从范围生成随机整数 [英] Generating random integer from a range

问题描述

我需要一个函数，它会在给定的范围内生成一个随机整数（包括边框值）。我没有不合理的质量/随机性要求，我有四个要求：

• 我需要它快。我的项目需要生成数百万（或有时甚至数千万）的随机数，我的当前函数已经被证明是一个瓶颈。


• 我需要它是相当均匀的rand（）完全正确）。


• 最小 - 最大范围可以是从< 0，1>到< -32727，32727&

我目前有以下C ++代码：

  output = min +（rand（）*（int）（max  -  min）/ RAND_MAX）
  ）的主要问题，其中最后一个值几乎从不返回。
 
 
 
所以我抓住了笔和纸，下面的公式（建立在（int）（n + 0.5）整数舍入技巧）：
 
 
 
  
 
 
 
但它仍然不能给我均匀分布。重复运行10000样本给我的比值为37:50:13的值为-1,0.1。
 
 
 
你能建议更好的公式吗？ （或甚至整个伪随机数生成函数）
解决方案
您尝试过：
  output = min +（rand（）％（int）（max  -  min + 1））
  
这仍然是偏向较低的数字，但比原来的版本要少得多。它也可以扩展它，以消除偏见。
 
I need a function which would generate a random integer in given range (including border values). I don't unreasonable quality/randomness requirements, I have four requirements:


I need it to be fast. My project needs to generate millions (or sometimes even tens of millions) of random numbers and my current generator function has proven to be a bottleneck.
I need it to be reasonably uniform (use of rand() is perfectly fine).
the min-max ranges can be anything from <0, 1> to <-32727, 32727>.
it has to be seedable.


I currently have following C++ code:
output = min + (rand() * (int)(max - min) / RAND_MAX)
The problem is, that it is not really uniform - max is returned only when rand() = RAND_MAX (for Visual C++ it is 1/32727). This is major issue for small ranges like <-1, 1>, where the last value is almost never returned.

So I grabbed pen and paper and came up with following formula (which builds on the (int)(n + 0.5) integer rounding trick):



But it still doesn't give me uniform distribution. Repeated runs with 10000 samples give me ratio of 37:50:13 for values values -1, 0. 1.

Could you please suggest better formula? (or even whole pseudo-random number generator function)
解决方案
Have you tried:
output = min + (rand() % (int)(max - min + 1))
This is still slightly biased towards lower numbers but much less so than your original version. It's also possible to extend it so that it removes the bias.

                        
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