Yeah, that was silly of me

True, but just in case ;)

Simple example: bimap (+ 10) (* 13) (7, 3) = (17, 39)

So, (bimap f g (a,b)) applies f to a, and g to b, while maintaining the tuple structure.

Another example, but with Either:

bimap (+ 10) (* 13) (Left 9)  = Left  19
bimap (+ 10) (* 13) (Right 2) = Right 26

Now, a clever trick is to apply join to bimap. It happens to be the case that join bimap applies the same function to both parts of a tuple:

join bimap (+2) (2,3) = (4,5)

The reason this works is due to the definition of join and the monad instance for functions (i.e. (->) r):

-- definitions
join x  :=  x >>= id
f >>= g :=  \r -> g (f r) r     -- (for functions only)

-- therefore:
join f  =  f >>= id
        =  \r -> id (f r) r
        =  \r -> f r r

-- specific to bimap:
join bimap = \r -> bimap r r

Thank you! It's so satisfying to lay things out his way :)

Same here. The random test fails with the undecypherable:

Falsifiable (after 4 tests and 6 shrinks): expected: Just 0 but got: Nothing "a" Positive {getPositive = 1} -1

What am I supposed to do with that? ...