5 ms·

That's what I'm used to as well, but in my experience a lot of programmers take fold and reduce to be synonyms. A monoidal reduce is much less "scary" than a ge

That's what I'm used to as well, but in my experience a lot of programmers take fold and reduce to be synonyms. A monoidal reduce is much less "scary" than a general fold. I suspect most programmers have never[1] heard the word monoid, let alone know what it means, and having to remember the meaning of a weird new word is enough to make most people dislike something compared to the simpler more familiar operations. [1]Or if they have, their only encounter with it is the "a monad is just a monoid in the category of endofunctors" meme.

sigbottle 12d ago
I do know what a monoid is, but a monad in the category of endofunctors is the scary word for me :sob:
- ndriscoll 11d ago
  It just means if you have some functor F (generic type with a well-behaved `map` function, like List), then you have a `flatten` operation F[F[_]] - > F[_], and like a monoidal product, it's associative. So if you have a triply nested List, you can flatten inside first or outside first. Also, like a monoid, it has an "identity" function wrap: A->F[A] (e.g. x -> [x]). Identity in the sense that "multiplying" (flattening) with wrap does nothing. i.e. wrap(flatten(x)) = flatten(wrap(x)) = x when those things make sense. So basically wrapping and flattening behave in a sane way. Flatten is your multiply, wrap is your multiplicative identity, and it's like a monoid if you squint.
  - adastra22 11d ago
    You have become the meme.
    - ndriscoll 11d ago
      "The meme" literally comes from a book that was offering it as an intuitive explanation of a long definition, assuming you know what a monoid is. All the laws and stuff boil down to "if you generalize the idea of a monoid a little bit, and if you have some functor+flatten+wrap forming a monoid, we call that a monad." If you don't know what that stuff means then obviously it's not for you, but if you do, then it's actually a concise way to give an intuition for "what (or why) it is," which is basically just that flatten is associative and wrap is neutral. Like if someone says they know about rings and modules, you might say that an ideal is just an R-submodule of R, which grants an interesting perspective and gives a quick, memorable definition. But if they don't know about modules, you might not give them that definition.
      - adastra22 10d ago
        The point is that the terminology surrounding this is impenetrable and non intuitive. Rather than acknowledge that, we get a lesson on category theory, which is missing the point.
        ndriscoll 10d ago
        The person I replied to said they know what monoid means, so they're familiar with algebra. The explanation is intuitive for someone familiar with undergraduate algebra (adapted to also assume some familiarity with programming and show how it connects). That's literally where the meme comes from, an intuitive remark from an introductory text on category theory. If you think it's impenetrable, it's not for you. You don't have the correct background, so ignore it.
        adastra22 9d ago
        It may have come from an old textbook, but its origin as a meme is a 2009 joke blog post that gets its humor from making fun of how obtuse and pedagogically inept that explanation is. When people reference it today, this is what they are referencing: https://james-iry.blogspot.com/2009/05/brief-incomplete-and-mostly-wrong.html?m=1 https://james-iry.blogspot.com/2009/05/brief-incomplete-and-...
        ndriscoll 9d ago
        I'm aware of the meme, but it's not obtuse or pedagogically inept. Like if someone says they're familiar with groups and Fourier transforms, but not what it means to say wavelets are the Fourier basis for the affine group, and I break that down, and you don't know what any of those words mean, that's not me giving an obtuse explanation of wavelets; that's you wandering into the wrong conversation.
- antonvs 11d ago
  Do you want to understand monads, or do you want to understand the original quote that the joke you referenced was based on? For the record, the original quote by Saunders Mac Lane is "a monad in X is just a monoid in the category of endofunctors of X, with product × replaced by composition of endofunctors and unit set by the identity endofunctor." That quote is a statement in category theory. The author probably never heard of, say, Haskell - he was a pure mathematician. You can't usefully express that quote in Haskell code. You can treat it as a kind of formal description of what monads are, and Haskell generally conforms to that. But in that context, the quote itself is essentially using category theory as a metalanguage, in the same sort of way as one might write a mathematical statement that captures the semantics of some programming language expression. That said, the quote can be handwavingly understood if you know what a monoid is, and that for monads, the identity object is the identity functor, its product is `join`[1] and its unit and multiplication satisfy the usual monoid laws. For a concrete example, consider this Haskell expression using the `Maybe` monad: do x <- Just 3 return (x + 1) That desugars to: Just 3 >>= \x -> Just (x + 1) Which we can desugar to an expression in terms of the monad's monoidal product, `join`, by substituting the definition of `>>=` in terms of `join`[1] to get: join (fmap (\x -> Just (x + 1)) (Just 3)) You can evaluate that in Haskell and you'll get `Just 4`, just like the original expression. So what happened there? The inner expression `fmap (\x -> Just (x + 1)) (Just 3)` applies the anonymous function to `Just 3` to get the double-wrapped `Just (Just 4)`. One of the `Just` wrappers is then eliminated with `join`. (Btw, the fact that we have a Maybe within a Maybe here is related to the fact "monads are monoids in the category of endofunctors" - a category that maps to itself. That's where that part of the quote comes from.) In this simple example, there's some unnecessary machinery - you can get the same result with `fmap (\x -> x + 1) (Just 3)`, without the extra `Just` wrapper or the `join` to eliminate it. But then you lose the ability to do things "in the monad": the anonymous function becomes just an ordinary function, it doesn't have access to the monadic wrapper. Many of the useful things that monads can do are because the wrapper is available in every function, so you can store state in it (Reader monad), create new wrapper instances with different state and pass those on (Writer and State monad), etc. --- [1] x >>= f = join (fmap f x)

sigbottle 12d ago

I do know what a monoid is, but a monad in the category of endofunctors is the scary word for me :sob:

ndriscoll 11d ago
It just means if you have some functor F (generic type with a well-behaved `map` function, like List), then you have a `flatten` operation F[F[_]] - > F[_], and like a monoidal product, it's associative. So if you have a triply nested List, you can flatten inside first or outside first. Also, like a monoid, it has an "identity" function wrap: A->F[A] (e.g. x -> [x]). Identity in the sense that "multiplying" (flattening) with wrap does nothing. i.e. wrap(flatten(x)) = flatten(wrap(x)) = x when those things make sense. So basically wrapping and flattening behave in a sane way. Flatten is your multiply, wrap is your multiplicative identity, and it's like a monoid if you squint.
- adastra22 11d ago
  You have become the meme.
  - ndriscoll 11d ago
    "The meme" literally comes from a book that was offering it as an intuitive explanation of a long definition, assuming you know what a monoid is. All the laws and stuff boil down to "if you generalize the idea of a monoid a little bit, and if you have some functor+flatten+wrap forming a monoid, we call that a monad." If you don't know what that stuff means then obviously it's not for you, but if you do, then it's actually a concise way to give an intuition for "what (or why) it is," which is basically just that flatten is associative and wrap is neutral. Like if someone says they know about rings and modules, you might say that an ideal is just an R-submodule of R, which grants an interesting perspective and gives a quick, memorable definition. But if they don't know about modules, you might not give them that definition.
    - adastra22 10d ago
      The point is that the terminology surrounding this is impenetrable and non intuitive. Rather than acknowledge that, we get a lesson on category theory, which is missing the point.
      - ndriscoll 10d ago
        The person I replied to said they know what monoid means, so they're familiar with algebra. The explanation is intuitive for someone familiar with undergraduate algebra (adapted to also assume some familiarity with programming and show how it connects). That's literally where the meme comes from, an intuitive remark from an introductory text on category theory. If you think it's impenetrable, it's not for you. You don't have the correct background, so ignore it.
        adastra22 9d ago
        It may have come from an old textbook, but its origin as a meme is a 2009 joke blog post that gets its humor from making fun of how obtuse and pedagogically inept that explanation is. When people reference it today, this is what they are referencing: https://james-iry.blogspot.com/2009/05/brief-incomplete-and-mostly-wrong.html?m=1 https://james-iry.blogspot.com/2009/05/brief-incomplete-and-...
        ndriscoll 9d ago
        I'm aware of the meme, but it's not obtuse or pedagogically inept. Like if someone says they're familiar with groups and Fourier transforms, but not what it means to say wavelets are the Fourier basis for the affine group, and I break that down, and you don't know what any of those words mean, that's not me giving an obtuse explanation of wavelets; that's you wandering into the wrong conversation.
antonvs 11d ago
Do you want to understand monads, or do you want to understand the original quote that the joke you referenced was based on? For the record, the original quote by Saunders Mac Lane is "a monad in X is just a monoid in the category of endofunctors of X, with product × replaced by composition of endofunctors and unit set by the identity endofunctor." That quote is a statement in category theory. The author probably never heard of, say, Haskell - he was a pure mathematician. You can't usefully express that quote in Haskell code. You can treat it as a kind of formal description of what monads are, and Haskell generally conforms to that. But in that context, the quote itself is essentially using category theory as a metalanguage, in the same sort of way as one might write a mathematical statement that captures the semantics of some programming language expression. That said, the quote can be handwavingly understood if you know what a monoid is, and that for monads, the identity object is the identity functor, its product is `join`[1] and its unit and multiplication satisfy the usual monoid laws. For a concrete example, consider this Haskell expression using the `Maybe` monad: do x <- Just 3 return (x + 1) That desugars to: Just 3 >>= \x -> Just (x + 1) Which we can desugar to an expression in terms of the monad's monoidal product, `join`, by substituting the definition of `>>=` in terms of `join`[1] to get: join (fmap (\x -> Just (x + 1)) (Just 3)) You can evaluate that in Haskell and you'll get `Just 4`, just like the original expression. So what happened there? The inner expression `fmap (\x -> Just (x + 1)) (Just 3)` applies the anonymous function to `Just 3` to get the double-wrapped `Just (Just 4)`. One of the `Just` wrappers is then eliminated with `join`. (Btw, the fact that we have a Maybe within a Maybe here is related to the fact "monads are monoids in the category of endofunctors" - a category that maps to itself. That's where that part of the quote comes from.) In this simple example, there's some unnecessary machinery - you can get the same result with `fmap (\x -> x + 1) (Just 3)`, without the extra `Just` wrapper or the `join` to eliminate it. But then you lose the ability to do things "in the monad": the anonymous function becomes just an ordinary function, it doesn't have access to the monadic wrapper. Many of the useful things that monads can do are because the wrapper is available in every function, so you can store state in it (Reader monad), create new wrapper instances with different state and pass those on (Writer and State monad), etc. --- [1] x >>= f = join (fmap f x)