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I have this problem for which I am struggling to find an efficient dynamic programming algorithm. Would be thankful for some help!!

Let $A = \{ a_1, a_2, ..., a_n \}$ be a set where $a_i \in \mathbb{N}$ for $i=1,...,n$.

The goal is to determine whether there exist two disjoint subsets $M,N \subset A$ such that the sum of all elements in $M$ is equal to exactly $\textit{twice}$ the sum of all elements in $N,$ and $M \not = \emptyset$ and $N \not = \emptyset.$

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    $\begingroup$ Let $t$ be the sum of all input values. I think your best bet will be an $O(nt^2)$-time, $O(t^2)$-space algorithm that generalises the usual Knapsack algorithm by looping through all items in some order, and generating all pairs of sums that can be formed from the first $i$ elements -- by adding the $i$-th item to either element of all pairs of sums $(a, b)$ that can be formed from the first $i-1$ items. At the end, look for pairs of the form $(a, 2a)$. This is only pseudopolynomial time, but I'm certain ordinary Knapsack can be reduced to this. $\endgroup$
    – j_random_hacker
    Apr 9 '20 at 0:58
  • $\begingroup$ Ins't the answer trivially yes? Pick $M=N=\emptyset$. $\endgroup$
    – Steven
    Apr 9 '20 at 16:00
  • $\begingroup$ That's my bad, both $M$ and $N$ must not be empty. Will add that in now. $\endgroup$
    – yellowsuzuki
    Apr 9 '20 at 16:13
  • $\begingroup$ Thank you both for your answers! $\endgroup$
    – yellowsuzuki
    Apr 9 '20 at 21:22
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Let $\sigma(S)$ denote the sum of all the elements in $S$ and $A_i = \{a_1, \dots, a_i\}$.

Given, $i=0,\dots,n$ and $w \in \mathbb{Z}$, define $OPT[i,w]$ as the mazimum cardinality of a subset $S \cup S' \subseteq A_i$ where $S \cap S' = \emptyset$ and $\sigma(S) - 2\sigma(S') = w$. If no possible choice for $S$ and $S'$ exists, then let $OPT[i,w]=-\infty$.

According to the above definition: $$ OPT[0, w] = \begin{cases} 0 & \text{if } w=0 \\ -\infty & \text{if } w \neq 0 \\ \end{cases} $$

For $i>0$: $$ OPT[i, w] = \max\{ OPT[i-1, w], 1 + OPT[i-1, w-a_i], 1+ OPT[i-1, w+2a_i] \}. $$

The answer to your problem is true iff $OPT[n, 0] > 0$. Each $OPT[i, w]$ can be computed in constant time (if you have already computed the values of $OPT[i-1, \cdot]$). Moreover, there are only $O(n)$ possible choices for $i$ and $O(t)$ sensible choices for $w$, where $t = \sum_{i=1}^n a_i$. This gives you a dynamic programming algorithm with time complexity $O(n t)$ and space complexity $O(t)$.

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  • $\begingroup$ That checks out!! Thanks a lot for the answer, appreciate it!! $\endgroup$
    – yellowsuzuki
    Apr 9 '20 at 21:22
  • $\begingroup$ Nice answer! You could use a single bit for each $OPT[i,w]$, and set a global "solution exists" flag when some $OPT[i, 0]$ can be achieved from some $OPT[i-1, x]$ with $x \ne 0$. $\endgroup$
    – j_random_hacker
    Apr 9 '20 at 21:51
  • $\begingroup$ @j_random_hacker, yep! My first write up used one bit per $OPT[i,w]$ but then I had the problem that $N=M=\emptyset$ would always be a solution. I thought of keeping two bits per $OPT[i,w]$ ("is there a solution?" and "is there a non-trivial solution?") but it felt "hacky" so I looked for a "natural" subproblem definition that would circumvent the problem altogether. $\endgroup$
    – Steven
    Apr 9 '20 at 22:04

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