Find all integer points at distance d from line segment (0, b)

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I'm reading a scientific paper on integer linear programming and trying to understand a specific part of it.

There is a point $ b \in \mathbb Z^m $ and a set $\mathcal S$ that consist of all points $x \in \mathbb Z^m$ for which there exists some $\alpha \in \mathbb R, \alpha \geq 0, \alpha \leq 1$ with
$\|x - \alpha \cdot b\|_{\infty} \leq 2m \cdot \Delta$
($\Delta \in \mathbb Z,$ so $2m \cdot \Delta $ is some constant integer value)

So in short, the set $\mathcal S$ contains integer points which are at some distance $d$ from the line segment connecting $0$ and $b$.

The paper says (right in the middle of pg 5) the following:

We now argue that $\mathcal |S| \leq (4m · \Delta + 1)^m ·\|b\|_1$ .
Let $ f \in \mathbb R^m$ be any point. Since $2m \cdot \Delta $ is an integer, the integer points at distance at most $2m \cdot \Delta $ from $f$ are contained in the set of integer points at distance at most $2m \cdot \Delta $ from $\lfloor f \rfloor $. Therefore, an upper bound on $\mathcal |S|$ is the number of different integer vectors that can be obtained by rounding a point on the line-segment $(0, b)$ times $ (4m · \Delta + 1)^m $ . The number of rounded integer points is at most $\|b\|_1$.

I've been trying to understand where the $ (4m · \Delta + 1)^m $ and $\|b\|_1$ come from, but to no avail.
Could you please give me any tips or suggestions about this particular variables or the problem in general?

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The key point here is the observation that the integer points at distance at most $k=2m\cdot\Delta$ from $f$ are contained in the set of integer points at distance at most $k$ from $\lfloor f \rfloor$. We can see this by noting that $|x_i-f_i| \le k$ is equivalent to $-k\le x_i-f_i \le k$. Since $x_i,k$ are integers, the inequality $x_i-f_i\le k \leftrightarrow x_i-k\le f_i$ is just the same as $x_i-k\le \lfloor f_i \rfloor$ (and similarly to the other inequality). With this key observation, to bound the number of integral points in the set $\|x - \alpha \cdot b\|_{\infty} \leq 2m \cdot \Delta$, we only need to bound the total number of integral points obtained from the sets $\|x - c\|_{\infty} \leq 2m \cdot \Delta$ where $c$ is an integral point on the line segment between $0$ and $b$. And how many such integral points $c$? There are at most $\|b\|_1$ such points.

Now, let's think about the set of points $x$ that $\|x - 0\|_{\infty} \leq k $. This is just the hypercube with length $k$ in $\mathbb{R}^m$ and it has exactly $(2k+1)^m$ integral points (for each integral point $(x_1,\ldots ,x_m)$ there are $2k+1$ ways to choose an integral value for each $x_i$).

Now for any integral point $b$, the set of points $x$ that $\|x - b\|_{\infty} \leq k $ is just the translation of hypercube with length $k$ in $\mathbb{R}^m$ by $b$ so it has exactly the same number of integral points in it, which is $(2k+1)^m$. This is the reason why they have $(4m · \Delta + 1)^m$ in their estimation.