Subset sum problem

In computer science, the subset sum problem is an important problem in complexity theory and cryptography. The problem is this: given a set of integers, does the sum of some non-empty subset equal exactly zero? For example, given the set { −7, −3, −2, 5, 8}, the answer is yes because the subset { −3, −2, 5} sums to zero. The problem is NP-complete.

An equivalent problem is this: given a set of integers and an integer s, does any non-empty subset sum to s? Subset sum can also be thought of as a special case of the knapsack problem. One interesting special case of subset sum is the partition problem, in which s is half of the sum of all elements in the set.

General discussion

The subset sum problem is a good introduction to the NP-complete class of problems. There are two reasons for this

• It is a decision and not an optimization problem
• It has a very simple formal definition and problem statement.

Although the subset sum problem is a decision problem, the cases when an approximate solution is sufficient have also been studied, in the field of approximations algorithms; one algorithm for the approximate version of the subset sum problem is given below.

Complexity

The complexity (difficulty of solution) of subset sum can be viewed as depending on two parameters, N, the number of decision variables, and P, the precision of the problem (stated as the number of binary place values that it takes to state the problem). (Note: here the letters N and P mean something different than what they mean in the NP class of problems.)

The complexity of the best known algorithms is exponential in the smaller of the two parameters N and P. Thus, the problem is most difficult if N and P are of the same order. It only becomes easy if either N or P becomes very small.

If N (the number of variables) is small, then an exhaustive search for the solution is practical. If P (the number of place values) is a small fixed number, then there are dynamic programming algorithms that can solve it exactly.

What is happening is that the problem becomes seemingly non-exponential when it is practical to count the entire solution space. There are two ways to count the solution space in the subset sum problem. One is to count the number of ways the variables can be combined. There are 2^N possible ways to combine the variables. However, with N = 10, there are only 1024 possible combinations to check. These can be counted easily with a branching search. The other way is to count all possible numerical values that the combinations can take. There are 2^P possible numerical sums. However, with P = 5 there are only 32 possible numerical values that the combinations can take. These can be counted easily with a dynamic programming algorithm. When N = P and both are large, then there is no aspect of the solution space that can be counted easily.

Efficient algorithms for both small N and small P cases are given below.

Exponential time algorithm

There are several ways to solve subset sum in time exponential in N. The most naïve algorithm would be to cycle through all subsets of N numbers and, for every one of them, check if the subset sums to the right number. The running time is of order O(2^NN), since there are 2^N subsets and, to check each subset, we need to sum at most N elements.

A better exponential time algorithm is known which runs in time O(2^N/2). The algorithm splits arbitrarily the N elements into two sets of N/2 each. For each of these two sets, it stores a list of the sums of all 2^N/2 possible subsets of its elements. Which list to be sorted, which would ordinarily take time O(2^N/2N). However, given a sorted list of sums for k elements, the list can be expanded to two sorted lists with the introduction of a (k + 1)st element, and these two sorted lists can be merged in time O(2^k). Thus, each list can be generated in sorted form in time O(2^N/2). Given the two sorted lists, the algorithm can check if an element of the first array and an element of the second array sum up to s in time O(2^N/2). To do that, the algorithm passes through the first array in decreasing order (starting at the largest element) and the second array in increasing order (starting at the smallest element). Whenever the sum of the current element in the first array and the current element in the second array is more than s, the algorithm moves to the next element in the first array. If it is less than s, the algorithm moves to the next element in the second array. If two elements with sum s are found, it stops. No algorithm with a faster worst-case running-time has ever been found since Horowitz and Sahni first published this algorithm in 1974. Not only this, but no algorithm that solves all instances of the subset sum problem that uses a different strategy than that of Horowitz and Sahni's algorithm has ever been found to run faster than the order of 2^N time in the worst-case scenario.

Pseudo-polynomial time dynamic programming solution

The problem can be solved as follows using dynamic programming. Suppose the sequence is

x₁, ..., x_n

and we wish to determine if there is a nonempty subset which sums to 0. Let N be the sum of the negative values and P the sum of the positive values. Define the boolean-valued function Q(i,s) to be the value (true or false) of

"there is a nonempty subset of x₁, ..., x_i which sums to s".

Thus, the solution to the problem is the value of Q(n,0).

Clearly, Q(i,s) = false if s < N or s > P so these values do not need to be stored or computed. Create an array to hold the values Q(i,s) for 1 ≤ i ≤ n and N ≤ s ≤ P.

The array can now be filled in using a simple recursion. Initially, for N ≤ s ≤ P, set

Q(1,s) := (x₁ == s).

Then, for i = 2, …, n, set

Q(i,s) := Q(i − 1,s) or (x_i = s) or Q(i − 1,s − x_i)   for N ≤ s ≤ P.

For each assignment, the values of Q on the right side are already known, either because they were stored in the table for the previous value of i or because Q(i − 1,s − x_i) = false if s − x_i < N or s − x_i > P. Therefore, the total number of arithmetic operations is O(n(P − N)). For example, if all the values are O(n^k) for some k, then the time required is O(n^k+2).

This algorithm is easily modified to return the subset with sum 0 if there is one.

This solution does not count as polynomial time in complexity theory because P − N is not polynomial in the size of the problem, which is the number of bits used to represent it. This algorithm is polynomial in the values of N and P, which are exponential in their numbers of bits.

A more general problem asks for a subset summing to a specified value (not necessarily 0). It can be solved by a simple modification of the algorithm above. For the case that each x_i is positive and bounded by the same constant, Pisinger found a linear time algorithm.

Polynomial time approximate algorithm

An approximate version of the subset sum would be: given a set of N numbers x₁, x₂, ..., x_N and a number s, output

• yes, if there is a subset that sums up to s;
• no, if there is no subset summing up to a number between (1 − c)s and s for some small c > 0;
• any answer, if there is a subset summing up to a number between (1 − c)s and s but no subset summing up to s.

If all numbers are non-negative, the approximate subset sum is solvable in time polynomial in N and 1/c.

The solution for subset sum also provides the solution for the original subset sum problem in the case where the numbers are small (again, for nonnegative numbers). If any sum of the numbers can be specified with at most P bits, then solving the problem approximately with c = 2^−P is equivalent to solving it exactly. Then, the polynomial time algorithm for approximate subset sum becomes an exact algorithm with running time polynomial in N and 2^P (i.e., exponential in P).

The algorithm for the approximate subset sum problem is as follows:

 initialize a list S to contain one element 0.
 for each i from 1 to N do
   let T be a list consisting of xi + y, for all y in S
   let U be the union of T and S
   sort U
   make S empty 
   let y be the smallest element of U 
   add y to S 
   for each element z of U in increasing order do
      //trim the list by eliminating numbers close to one another
      //and throw out elements greater than s
     if y + cs/N < z ≤ s, set y = z and add z to S 
 if S contains a number between (1 − c)s and s, output yes, otherwise no

The algorithm is polynomial time because the lists S, T and U always remain of size polynomial in N and 1/c and, as long as they are of polynomial size, all operations on them can be done in polynomial time. The size of lists is kept polynomial by the trimming step, in which we only include a number z into S if it is greater than the previous one by cs/N and not greater than s.

This step ensures that each element in S is smaller than the next one by at least cs/N and do not contain elements greater than s. Any list with that property consists of no more than N/c elements.

The algorithm is correct because each step introduces an additive error of at most cs/N and N steps together introduce the error of at most cs.

Further reading

• Cormen, Thomas H.; Leiserson, Charles E.; Rivest, Ronald L.; Stein, Clifford (2001) [1990]. "35.5: The subset-sum problem". Introduction to Algorithms (2nd ed.). MIT Press and McGraw-Hill. ISBN 0-262-03293-7.
• Michael R. Garey and David S. Johnson (1979). Computers and Intractability: A Guide to the Theory of NP-Completeness. W.H. Freeman. ISBN 0-7167-1045-5. A3.2: SP13, pg.223.