| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Given a 2-d convex polygon with N corners, what's the max number of diagonal cross points you can get (there are N(N-3)/2 diagonals). A ONE LINE reply + proof is required...
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 311.1 | RAINBO::GRANT | Tue Jun 25 1985 16:43 | 1 | ||
For any 4 corners, a max. of 1 D.C.P. For N>=4, this is C(N,4) = N!/(24*(N-4)!) | |||||
| 311.2 | LATOUR::JMUNZER | Fri Jun 28 1985 14:09 | 5 | ||
...because if the points are 123...N, a DCP is where w-y hits x-z: 1 <= w < x < y < z <= N [Much too long-winded.] | |||||