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Tejas Dinkar
October 10, 2014
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Monads you've already put in production (without knowing it)
Tejas Dinkar
October 10, 2014
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Transcript
Monads you are already using in prod Tejas Dinkar nilenso
about.me • Hi, I’m Tejas • Nilenso: Partner • twitter:
tdinkar • github: gja
Serious Pony
Online Abuse
Trouble at the Koolaid Point http://seriouspony.com/trouble-at-the-koolaid-point/ https://storify.com/adriarichards/telling-my-troll-story-because- kathy-sierra-left-t
If you think you understand Monads, you don't understand Monads.
None
This talk is inaccurate and will make a mathematician cry
None
Goal of this talk For you to say “Oh yeah,
I’ve used that hack”
None
Monads • Programmable Semicolons • Used to hide plumbing away
from you • You can say Monads in almost any sentence and people will think you are smart
None
Values Value
Monads Value Box
Mysore Masala Monad M onad Value
Monads Value Box
Monads • Monads define two functions • return takes a
value and puts it in a box • bind takes a box & function f, returning f(value) • it is expected that the function returns a box
Value Value Another Value Value Function return bind
Our Function Signatures Value f(value)
Some math (√4) + 5
Some math (√4) + 5 3 or 7!
Value 4
Monad [4]
[alive, dead]
ruby! x = [1, 2, 3] y = x.map {
|x| x + 1 } # y = [2, 3, 4]
return Value Value return
return def m_return(x) [x] end # m_return(4) => [4]
The functions Value f(value)
Square Root fn def sqrt(x) s = Math.sqrt(x) [s, -s]
end # sqrt(4) => [2, -2]
Increment Fn def inc_5(x) [x + 5] end # inc_5(1)
=> [6]
Bind Functions Another Value Value Function bind
Bind Function x = m_return(4) y = x.????? { |p|
sqrt(p) } # I want [-2, 2]
Bind Function x = m_return(4) y = x.map {|p| sqrt(p)
} # y => [[2, -2]] # ^—— Box in a box?
Bind Function x = m_return(4) y = x.mapcat {|p| sqrt(p)
} # y => [2, -2]
Putting it together m_return(4) .mapcat {|p| sqrt(p)} .mapcat {|p| inc_5(p)}
# => [3, 7]
You have invented the List Monad, used to model non-determinism
Congrats
Turtles all the way down
A small constraint • Let’s do a bit of a
self imposed constraint on this • Functions must return either 0 or 1 elements • (we’ll only model positive integers here)
return - stays the same
bind - stays the same x = m_return(4) y =
x.mapcat { |p| inc_5(p) } # y => 9
Square Root Fn def sqrt(x) if (x < 0) return
[] #error else [Math.sqrt(x)] end end # sqrt(4) => [2] # sqrt(-1) => []
Describe in English There is a list passed to each
step Maybe this list has just one element, or Maybe it has none
None
The Maybe Monad • The intent is to short circuit
computation • The value of the `box’ is None, or Just(Value) • You can think of it as a type-safe nil / null
try def try(x, f) if x == nil return f(x)
else return nil end end # 4.try { |x| x + 5 } => 9 # nil.try {|x| x + 5 } => nil
None
Let’s start over • The Monad Laws • Left Identity
• Right Identity • Associativity
Left Identity m_return(a).bind(f) == f(a)
Right Identity m.bind(m_return) == m
Associativity m.bind(f).bind(g) == m.bind(x -> f(x).bind(g))
Store Computation
The State Monad • Rest of the world - State
Machine (sorta) • The value inside the box f(state) => [r new-state] • Particularly useful in pure languages like Haskell • Let’s build a stack
The functions Value f(value)
The functions (f(value) state) [new-value, new-state]
push def push(val) lambda { |state| new_state = state.push(val) [value,
new_state] } end
pop def pop() lambda { |state| val = state.pop() [val,
state] } end
def double_top() lambda { |state| top = state.pop() [2 *
top, state.push(2*top)] } end double_top
return def m_return(x) lambda { |state| [x, state] } end
bind def bind(mv, f) lambda { |state| v, temp_state =
mv(state) state_fn = f(v) state_fn(temp_state) } end
example # Not working code ! m_return(4) .bind(a -> push(a))
.bind(b -> push(b + 1)) .bind(c -> double_top()) .bind(d -> sum_top2()) .bind(e -> pop())
None
Associativity m.bind(f).bind(g) == m.bind(x => f(x).bind(g))
turn this # Not working code ! m_return(4) .bind(a ->
push(a)) .bind(b -> push(b + 1)) .bind(c -> double_top()) .bind(d -> sum_top2()) .bind(e -> pop())
into this m_return(4) .bind(a -> push(a) .bind(b -> push(b +
1) .bind(c -> double_top() .bind(d -> sum_top() .bind(e -> pop())))))
done with ruby
imagine # Not working code state_monad { a <- m_return(4)
b <- push(a) c <- push(b + 1) d <- double_top() e <- sum_top2() pop() }
Back to List m_return(4) .mapcat {|p| sqrt(p)} .mapcat {|p| inc_5(p)}
# => [3, 7]
Back to List m_return(4) .mapcat {|a| sqrt(a) .mapcat {|b| inc_5(b)}}
# => [3, 7]
Back to List list_monad { a <- m_return(4) b <-
sqrt(a) c <- inc_5(b) c }
On to Clojure • this is an example from clojure.net
• the state is a vector containing every function we’ve called so far
(defn inc-s [x] (fn [state] [(inc x) (conj state :inc)]))
in clojure (defn inc-s [x] (fn [state] [(inc x) (conj
state :inc)])) (defn do-things [x] (domonad state-m [a (inc-s x) b (double-s a) c (dec-s b) d (dec-s c)] d)) ! ((do-things 7) []) => [14 [:inc :double :dec :dec]]
state monad in Clojure (defmonad state-m "Monad describing stateful computations.
The monadic values have the structure (fn [old-state] [result new-state])." [m-result (fn m-result-state [v] (fn [s] [v s])) m-bind (fn m-bind-state [mv f] (fn [s] (let [[v ss] (mv s)] ((f v) ss)))) ])
state monad in Haskell inc = state (\st -> let
st' = st +1 in (st’,st')) inc3 = do x <- inc y <- inc z <- inc return z
Finally, IO
IOMonad • rand-int(100) is non deterministic !
ay-yo
IOMonad • rand-int(100) is non deterministic • rand-int(100, seed =
42) is deterministic • monadic value: f(world) => [value, world-after-io]
IOMonad • puts() just `appends to a buffer’ in the
real world • How does gets() return different strings? • gets() returns a fixed value based on the `world’
Image Credits http://www.myfoodarama.com/2010/11/masala- dosa.html http://www.clojure.net/2012/02/10/State/ http://www.cafepress.com/ +no_place_like_home_ruby_slippers_3x5_area_rug, 796646161 http://www.netizens-stalbans.co.uk/installs-and- upgrades.html.htm
http://www.hpcorporategroup.com/what-is-the-life- box.html
Thank You MANY QUESTIONS? VERY MONAD SO FUNCTIONAL Y NO
CLOJURE?
[email protected]
@tdinkar WOW WOW WOW MUCH EASY SUPER SIMPLE