Universal Dungeon Maker : Dark Lord (by GameCoaster)

Hi, All! :>

The following is a guide to break through the highest difficulty levels with Dark Lord Emma.


If you look at the video first, it is possible without war horn and electrical short.


Next, look at the following image.


1) As you can see, there are no electrical short & war horn.
Therefore, it is liberated from the pressure to make relatively good traps and battle rooms.

2) If only a few elite monsters exist, that's enough. You do not need many monsters.

3) Emma's first skill is quite useful. And it is only required one cost.

4) Let's maximize the effect of Slow by arranging Icebolts on the first line.
If you have a Relic Pearl Shell, it is even better. Because it enhances the effect of Slow.


Okay. Let's look at the following conditions to achieve this.

1) First, there must be monsters & facility succession.

2) Relic: Pearl Shell, Mana recovery artifacts, 25% regeneration of the damage received, and frenzy related artifacts.

3) The tail decoration of Emma, an artifact dedicated to Emma, is essential. Because this strategy is for Emma users.(If the enemy is Fragile state, the Emma skill Fury will exert a great deal of power.)

4) Super Monsters are required. They are White Fang, Suparna, and Dryanid.(+ Death Knight)

5) Required skills of Dark Lord Emma : Parry Weapon, Foresight, Executioner, Mystic One, Earthquake.


*Strategic placement in the Dungeon.

1) Let's get hatcheries early and deploy them.

2) As you expand the dungeon, install an acceleration room in front of the Dark Lord room.

3) Place Panics and Icebolts referring to the screenshot above.
(They do not need level up, but let's upgrade.)

4) The role of the hatchery is to level up Dark Lord Emma.
If you create monsters, give them to Emma. She will be incredibly strong.


That's all. :>

Emma's massive attack will defeat most enemies!



My favorite Dark Lord is Emma. Emma is love. :>

Dungeon Maker : Dark Lord, Emma (by migelhososi)

 

In the current version, a total of 34 Heroes invade your dungeon. Who is the most powerful of them? Maybe, Ang...?!

 

Yeah, I hardly found them.

But someday, reviews and videos of this game will come out.
Maybe...? Lol

"Wait and Hope"

 

The Study of Hero Movements. (by @Quoad)


There is a Japanese user who examined it.
(https://twitter.com/tp_kuronekotei/status/1002457540515414017)


The numbers on the far left are Day.

The table on the left is where the hero arrives after 100 steps.
The table on the right shows the number of times the hero has touched that tile during 100 steps (probability)

It is quite amazing and interesting.
His(tp_kuronekotei) calculation results are quite similar to your(@Quoad) calculation results.





The Study of Hero Movements. (by @Quoad)

@Quoad said :

Has anyone calculated the trap worth / footfall of various paths / squares?
I’m sure there’s a simpler way of doing it than what I’ve begun to bodge out.
Regrettably, I’m not a mathematician and am not entirely sure what to google.

(The relationships are predictable and constant, though. Every path entering a corner has only 1 way out. Every path entering a side square has 2 exits. Every path entering a central square has 3 exits. So it should be fairly straightforward to formalise / calculate.)



@Quoad said :
And each square has regular patterns.
This should be totally doable, but I’m not sure my head’s up to it #rofl
Corners are clearly the least valuable. (100% or what they receive goes on to the next square, which also receives input from others).
Sides are clearly the next least valuable.
But by how much?

@deresi said :
I was simply understanding heroes' steps.
IMO, The rule of the heroes' steps...
- Heroes do not turn 180 degrees
- Heroes enter the boss room when coming in front of the boss room
- Heroes' steps on all routes with equal probability
But your analysis is quite interesting.

@Quoad said :
Yeah. Those rules. That’s what I’m trying to implement.
It should be a fairly straightforward repeating pattern. Just trying to puzzle out the paths does my head in. (And I don’t think it’s necessary, tbf - it should be possible to work it out inductively, whilst I’m bodging away at trial and error. Which swiftly becomes complex.)

@Quoad said :
The first four moves #rofl
Last pic is the room numbering scheme.



Edit: +1


Edit: and as a proportion of all moves taken....

Square 1 (entry square) receives

• 24.4% of ALL contacts.

The other central squares (in order) receive:

• 13.33%
• 8.89%
• 0.74%
• 0.25% of all contacts.

The sides (in order) each receive:

• 10.37%
• 5.18%
• 2.84% of all contacts.

And the first corners receive

• 7.78%.

I’m sure there are proportional relationships between those that would stabilise / tend towards stability over turns. And repeat visits to squares (vs first visits) might be significant. (Remove the first turn, and square 1 only receives 5.56% of visits).

I need to think about that tho!


@Quoad said :
I believe some aspects of this may be relevant, though a toddler’s bedtime demands immediate attention: https://en.m.wikipedia.org/wiki/Random_walk
I also suspect that the precipitous drop between the 2nd and 3rd central tile is an artefact of the number of steps i modelled / stopped at. I think the gap would diminish if there were +2. (I may be wrong).

@Quoad said :
Actually, yeah. I’m quite confident that the picture is incomplete, and the rate of drop off is artificially high. The first non-starting central square (#4) has ⅓ of that-turn steps the first time its encountered. That’s doubled (well, +⅓) the next time it’s visited.
#7 has 1/9 of that-turn steps the first time its encountered. That’s quadrupled (well, +1/3) next time it’s visited.
That’s... interesting!
Edit: I wonder if #10 also gains +1/3 on a second visit :hmmmmm:

@Quoad said :
Oh, hello.
And the %s for each first contact with each central square are ⅓, 1/9, 1/27, and 1/81.
That’s entirely predictable, bc each first contact arises only from the previous central square. And each path from a central square forks into 3.
Hmmm.

@Quoad said :
I’m also convinced that the number branches are important.
Looking at the path diagram:
3,1,2,2
3,1,2,3
3,3,2,1
3,3,2,2
3,3,3,2
3,3,3,3
3,3,3,2... (herein in its a mirror image of the first 5 paths)
And the cell numbers appear predictably (they’re also mirror images of each other from top to bottom, if you look at the final branch).
And the paths give you the %s. (1/3/1/2/2 = 1/12 = hence this being the rate for 7 and 11 for the top branch).
HMMMM.

@Quoad said :
I have modeled +1 turn, which does indeed suggest that the lower rates for subsequent central squares was an artefact of the turn I stopped at.

Turn 6 is the first 'cash out' move - when individuals hit C6, and so (effectively) exit the dungeon.

I am surprised to see that C2 looks almost as promising (after 6 moves, lol) as C1. I'm unsure what would happen over subsequent turns.

At 6 turns, my brain has begun to melt - because I'm having to figure out the proportion of each cell's value that comes from which direction, and (consequently) which can go elsewhere. Which is doing my head in for central squares. I'm not sure if this is actually complicated, or if I'm making it unnecessarily difficult by getting confused. (And once I've done it once, I think it should be straightforward from thereon in - it's just puzzling it out that once. Which I may well've already done, in one of these squares. C2?! HMMM).



@Quoad said :
BY the 7th move, the 4th central square will receive:

From C3:
One third of C3's input from C2;
one third of C3's input from T3;
and one third of C3's input from B3.

From T4:
Half of T4's input from T5
Half of T4's input from T3

From B4:
Half of B4's input from B5
Half of B4's input from B3

From C5:
One third of C5's input from B5
One third of C5's input from T5
And nothing C5's input from C6, because C6 GIVES NOTHING BACk.

OK. Er. Yeah.

THE SAME formula applies for the central 2 central squares (C3 and C4), I think.

Corners create special conditions.

Soooo...

for any established move after the ball is rolling, C2 will receive...

From C1:
Half of C1's input from T1
Half of C1's input from T2

From T2:
Half of T2's input from T1
Half of T2's input from T3

From B2:
Half of B2's input from B1
Half of B2's input from B3

From C3:
One third of C3's input from B3
One third of C3's input from T3
One third of C3's input from C4

Christ.

Edit: so, ok, the only difference in that appears to be the role of C1. Which (for a single path) has no C0 feeding into it.

I thought this would be easy, once I cracked it. Once cells have the capacity to receive 2nd or 3rd visits, it becomes a lot more complex!

@Quoad Said :

After posing this question elsewhere, I think someone has very kindly modelled 1,000,000 heroes, each taking up to 100 steps.

I think it’s been done correctly - ie, all enter in square 1, and exit if they touch square 16 (column 6 middle). Backtracking / 180 degree turns not allowed.

Central squares are 1, 4, 7, 10, 13, 16. The others are top and bottoms in their relevant columns (so 2 & 3 are C1, 5 & 6 are C 2...). 2, 3, 17 and 18 are the corners.

Long story short - the central squares in columns 1 and 2 are nearly identical. Corners are poor, for any measure of relative footfall. Footfall diminishes as heroes progress.

This is only really relevant for traps tho, and even then it’s complicated to work out how to apply it - particularly when using traps whose damage stacks.