# List the smallest $k$ subset sums

Given a set of positive reals $\set{x_1,\ldots,x_n}$ where $x_1<x_2<\ldots<x_n$, find the smallest $k$ subset sums.

We can assume $n\leq k$, because we do not have to read $x_j$ if $j>k$.

Torsten Gross and Nils Blüthgen posted a $O(k^2)$ time solution on arXiv.

We show a $O(k\log k)$ time algorithm, which is optimal if we want to output the numbers in order.

We list the sums one by one by maintaining a priority queue of sums. We start with the empty set. Assume that we added the sum induced by $I\subset [n]$ (that is, $\sum_{i\in I} x_i$) into the output, let $j=1+\max I$. Now we can consider two possibilities by extending the current solution: the sum induced by $I\cup \set{j}$ or the sum induced by $I\cup \set{k}$ where $k>j$. We will add both possibilities to the queue so that one can inspect them later. We can avoid storing the sets, only the values are required.

Here is a python implementation.

```
def first_k_subset_sums(x,k):
n = len(x)
h = []
output = [0] # need to account for the empty set
heapq.heappush(h,(x[0],0))
while h and len(output)<k:
(u,b) = heapq.heappop(h)
output.append(u)
if b+1<n:
heapq.heappush(h,(u+x[b+1],b+1))
heapq.heappush(h,((u-x[b])+x[b+1],b+1))
return output
```

If we want to output the sets themselves, not just the values, does the running time change? If a set $I$ is in the output, then all subsets of $I$ must also be in the output. Hence the largest set we can ever output has size $O(\log k)$. Therefore the total output length is at most $O(k\log k)$.

This is also a lower bound. Consider when $x_i=2^i$, then we will output all subsets of $\set{x_1,\ldots,x_{\log k}}$, and we know that $\sum_{i=1}^{\log k} i{\log k\choose i} = \Omega(\log k)$.

If we don't have to list the smallest $k$ subset sum values in order, then $O(k)$ is possible, see this mathoverflow answer by David Eppstein.

If we are interested in the smallest $k$ *distinct* subset sum. I don't know of any algorithm that performs better than $O(nk)$, even if we know that $n=\Omega(k)$.

# Acknowledgements

I would like to thank Tana Wattanawaroon for helpful discussions and taking an interest in this problem.