Feed Icon RSS 1.0 XML Feed available

Sums of numbers that add to X

Date: 10-Oct-2004/17:45

Tags: , ,

Characters: (none)

I saw the following coding problem:
Write a function that takes 4 integers as input and returns an integer. The four input values will each range from 1-10, and the function should calculate the number of combinations of input that add to 15 and return that number of combinations.
For example, if the four input values were labeled A-D and had the values A=5, B=5, C=5 and D=10 there would be 4 combinations adding to 15.
  • A+D = 15
  • B+D = 15
  • C+D = 15
  • A+B+C = 15
So the desired output would be: 4
Please describe how you would modify your solution to accept N input values (instead of just four).
My first try turned out to be slower than a solution from an experienced C++ programmer who'd leveraged the C++ standard library algorithms. Here's how he did it:
class Knapsack
  Knapsack (
    std::vector::const_iterator beg,
    std::vector::const_iterator end,
    int sum
    : _pieces (beg, end)
    , _sum (sum)
        // precondition: input was sorted

  int UniqueCombinations ();

  std::vector _pieces;
  int _sum;

// Each combination is counted only once
// Hint: If a pick contains several equal items in a row,
// they were all picked from the front of a run of
// equal items in the (sorted) input (see the skipping below)
int Knapsack::UniqueCombinations ()
  int count = 0;
  std::vector::const_iterator it = _pieces.begin ();
  std::vector::const_iterator end = _pieces.end ();
  assert (it != end);    
  do {
    int curItem = *it;
    int newSum = _sum - curItem;

    if (newSum == 0) {
    } else if (newSum > 0 && it != end) {
     Knapsack smallerKnapsack (it, end, newSum);
     count += smallerKnapsack.UniqueCombinations ();

    // skip all equal value items
    while (it != end && *it == curItem) {
  } while (it != end);
  return count;

int FillKnapsack (int const * arr, int size, int sum)
  if (size == 0)
    std::cout << "0 combinations: empty array" << std::endl;
    return 0;

    std::vector data (arr, arr + size);
    // sort in decreasing order
    std::sort (data.begin (), data.end (), std::greater ());
    Knapsack knapsack (data.begin (), data.end (), sum);
    int comb = knapsack.UniqueCombinations ();

    std::cout << comb << " combination(s)" << std::endl;
    return comb;
  catch (std::bad_alloc)
    std::cerr << "Out of memory" << std::endl;
  catch (...)
    std::cerr << "Unknown error" << std::endl;
  return 0;
Not to be outdone, I started wondering if there was a way to do it even faster. Since the values in the array are bounded, I decided to immediately transform the input into a histogram telling how many of each number appeared. Then I would do a brute force march through all the histograms that were "recipes" adding to the target sum, and add the appropriate number of sums for each match. So in the sample case:
  • A = 5
  • B = 5
  • C = 5
  • D = 10
I'd quickly transform it to mean:
[0,0,0,0,3,0,0,0,0,1] // three fives and one 10
Then, I'd generate the exhaustive list of "recipes" that sum to 15:
[15,0,0,0,0,0,0,0,0,0] // fifteen ones
[13,1,0,0,0,0,0,0,0,0] // thirteen ones and one two
[12,0,1,0,0,0,0,0,0,0] // twelve ones and one three
/* ... */
[0,0,0,0,1,0,0,0,0,1] // one ten and one five
From there it was just a matter of seeing how many times a recipe could be applied on the input. That's relatively straightforward combinatorics, but my example cases on large data sets overflowed my naive algorithm for combinations. It was a hard bug to track down, but once I replaced it with a better routine off the web I had a blazing fast algorithm. Its performance is determined by the range of the numbers and the value of the sum, not the size of the input array.
UPDATE 17-Jan-2014 The code below was written 10 years ago; when I wasn't particularly comfortable with the C++ collection classes.
// how many combinations of k objects can you choose from a set of n objects ?
// this method avoids overflow that you hit with:
//    factorial(n) / (factorial(k) * factorial(n-k))
// see http://www.cs.sjsu.edu/faculty/fecteau/cs140/combPerm.html
unsigned int combinations(unsigned int r, unsigned int n)
  assert(n >= k);
  unsigned int k = r < n-r ? r : n-r;
  unsigned int answer = 1;
  unsigned int multiplier = n;
  unsigned int divisor = 1;
  while (divisor <= k) {
    answer = (answer * multiplier) / divisor; // this will evenly divide 
  return answer;

// SumsRecipe
// Given a "recipe" for producing a certain sum, tell how many ways that recipe
// can be achieved given the set of numbers indicated by Nums.  Both the recipe
// and the input list are set up so that array[j] tells you how many j's you have
// (or in the case of the recipe, how many j's are needed)
unsigned int SumsRecipe(
  const unsigned int Nums[],
  const unsigned int Recipe[],
  unsigned int Max
) {
  unsigned int Sums = 1;
  for (unsigned int j = 1; j <= Max; j++)
    if (Nums[j] < Recipe[j])
      return 0;
    if (Recipe[j] != 0)
      Sums *= combinations(Recipe[j], Nums[j]);
  return Sums;

// EnumRecipe
// Class which provides a somewhat brute-force method of enumerating over all
// the possible "Recipes" for making the value X as a sum of numbers 1-Max.
// It really just ticks through the recipes like an odometer, trying:
//   1 one, then 2 ones, then 3 ones...
// When it hits the maximum number of ones you can have it then tries:
//   1 two with 0 ones, then 1 two with 1 one, then 1 two with 2 ones...
// Although it might seem like this is a terrible way to generate these
// recipes, there are a couple of "saving graces".  We know
// we don't have more than Length numbers available, and that the total
// can't be greater than X...so if we hit an excess of either of these
// conditions we can fast-forward the "odometer" to the next setting
// which zeroes one of the wheels.
class EnumRecipe
  unsigned int m_X; // Target value for our recipe's sum
  unsigned int m_Length; // Most numbers that could occur in a recipe
  unsigned int m_Max; // Maximum value of any given number
  unsigned int m_Used; // Numbers used in this recipe
  unsigned int m_Total; // Current running total sum
  bool m_First; // First time calling the enumerator?
  const unsigned int* m_Nums; // How many of each number is available in the input?

    unsigned int X,  
    const unsigned int* Nums, 
    unsigned int Length, 
    unsigned int Max
  ) :
    // Nothing to do

  bool NextRecipe(unsigned int Recipe[])
    if (m_First) {
      // since this is the first call, start by zeroing out
      // the recipe that the user passed in...
      for (unsigned int j = 0; j <= m_Max; j++)
        Recipe[j] = 0;
      m_First = false;
      // Assert(m_Total == 0);
      // Assert(m_Used == 0);

    unsigned int Wheel = 1;
    while (true) {
      // Will increasing the current "Wheel" lead to a
      // "Recipe" that uses more values than we have as
      // input, or one that the recipe's summation will 
      // exceed our target X value?
      if ((m_Used < m_Length) && 
          (m_Total < m_X) && 
          (m_Nums[Wheel] >= Recipe[Wheel])
      ) {
        m_Total += Wheel;

        if (m_Total == m_X)
          return true;

        // as on an odometer, each time we successfully
        // increment one of the wheels, the next wheel
        // we'll try to increment will be the first again
        Wheel = 1;

      } else {
        // We reached the maximum value for this "wheel"
        // Is it the last one?  If so, we're done.
        if (Wheel == m_Max)
          return false;

        // Zero this wheel and update our state variables.
        // This intermediate state is one that we've visited
        // before, so no need to test total against m_X
        m_Total -= (Wheel * Recipe[Wheel]);
        m_Used -= Recipe[Wheel];
        Recipe[Wheel] = 0;

    return true;

  virtual ~EnumRecipe()
    // Nothing to do

// SumsThatAddToXSpecialized
// Specialized version of code which answers the "how many ways
// are there to achieve the sum of X using the integers in Values[]".
// This case is specialized because it assumes that the
// values in the array range from 1..Max.  What's different about
// this particular approach is that the performance is not a
// function of the length of the input array.
unsigned int SumsThatAddToXSpecialized(
    unsigned int X, 
    const unsigned int Values[], 
    unsigned int Length,
    unsigned int Max
) {
  unsigned int Nums[Max+1];
  for (unsigned int j = 0; j <= Max; j++) {
    Nums[j] = 0;
  for (unsigned int i = 0; i < Length; i++) {
    // Assert((Values[i] <= Max) && (Values[i] > 0));

  unsigned int Sums = 0;

  unsigned int Recipe[Max+1];
  EnumRecipe er (X, Nums, Length, Max);
  while (er.NextRecipe(Recipe)) {
    Sums += SumsRecipe(Nums, Recipe, Max);
  return Sums;
It's fun to see something like this run so fast and still be correct!
Business Card from SXSW
Copyright (c) 2007-2018 hostilefork.com

Project names and graphic designs are All Rights Reserved, unless otherwise noted. Software codebases are governed by licenses included in their distributions. Posts on blog.hostilefork.com are licensed under the Creative Commons BY-NC-SA 4.0 license, and may be excerpted or adapted under the terms of that license for noncommercial purposes.