Updated SM2 MFMM (2) (markdown)

SM2-MFMM-(2).md

@@ -379,4 +379,84 @@ $t_0=0+0$
 	ADOXQ t0, acc0
 ```
 乘法: 5  
-加法：9  
+加法：9  
+
+### 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  
+```asm
+	// First reduction step, [ord3, ord2, ord1, ord0] = [1, -0x100000000, -1, ord1, ord0]
+	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)
+	ADCQ $0, DX                 // DX = carry1 + H(t0 * ord0)
+	MOVQ DX, t1                 // t1 = carry1 + H(t0 * ord0)
+	MOVQ t0, acc0               // acc0 =  t0
+
+	MOVQ p256ord<>+0x08(SB), AX
+	MULQ t0
+	ADDQ t1, acc1               // (carry2, acc1) = acc1 + t1
+	ADCQ $0, DX                 // DX = carry2 + H(t0*ord1)
+
+	ADDQ AX, acc1               // (carry3, acc1) = acc1 + t1 + L(t0*ord1)
+	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  
+移位：2  
+加法：8  
+减法：3  
+
+**使用MULXQ**:  
+```asm
+	// First reduction step, [ord3, ord2, ord1, ord0] = [1, -0x100000000, -1, ord1, ord0]
+	MOVQ acc0, DX
+	MULXQ p256ordK0<>(SB), t0, AX
+	// 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 t0, DX                 // Y = t0 = (k0 * acc0) mod 2^64
+	MULXQ p256ord<>+0x00(SB), AX, t1
+	ADDQ AX, acc0               // (carry1, acc0) = acc0 + L(t0 * ord0)
+	ADCQ t1, acc1               // (carry2, acc1) = acc1 + H(t0 * ord0) + carry1
+	MOVQ t0, acc0               // acc0 = t0 
+
+	MULXQ p256ord<>+0x08(SB), AX, t1
+	ADCQ $0, t1                 // t1 = carry2 + H(t0*ord1)
+	ADDQ AX, acc1               // (carry3, acc1) = acc1 + L(t0*ord1)
+	ADCQ t1, acc2               // (carry4, acc2) = acc2 + t1 + carry3
+	ADCQ $0, acc3               // (carry5, acc3) = acc3 + carry4
+	ADCQ $0, acc0               //           acc0 = t0 + carry5 
+	// calculate the negative part: [acc0, acc3, acc2, acc1] - [0, 0x100000000, 1, 0] * t0
+	MOVQ t0, AX
+	//MOVQ t0, DX              // This is not required due to t0=DX already
+	SHLQ $32, AX
+	SHRQ $32, DX
+
+	SUBQ t0, acc2
+	SBBQ AX, acc3
+	SBBQ DX, acc0
+```
+乘法: 3  
+移位：2  
+加法：7  
+减法：3  
+
+看来在支持**MULXQ/ADCXQ/ADOXQ**的情况下，使用方案一更好！
+
+