Talk:Wandering monsters

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Since I was still in a BHY ascension when the Dia De Los Muertos Burrachos came around, I took the opportunity to do some spading on the unresolved issue mentioned at the end of this article.

To see what happens when two wandering monsters are slated for the next round (i.e. have had their timers exceeded) I waited until I hit a Dia De Los Muertos Burrachos monster (forgot to write down which one), and then adventured in the cake-shaped arena for 35 turns. According to the wiki page, this should avoid tripping either adventure, while simultaneously running out both timers (25-35 for corpses, and 15-20 for bees).

My next adventure was in the spooky forest (chosen arbitrarily), and I faced a Moneybee. I adventured again in the same spot, and got La Novia Cadaver. This makes it seem like the wandering monster adventures queue up, at least between corpses and bees. If they queue in the order that their timers expire then bees should always be first.

I then ran the experiment two more times. The first time was duplicated exactly, except that I switched to the black forest for my two adventure.php adventures (to avoid the chance of a semi-rare preempting the wandering monsters). Immediately following the last post, I ran the 35 adventures in the arena and then fought a Mumblebee and El Novio Cadaver.

The second time, I accidentally overdrank (for some reason thinking that a corpse island iced tea would only add 4 drunkenness), and ended up falling down drunk, but with enough adventures remaining. Since the wiki said that wandering monsters can be fought in a drunken stupor, I spent 35 turns at the Institute for Canadian Studies, after which I went back to the black forest, and lo and behold I fought another Mumblebee, with El Padre Cadaver right on its heels in the following adventure.

All of this seems to be perfectly consistent with the idea that wandering monsters (at least corpses and bees) queue up in the order that their timers expire, and then dequeue in a first-in first-out manner in successive adventure.php loads.

--Zemrude 15:29, 8 August 2011 (CEST)

  • Nice method. You know, on a Muertos Burrachos day, under BHY, after defeating your nemesis in the dark and dank cave, well, you can have three sets of wandering monsters following you. Too bad the Obtuse Angel has a 'b', or you could add a romantic arrow. --Club (#66669) (Talk) 17:21, 8 August 2011 (CEST)
    • Thanks! It's a shame I didn't think of that earlier, I had already beaten all the wandering monsters from the nemesis quest this time around. Speaking of methods, I'm new to editing this wiki, and I'm not sure if this consitutes enough of a spade to stick something on the page, or if I should keep it in the discussion until someone else has had a chance to replicate? (which they could manage with BHY and the nemesis quest, per your awesome insight)--Zemrude 23:55, 8 August 2011 (CEST)
      • I think this rates as talk page level spading. Some players are much more rigorous. I'm usually not one of those players. --Club (#66669) (Talk) 01:37, 9 August 2011 (CEST)
      • I don't think it quite answers the priority question, but the queueing up is at least good info. It should be hypothetically possible for two (or more) wandering monsters to occur on the same turn, without having forced them to do so (just simply that both of their windows are active simultaneously from normal adventuring). Does one of them always take priority? For that matter, exactly when is it the game decides when you get the wandering monster? Random roll made on every adventure within a window, guaranteed at the end? Rolled immediately after the previous wandering monster? If it rolls them ahead of time, what happens when it rolls up a different wandering monster that could normally occur on that turn? Is that turn impossible to roll up? Or would it bump the previous one back/ahead a turn (and which one/how so)? These are hard to answer questions, since the amount of data needed is well beyond what a single person can produce without a very large number of multis (and the ability to run them all somehow). Some may be impossible to distinguish from other possibilities, others may be distinguishable but require tens of thousands of data points to verify. In any case, you can check [kolspading.com kolspading]. I'm fairly sure there's been a thread about this before that hasn't seen action lately, and you can always add in your observations to help out and compare them with others. --Flargen 05:43, 9 August 2011 (CEST)
  • Apparently "queueing up" is no longer what happens, or is not what happens in all cases. I faxed a cement_penguin and winked at it with my Reanimator on El Día. As soon as I winked at it (while still in the combat), I set a /timer for 25 turns. I fought rats in the Tavern Cellar until the timer expired, then brought out my Happy Medium and went to the Valley of Rofl. On the first turn I got an El Día monster. On the next 5 or 6 turns, before giving up in frustration, I got only regular Valley monsters. The reanimated penguin did not show up at all.

    I went back to the tavern, and killed about 10 more rats, so that I was somewhere near the start of the next Reanimator interval. Brought out the medium again, then adventured in the Valley for a few turns, and this time the reanimated penguin showed up. --Greycat (talk) 16:02, 14 December 2013 (UTC)
    • The third reanimated penguin did eventually show up. I wasn't counting turns or anything, but I went back to the Valley later with the Goth Kid to flush out another El Día monster or maybe a crayon shaving, and got a penguin instead. --Greycat (talk) 16:37, 14 December 2013 (UTC)

Delay?

Wandering monsters, from what I'm understanding here, don't take, use or otherwise reference delay(), right? So I couldn't use free runaways or hipster fights to decrement the counter while avoiding use of turns, correct?--Hastifer 23:56, 19 January 2012 (CET)

Almost certainly they do not use delay(), especially since delay() is always 5. --PsyMar 16:59, 15 December 2012 (CET)