Anonymous at Sun, 16 Feb 2025 09:50:38 UTC No. 16587710

Simply claim that you cannot know whether it's true or false until it is destructively observed, at which point it's neither true or false, it only *was* true or false, before you destroyed it, you monster.

Anonymous at Sun, 16 Feb 2025 16:40:47 UTC No. 16588046

>>16587665
I need the help chat I have like 16 like laters and most of the time most of those 16 get uptoo like 16*47 or more.

I can't manually make that as conplex as posible when I start doing the logic gates and such.
I need a trick of some sort.

Anonymous at Sun, 16 Feb 2025 16:55:45 UTC No. 16588058

>>16587665
https://en.wikipedia.org/wiki/Boolean_satisfiability_problem

Anonymous at Sun, 16 Feb 2025 17:59:53 UTC No. 16588151

>>16588058
I got some good ideas from that thanks.

But like whats your opinion about it anon, if you have a list of like 1200 bolean letters how do you make an equation as complicated as posible yourself.

Like I might use like something similar to the DPLL algorith you gave me but with formulas that sesignate entire chunks and then use a different number showing system when I have to get the full thing by hand.

A lot of the algorithsm are about simplyfing the equation but I am unsure if thier is any work on how to make them as large as posible.

🗑️ Anonymous at Sun, 16 Feb 2025 22:37:36 UTC No. 16588522

>>16587710
If you don't know whether it is a tulpa, then it is a tulpa.

Anonymous at Sun, 16 Feb 2025 22:53:09 UTC No. 16588534

>>16587665
What the fuck do you mean by complex as possible and what kind of a project would require this?

Anonymous at Sun, 16 Feb 2025 23:06:56 UTC No. 16588542

>>16587665
you cant

Anonymous at Mon, 17 Feb 2025 00:41:14 UTC No. 16588606

>>16588534
>>>16587665 (OP) #
>What the fuck do you mean by complex as possible
The oposite of simplyfying.
Like something like this
X+y'
Turns into this

X+y'*x''+y'*(X*1)+y'*x''+(y*0)*x+(y*0)+(x*1)+(y*0).
Hopefully this makes it clear.

>what kind of a project would require this?

Is a secret

Anonymous at Mon, 17 Feb 2025 00:42:24 UTC No. 16588608

>>16588542
Why not

Anonymous at Mon, 17 Feb 2025 01:47:10 UTC No. 16588646

>>16588606
>The oposite of simplyfying
https://youtu.be/k0qmkQGqpM8

the closest thing I can think of that matches your “description” is the concept of a free object in algebra and category theory. It allows you to construct an arbitrary algebraic structure from a generating set and you can prove that all algebric objects are quotient object of some free object. For example, every commutative ring is a quotient ring of the polynomial ring of integers. Another example is that every group is a quotient group of some free group.
https://en.wikipedia.org/wiki/Free_object

Anonymous at Mon, 17 Feb 2025 03:45:44 UTC No. 16588737

>>16588646
Alright so , looking into it this qoute below seems to tell me that it might work out with some modifications or at least some though about it.
But like doesn't it change the meaning if applied unto an operation? , like if X and Y in theory need to mean a specific value.
Like y always has to be 0?

"The idea of 'free' is that you have some setSS(finite or infinite), and you want to create some object (group, module, algebra, etc.) which is both generated by the elements in S, and there are also no restrictions/relations between the elements in S.

ttakeS={x,y}S={x,y}for example. The free group generated by S would containxxandyy, but it would also have to containxyxy,x−1x−1,y−1y−1,


One consequence of this is that to

describe any homomorphism from the free object generated bySSto some other object, it's enough to say where the elements ofSSgo AND there are no restrictions on where the elements ofSScan go. So for instance the complex numbers are not a free real algebra

:CCis generated (as anR−R−algebra) by11andii, but any homomorphismCACAmust sendiito some element that squares to−1−1. And it's important to know that the notion of 'free' depends on the category being considered. The polynomial ringR[x,y]R[x,y]is a free abelianR−R−algebra, but not a freeR−R−algebra.


This last property (that any homomorphism is uniquely described by an arbitrary choice of where the elements ofSSget sent) is exactly what Thomas Andrews means when he says that 'free' is an adjoint of a forgetful functor"

Anonymous at Mon, 17 Feb 2025 03:53:07 UTC No. 16588745

>>16588737
>ttakeS={x,y}S={x,y}for example. The free group generated by S would containxxandyy, but it would also have to containxyxy,x−1x−1,y−1y−1,
It contains all kinds of elements. The free group F({x}) is already the integers, which are infinite. Two or more elements makes it nonabelian. Btw, this gives you a simple proof that every abelian group is some direct sum of cyclic groups. Free groups are wild and poorly understood. But I guess this matches your definition of “complicated”. And I don’t see a reason this wouldn’t work with booleans. Just find the appropriate adjoint functor and voila.

Anonymous at Mon, 17 Feb 2025 04:26:20 UTC No. 16588765

>>16588745
It should work baring I understood it , but I wanted to ask just in case.

Anonymous at Mon, 17 Feb 2025 04:48:45 UTC No. 16588793

>>16588765
You seem to know your shit when it comes to abstract algebra and even category theory, so I trust you can make this work. Good luck.

Anonymous at Mon, 17 Feb 2025 06:03:25 UTC No. 16588849

>>16588793
Thank you for all the help and encoregement anon.
You are a nice fella.

Anonymous at Mon, 17 Feb 2025 06:39:01 UTC No. 16588881

>>16588606
>Is a secret
I think I know what it is. I did not bother because questionable payoff.
I wont give away what it probably is. Good luck with it though.