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Re^7: Derangements iterator (callbacks)

by Anonymous Monk
on Jan 02, 2006 at 19:27 UTC ( #520432=note: print w/replies, xml ) Need Help??


in reply to Re^6: Derangements iterator (callbacks)
in thread Derangements iterator

#!/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, $xs) = @_; my($head, @tail) = @$xs; $head ? foldl_rec($f, $f->($z,$head), \@tail) : $z; } # "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, $xs) = @_; my($head, @tail) = @$xs; $head ? $f->($head,foldr_rec($f, $z, \@tail)) : $z; } $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

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