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If you know any linear algebra here's an example that has nothing to do with side effects: Let X be any set. Let mX be the set of formal linear combinations of

If you know any linear algebra here's an example that has nothing to do with side effects: Let X be any set. Let mX be the set of formal linear combinations of elements of X (that is, expressions like 3x₁+2x₂ where x₁ and x₂ are elements of X). Then m is a monad. fmap is variable substitution. Ex. fmap [x₁->y₁, x₂->y₂] (3x₁ + 2x₂) = (3y₁ + 2y₂). join is simplification of nested expressions. Ex. join (3(2x₁) + 2(x₁ + 2x₂)) = (8x₁ + 4x₂). >>= is substitution and simplification. Ex. (3x₁ + 2x₂) >>= [x₁->3y₂, x₁->-y₂] = (3(3y₂) + 2(-y₂)) = 7y₂. A Kleisli arrow X->mY is a system of linear equations, ie. a matrix x₁ = 3y₁ + 6y₂ x₂ = y₁ - 2y₂ >=> is composition of linear systems, ie. matrix multiplication. Whereas in IO you think of f >=> g as meaning "first f, then g", here f >=> g is the matrix product fg, ie. "first g, then f".

ggm 7y ago
You can do it, or explain it but you can't do both: proof see above
perl4ever 7y ago
I was like if m in mX is a monad, how is m not a function? Or if it is a function, how is it different from functions in general? Googling, I found the following helpful comment: "...while addition and multiplication are both monoids over the positive natural numbers, a monad is a monoid object in a category of endofunctors: return is the unit, and join is the binary operation. It couldn't be more simple. If that confuses you, it might be helpful to see a Monad as a lax functor from a terminal bicategory"
- perl4ever 7y ago
  Wait, I think I may have found a good explanation: "A few months ago Brent Yorgey complained about a certain class of tutorials which present monads by explaining how monads are like burritos. At first I thought the choice of burritos was only a facetious reference to the peculiar and sometimes strained analogies these tutorials make. But then I realized that monads are like burritos." https://blog.plover.com/prog/burritos.html https://blog.plover.com/prog/burritos.html

ggm 7y ago

You can do it, or explain it but you can't do both: proof see above

perl4ever 7y ago

I was like if m in mX is a monad, how is m not a function? Or if it is a function, how is it different from functions in general? Googling, I found the following helpful comment: "...while addition and multiplication are both monoids over the positive natural numbers, a monad is a monoid object in a category of endofunctors: return is the unit, and join is the binary operation. It couldn't be more simple. If that confuses you, it might be helpful to see a Monad as a lax functor from a terminal bicategory"

perl4ever 7y ago
Wait, I think I may have found a good explanation: "A few months ago Brent Yorgey complained about a certain class of tutorials which present monads by explaining how monads are like burritos. At first I thought the choice of burritos was only a facetious reference to the peculiar and sometimes strained analogies these tutorials make. But then I realized that monads are like burritos." https://blog.plover.com/prog/burritos.html https://blog.plover.com/prog/burritos.html