A Complete(ish) Map of DMS (Wanted: A myth buster)

[-Shiny_Tsukkomi-]

petiele

-pic-
Screenie of the room you're missing.

Thanks! biggrin

Now to check if the dimensions fit the map...meh, I'll do that tomorrow or somethin'. *tired*

[MeMyselfAndEye]

as far as the math bit goes, rule of thumb for probability is (options)^(possibilities)
That's why even if you flip a coin 10 million times you will only get a *50:50 chance

*caveat in this example: it has been shown that the "face" portion of a coin slightly increases the chance of "tails" when being tossed, but that's a physical property of the coin and has nothing to do with the fair statistics of it.

so i suppose it would be (20)^(#of unique rooms possible) if you only look at ym, added to (20)^(#of unique rooms possible). Now if each of the possible rooms could only be used once, i think the equation is then altered by adding factorials to the powers to show each room can only be chosen once so
(20!)^(#of unique rooms possible)

example:
repeatable: 20!^20=20^20+19^20+...+1^20

not repeatable: 20!^20!= 20^20+19^19+...+1^1

now i'm not a Mathematician by any means but from what i remember from Calculus 2 that should be correct (please do inform me if i am wrong). On a related note i can almost guarantee the first option is correct for this circumstance because I've experience multiple dead ends in a single ym before.

[Zzyli]

MeMyselfAndEye

as far as the math bit goes, rule of thumb for probability is (options)^(possibilities)
That's why even if you flip a coin 10 million times you will only get a *50:50 chance

*caveat in this example: it has been shown that the "face" portion of a coin slightly increases the chance of "tails" when being tossed, but that's a physical property of the coin and has nothing to do with the fair statistics of it.

so i suppose it would be (20)^(#of unique rooms possible) if you only look at ym, added to (20)^(#of unique rooms possible). Now if each of the possible rooms could only be used once, i think the equation is then altered by adding factorials to the powers to show each room can only be chosen once so
(20!)^(#of unique rooms possible)

example:
repeatable: 20!^20=20^20+19^20+...+1^20

not repeatable: 20!^20!= 20^20+19^19+...+1^1

now i'm not a Mathematician by any means but from what i remember from Calculus 2 that should be correct (please do inform me if i am wrong). On a related note i can almost guarantee the first option is correct for this circumstance because I've experience multiple dead ends in a single ym before.

Zzyli - Zzy as in Xylophone, Li as in leaf

The thing that complicates the issue a bit is it's about the permutations of entrances & exits of the room. Would you consider a map where room 4 joins to room 5 on the left the same as where it joins on the right?

We have
1 Ingress
1 Egress
1 1 door room
10 2 door rooms
9 3 door rooms
1 4 door room

Thus 54 Apertures in total. Some are excluded from joining to others by virtue of them being in the same room, also the Ingress & the Egress can't join. I'm not sure exactly how to work it from here... I'm still thinking about it. Can we start with the 4 door room & randomly assign rooms from there? Can we then safely move through the number of doors & assign them randomly? How do we exclude impossible permutations?

Accentuate the positive and eliminate the negative...

[Runic Bulwark]

-Shiny_Tsukkomi-

MeMyselfAndEye

awe sad, i hit a dead end yesterday in GR with my attunement crew. I would have screenied if i had known (only just found this thread today)

gonk

Bummer! The elusive room remains elusive... gonk

eek So it does exist....

[User 9900289]

le bump

[Nanodeath]

-Shiny_Tsukkomi-

I'm not posting that fast...:c

*last reserve*

Lol if I knew a good way to map the mazes by hand, I'd write a program to do it. xd
Hehe I had some ideas, but they seemed overcomplicated. rofl

[Nanodeath]

-Shiny_Tsukkomi-

Are there any good mathematical people out there who are capable of calculating how many possible mazes can be generated from the conditions given stated above? I have NO idea how to calculate that.

I bet the answer requires some scientific notation...


Any suggestions and feedback are appreciated. ^^

After much brain bewilderment, I found your answer.
There are 54 doors total in yellow maze (10 2-door rooms, 9 3-door rooms, 1 4-door room, and 3 1 door rooms (entrance and exit and DE)).

Therefore if I keep picking a random door and connecting it with another random door, there are. 54! Possible permutations of doors as I connect them. However, the order in which I pair them off doesn't matter, so I'll divide by the number of possible permutations of pairs of doors, which is 27!.. Also you need to divide by 2^(number of pairs) since direction of each connection is irrelevant

So for YM it's (54!/27!)/(2^27) Possible mazes.
Same logic for GRs gives (52!/26!)/(2^26) Possible mazes.

Lol also, for every set of identical rooms, divide the result by n!, where n is the number of identical rooms. (Edit: Wait a second.... I smell this being invalid. rofl with 2 unique 2-door rooms, my formula evaluates to 3, which is correct.. but if the rooms are identical, this would reduce it to 1.5, but it should still be 3 @_@ what I do wrong? crying )

EDIT: CRAP! I FORGOT ALL THE ROOMS HAVE TO BE CONNECTED IN 1 GROUP... crying Lol this count is wrong... back to square one. crying

[Mayakashi Minshuu]

Been looking for maps since I started playing DMS.
Thank you.

[Nanodeath]

Zzyli

MeMyselfAndEye

as far as the math bit goes, rule of thumb for probability is (options)^(possibilities)
That's why even if you flip a coin 10 million times you will only get a *50:50 chance

*caveat in this example: it has been shown that the "face" portion of a coin slightly increases the chance of "tails" when being tossed, but that's a physical property of the coin and has nothing to do with the fair statistics of it.

so i suppose it would be (20)^(#of unique rooms possible) if you only look at ym, added to (20)^(#of unique rooms possible). Now if each of the possible rooms could only be used once, i think the equation is then altered by adding factorials to the powers to show each room can only be chosen once so
(20!)^(#of unique rooms possible)

example:
repeatable: 20!^20=20^20+19^20+...+1^20

not repeatable: 20!^20!= 20^20+19^19+...+1^1

now i'm not a Mathematician by any means but from what i remember from Calculus 2 that should be correct (please do inform me if i am wrong). On a related note i can almost guarantee the first option is correct for this circumstance because I've experience multiple dead ends in a single ym before.

Zzyli - Zzy as in Xylophone, Li as in leaf

The thing that complicates the issue a bit is it's about the permutations of entrances & exits of the room. Would you consider a map where room 4 joins to room 5 on the left the same as where it joins on the right?

We have
1 Ingress
1 Egress
1 1 door room
10 2 door rooms
9 3 door rooms
1 4 door room

Thus 54 Apertures in total. Some are excluded from joining to others by virtue of them being in the same room, also the Ingress & the Egress can't join. I'm not sure exactly how to work it from here... I'm still thinking about it. Can we start with the 4 door room & randomly assign rooms from there? Can we then safely move through the number of doors & assign them randomly? How do we exclude impossible permutations?

Accentuate the positive and eliminate the negative...

Well based on my math, the number of ways to connect 54 apertures is 27! (and I did doublecheck my formula on some test mazes), but yeah.. idk how to count impossible permutations yet. >:[ -will figure it out very soon, or just count them..- rofl

[Zzyli]

Hashtable

Well based on my math, the number of ways to connect 54 apertures is 27! (and I did doublecheck my formula on some test mazes), but yeah.. idk how to count impossible permutations yet. >:[ -will figure it out very soon, or just count them..- rofl

Zzyli - Zzy as in Xylophone, Li as in leaf

I don't know if it's easier to just brute-force the possible options for each door & then combine them in some kind of 'box expression' probability (probably only I call it that...). I've been mulling it over a bit.

I think I may brute-force the combinatorics for each room & see if we can put our heads together on constructing a probability model after that.

Accentuate the positive and eliminate the negative...

[Nanodeath]

Zzyli

Hashtable

Well based on my math, the number of ways to connect 54 apertures is 27! (and I did doublecheck my formula on some test mazes), but yeah.. idk how to count impossible permutations yet. >:[ -will figure it out very soon, or just count them..- rofl

Zzyli - Zzy as in X[/ocolor]ylophone, Li as in leaf

I don't know if it's easier to just brute-force the possible options for each door & then combine them in some kind of 'box expression' probability (probably only I call it that...). I've been mulling it over a bit.

I think I may brute-force the combinatorics for each room & see if we can put our heads together on constructing a probability model after that.

Accentuate the positive and eliminate the negative...

To count impossible permutations:
Assume 2 sets of rooms that aren't connected.

Then for each of these two sets use the formula I stated earlier, and multiply these 2 numbers together to find number of impossible permutations for these 2 sets. (Odd number of doors = 0 possibilities

Add all the results up and subtract from the total.

Lol I'm sleepy, but I'm sure this is easier than brute forcing the whole maze. I'll get back on that. surprised

Edit: wait.. if there are 4 groups of nonconnected rooms a, b, c, and d, then you would need to make sure that groupings of ABC and D, AB and CD, A and BCD, etc weren't counted seperately... confused Hmmmm.... e_e

[Nanodeath]

Zzyli

Hashtable

Well based on my math, the number of ways to connect 54 apertures is 27! (and I did doublecheck my formula on some test mazes), but yeah.. idk how to count impossible permutations yet. >:[ -will figure it out very soon, or just count them..- rofl

Zzyli - Zzy as in Xylophone, Li as in leaf

I don't know if it's easier to just brute-force the possible options for each door & then combine them in some kind of 'box expression' probability (probably only I call it that...). I've been mulling it over a bit.

I think I may brute-force the combinatorics for each room & see if we can put our heads together on constructing a probability model after that.

Accentuate the positive and eliminate the negative...

Had to edit both of my posts because of an absurd error. :rouge: I shouldn't do math half asleep. xp

Lol the 27! Was an invalid simplification. rofl

Hehe anyways, we really should figure this out -interest sparked-

[Nanodeath]

iMuSic_Nvr_sLeEpS

-Shiny_Tsukkomi-

MeMyselfAndEye

awe sad, i hit a dead end yesterday in GR with my attunement crew. I would have screenied if i had known (only just found this thread today)

gonk

Bummer! The elusive room remains elusive... gonk

eek So it does exist....

If the numbers of rooms Shiny stated for GRs:
11 2-door rooms
8 3-door rooms
1 4-door room

Are always the same, then it would be impossible to add a dead end room. confused

Maybe it changes? surprised

[Runic Bulwark]

Hashtable

iMuSic_Nvr_sLeEpS

-Shiny_Tsukkomi-

MeMyselfAndEye

awe sad, i hit a dead end yesterday in GR with my attunement crew. I would have screenied if i had known (only just found this thread today)

gonk

Bummer! The elusive room remains elusive... gonk

eek So it does exist....

If the numbers of rooms Shiny stated for GRs:
11 2-door rooms
8 3-door rooms
1 4-door room

Are always the same, then it would be impossible to add a dead end room. confused

Maybe it changes? surprised

Hmm perhaps a glitch

[-Shiny_Tsukkomi-]

iMuSic_Nvr_sLeEpS

Hashtable

iMuSic_Nvr_sLeEpS

-Shiny_Tsukkomi-

MeMyselfAndEye

awe sad, i hit a dead end yesterday in GR with my attunement crew. I would have screenied if i had known (only just found this thread today)

gonk

Bummer! The elusive room remains elusive... gonk

eek So it does exist....

If the numbers of rooms Shiny stated for GRs:
11 2-door rooms
8 3-door rooms
1 4-door room

Are always the same, then it would be impossible to add a dead end room. confused

Maybe it changes? surprised

Hmm perhaps a glitch

A glitch from beta DMS? Sounds cool...xD

In the event that a dead end would exist in a generated GR maze, they would have to add in another 2-way room.

*dying to see this rumored dead end*