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Square, Circle, Tri then Star for me

Btw there are some leftover monster after u flip the switch , couldn't get past the barrier behind it

< Message edited by yjjoel26 -- 3/24/2020 16:27:19 >

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It goes least to most blinky. The first lever will have one static light with no blinking. The second will have a blinky-static-blinky pattern. The third will be blinky-blinky-static. Finally, the fourth will be blinky-no light-blinky. I hope this helps!

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Yeah, I followed the same pattern and it worked for some reason (though the shapes seems to be randomized judging from the other replies). I still don't get why "blink-on-blink" is somehow "less blinky" than "blink-blink-on" though. This seems like a bad puzzle overall - I could have sworn in the beginning that the answer had something to do with superpositioning the lights and/or some switches overriding others, but nope...the answer is apparently "go from least blinky to most blinky" lolwut?

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My issue with the puzzle is that what is considered "more" or "less" blinks. Most people worked out that the sequence was going to be either most blinks to least blinks or vice versa. But to me blink-light-blink is more "blinky" that blink-no light-blink. So I kept getting it wrong. I think speed of the lights flashing should be changed so they are more distinct for each other (so faster light flashes for more "blinky").

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I don't see how it could be a binary puzzle is there are 3 states: flash, off, on. That is trinary.

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These are what I had:

circle - x1x
square - xx1
star - x0x
triangle-001

0=off
1=on
x=blinking

triangle(1)-->circle(2)-->square(3)-->star(54)

circle=treat both blinking states as 0
square=from left to right, first is 0 and second is 1

The blinking ones can be treated as either 0 or 1.

Credit goes to Beeeen#6369 from Discord.

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Glad to see the solution, although I have to agree it sounds like it's going to be more to do with binary (or trinary?) systems than just relying on the blinks/frequency, but I'm bad at binary math xD

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Each pattern represents the last 3 digits for 1, 2, 3, and 4 in binary. It works when you keep in mind with what we learned; blanks are zero, statics are one and blinking can be either or.

1 is 001 or Blank, Blank, Static
2 is 010 or Blinking, Static, Blinking
3 is 011 or Blinking, Blinking, Static
4 is 100 or Blinking, Blank, Blinking

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You mean the room with the leftovers and the invisible wall you can't get past?

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That's the one.

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Although blinking can either be 0 or 1 but at the same time it isn't e.g. each blinking is specifically either 0 or 1 but which it is is hidden from you and you just have to figure it out - basically ignore the blinking and treat the rest as binary.

For a further breakdown of StarAngel's post, bear in mind it is showing the sequence so the numbers are 1 - 4:

Blank, Blank, Static - 1, nothing hidden
Blinking, Static, Blinking - 2 or 3 (or 6/7 but capped at 4)
Blinking, Blinking, Static - 1 or 3 (or ...), cannot be 1 so 3
Blinking, Blank, Blinking - 1 or 4 (or ...), cannot be 1 so 4

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Is anyone able to go to it again after finding it for the first time? I can't seem to reproduce it even though I followed the same procedure - find the correct lever sequence, go to the lever in the other room then go back to the previous screen. It would be nice to for someone to post a screenshot in here for posterity.

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The secret room isnt lever dependent, I believe! Poke around near the star room and keep an eye on the reflections in the floor for clues.

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I'm posting a screenshot of the secret room for archiving purposes (this war is supposed to be about an archive after all :P): https://i.imgur.com/t2FGeKD.png