如何创建不均匀的范围数随机函数？ [英] How to create uneven range number random function?

问题描述

我们知道经典范围随机函数是这样的：

  public static final int random（final int min，final int max）{
 Random rand = new Random（）; 
返回min + rand.nextInt（max  -  min + 1）; // +1包含最多
} 
  

我想创建用于生成的算法函数在1..10之间的范围内随机编号，但不一致的可能性如下：

1）1,2,3 - > 3/6（1/2）

2 ）4,5,6,7 - > 1/6

3）8,9,10 - > 2/6（1/3）

上述意味着该函数有1/2的机会返回1到3之间的数字，1/6的机会返回4到7之间的数字，以及1/3的机会返回8到10之间的数字。 / p>

任何人都知道算法吗？

更新：

实际上，1..10之间的范围仅作为示例。我想要创建的函数适用于任何数字范围，例如：1..10000，但规则仍然相同：顶部范围为3/6（30％部分），中间范围为1/6（下一个） 40％部分），底部范围2/6（最后30％部分）。

解决方案

使用上述算法：

  int temp = random（0,5）; 
 if（temp< = 2）{
 return random（1,3）; 
}否则if（temp< = 3）{
 return random（4,7）; 
} else {
 return random（8,10）; 
} 
  

这应该可以解决问题。

编辑：根据您的评论中的要求：

  int first_lo = 1，first_hi = 3000; // 1/2在[first_lo，first_hi]中选择数字的机会
 int second_lo = 3001，second_hi = 7000; // 1/6机会在[second_lo，second_hi]中选择一个数字
 int third_lo = 7001，third_hi = 10000; // 1/3机会在[third_lo，third_hi]中选择一个数字
 int第二个
 int temp = random（0,5）; 
 if（temp< = 2）{
 return random（first_lo，first_hi）; 
}否则if（temp< = 3）{
 return random（second_lo，second_hi）; 
} else {
 return random（third_lo，third_hi）; 
} 
  

We know that the classic range random function is like this:

public static final int random(final int min, final int max) {
    Random rand = new Random();
    return min + rand.nextInt(max - min + 1);  // +1 for including the max
}

I want to create algorithm function for generating number randomly at range between 1..10, but with uneven possibilities like:
1) 1,2,3 -> 3/6 (1/2)
2) 4,5,6,7 -> 1/6
3) 8,9,10 -> 2/6 (1/3)

Above means the function has 1/2 chance to return number between 1 and 3, 1/6 chance to return number between 4 and 7, and 1/3 chance to return number between 8 and 10.

Anyone know the algorithm?

UPDATE:
Actually the range between 1..10 is just served as an example. The function that I want to create would apply for any range of numbers, such as: 1..10000, but the rule is still same: 3/6 for top range (30% portion), 1/6 for middle range (next 40% portion), and 2/6 for bottom range (last 30% portion).

解决方案

Use the above algorithm:

int temp = random(0,5);
if (temp <= 2) {
  return random(1,3);
} else if (temp <= 3) {
 return random(4,7);
} else  {
 return random(8,10);
}

This should do the trick.

EDIT: As requested in your comment:

int first_lo = 1, first_hi = 3000; // 1/2 chance to choose a number in [first_lo, first_hi]
int second_lo = 3001, second_hi = 7000; // 1/6 chance to choose a number in [second_lo, second_hi] 
int third_lo = 7001, third_hi = 10000;// 1/3 chance to choose a number in [third_lo, third_hi] 
int second
int temp = random(0,5);
if (temp <= 2) {
  return random(first_lo,first_hi);
} else if (temp <= 3) {
 return random(second_lo,second_hi);
} else  {
 return random(third_lo,third_hi);
}

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