java – cpu的矩阵访问和乘法优化

2023年2月3日 209次阅读

我在
java中制作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？我要实施的是：

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.

主要目的是具有(8的倍数,8的倍数)大小的矩阵并且应用具有多个线程的avx内在函数(每个胎面同时在一组上工作).

我只实现了矢量*矢量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),当在包装器中启用多线程时(8x – >几乎使用),它也会停留在内存带宽上限20GB / s接近我的ddr3的限制)已经尝试过opencl它对cpu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 – 对角线和该块之间的矩阵乘法之间进行更多更新.

最佳答案 一些库使用
Expression Templates来实现非常特定的优化函数的应用,用于级联矩阵运算.

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

这使得语句的连接成为可能：

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

在这种情况下,您将需要3个类：

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,以便能够为任何平均值的任何计算都有一个特殊的函数.

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