Calculate all series combinations

Question

I have a list of items, and each item has a quantity.

var items = {
    1: 12,   // we have 12 x item1
    2: 1,    // we have 1 x item2
    3: 1,
    4: 7,
    5: 2,
    6: 2
};

Alternatively, this can be considered as:

var items = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
             2, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6];

How would you like to get a list of each combination of these elements, bearing in mind that order is completely unimportant (and therefore [1,2,3] == [3,2,1]), and that not every element should exist as a result.

I assume the result might look like this:

[1]
[1, 1]
[1, 2]
[1, 3]
...

or even better:

{1 : 1}          // 1 x item1
{1 : 2}          // 2 x item1 
{1 : 1, 2 : 1}   // 1 x item1, 1 x item2
{1 : 1, 3 : 1}
....

Answer 1

I assume that the quantity of each element is limited.

: 1, . , 1s 2 . , , 2 . 2s , 3. ...

.

---

, ... - ;-) , - " ".

javaScript:

var limits = [1, 3, 5, 2];

function out(arr){
  var text = '';
  for (var i=0; i < arr.length; i++){
    text += arr[i] + '.'
  }
  var log = document.getElementById('log');
  var p = document.createElement('p');
  log.appendChild(p);
  p.innerHTML = '<span>' + text + '</span>';
}

function generateNextSet(set){
  for (var i = 0; i < set.length; i++){
    var amount = set[i];
    if (amount + 1 > limits[i]){
      set[i] = 0;
    } else {
      set[i] = amount + 1;
      return set;
    }
  }
  return false;
}

function generateSets(){
  var initial_set = [0, 0, 0, 0]
  var set = generateNextSet(initial_set);
  out(set);
  while (set = generateNextSet(set)) {
    out(set);
  }
};

div "log" - generateSets(), .

Answer 2

: , , :)

---

, ;

var items {
    1 : 1,
    2 : 1,
    4 : 1,
    8 : 1,
    16: 1,
    32: 1
};

, :

• 0 ([] - )
• 1 ([1])
• 2 ([2])
• 3 ([2, 1])
• 4 ([4])
• .

, , , mixed-radix.

, 0 MAX. , , , .

var items = {
    1: 12,   // we have 12 x item1
    2: 1,    // we have 1 x item2
    3: 1,
    4: 7,
    5: 2,
    6: 2
};

var counter = {
    1: 0,
    2: 0,
    3: 0,
    4: 0,
    5: 0,
    6: 0
};

function increment(digit) {
    if (digit > 6) {
        return false;
    }

    var value = counter[digit] + 1;

    if (value > items[digit]) {
        counter[digit] = 0;
        return increment(digit + 1);
    }

    counter[digit] = value;

    return true;
}

while (increment(1)) {
    var set = [];

    for (var digit in counter) {
        var value = counter[digit];

        for (var i = 0; i < value; i++) {
            set.push(digit);
        }
    }

    document.write("<div>" + set + "</div>");
}

:

1
1,1
1,1,1

---- snip ----

2
1,2
1,1,2

---- big snip ----

1,1,1,1,1,1,1,1,2,3,4,4,4,4,4,4,4,5,5,6,6
1,1,1,1,1,1,1,1,1,2,3,4,4,4,4,4,4,4,5,5,6,6
1,1,1,1,1,1,1,1,1,1,2,3,4,4,4,4,4,4,4,5,5,6,6
1,1,1,1,1,1,1,1,1,1,1,2,3,4,4,4,4,4,4,4,5,5,6,6
1,1,1,1,1,1,1,1,1,1,1,1,2,3,4,4,4,4,4,4,4,5,5,6,6

Answer 3

.

, n 1, : [5,6] → [5,5,6], [5,6,6], [5,5, 6,6].

[]
[1] -> [1,1], [1,1,1] etc
  [1,2] -> [1,1,2], ...
  [1,3] -> [1,1,3]
  [1,4] -> [1,1,4], ...., [1,4,4], -- all combinations of all multi quantity
[2]
[3]
[4] -> [4,4], [4,4,4] etc
[5] -> [5,5]
[6] -> [6,6]

Etc...

():

Combinations: {N -> N} -> [[N]]
Combinations(s) == CombinationsX(s, [])

CombinationsX: {N -> N} X [N] -> [[N]]
Combinationsx(s, g) ==
  if s = {} then return []
  else
    {a -> b} = hd s
    ts = tl s
    res = Combinationsx(ts, g) 
    for q in 1..b do
      g = g + [a]
      res = res ++ Combinationsx(ts, g)
    return res      

Answer 4

A good resource for is the algorithm described by Kenneth Rosen in Discrete Mathematics and its applications . Many problems can use this general algorithm, and it's nice to have it in your toolbox.