Difference between Monad and Applicative in Haskell

Question

I just read the following from typeclassopedia about the difference between Monad and Applicative. I can understand that there is no join

Answer 1

Just 1 describes a "computation", whose "result" is 1. Nothing describes a computation which produces no results.

The difference between a Monad and an Applicative is that in the Monad there's a choice. The key distinction of Monads is the ability to choose between different paths in computation (not just break out early). Depending on a value produced by a previous step in computation, the rest of computation structure can change.

Here's what this means. In the monadic chain

return 42            >>= (\x ->
if x == 1
   then
        return (x+1) 
   else 
        return (x-1) >>= (\y -> 
        return (1/y)     ))

the if chooses what computation to construct.

In case of Applicative, in

pure (1/) <*> ( pure (+(-1)) <*> pure 1 )

all the functions work "inside" computations, there's no chance to break up a chain. Each function just transforms a value it's fed. The "shape" of the computation structure is entirely "on the outside" from the functions' point of view.

A function could return a special value to indicate failure, but it can't cause next steps in the computation to be skipped. They all will have to process the special value in a special way too. The shape of the computation can not be changed according to received value.

With monads, the functions themselves construct computations to their choosing.