Exercise 4 State & IO Higher Kinds More Examples Administrivia
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Software System Design and Implementation
Functors, Applicatives, and Monads Practice
Curtis Millar
CSE, UNSW (and Data61) 15 July 2020
Exercise 4 State & IO Higher Kinds More Examples Administrivia
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Capitalise all characters in the input file.
Exercise 4
Exercise 4 State & IO Higher Kinds More Examples Administrivia
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Capitalise all characters in the input file. Sum all the numbers in the input file.
Exercise 4
Exercise 4
State & IO Higher Kinds More Examples Administrivia
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Capitalise all characters in the input file. Sum all the numbers in the input file. Implement a guessing game AI.
Exercise 4
Exercise 4 State & IO Higher Kinds More Examples Administrivia
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State & IO
Week 5 covered State and IO and you¡¯ve had a few weeks to work with them. Do you have any questions?
Exercise 4 State & IO Higher Kinds More Examples Administrivia
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Functors, Applicatives, Monads
Consider higher-kinded types of kind * -> * that contain or produce their argument type.
Exercise 4
State & IO Higher Kinds More Examples Administrivia
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Functors, Applicatives, Monads
Consider higher-kinded types of kind * -> * that contain or produce their argument type.
Functor lets us use a pure function to map between the higher-kinded type applied to different concrete types.
Exercise 4
State & IO Higher Kinds More Examples Administrivia
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Functors, Applicatives, Monads
Consider higher-kinded types of kind * -> * that contain or produce their argument type.
Functor lets us use a pure function to map between the higher-kinded type applied to different concrete types.
Applicative lets us apply a n-ary function in the context of the higher-kinded type.
Monad lets us sequentially compose functions that return values in the higher-kinded type.
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class Functor f where
fmap :: (a -> b) -> f a -> f b
The functor type class must obey two laws:
Functors
Exercise 4 State & IO Higher Kinds More Examples Administrivia
class Functor f where
fmap :: (a -> b) -> f a -> f b
The functor type class must obey two laws:
Functor Laws
1 fmap id == id
2 fmap f . fmap g == fmap (f . g)
Functors
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Exercise 4 State & IO Higher Kinds More Examples Administrivia
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Applicatives
class Functor f => Applicative f where
pure :: a -> f a
(<*>) :: f (a -> b) -> f a -> f b
The functor type class must obey four additional laws:
Exercise 4 State & IO Higher Kinds More Examples Administrivia
Applicatives
class Functor f => Applicative f where
pure :: a -> f a
(<*>) :: f (a -> b) -> f a -> f b
The functor type class must obey four additional laws:
Applicative Laws
1 pure id <*> v = v (Identity)
2 pure f <*> pure x = pure (f x) (Homomorphism)
3 u <*> pure y = pure ($ y) <*> u (Interchange)
4 pure(.) <*>u<*>v<*>w=u<*>(v<*>w) (Composition)
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Exercise 4 State & IO Higher Kinds More Examples Administrivia
Alternative Applicative
It is possible to express Applicative equivalently as:
class Functor f => App f where
pure :: a -> f a
tuple :: f a -> f b -> f (a,b)
Example (Alternative Applicative)
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Exercise 4 State & IO Higher Kinds More Examples Administrivia
Alternative Applicative
It is possible to express Applicative equivalently as:
class Functor f => App f where
pure :: a -> f a
tuple :: f a -> f b -> f (a,b)
Example (Alternative Applicative)
1 Using tuple, fmap and pure, let¡¯s implement <*>.
2 And, using <*>, fmap and pure, let¡¯s implement tuple.
done in Haskell.
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Exercise 4 State & IO Higher Kinds More Examples Administrivia
Alternative Applicative
It is possible to express Applicative equivalently as:
class Functor f => App f where
pure :: a -> f a
tuple :: f a -> f b -> f (a,b)
Example (Alternative Applicative)
1 Using tuple, fmap and pure, let¡¯s implement <*>.
2 And, using <*>, fmap and pure, let¡¯s implement tuple.
done in Haskell.
Proof exercise: Prove that tuple obeys the applicative laws.
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Exercise 4 State & IO Higher Kinds More Examples Administrivia
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Monads
class Applicative m => Monad m where
(>>=) :: m a -> (a -> m b) -> m b
We can define a composition operator with (>>=):
Exercise 4 State & IO Higher Kinds More Examples Administrivia
Monads
class Applicative m => Monad m where
(>>=) :: m a -> (a -> m b) -> m b
We can define a composition operator with (>>=):
(<=<) :: (b -> m c) -> (a -> m b) -> (a -> m c)
(f <=< g) x = g x >>= f
The monad type class must obey three additional laws:
Monad Laws
1 f <=< (g <=< x) == (f <=< g) <=< x (associativity)
2 pure <=< f == f (left identity)
3 f <=< pure == f (right identity)
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Exercise 4 State & IO Higher Kinds More Examples Administrivia
Alternative Monad
It is possible to express Monad equivalently as: class Applicative m => Mon m where
join :: m (m a) -> m a Example (Alternative Monad)
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Exercise 4 State & IO Higher Kinds More Examples Administrivia
Alternative Monad
It is possible to express Monad equivalently as: class Applicative m => Mon m where
join :: m (m a) -> m a
Example (Alternative Monad)
1 Using join and fmap, let¡¯s implement >>=.
2 And, using >>= let¡¯s implement join.
done in Haskell.
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Exercise 4 State & IO Higher Kinds More Examples Administrivia
data Tree a
= Leaf
| Node a (Tree a) (Tree a)
deriving (Show)
Example (Tree Example)
Show that Tree is an Applicative instance. done in Haskell.
Tree Example
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Exercise 4 State & IO Higher Kinds More Examples Administrivia
data Tree a
= Leaf
| Node a (Tree a) (Tree a)
deriving (Show)
Example (Tree Example)
Show that Tree is an Applicative instance. done in Haskell.
Note that Tree is not a Monad instance.
Tree Example
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Exercise 4 State & IO Higher Kinds More Examples Administrivia
Formulas Example
data Formula v = Var v
| And (Formula v) (Formula v)
| Or (Formula v) (Formula v)
| Not (Formula v)
| Constant Bool
deriving (Eq,Show)
Example (Formulas Example)
Show that Formula is a Monad instance. done in Haskell.
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Homework
Exercise 4 State & IO Higher Kinds More Examples Administrivia
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Homework
1 Week 5¡¯s quiz is due on Friday. Make sure you submit your answers.
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2
Week 5¡¯s quiz is due on Friday. Make sure you submit your answers.
The fifth programming exercise is due by the start of my next lecture (in 7 days).
Homework
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2 3
Week 5¡¯s quiz is due on Friday. Make sure you submit your answers.
The fifth programming exercise is due by the start of my next lecture (in 7 days). This week¡¯s quiz is also up, it¡¯s due Friday week (in 9 days).
Homework
Exercise 4 State & IO Higher Kinds More Examples Administrivia
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Consultations
Consultations will be made on request. Ask on piazza or email cs3141@cse.unsw.edu.au.
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Consultations
Consultations will be made on request. Ask on piazza or email cs3141@cse.unsw.edu.au.
If there is a consultation it will be announced on Piazza with a link a room number for Hopper.
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Consultations
Consultations will be made on request. Ask on piazza or email cs3141@cse.unsw.edu.au.
If there is a consultation it will be announced on Piazza with a link a room number for Hopper.
Will be in the Thursday lecture slot, 9am to 11am on Blackboard Collaborate.
Exercise 4
State & IO Higher Kinds More Examples Administrivia
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Consultations
Consultations will be made on request. Ask on piazza or email cs3141@cse.unsw.edu.au.
If there is a consultation it will be announced on Piazza with a link a room number for Hopper.
Will be in the Thursday lecture slot, 9am to 11am on Blackboard Collaborate.
Make sure to join the queue on Hopper. Be ready to share your screen with REPL (ghci or stack repl) and editor set up.