Timeline for "Critter Tracking: When does it cross its own path?"
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 2, 2015 at 14:18 | comment | added | WernerCD | Added Version #2, which includes some things from this answer (StopWatch, Action list) | |
| Jul 1, 2015 at 8:50 | comment | added | outoftime | @WernerCD I went a sleep and when wake up, Rick Davin finished his version first. | |
| Jun 30, 2015 at 22:57 | comment | added | WernerCD | Otherwise, I do like the Action part (I've used Actions in passing before). I'll have to see if i can wrap my head around it and work it in. | |
| Jun 30, 2015 at 22:55 | comment | added | WernerCD | I was going to ask how this tracked the points the critter walked to determine when if crossed it's path (which is why my answer has Dict<Int,List<int>> - Dict<int, is X, List<int>> is Y and they both are required to "remember" where we've traveled. If I've put 4 in A[0], that means I go north 4 spaces: (0,0),(0,1),(0,2),(0,3),(0,4). Path[0] would then be a list of 0...4. Later, if I stop on (0,3) (not an end point, but a part of the path), that counts as crossing the path. | |
| Jun 30, 2015 at 22:33 | comment | added | Rick Davin | I can forgive the Java style answer for a C# question, but your solve method returns the wrong result for the example input set of { 1, 3, 2, 5, 4, 4, 6, 3, 2 }. It's not a question of whether the ending points lands on a previous point but rather if the segment (or vector) between the current point and previous point would collide with any segments formed by the previous coordinates. | |
| Jun 30, 2015 at 21:44 | history | answered | outoftime | CC BY-SA 3.0 |