This Streamer got hurt from Speedrunning Super Mario 64 | Sophomore 16 Star Invitational 

You’re absolutely correct about BoxBox = Box^2. No idea what Simply was thinking

Your commentary was goated

Thank you for finally clarifying that box^2 = boxbox, I really lost my mind when people we're trying to "prove" that it's not correct

The mistakes made towards the beginning and end of the runs were definitely a factor.

@@rud5101 the begining and ending half specifically

I feel like it really comes down to who has the fastest movement with the less mistakes during the begging, middle and end. I don’t now why but i also feel that having good bowser throws is a positive

So basically who played better. Thanks i wouldn't guess

@@thevirtuoso83621 year later I hope you are not still this dense

Thats what I'm sayin

Literally was I was thinking too

It is not know that the box in his name is a physical box

ergo XD

A playlist for a playlist pagman

Especially an unedited one.

Yes I’ve been waiting for sophomore 16 playlist as well

Try using your other hand to give your preferred hand a rest during all of that "speedrunning"

Check out simply’s video on his hand problems, his were insanely bad but were actually all in his mental and once he realized it they went back to normal

Also, if you consider "boxbox" as a string in a formal language, then given a a string, S, S^n is written as S repeated n times. Therefore if S = "box" then S^2 = SS = "box"^2 = "boxbox". Therefore, box^2 = boxbox Q.E.D.

why do you assume b, o, and x lie in a commutative ring? You realize that in many sets, multiplication is non commutative, right?

bro

The letters aren't variables though box is a whole word

easier: Let y = box Then boxbox = yy = y^2 = (box)^2

You almost never omit the operator * and all other explicit delimiters (e.g. parentheses aroud (box) like you did) when dealing with multi letter variables. To see why this becomes ambiguous: is boxbox = box * box, b * o * x * b * o * x, its own variable, or some other wild combination? (box)(box) = box * box by associative (and multiplicative identity) property, yes, but writing it as boxbox is just being confusing even if you know that box is a variable beforehand. Put differently, omitting the operator is convention to make it easier for humans to read and write, not a math property. Where this operator IS still omitted with multiple letters is **string concatenation**, which proves my point that it should be 2*box. Quotes around string of text just for clarity (unambiguous if it wasn't inline in a sentence): you take "box" || "box" = "boxbox", which if you repeatedly concatenate the **same** string together can be represented as "boxbox" = 2*"box" = "box" || "box" (Tu-box, if you will). Simply's right here before he gave up. This is true with both literal strings and variables: for example, if we knew: y = box where y is a variable and {b, o, x} are all symbols, then: yy = y || y = boxbox (still unambiguous with explicitly defined alphabet symbols/terminals, variables, and grammar/parsing rules... even recursively :)

Also, if you treat box as a variable number it's all meaningless anyway. if box = 0,1 then box^2 = 0,1 (not null string or "box"), and unless you accept the possibility that box = 2^1/2 (which is just irrational :) you can never get box^2 = 2 with standard arithmetic, let alone "boxbox". Alternatively box = box * box = box * box * box * ... if and only if box = 0,1, but 1) this doesn't generalize to unconstrained variable so dom is wrong that box^2 = boxbox in general, and 2) see above about ambiguity of omitting *

I agree that "box" is a single variable but one step further: "boxbox" only has 2 "box"es if you count them, which would imply it's 2*"box" (Tu-box, if you will). Here the implicit operator is repeated concatenation (x*"text" as """adding""" same string of text to itself x times to make "texttext...text"). The only way "boxbox" could be "box^2" is if you treat the implicit operator as multiplication (e.g. writing x * y as xy) where the variable name is "box". But since there are only 2 boxes and not "box"^2 "box"es displayed in the final result, and this operator isnt ever implicit with multi-letter variables, this feels much less intuitive.

Editor gonna get a write up for this

Yep, saying the answer is always BoxBox is the option select here. No matter what you're right.

If you consider BOX one variable, then you can assume BOXBOX is the same as BOX*BOX, the multiplication is implied, same as when you see 2x or 2yz, it's implied that 2x = 2*x and 2yz = 2*y*z... so, BOXBOX is not BOX + BOX, it's actually BOX*BOX, or BOX^2

@@ethanimus9241 uhmm… actually 🤓☝️ saying boxbox does imply that it is box+box but if we say box(box) it does imply that is is box•box using the distributive property 😎☝️

WTF?

If I have two values, x and y that multiplied together equal z, you can write it as xy=z. This is because similarly to coefficients like 3x, it is implied the two values are multiplied together. If x and y in this example both happen to be the same value (say 2) then we could right it instead as xy=4. Since x=y you can substitute x=y into the equation and get xx=4 or x^2=4. Now if we consider box a variable which is how most people seem to interpret it including yourself (" If box is represented by"). boxbox would imply box x box which is just box^2 No one in their sane mind writes xy to define a value that is equivalent to 10x + 1y like you did with the 22 example.

@@bugsybro3011 You're right it would be insanity writing it as 22. That was in reference to what Dom said in the video. Simply said (paraphrasing) "1x1 is 1" and dom said "yeah but what is 2x2?". Dom was trying to use 2 to represent Box in that scenario to show box squared is boxbox but for box squared to be boxbox in that scenario then 2x2 would have to equal 22.

This is 2 years later but thank you for saying this, it was driving me crazy how Dom was trying to rationalize Boxbox.

If Box is 1 or 0 it actually could be true

box != box^3, therefor your math is wrong an shame on you're family

box^2 = (edge^3)^2 = edge^6

Best worst commentary this match

in that case box^2 = box which also = boxbox because boxbox = box

But if you represent BOX with any number other than 1 then it has to be box^2

Because he is commenting it duh