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Because it may allow us to use dynamic programming to calculate Q(m, n) and because some new quadrilaterals can be easily generated simply by moving edges and v
by marcodiego 5y ago
Because it may allow us to use dynamic programming to calculate Q(m, n) and because some new quadrilaterals can be easily generated simply by moving edges and vertices to new spaces when m or n is increased. So, it seems natural for me to try to write Q(m + 1, n) as a function of Q(m, n).
Also, note that Q(m, n) = Q(n, m), so if we can calculate Q(m + 1, n) as a function of Q(m, n) we can also do it for Q(m, n + 1). Calculating Q(m + 1, n) as a function of Q(m, n) doesn't seem complicated if it weren't by the rules "no straight angles and does not self-intersect". Maybe it can be done with some combinatorics, but seems beyond my skill. Also, expressing it in terms of combinations may also simplify calculation of rest of division.
If such relation can be found, I think the problem may be easily solved.
- deleted 5y ago[deleted]
- marcodiego 5y agoThis made me think of... U(m, n) as the number of unique forms of Q(m, n). By unique forms, I mean forms that can't be obtained by simply translating previously found forms. Maybe it is easy to calculate Q(m, n) as a function of U(m, n) and maybe calculating U(m, n) as a function of U(m - 1, n) is a bit easier than calculating Q(m - 1, n).