将数字拆分为 4 个随机数 [英] Split number into 4 random numbers
问题描述
我想将 10 拆分成 4 个随机数的数组，但都不能是 0 或高于 4.例如 [1,2,3,4]、[1,4,4,1] 或 [4,2,3,1]代码>.
I want to split 10 into an array of 4 random numbers, but neither can be 0 or higher than 4. For example [1,2,3,4], [1,4,4,1] or [4,2,3,1].
我认为这是一个简单的问题，但出于某种原因，我想不出如何做到这一点.如果有人有一些非常有帮助的说明！
I think it's an easy question, but for some reason I can't think of how to do this. If someone has some instruction that would be very helpful!
这是我现在拥有的代码，但我也生成了 10 以下的总数:
This is the code I have now, but I generates also a total number under 10:
let formation = [];
let total = 0;
for (let i = 0; i < 4; i ++) {
if (total < 9) {
formation[i] = Math.floor(Math.random() * 4) + 1;
} else {
formation[i] = 1;
}
}
推荐答案
您可以创建所有可能的组合并选择一个随机数组.
You could create all possible combinations and pick a random array.
function get4() {
function iter(temp) {
return function (v) {
var t = temp.concat(v);
if (t.length === 4) {
if (t.reduce(add) === 10) {
result.push(t);
}
return;
}
values.forEach(iter(t));
};
}
const
add = (a, b) => a + b,
values = [1, 2, 3, 4],
result = [];
values.forEach(iter([]));
return result;
}
console.log(get4().map(a => a.join(' ')));.as-console-wrapper { max-height: 100% !important; top: 0; }它的工作原理是使用随机值的因子和偏移量，基于实际总和、索引、下一个索引所需的最小总和以及最大总和.
It works by using a factor for the random value and an offset, based on the actual sum, index, minimum sum which is needed for the next index, and the maximum sum.
偏移量通常是最小和，或者和和最大和之差的较大值.为得到因子，取三个值作为乘以随机值的最小值.
The offset is usually the minimum sum, or the greater value of the difference of sum and maximum sum. For getting the factor, three values are taken for the minimum for multiplying the random value.
该表说明了总和的所有可能值和所需的迭代，基于给定的值和获取所有值的迭代.
The table illustrates all possible values of the sum and the needed iterations, based on a given value and the iteration for getting all values.
一开始，总和是小部分分布的值.结果是剩余和为 14 ... 10 的第二个块，因为可以取值 1 ... 5.第三轮遵循相同的规则.最后，将剩余的总和作为该值的偏移量.
At the beginning the sum is the value for distribution in small parts. The result is the second block with a rest sum of 14 ... 10, because it is possible to take a value of 1 ... 5. The third round follows the same rules. At the end, the leftover sum is taken as offset for the value.
具有 1、...、5 值和 5 元素且总和为 15 的示例以及所有可能性:
An example with 1, ..., 5 values and 5 elements with a sum of 15 and all possibilities:
min: 1
max: 5
length: 5
sum: 15
smin = (length - index - 1) * min
smax = (length - index - 1) * max
offset = Math.max(sum - smax, min)
random = 1 + Math.min(sum - offset, max - offset, sum - smin - min)
index sum sum min sum max random offset
------- ------- ------- ------- ------- -------
_ 0 15 4 20 5 1
1 14 3 15 5 1
1 13 3 15 5 1
1 12 3 15 5 1
1 11 3 15 5 1
_ 1 10 3 15 5 1
2 13 2 10 3 3
2 12 2 10 4 2
2 11 2 10 5 1
2 10 2 10 5 1
2 9 2 10 5 1
2 8 2 10 5 1
2 7 2 10 5 1
2 6 2 10 4 1
_ 2 5 2 10 3 1
3 10 1 5 1 5
3 9 1 5 2 4
3 8 1 5 3 3
3 7 1 5 4 2
3 6 1 5 5 1
3 5 1 5 4 1
3 4 1 5 3 1
3 3 1 5 2 1
_ 3 2 1 5 1 1
4 5 0 0 1 5
4 4 0 0 1 4
4 3 0 0 1 3
4 2 0 0 1 2
4 1 0 0 1 1
示例代码采用目标 1, ..., 4 长度为 4 部分和 的总和10.
The example code takes the target 1, ..., 4 with a length of 4 parts and a sum of 10.
function getRandom(min, max, length, sum) {
return Array.from(
{ length },
(_, i) => {
var smin = (length - i - 1) * min,
smax = (length - i - 1) * max,
offset = Math.max(sum - smax, min),
random = 1 + Math.min(sum - offset, max - offset, sum - smin - min),
value = Math.floor(Math.random() * random + offset);
sum -= value;
return value;
}
);
}
console.log(Array.from({ length: 10 }, _ => getRandom(1, 4, 4, 10).join(' ')));.as-console-wrapper { max-height: 100% !important; top: 0; }这篇关于将数字拆分为 4 个随机数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
