Talk:Bring to another

This seems like another esolang in which its owner forgot about how it works. This is the same thing that happened to Burn mariomakercalculator↑↑↑↑↑↑↑↑↑↑↑↑↑↑ 15:49, 28 December 2025 (BRZ)

It infact literally is! --Yayimhere2(school) (talk) 07:38, 31 December 2025 (UTC)

User:A()'s theory

It's possible that all numbers get store in a stack when unused. Ex:

105 32 104 -> 104
               32
              105

And ';' prints the most recent value on the stack and pops it off. Also '(' , ')' could represent an infinte loop possibly... --(this comment by A() at 23:59, 30 December 2025‎ UTC; please sign your comments with ~~~~)

Idk maybe --Yayimhere2(school) (talk) 07:38, 31 December 2025 (UTC)

User:↑'s theories

In the FizzBuzz example, i note $. It's the identifier for the counter i think.

And get the snippet:

($)/([3][5][0]) 0 000 =

which has slots 3, 5 and 0. 0 is probably empty slot. And get the part:

0 000 =

Which is probably the modification done if the result is 0. So 3 turns into 0, 5 turns into 000, and 15 is probably 0000 (000 + 0). mariomakercalculator↑↑↑↑↑↑↑↑↑↑↑↑↑↑ 11:26, 31 December 2025 (BRZ)

By this theory, it doesnt seem to take account for looping. --Yayimhere2(school) (talk) 15:14, 31 December 2025 (UTC)

I probably forgot to mention that when operating on the counter (or any variable that has more than 1 value), it operates on all the values. mariomakercalculator↑↑↑↑↑↑↑↑↑↑↑↑↑↑ 13:07, 31 December 2025 (BRZ)

It still, by that theory, doesnt account for the looping. and that statement is a big one which you have to manage in the 110 and everything else. --Yayimhere2(school) (talk) 16:09, 31 December 2025 (UTC)

Some Theories About Syntax User:A()

Now I am most likely being very stupid but, It does seem the syntax of:

(value1)/(value(s))

then '=' prints if it equals some other value:

(value1)/(value(s)) value <jump to beginning if 0> =

It has something to do with lambda calculus, so maybe 000 takes in = as a param. This whole program can be encapsulated into a function by value1. If we can make a program with this syntax style that has two inputs and outputs the second one we can prove this theory. A() (talk) 19:16, 31 December 2025 (UTC)

However, does this work with the other programs. --Yayimhere2(school) (talk) 20:03, 31 December 2025 (UTC)

((λ)/([x][y])([y]))(x')(y') -> (y')

This seems closeA() (talk) 20:17, 31 December 2025 (UTC)

That should returns x though? --Yayimhere2(school) (talk) 20:48, 31 December 2025 (UTC)

Wait, so [y] omits y? -- A() Being stupid A() (talk) 20:59, 31 December 2025 (UTC)
Like, the k combinator does k x y = x --Yayimhere2(school) (talk) 21:00, 31 December 2025 (UTC)

((λ)/([x][y])([x]))(x')(y') -> (y')

Think this is correct. A() (talk) 21:02, 31 December 2025 (UTC)
But also, what is it? what is that little diagram? --Yayimhere2(school) (talk) 21:02, 31 December 2025 (UTC) '-> [y']' shows it returns the second value. This was an attempt at making the false Boolean A() (talk) 21:06, 31 December 2025 (UTC)
which one? --Yayimhere2(school) (talk) 21:07, 31 December 2025 (UTC)

The Y value -- A() (talk) 21:10, 31 December 2025 (UTC)

so it returns Y, the opposite of what k does???? --Yayimhere2(school) (talk) 21:11, 31 December 2025 (UTC)

Yes exactly! -- A() (talk) 21:12, 31 December 2025 (UTC)

If you theory predicts that, then its, quite literally, wrong --Yayimhere2(school) (talk) 21:13, 31 December 2025 (UTC)

Oh well, at least I tried. -- A() (talk) 21:14, 31 December 2025 (UTC)

After thinking deeply, i think it's more like this:

(λ)/(y)?[Aₙ.y](f)

-- A() (talk) 20:24, 2 January 2026 (UTC)
if this still implies the second argument is returned, then its incorrect. if the opposite, then the lambda calculus related program agrees. --Yayimhere2(school) (talk) 20:32, 2 January 2026 (UTC)

There is not enough information about the lambda syntax to make a good theory. It's possible that if I got enough information that I could a better theory. I don't why the Substack is there, maybe the ? could suggest a possible return function. Using the information I already know about this language, the only thing I can assume is that the ( < name > ) is the name of the lambda calculus expression. That may be wrong of course but how can I be certain with little information? Maybe if I wait long enough, the answer might reveal itself. Only time can tell. -- A() (talk) 20:45, 2 January 2026 (UTC)
perhaps not. Though I think when enough info has been found about the other programs, that knowledge can be applied to this specific case. --Yayimhere2(school) (talk) 08:26, 3 January 2026 (UTC)
You perhaps want to look at this language again, because I found a little batch of two program recently, and ive added them to the page! --Yayimhere2(school) (talk) 08:47, 3 January 2026 (UTC)

Dissection

(x) # search for input x
/[  # Encapsulate func
  [ₙ]
  (λ) # func (I am not sure)
  A*  # return if found (I am not sure about this either)
  [x] # output x (Substack? Substack of x?)
  (?) # possible if statement
]?

It is obvious that the (x)/ means input 'x' or something. -- A() (talk) 16:39, 3 January 2026 (UTC)

That kinda does make sense, because then lambda is passed to x in the original program, which then allows it to be an abstraction OF a lambda. --Yayimhere2(school) (talk) 17:09, 3 January 2026 (UTC)

(the (x)/ part specifically)

Information I've gathered

1. (x)/ Probably means something like input 'x'
2. (λ)A*[x] Probably means output 'x' or something. It's analogous to [Aₙ.x]
3. It's possible that I need two inputs for the Choose2nd functions to work

Ex:

(y,x)/[[ₙ](λ)A*[y](?)]?
or 
(x,y)/[[ₙ](λ)A*[y](?)]?

Depends on which direction it registers input. I am possibly wrong, but I am getting closer.-- A() (talk) 21:33, 3 January 2026 (UTC)

Yea seems to make sense. You could go check out what aadenboy wrote if you haven't. I think A represents when another argument is taken but it doesnt appear in the lambda. Maybe? There is a chance * is concatenation(im a big fan of underload, and propably knew of it at the time). I think I'd look something like:

(y)(x)/[[ₙ](λ)A*[y](?)]?

Because underload, and also, reverse Polish notation(because it's stack based, atleast it most likely is). Also, I think the reason this doesnt have a name is because it gets applied? like because, theres another ?. Maybe idk. --Yayimhere2(school) (talk) 08:06, 4 January 2026 (UTC)

aadenboy's part in this

loose observation, [n][n][n]... might be a tuple, and (...) in general is an argument?

($)/([3][5][0])

(     - argument 1
  $     - accumulator?
)     - end of argument 1
/     - apply?
(     - argument 2
  [3]   - tuple value 1
  [5]   - tuple value 2
  [0]   - tuple value 3
)     - end of argument 2

$ -> [3] = $ mod 3
$ -> [5] = $ mod 5
$ -> [0] = $ mod 0 (this doesn't make sense, but whatever. why is there a [0] to begin with, actually?)
returns [result][result][result]

—aadenboy (talk|contribs) 00:31, 4 January 2026 (UTC)

Yea seems reasonable. Perhaps mod 0 is a special case, of some sort. --Yayimhere2(school) (talk) 08:01, 4 January 2026 (UTC)

Perhaps Mod 0 is the identity.

Makes senseA() (talk) 18:25, 4 January 2026 (UTC)

Truth machine

Ive looked at the truth machine:

[&&] ; take input, and push as a substack(??)
-() ; repeat *something*( the () ) - times. Perhaps - is the top of the topmost substack
{ start loop
  " duplicate
  % pop and print
} end loop
% print if 0

Now, my theory is, () applies to the loop, because the loop applies to the op of stack not top of substack. Now, the - references the top of stack, so if its zero, the loop is skipped, else, the loop runs once, but the loop is infinite anyways so it runs forever. --Yayimhere2(school) (talk) 11:06, 4 January 2026 (UTC)

This is also congruent with &&{"%}%, because here no substacks are used. --Yayimhere2(school) (talk) 11:12, 4 January 2026 (UTC)