Different monsters on the same tile? 0.19

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Big-J-Q
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Different monsters on the same tile? 0.19

Post by Big-J-Q »

I don't think this is necessarily a bug, but in DM, on the Scorpion level, I had a Beholder and a Scorpion on the same tile, and it looked quite silly!:)

Are different kinds of monsters allowed to be on the same tile? Although this was a special case, since it occured in the place where the monsters are trapped behind teleporters (just before the place where you get the boots of speed). A nearby scorpion must have found its way to the teleport place, and accidentally teleported to a same tile with a beholder. Too bad I don't have a screenshot! :wink:

How would this be handled in DM?
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George Gilbert
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Re: Different monsters on the same tile? 0.19

Post by George Gilbert »

What you've seen is a bug, but the general case of having different types of monsters on a tile is not.

DM only allowed a single type of monster on a tile (actually it was even more restrictive than that) whereas RTC is meant to allow anything that will fit onto a tile.

As the scorpion takes up all of a tile, then no other monsters should be allowed on the tile with it (this is the bug). It is however quite possible (and not a bug) to have, say, a skeleton and a beholder (and a screamer and a mummy) one in each of the four corners of a tile.
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beowuuf
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Re: Different monsters on the same tile? 0.19

Post by beowuuf »

In DM monsters were queued if teleported onto the same square, am I right in thinking in RTC this isn't the case ,and they are just placed on?
I've noticed the monster crush on level 10 too, but only at the point where all the creatures are trapped behind teleporters in the red cross room
Was unable to (permenantly) kill off ian_scho (Haynuus), Ameena, oh_brother (Westian), money (Falkor), raixel (Petal) and Lord_Bones (Aurek) in the DM D&D game Time's Champions!

CONGRATULATIONS TO THOSE WHO MADE THE GAME WHAT IT WAS - GREAT!
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George Gilbert
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Re: Different monsters on the same tile? 0.19

Post by George Gilbert »

Nearly - DM does keep monsters in limbo until their teleport destination is free, and RTC doesn't. RTC doesn't however just plonk the monster down regardless, it tries to move the monster already on the tile to the neighbouring ones first; and only if that fails does it add them on top of each other. I'm not sure what the "best" algorithm here should be. Any votes?
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beowuuf
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Re: Different monsters on the same tile? 0.19

Post by beowuuf »

Can you have the failing option as DM's limbo method?
Otherwise can't you have monsters placed in the walls too? In DM they could walk out of a wall, is it the same with RTC?
Then it doesn't matter if alot of other monsters are placed on the same tile, you can't see it, I'm guessing only one acceptable group can then walk out at a time afterwards.
Was unable to (permenantly) kill off ian_scho (Haynuus), Ameena, oh_brother (Westian), money (Falkor), raixel (Petal) and Lord_Bones (Aurek) in the DM D&D game Time's Champions!

CONGRATULATIONS TO THOSE WHO MADE THE GAME WHAT IT WAS - GREAT!
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Lunever
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Re: Different monsters on the same tile? 0.19

Post by Lunever »

I'd prefer the DM-limbo-method, but only, if it is not possible to move the monsters already occupying the tile to a adjacent tile. I do not think, that placing them upon walls is a good thing, cause there may be some places, where the monsters might end up in an area of the dungeon where they were not intended to be.
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George Gilbert
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Re: Different monsters on the same tile? 0.19

Post by George Gilbert »

OK - I've altered the code to make it much more unlikely that monsters will appear on top of each other - now small monsters will try to land on each of the sub-tiles too, not just the part of the tile first intended.

To avoid the chances of monsters (and more importantly any items they're carrying) being stuck in limbo (you could easily imagine a situation where a tile would never become free) as a last resort the monster will get dumped on the intended destination tile. Within a few seconds it should sort itself out though.

This fix will be in V0.20
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