HiLander | actually, no my suggestion doesn't work go with Dr_Ford's |

thermoplyae | #math has spoken |

action | Jan` kowtows |

Jan` | kowtows |

action | jamisnemo tries that order.. |

jamisnemo | tries that order.. |

HiLander | (i forgot that axis-aligned makes it the product of intervals, and the intersection is the product of the intersections) |

thermoplyae | (actually, y' = x^7 y^2 - y/x) |

hagabaka | "determine the segment of the y-axis where both rectangles exist, and determine the ranges upon the x-axis for this. if both ranges intersect at all, the rectangles intersect." |

jamisnemo | yea that y gets stuck otherwise as y/x er.. yea alright i see :p thank you! |

hagabaka | do you mean the the x-projections of both rectangles overlap, and y-projections also overlap? |

thermoplyae | I'm not sure I see, I was just pointing out an algebra mistake |

hagabaka | i don't think that's necessary for the rectangles to intersect |

thermoplyae | If you get it, then great |

Jan` | I think I sort of do |

action | Jan` does some thinking |

Jan` | does some thinking |

thermoplyae | hagabaka: It's necessary, and sufficient if the rectangles are axis-aligned |

Solak | Hello. |

Dr_Ford | it is. their x-position-segment may be the same exact thing, and one may be above the other, completely without collision. that's because their y-position-range doesn't intersect |

hagabaka | oh, i meant not sufficient so they are axis-aligned? |

thermoplyae | yes |

hagabaka | hmm |

HiLander | if they're not axis-aligned, it's insufficient |

Jan` | they're axis aligned. |

jamisnemo | it's not separable but it's doable :p |

Jan` | If they weren't, life would get very difficult. |

jamisnemo | y'=f(x,y) |

Dr_Ford | yah, for axis aligned rectangle collision detection, I'm sure there's trig involved |

jamisnemo | I'm just losing my mind |

Jan` | you mean for axis non-aligned |

thermoplyae | Is it sufficient if only one is axis-aligned? |

TRWBW | well you can always do it as polygons, based on edge intersections |

jamisnemo | y=(x^8y^2)/8 - ln(x)*y ugh |

thermoplyae | I think so. If so, then you can construct a rotation matrix to rotate one into axis alignment and then apply the easy method above |