This is the best lambda calculus video I've watched of all. Thank you!

Obviously understanding the value of lambda calculus is in exploring how robust it is on an abstract level, but it would be great to see this in an applied context as well! Thank you for your content!

Thank you so much, this was such a great explanation

Yesterday I heard of Lambda Calculus for the first time. It is something that is right up my street and I look forward to understanding it soon. However, for now, I am encountering barrage after barrage of geeky drivel from one geek back to himself and his fellow geeks who are entertaining one another with the notion that they have explained this thing clearly and adequately.

Multiplication without recursion was crazy to see. Really showcases the possibilities of lambda calculus. Great video, any chance you could increase your volume slightly?

Haskell is goated

Great video, by far the best explanation I've seen on the topic! My only critique is that the animations were a bit too fast, especially in the part about basic arithmetic operations

@blueoxygenurawr404

Ай бұрын

ikrr speedy gonazalez

Extremely interesting way of computing already curious to learn more about catergory theory!

Great video! well explained

thanks for the video! I find the topic so interesting!

@anakimluke

Жыл бұрын

ps. your next video is on category theory!!?? even more interested now!!

I finally understand the y-combinator now.

Very good video!

Can you enumerate each of the lambda calculus functions in order of the number of bits of information that it takes to represent their function?

Good one thanks

3:05 does the function return itself (i.e. return a function), or the value it received?

@raphaeld9270

5 ай бұрын

It is an error in the video it seems : `f(x) =x ` is the Identity function and just returns the input value as its output. 12:26 is closer to a function that "returns itself".

@LambdaJack

2 ай бұрын

Which function!? A composition specialized to it's recursive specialization is it's recursive specialization. A better name than Y could be "recursive", as in "(recursive compositio)". "id" as composition does not _specify_ at all what should be done, so it could be anything. As a Degenerate Case a Composition can take the Self-Reference and do nothing with it.

@beantown_billy2405

2 ай бұрын

@@LambdaJack The function at 3:05, f(x)=x

Great video. I wished you putting a little more emphasis on, explicit putting parentheses on lambda abstraction vs no explicit parentheses. For example: Lambda x.x y Is different then: (Lambda x.x) y Sorry if bad English.

@TonyZhang01

Жыл бұрын

Definitely. I hoped that the use of parentheses was pretty clear based on the examples and what I wrote, and also because it is used the same way as they are normally used, so if I focused on talking about it it might seem more confusing than it actually is

Instead of using church encodings a great alternative is PCF xx

有沒有中文的

First!

@incognitohacks4850

Жыл бұрын

Ackshually

@alonelyphoenix8942

11 ай бұрын

Why

λ Beautiful λ

Second

chinz

I am happy that the Romans did not invent lambda Calculus -- otherwise buildings worldwide must bear the mark of Church Encoding f(f(f(f(f(f(f(f(f(f(fx))))))))))))

The quality of information Is exelent but the way how It Is explained Is really low level. Really Fast to have time to be understand. Sonetimes It looks that It Is an automatic voice that speak. I really Belive that this video Is superwow for the content but It Is terribly hard to really understand the content.

You say that the full definition of the add function is 🐑fxmn. (m f) (n f x). Sorry, I haven’t found a way to input the lambda character on an iPad, so I used the lamb emoji. How can I walk through the steps to apply that function? Is it like this? 🐑fxmn. (m f) (n f x) f x 2 3 (2 f) (3 f x) What do I do from here?

@MarkVolkmann

2 ай бұрын

I think I found an answer. (λfxmn. (m f) (n f x)) 2 3 -- Substitute 2 for m and 3 for n. λfx. (2 f) (3 f x) -- Substiture the λ terms for 2 and 3. λfx. (λfx.f (f x) f) (λfx.f (f (f x)) f x) -- Apply the arguments f and x in the last term. λfx. (λfx.f (f x) f) (f (f (f x))) -- Apply the argument f in the last term. λfx. (λx.f (f x)) (f (f (f x))) -- Apply the argument `(f (f (f x)))` to the function on its left. λfx. f (f (f (f (f x)))) -- This is the definition of the number 5.

@MarkVolkmann

2 ай бұрын

I haven’t been able to work out a similar set of steps to demonstrate multiplication though. I could use help with that.