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Re^5: Derangements iterator (callbacks)

 on Jan 01, 2006 at 18:21 UTC ( #520270=note: print w/replies, xml ) Need Help??

in reply to Re^4: Derangements iterator (callbacks)

From enumerators to cursors: turning the left fold inside out...
The mechanical inversion procedure presented in * had a catch: it relies on shift/reset (or call/cc plus a mutable cell, which is the same thing). How can we do such an inversion in Haskell? We can introduce a right fold enumerator, which is more amenable to such transformations. Or we can use a continuation monad and emulate shift/reset. The present article demonstrates the third approach: a non-recursive left-fold. We argue that such a left fold is the best interface for a collection. Indeed, given the non-recursive left-fold we can:
• instantiate it into the ordinary left fold
• instantiate in into a stream
If we turn two enumerators into streams, we can *safely* interleave these streams.

We should point out that the relation between the left fold, the non-recursive left fold and the stream is deep. The ordinary, recursive left fold is the fix point of the non-recursive one. On the other hand, the instantiation of the non-recursive left fold as a stream, as we shall see, effectively captures a continuation in a monadic action. We see once again that call/cc and Y are indeed two sides of the same coin **.

The rest of the article demonstrates the inversion procedure. The procedure is generic, as evidenced by its polymorphic type. We illustrate the technique on an example of a file considered a collection of characters. Haskell provides a stream interface to that collection: hGetChar. We implement a left fold enumerator. We then turn that enumerator back to a stream: we implement a function 'myhgetchar' _only_ in terms of the left fold enumerator. The approach is general and uses no monadic heavy lifting.

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Re^6: Derangements iterator (callbacks)
by demerphq (Chancellor) on Jan 02, 2006 at 12:33 UTC

Well, seeing as apparently you understand this article maybe you could give us an example how to turn foreach (@list){ ... } into a cursor based approach. Since perl doesnt support first class continuations I guess you will need to implement this "left fold" operator. Which in itself would be pretty interesting. Actually even explaining in normal english (unlike the functional jargon gobbly-gook that the article uses) what this "left fold" operator does would be nice.

---
\$world=~s/war/peace/g

```#!/usr/bin/perl -w

# A non-recursive left fold (foldl), taken from Language::Functional
sub foldl(&\$\$) {
my(\$f, \$z, \$xs) = @_;
map { \$z = \$f->(\$z, \$_) } @{\$xs};
return \$z;
}

# Recursive foldl
sub foldl_rec {
my(\$f, \$z, \$xs) = @_;
}

# "Fold" is the universal list traversal function. Also known as
# "reduce" (see List::Util) and "accumulate" (C++ STL).  Any function
# you write that munges lists (map, grep, etc.) can be rewritten in
# terms of a fold.  It essentially takes a list and replaces each "con
+s"
# constructor with a function.  Stated another way, if you have a list
# @a = (1, 2, 3, 4), fold will replace the commas with another functio
+n
# of your choosing.  Let's say you want the sum of the elements in @a.
# Replace the commas with a '+' sign, (1 + 2 + 3 + 4).  Easy isn't it?
# You might write it as...

\$s = foldl(sub{ \$_[0] + \$_[1] }, 0, [1..4]);
print "sum = \$s\n";  # 10

# ...in addition to providing the function and the list, you supply an
# initial value to start out with.  In the case of \$sum above, we use
# 0.  If you want the product of the elements in the list you can chan
+ge
# to...

\$p = foldl(sub{ \$_[0] * \$_[1] }, 1, [1..4]);
print "product = \$p\n";  # 24

# The "left" portion comes into play because we start at the left end
+of
# the list and work towards the right.  The actual sum that is
# calculated is (((((0+1)+2)+3)+4).  It only makes a difference when t
+he
# function used isn't associative.  Subtraction is an example...

\$l = foldl(sub{ \$_[0] - \$_[1] }, 0, [1..4]);
print "left fold subtraction = \$l\n";  # ((((0-1)-2)-3)-4) == -10

# Recursive foldr
sub foldr_rec {
my(\$f, \$z, \$xs) = @_;
}

\$r = foldr_rec(sub{ \$_[0] - \$_[1] }, 0, [1..4]);
print "right fold subtraction = \$r\n";  # (1-(2-(3-(4-0)))) == -2

# The dual of "fold" is the universal list creation function, "unfold"
+.
# See more unfold in action...
#
#    http://use.perl.org/~Greg%20Buchholz/journal/26747

Actually even explaining in normal english (unlike the functional jargon gobbly-gook that the article uses) what this "left fold" operator does would be nice.
Another explaination of the original paper.

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