Question

我在java中做了一些内在优化的矩阵包装器（在JNI的帮助下）。需要肯定这一点，你能给出关于矩阵优化的一些提示吗？我要实现的是：

Matrix可以表示为四组缓冲区/数组，一组用于水平访问，一组用于垂直访问，一组用于对角访问，命令缓冲区仅在需要时计算矩阵元素。这是一个例子。

Matrix signature: 

0  1  2  3  

4  5  6  7

8  9  1  3

3  5  2  9

First(hroizontal) set: 
horSet[0]={0,1,2,3} horSet[1]={4,5,6,7} horSet[2]={8,9,1,3} horSet[3]={3,5,2,9}

Second(vertical) set:
verSet[0]={0,4,8,3} verSet[1]={1,5,9,5} verSet[2]={2,6,1,2} verSet[3]={3,7,3,9}

Third(optional) a diagonal set:
diagS={0,5,1,9} //just in case some calculation needs this

Fourth(calcuation list, in a "one calculation one data" fashion) set:
calc={0,2,1,3,2,5} --->0 means multiply by the next element
                       1 means add the next element
                       2 means divide by the next element
                       so this list means
                       ( (a[i]*2)+3 ) / 5  when only a[i] is needed.
Example for fourth set: 
A.mult(2),   A.sum(3),  A.div(5), A.mult(B)
(to list)   (to list)  (to list) (calculate *+/ just in time when A is needed )
 so only one memory access for four operations.
 loop start
 a[i] = b[i] * ( ( a[i]*2) +3 ) / 5  only for A.mult(B)
 loop end

如上所示，当需要访问列元素时，第二组提供连续访问。没有飞跃。第一组水平访问也达到了同样的效果。

这应该让一些事情变得更容易，有些事情更难：

 Easier: 
 **Matrix transpozing operation. 
 Just swapping the pointers horSet[x] and verSet[x] is enough.

 **Matrix * Matrix multiplication.
 One matrix gives one of its horizontal set and other matrix gives vertical buffer.
 Dot product of these must be highly parallelizable for intrinsics/multithreading.
 If the multiplication order is inverse, then horizontal and verticals are switched.

 **Matrix * vector multiplication.
 Same as above, just a vector can be taken as horizontal or vertical freely.

 Harder:
 ** Doubling memory requirement is bad for many cases.
 ** Initializing a matrix takes longer.
 ** When a matrix is multiplied from left, needs an update vertical-->horizontal
 sets if its going to be multiplied from right after.(same for opposite)
 (if a tranposition is taken between, this does not count)


 Neutral:
 ** Same matrix can be multiplied with two other matrices to get two different
 results such as A=A*B(saved in horizontal sets)   A=C*A(saved in vertical sets)
 then A=A*A gives   A*B*C*A(in horizontal) and C*A*A*B (in vertical) without
 copying A. 

 ** If a matrix always multiplied from left or always from right, every access
 and multiplication will not need update and be contiguous on ram.

 ** Only using horizontals before transpozing, only using verticals after, 
 should not break any rules.

主要目的是使用矩阵（8的倍数，8的倍数）并应用具有多个线程的avx内在函数（每个步骤同时在一个集合上工作）。

我只实现了vector * vector dotproduct。 如果你掌握了编程方法，我会进入这个方向。

我写的dotproduct（带内在函数）比循环展开的版本快6倍（这是乘法的两倍快），当在包装器中启用多线程时，它也会停留在内存带宽上限（8x - ＆gt;使用接近我的ddr3限制的接近20GB / s）已经尝试过opencl，它对于cpu来说有点慢，但对于gpu来说非常好。

谢谢。

编辑：“块矩阵”缓冲区如何执行？当乘以大矩阵时，小补丁以特殊方式相乘，缓存可能用于减少主存储器访问。但是，这需要在垂直 - 水平 - 对角线和此块之间的矩阵乘法之间进行更多更新。

Answer 1

这实际上相当于缓存转置。听起来你打算急切地这样做;我只是在需要时才计算换位，并在需要时再次记住它。这样，如果你从未需要它，那么它永远不会被计算出来。

Answer 2

有几个库使用Expression Templates来为一系列矩阵运算启用非常具体的优化函数。

The C++ Programming Lanuage还有一章关于“融合操作”（29.5.4，第4版）。

这样可以连接语句la：

M = A*B.transp(); // where M, A, B are matrices

在这种情况下，您需要有3个课程：

class Matrix;

class Transposed
{
public:
  Transposed(Matrix &matrix) : m_matrix(matrix) {}
  Matrix & obj (void) { return m_matrix; }
private:
  Matrix & m_matrix;
};

class MatrixMatrixMulTransPosed
{
public:
  MatrixMatrixMulTransPosed(Matrix &matrix, Transposed &trans) 
    : m_matrix(matrix), m_transposed(trans.obj()) {}
  Matrix & matrix (void) { return m_matrix; }
  Matrix & transposed (void) { return m_transposed; }
private:
  Matrix & m_matrix;
  Matrix & m_transposed;
};

class Matrix
{
  public:
    MatrixMatrixMulTransPosed operator* (Transposed &rhs)
    { 
      return MatrixMatrixMulTransPosed(*this, rhs); 
    }

    Matrix& operator= (MatrixMatrixMulTransPosed &mmtrans)
    {
      // Actual computation goes here and is stored in this.
      // using mmtrans.matrix() and mmtrans.transposed()
    }
};

你可以推进这个概念，以便能够为每个任何平均值的计算都设置一个特殊的函数。

cpu的矩阵访问和乘法优化

2 个答案: