explaining the problem, and why is not a monad

biblio.bib

@@ -12,4 +12,12 @@
   author = {François Pottier},
   title = {the {Inferno} library: \url{https://gitlab.inria.fr/fpottier/inferno}},
   year = 2014,
-}
+}
+
+@techreport{composing-monads,
+  title = {Composing Monads},
+  author = {Mark P. Jones and Luc Duponcheel},
+  institution ={Yale University},
+  year = {1993},
+  number = {YALEU/DCS/RR-1004},
+}

workshop.tex

@@ -133,25 +133,33 @@ concluded by a final ``map'' that takes all the witnesses of the
 constraint-solving pieces, and combines them into a final witness for
 the inferred term.
 
-\subsection{Our sufferings} Now consider extending pair to handle not
+\subsection{The problem} Consider extending this code to handle not
 just binary pairs, but n-ary tuples \lstinline{Tuple us}. Instead of
 a statically-known number of inference variables (2), we want to
-generate one for each subterm of the list. How would you program this
-with the proposed API?
-
-There would be a standard solution if \lstinline{'a co} was an
-inference/elaboration \emph{monad}, with a fresh-inference-variable
-operation \lstinline{exist : (variable * F.ty) co}. We would write
-something like, with a generic combinator %
-\lstinline{traverse : 'a co list -> 'a list co}:
+generate one for each subterm of the list, to build the result
+type. How would you program this with the proposed API?
+
+If \lstinline{'a co} was an inference/elaboration \emph{monad}, it
+would be natural to decompose the problem with a function of type
+\lstinline{'a list -> Infer.variable list co}, or possibly
+\lstinline{mapM : ('a -> 'b co) -> 'a list -> 'b list co},
+and then \lstinline{bind} the result to continue. For example:
+
 \begin{lstlisting}
 | ML.Tuple us ->
-  traverse (List.map (fun _ -> exists) us) >>= fun vs ->
-  List.mapM2 (fun (v, ty) u -> hastype u w ^& return ty) vs us >>= fun utys ->
+  let on_each u =
+    exist (fun v -> v ^& hastype u v)
+    <$$> fun (ty, (v, u')) -> (v, (u', ty)) in
+  mapM on_each us >>= fun vutys ->
+  let vs, utys = List.split vutys in
   w --- Infer.product vs ^^
   return (ML.tuple utys)
 \end{lstlisting}
-Instead, we have to write code in continuation-passing style:
+%
+But \lstinline{'a co} is \emph{not} a monad for a good reason -- we
+explain this in the next section. Instead, we have to write code in
+continuation-passing style:
+%
 \begin{lstlisting}
   let rec traverse (us : 'a list) (k : variable list -> 'r co) : ((F.term * F.ty) list * 'r) co =
     match us with
@@ -161,13 +169,77 @@ Instead, we have to write code in continuation-passing style:
         traverse us (fun vs ->
           hastype u v ^&
           k (v :: vs)
-      )) <$$> fun (ty, u', (utys, rs)) -> ((u', ty) :: utys, r)
+      )) <$$> fun (ty, (u', (utys, rs))) -> ((u', ty) :: utys, r)
   in
   traverse us (fun vs ->
     w --- Infer.product vs
   ) <$$> fun (utys, ()) -> F.Tuple (List.combine u's tys)
 \end{lstlisting}
 
+Note: \lstinline{traverse} is called with a single continuation; it is
+possible to inline this continuation in the empty-list case, and get
+a simpler, less generic function, that could not be reused
+elsewhere. We wish to discuss the reusable presentation in this document;
+the API design improvements also help in the non-reusable case.
+
+There are two difficulties with this API and the code we had to write:
+\begin{enumerate}
+\item It is much harder to figure out how to write this code than in the monadic
+  version, which allows a natural problem decomposition. Programmers may not know
+  at all how to generate dynamically many inference variables, and/or when a CPS
+  is necessary and how to achieve it.
+\item The code, in either versions, is hard to read. In particular,
+  values are \emph{named} far from where they are \emph{expressed} in
+  the program; notice for example how \lstinline{u'} binds the witness
+  of \lstinline{hastype u v}, but is placed farther away in the
+  program. When we explain this code, we say that the
+  \lstinline{Infer.ty} result of \lstinline{exists} or the
+  \lstinline{(F.term * F.ty) list} result of \lstinline{traverse} is
+  ``pushed on the stack''; we would rather avoid this particular style
+  of point-free programming.
+\end{enumerate}
+
+Our contribution is to identify, explain this problem, and propose
+a partial solution: a systematic approach to programming with
+``binders'' that can be explained and documented, and a different API
+that lets us name terms in a more natural, local way -- at the cost of
+``unpacking'' operations later on, as we will explain.
+
+While our work is specific to Inferno (and OCaml), we believe that
+this API problem arises in the more general case of an applicative
+functor with \emph{quantifiers}, as \lstinline{exist} in our example.
+
+\subsection{Not a monad} At the cost of our page limit, let us explain
+why the \lstinline{'a co} type of Inferno cannot be given the
+structure of a monad.
+
+The witness \lstinline{'a} is built with knowledge of the
+\emph{global} solution to the inference constraints
+generated. Consider the applicative combinator%
+\lstinline{pair : 'a co -> 'b co -> ('a * 'b) co}; both arguments
+generate a part of the constraint, but the witnesses of \lstinline{'a}
+and \lstinline{'b} can only be computed once the constraints on
+\emph{both} sides are solved. For example, when inferring the ML term
+\lstinline{(x, x + 1)}, the type inferred for the first component is
+determined by unifications coming from the constraint of the second.
+
+Implementing \lstinline{bind : 'a co -> ('a -> 'b co) -> 'b co}
+would require building the witness for \lstinline{'a} before the
+constraint arising from \lstinline{'b co} is known; this cannot be
+done if \lstinline{'a} requires the final solution.
+
+If you are abstractly inclined: internally \lstinline{'a co} is
+defined as \lstinline{raw_constraint * (solution -> 'a)} it is the
+composition of a ``writer'' monad that generate constraints and
+a ``reader'' monad consuming the solution. For the composition
+$F \circ G$ of two monads to itself be a monad, a sufficient
+condition\citep{composing-monads} is that they commute:
+$(F \circ G) \to (G \circ F)$ -- this suffices to define
+$\mathtt{join} : (F \circ G \circ F \circ G) \to F \circ G$ from the
+$\mathtt{join}$ operation of $F$ and $G$. Commuting the reader with
+the writer would require reading the solution before writing the
+constraint; this is not possible if we want the final solution.
+
 \bibliography{biblio}
 
 \end{document}