So basically this removes the “|” and it would just be how1what2when3where4?
For this, we could have it run a check 50 times, and then randomInt will decrease by 1 everytime it returns 0
Oh wait , that would just leave us with the last word…
Yep!
It’d be more like 0how1what2when3where though. Starting from 0 gives us 1 more number before we reach triple digits, which probably won’t ever be relevant, but still.
We could have the property name be “lib132”, where lib(1)01 is the library number, and lib1(32) is the amount of words in that library.
Give me a bit to make the code
nevermind it wont let me put the block in cus its
green ![]()
Late to the party but this is my mojo ![]()
where is our starting point exactly what’s gonna be our first approach to test this and further on investigate upon this?
Okay I will find all the posts for you to read so you know where we are at.
https://forum.creative.gimkit.com/t/pseudo-ai-with-rng-how-do-we-do-it/185827/30?u=suspnded_arr0w_shaft
https://forum.creative.gimkit.com/t/pseudo-ai-with-rng-how-do-we-do-it/185827/31?u=suspnded_arr0w_shaft
https://forum.creative.gimkit.com/t/pseudo-ai-with-rng-how-do-we-do-it/185827/41?u=suspnded_arr0w_shaft
https://forum.creative.gimkit.com/t/pseudo-ai-with-rng-how-do-we-do-it/185827/73?u=suspnded_arr0w_shaft
https://forum.creative.gimkit.com/t/pseudo-ai-with-rng-how-do-we-do-it/185827/87?u=suspnded_arr0w_shaft
I didn’t link them because it was really cluttered…
Testing f@iled. Please can you explain why?
Wolf.rising as my ADHD will not let me go to each and every link suspended gave
me can you just summarize what you guys understand rn
100TH POST ![]()
@Gimkitsuggestor
What we have right now is just a giant random number-generating machine.
Basically, we set how as the only response for the AI.
BUT
We want a more linguistic approach, so instead of just printing “how”, it finds related words in a library corresponding to “how” (001-001-001-001)
example:
BUT
@some_kid found a much more compact and storage-efficient way to store a library. Check the below quote.
What we are working on is finding the max integer, but this new setup could technically increase our word count by a whopping 256x. (this is assuming each library contains “a” for each word, which would be silly, but this is all still theoretical so…)
Technically, the setup has less than 256 entries, because after 9, our entries need more than 1 digit for the index.
We’re storing our libraries in text properties as their default value. Libraries look like this:
0and1additionally2also3in addition4secondly and so on.
@Gimkitsuggestor
To put this into perspective, our previous max word count was 4096.
The average length of a word is 5.1 characters so lets plug that in.
512 / 5.1 = 100.392156863
100.392156863 x 4096 = 411206.27451
411206 max word count
The OED only has ~600000 words. Thats 68% of the english dictionary we could use!
Disclaimer: Number is not accurate AT ALL
This is because:
- as @some_kid stated, after 9, we use double-digits
- Not all words are 5.1 letters long…
512 / 5.1 would be the number of words we can have per property, not taking into account the numbers we use to index them. Instead of multiplying that by our previous max word count, we’d multiply it by the number of properties we have available to use, which would likely be ~124 considering we use properties for more than one thing.
So, taking into account our property limit and our indexes, we can store ~9,069 words.
I got this figure with 512 / 7 * 124. The lack of a .1 on the end of 7 is due to the 10 single-digit indexes we get per library.
Consider that, in order to achieve this figure, we have to use every space in every library, which likely wouldn’t be feasible due to each library only containing a specific type of thing, like superheroes or Gimkit device names.
Okay so I am really good at math trust.
I might be able to work on the AI later.
Please update me with any ideas!
NOOOOOOOOOOOO @kormorant GOT SUSPENDED I MIGHT ACTUALLY CRY ![]()
We really needed korm banned thanks jeffy they was creating lots of dissent and clutter and ruining the forums
literally one of the people who shouldve been made TL4 is now temp-banned ![]()
lol wait korm is unsuspended on my birthday.
My birthday present is korm ![]()
@some_kid I think I found a way to get the length of a library.
![]()
I feel like this is too simple though, so correct me if i am wrong.
When I said the length of the library, I meant determining how many entries are in it.
Sorry if what I said was confusing!
Hmmmmm
So what if we have a loop to count the numbers in the property starting at 1, and ending when it returns 0/false
Then that will broadcast to the library searcher, and tell it what to roll
That does work. You’d just have to set up a property so that the library searcher can get the information from the loop.

