在CUDA内核中的均匀分布的伪随机整数 [英] Uniformly distributed pseudorandom integers inside CUDA kernel

问题描述

如何在内核中生成均匀分布的伪随机整数？据我所知，Curand Api允许使用泊松离散分布，但不一致。

解决方案

：

1）使用curand_uniform从均匀分布中获取随机浮点数，然后将其映射到整数区间：

  float randu_f = curand_uniform（& localState）; 
 randu_f * =（B-A + 0.999999）; //你不应该使用（B-A + 1）* 
 randu_f + = A; 
 int randu_int = __float2int_rz（randu_f）; 
 

__ float2int_rz将舍入到零模式中的单精度浮点值x转换为有符号整数。

* curand_uniform返回在0.0和1.0之间均匀分布的伪随机浮点数序列。它可以从0.0返回到1.0，其中包括1.0并且排除0.0。
您应该使用largest_float_before_1或小于1，因为有一个小机会你将随机1，并且你可以超出范围。我也没有检查是否largest_float_before_1和GPU上的浮点运算保证不超过定义的边界。

2）调用curand返回一个伪随机数序列： / p>

  int randu_int = A + curand（& localState）％（BA）;但是，模数在GPU上非常昂贵，方法1更快。
 
 
 
 
 
 $ b < b
How can I generate uniformly distributed pseudorandom integers within a kernel? As far as I know Curand Api allows to use poisson discrete distribution, but not uniform.
解决方案
I suggest two options within a Kernel:

1) using curand_uniform to obtain a random floating point number from a uniform distribution, then map it to integer interval:
float randu_f = curand_uniform(&localState);
randu_f *= (B-A+0.999999); // You should not use (B-A+1)*
randu_f += A;
int randu_int = __float2int_rz(randu_f);
__float2int_rz Convert the single-precision floating point value x to a signed integer in round-towards-zero mode.

*curand_uniform returns a sequence of pseudorandom floats uniformly distributed between 0.0 and 1.0. It may return from 0.0 to 1.0, where 1.0 is included and 0.0 is excluded.
You should use biggest_float_before_1 or a little less than 1, because there is a small chance You will random 1, and You can get out of bounds. I didn't also check does biggest_float_before_1 and floating-point operations on GPU guarantee not to exceed from defined bounds.

2) calling curand returns a sequence of pseudorandom numbers:
int randu_int = A + curand(&localState) % (B-A);
However, modulo is very expensive on GPU and method 1 is faster.

                        
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