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Updated SM2 WWMM (2) (markdown)
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@ -452,50 +452,11 @@ $T_3=t_7 \ast 2^{448} + t_6 \ast 2^{384} + t_5 \ast 2^{320} + (t_4+Y-Y>>32) \ast
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加法:8
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减法:3
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**使用MULXQ**:
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```asm
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// First reduction step, [ord3, ord2, ord1, ord0] = [1, -0x100000000, -1, ord1, ord0]
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MOVQ acc0, DX
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MULXQ p256ordK0<>(SB), t0, AX
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// 处理第一个加法,以便释放acc0
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MOVQ t0, DX // Y = t0 = (k0 * acc0) mod 2^64
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MULXQ p256ord<>+0x00(SB), AX, t1
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ADDQ AX, acc0 // (carry1, acc0) = acc0 + L(t0 * ord0),acc0 可以被释放了。
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ADCQ t1, acc1 // (carry2, acc1) = acc1 + H(t0 * ord0) + carry1
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// calculate the negative part: [t0, acc3, acc2, acc1] - [0, 0x100000000, 1, 0] * t0
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// [t0 - t0>>32, acc3-t0<<32, acc2-t0, acc1]
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// 处理减法
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MOVQ t0, acc0 // acc0 = t0
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MOVQ t0, AX
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SHLQ $32, AX
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SHRQ $32, DX
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SUBQ t0, acc2
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SBBQ AX, acc3
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SBBQ DX, acc0
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// 处理加法
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MOVQ t0, DX
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MULXQ p256ord<>+0x08(SB), AX, t1
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ADCQ $0, t1 // t1 = carry2 + H(t0*ord1)
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ADDQ AX, acc1 // (carry3, acc1) = acc1 + L(t0*ord1)
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ADCQ t1, acc2 // (carry4, acc2) = acc2 + t1 + carry3
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ADCQ $0, acc3 // (carry5, acc3) = acc3 + carry4
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ADCQ $0, acc0 // acc0 = t0 + carry5
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```
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乘法: 3
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移位:2
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加法:7
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减法:3
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| 方案 | 乘法 | 移位 | 加法 | 减法 |
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| ----------- | ----------- | ----------- | ----------- | ----------- |
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| 方案一 | 5 | 0 | 14 | 0 |
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| 方案一(MULX/ADCX/ADOX) | 5 | 0 | 9 | 0 |
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| 方案二 | 3 | 2 | 8 | 3 |
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| 方案二(MULX) | 3 | 2 | 7 | 3 |
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看来在支持**MULXQ/ADCXQ/ADOXQ**的情况下,使用方案一(MULX/ADCX/ADOX)更好!
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@ -638,49 +599,10 @@ $T_3=(t_4+Y-Y>>32) \ast 2^{256}+(t_3 - Y<<32) \ast 2^{192} + (t_2 - Y) \ast 2^{1
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加法:9
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减法:3
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**使用MULXQ**:
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```asm
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// First reduction step
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MOVQ acc0, DX
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MULXQ p256ordK0<>(SB), t0, AX
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// 处理第一个加法,以便释放acc0
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MOVQ t0, DX
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MULXQ p256ord<>+0x00(SB), AX, BX
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ADDQ AX, acc0 // acc0 可以被释放了
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ADCQ BX, acc1
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// 开始处理减法
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MOVQ t0, acc0
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MOVQ t0, AX
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SHLQ $32, AX
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SHRQ $32, DX
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SUBQ t0, acc2
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SBBQ AX, acc3
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SBBQ DX, acc0
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// 处理加法
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MOVQ t0, DX
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MULXQ p256ord<>+0x08(SB), AX, BX
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ADCQ $0, BX
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ADDQ AX, acc1
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ADCQ BX, acc2
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ADCQ $0, acc3
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ADCQ acc0, acc4
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ADCQ $0, acc5
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```
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乘法: 3
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移位:2
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加法:8
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减法:3
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| 方案 | 乘法 | 移位 | 加法 | 减法 |
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| ----------- | ----------- | ----------- | ----------- | ----------- |
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| 方案一 | 5 | 0 | 15 | 0 |
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| 方案一(MULX/ADCX/ADOX) | 5 | 0 | 10 | 0 |
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| 方案二 | 3 | 2 | 9 | 3 |
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| 方案二(MULX) | 3 | 2 | 8 | 3 |
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看来在支持**MULXQ/ADCXQ/ADOXQ**的情况下,使用方案一(MULX/ADCX/ADOX)更好!
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