Domination in Nice Lattices

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Let an integer vector be nice when it has only two nonzero components, which sum to zero. So (0, 0, 3, 0, -3) and (-1, 0, 1, 0, 0) are examples of nice vectors in $n=5$ dimensions.

Call a lattice nice if it is of the form $\mathbb{Z}$-span({$v_1, v_2, \dotsc, v_m$}), where all $v_i$ are nice. (Note: the $v_i$ are not necessarily linearly independent so $m$ could be larger than the dimension; although WOLOG $m \le \tbinom{n}{2}$.)

Is the following decision problem in P?

• INPUT: a nice lattice and a vector $x \in \mathbb{Z}^n$
• QUESTION: does the lattice contain a $y$ such that $y_i \ge x_i$ for all $i=1, \dotsc, n$?

Motivation and background:

• in general lattices, the problem is NP-complete (via the unbounded knapsack problem)
• if this problem lies in P, one can solve an interesting more general problem

A possibly interesting partial result would be to demonstrate any useful structure for nice lattices!

(I posted a flow formulation of the problem on cstheory)

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Trivial comment.  If we take $v_1=(1,-1,0, \dots, 0), v_2=(0,1,-1,0, \dots, 0), \dots, v_{n-1}=(0, \dots, 1,-1)$, then the $\mathbb{Z}$-span of all the $v_i$'s is in fact all integer vectors whose coordinates sum to zero.   – Tony Huynh Feb 22 2011 at 14:53
Not-as-trivial comment:  There are at most finitely many vectors y which are both nice and greater than x (the sum of whose components must be at most 0).  This can be transformed into finding integer solutions to a system of linear inequalities; if the transform went the other way, I think it would show the problem to be NP-complete.  Gerhard "Ask Me About System Design" Paseman, 2011.02.22 – Gerhard Paseman Feb 22 2011 at 17:16
Of course, there is the brute force approach: if the sum of the coefficients of x is -r, try the O(r^(n-1)) nice vectors above x and see which are in the desired lattice.  Gerhard "Is There A Better Way" Paseman, 2011.02.22 – Gerhard Paseman Feb 22 2011 at 17:23
Yes, your comments are valid. Tony, more generally than what you state, the problem becomes poly-time solvable if all of the nonzero entries form a chain under division (e.g., all 1s, 2s, 6s, 30s...) – Dave Pritchard Feb 24 2011 at 11:47
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