +       std     T1, 8(RP)
> > +       std     T2, 16(RP)
> > +
> > +       blr
> > +EPILOGUE(_nettle_ecc_secp192r1_modp)
> 
> Regards,
> /Niels
> 

Amitay.
-- 

Good literature is about Love and War.
Junk Fiction is about Sex and Violence.
_______________________________________________
nettle-bugs mailing list
[email protected]
http://lists.lysator.liu.se/mailman/listinfo/nettle-bugs

Reply via email to