Divide the digit into four separate random figures

I am looking to divide the number 10 into an array of 4 random numbers, where each number is between 1 and 4, excluding 0. Examples of valid arrays include [1,2,3,4], [1,4,4,1], or [4,2,3,1].

While I believe this might be a simple task, I'm having trouble figuring out the solution. Any guidance on how to achieve this would be greatly appreciated!

Edit: Here is the code that I currently have, but it sometimes generates a total sum greater than 10:

  let arrangement = [];
  let total = 0;

   for (let i = 0; i < 4; i ++) {
    if (total < 9) {
      arrangement[i] = Math.floor(Math.random() * 4) + 1; 
    } else {
      arrangement[i] = 1;
    }
  }
javascriptarraysalgorithmmathrandom

Answer №1

To generate random arrays, you have the option of creating all possible combinations and then selecting a random array from them.

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; }

A unique approach for obtaining random values without relying on a predefined list of combinations

This technique involves using a factor for the random value and an offset, determined by factors including actual sum, index, minimum required sum for next index, and maximum sum.

The offset typically equals to either the minimum sum or the greater value between the difference of total sum and maximum sum. To calculate the factor, three values are considered as multiplicands for the random value.

The table showcases various sums and corresponding iterations needed based on specified values and iteration count for generating all values.

Initially, the sum represents the value distributed in smaller units. The resulting block follows with a remaining sum ranging from 14 to 10, achievable through values from 1 to 5. Subsequent rounds adhere to the same guidelines. Ultimately, any leftover sum is utilized as an offset for determining the value.

An instance featuring values 1 through 5, 5 elements totaling 15, and exhaustive options:

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

In the provided example code, select numbers ranging from 1 to 4, with 4 parts and total 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; }

Answer №2

While arriving a little late to the party, I've thoroughly enjoyed tackling this task. My approach is designed to be efficient even with large values and doesn't rely on luck to find a random match. It's compact, unbiased, and only creates the necessary partitions.

As long as the value of max isn't overly limiting, this method should work smoothly.

const len = 4;
const total = 10;
const max = 4;

let arr = new Array(len);
let sum = 0;
do {
  // generating random numbers
  for (let i = 0; i < len; i++) {
    arr[i] = Math.random();
  }
  // calculating total of random numbers
  sum = arr.reduce((acc, val) => acc + val, 0);
  // determining scale to use on the numbers
  const scale = (total - len) / sum;
  // scaling the array
  arr = arr.map(val => Math.min(max, Math.round(val * scale) + 1));
  // re-calculating the sum
  sum = arr.reduce((acc, val) => acc + val, 0);
  // looping if sum is not exactly the expected total due to scale rounding effects
} while (sum - total);

console.log(arr);

Answer №3

The most straightforward approach is to use brute force.

  1. Begin by creating a while loop for your calculations
  2. Inside the loop, generate an empty array and populate it with random values until the desired length is reached
  3. Check if the sum of the array equals your target value; if so, terminate the loop

This process will continue executing until a solution is achieved.

However, there are two factors to keep in mind:

  1. You can easily verify if a solution is feasible by ensuring that multiplying the length-of-array by minimum-value does not exceed the sum, and multiplying it by maximum-value does not fall below the sum.
  2. A loop based on random conditions could potentially run indefinitely, so setting a maximum number of iterations may be advisable.

Both considerations are addressed in the following code snippet:

function generateRandomNumber(max, min) {
  while (true) {
    var r = Math.round(Math.random() * max);
    if (r >= min) {
      return r;
    }
  }
}

function splitNumberIntoComponentsInRange(numberToSplit, numberOfSplits, maxValue, minValue, onUpdate) {
  if (minValue === void 0) {
    minValue = 1;
  }
  //Ensuring a possible result
  if (maxValue * numberOfSplits < numberToSplit || minValue * numberOfSplits > numberToSplit) {
    return new Promise(function(resolve, reject) {
      resolve(false);
    });
  }
  //Initializing return array
  var arr = [];
  var accumulator = 0;
  while (arr.length < numberOfSplits) {
    var val = generateRandomNumber(Math.floor(numberToSplit / numberOfSplits), minValue);
    accumulator += val;
    arr.push(val);
  }
  return new Promise(function(resolve, reject) {
    function executeTest() {
      var d = Date.now();
      var localMaxValue = Math.min(maxValue, Math.ceil((numberToSplit - accumulator) / 4));
      //Combination loop
      while (accumulator < numberToSplit && Date.now() - d < 17) {
        var index = Math.round(Math.random() * (arr.length - 1));
        if (arr[index] >= maxValue) {
          continue;
        }
        var r = generateRandomNumber(localMaxValue, minValue);
        while (arr[index] + r > maxValue || accumulator + r > numberToSplit) {
          if (Date.now() - d >= 17) {
            break;
          }
          r = generateRandomNumber(localMaxValue, minValue);
        }
        if (arr[index] + r > maxValue || accumulator + r > numberToSplit) {
          continue;
        }
        arr[index] += r;
        accumulator += r;
      }
      if (accumulator < numberToSplit) {
        if (onUpdate !== void 0) {
          onUpdate(arr);
        }
        requestAnimationFrame(executeTest);
      } else {
        resolve(arr);
      }
    }
    executeTest();
  });
}
//TEST
var table = document.body.appendChild(document.createElement('table'));
table.innerHTML = "<thead><tr><th>Number to split</th><th>Number of splits</th><th>Max value</th><th>Min value</th><th>Run</th></tr></thead>" +
  "<tbody><tr><th><input id=\"number-to-split\" value=\"10\" type=\"number\" min=\"1\"/></th><th><input id=\"number-of-splits\" value=\"4\" type=\"number\" min=\"1\"/></th><th><input id=\"max-value\" type=\"number\" min=\"1\" value=\"4\"/></th><th><input id=\"min-value\" type=\"number\" min=\"1\" value=\"1\"/></th><th><input id=\"run\" type=\"button\" value=\"Run\"/></th></tr></tbody>";
var output = document.body.appendChild(document.createElement('pre'));
output.style.overflowX = "scroll";
document.getElementById("run").onclick = function() {
  splitNumberIntoComponentsInRange(parseInt(document.getElementById("number-to-split").value, 10), parseInt(document.getElementById("number-of-splits").value, 10), parseInt(document.getElementById("max-value").value, 10), parseInt(document.getElementById("min-value").value, 10))
    .then(function(data) {
      if (data !== false) {
        output.textContent += data.join("\t") + '\n';
      } else {
        output.textContent += 'Invalid data\n';
      }
    });
};

EDIT 1 - Handling Large Calculations

By utilizing requestAnimationFrame and Promises, the code now operates asynchronously, enabling longer computation times without disrupting the user.

I have also modified the random function to adjust based on the remaining range, significantly reducing the total calculations needed for larger numbers.

Answer №4

To solve this problem, you will first need to calculate the partitions of the number 10 (refer to this link for more information). Once you have determined all the partitions, apply your conditions on the resulting set.

// Here is a code snippet to generate partitions
// Source: https://gist.github.com/k-hamada/8aa85ac9b334fb89ac4f

function* partitions(n) {

    if (n <= 0) throw new Error('positive integer only');
    yield [n];

    var x = new Array(n);
    x[0] = n;
    for (var i = 1; i < n; i++) x[i] = 1;

    var m = 0, h = 0, r, t;
    while (x[0] != 1) {
        if (x[h] == 2) {
            m += 1;
            x[h] = 1;
            h -= 1;
        } else {
            r = x[h] - 1;
            x[h] = r;

            t = m - h + 1;
            while (t >= r) {
                h += 1;
                x[h] = r;
                t -= r;
            }
            m = h + (t !== 0 ? 1 : 0);
            if (t > 1) {
                h += 1;
                x[h] = t;
            }
        }
        yield x.slice(0, m + 1);
    }
}

results = [];
// Retrieve all possible partitions for the given number
for (var partition of partitions(10)) {
    // Apply the specified conditions (4 numbers in each partition, none greater than 4)
    if(partition.length != 4 || partition.some((x) => x > 4)) continue;
    results.push(partition);
}
console.log(results);

Answer №5

Given the following scenario:

If you have a collection of n positive numbers that add up to S, at least one number in the collection will be less than the average S divided by n (S/n)

and if you require a result set containing exactly 4 numbers,

You can implement the algorithm described below:

  1. Choose a random number from the range [1, floor(S/n)]. For example, if floor(10/4) = 2, select a random number between [1,2]. Let this be x1.
  2. Select another random number from the range [1, floor((S - x1)/(n - 1))], and denote it as x2.
  3. Pick yet another random number from the range [1, floor((S - x1 - x2)/(n - 2))].
  4. Repeat this process until you obtain x(n-1).
  5. Calculate the final number by subtracting x1, x2 .... x(n-1) from S.

Finally, enhance the above algorithm by adding a condition that sets an upper limit for the random numbers.

In n steps, you will have your collection.

    function getRandomInt(min, max) {
       return Math.floor(Math.random() * (max - min + 1)) + min;
    }

    function getRandomCollection(min, max, length, sum) {
        var collection = [];
        var leftSum = sum - (min - 1);

        for(var i = 0; i < length - 1; i++) {
             var number = getRandomInt(min, Math.min(Math.ceil(leftSum/(length - i)), max));
             leftSum -= number;
             collection.push(number);
        }
        leftSum += min - 1;
        while(leftSum > max) {
             var randomIndex = Math.floor(Math.random() * collection.length);
             if(collection[randomIndex] < max) {
                  collection[randomIndex]++;
                  leftSum--;
             }
        }
        
        collection.push(leftSum);
        return collection;
    }
    console.log(getRandomCollection(1, 4, 4, 10).join(' + ') + ' = 10');
    console.log(getRandomCollection(3, 20, 10, 100).join(' + ') + ' = 100');

Reference

Check out my answer using the same algorithm for a different question

Answer №6

Efficient yet with a bias and unpredictable ending

function createPartition(sum, len, min, max) {
    const array = Array(len).fill(min)
    while (array.reduce((acc,val)=>acc+val) < sum) {
        const index = Math.random()*len|0
        if (array[index] < max) array[index]++
    }
    return array
}

console.log(Array(10).fill().map(_=>createPartition(10, 4, 1, 4).join(' ')))
.as-console-wrapper { max-height: 100% !important; top: 0; }

The while loop has the potential to run infinitely with a very low probability. To avoid this issue, you could maintain another array of "valid indexes" and remove elements from it as values reach their maximum.

Answer №7

As a middle school math teacher who unexpectedly came across this page, I have been pondering the effectiveness of the following approach:

numbers = [1, 1, 1, 1];
completed = [0, 0, 0, 0];
for (index = 0; index < 6; index++) {
    do {
        current = Math.floor(Math.random() * 4);
    } while (completed[current] == 1);
    numbers[current] = numbers[current] + 1;
    if (numbers[current] == 4) {
        completed[current] = 1;
    }
}

Answer №8

Here is a simple way to generate a random number between 1 and 4:

You can adjust this code snippet within a function to suit your specific requirements for generating arrays.

Math.floor(Math.random() * 4) + 1 


var randomNumber = Math.floor(Math.random() * 4) + 1 ;
console.log(randomNumber);

Answer №9

This task proved to be too simple.

var data = null;
while(true) {
    var sum = 0;
    var expectedSum = 10;
    data = [];
    while(expectedSum !== sum) {
        //var item = Math.floor(Math.random() * 9) + 1;
        var item = Math.floor(Math.random() * 4) + 1;
        if(item + sum > expectedSum) {
            continue;
        }
        sum += item;
        data.push(item);
    }
    if(data.length === 4) {
        break;
    } else {
        console.log('false iteration')
    }
}
console.log(data);

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