View difference between Paste ID: uN1vDUcR and qkwXkwLr

   // replace calls of returned functions by .apply for clarity
-// difference between foldLeft and foldRight is associativity:
+  // these are our calls to g
-foldLeft(List(1,2,3,4), 0)((a,b) => a + b) = ((((0+1)+2)+3)+4)
+  foldLeftViaFoldRight(List(1,2), 0)(f)
-foldRight(List(1,2,3,4), 0)((b, a) => b + a) = (1+(2+(3+(4+0))))
+
   foldRight(List(1,2), b => b)((a,g) => b => g(f(b, a)))(0)
-//So to invert associativity, we build up a function, starting with identity.
+  (b1 => foldRight(List(2), b => b)((a,g) => b => g(f(b, a))).apply(f(b1, 1)))(0) // (1)
 
-//added types on foldRight, note that foldRight's B is not foldLeft's B, but a function B => B:
+  // look at inner fold right separately:
-def foldLeftViaFoldRight[A,B](l: List[A], z: B)(f: (B,A) => B): B =
+  foldRight(List(2), b => b)((a,g) => b => g(f(b, a)))
-    foldRight[A, B => B](l, (b:B) => b)((a,g) => // g is foldRight's accumulator
+  b2 => foldRight(Nil, b => b)((a,g) => b => g(f(b, a))).apply(f(b2, 2))
-			b => g(f(b,a)) // we return a new function that applies f first, then our accumulated function up to now
+  b2 => (b => b).apply(f(b2, 2)) // we reach zero element, identity
-    )(z)                   // the result of our foldRight is B => B, we call it with the zero element to get the result
+  b2 => f(b2, 2))
 
   // insert into (1)
   (b1 => (b2 => f(b2, 2)).apply(f(b1, 1)))(0)
   (b2 => f(b2, 2)).apply(f(0, 1))) // apply the function from foldRight to zero value
   f(f(0, 1), 2))