I checked my data and found out, that any combinations of
B1 B2 D in Door 1 2 3
B2 D DT in Door 2 3 4
D DT RT in Door 3 4 5
never occured so far.
And so the counter part of
DT RT S in Door 4 5 6 (Counter of 1 2 3)
RT S B1 in Door 5 6 1 (Counter of 2 3 4)
S B1 B2 in Door 6 1 2 (Counter of 3 4 5)
did neither ofc
Same with
B1 B2 D DT never been in 1 2 3 4
B2 D DT RT never been in 2 3 4 5
D DT RT S never been in 3 4 5 6
DT RT S B1 never been in 4 5 6 1
RT S B1 B2 never been in 5 6 1 2
S B1 B2 D never been in 6 1 2 3
in any combination
4 Likes
Green: 0%
Non-trap in door 4
Non-trap in door 5
Interesting fact:2 & 3 if B2 is in 5 so far
1 Like
AMT
September 27, 2022, 1:30am
509
CaptainAwesome:
B1 B2 D in Door 1 2 3
B2 D DT in Door 2 3 4
D DT RT in Door 3 4 5
Isn’t that because D can never be on 3? (…and B1 on 1, etc)
WhoDat
September 27, 2022, 2:55am
510
Thank you for sharing your analysis but let me understand the last part. Why B2 cannot be in Room 6? because if the RT falls in Room 4 and DT falls in Room 5 then B2 will be in Room 6. is it possible to have B2 in Room 6? I checked the screenshots and never found the two traps in Room 4 and 5 as a combination as I stated. I would like to understand the logic behind it please. Thanks
TbIKBA:
I can attempt to answer that.
4RT/5DT can not occur as it would create a closed loop leaving 6 in its own thread.
4RT, check 5 → 5DT would refer back to check 4 etc.
The way they have programmed it, all 6 must create a full loop without any self-contained closed loops.
Another way to check would be to look backwards:
6B2 → 2B1 → 1B3 → 3S → 6… is already B2
Full loop without 4/5. This version isn’t very practical while going through the dungeon, but after the fact its easy to learn/figure out.
Just to clarify, I’m not saying 6B2 can never happen, it just can never happen (at this time) when the combination is 1B3, 2B1, 3S.
3 Likes
TbIKBA
September 27, 2022, 3:23am
512
I myself do not yet understand the logic why it is so coded. But the algorithm works in such a way that after each opening of the door, the “diagonal” is saved
EDIT: We can continue the logic. Open door 4 and there will be a RT. Then the “diagonal” will be rebuilt and a cross will become in 5DT
2 Likes
You might have missed
means B1 (in 2 or 3) and B2 (in 1 or 3) and D (in 1 or 2) can never happen.
i.e.
It fits with the explanation of @TbIKBA
TbIKBA:
We have already understood behind which door certain results cannot be
1
2
3
4
5
6
B1
x
B2
x
B3
x
DT
x
RT
x
S
x
And if we open any door, then the diagonally symmetrical door also becomes with a cross
1
2
3
4
5
6
B1
x
x
B2
x
B3
o
x
DT
x
RT
x
S
x
I went on to figure out why there couldn’t be three bosses in the first three doors. And I understood the logic that can be followed for any opening doors. I’ll use the example of the two-horns in doors 1-2-3 to show why they can’t be there. This applies to other combinations as well.
1
2
3
4
5
6
B1
x
x
B2
x
x
B3
o
x
x
x
x
x
DT
x
x
RT
x
x
S
x
x
I decided to remove the extra row and column so that they do not distract attention. It turned out that the diagonal line was preserved if the lines were rearranged in the right order. That is, in each row and column there is one cross.
2
3
4
5
6
B1
x
B2
x
DT
x
RT
x
S
x
Let’s continue, in room 2 we will put B1. And remove the extra row and column (I will skip the step with setting “o” and “x”)
The result is a 4x4 table in which you need to complete the diagonal so that the original logic continues to work.
3
4
5
6
B2
х
DT
x
RT
x
S
x
And now we see that there can no longer be a boss in room 3. Again, this works for other combinations that I have tested. I think we can continue this logic for the 3x3 table, but by then we should already have a perfect run
and stretches to the 4- and 5-strings i. e. B1 B2 D DT RT can not be in any combination of 12345. It would leave S alone in 6.
So there must be a closed loop of 6 in all rooms, but never a loop of less then 6 in their belonging rooms.
3 Likes
AMT
September 27, 2022, 4:49am
514
1 Like
I also works for non-adjacent rooms
B1 in 3 & D in 4 exclude DT in 1
If you want to go through the data by yourself
B1
B2
D
DT
RT
S
5
4
6
1
3
2
2
6
4
5
1
3
4
3
1
6
2
5
3
4
5
6
2
1
5
3
6
1
2
4
4
3
1
6
2
5
4
6
2
5
3
1
4
1
2
6
3
5
4
1
5
3
6
2
5
3
6
1
2
4
5
4
1
3
6
2
6
3
5
1
4
2
5
3
4
1
6
2
4
3
6
5
2
1
4
6
5
3
2
1
5
1
4
6
3
2
3
1
5
6
4
2
4
1
5
3
6
2
6
1
4
5
2
3
2
3
4
5
6
1
5
1
2
6
4
3
4
1
2
5
6
3
2
3
4
5
6
1
3
4
6
5
1
2
4
6
2
3
1
5
4
6
1
2
3
5
4
3
1
6
2
5
2
6
4
5
1
3
4
1
6
5
3
2
5
3
6
1
2
4
4
1
5
6
2
3
2
5
4
6
3
1
4
3
6
2
1
5
5
4
2
6
3
1
4
6
5
2
1
3
6
4
5
1
2
3
6
1
4
5
2
3
3
1
6
2
4
5
2
3
4
6
1
5
4
6
5
2
1
3
5
6
2
1
3
4
5
3
1
6
4
2
6
5
4
1
3
2
2
6
5
3
1
4
4
6
5
3
2
1
5
3
1
2
6
4
4
5
6
3
1
2
3
5
6
1
4
2
5
6
2
3
4
1
2
5
4
6
3
1
6
3
4
1
2
5
6
4
5
3
1
2
6
4
1
5
3
2
3
5
2
1
6
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4
1
5
6
2
3
6
5
1
3
4
2
4
1
5
6
2
3
5
3
1
6
4
2
6
4
5
1
2
3
4
6
2
5
3
1
6
1
2
5
3
4
5
3
4
6
2
1
5
2
1
6
4
3
6
5
2
1
4
3
4
5
1
6
3
2
2
6
5
1
4
3
6
1
5
2
4
3
2
4
1
5
6
3
5
6
4
1
2
3
4
3
6
5
2
1
5
4
1
6
2
3
6
4
1
5
3
2
4
1
6
3
2
5
6
4
1
5
3
2
5
1
2
6
4
3
4
3
1
6
2
5
4
6
5
2
1
3
5
4
1
3
6
2
4
1
5
3
6
2
2
3
4
6
1
5
6
4
2
1
3
5
3
1
6
2
4
5
3
6
5
1
2
4
4
6
1
5
2
3
5
4
2
1
6
3
3
6
2
1
4
5
3
1
5
6
4
2
3
5
2
1
6
4
4
3
5
6
1
2
2
4
5
3
6
1
5
3
1
2
6
4
5
1
4
6
3
2
5
6
1
3
2
4
6
1
5
2
4
3
2
6
4
1
3
5
3
5
6
2
1
4
2
6
4
1
3
5
4
6
2
3
1
5
5
4
6
3
2
1
2
6
1
3
4
5
5
6
1
2
4
3
6
4
2
1
3
5
6
3
1
5
2
4
4
1
2
6
3
5
2
3
6
5
1
4
5
6
4
2
3
1
6
1
4
5
2
3
4
1
2
5
6
3
2
3
5
6
4
1
4
5
1
2
6
3
4
5
1
2
6
3
4
3
1
5
6
2
4
1
6
5
3
2
3
5
6
1
4
2
3
1
4
6
2
5
4
1
5
3
6
2
5
1
6
2
3
4
2
6
4
5
1
3
6
4
1
3
2
5
6
4
1
3
2
5
4
3
1
6
2
5
3
4
6
1
2
5
5
6
4
2
3
1
2
6
5
3
1
4
6
1
4
5
2
3
6
4
1
5
3
2
3
5
2
6
4
1
3
4
2
6
1
5
6
1
5
3
2
4
5
6
1
2
4
3
4
6
1
2
3
5
5
3
1
2
6
4
2
6
4
5
1
3
6
5
4
1
3
2
4
3
6
2
1
5
6
5
4
1
3
2
5
1
4
2
6
3
5
6
1
2
4
3
2
4
1
5
6
3
5
6
1
2
4
3
3
1
5
6
4
2
6
4
2
5
1
3
6
5
1
3
4
2
6
1
4
5
2
3
4
6
1
5
2
3
2
3
5
1
6
4
2
4
5
6
1
3
3
6
4
5
2
1
5
4
6
3
2
1
2
6
4
1
3
5
4
5
2
6
1
3
6
3
5
1
4
2
2
5
1
3
6
4
6
5
1
3
4
2
2
3
4
6
1
5
5
4
2
6
3
1
5
4
2
1
6
3
3
5
6
2
1
4
4
5
6
3
1
2
3
6
5
2
4
1
4
1
2
5
6
3
6
1
2
3
4
5
3
5
4
6
1
2
2
6
1
5
3
4
3
4
6
5
1
2
3
6
4
2
1
5
3
5
6
2
1
4
4
1
2
5
6
3
4
6
5
3
2
1
5
1
2
3
6
4
5
1
2
6
4
3
3
5
4
2
6
1
5
4
1
3
6
2
5
4
2
6
3
1
2
6
4
5
1
3
3
6
2
5
1
4
5
3
6
2
4
1
6
3
5
2
1
4
3
1
5
2
6
4
6
3
4
1
2
5
4
3
1
5
6
2
4
1
6
3
2
5
4
1
2
6
3
5
4
6
5
3
2
1
2
3
4
6
1
5
2
6
5
1
4
3
2
6
1
3
4
5
5
3
1
6
4
2
3
5
6
2
1
4
3
5
4
2
6
1
5
6
1
3
2
4
4
1
5
6
2
3
2
3
6
5
1
4
3
4
5
1
6
2
3
6
4
5
2
1
5
4
2
1
6
3
5
6
4
2
3
1
3
5
4
2
6
1
5
3
1
6
4
2
2
3
4
6
1
5
2
5
4
1
6
3
5
6
2
1
3
4
5
4
6
1
3
2
4
1
2
5
6
3
6
5
1
3
4
2
6
3
4
5
1
2
5
6
4
1
2
3
6
5
2
3
1
4
6
1
5
2
4
3
6
5
2
1
4
3
5
1
6
2
3
4
5
4
2
1
6
3
4
1
5
6
2
3
6
1
2
5
3
4
4
5
2
3
6
1
3
4
2
5
6
1
3
5
6
2
1
4
3
6
2
1
4
5
2
6
5
1
4
3
4
5
2
6
1
3
3
1
6
2
4
5
2
6
5
3
1
4
5
3
1
2
6
4
5
6
1
3
2
4
5
6
1
2
4
3
2
4
1
6
3
5
3
5
6
1
4
2
3
6
2
5
1
4
3
4
2
6
1
5
2
4
1
5
6
3
4
1
5
6
2
3
5
4
6
1
3
2
3
1
5
6
4
2
6
5
4
2
1
3
6
4
1
3
2
5
3
1
4
5
6
2
3
6
2
1
4
5
4
1
6
3
2
5
3
6
5
2
4
1
6
1
2
5
3
4
3
6
2
1
4
5
5
3
6
2
4
1
2
5
4
6
3
1
4
1
6
5
3
2
5
3
4
1
6
2
5
6
4
2
3
1
5
4
1
3
6
2
2
4
6
3
1
5
2
5
6
1
3
4
6
3
4
1
2
5
4
3
6
2
1
5
6
3
1
5
2
4
4
5
1
6
3
2
2
5
6
3
4
1
3
6
2
1
4
5
6
3
5
2
1
4
3
5
2
1
6
4
2
3
6
1
4
5
2
5
4
6
3
1
2
4
5
3
6
1
3
4
6
5
1
2
4
1
6
5
3
2
3
6
5
2
1
4
3
5
2
6
4
1
5
3
4
6
2
1
4
5
1
6
3
2
5
4
2
1
6
3
2
5
1
6
4
3
6
4
2
1
3
5
3
5
4
6
1
2
3
6
4
2
1
5
5
6
4
1
2
3
6
5
4
1
3
2
4
6
1
5
2
3
2
3
5
1
6
4
4
6
2
3
1
5
4
6
1
2
3
5
3
6
2
1
4
5
3
4
6
1
2
5
4
3
1
6
2
5
This leads to the following conclusion
Door 1 can NOT contain a DT anymore since B1 in 3 and D in 4 exclude DT in 1
It’s way easier to understand if we replace
Door 1 can NOT contain a 4 anymore since 1 in 3 and 3 in 4 exclude 4 in 1.
Another example from my layout today
Opening 4 (B2 aka 2) followed by opening 2 (RT aka 5) excludes DT aka 4 in 5
Additional Example
Door 6 can not have a DT
It doesn’t help which door should be your 2nd pick, but it helps with your 3rd and following picks.
2 Likes
NeilTur
September 27, 2022, 6:46am
516
TheIdleOne:
Just to clarify, I’m not saying 6B2 can never happen, it just can never happen (at this time) when the combination is 1B3, 2B1, 3S.
Does that mean 6D cannot be when 1B2, 2B1, 3S as well?
1B2 → 2B1 shouldn’t happen as well, but 3S → 6D is also a closed loop
3 Likes
AMT
September 27, 2022, 7:06am
519
3 Likes
lmfao.
The good news: I had a good feeling it wasn’t 4,5.
The bad news: I thought I might have been back on the 4 track, but in reality Im still on the 5 track. 4 was a cursed trap, run over.
1DT
Tracked Run Summary: Day 26
Brown: 3,5,6 (4 was an altar) 3,4,6 (2 was a diamond bonus) 4,5,6 (3 was an altar) 3,5,6 (4 was an altar) 3,4,6 (5 was descending stairs to level 3) Purple: 3,5,6 (4 was an altar) Diamond: 4,5,6 (3 was a diamond bonus) Blue: 3,5,6 (4 was an altar) Diamond: 1,4,5 (2 was an altar) Blue: 2,4,6 (5 was ascending stairs)
1 Like
45
But I could figure out, that 6 can not be RT
And from here on...
3 can not be RT
Followed by...
2 can not be RT.
1 Like
AMT
September 27, 2022, 7:27am
523
Exactly like yesterday’s, except the descending stairs instead of ascending
2 Likes
Day 35. 4/35.
Used the tool. I still had better chance of getting a perfect run, than trying blindly, but no luck today.
546132
1 Like
Shaylynn:
I took your first 3 doors and checked my data.
1 Like
4th door 50%, 0, 33% should always actually be 50%, 0, 50%
2 Likes
