# Lambda Calculus - Evaluating Expressions #4

# Lambda Calculus - Evaluating Expressions #4

exfalso + 1 comment These questions are so pointless, they put off people who aren't familiar with church encodings, and piss off people who are. Booo

satyajitghana191 + 0 comments exactly, lmao

madspbuch + 0 comments Yep, These questions are pointless. Without the knowledge that we are talking about church numerals, it is just combinators.

alexkorotkikh + 1 comment I'm not very good in lambda calculus :( can somebody point me to some resources I can read and understand how result can be integer if there's no numeric input for lambda expression.

madspbuch + 0 comments Look up church numerals :-) (as stated otherwise in this thread, the questions are pointless without this piece of intermation.)

peterb3 + 0 comments The answer is "2", and this is an absolutely terrible series of problems that barely explains why anyone should care. Even knowing what I know about Church numerals I feel that all these questions are doing is irritating the user.

scott_sedgwick + 1 comment Hi alexkorotkikh,

I don't pretend to deeply understand this, but basically all those things were Church numerals.

IIRC, there is a chapter devoted to them in SICP too.

Scott.

alexkorotkikh + 0 comments Thanks a lot, Scott! Will check and try again :)

xdavidliu + 0 comments the last question Evaluating Expressions #3, was changed to the more reasonable "what Church numeral does this expression correspond to", but this question still has the nonsensical "compute the integer value". The change should be made here as well.

dMike + 0 comments I've made it this far, put the correct answers in the box, but I have no idea what I'm doing :/

sughan100 + 1 comment lamda X.lamda Y.X(X(Y))=2

Akeempositive + 2 comments Can you explain, please?

zhiwangshi28 + 0 comments 0 = Î»f.Î»x.x

1 = Î»f.Î»x.f x

2 = Î»f.Î»x.f (f x)

in this problem, replace f with x and x with y

khairul + 0 comments The way I think of church numerals is as follows. 0 is defined as just your input so x 1 is defined as applying a generic function once to your generic input so f(x) 2 is f(f(x)) and so on. In lambda calculus this 2 would be expressed as Î»f.Î»x.f f x.

No more comments

Sort 9 Discussions, By:

Please Login in order to post a comment