快速 SSE 阈值算法

Question

我正在尝试使用 SSE 提出一个非常快速的阈值算法来替换它：

uint8_t *pSrc, *pDst;

// Assume pSrc and pDst point to valid data

// Handle left edge
*pDst++ = *pSrc++;

// Likeness filter
for (uint32_t k = 2; k < width; k++, pSrc++, pDst++)
    if ((*pDst - *pSrc) * (*pDst - *pSrc) > 100 /*THRESHOLD_SQUARED*/) {
        *pDst = *pSrc;
    }
}

// Handle right edge
*pDst++ = *pSrc++;

到目前为止，我有这个：

const uint8_t THRESHOLD = 10;

__attribute__((aligned (16))) static const uint8_t mask[16] = {
    THRESHOLD, THRESHOLD, THRESHOLD, THRESHOLD,
    THRESHOLD, THRESHOLD, THRESHOLD, THRESHOLD,
    THRESHOLD, THRESHOLD, THRESHOLD, THRESHOLD,
    THRESHOLD, THRESHOLD, THRESHOLD, THRESHOLD
};

__m128i xmm1, xmm3, xmm4, xmm5, xmm6, xmm7, xmm8, xmm9;
xmm1 = _mm_load_si128((__m128i const *)mask);
xmm6 = _mm_setzero_si128();

uint8_t *pSrc, *pDst;

// Assume pSrc and pDst point to valid data

// I have other code with another mask for the first 16 entries

for (uint32_t k = 16; k < (width - 16); k += 16, pSrc += 16, pDst += 16) {
    xmm3 = _mm_load_si128((__m128i const *)pDst);
    xmm4 = _mm_load_si128((__m128i const *)pSrc);
    xmm5 = _mm_unpacklo_epi8(xmm3, xmm6);
    xmm7 = _mm_unpackhi_epi8(xmm3, xmm6);
    xmm8 = _mm_unpacklo_epi8(xmm4, xmm6);
    xmm9 = _mm_unpackhi_epi8(xmm4, xmm6);
    xmm5 = _mm_sub_epi16(xmm5, xmm8);
    xmm7 = _mm_sub_epi16(xmm7, xmm9);
    xmm5 = _mm_abs_epi16(xmm5);
    xmm7 = _mm_abs_epi16(xmm7);
    xmm5 = _mm_packs_epi16(xmm5, xmm7);
    xmm5 = _mm_cmpgt_epi8(xmm5, xmm1);
    xmm3 = _mm_blendv_epi8(xmm3, xmm4, xmm5);
    _mm_store_si128((__m128i *)pDst, xmm3);
}

// I have other code with another mask for the last 16 entries

我有想法使用另一种算法来处理两个值之差的绝对值（主要是留在U8（uchar）空间）：

a' = a >> 1;
b' = b >> 1;
diff = (abs(sub(a' - b')) << 1) + ((a ^ b) & 1);

这需要 8 条 SSE 指令，而不是上面的 9 条（不包括编译器生成的任何额外寄存器移动），但我不确定它是否由于依赖延迟而更快。

是否有其他 SSE 专家有更好的建议（最高使用 SSE 4.2）？

更新 1 - 感谢 Yves 的建议！

const uint8_t THRESHOLD = 10;

__attribute__((aligned (16))) static const uint8_t mask[16] = {
    THRESHOLD, THRESHOLD, THRESHOLD, THRESHOLD,
    THRESHOLD, THRESHOLD, THRESHOLD, THRESHOLD,
    THRESHOLD, THRESHOLD, THRESHOLD, THRESHOLD,
    THRESHOLD, THRESHOLD, THRESHOLD, THRESHOLD
};

__m128i xmm1, xmm3, xmm4, xmm5, xmm6, xmm7;
xmm1 = _mm_load_si128((__m128i const *)mask);
xmm6 = _mm_setzero_si128();

uint8_t *pSrc, *pDst;

// Assume pSrc and pDst point to valid data

// I have other code with another mask for the first 16 entries

for (uint32_t k = 16; k < (width - 16); k += 16, pSrc += 16, pDst += 16) {
    xmm3 = _mm_load_si128((__m128i const *)pDst);
    xmm4 = _mm_load_si128((__m128i const *)pSrc);
    xmm5 = _mm_subs_epu8(xmm3, xmm4);
    xmm7 = _mm_subs_epu8(xmm4, xmm3);
    xmm5 = _mm_adds_epu8(xmm5, xmm7);
    xmm5 = _mm_subs_epu8(xmm5, xmm1);
    xmm5 = _mm_cmpeq_epi8(xmm5, xmm6);
    xmm4 = _mm_blendv_epi8(xmm4, xmm3, xmm5);
    _mm_store_si128((__m128i *)pDst, xmm4);
}

// I have other code with another mask for the last 16 entries

Answer 1

Simd 库中有一些有用的函数：

inline __m128i Combine(__m128i mask, __m128i positive, __m128i negative)
{
    return _mm_or_si128(_mm_and_si128(mask, positive), _mm_andnot_si128(mask, negative));
}

inline __m128i AbsDifferenceU8(__m128i a, __m128i b)
{
    return _mm_sub_epi8(_mm_max_epu8(a, b), _mm_min_epu8(a, b));
}

inline __m128i LesserOrEqual8u(__m128i a, __m128i b)
{
    return _mm_cmpeq_epi8(_mm_min_epu8(a, b), a);
}

所以 SSE2 优化将如下所示：

__m128i t = _mm_set1_epi8(threshold);
for (uint32_t k = 16; k < width - 16; pSrc += 16, pDst += 16)
{
    __m128i src = _mm_load_si128((__m128i*)pSrc);
    __m128i dst = _mm_load_si128((__m128i*)pDst);
    __m128i mask = LesserOrEqual8u(AbsDifferenceU8(src, dst), t);
    _mm_strore_si128((__m128i*)pDst, Combine(mask, dst, src);
}

Answer 2

有一种有效的替代方法可以利用算术饱和度来计算绝对差。

确实，饱和减法计算A - B = Max(A - B, 0)，所以|A-B| = (A - B) + (B - A)。

Diff= _mm_adds_epu8(_mm_subs_epu8(A, B), _mm_subs_epu8(B, A));

总和不会饱和。这样，您可以保持 16 x 8 位无符号并获得最大吞吐量。