在具有平均“x"的2个数字之间获取“n"随机值 [英] Get `n` random values between 2 numbers having average `x`

问题描述

我想得到 n 个介于 1 到 5(包括两者)之间的随机数(例如 n=16)(整数)，因此平均值为 x.

I want to get n random numbers(e.g n=16)(whole numbers) between 1 to 5(including both) so that average is x.

x 可以是 (1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5) 之间的任何值.

x can be any value between (1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5).

我正在使用 PHP.

例如假设我的平均 x=3.

e.g. Suppose I have average x= 3.

然后需要 16 个介于 1 到 5 之间的整数(包括两者).比如 (1,5,3,3,3,3,2,4,2,4,1,5,1,5,3,3)

Then required 16 whole numbers between 1 to 5(including both). like (1,5,3,3,3,3,2,4,2,4,1,5,1,5,3,3)

更新:

如果 x=3.5 表示 16 个数字的平均值应该在 3.5 到 4 之间.
如果 x=4 表示 16 个数字的平均值应该在 4 到 4.5 之间
如果 x=5 表示所有数字都是 5

if x=3.5 means average of 16 numbers should be between 3.5 to 4.
and if x=4 means average of 16 numbers should be between 4 to 4.5
and if x=5 means all numbers are 5

推荐答案

这个答案允许目标平均值的任何值(无论 n 是奇数还是偶数)，并避免使用递归来优化性能.

This answer allows any value for the target average (regardless of whether n is odd or even) and avoids the use of recursion to optimize performance.

函数

function getRandomNumbersWithAverage($target_average, $n, $min=1, $max=5)
{

  if($min>$max) list($min, $max) = array($max, $min);
  if($target_average<$min || $target_average>$max) return false;
  else if($target_average==$min) return array_fill(0, $n, $min);
  else if($target_average==$max) return array_fill(0, $n, $max);

  if($n<1) return false;
  else if($n==1) return array($target_average);
  else
  {
    $numbers = array();
    for($i=0;$i<$n;$i++)
    {
      $sum = array_sum($numbers);
      $average = $i ? $sum/($i+1) : ($min+$max)/2;
      $contrived_number = $target_average*($i+1) - $sum;
      // Last one must be contrived
      if($i==$n-1) $new_number = ceil($contrived_number); // Round up
      else
      {
        // The tolerance gets smaller with iteration
        $tolerance = ($max-$min)*(1-($i/($n-1)));
        $temp_min = ($contrived_number-$tolerance);
        if($temp_min<$min) $temp_min = $min;
        $temp_max = ($contrived_number+$tolerance);
        if($temp_max>$max) $temp_max = $max;
        $new_number = mt_rand($temp_min, $temp_max);
      }
      if($new_number==0) $new_number = 0; // Handle -0
      $numbers[] = $new_number;
    }
    // Since the numbers get more contrived towards the end, it might be nice to shuffle
    shuffle($numbers);
    return $numbers;
  }
}

示例输出:

getRandomNumbersWithAverage(1, 12)

produced the numbers: (1,1,1,1,1,1,1,1,1,1,1,1) having an average of: 1


getRandomNumbersWithAverage(1.1, 13)

produced the numbers: (1,1,1,1,1,1,1,4,1,1,1,0,1) having an average of: 1.1538461538462


getRandomNumbersWithAverage(2.7, 14)

produced the numbers: (3,3,2,5,1,2,4,3,3,2,3,3,3,1) having an average of: 2.7142857142857


getRandomNumbersWithAverage(2.7, 15)

produced the numbers: (3,3,4,3,4,2,1,1,3,2,4,1,5,1,4) having an average of: 2.7333333333333


getRandomNumbersWithAverage(3.5, 16)

produced the numbers: (5,5,4,3,1,5,5,1,2,5,3,3,4,4,4,2) having an average of: 3.5


getRandomNumbersWithAverage(3.5, 17)

produced the numbers: (5,2,3,5,4,1,2,3,5,4,5,4,2,3,5,3,4) having an average of: 3.5294117647059


getRandomNumbersWithAverage(4, 18)

produced the numbers: (3,5,5,3,5,5,3,4,4,4,5,2,5,1,5,4,5,4) having an average of: 4


getRandomNumbersWithAverage(4.9, 19)

produced the numbers: (5,5,5,5,7,5,5,5,5,6,5,3,5,5,3,5,5,5,5) having an average of: 4.9473684210526


getRandomNumbersWithAverage(5, 20)

produced the numbers: (5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5) having an average of: 5


getRandomNumbersWithAverage(0.5, 10)

does not produce numbers


getRandomNumbersWithAverage(0, 9)

does not produce numbers


getRandomNumbersWithAverage(-1, 8)

does not produce numbers


getRandomNumbersWithAverage(5.5, 7)

does not produce numbers


getRandomNumbersWithAverage(6, 6)

does not produce numbers


getRandomNumbersWithAverage(6, 5, 1, 7)

produced the numbers: (7,7,2,7,7) having an average of: 6


getRandomNumbersWithAverage(6, 5, 1, 6)

produced the numbers: (6,6,6,6,6) having an average of: 6


getRandomNumbersWithAverage(3, 1)

produced the numbers: (3) having an average of: 3

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