# Subset sum problem with many divisibility conditions

Let $S$ be a set of natural numbers. We consider $S$ under the divisibility partial order, i.e. $s_1 \leq s_2 \iff s_1 \mid s_2$. Let

$\qquad \displaystyle \alpha(S) = \max \{|V| \mid V\subseteq S, V$ an antichain$\}$.

If we consider the subset sum problem where the multiset of numbers are in $S$ , what can we say about about the complexity of the problem related to $\alpha(S)$? It is simple to see if $\alpha(S)=1$, then the problem is easy. Note it is easy even for the harder knapsack problem when $\alpha(S)=1$$\dagger. \dagger Solving sequential knapsack problems by M. Hartmann and T. Olmstead (1993) • Instead of "relation", I suggest using the terms "partial order". Also, on minimal thought, the Frobenius coin problem might be relevant (of course, not sure, though) – Aryabhata Jul 12 '13 at 0:58 ## 1 Answer This problem can be solved in polynomial time using linear programming, and this is actually true for any partial order (S,\le). By the way, we can prove by induction that for any finite partial order set (S,\le), there exists a finite set S'\subseteq\mathbb{N} and a bijection f:S\rightarrow S', such that for all s_1,s_2\in S, s_1\le s_2 \Leftrightarrow f(s_1) | f(s_2). Let \mathcal{C} be the set formed by the chains in S. Remind that C is a chain iff for all v,v' in C, v\le v' or v'\le v Now create a boolean variable x_v for each v\in S, and a boolean variable y_C for each chain C. We can write the following linear program (P) for our problem :$$ \begin{split} \text{Max} \displaystyle\sum\limits_{v\in S} x_v \\ \text{subject to} \displaystyle\sum\limits_{v\in C} &x_v \le 1, \forall C\in\mathcal{C}\\ &x_v \in \{0,1\}, v\in S \end{split} $$and its dual (D) :$$ \begin{split} \text{Min} \displaystyle\sum\limits_{C\in \mathcal{C}} y_C\\ \text{subject to} \displaystyle\sum\limits_{C:v\in C} &y_C \ge 1, \forall v\in S\\ &y_C \in \{0,1\}, C\in \mathcal{C} \end{split} $$Then the problem of finding the minimum cover of an ordered set by chains is the dual of our problem. Dilworth's theorem states that There exists an antichain A, and a partition of the order into a family P of chains, such that the number of chains in the partition equals the cardinality of A which means that the optimal solution of these two problems match : Opt(P)=Opt(D) Let (P^*) (resp. (D^*)) be the relaxation of (P) (resp. (D)) i.e. the same linear program where all constraints x_v\in\{0,1\} (resp. y_C\in\{0,1\}) are replaced by x_v\in [0,1] (resp. y_C\in [0,1]). Let Opt(P^*) and Opt(D^*) be their optimal solutions. Since \{0,1\}\subseteq [0,1] we have :$$ Opt(P)\le Opt(P^*) \text{ and } Opt(D^*)\le Opt(D) $$and weak duality theorem establishes that Opt(P^*)\le Opt(D^*) then by putting everything together we have :$$ Opt(P)= Opt(P^*)=Opt(D^*)=Opt(D)$$Then, using Ellipsoid method, we can compute$Opt(P^*)$($=Opt(P)$) in polynomial time. There are an exponential number of constraints but there exists a polynomial time separation oracle. Indeed given a solution$X$, we can enumerate all couples$s_1,s_2\in X$and check if$s_1\le s_2$or$s_2\le s_1$, and therefore decide in polynomial time whether$X$is feasible or otherwise the constraint associated to the chain$\{v_1,v_2\}$is violated. • Ellispoid method works whatever the number of constraints, if we have (1) a polynomial number of variables and (2) a separation oracle which given any solution$x$decides in polynomial time whether$x$is feasible or find a constraints violated by$x\$. I recommend reading [www-math.mit.edu/~goemans/18433S09/ellipsoid.pdf], wikipedia is not very clear on this point – Mathieu Mari May 25 '18 at 18:01
• Thanks for explaining why the exponential number of constraints is not a problem, and the relevance of duality. Very nice! – D.W. May 25 '18 at 20:12