arm - NEON 边境检查

Question

引用我之前关于边境检查条件的问题 - Border check in image processing ？ 我正在使用 neon 编写代码进行边界检查。编写代码时遇到以下问题:

输入:

 --------------------------------
|221 220 228 223 230 233 234 235 ..
|71  73  70  78  92  130 141 143 ..
|

要求:

  -1 * v_m1_m1 + 0 * v_m1_0 + 1 * v_m1_p1
  -1 * v_0_m1  + 0 * v_0_0  + 1 * v_0_p1    --> v_out
  -1 * v_p1_m1 + 0 * v_p1_0 + 1 * v_p1_p1

伪代码:

for i = 0 to nrows - 1
        // init row pointers
        p_row_m1 = src + src_width * MAX(i-1, 0);           // pointing to minus1 row
        p_row_0  = src + src_width * i;                     // pointing to current row
        p_row_p1 = src + src_width * MIN(i+1, src_width-1); // pointing to plus1 row

        v_m1_m1 = vdupq_n_u32(p_row_m1[0]);   // fill left vector from src[i-1][0]
        v_0_m1  = vdupq_n_u32(p_row_0[0]);    // fill left vector from src[i][0]
        v_p1_m1 = vdupq_n_u32(p_row_p1[0]);   // fill left vector from src[i+1][0]

        v_m1_0 = vld1q_u32(&p_row_m1[0]);   // load center vector from src[i-1][0..7]
        v_0_0  = vld1q_u32(&p_row_0[0]);    // load center vector from src[i][0..7]
        v_p1_0 = vld1q_u32(&p_row_p1[0]);   // load center vector from src[i+1][0..7]

        for j = 0 to (ncols - 4) step 4         // assuming 4 elements per SIMD vector

            v_m1_p1  = vld1q_u32(&p_row_m1[j+4]);   // load right vector from src[i-1][0..7]
            v_0_p1   = vld1q_u32(&p_row_0[j+4]);    // load right vector from src[i][0..7]
            v_p1_p1  = vld1q_u32(&p_row_p1[j+4]);   // load right vector from src[i+1][0..7]
    //
    // you now have a 3x3 arrangement of vectors on which
    // you can perform a neighbourhood operation and generate
    // 16 output pixels for the current iteration:
    //
    //    v_m1_m1  v_m1_0  v_m1_p1
    //    v_0_m1   v_0_0   v_0_p1
    //    v_p1_m1  v_p1_0  v_p1_p1
    //
    //               |
    //               V
    //
    //              v_out
    vst1q_s32(v_out, &image_out[i][j])      // store output vector at image_out[i][j..j+15]
    // shuffle vectors so that we can use them on next iteration
    v_m1_m1 = v_m1_0
    v_m1_0  = v_m1_p1

    v_0_m1  = v_0_0 
    v_0_0   = v_0_p1

    v_p1_m1 = v_p1_0
    v_p1_0  = v_p1_p1

      end_for
      // for final iteration we need to handle right edge pixels...
      v_m1_p1 = vdupq_n_u32(p_row_m1[ncols-1])     // fill right vector from image[i-1][ncols-1]
      v_0_p1  = vdupq_n_u32(p_row_0[ncols-1])       // fill right vector from image[i][ncols-1]
      v_p1_p1 = vdupq_n_u32(p_row_p1[ncols-1])     // fill right vector from image[i+1][ncols-1]
      // calculate v_out as above
      vst1q_s32(v_out, &image_out[i][j])        // store output vector at image_out[i][ncols_16..ncols-1]
    end_for

在 C 代码中，如果我想计算 221 (@i,j =0,0) ，带边框，它看起来像:

 221  221 220
      -----------
 221 |221 220
     |
 71  |71  73

在 NEON 代码中，如果我想计算 221 (@i,j =0,0) ，带有边框，它看起来像:

221 221 221 221     221 220 228 223   230 233 234 235
                    -------------------------------
221 221 221 221    |221 220 228 223   230 233 234 235
                   |
71  71  71  71     |71  73  70  78    92  130 141 143

计算 221: 在 neon 代码中，邻居如下所示，根据要求，它出现了错误:

v_m1_m1.0  v_m1_0.0  v_m1_p1.0
v_0_m1.0   v_0_0.0   v_0_p1.0
v_p1_m1.0  v_p1_0.0  v_p1_p1.0

221  221  230
     ---
221 |221| 230
     ---
71   71   92

如何使用通用代码逻辑处理边框？

Answer 1

看来您需要了解如何通过邻域运算实现 SIMD 内核。这是重要的部分:

//
// you now have a 3x3 arrangement of vectors on which
// you can perform a neighbourhood operation and generate
// 4 output pixels for the current iteration:
//
//    v_m1_m1  v_m1_0  v_m1_p1
//    v_0_m1   v_0_0   v_0_p1
//    v_p1_m1  v_p1_0  v_p1_p1
//
//               |
//               V
//
//              v_out

假设您想在十字形内核上实现求和，即:

y[i][j] = x[i][j] + x[i-1][j] + x[i+1][j] + x[i][j-1] + x[i][j+1];

SIMD 中的伪代码为:

// sum vertically: x[i][j] + x[i-1][j] + x[i+1][j]
v_out = v_m1_0;
v_out = v_out + v_0_0;      // vaddq
v_out = v_out + v_p1_0;     // vaddq

// add the x[i][j-1] components
v_temp = v_0_m1:v_0_0 >> 1; // vextq - use this to get a right-shifted vector
v_out = v_out + v_temp;     // vaddq

// add the x[i][j+1] components
v_temp = v_0_0:v_0_p1 << 1; // vextq - use this to get a left-shifted vector
v_out = v_out + v_temp;     // vaddq

此时v_out现在包含输出元素y[i][j]..y[i][j+3]的四个总和。换句话说，我们评估了一个内核中的四个输出点。现在，我们将所有向量向左移动 1，并为右侧列加载 3 个新向量，然后使用 j += 4 再次执行此操作。如果您查看上一个问题的原始伪代码，您会发现边界情况都是通过用边缘值填充向量来处理的。