- ta tb tc - . . . -_|__ _|__ _|__ -| | | | | | -1 . . 3 1 . - _|__ _|__ _|__ - | | | | | | - 2 3 1 2 3 2 - -let ta = Node (Leaf 1, Node (Leaf 2, Leaf 3));; -let tb = Node (Node (Leaf 1, Leaf 2), Leaf 3);; -let tc = Node (Leaf 1, Node (Leaf 3, Leaf 2));; -- -So `ta` and `tb` are different trees that have the same fringe, but -`ta` and `tc` are not. - -The simplest solution is to map each tree to a list of its leaves, -then compare the lists. But because we will have computed the entire -fringe before starting the comparison, if the fringes differ in an -early position, we've wasted our time examining the rest of the trees. - -The second solution was to use tree zippers and mutable state to -simulate coroutines (see [[coroutines and aborts]]). In that -solution, we pulled the zipper on the first tree until we found the -next leaf, then stored the zipper structure in the mutable variable -while we turned our attention to the other tree. Because we stopped -as soon as we find the first mismatched leaf, this solution does not -have the flaw just mentioned of the solution that maps both trees to a -list of leaves before beginning comparison. - -Since zippers are just continuations reified, we expect that the -solution in terms of zippers can be reworked using continuations, and -this is indeed the case. Before we can arrive at a solution, however, -we must define a data structure called a stream: - - type 'a stream = End | Next of 'a * (unit -> 'a stream);; - -A stream is like a list in that it contains a series of objects (all -of the same type, here, type `'a`). The first object in the stream -corresponds to the head of a list, which we pair with a stream -representing the rest of a the list. There is a special stream called -`End` that represents a stream that contains no (more) elements, -analogous to the empty list `[]`. - -Actually, we pair each element not with a stream, but with a thunked -stream, that is, a function from the unit type to streams. The idea -is that the next element in the stream is not computed until we forced -the thunk by applying it to the unit: - -

-# let rec make_int_stream i = Next (i, fun () -> make_int_stream (i + 1));; -val make_int_stream : int -> int stream =- -You can think of `int_stream` as a functional object that provides -access to an infinite sequence of integers, one at a time. It's as if -we had written `[1;2;...]` where `...` meant "continue indefinitely". - -So, with streams in hand, we need only rewrite our continuation tree -monadizer so that instead of mapping trees to lists, it maps them to -streams. Instead of - - # tree_monadize (fun a k -> a :: k a) t1 (fun t -> []);; - - : int list = [2; 3; 5; 7; 11] - -as above, we have - - # tree_monadize (fun i k -> Next (i, fun () -> k ())) t1 (fun _ -> End);; - - : int stream = Next (2,-# let int_stream = make_int_stream 1;; -val int_stream : int stream = Next (1, ) (* First element: 1 *) -# match int_stream with Next (i, rest) -> rest;; -- : unit -> int stream = (* Rest: a thunk *) - -(* Force the thunk to compute the second element *) -# (match int_stream with Next (i, rest) -> rest) ();; -- : int stream = Next (2, ) -

-let rec compare_streams stream1 stream2 = - match stream1, stream2 with - | End, End -> true (* Done! Fringes match. *) - | Next (next1, rest1), Next (next2, rest2) when next1 = next2 -> compare_streams (rest1 ()) (rest2 ()) - | _ -> false;; - -let same_fringe t1 t2 = - let stream1 = tree_monadize (fun i k -> Next (i, fun () -> k ())) t1 (fun _ -> End) in - let stream2 = tree_monadize (fun i k -> Next (i, fun () -> k ())) t2 (fun _ -> End) in - compare_streams stream1 stream2;; -- -Notice the forcing of the thunks in the recursive call to -`compare_streams`. So indeed: - -

-# same_fringe ta tb;; -- : bool = true -# same_fringe ta tc;; -- : bool = false -- -Now, this implementation is a bit silly, since in order to convert the -trees to leaf streams, our tree_monadizer function has to visit every -node in the tree. But if we needed to compare each tree to a large -set of other trees, we could arrange to monadize each tree only once, -and then run compare_streams on the monadized trees. - -By the way, what if you have reason to believe that the fringes of -your trees are more likely to differ near the right edge than the left -edge? If we reverse evaluation order in the tree_monadizer function, -as shown above when we replaced leaves with their ordinal position, -then the resulting streams would produce leaves from the right to the -left. - The idea of using continuations to characterize natural language meaning ------------------------------------------------------------------------ diff --git a/using_continuations_to_solve_same_fringe.mdwn b/using_continuations_to_solve_same_fringe.mdwn new file mode 100644 index 00000000..b65add60 --- /dev/null +++ b/using_continuations_to_solve_same_fringe.mdwn @@ -0,0 +1,130 @@ +Using continuations to solve the same fringe problem +---------------------------------------------------- + +We've seen two solutions to the same fringe problem so far. +The problem, recall, is to take two trees and decide whether they have +the same leaves in the same order. + +

+ ta tb tc + . . . +_|__ _|__ _|__ +| | | | | | +1 . . 3 1 . + _|__ _|__ _|__ + | | | | | | + 2 3 1 2 3 2 + +let ta = Node (Leaf 1, Node (Leaf 2, Leaf 3));; +let tb = Node (Node (Leaf 1, Leaf 2), Leaf 3);; +let tc = Node (Leaf 1, Node (Leaf 3, Leaf 2));; ++ +So `ta` and `tb` are different trees that have the same fringe, but +`ta` and `tc` are not. + +The simplest solution is to map each tree to a list of its leaves, +then compare the lists. But because we will have computed the entire +fringe before starting the comparison, if the fringes differ in an +early position, we've wasted our time examining the rest of the trees. + +The second solution was to use tree zippers and mutable state to +simulate coroutines (see [[coroutines and aborts]]). In that +solution, we pulled the zipper on the first tree until we found the +next leaf, then stored the zipper structure in the mutable variable +while we turned our attention to the other tree. Because we stopped +as soon as we find the first mismatched leaf, this solution does not +have the flaw just mentioned of the solution that maps both trees to a +list of leaves before beginning comparison. + +Since zippers are just continuations reified, we expect that the +solution in terms of zippers can be reworked using continuations, and +this is indeed the case. Before we can arrive at a solution, however, +we must define a data structure called a stream: + + type 'a stream = End | Next of 'a * (unit -> 'a stream);; + +A stream is like a list in that it contains a series of objects (all +of the same type, here, type `'a`). The first object in the stream +corresponds to the head of a list, which we pair with a stream +representing the rest of a the list. There is a special stream called +`End` that represents a stream that contains no (more) elements, +analogous to the empty list `[]`. + +Actually, we pair each element not with a stream, but with a thunked +stream, that is, a function from the unit type to streams. The idea +is that the next element in the stream is not computed until we forced +the thunk by applying it to the unit: + +

+# let rec make_int_stream i = Next (i, fun () -> make_int_stream (i + 1));; +val make_int_stream : int -> int stream =+ +You can think of `int_stream` as a functional object that provides +access to an infinite sequence of integers, one at a time. It's as if +we had written `[1;2;...]` where `...` meant "continue indefinitely". + +So, with streams in hand, we need only rewrite our continuation tree +monadizer so that instead of mapping trees to lists, it maps them to +streams. Instead of + + # tree_monadize (fun a k -> a :: k a) t1 (fun t -> []);; + - : int list = [2; 3; 5; 7; 11] + +as above, we have + + # tree_monadize (fun i k -> Next (i, fun () -> k ())) t1 (fun _ -> End);; + - : int stream = Next (2,+# let int_stream = make_int_stream 1;; +val int_stream : int stream = Next (1, ) (* First element: 1 *) +# match int_stream with Next (i, rest) -> rest;; +- : unit -> int stream = (* Rest: a thunk *) + +(* Force the thunk to compute the second element *) +# (match int_stream with Next (i, rest) -> rest) ();; +- : int stream = Next (2, ) +

+let rec compare_streams stream1 stream2 = + match stream1, stream2 with + | End, End -> true (* Done! Fringes match. *) + | Next (next1, rest1), Next (next2, rest2) when next1 = next2 -> compare_streams (rest1 ()) (rest2 ()) + | _ -> false;; + +let same_fringe t1 t2 = + let stream1 = tree_monadize (fun i k -> Next (i, fun () -> k ())) t1 (fun _ -> End) in + let stream2 = tree_monadize (fun i k -> Next (i, fun () -> k ())) t2 (fun _ -> End) in + compare_streams stream1 stream2;; ++ +Notice the forcing of the thunks in the recursive call to +`compare_streams`. So indeed: + +

+# same_fringe ta tb;; +- : bool = true +# same_fringe ta tc;; +- : bool = false ++ +Now, this implementation is a bit silly, since in order to convert the +trees to leaf streams, our tree_monadizer function has to visit every +node in the tree. But if we needed to compare each tree to a large +set of other trees, we could arrange to monadize each tree only once, +and then run compare_streams on the monadized trees. + +By the way, what if you have reason to believe that the fringes of +your trees are more likely to differ near the right edge than the left +edge? If we reverse evaluation order in the tree_monadizer function, +as shown above when we replaced leaves with their ordinal position, +then the resulting streams would produce leaves from the right to the +left. + -- 2.11.0