PyOpenCL - 英特尔和NVidia

Question

我正在尝试运行基于高斯脉冲传播的模拟。我在我的Windows桌面与i5 4590＆amp;之间进行交叉开发。 GTX 970（最新驱动程序）和我2015年初的macbook air。

当我运行我的主代码时，我的桌面上得不到任何不错的结果（计算分歧），但在我的Mac上，结果似乎没问题。

为了进一步研究这个问题，我试图运行简单的高斯传播。 macbook上的结果或多或少都没问题，而在桌面上则是一团糟。

我在两台机器上运行相同的代码，两者都有相同的python（2.7.10）分发版和相应的模块。

这是我的代码

import scipy as sp
import pyopencl as cl
import matplotlib.pyplot as plot

ctx = cl.create_some_context()
queue = cl.CommandQueue(ctx)
MF = cl.mem_flags

dx = 0.01
X = sp.arange(0.0, 100, dx)
N = len(X)

A_h = (sp.exp(-(X-50)**2/10.)*sp.exp(-1j*1000*X)).astype(sp.complex64)
A_d = cl.Buffer(ctx, MF.READ_WRITE | MF.COPY_HOST_PTR, hostbuf=A_h)

plot.figure()
plot.plot(X, abs(A_h) ** 2 / (abs(A_h) ** 2).max())

Source = """
    #define complex_ctr(x, y) (float2)(x, y)
    #define complex_mul(a, b) complex_ctr(mad(-(a).y, (b).y, (a).x * (b).x), mad((a).y, (b).x, (a).x * (b).y))
    #define complex_unit (float2)(0, 1)

    __kernel void propagate(__global float2 *A){
        const int gid_x = get_global_id(0);
        float EPS = 0.1f;
        A[gid_x] = A[gid_x] + EPS*complex_mul((A[gid_x+1] + A[gid_x-1]), complex_unit);
    }
"""
prg = cl.Program(ctx, Source).build()
for i in range(3000):
    print i
    event = prg.propagate(queue, (N,), None, A_d)
    event.wait()
cl.enqueue_copy(queue, A_h, A_d)

plot.plot(X, abs(A_h) ** 2 / (abs(A_h) ** 2).max())

plot.show()

以下是结果

桌面结果：

Mac结果：

绿线对应于传播后的高斯线，蓝线是初始高斯

NVidia方面可能导致此问题的原因是什么？我想我错过了防​​止这种情况发生的关键步骤，并且由于运气而在mac上运行

修改的

这是我根据用户建议工作的最终代码

import scipy as sp
import pyopencl as cl
import matplotlib.pyplot as plot

ctx = cl.create_some_context()
queue = cl.CommandQueue(ctx)
MF = cl.mem_flags

dx = sp.float32(0.001)
X = sp.arange(0.0, 100, dx).astype(sp.float32)
N = len(X)

A_h = (sp.exp(-(X-50)**2/10.)*sp.exp(1j*1000*X)).astype(sp.complex64)
A_d = cl.Buffer(ctx, MF.READ_WRITE | MF.COPY_HOST_PTR, hostbuf=A_h)
B_d = cl.Buffer(ctx, MF.READ_WRITE | MF.COPY_HOST_PTR, hostbuf=A_h)

plot.figure()
plot.plot(X, abs(A_h) ** 2 / (abs(A_h) ** 2).max())

Source = """
    #define complex_ctr(x, y) (float2)(x, y)
    #define complex_mul(a, b) complex_ctr((a).x * (b).x - (a).y * (b).y, (a).x * (b).y + (a).y * (b).x)
    #define complex_unit (float2)(0, 1)

    __kernel void propagate(__global float2 *A){
        const int gid_x = get_global_id(0);
        float EPS = 0.1f;
        float2 a1, a2;
        a1 = A[gid_x-1];
        a2 = A[gid_x+1];
        barrier(CLK_GLOBAL_MEM_FENCE);
        A[gid_x] += EPS*complex_mul((a1 + a2), complex_unit);
    }
"""

prg = cl.Program(ctx, Source).build()
for i in range(12000):
    print i
    evolve = prg.propagate(queue, (N,), None, A_d)
    evolve.wait()
cl.enqueue_copy(queue, A_h, A_d)

plot.plot(X, abs(A_h) ** 2)

plot.show()

Answer 1

编辑：哦，只是阅读@talonmies评论，它与此相同。

此代码在OpenCL中不安全，它有datarace问题：

A[gid_x] = A[gid_x] + EPS*complex_mul((A[gid_x+1] + A[gid_x-1]), complex_unit);

每个工作项x都使用x+1和x-1。根据工作项目的时间表，结果会有所不同。

使用2个缓冲区，从A读取，写入B，easy：

import scipy as sp
import pyopencl as cl
import matplotlib.pyplot as plot

ctx = cl.create_some_context()
queue = cl.CommandQueue(ctx)
MF = cl.mem_flags

dx = 0.01
X = sp.arange(0.0, 100, dx)
N = len(X)

A_h = (sp.exp(-(X-50)**2/10.)*sp.exp(-1j*1000*X)).astype(sp.complex64)
A_d = cl.Buffer(ctx, MF.READ_WRITE | MF.COPY_HOST_PTR, hostbuf=A_h)
B_d = cl.Buffer(ctx, MF.READ_WRITE)

plot.figure()
plot.plot(X, abs(A_h) ** 2 / (abs(A_h) ** 2).max())

Source = """
    #define complex_ctr(x, y) (float2)(x, y)
    #define complex_mul(a, b) complex_ctr(mad(-(a).y, (b).y, (a).x * (b).x), mad((a).y, (b).x, (a).x * (b).y))
    #define complex_unit (float2)(0, 1)

    __kernel void propagate(__global float2 *A, __global float2 *B){
        const int gid_x = get_global_id(0);
        float EPS = 0.1f;
        B[gid_x] = A[gid_x] + EPS*complex_mul((A[gid_x+1] + A[gid_x-1]), complex_unit);
    }
"""
prg = cl.Program(ctx, Source).build()
for i in range(3000):
    print i
    event = prg.propagate(queue, (N,), None, A_d, B_d)
    A_d, B_d = B_d, A_d #Swap buffers, so A always has results
    #event.wait() #You don't need this, this is slowing terribly the execution, enqueue_copy already waits
cl.enqueue_copy(queue, A_h, A_d)

plot.plot(X, abs(A_h) ** 2 / (abs(A_h) ** 2).max())

plot.show()