Longest increasing subsequence: Difference between revisions
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sub lis(@S) { |
sub lis(@S) { |
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my $iter = @S[*-1][*-1]; |
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@S[*-1][*-1], *.value ...^ !*.defined |
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); |
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$iter.=value; |
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} |
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} |
} |
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say lis patience(<3 2 6 4 5 1>); |
say lis patience(<3 2 6 4 5 1>); |
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say lis patience(<0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15>».Int); |
say lis patience(<0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15>».Int); |
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#say patience(<7 2 8 1 3 4 10 6 9 5>».Int);</lang> |
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{{out}} |
{{out}} |
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<pre>2 4 5 |
<pre>2 4 5 |
Revision as of 08:53, 18 August 2013
You are encouraged to solve this task according to the task description, using any language you may know.
Calculate and show here a longest increasing subsequence of the list:
And of the list:
Note that a list may have more than one subsequence that is of the maximum length.
- Ref
- Dynamic Programming #1: Longest Increasing Subsequence on Youtube
- An efficient solution can be based on Patience sorting.
Common Lisp
Common Lisp: Using the method in the video
Slower and more memory usage compared to the patience sort method. <lang lisp>(defun longest-increasing-subseq (list)
(let ((subseqs nil)) (dolist (item list) (let ((longest-so-far (longest-list-in-lists (remove-if-not #'(lambda (l) (> item (car l))) subseqs))))
(push (cons item longest-so-far) subseqs)))
(reverse (longest-list-in-lists subseqs))))
(defun longest-list-in-lists (lists)
(let ((longest nil)
(longest-len 0))
(dolist (list lists) (let ((len (length list)))
(when (> len longest-len) (setf longest list longest-len len))))
longest))
(dolist (l (list (list 3 2 6 4 5 1) (list 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15)))
(format t "~A~%" (longest-increasing-subseq l))))</lang>
- Output:
(2 4 5) (0 2 6 9 11 15)
Common Lisp: Using the Patience Sort approach
This is 5 times faster and and uses a third of the memory compared to the approach in the video. <lang lisp>(defun lis-patience-sort (input-list)
(let ((piles nil)) (dolist (item input-list) (setf piles (insert-item item piles))) (reverse (caar (last piles)))))
(defun insert-item (item piles)
(let ((inserted nil)) (loop for pile in piles
and prev = nil then (car pile) and i from 0 do (when (and (not inserted) (<= item (caar pile))) (setf inserted t (elt piles i) (push (cons item prev) (elt piles i)))))
(if inserted
piles (append piles (list (list (cons item (caar (last piles)))))))))
(dolist (l (list (list 3 2 6 4 5 1) (list 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15)))
(format t "~A~%" (lis-patience-sort l)))</lang>
- Output:
(2 4 5) (0 2 6 9 11 15)
D
From the second Python entry, using the Patience sorting method. <lang d>import std.stdio, std.algorithm, std.array;
/// Return one of the Longest Increasing Subsequence of /// items using patience sorting. T[] lis(T)(in T[] items) pure nothrow if (__traits(compiles, T.init < T.init)) {
if (items.empty) return null;
static struct Node { T val; Node* back; } auto pile = [[new Node(items[0])]];
OUTER: foreach (immutable di; items[1 .. $]) { foreach (immutable j, ref pj; pile) if (pj[$ - 1].val > di) { pj ~= new Node(di, j ? pile[j - 1][$ - 1] : null); continue OUTER; } pile ~= [new Node(di, pile[$ - 1][$ - 1])]; }
T[] result; for (auto ptr = pile[$ - 1][$ - 1]; ptr != null; ptr = ptr.back) result ~= ptr.val; result.reverse(); return result;
}
void main() {
foreach (d; [[3,2,6,4,5,1], [0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]]) d.writeln;
}</lang>
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Déjà Vu
<lang dejavu>in-pair: if = :nil dup: false drop else: @in-pair &> swap &< dup
get-last lst: get-from lst -- len lst
lis-sub pile i di: for j range 0 -- len pile: local :pj get-from pile j if > &< get-last pj di: push-to pj & di if j get-last get-from pile -- j :nil return push-to pile [ & di get-last get-last pile ]
lis d: local :pile [ [ & get-from d 0 :nil ] ] for i range 1 -- len d: lis-sub pile i get-from d i [ for in-pair get-last get-last pile ]
. lis [ 3 2 6 4 5 1 ] . lis [ 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 ] </lang>
- Output:
[ 2 4 5 ] [ 0 2 6 9 11 15 ]
Icon and Unicon
The following works in both languages:
<lang unicon>procedure main(A)
every writes((!lis(A)||" ") | "\n")
end
procedure lis(A)
r := [A[1]] | fail every (put(pt := [], [v := !A]), p := !pt) do if put(p, p[-1] < v) then r := (*p > *r, p) else p[-1] := (p[-2] < v) return r
end</lang>
Sample runs:
->lis 3 2 6 4 5 1 3 4 5 ->lis 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 0 4 6 9 11 15 ->
Java
A solution based on patience sorting, except that it is not necessary to keep the whole pile, only the top (in solitaire, bottom) of the pile, along with pointers from each "card" to the top of its "previous" pile. <lang java>import java.util.*;
public class LIS {
public static <E extends Comparable<? super E>> List<E> lis(List<E> n) { List<Node<E>> pileTops = new ArrayList<Node<E>>(); // sort into piles for (E x : n) {
Node<E> node = new Node<E>(); node.value = x;
int i = Collections.binarySearch(pileTops, node); if (i < 0) i = ~i;
if (i != 0) node.pointer = pileTops.get(i-1);
if (i != pileTops.size()) pileTops.set(i, node); else pileTops.add(node); }
// extract LIS from nodes List<E> result = new ArrayList<E>(); for (Node<E> node = pileTops.get(pileTops.size()-1); node != null; node = node.pointer) result.add(node.value); Collections.reverse(result); return result;
}
private static class Node<E extends Comparable<? super E>> implements Comparable<Node<E>> {
public E value; public Node<E> pointer;
public int compareTo(Node<E> y) { return value.compareTo(y.value); } }
public static void main(String[] args) {
List<Integer> d = Arrays.asList(3,2,6,4,5,1); System.out.printf("an L.I.S. of %s is %s\n", d, lis(d));
d = Arrays.asList(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15);
System.out.printf("an L.I.S. of %s is %s\n", d, lis(d));
}
}</lang>
- Output:
an L.I.S. of [3, 2, 6, 4, 5, 1] is [2, 4, 5] an L.I.S. of [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] is [0, 2, 6, 9, 11, 15]
Perl
<lang Perl>sub lis {
my @l = map [], 1 .. @_; push @{$l[0]}, +$_[0]; for my $i (1 .. @_-1) { for my $j (0 .. $i - 1) { if ($_[$j] < $_[$i] and @{$l[$i]} < @{$l[$j]} + 1) { $l[$i] = [ @{$l[$j]} ]; } } push @{$l[$i]}, $_[$i]; } my ($max, $l) = 0, []; for (@l) { ($max, $l) = (scalar(@$_), $_) if @$_ > $max; } return @$l;
}
print join ' ', lis 3, 2, 6, 4, 5, 1; print join ' ', lis 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15; </lang>
- Output:
2 4 5 0 2 6 9 11 15
Perl 6
Dynamic programming
Straight-forward implementation of the algorithm described in the video.
<lang Perl 6>sub lis(@d) {
my @l = [].item xx @d; @l[0].push: @d[0]; for 1 ..^ @d -> $i { for ^$i -> $j { if @d[$j] < @d[$i] && @l[$i].elems < @l[$j].elems + 1 { @l[$i] = [ @l[$j][] ] } } @l[$i].push: @d[$i]; } return max :by(*.elems), @l;
}
say lis([3,2,6,4,5,1]); say lis([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]);</lang>
- Output:
2 4 5 0 2 6 9 11 15
Patience sorting
<lang Perl 6>sub patience(@d is copy) {
my @S = [@d.shift() => Mu].item; for @d -> $card { if defined my $i = first { @S[$_][*-1].key > $card }, ^@S { @S[$i].push: $card => $i ?? @S[$i-1][*-1] !! Mu; } else { @S.push: [ $card => @S ?? @S[*-1][*-1] !! Mu ].item } } return @S;
}
sub lis(@S) {
reverse map *.key, (
@S[*-1][*-1], *.value ...^ !*.defined
);
}
say lis patience(<3 2 6 4 5 1>);
say lis patience(<0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15>».Int);
- say patience(<7 2 8 1 3 4 10 6 9 5>».Int);</lang>
- Output:
2 4 5 0 2 6 9 11 15
Python
Python: Method from video
<lang python>def longest_increasing_subsequence(d):
'Return one of the L.I.S. of list d' l = [] for i in range(len(d)): l.append(max([l[j] for j in range(i) if l[j][-1] < d[i]] or [[]], key=len) + [d[i]]) return max(l, key=len)
if __name__ == '__main__':
for d in [[3,2,6,4,5,1], [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]]: print('a L.I.S. of %s is %s' % (d, longest_increasing_subsequence(d)))</lang>
- Output:
a L.I.S. of [3, 2, 6, 4, 5, 1] is [3, 4, 5] a L.I.S. of [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] is [0, 4, 6, 9, 13, 15]
Python: Patience sorting method
<lang python>from collections import namedtuple
class Node(namedtuple('Node_', 'val back')):
def __iter__(self): while self is not None: yield self.val self = self.back
def lis(d):
"""Return one of the L.I.S. of list d using patience sorting.""" if not d: return [] pile = [[Node(d[0], None)]] for di in d[1:]: for j, pj in enumerate(pile): if pj[-1].val > di: pj.append(Node(di, None if not j else pile[j - 1][-1])) break else: pile.append([Node(di, pile[-1][-1])])
return [val for val in pile[-1][-1]] [::-1]
if __name__ == '__main__':
for d in [[3,2,6,4,5,1], [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]]: print('a L.I.S. of %s is %s' % (d, lis(d)))</lang>
- Output:
a L.I.S. of [3, 2, 6, 4, 5, 1] is [2, 4, 5] a L.I.S. of [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] is [0, 2, 6, 9, 11, 15]
Swym
Based on the Python video solution. Interpreter at [[1]] <lang swym>Array.'lis' {
'stems' = Number.Array.mutableArray[ [] ]
forEach(this) 'value'-> { 'bestStem' = stems.where{==[] || .last < value}.max{.length}
stems.push( bestStem + [value] ) }
return stems.max{.length}
}
[3,2,6,4,5,1].lis.trace [0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15].lis.trace</lang>
- Output:
[3,4,5] [0,4,6,9,13,15]
Tcl
<lang tcl>package require Tcl 8.6
proc longestIncreasingSubsequence {sequence} {
# Get the increasing subsequences (and their lengths) set subseq [list 1 [lindex $sequence 0]] foreach value $sequence {
set max {} foreach {len item} $subseq { if {[lindex $item end] < $value} { if {[llength [lappend item $value]] > [llength $max]} { set max $item } } elseif {![llength $max]} { set max [list $value] } } lappend subseq [llength $max] $max
} # Pick the longest subsequence; -stride requires Tcl 8.6 return [lindex [lsort -stride 2 -index 0 $subseq] end]
}</lang> Demonstrating: <lang tcl>puts [longestIncreasingSubsequence {3 2 6 4 5 1}] puts [longestIncreasingSubsequence {0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15}]</lang>
- Output:
3 4 5 0 4 6 9 13 15