cpu的java矩阵访问与乘法优化
我正在用java(借助JNI)制作一些内部优化的矩阵包装器。需要肯定这一点,你能给出一些关于矩阵优化的提示吗我要实施的是:
矩阵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,以及一个仅在需要时计算矩阵元素的命令缓冲区。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
如上所述,当需要访问列元素时,第二个集合提供连续访问。没有跳跃。第一套水平通道也实现了同样的效果
这会让一些事情变得更容易一些事情变得更难:
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.
主要目的是拥有一个(8的倍数,8的倍数)大小的矩阵,并使用多线程应用avx Intrinsic(每个踏板同时在一组上工作)
我只得到了向量*向量点积如果编程大师们给出指导,我将对此进行探讨
我写的dotproduct(使用内部函数)比循环展开版本快6倍(它是一个一个乘法的两倍),当包装器中启用多线程时(8x-->;使用接近我的ddr3限制的近20GB/s),它已经尝试了opencl,并且cpu速度有点慢,但对gpu来说很棒
谢谢
编辑:一个“块矩阵”缓冲区将如何执行?当将大矩阵相乘时,小的补丁会以一种特殊的方式相乘,缓存可能用于减少主内存访问。但这需要在垂直-水平对角线和该块之间的矩阵乘法之间进行更多更新
# 1 楼答案
有几个库使用Expression Templates来支持对矩阵操作的级联应用非常具体、优化的函数
The C++ Programming Lanuage还有一个关于“融合操作”的简短章节(29.5.4,第四版)
这使得语句能够串联起来:
在这种情况下,您需要有3个类:
你可以改进这个概念,使它能够为每一个计算提供一个特殊的函数,无论用什么方法,它都是至关重要的