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Updated SM2 WWMM (2) (markdown)

Sun Yimin 2024-02-24 13:49:14 +08:00
parent 3dbe3ed846
commit 9bec042da7
1 changed files with 25 additions and 26 deletions
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SM2-WWMM-(2).md

@ -399,7 +399,6 @@ $t_0=0+0$
### 方案二：(移位、加法、减法)
因为Order素数不是MFMM，所以这个方案其实优势不大、甚至没有优势，尤其是在使用**使用MULXQ/ADCXQ/ADOXQ**的情况下。
移位针对 $O_3$ $O_2$ 乘法
这里加法没有溢出风险：Y不可能是0xFFFFFFFFFFFFFFFF。
$T_2=T_1 \ast O=Y \ast O= Y \ast 2^{256}-(Y \ast 2^{32}) \ast 2^{192} - Y \ast 2^{128} + (Y \ast O_1) \ast 2^{64} + (Y \ast O_0)$
@ -407,12 +406,11 @@ $T_3=T + T_2=t_7 \ast 2^{448} + t_6 \ast 2^{384} + t_5 \ast 2^{320} + t_4 \ast 2
$T_3=t_7 \ast 2^{448} + t_6 \ast 2^{384} + t_5 \ast 2^{320} + (t_4+Y) \ast 2^{256}+(t_3 - Y \ast 2^{32}) \ast 2^{192} + (t_2 - Y) \ast 2^{128} + (t_1 + Y \ast O_1) \ast 2^{64} + (t_0 + Y \ast O_0) $
```asm
// First reduction step, [ord3, ord2, ord1, ord0] = [1, -0x100000000, -1, ord1, ord0]
// First reduction step
MOVQ acc0, AX
MULQ p256ordK0<>(SB)
MOVQ AX, t0 // Y = t0 = (k0 * acc0) mod 2^64
// calculate the positive part first: [1, 0, 0, ord1, ord0] * t0 + [0, acc3, acc2, acc1, acc0]
// the result is [acc0, acc3, acc2, acc1], last lowest limb is dropped.
MOVQ p256ord<>+0x00(SB), AX
MULQ t0
ADDQ AX, acc0 // (carry1, acc0) = acc0 + L(t0 * ord0)
@ -420,6 +418,16 @@ $T_3=t_7 \ast 2^{448} + t_6 \ast 2^{384} + t_5 \ast 2^{320} + (t_4+Y) \ast 2^{25
MOVQ DX, t1 // t1 = carry1 + H(t0 * ord0)
MOVQ t0, acc0 // acc0 = t0
// calculate the negative part: [acc0, acc3, acc2, acc1] - [0, 0x100000000, 1, 0] * t0
MOVQ t0, AX
MOVQ t0, DX
SHLQ $32, AX
SHRQ $32, DX
SUBQ t0, acc2
SBBQ AX, acc3
SBBQ DX, acc0
MOVQ p256ord<>+0x08(SB), AX
MULQ t0
ADDQ t1, acc1 // (carry2, acc1) = acc1 + t1
@ -429,15 +437,6 @@ $T_3=t_7 \ast 2^{448} + t_6 \ast 2^{384} + t_5 \ast 2^{320} + (t_4+Y) \ast 2^{25
ADCQ DX, acc2
ADCQ $0, acc3
ADCQ $0, acc0
// calculate the negative part: [acc0, acc3, acc2, acc1] - [0, 0x100000000, 1, 0] * t0
MOVQ t0, AX
MOVQ t0, DX
SHLQ $32, AX
SHRQ $32, DX
SUBQ t0, acc2
SBBQ AX, acc3
SBBQ DX, acc0
```
乘法: 3
@ -611,16 +610,7 @@ $t_5=t_5 - 0$
ADCQ $0, DX
MOVQ DX, BX
MOVQ p256ord<>+0x08(SB), AX
MULQ t0
ADDQ BX, acc1
ADCQ $0, DX
ADDQ AX, acc1
ADCQ DX, acc2
ADCQ $0, acc3
ADCQ t0, acc4
ADCQ $0, acc5
MOVQ t0, acc0
MOVQ t0, AX
MOVQ t0, DX
SHLQ $32, AX
@ -628,8 +618,17 @@ $t_5=t_5 - 0$
SUBQ t0, acc2
SBBQ AX, acc3
SBBQ DX, acc4
SBBQ $0, acc5
SBBQ DX, acc0
MOVQ p256ord<>+0x08(SB), AX
MULQ t0
ADDQ BX, acc1
ADCQ $0, DX
ADDQ AX, acc1
ADCQ DX, acc2
ADCQ $0, acc3
ADCQ acc0, acc4
ADCQ $0, acc5
```
乘法: 3
移位：2
@ -674,7 +673,7 @@ $t_5=t_5 - 0$
| ----------- | ----------- | ----------- | ----------- | ----------- |
| 方案一 | 5 | 0 | 15 | 0 |
| 方案一（MULX/ADCX/ADOX） | 5 | 0 | 10 | 0 |
| 方案二 | 3 | 2 | 9 | 4 |
| 方案二 | 3 | 2 | 9 | 3 |
| 方案二（MULX） | 3 | 2 | 8 | 4 |
看来在支持**MULXQ/ADCXQ/ADOXQ**的情况下，使用方案一（MULX/ADCX/ADOX）更好！