Introduction to lambda expression & lambda calculus

Introduction to lambda expression & lambda calculus

• 1.
  Requirements • Common Knowledgeof Programming • Maybe Python/Ruby • Recursion
• 2.
  LAMBDA INTRO KIM LEO
• 3.
  λ
• 4.
  lambda expression def square(x): returnx*x => square = lambda x: x*x e.g.: (lambda x: x*x)(8) => 8*8 => 64 But how could make use of LAMBDA?
• 5.
  lambda expression alist: [8,5, 2, 4, 62, 4, 6] anOp: blablabla apply the operation to the list. like: for i = 0 to len(alist): anOp( alist[i] ) or: foreach i in alist: anOp( i )
• 6.
  Implement apply_square_for_each(alist): foreach i inalist: square( i ); apply_plusplus7_for_each(alist): foreach i in alist: plusplus7( i ); apply_for_each(alist, op): case when op == "square": foreach i in alist: square( i ) .....
• 7.
  lambda expression apply_for_each(alist, alambda): foreachi in alist: alambda( i ) apply_for_each([1,2,3], square) apply_for_each([ ... ], anything_you_want_applied_to) apply_for_each => map/each
• 8.
  General LISt Processing Map： [x,y,z] -> map(op,[x,y,z]) => [op(x),op(y),op(z)] Reduce ： [x,y,z] -> reduce(op,[x,y,z],v) => op(op(op(v,x),y),z) Filter ： [x,y,z] -> filter(op,[x,y,z]) => a subset of the list for each op(i) => true
• 9.
  more.. each, find, every/all,any/some, invoke, where, sort, group/groupBy, countBy .... (underscore.js/Ruby Enumerable Module) var all_of_us def lonelyboy def lonelygirl zip( filter( lonelyboy, all_of_us), filter( lonelygirl, all_of_us) )
• 10.
  Lambda Calculus lambda x:x lambda x,y: y(x) (lambda x:x)(1) 1 xx.λ xyyx ..λλ 1).( xxλ xfxf ..λλ
• 11.
  Recursion Y := λg.(λx.g(x x)) (λx.g (x x)) Y f => λg.(λx.g (x x)) (λx.g (x x)) f => (λx.f (x x)) (λx.f (x x)) => f ((λx.f (x x)) (λx.f (x x))) => f (Y f)
• 12.
  SKI Combinator I :=λx.x K := λx.λy.x S := λx.λy.λz.x z (y z) OR: Ix = x Kxy = x Sxyz = xz(yz)
• 13.
  self-application SIIα = Iα(Iα)= αα SII(SII) = I(SII)(I(SII)) = I(SII)(SII) = SII(SII)
• 14.
  reverse S(K(SI))K αβ → K(SI)α(Kα)β → SI(Kα) β → Iβ(Kαβ) → Iβα βα
• 15.
  Y in SKI Y= S (K (S I I)) (S (S (K S) K) (K (S I I))) Y' = S S K (S (K (S S (S (S S K)))) K)