Find All Combinations of a List of Numbers with a Given Sum

Find all combinations of a list of numbers with a given sum

You could use itertools to iterate through every combination of every possible size, and filter out everything that doesn't sum to 10:

import itertools

numbers = [1, 2, 3, 7, 7, 9, 10]
target = 10

result = [seq for i in range(len(numbers), 0, -1)
          for seq in itertools.combinations(numbers, i)
          if sum(seq) == target]

print(result)

Result:

[(1, 2, 7), (1, 2, 7), (1, 9), (3, 7), (3, 7), (10,)]

Unfortunately this is something like O(2^N) complexity, so it isn't suitable for input lists larger than, say, 20 elements.

How to get all combination for a given sum with a given number of elements

A very simple approach is to use itertools.combinations to generate all the possible combinations, and select those that produce the desired sum.

>>> import itertools
>>> numbers = [0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9]
>>> target_sum = 1
>>> number = 3
>>> [c for c in itertools.combinations(numbers, number) if sum(c) == target_sum]
[(0.1, 0.2, 0.7), (0.1, 0.3, 0.6), (0.1, 0.4, 0.5), (0.2, 0.3, 0.5)]

If you want all the possible different orderings as distinct solutions, use itertools.permutations instead:

>>> [c for c in itertools.permutations(numbers, number) if sum(c) == target_sum]
[(0.1, 0.2, 0.7), (0.1, 0.3, 0.6), (0.1, 0.4, 0.5), (0.1, 0.5, 0.4), (0.1, 0.6, 0.3), (0.1, 0.7, 0.2), (0.2, 0.1, 0.7), (0.2, 0.3, 0.5), (0.2, 0.5, 0.3), (0.3, 0.1, 0.6), (0.3, 0.2, 0.5), (0.3, 0.5, 0.2), (0.4, 0.1, 0.5), (0.4, 0.5, 0.1), (0.5, 0.1, 0.4), (0.5, 0.2, 0.3), (0.5, 0.3, 0.2), (0.5, 0.4, 0.1), (0.6, 0.1, 0.3), (0.7, 0.1, 0.2)]

If you want to be able to use each element more than once, use itertools.combinations_with_replacement:

>>> [c for c in itertools.combinations_with_replacement(numbers, number) if sum(c) == target_sum]
[(0.1, 0.1, 0.8), (0.1, 0.2, 0.7), (0.1, 0.3, 0.6), (0.1, 0.4, 0.5), (0.2, 0.2, 0.6), (0.2, 0.3, 0.5), (0.2, 0.4, 0.4), (0.3, 0.3, 0.4)]

If you want to use elements multiple times and get all the possible orderings, use itertools.product:

>>> [c for c in itertools.product(numbers, repeat=number) if sum(c) == target_sum]
[(0.1, 0.1, 0.8), (0.1, 0.2, 0.7), (0.1, 0.3, 0.6), (0.1, 0.4, 0.5), (0.1, 0.5, 0.4), (0.1, 0.6, 0.3), (0.1, 0.7, 0.2), (0.1, 0.8, 0.1), (0.2, 0.1, 0.7), (0.2, 0.2, 0.6), (0.2, 0.3, 0.5), (0.2, 0.4, 0.4), (0.2, 0.5, 0.3), (0.2, 0.6, 0.2), (0.3, 0.1, 0.6), (0.3, 0.2, 0.5), (0.3, 0.3, 0.4), (0.3, 0.4, 0.3), (0.3, 0.5, 0.2), (0.4, 0.1, 0.5), (0.4, 0.2, 0.4), (0.4, 0.3, 0.3), (0.4, 0.4, 0.2), (0.4, 0.5, 0.1), (0.5, 0.1, 0.4), (0.5, 0.2, 0.3), (0.5, 0.3, 0.2), (0.5, 0.4, 0.1), (0.6, 0.1, 0.3), (0.6, 0.2, 0.2), (0.7, 0.1, 0.2), (0.8, 0.1, 0.1)]

Finding all possible combinations of numbers to reach a given sum

This problem can be solved with a recursive combinations of all possible sums filtering out those that reach the target. Here is the algorithm in Python:

def subset_sum(numbers, target, partial=[]):
    s = sum(partial)

    # check if the partial sum is equals to target
    if s == target: 
        print "sum(%s)=%s" % (partial, target)
    if s >= target:
        return  # if we reach the number why bother to continue
    
    for i in range(len(numbers)):
        n = numbers[i]
        remaining = numbers[i+1:]
        subset_sum(remaining, target, partial + [n]) 
   

if __name__ == "__main__":
    subset_sum([3,9,8,4,5,7,10],15)

    #Outputs:
    #sum([3, 8, 4])=15
    #sum([3, 5, 7])=15
    #sum([8, 7])=15
    #sum([5, 10])=15

This type of algorithms are very well explained in the following Stanford's Abstract Programming lecture - this video is very recommendable to understand how recursion works to generate permutations of solutions.

Edit

The above as a generator function, making it a bit more useful. Requires Python 3.3+ because of yield from.

def subset_sum(numbers, target, partial=[], partial_sum=0):
    if partial_sum == target:
        yield partial
    if partial_sum >= target:
        return
    for i, n in enumerate(numbers):
        remaining = numbers[i + 1:]
        yield from subset_sum(remaining, target, partial + [n], partial_sum + n)

Here is the Java version of the same algorithm:

package tmp;

import java.util.ArrayList;
import java.util.Arrays;

class SumSet {
    static void sum_up_recursive(ArrayList<Integer> numbers, int target, ArrayList<Integer> partial) {
       int s = 0;
       for (int x: partial) s += x;
       if (s == target)
            System.out.println("sum("+Arrays.toString(partial.toArray())+")="+target);
       if (s >= target)
            return;
       for(int i=0;i<numbers.size();i++) {
             ArrayList<Integer> remaining = new ArrayList<Integer>();
             int n = numbers.get(i);
             for (int j=i+1; j<numbers.size();j++) remaining.add(numbers.get(j));
             ArrayList<Integer> partial_rec = new ArrayList<Integer>(partial);
             partial_rec.add(n);
             sum_up_recursive(remaining,target,partial_rec);
       }
    }
    static void sum_up(ArrayList<Integer> numbers, int target) {
        sum_up_recursive(numbers,target,new ArrayList<Integer>());
    }
    public static void main(String args[]) {
        Integer[] numbers = {3,9,8,4,5,7,10};
        int target = 15;
        sum_up(new ArrayList<Integer>(Arrays.asList(numbers)),target);
    }
}

It is exactly the same heuristic. My Java is a bit rusty but I think is easy to understand.

C# conversion of Java solution: (by @JeremyThompson)

public static void Main(string[] args)
{
    List<int> numbers = new List<int>() { 3, 9, 8, 4, 5, 7, 10 };
    int target = 15;
    sum_up(numbers, target);
}

private static void sum_up(List<int> numbers, int target)
{
    sum_up_recursive(numbers, target, new List<int>());
}

private static void sum_up_recursive(List<int> numbers, int target, List<int> partial)
{
    int s = 0;
    foreach (int x in partial) s += x;

    if (s == target)
        Console.WriteLine("sum(" + string.Join(",", partial.ToArray()) + ")=" + target);

    if (s >= target)
        return;

    for (int i = 0; i < numbers.Count; i++)
    {
        List<int> remaining = new List<int>();
        int n = numbers[i];
        for (int j = i + 1; j < numbers.Count; j++) remaining.Add(numbers[j]);

        List<int> partial_rec = new List<int>(partial);
        partial_rec.Add(n);
        sum_up_recursive(remaining, target, partial_rec);
    }
}

Ruby solution: (by @emaillenin)

def subset_sum(numbers, target, partial=[])
  s = partial.inject 0, :+
# check if the partial sum is equals to target

  puts "sum(#{partial})=#{target}" if s == target

  return if s >= target # if we reach the number why bother to continue

  (0..(numbers.length - 1)).each do |i|
    n = numbers[i]
    remaining = numbers.drop(i+1)
    subset_sum(remaining, target, partial + [n])
  end
end

subset_sum([3,9,8,4,5,7,10],15)

Edit: complexity discussion

As others mention this is an NP-hard problem. It can be solved in exponential time O(2^n), for instance for n=10 there will be 1024 possible solutions. If the targets you are trying to reach are in a low range then this algorithm works. So for instance:

subset_sum([1,2,3,4,5,6,7,8,9,10],100000) generates 1024 branches because the target never gets to filter out possible solutions.

On the other hand subset_sum([1,2,3,4,5,6,7,8,9,10],10) generates only 175 branches, because the target to reach 10 gets to filter out many combinations.

If N and Target are big numbers one should move into an approximate version of the solution.

Given a list of numbers, how do you create all the combinations of sums and return a list of those sum

Using itertools.permutations:

>>> import itertools
>>> nums = [1, 2, 3]
>>> [sum(p) for n in range(1, len(nums)+1) for p in itertools.permutations(nums, n)]
[1, 2, 3, 3, 4, 3, 5, 4, 5, 6, 6, 6, 6, 6, 6]

If you want to see exactly what permutations is doing rather than just seeing the final sums, you can do fun things with map and join:

>>> [f"{'+'.join(map(str, p))}={sum(p)}" for n in range(1, len(nums)+1) for p in itertools.permutations(nums, n)]
['1=1', '2=2', '3=3', '1+2=3', '1+3=4', '2+1=3', '2+3=5', '3+1=4', '3+2=5', '1+2+3=6', '1+3+2=6', '2+1+3=6', '2+3+1=6', '3+1+2=6', '3+2+1=6']

Generating all combinations of list of numbers where the sum of the combination is = N

A simple k-combination of a finite set S is a subset of k distinct elements of S. Specifying a subset does not arrange them in a particular order.

You can use the CombinatoricsLib. CombinatoricsLib is a java library for generating combinatorial objects. https://code.google.com/p/combinatoricslib/

Using this:

    public static void main(String[] args) {

           // Create the initial vector
           ICombinatoricsVector<Integer> initialVector = Factory.createVector(
              new Integer[]  {1, 5, 4, 3, 6, 9, 7, 4, 3, 8, 2, 1, 6, 3, 7} );          

           int subsetMaxSize = 5;
           int upperLimit = 10;  
           int lowerLimit = 8;
           for(int i = 1; i <= subsetMaxSize; i++)
           {
               Generator<Integer> gen = Factory.createSimpleCombinationGenerator(initialVector, i);
               for (ICombinatoricsVector<Integer> combination : gen)
               {
                   int sum = vectorSum(combination);
                   if(validateSum(sum, lowerLimit, upperLimit))
                       printVector(combination);
               }   
           }
    }

    public static boolean validateSum(Integer value, Integer lowerLimit, Integer upperLimit)
    {
        if(value <= upperLimit && value > lowerLimit)   
            return true;
        return false;           
    }

    public static Integer vectorSum(ICombinatoricsVector<Integer> vect)
    {
        Integer sum = 0;    
        for(int i = 0; i < vect.getSize(); i++) 
            sum += vect.getValue(i);
        return sum;
    }

    public static void printVector(ICombinatoricsVector<Integer> vect)
    {
        String output = ""; 
        for(int i = 0; i < vect.getSize(); i++)
            output += vect.getValue(i) + ", ";
        System.out.println(output);
    }

will return the output

9, 
1, 9, 
1, 8, 
5, 4, 
5, 4, 
4, 6, 
4, 6, 
3, 6, 
3, 7, 
3, 6, 
3, 7, 
6, 4, 
6, 3, 
6, 3, 
9, 1, 
7, 3, 
7, 2, 
7, 3, 
4, 6, 
3, 6, 
3, 7, 
8, 2, 
8, 1, 
2, 7, 
6, 3, 
3, 7, 
1, 5, 4, 
1, 5, 3, 
1, 5, 4, 
1, 5, 3, 
1, 5, 3, 
1, 4, 4, 
1, 3, 6, 
1, 3, 6, 
1, 6, 3, 
1, 6, 2, 
1, 6, 3, 
1, 7, 2, 
1, 7, 1, 
1, 3, 6, 
1, 8, 1, 
1, 2, 6, 
1, 2, 7, 
1, 1, 7, 
1, 6, 3, 
5, 4, 1, 
5, 3, 2, 
5, 3, 1, 
5, 4, 1, 
5, 3, 2, 
5, 3, 1, 
5, 2, 3, 
5, 1, 3, 
4, 3, 3, 
4, 3, 2, 
4, 3, 3, 
4, 4, 2, 
4, 4, 1, 
4, 3, 2, 
4, 3, 3, 
4, 2, 3, 
3, 6, 1, 
3, 4, 3, 
3, 4, 2, 
3, 4, 3, 
3, 3, 3, 
3, 1, 6, 
6, 3, 1, 
6, 2, 1, 
6, 1, 3, 
7, 2, 1, 
4, 3, 2, 
4, 3, 3, 
4, 2, 3, 
3, 1, 6, 
2, 1, 6, 
2, 1, 7, 
1, 6, 3, 
1, 5, 3, 1, 
1, 5, 3, 1, 
1, 5, 2, 1, 
1, 5, 1, 3, 
1, 4, 3, 2, 
1, 4, 3, 1, 
1, 4, 4, 1, 
1, 4, 3, 2, 
1, 4, 3, 1, 
1, 4, 2, 3, 
1, 4, 1, 3, 
1, 3, 4, 2, 
1, 3, 4, 1, 
1, 3, 3, 2, 
1, 3, 3, 3, 
1, 3, 2, 3, 
1, 6, 2, 1, 
1, 4, 3, 2, 
1, 4, 3, 1, 
1, 4, 2, 3, 
1, 4, 1, 3, 
1, 3, 2, 3, 
1, 2, 1, 6, 
4, 3, 2, 1, 
4, 3, 2, 1, 
4, 2, 1, 3, 
3, 4, 2, 1, 
3, 3, 2, 1, 
3, 3, 1, 3, 
3, 2, 1, 3, 
4, 3, 2, 1, 
4, 2, 1, 3, 
3, 2, 1, 3, 
1, 3, 3, 2, 1, 
1, 3, 2, 1, 3, 
1, 3, 2, 1, 3, 

Finding all possible combinations of numbers of an array to reach a given sum

Took me while to code this. So it's basically brute force. I recursively (backtracking) generate all possible expression with the operators given and then evaluate them. Note these are just infix expression(s).

Now this is a very slow solution. There are several optimization one can do here.

vector<string> allCombinations(vector<int> &arr, int k)
{
    int n = (int)arr.size();
    string operators = "+-*";
    vector<string> ans;
    // To check precedence of operators
    auto prec = [&](char op) -> int
    {
        if (op == '*' or op == '/') return 2;
        if (op == '+' or op == '-') return 1;
        return -1;
    };
    // For infix evaluation (kindof a helper function)
    auto compute = [&](int v1, char op, int v2) -> int
    {
        if (op == '+') return v1 + v2;
        if (op == '-') return v1 - v2;
        if (op == '*') return v1 * v2;
        if (op == '/') return v1 / v2;
        assert(false);
        return INT_MAX;
    };
    // Main infix evaluation function
    auto evaluate = [&](string s) -> int
    {
        int len = (int)s.size();
        // vector is being used as a STACK
        vector<int> val;
        vector<char> ops;
        for (int i = 0; i < len; i++)
        {
            char curr = s[i];
            if (curr == ' ') continue;
            if (isdigit(curr))
            {
                int v = 0;
                while (i < len and isdigit(s[i])) v = 10 * v + (s[i++] - '0');
                val.push_back(v);
                i--;
            }
            else
            {
                while (!ops.empty() and prec(curr) <= prec(ops.back()))
                {
                    int v1 = val.back();
                    val.pop_back();
                    int v2 = val.back();
                    val.pop_back();
                    char op = ops.back();
                    ops.pop_back();
                    val.push_back(compute(v2, op, v1));
                }
                ops.push_back(curr);
            }
        }
        while (!ops.empty())
        {
            int v1 = val.back();
            val.pop_back();
            int v2 = val.back();
            val.pop_back();
            char op = ops.back();
            ops.pop_back();
            val.push_back(compute(v2, op, v1));
        }
        return val.back();
    };
    // Generates all expression possible
    function<void(int, string&)> generate = [&](int i, string &s) -> void
    {
        s += to_string(arr[i]);
        if (i == n - 1)
        {
            if (evaluate(s) == k) ans.push_back(s);
            // Backtrack
            s.pop_back();
            return;
        }
        for (char &ops : operators) 
        {
            s.push_back(ops);
            generate(i + 1, s);
            // Backtrack
            s.pop_back();
        }
        // Backtrack
        s.pop_back();
    };
    string s;
    // Try all combinations
    sort(arr.begin(), arr.end());
    do
    {
        generate(0, s);
    } while (next_permutation(arr.begin(), arr.end()));
    return ans;
}

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