Subset sum problem

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 ${\displaystyle O(2^{n}\cdot n)}$, since there are ${\displaystyle 2^{n}}$ subsets and, to check each subset, we need to sum at most n elements.

The algorithm can be implemented by depth-first search of a binary tree: each level in the tree corresponds to an input number; the left branch corresponds to excluding the number from the set, and the right branch corresponds to including the number (hence the name Inclusion-Exclusion). The memory required is ${\displaystyle O(n)}$. The run-time can be improved by several heuristics:[5]

• Process the input numbers in descending order.
• If the integers included in a given node exceed the sum of the best subset found so far, the node is pruned.
• If the integers included in a given node, plus all remaining integers, are less than the sum of the best subset found so far, the node is pruned.

In 1974, Horowitz and Sahni[6] published a faster exponential-time algorithm, which runs in time ${\displaystyle O(2^{n/2}\cdot (n/2))}$, but requires much more space - ${\displaystyle O(2^{n/2})}$. The algorithm splits arbitrarily the n elements into two sets of ${\displaystyle n/2}$ each. For each of these two sets, it stores a list of the sums of all ${\displaystyle 2^{n/2}}$ possible subsets of its elements. Each of these two lists is then sorted. Using even the fastest comparison sorting algorithm, Mergesort for this step would take time ${\displaystyle O(2^{n/2}n)}$. However, given a sorted list of sums for ${\displaystyle k}$ elements, the list can be expanded to two sorted lists with the introduction of a (${\displaystyle k+1}$)th element, and these two sorted lists can be merged in time ${\displaystyle O(2^{k})}$. Thus, each list can be generated in sorted form in time ${\displaystyle O(2^{n/2})}$. 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 T in time ${\displaystyle O(2^{n/2})}$. 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 T, the algorithm moves to the next element in the first array. If it is less than T, the algorithm moves to the next element in the second array. If two elements that sum to T are found, it stops. (The sub-problem for two elements sum is known as two-sum.[7])

In 1981, Schroeppel and Shamir presented an algorithm[8] based on Horowitz and Sanhi, that requires similar runtime - ${\displaystyle O(2^{n/2}\cdot (n/4))}$, much less space - ${\displaystyle O(2^{n/4})}$. Rather than generating and storing all subsets of n/2 elements in advance, they partition the elements into 4 sets of n/4 elements each, and generate subsets of n/2 element pairs dynamically using a min heap, which yields the above time and space complexities since this can be done in ${\displaystyle O(k^{2}\log(k))}$ and space ${\displaystyle O(k)}$ given 4 lists of length k.

Due to space requirements, the HS algorithm is practical for up to about 50 integers, and the SS algorithm is practical for up to 100 integers.[5]

In 2010, Howgrave-Graham and Joux[9] presented a probabilistic algorithm that runs faster than all previous ones - in time ${\displaystyle O(2^{.337n})}$ using space ${\displaystyle O(2^{.256n})}$. It solves only the decision problem, cannot prove there is no solution for a given sum, and does not return the subset sum closest to T.

The techniques of Howgrave-Graham and Joux were subsequently extended [10] bringing the time-complexity to ${\displaystyle O(2^{.291n})}$.

The following very simple algorithm has an approximation ratio of 1/2:[16]

• Order the inputs by descending value;
• Put the next-largest input into the subset, as long as it fits there.

When this algorithm terminates, either all inputs are in the subset (which is obviously optimal), or there is an input that does not fit. The first such input is smaller than all previous inputs that are in the subset and the sum of inputs in the subset is more than T/2 otherwise the input also is less than T/2 and it would fit in the set. Such a sum greater than T/2 is obviously more than OPT/2.

The following algorithm attains, for every ${\displaystyle \epsilon >0}$, an approximation ratio of ${\displaystyle (1-\epsilon )}$. Its run time is polynomial in n and ${\displaystyle 1/\epsilon }$. Recall that n is the number of inputs and T is the upper bound to the subset sum.

initialize a list L to contain one element 0.

for each i from 1 to n do
    let Ui be a list containing all elements y in L, and all sums xi + y for all y in L.
    sort Ui in ascending order
    make L empty 
    let y be the smallest element of Ui
    add y to L
    for each element z of Ui in increasing order do
        // Trim the list by eliminating numbers close to one another
        // and throw out elements greater than the target sum T.
        if y +  ε T/n < z ≤ T then
            y = z
            add z to L

return the largest element in L.

Note that without the trimming step (the inner "for each" loop), the list L would contain the sums of all ${\displaystyle 2^{n}}$ subsets of inputs. The trimming step does two things:

• It ensures that all sums remaining in L are below T, so they are feasible solutions to the subset-sum problem.
• It ensures that the list L is "sparse", that is, the difference between each two consecutive partial-sums is at least ${\displaystyle \epsilon T/n}$.

These properties together guarantee that the list L contains no more than ${\displaystyle n/\epsilon }$ elements; therefore the run-time is polynomial in ${\displaystyle n/\epsilon }$.

When the algorithm ends, if the optimal sum is in L, then it is returned and we are done. Otherwise, it must have been removed in a previous trimming step. Each trimming step introduces an additive error of at most ${\displaystyle \epsilon T/n}$, so n steps together introduce an error of at most ${\displaystyle \epsilon T}$. Therefore, the returned solution is at least ${\displaystyle {\text{OPT}}-\epsilon T}$ which is at least ${\displaystyle (1-\epsilon ){\text{OPT}}}$ .

The above algorithm provides an exact solution to SSP in the case that the input numbers are small (and non-negative). If any sum of the numbers can be specified with at most P bits, then solving the problem approximately with ${\displaystyle \epsilon =2^{-P}}$ 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 ${\displaystyle 2^{P}}$ (i.e., exponential in P).

Kellerer, Mansini, Pferschy and Speranza[17] and Kellerer, Pferschy and Pisinger[18] present other FPTAS-s for subset sum.