hint request: counting the elements in a subset of a numerical semigroup

Question

I'm working on a problem for an online judge and I'm stuck. I'd like a nudge in the right direction (not an outright solution, please), relevant references, theorems, etc. After reading through relevant entries here and on the web, I believe the correct mathematical statement of the problem is to count the exact number of elements in a subset of a numerical semigroup $\lbrace a, b \rbrace$ (the set of all linear combinations of $a,b$ with non-negative integer coefficients). Specifically, to count the element of set $S$: \begin{equation} S = \lbrace x \vert x \in \lbrace a,b \rbrace, X_{min} \le x \le X_{max} \rbrace \end{equation} The range of possible values is: $1 \le a, b, X_{min}, X_{max} \le \approx 2^{63}$. With these sizes, only a closed form $O(1)$ solution will do. I already have a solution for the cases: $a=b$ (trivial) and $b = a + 1$ (not trivial but not too challenging, interesting case). From the answers to this question: Determine the Set of a Sum of Numbers, I suspect some sort of generalization of the gap counting formula for the case where $gcd(a,b)=1$: $gaps=(a-1)(b-1)/2$, to be able to quickly calculate the number of gaps below an arbitrary $x$ is the key for answering this question.

What I've noticed so far: I've investigated trying to find a pattern for the gaps between successive multiples of $a$ (assuming $a<b$). I've noticed that each multiple of $b$ occupies one of the $(a-1)$ positions between multiples of $a$, and that the $(a-1)$th multiple of $b$ fills that last gap. The first $b$ fills position $b_a$ and the next multiple of b occupies $(b_a + (b-a)_a)_a$. But I haven't been able to generalize anything from that yet.

(My math background is undergraduate level calculus (2 years), linear algebra/matrix theory, discrete math (for CS students), and a real analysis course, but no number theory.) Thanks for any hints / pointers!

Answer 1

Your problem is a specific case of counting lattice points (vectors all of whose coordinates are integers) inside of a polytope (a region in $n$-space which is the intersection of a finite number of half spaces -- sets of vectors $x$ satisfying $a \cdot x \le b$, where $a$ is some non-zero vector and $b$ is a scalar). In your case you're in dimension 2 which makes some things simpler. The ingredients are a method to calculate the number of lattice points on a line segment, and in the interior of a right triangle). You can find the details in "A Simple Algorithm for Lattice Point Counting in Rational Polygons: by Hiroki Yanagisawa in http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.85.5275&rep=rep1&type=pdf . Here is the idea: If $a,b,c$ are positive integers define $N(a,b,c)$ to be the number of lattice points in $\{ x \in \mathbb{R}^2 | a x + by \le c \}$. The first observations is that $N(a,b,c) = N(b,a,c)$ and that $N(a,a,c) = \lfloor c/a \rfloor (\lfloor c/a \rfloor - 1)/2$, and if $a > b$ let $m = \lfloor c/a \rfloor, h = \lfloor (c-am)/b\rfloor, k = \lfloor (a-1)/b\rfloor, c' = c - b(km + h)$. Then $N(a,b,c) = N(a-bk,b,c') + km(m-1)/2 + m h$. The latter comes from dividing the triangle into regions some of which have numbers of lattice points that are easy to calculate. This gives a recursive algorithm (something like the Euclidean algorithm) which runs in time $O(\max(\log(a),\log(b),\log(c)))$.

Answer 2

You can reduce the general problem to the problem for $\gcd(a,b)=1$ using

$$ \begin{align} a&\to a/g\;,\\ b&\to b/g\;,\\ X_{\text{min}}&\to\lceil X_{\text{min}}/g\rceil\;,\\ X_{\text{max}}&\to\lfloor X_{\text{max}}/g\rfloor\;, \end{align} $$

where $g=\gcd(a,b)$. Then you can find a solution to $ma+nb=1$ (with $m$ or $n$ negative), use it to represent all integers in $[1,ab]$ as linear combinations of $a$ and $b$, and add a sufficient multiple of $a$ or $b$ to make all coefficients nonnegative. All numbers in and beyond the resulting interval are in the semigroup, and you only have to deal with the cases below the interval.