[Solved]: Overflow safe summation

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Problem Detail: Suppose I am given $n$ fixed width integers (i.e. they fit in a register of width $w$), $a_1, a_2, dots a_n$ such that their sum $a_1 + a_2 + dots + a_n = S$ also fits in a register of width $w$. It seems to me that we can always permute the numbers to $b_1, b_2, dots b_n$ such that each prefix sum $S_i = b_1 + b_2 + dots + b_i$ also fits in a register of width $w$. Basically, the motivation is to compute the sum $S = S_n$ on fixed width register machines without having to worry about integer overflows at any intermediate stage. Is there a fast (preferably linear time) algorithm to find such a permutation (assuming the $a_i$ are given as an input array)? (or say if such a permutation does not exist).

Asked By : Aryabhata

Answered By : Dave Clarke

Strategy
The following linear-time algorithm adopts the strategy of hovering around $0$, by choosing either positive or negative numbers based on the sign of the partial sum. It preprocesses the list of numbers; it computes the permutation of the input on-the-fly, while performing the addition. Algorithm

  1. Partition $a_1, ldots, a_n$ into a two lists, the positive elements $P$ and the negative elements $M$. Zeros can be filtered out.
  2. Let $Sum=0$.
  3. While both lists are non-empty
  4. $~~~~~~$If $Sum>0$ { $Sum:=Sum+text{head}(M)$; $M:=text{tail}(M)$; }
  5. $~~~~~~$else { $Sum:=Sum+text{head}(P)$; $P:=text{tail}(P)$; }
  6. When one of the two lists becomes empty, add the rest of the remaining list to $S$.

Correctness
Correctness can be established using a straightforward inductive argument on the length of the list of numbers. First, prove that if $a_1, ldots, a_n$ are all positive (or all negative), and their sum does not cause overflow, then nor do the prefix sums. This is straightforward. Second, prove that $Sum$ is within bounds is an invariant of the loop of the algorithm. Clearly, this is true upon entry into the loop, as $Sum=0$. Now, if $Sum>0$, adding a negative number that is within bounds to $Sum$ does not cause $Sum$ to go out of bounds. Similarly, when $Sumle0$ adding a positive number that is within bounds to sum does not cause $Sum$ to go out of bounds. Thus upon exiting the loop, $Sum$ is within bounds. Now, the first result can be applied, and together these are sufficient to prove that the sum never goes out of bounds.

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