Monoid and Monad

Quora: What is the difference between monoid and monad? by Mort Yao https://www.quora.com/What-is-the-difference-between-monoid-and-monad/answer/Mort-Yao?share=41115848&srid=ZyHj

A well-said, perhaps the briefest ever answer is: monad is just a monoid in the category of endofunctors.

monoid is defined as an algebraic structure (generally, a set) M with a binary operation (multiplication) • : M × M → M and an identity element (unit) η : 1 → M, following two axioms:

i. Associativity
∀ a, b, c ∈ M, (a • b) • c = a • (b • c)

ii. Identity
∃ e ∈ M ∀ a ∈ M, e • a = a • e = a

When specifying an endofunctor T : X → X (which is a functor that maps a category to itself) as the set M, the Cartesian product of two sets is just the composition of two endofunctors; what you get from here is a monad, with these two natural transformations:

1. The binary operation is just a functor composition μ : T × T → T
2. The identity element is just an identity endofunctor η : I → T

Satisfied the monoid axioms (i. & ii.), a monad can be seen as a monoid which is an endofunctor together with two natural transformations. 

The name “monad” came from “monoid” and “triad”, which indicated that it is a triple (1 functor + 2 trasformations), monoidic algebraic structure.



In other words, monoid is a more general, abstract term. When applying it to the category of endofunctors, we have a monad.

https://www.quora.com/What-is-the-difference-between-monoid-and-monad/answer/Bartosz-Milewski?share=e4dc6c63&srid=ZyHj

State Monad: Milewski

Don’t fear the Monad

Brian Beckman: 

You can understand Monad without too much Category Theory.

Functional Programming = using functions to compose from small functions to very complex software (eg. Nuclear system, driverless car software…).

Advantages of Functional Programming:

  • Strong Types Safety: detect bugs at compile time. 
  • Data Protection thru Immutability: Share data safely in Concurrent / Parallel processing.
  • Software ‘Componentisation’  ie Modularity : Each function always returns the same result, ease of software reliability testing.

Each “small” function is a Monoid.
f : a -> a (from input of type ‘a‘ , returns type ‘a’)
g: a -> a

compose h from f & g : (strong TYPING !!)
h = f。g : a -> a

[Note] : Object in Category, usually called  Type in Haskell, eg. ‘a’ = Integer)

You already know a Monoid (or Category in general) : eg Clock

  1. Objects: 1 2 3 …12 (hours)
  2. Arrow (Morphism): rule “+”: 
    • 7 + 10 = 17 mod 12 = 5
  3. Law of Associativity:
    x + (y + z) = (x + y) + z
  4. Identity (or “Unit”):  (“12”):
    x + 12 = 12 + x = x

More general than Monoid is a “Monoidal” Category where: (instead of only single object ‘a’, now more “a b c…”)
f : a -> b
g: b -> c
h = f。g : a -> c

Function under composition Associative rule and with an Identity => Monoid


Monad (M): a  way to manage  the side-effects (I/O, exception , SQL Database, etc) within the Functional Programming way like monoidal categories: ie associative composition, identity.

Remark: For the last 60 years in Software, there have been 2 camps: 

  1. Bottom-Up Design: from hardware foundation,  build performance-based languages: Fortran, C, C++, C#, Java…
  2. Top-Down Design: from Mathematics foundation, build functional languages (Lambda-Calculus, Lisp, Algo, Smalltalk, Haskell…). 
  3. F# (Microsoft) is the middle-ground between 1 & 2.

Ref: What is a Monad ?

Monad = chaining operations with binding “>>=”

  • Possible use: allows to write mini-language, parser…

BM Category Theory 10: Monad & Monoid

10.1 Monad
Analogy

Function : compose “.“, Id

Monad: compose “>>=“, return

Imperative (with side effects eg. state, I/O, exception ) to Pure function by hiding or embellishment in Pure function but return “embellished” result.

10. 2 Monoid

Monoid in category of endo functors = Monad

Ref Book : 

What is the significance of monoids in category theory? by Bartosz Milewski 

BM Category Theory 3 & 4 Monoid, Kleisli Category (Monad), Terminal /Initial Object, Product… Free Monoid 

[Continued from 1.1 to 2.2]

3.1 Monoid M (m, m)

Same meaning in Category as in Set: Only  1 object, Associative, Identity

Thin / Thick Category:

  • “Thin” with only 1 arrow between 2 objects; 
  • “Thick” with many arrows between 2 objects.

Arrow : relation between 2 objects. We don’t care what an arrow actually is (may be total / partial order relations like = or \leq , or any relation), just treat arrow abstractly.

Note: Category Theory’s “Abstract Nonsense” is like Buddhism “空即色, 色即空” (Form = Emptiness).

Example of Monoid: String Concatenation: identity = Null string.

Strong Typing: function f calls function g, the type of the output of f must match with the type of the input of g.

Weak Typing:   no need to match type. eg. Monoid.

Category induces a Hom-Set: (Set of “Arrows”, aka Homomorphism 同态, which preserves structure after the “Arrow”)

  • C (a, b) : a -> b
  • C (a,a) for Monoid : a -> a

3.2 Kleisli Category (Monad) 

Morphism between a, b :
\boxed{a \to (b, \text {string})}

In brief, embellish the returned output b with string. 

In the example of Kleisli Category, using Monoid (single object) with embellishing string => Monad (covered in later videos).

4.1 Monad (one definition) : Kleisli

Set and Category SET: 

  • Set has elements and function between sets.
  • Category SET has NO element, only arrows (morphisms)

Universal Construction: define relations between Objects.

Set is rich with functions, except 1 case: There is no function from a non-empty set to Void (because a function must have an image in the co-domain, but Void has no element as the Image.)

Terminal Object : all other objects (each with UNIQUE arrow) point to it. [Denoted as () or Unit]

\boxed { \forall a, \exists f :: a \to ()}

\boxed { \forall a, f :: a \to (), g :: a \to () \implies f = g } (Uniqueness of arrow)

Note: Terminal Object (T.O.) is the largest object. There is no T.O. in Natural numbers.

Initial Object: reverse of Terminal object. Outgoing arrow from Void.

absurd :: Void -> a

Category “Equality”:

  • 2 objects are isomorphic (exists an inverse arrow), but we don’t say they are equal;
  • Arrows can be equal by associativity :  f \circ (g \circ h) = (f \circ g) \circ h

Terminal Object (Similarly, Initial Object) is unique (up to ONLY ONE Isomorphism). 

Assume both Ta and Tb are Terminal Objects. Prove Ta = Tb by showing there is only one isomorphism.

Note1: example of 2 isomorphisms between 2 Objects : (0, 1) and (T, F) => 

  • 1st isomorphism : (1 ~ T), (0 ~ F)
  • 2nd isomorphism: (1 ~ F), (0 ~ T)

Note2: There are outgoing arrows from T.O. => generalising.

4.2 Products

C^{op} : also a Category with all arrows in C reversed.

Products (Cartesian) of 2 objects (a,b): by Universal Construction.

p and q are (good) projections of c.
p’ and q’ are (another) projections of c’.

c is the product of (a, b) if 
there is a UNIQUE  isomorphism m,
\boxed {m :: c' \to c }
\boxed {p' = p  \circ m}
\boxed{q' = q  \circ m}

II 3.2 Free Monoid

Monoid : with a Set of 2 generators (a, b)

  • Identity: e*a = a*e = a; 
  • Associativity

Free Monoid = {e, a, b, ab, ba, aab, abb, ….}, notice it is generated by the set {a, b} and thus not a Monoid (noted as Mon).

We can create a Functor U (aka  Forgetting function) between Monoid and its underlying Set: 

\boxed { U :: Mon \to Set }

Minggatu-Catalan Number

Minggatu-Catalan Number

image

First discovered by the Chinese Mathematician Minggatu 明安图 (清 康熙, 1730), later by the French (né Belgian) École Polytechnique mathematician Eugène Charles Catalan (1814 – 1894).

Proof: Consider ways of making sums with {0, 1, 2, 3, 4…} and +, ( , ).

0 = (0)
1 way for 0
\boxed {C_{0} = 1}

0+1 =(0+1) :
1 way for 1 \boxed {C_{1} = 1}

0 +1 +2 = ((0+(1+2)) = ((0+1)+2):
2 ways for 2 \boxed {C_{2} = 2}

0+1+2+3 = (0+(1+(2+3))) = (0+((1+2)+3))= ((0+1)+(2+3)) = ((0+(1+2))+3) = (((0+1)+2)+3) :
5 ways for 3 \boxed {C_{3} = 5}

0+1+2+3+…+n = ?
? ways for n \boxed {C_{n} = ?}

Try finding the pattern for C_{4}:

Let A represents {0, 1, 2, 3, 4}.
There are 4 cases:

Case 1:
(\underbrace{(A)}_{C_{0}} + \underbrace{(A+ A+ A+ A)} _{C_{3}})

Case 2:
( \underbrace{ (A+A ) }_{C_{1}} + \underbrace{ (A+ A+ A) }_{C_{2}} )

Case 3:
( \underbrace{ (A+A+A ) }_{C_{2}} + \underbrace{ ( A+ A) }_{C_{1}})

Case 4:
( \underbrace{ (A+A+A+A) }_{C_{3}} + \underbrace{ (A) }_{C_{0}} )

C_{4} = C_{0}.C_{3} + C_{1}.C_{2} + C_{2}.C_{1}+ C_{3}.C_{0}

Generalise:
C_{n+1} = C_{0}.C_{n-0} + C_{1}.C_{n-1} + C_{2}.C_{n-2} + ...+ C_{k}.C_{n-k} + ...+ C_{n-0}.C_{0}

Recurrence Sequence of C_{n}:
\boxed{ \begin{cases} C_{0} =1, \\ C_{n+1} = \displaystyle\sum_{k=0}^{n}C_{k}C_{n-k} , & \text{( } n \geq {0} \text {)} \end{cases} }

image

Step 1: Generating Function C(x) :
Let C(x) =C_{0} + C_{1}x+ C_{2}x^2 + ...+ C_{n}x^n

\boxed {\displaystyle C(x) = \sum_{n=0}^{\infty} C_{n} x^n }

C(x)^{2} = (C_{0}C_{0})+ (C_{0}C_ {1} + C_{1}C_{0})x+ (C_{0}C_{2}+ C_{1}C_{1}+ C_{2}C_{0} ) x^2 + ...

The coefficient for x^{n}: \displaystyle \sum_{k=0}^{n}C_{k}C_{n-k}

\displaystyle C(x)^2 = \sum_{n=0}^{\infty} \underbrace{ \left (\sum_{k=0}^{n}C_{k}C_{n-k} \right)}_{\text {coeff of }x^{n}} x^n = \sum_{n=0}^{\infty} \underbrace{ C_{n+1}}_{\text {recurrence}} x^n

\displaystyle x.C(x)^{2} =x. \sum_{n=0}^{\infty} C_{n+1} x^n = \sum_{n=0}^{\infty} C_{n+1} x^{n +1}
Change (n+1) to n by adjusting initial value from n= 0 to 1:
\displaystyle x.C(x)^{2} = \sum_{n=1}^{\infty} C_{n} x^n

Reset the initial value from n=1 to 0:
\displaystyle x.C(x)^{2} = \sum_{n=0}^{\infty} C_{n} x^n - C_{0}
We have seen
C_{0}= 1 and
\displaystyle C(x) = \sum_{n=0}^{\infty} C_{n} x^n

\displaystyle x.C(x)^{2} - C(x) +1 = 0
This is the quadratic equation on C (x):

C(x)= \frac{1 \pm \sqrt{1-4x} }{2x}

2x.C(x)= 1 \pm \sqrt{1-4x}

Case: 2x.C(x)= 1 + \sqrt{1-4x}
When x \rightarrow 0
2x.C(x)= 0
1 + \sqrt{1-4x}= 1+1=2
Impossible !

Case: 2x.C(x)= 1 - \sqrt{1-4x}
When x \rightarrow 0
2x.C(x)= 0
1 - \sqrt{1-4x}= 1- 1=0

Step 2: The closed form for the generating function C(x) :

\boxed { C(x)= \frac{1 - \sqrt{1-4x} }{2x} }

Step 3: The closed form for C_{n}

We expand \sqrt{1-4x} into power series:

Let \displaystyle \sqrt{1-4x}= K(x)= \sum_{k=0}^{\infty}K_{k}x^{k}

2xC(x)= 1 - \sqrt{1-4x}
\displaystyle 2xC(x)= 1 - \sum_{k=0}^{\infty}K_{k}x^{k}

\displaystyle 2x\sum_{k=0}^ {\infty}C_{k}x^{k} = 1 - \sum_{k=0}^{\infty}K_{k}x^{k}

Move 2x inside the left sum:
\displaystyle \sum_{k=0}^ {\infty}2C_{k}x^{k+1} = 1 - \sum_{k=0}^{\infty}K_{k}x^{k}

Synchronise the indices k,
\displaystyle \sum_{k=1}^ {\infty}2C_{k-1}x^{k} = 1 - K_{0} - \sum_{k=1}^{\infty}K_{k}x^{k}

\displaystyle \sum_{k=1}^ {\infty}2C_{k-1}x^{k} + \sum_{k=1}^{\infty}K_{k}x^{k} = 1 - K_{0}

\displaystyle \sum_{k=1}^ {\infty} (2C_{k-1} + K_{k})x^{k} = 1 - K_{0}

We compare the coefficient of each term x^{k} :

x^{0}: 0 = 1- K_{0}
x^{1}: 2C_{0} + K_{1} = 0
x^{2}: 2C_{1} + K_{2} = 0

x^{n}: 2C_{n} + K_{n+1} = 0

\boxed{ \begin{cases} K_{0} =1, \\ C_{n} = \displaystyle - \frac {K_{n+1}}{2} , & \text{( } n \geq {0} \text {)} \end{cases} }

We need to express K_{n+1} in term of n only.

Go back to the definition of K(x):
\displaystyle K(x)= \sum_{k=0}^{\infty}K_{k}x^{k}

\displaystyle K(x)= K_{0}+K_{1}x + K_{2}x^{2} + ...+ K_{n}x^{n}+...

Differentiate:
\displaystyle K'(x)=1.K_{1}x + 2K_{2}x^{1} + ...+ nK_{n}x^{n-1}+...

\displaystyle K'(0)= 1K_{1}

Differentiate 2nd time:
\displaystyle K''(x)=2.1K_{2} + ...+ n. (n-1)K_{n}x^{n-2}+...

\displaystyle K''(0)= 2.1K_{2}

Continue …

K^{(n)}(x)= n (n-1)(n-2)...2.1K_{n} + ...

K^{(n)}(0)= n! K_{n}

\boxed { K_{n} = \frac { K^{(n)}(0)}{n!} = \frac { K^{(n)}(0)}{n^{\underline {n}}} }

Note: See blog on “Falling Factorial”: x^{\underline {n}}

Next step, we shall find K^{(n)}(0) in term of n
Recall
K(x) = (1-4x)^{\frac {1}{2}}

K'(x) = -2.(1-4x)^{-\frac {1}{2}}
K''(x) = -2.2(1-4x)^{-\frac {3}{2}}
K'''(x) = -2.4.3(1-4x)^{-\frac {5}{2}}
K''''(x) = -2.6.5.4(1-4x)^{-\frac {7}{2}}

K^{(n)}(x) = -2.(2n-2)^{\underline {n-1}}(1-4x)^{-\frac {2n-1}{2}}

K^{(n+1)}(x) = -2.(2n)^{\underline {n}}(1-4x)^{-\frac {2n+1}{2}}

\boxed { K^{(n+1)}(0) = -2.(2n)^{\underline {n}} }

Also from above, we have seen:
\boxed { K_{n} = \frac { K^{(n)}(0)}{n^{\underline {n}}} }

Change n to (n+1):
K_{n+1} = \frac { K^{(n+1)}(0)}{(n+1)^{\underline {n+1}}}

K_{n+1} = \frac { -2.(2n)^{\underline {n}} } {(n+1)^{\underline {n+1}}}
From step 3 above, we also know:
C_{n} = - \frac {K_{n+1}} {2}

C_{n} = \frac { (2n)^{\underline {n}} } {(n+1)^{\underline {n+1}}} = \frac {1}{n+1} \frac { (2n)^{\underline {n}} } {(n)^{\underline {n}}}
since (n+1)^{\underline {n+1}} = (n+1)!= (n+1).n! = (n+1). (n)^{\underline {n}}
and
\frac { (2n)^{\underline {n}}} {(n)^{\underline {n}}} = \binom {2n}{n}  (by definition of Falling Factorial)

\boxed {\displaystyle C_{n} = \frac {1}{n+1} \binom {2n}{n}  } [QED]

image

Ref:

1. History:
image

2. Proof of Minggatu Catalan numbers

3. Monoid

4. Kleene Star

5. Ref:
Math Girls by Hiroshi Yuki et al.
http://www.amazon.co.uk/dp/0983951306/ref=cm_sw_r_udp_awd_UKSvtb1AXNWCE