# Lambda Calculus - Reductions #1

# Lambda Calculus - Reductions #1

nixphix + 6 comments If you are new to lambda calculus following video & links will help

kincaidsavoie + 0 comments Thank you!

mattklein999 + 0 comments Yes, thank you! That PDF is a perfect introduction!

redemption + 1 comment Thanks for the PDF!!!

- AR
arafaliy + 0 comments [deleted]

- AR
arafaliy + 0 comments Thanks :)

wu_ct + 1 comment There is a version 2 of the origin PDF tutorial which has been fixed some errors and added more illustrations. Here is the link: http://www.inf.fu-berlin.de/inst/ag-ki/rojas_home/documents/tutorials/lambda.pdf

Kaan_Aydinay + 0 comments Thanks very much !!

Risen_Again + 0 comments I started coding a lexical parser for this. Then read a comment that we just have to write plain text answer for the given expression. Facepalm.

kkweon + 0 comments (Function)argument e.g (λx.(+1)x)3 = (λx.+1 x)3 = (+ 1 3) = 4

((λx.(x y))(λz.z)) = (λx.(x y))(λz.z) = ((λz.z)y) = (λz.z)y = y

danthemango + 0 comments I think the question should mention that it is looking for the result of the reduction. I thought it was looking for some kind of input to the lambda expression.

- AS
sethia_arun + 0 comments I liked this content http://www.nyu.edu/projects/barker/Lambda/

Zhomart + 1 comment Hi, I still don't understand why application can be written in the statement of the function?

E.g.

`λx.(x y)`

Shouldn't it be written as

`(λx.x)y`

?- 井ま
g2198207 + 0 comments λx.(x y) is a function that applies y to its first argument [maybe clearer this way: (λx.x y) ]

(λx.x)y is just y applied to the identity function, reduces to y

shekhar_kotekar + 1 comment Hi,

We are supposed to write Scala/Java/ code for this problem or we are supposed to write steps to reduce that expression?

I cannot see any language related drop down box for this problem. Here is the screenshot of it.

wisn98 + 1 comment We are not writing code here. Just put your answer on the text editor and submit it.

shekhar_kotekar + 1 comment Okay.THanks.

wisn98 + 0 comments Yes, you're welcome :D

No more comments

Sort 7 Discussions, By:

Please Login in order to post a comment