3 ms·
Note: Q(m + 1, n) = each valid quad of Q(m, n) moving an edge to another possible one in the new line + each valid quad of Q(m, n) moving a vertex to the
by marcodiego 5y ago
Note: Q(m + 1, n) =
each valid quad of Q(m, n) moving an edge to another possible one in the new line +
each valid quad of Q(m, n) moving a vertex to the new line +
the same thing moving each unique valid quad of Q(m, n) to the new line.
The number of new possible edges on the new line is n * (n - 1) / 2 . The problem is: moving each edge of each quad that can be generated in Q(m, n) to the new possible edges may generate quads that break the rules.
The number of new possible vertices on the new line is (m + 1) . The problem is: moving each vertex of each quad that can be generated in Q(m, n) to the new possible vertices may generate quads that break the rules.
Finding a way to calculate both terms looks like good progress.
- tylerhou 5y agoIt might be easier to define f(m, n, v) where f(m, n, v) is the number of v-vertex shapes that can fit within a m-n square with exactly one vertex in the top left hand corner. Then Q(m, n) = \sum_{i \in m, j \in n} f(m, n, 4). f also lends itself to a much easier recurrence.
- marcodiego 5y agoFor me, it seems easy if we don't have to care about the imposed restrictions: no crossing and no straight angles. Hmmmm.... New ideas coming... Maybe we don't have to care about crossings: any crossing can be solved by reordering the vertices, so, all we have to care about is the number unique vertices positions! I still don't know how to avoid straight angles though.
- tylerhou 5y agoIf you didn’t care about the restrictions, then the answer is N*M choose 4.
- marcodiego 5y agoI think I've made some progress... When one line is added to the problem, it allows us to put 1 or 2 vertices there, so... Consider U(m, n) as the number of new unique forms possible for Q(m, n). I'm not sure, but it looks like: U(1, 1) = 1 U(m + 1, n) = U(m, n) * (n + (n+1)*n/2) - X U(m, n + 1) = U(m, n) * (m + (m+1)*m/2) - X Where X is the number of invalid forms generated adding one or two vertices in the new line because of straight angles in diagonals. I'm considering only diagonals, because adding only one or two vertices in the new line will not align 3 vertices horizontally or vertically if all "previous forms" were valid. If we can find X, then, I think Q is given by: Q(m, n) = m*n*U(1, 1) + (m-1)*(n-1)*U(2, 2) + (m-2)*(n-2)*U(3, 3) +...+ 1*1*U(m, n)
- tylerhou 5y ago> If we can find X Almost certainly we need to add terms into the DP state to compute X; unfortunately adding just the number of vertices doesn’t work as some triangles have two edges in which a point can be inserted to create a quadrilateral, while others only have one. The difficulty is finding those terms, though…
- marcodiego 5y agoWhat do you mean by "the DP state"? Also, I think my previous formula is wrong. Maybe it is the sum of: (m - i + 1)*(n - j + 1)*U(i, j) With i, j in [1..m]. But maybe it is still wrong because U(i, j) will re-count combinations from U(i - 1, j) and U(i, j - 1)... Argh! This is complicated! I think I'll have to revisit it if I ever have a good idea.
- tylerhou 5y agoBy DP state, I mean the arguments to the recursive function (for U, the state is the tuple (i, j)). For example, if you wanted to use DP to compute all-pairs-shortest-paths (Floyd-Warshall), you add a third parameter to the DP state ((f(u, v, k)).
- marcodiego 5y agoSure! Dynamic programming, right?