How many demons?

[Milkeno]

There are a certain number of demons in a doom map. for any three demons, there are two who aren't of the same enemy type. for any four demons, there will always be two who are of the same enemy type. what is the maximum number of demons in the map?

Edited April 11, 2024 by Milkeno

[Redneckerz]

Depends on the map.

[Murdoch]

... what?

[Milkeno]

2 hours ago, Redneckerz said:

Depends on the map.

Think about this carefully, for any four demons, there will always be two who are of the same enemy type and for any three demons, there are two who aren't of the same enemy type. You have to fulfill both of these you can't insert just any number into one of them.

[Cilian]

is this a math problem?

[SleepyVelvet]

After your typo fix and assuming we're allowing for Icon of Sins and Wolfenstein SS, I believe the answer is

6

Edited April 11, 2024 by SleepyVelvet

[bejiitas_wrath]

Maybe we need a program to calculate this formula for any map.

[OliveTree]

are there just two who are the same demon out of every set of four? or are there at least two who are the same demon. likewise, are there just two demons who aren't the same as each other in every set of three, or are at least two demons out of every set of three different from each other. These produce different results.

[kwc]

Remember to visualize the complex problems, and relax…
 

At 7:30 a.m, a portal to hell traveling 60 light years an hour leaves Phobos bound for Deimos, 520 light years away. At the same time, a portal traveling 30 light years an hour and carrying 40 demons leaves Deimos bound for Phobos. It's eight Mancubi long and always carries the same number of demons on each trip. An hour later, a number of demons equal to half the number of minutes past the hour arrive but three times as many plus six get on. At the second teleporter on Deimos, half the demons plus two arrive but twice as many enter the teleporter as had entered the first teleporter on Phobos.

Edited April 11, 2024 by kwc

[Shepardus]

One, because you are the demons.

Spoiler

6 (3 types, 2 of each type)

 

[Osmosis Bones]

I would've aced my math exams if the questions were all Doom themed

[Redneckerz]

2 hours ago, Milkeno said:

Think about this carefully, for any four demons, there will always be two who are of the same enemy type and for any three demons, there are two who aren't of the same enemy type. You have to fulfill both of these you can't insert just any number into one of them.

I thought about it carefully. It depends on the map.

 

Spoiler

I also suck at math and i don't condone math related trivia on the morning.

 

[Chernobog The Exalted]

2 hours ago, kwc said:

It's eight Mancubi long

Americans will use anything but the metric system

[Chernobog The Exalted]

also, what even is this question? I don't understand, mortals confuse me sometimes

[ObserverOfTime]

Let’s analyze the problem step by step:

For any three demons, there are two who aren’t of the same enemy type. This means that there are at least two different enemy types among these three demons.

 

For any four demons, there will always be two who are of the same enemy type. This implies that there must be at least one pair of demons with the same enemy type.

 

To maximize the number of demons, we want to minimize the number of different enemy types. Let’s consider the following scenarios:

If we have only one enemy type, we can have at most three demons (since any four would violate the second condition).

If we have two enemy types, we can have at most two demons of each type (total of four demons).

If we have three enemy types, we can have at most one demon of each type (total of three demons).

 

Therefore, the maximum number of demons in the map is four. This can be achieved by having two demons of one type and two demons of another type. Any additional demon would violate the second condition.

 

Hence, the answer is four demons.

 

Edit: Only now I realize I probably did your math homework for you lmao

 

Edited April 11, 2024 by ObserverOfTime

[TheHambourgeois]

Two barons of hell stand at a fork in the road, one of them always tells the truth, one of them always lies. One path leads to a regular exit, one path leads to a death exit. What question can you ask them that will allow you to carry your inventory to the next map?

[Milkeno]

16 hours ago, Shepardus said:

One, because you are the demons.

  Reveal hidden contents

6 (3 types, 2 of each type)

 

I actually got 5 but I think that's due to my faulty programing
 

 

17 hours ago, bejiitas_wrath said:

Maybe we need a program to calculate this formula for any map.

I wrote mine in zscript >.> it outputs to the ammo count

[Maximum Matt]

And of course, the answers are completely nullified in Nightmare mode

[Insaneprophet]

A cyber demon exits a bathroom stall and tells doomguy he left a bfg in ther for him

Doom guy says "bull shit"

[LoatharMDPhD]

22 hours ago, kwc said:

Remember to visualize the complex problems, and relax…

...

// insert image of the Rock saying "semper fi, motherfucker..."

[LoatharMDPhD]

// isn't this am I am legend type affair where doomguy is the demon and all the monsters are the regular folk?

[ReaperAA]

Lets look at the conditions:

• 1st condition says that for any 3 demons, there must be 2 who are different from each other. Thus we can't have all 3 demons be of same type. This means that there can be a maximum of 2 of each demon type in the map.
• 2nd condition states that for any 4 demons, there must be 2 demons who are of the same type. Thus there can only be a maximum of 3 types of demons in the map, since adding a 4th type would violate the rule.

 

Thus we get the equation:

Number of demons per type x Number of types = 2 x 3 = 6

 

The answer is 6.

Edited April 12, 2024 by ReaperAA

[Doom-X-Machina]

The formula is sound until 1 of those demons happens to be a Pain Elemental.

[realjohnmadden]

4 hours ago, Doom-X-Machina said:

The formula is sound until 1 of those demons happens to be a Pain Elemental.