Longest increasing subsequence: Difference between revisions
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{{trans|Python}}
<
V n = x.len
V P = [0] * n
Line 47:
L(d) [[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 #. is #.’.format(d, longest_increasing_subsequence(d)))</
{{out}}
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=={{header|360 Assembly}}==
{{trans|VBScript}}
<
LNGINSQ CSECT
USING LNGINSQ,R13 base register
Line 177:
XDEC DS CL12 temp for xdeco
YREGS
END LNGINSQ</
{{out}}
<pre>
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{{Trans|Phix}} … modified to return ''multiple'' co-longest sequences where found. It's not clear how equal values should be treated. Here the behaviour happens to be as in the demo code at the end.
<
script o
property inputList : aList
Line 251:
set end of output to {finds:longestIncreasingSubsequences(input's contents)}
end repeat
return output</
{{output}}
<
=={{header|Arturo}}==
<
l: new [[]]
loop d 'num [
Line 284:
] 'seq [
print ["LIS of" seq "=>" lis seq]
]</
{{out}}
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=={{header|AutoHotkey}}==
<
for k, v in Lists {
Line 313:
}
return, D
}</
'''Output:'''
<pre>3, 4, 5
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=={{header|C}}==
Using an array that doubles as linked list (more like reversed trees really). O(n) memory and O(n<sup>2</sup>) runtime.
<
#include <stdlib.h>
Line 363:
lis(y, sizeof(y) / sizeof(int));
return 0;
}</
{{out}}
<pre>
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===Recursive===
{{works with|C sharp|6}}
<
using System.Collections;
using System.Collections.Generic;
Line 420:
IEnumerator IEnumerable.GetEnumerator() => GetEnumerator();
}
}</
===Patience sorting===
{{works with|C sharp|7}}
<
{
public static T[] Find<T>(IList<T> values, IComparer<T> comparer = null) {
Line 449:
Console.WriteLine(string.Join(",", LIS.Find(new [] { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 })));
}
}</
{{out}}
<pre>
Line 458:
Patience sorting
=== C++11 ===
<
#include <list>
#include <algorithm>
Line 523:
show_lis(std::list<int> { 3, 2, 6, 4, 5, 1 });
show_lis(std::vector<int> { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 });
}</
=== C++98 ===
<
#include <list>
#include <algorithm>
Line 607:
show_lis(vec1);
show_lis(vec2);
}</
{{out}}
<pre>2 4 5
Line 617:
The combination is done using ''cons'', so what gets put on a pile is a list -- a descending subsequence.
<
(let [[les gts] (->> piles (split-with #(<= (ffirst %) card)))
newelem (cons card (->> les last first))
Line 628:
(println (a-longest [3 2 6 4 5 1]))
(println (a-longest [0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15]))</
{{out}}
<syntaxhighlight lang="text">(2 4 5)
(0 2 6 9 11 15)</
=={{header|Common Lisp}}==
===Common Lisp: Using the method in the video===
Slower and more memory usage compared to the patience sort method.
<
(let ((subseqs nil))
(dolist (item list)
Line 655:
(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))))</
{{out}}
<pre>(2 4 5)
Line 661:
===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.
<
(let ((piles nil))
(dolist (item input-list)
Line 683:
(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)))</
{{out}}
<pre>(2 4 5)
Line 689:
===Common Lisp: Using the Patience Sort approach (alternative)===
This is a different version of the code above.
<
(multiple-value-bind
(i prev)
Line 708:
(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-inc-seq l)))</
{{out}}
<pre>(2 4 5)
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{{trans|Haskell}}
Uses the second powerSet function from the Power Set Task.
<
T[] lis(T)(T[] items) pure nothrow {
Line 731:
[3, 2, 6, 4, 5, 1].lis.writeln;
[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15].lis.writeln;
}</
{{out}}
<pre>[2, 4, 5]
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{{trans|Python}}
From the second Python entry, using the Patience sorting method.
<
/// Return one of the Longest Increasing Subsequence of
Line 776:
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]])
d.lis.writeln;
}</
The output is the same.
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{{trans|Java}}
With some more optimizations.
<
T[] lis(T)(in T[] items) pure nothrow
Line 831:
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]])
d.writeln;
}</
The output is the same.
=={{header|Déjà Vu}}==
{{trans|Python}}
<
if = :nil dup:
false drop
Line 861:
!. lis [ 3 2 6 4 5 1 ]
!. lis [ 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 ]
</syntaxhighlight>
{{out}}
<pre>[ 2 4 5 ]
[ 0 2 6 9 11 15 ]</pre>
=={{header|EasyLang}}==
{{trans|Ring}}
<syntaxhighlight>
func[] lis x[] .
n = len x[]
len p[] n
len m[] n
for i to n
lo = 1
hi = lng
while lo <= hi
mid = (lo + hi) div 2
if x[m[mid]] < x[i]
lo = mid + 1
else
hi = mid - 1
.
.
if lo > 1
p[i] = m[lo - 1]
.
m[lo] = i
if lo > lng
lng = lo
.
.
len res[] lng
if lng > 0
k = m[lng]
for i = lng downto 1
res[i] = x[k]
k = p[k]
.
.
return res[]
.
tests[][] = [ [ 3 2 6 4 5 1 ] [ 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 ] ]
for x to len tests[][]
print lis tests[x][]
.
</syntaxhighlight>
{{out}}
<pre>
[ 2 4 5 ]
[ 0 2 6 9 11 15 ]
</pre>
=={{header|Elixir}}==
Line 871 ⟶ 918:
===Naive version===
very slow
<
# Naive implementation
def lis(l) do
Line 889 ⟶ 936:
IO.inspect Longest_increasing_subsequence.lis([3,2,6,4,5,1])
IO.inspect Longest_increasing_subsequence.lis([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15])</
{{out}}
Line 898 ⟶ 945:
===Patience sort version===
<
# Patience sort implementation
def patience_lis(l), do: patience_lis(l, [])
Line 925 ⟶ 972:
IO.inspect Longest_increasing_subsequence.patience_lis([3,2,6,4,5,1])
IO.inspect Longest_increasing_subsequence.patience_lis([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15])</
{{out}}
Line 949 ⟶ 996:
Function ''memo'' and ''patience2'' by [https://www.linkedin.com/in/find-roman/ Roman Rabinovich].
<
-module(longest_increasing_subsequence).
Line 1,159 ⟶ 1,206:
% **************************************************
</syntaxhighlight>
Output naive:
Line 1,184 ⟶ 1,231:
[0,2,6,9,11,15]
</pre>
=={{header|FreeBASIC}}==
<syntaxhighlight lang="vb">Sub Lis(arr() As Integer)
Dim As Integer lb = Lbound(arr), ub = Ubound(arr)
Dim As Integer i, lo, hi, mitad, newl, l = 0
Dim As Integer p(ub), m(ub)
For i = lb To ub
lo = 1
hi = l
Do While lo <= hi
mitad = Int((lo+hi)/2)
If arr(m(mitad)) < arr(i) Then
lo = mitad + 1
Else
hi = mitad - 1
End If
Loop
newl = lo
p(i) = m(newl-1)
m(newl) = i
If newL > l Then l = newl
Next i
Dim As Integer res(l)
Dim As Integer k = m(l)
For i = l-1 To 0 Step - 1
res(i) = arr(k)
k = p(k)
Next i
For i = Lbound(res) To Ubound(res)-1
Print res(i); " ";
Next i
End Sub
Dim As Integer arrA(5) => {3,2,6,4,5,1}
Lis(arrA())
Print
Dim As Integer arrB(15) => {0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15}
Lis(arrB())
Sleep</syntaxhighlight>
{{out}}
<pre>2 4 5
0 2 6 9 11 15</pre>
=={{header|Go}}==
Patience sorting
<
import (
Line 1,229 ⟶ 1,322:
fmt.Printf("an L.I.S. of %v is %v\n", d, lis(d))
}
}</
{{out}}
Line 1,237 ⟶ 1,330:
=={{header|Haskell}}==
===Naive implementation===
<
import Data.List ( maximumBy, subsequences )
import Data.List.Ordered ( isSorted, nub )
Line 1,248 ⟶ 1,341:
print $ lis [3,2,6,4,5,1]
print $ lis [0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]
print $ lis [1,1,1,1]</
{{out}}
Line 1,256 ⟶ 1,349:
===Patience sorting===
<
module Main (main, lis) where
Line 1,307 ⟶ 1,400:
print $ lis [3, 2, 6, 4, 5, 1]
print $ lis [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]
print $ lis [1, 1, 1, 1]</
{{out}}
Line 1,318 ⟶ 1,411:
The following works in both languages:
<
every writes((!lis(A)||" ") | "\n")
end
Line 1,328 ⟶ 1,421:
else p[-1] := (p[-2] < v)
return r
end</
Sample runs:
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These examples are simple enough for brute force to be reasonable:
<
longestinc=: ] #~ [: (#~ ([: (= >./) +/"1)) #:@I.@increasing</
In other words: consider all 2^n bitmasks of length n, and select those which strictly select increasing sequences. Find the length of the longest of these and use the masks of that length to select from the original sequence.
Line 1,350 ⟶ 1,443:
Example use:
<syntaxhighlight lang="j">
longestinc 3,2,6,4,5,1
2 4 5
Line 1,358 ⟶ 1,451:
0 2 6 9 13 15
0 4 6 9 11 15
0 4 6 9 13 15</
=={{header|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.
<
public class LIS {
Line 1,401 ⟶ 1,494:
System.out.printf("an L.I.S. of %s is %s\n", d, lis(d));
}
}</
{{out}}
Line 1,408 ⟶ 1,501:
=={{header|JavaScript}}==
<
if (input.length === 0) {
return [];
Line 1,441 ⟶ 1,534:
console.log(getLongestIncreasingSubsequence([0, 7, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]));
console.log(getLongestIncreasingSubsequence([3, 2, 6, 4, 5, 1]));
</syntaxhighlight>
{{out}}
Line 1,450 ⟶ 1,543:
===Patience sorting===
<
if (input.length === 0) {
return 0;
Line 1,489 ⟶ 1,582:
console.log(getLongestIncreasingSubsequence([0, 7, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]));
console.log(getLongestIncreasingSubsequence([3, 2, 6, 4, 5, 1]));
</syntaxhighlight>
{{out}}
Line 1,504 ⟶ 1,597:
Recent versions of jq have functions that obviate the need for the two generic functions defined in this subsection.
<
def _until:
if cond then . else (update | _until) end;
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else .[0] = $mid + 1
end )
| .[0];</
'''lis:'''
<
# Helper function:
Line 1,539 ⟶ 1,632:
)
| .[length - 1]
| reverse( recurse(.back) | .val ) ; </
'''Examples:'''
<
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]
) | lis</
{{out}}
<
[2,4,5]
[0,2,6,9,11,15]
</syntaxhighlight>
=={{header|Julia}}==
{{works with|Julia|0.6}}
<
function lis(arr::Vector)
if length(arr) == 0 return copy(arr) end
Line 1,574 ⟶ 1,667:
@show lis([3, 2, 6, 4, 5, 1])
@show lis([0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15])</
{{out}}
Line 1,582 ⟶ 1,675:
=={{header|Kotlin}}==
Uses the algorithm in the Wikipedia L.I.S. article:
<
fun longestIncreasingSubsequence(x: IntArray): IntArray =
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)
lists.forEach { println(longestIncreasingSubsequence(it).asList()) }
}</
{{out}}
Line 1,631 ⟶ 1,724:
=={{header|Lua}}==
<
local piles = { { {table.remove(seq, 1), nil} } }
while #seq>0 do
Line 1,657 ⟶ 1,750:
buildLIS({3,2,6,4,5,1})
buildLIS({0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15})
</syntaxhighlight>
{{out}}
Line 1,668 ⟶ 1,761:
stack:=stackitem(L(i)), ! stack(L(j)) returns a refence to a new stack object, with the first item on L(i) (which is a reference to stack object) and merge using ! the copy of L(j) stack.
<syntaxhighlight lang="m2000 interpreter">
Module LIS_example {
Function LIS {
Line 1,706 ⟶ 1,799:
}
LIS_example
</syntaxhighlight>
===Using arrays in an array===
<syntaxhighlight lang="m2000 interpreter">
Module LIS_example {
Function LIS {
Line 1,749 ⟶ 1,842:
}
LIS_example
</syntaxhighlight>
{{out}}
Line 1,760 ⟶ 1,853:
=={{header|Maple}}==
<
LIS := proc(L)
local i, j;
Line 1,785 ⟶ 1,878:
return output[index];
end proc:</
Alternatively, output the longest subsequence using built-in command max:
<
output[i];</
<
M := [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15];
LIS(L);
LIS(M);</
{{out}}
<pre>
Line 1,803 ⟶ 1,896:
=={{header|Mathematica}}/{{header|Wolfram Language}}==
Although undocumented, Mathematica has the function LongestAscendingSequence which exactly does what the Task asks for:
<
{{out}}
<pre>{{2,4,5},{0,2,6,9,11,15}}</pre>
Line 1,809 ⟶ 1,902:
=={{header|Nim}}==
{{trans|Python}}
<
var l: seq[seq[T]]
for i in 0 .. d.high:
Line 1,822 ⟶ 1,915:
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]]:
echo "A L.I.S. of ", d, " is ", longestIncreasingSubsequence(d)</
{{out}}
<pre>A L.I.S. of @[3, 2, 6, 4, 5, 1] is @[3, 4, 5]
Line 1,829 ⟶ 1,922:
=={{header|Objective-C}}==
Patience sorting
<
@interface Node : NSObject {
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}
return 0;
}</
{{out}}
<pre>an L.I.S. of (
Line 1,923 ⟶ 2,016:
=={{header|OCaml}}==
===Naïve implementation===
<
then x
else acc) [] l
Line 1,945 ⟶ 2,038:
in
List.map (fun x -> print_endline (String.concat " " (List.map string_of_int
(lis x)))) sequences</
{{out}}
<pre>
Line 1,953 ⟶ 2,046:
===Patience sorting===
<
let pile_tops = Array.make (List.length list) [] in
let bsearch_piles x len =
Line 1,974 ⟶ 2,067:
in
let len = List.fold_left f 0 list in
List.rev pile_tops.(len-1)</
Usage:
<pre># lis compare [3; 2; 6; 4; 5; 1];;
Line 1,980 ⟶ 2,073:
# lis compare [0; 8; 4; 12; 2; 10; 6; 14; 1; 9; 5; 13; 3; 11; 7; 15];;
- : int list = [0; 2; 6; 9; 11; 15]</pre>
=={{header|Pascal}}==
{{works with|FPC}}
O(NLogN) version.
<syntaxhighlight lang="pascal">
program LisDemo;
{$mode objfpc}{$h+}
uses
SysUtils;
function Lis(const A: array of Integer): specialize TArray<Integer>;
var
TailIndex: array of Integer;
function CeilIndex(Value, R: Integer): Integer;
var
L, M: Integer;
begin
L := 0;
while L < R do begin
{$PUSH}{$Q-}{$R-}M := (L + R) shr 1;{$POP}
if A[TailIndex[M]] < Value then L := M + 1
else R := M;
end;
Result := R;
end;
var
I, J, Len: Integer;
Parents: array of Integer;
begin
Result := nil;
if Length(A) = 0 then exit;
SetLength(TailIndex, Length(A));
SetLength(Parents, Length(A));
Len := 1;
for I := 1 to High(A) do
if A[I] < A[TailIndex[0]] then
TailIndex[0] := I
else
if A[TailIndex[Len-1]] < A[I] then begin
Parents[I] := TailIndex[Len - 1];
TailIndex[Len] := I;
Inc(Len);
end else begin
J := CeilIndex(A[I], Len - 1);
Parents[I] := TailIndex[J - 1];
TailIndex[J] := I;
end;
if Len < 2 then exit([A[0]]);
SetLength(Result, Len);
J := TailIndex[Len - 1];
for I := Len - 1 downto 0 do begin
Result[I] := A[J];
J := Parents[J];
end;
end;
procedure PrintArray(const A: array of Integer);
var
I: SizeInt;
begin
Write('[');
for I := 0 to High(A) - 1 do
Write(A[I], ', ');
WriteLn(A[High(A)], ']');
end;
begin
PrintArray(Lis([3, 2, 6, 4, 5, 1]));
PrintArray(Lis([0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]));
PrintArray(Lis([1, 1, 1, 1, 1, 0]));
end.
</syntaxhighlight>
{{out}}
<pre>
[2, 4, 5]
[0, 2, 6, 9, 11, 15]
[1]
</pre>
=={{header|Perl}}==
===Dynamic programming===
{{trans|Raku}}
<
sub lis {
Line 2,005 ⟶ 2,176:
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;</
{{out}}
<pre>2 4 5
Line 2,011 ⟶ 2,182:
===Patience sorting===
<
my @pileTops;
# sort into piles
Line 2,044 ⟶ 2,215:
print "an L.I.S. of [@d] is [@lis]\n";
}</
{{out}}
<pre>an L.I.S. of [3 2 6 4 5 1] is [2 4 5]
Line 2,051 ⟶ 2,222:
=={{header|Phix}}==
Using the Wikipedia algorithm (converted to 1-based indexing)
<!--<
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">lis</span><span style="color: #0000FF;">(</span><span style="color: #004080;">sequence</span> <span style="color: #000000;">x</span><span style="color: #0000FF;">,</span> <span style="color: #004080;">integer</span> <span style="color: #000000;">n</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #000000;">x</span><span style="color: #0000FF;">))</span>
Line 2,090 ⟶ 2,261:
<span style="color: #0000FF;">?</span><span style="color: #000000;">lis</span><span style="color: #0000FF;">(</span><span style="color: #000000;">tests</span><span style="color: #0000FF;">[</span><span style="color: #000000;">i</span><span style="color: #0000FF;">])</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<!--</
{{out}}
<pre>
Line 2,099 ⟶ 2,270:
=={{header|PHP}}==
Patience sorting
<
class Node {
public $val;
Line 2,135 ⟶ 2,306:
print_r(lis(array(3, 2, 6, 4, 5, 1)));
print_r(lis(array(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15)));
?></
{{out}}
<pre>Array
Line 2,154 ⟶ 2,325:
=={{header|Picat}}==
===Mode-directed tabling===
{{trans|Prolog}}
<syntaxhighlight lang="picat">table(+,+,max)
lis_mode(In, Out,OutLen) =>
one_is(In, [], Is),
Line 2,197 ⟶ 2,337:
( Current = [], one_is(T, [H], Final));
( Current = [H1|_], H1 @< H, one_is(T, [H|Current], Final));
one_is(T, Current, Final).</syntaxhighlight>
===Constraint modelling approach===
For larger instances, the sat solver is generally faster than the cp solver.
<syntaxhighlight lang="picat">lis_cp(S, Res,Z) =>
Len = S.len,
X = new_list(Len),
Line 2,221 ⟶ 2,360:
foreach(I in 1..N, J in I+1..N)
(A[I] #!= 0 #/\ A[J] #!= 0) #=> (A[I] #< A[J])
end.</
===Test===
<syntaxhighlight lang="picat">import sat. % for lis_cp
% import cp. % Slower than sat on larger instances.
go =>
nolog,
Tests = [
[3,2,6,4,5,1],
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15],
[1,1,1,1],
[4,65,2,-31,0,99,83,782,1]
],
Funs = [lis_mode, lis_cp],
foreach(Fun in Funs)
println(fun=Fun),
foreach(Test in Tests)
call(Fun,Test,Lis,Len),
printf("%w: LIS=%w (len=%d)\n",Test, Lis,Len)
end,
nl,
end,
nl.</syntaxhighlight>
{{out}}
<pre>[3,2,6,4,5,1]: LIS=[3,4,5] (len=3)
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]: LIS=[0,4,6,9,13,15] (len=6)
Line 2,229 ⟶ 2,392:
[4,65,2,-31,0,99,83,782,1]: LIS=[4,65,99,782] (len=4)</pre>
The mode directed tabling tends to be the fastest of the two methods.
=={{header|PicoLisp}}==
Adapted patience sorting approach:
<
(let (D NIL R NIL)
(for I Lst
Line 2,244 ⟶ 2,408:
(T (when R (queue 'D (car R)))
(push 'R I) ) ) )
(flip R) ) )</
Original recursive glutton:
<
(let N (pop 'L)
(maxi length
Line 2,267 ⟶ 2,431:
(test (-31 0 83 782)
(glutton (4 65 2 -31 0 99 83 782 1)) )</
=={{header|PowerShell}}==
{{works with|PowerShell|2}}
<
{
If ( $A.Count -lt 2 ) { return $A }
Line 2,318 ⟶ 2,482:
# Return the series (reversed into the correct order)
return $S[$Last..0]
}</
<
( Get-LongestSubsequence 0, 8, 4, 12, 2, 10, 6, 16, 14, 1, 9, 5, 13, 3, 11, 7, 15 ) -join ', '</
{{out}}
<pre>2, 4, 5
Line 2,330 ⟶ 2,494:
<
% we ask Prolog to find the longest sequence
aggregate(max(N,Is), (one_is(In, [], Is), length(Is, N)), max(_, Res)),
Line 2,344 ⟶ 2,508:
( Current = [H1 | _], H1 < H, one_is(T, [H | Current], Final));
one_is(T, Current, Final).
</syntaxhighlight>
Prolog finds the first longest subsequence
<pre> ?- lis([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15], Out).
Line 2,356 ⟶ 2,520:
===Python: O(nlogn) Method from Wikipedia's LIS Article[https://en.wikipedia.org/wiki/Longest_increasing_subsequence#Efficient_algorithms]===
<
"""Returns the Longest Increasing Subsequence in the Given List/Array"""
N = len(X)
Line 2,388 ⟶ 2,552:
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)))</
{{out}}
Line 2,395 ⟶ 2,559:
===Python: Method from video===
<
'Return one of the L.I.S. of list d'
l = []
Line 2,405 ⟶ 2,569:
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)))</
{{out}}
Line 2,412 ⟶ 2,576:
===Python: Patience sorting method===
<
from functools import total_ordering
from bisect import bisect_left
Line 2,440 ⟶ 2,604:
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)))</
{{out}}
Line 2,448 ⟶ 2,612:
=={{header|Racket}}==
Patience sorting. The program saves only the top card of each pile, with a link (cons) to the top of the previous pile at the time it was inserted. It uses binary search to find the correct pile.
<
(require data/gvector)
Line 2,474 ⟶ 2,638:
(if (<= item (car (gvector-ref piles middle)))
(loop first middle)
(loop (add1 middle) last)))))])))</
{{out}}
<pre>'(2 4 5)
Line 2,485 ⟶ 2,649:
Straight-forward implementation of the algorithm described in the video.
<syntaxhighlight lang="raku"
my @l = [].item xx @d;
@l[0].push: @d[0];
Line 2,500 ⟶ 2,664:
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]);</
{{out}}
<pre>[2 4 5]
Line 2,506 ⟶ 2,670:
===Patience sorting===
<syntaxhighlight lang="raku"
my @S = [@deck.shift() => Nil].item;
for @deck -> $card {
Line 2,521 ⟶ 2,685:
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>;</
{{out}}
<pre>[2 4 5]
Line 2,528 ⟶ 2,692:
=={{header|REXX}}==
{{trans|VBScript}}
<
$.=; $.1= 3 2 6 4 5 1 /*define the 1st list to be examined. */
$.2= 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 /* " " 2nd " " " " */
Line 2,560 ⟶ 2,724:
do L; $= @.k $; k= p.k /*perform this DO loop L times. */
end /*i*/
return strip($) /*the result has an extra leading blank*/</
{{out|output|text= when using the internal default input:}}
<pre>
Line 2,571 ⟶ 2,735:
=={{header|Ring}}==
<
# Project : Longest increasing subsequence
Line 2,630 ⟶ 2,794:
see svect
see "}" + nl
</syntaxhighlight>
Output:
<pre>
Line 2,639 ⟶ 2,803:
=={{header|Ruby}}==
Patience sorting
<
def lis(n)
Line 2,671 ⟶ 2,835:
p lis([3, 2, 6, 4, 5, 1])
p lis([0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15])</
{{out}}
<pre>[2, 4, 5]
Line 2,678 ⟶ 2,842:
=={{header|Rust}}==
<syntaxhighlight lang="rust">
fn lis(x: &[i32])-> Vec<i32> {
let n = x.len();
Line 2,723 ⟶ 2,887:
let list = vec![0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15];
println!("{:?}", lis(&list));
}</
{{out}}
Line 2,732 ⟶ 2,896:
===Patience sorting===
{{Out}}See it in running in your browser by [https://scalafiddle.io/sf/Wx8DsUO/1 ScalaFiddle (JavaScript)] or by [https://scastie.scala-lang.org/FtLHeaAwSrO6VXVOTTZ7FQ Scastie (JVM)].
<
val tests = Map(
"3,2,6,4,5,1" -> Seq("2,4,5", "3,4,5"),
Line 2,764 ⟶ 2,928:
allLongests.forall(lis => expect.contains(lis.mkString(",")))
})
}</
{{Out}}
<pre>3,2,6,4,5,1 has 2 longest increasing subsequences, e.g. 2,4,5
Line 2,770 ⟶ 2,934:
===Brute force solution===
<
def isSorted(l:List[Int])(f: (Int, Int) => Boolean) = l.view.zip(l.tail).forall(x => f(x._1,x._2))
def sequence(set: List[Int])(f: (Int, Int) => Boolean) = powerset(set).filter(_.nonEmpty).filter(x => isSorted(x)(f)).toList.maxBy(_.length)
sequence(set)(_<_)
sequence(set)(_>_)</
=={{header|Scheme}}==
Patience sorting
<
(define pile-tops (make-vector (length lst)))
(define (bsearch-piles x len)
Line 2,802 ⟶ 2,966:
(display (lis < '(3 2 6 4 5 1))) (newline)
(display (lis < '(0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15))) (newline)</
{{out}}
Line 2,810 ⟶ 2,974:
=={{header|Sidef}}==
Dynamic programming:
<
var l = a.len.of { [] }
l[0] << a[0]
Line 2,825 ⟶ 2,989:
say lis(%i<3 2 6 4 5 1>)
say lis(%i<0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15>)</
Patience sorting:
<
var pileTops = []
deck.each { |x|
Line 2,854 ⟶ 3,018:
say lis(%i<3 2 6 4 5 1>)
say lis(%i<0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15>)</
{{out}}
Line 2,865 ⟶ 3,029:
Patience sorting
{{works with|SML/NJ}}
<
let
val pile_tops = DynamicArray.array (length n, [])
Line 2,895 ⟶ 3,059:
app f n;
rev (DynamicArray.sub (pile_tops, DynamicArray.bound pile_tops))
end</
Usage:
<pre>- lis Int.compare [3, 2, 6, 4, 5, 1];
Line 2,904 ⟶ 3,068:
=={{header|Swift}}==
<
extension Array where Element: Comparable {
Line 2,951 ⟶ 3,115:
print("\(l1) = \(l1.longestIncreasingSubsequence())")
print("\(l2) = \(l2.longestIncreasingSubsequence())")</
{{out}}
Line 2,961 ⟶ 3,125:
{{trans|Python}}
Based on the Python video solution. Interpreter at [[http://cheersgames.com/swym/SwymInterpreter.html?Array.%27lis%27%0A%7B%0A%20%20%27stems%27%20%3D%20Number.Array.mutableArray%5B%20%5B%5D%20%5D%0A%20%0A%20%20forEach%28this%29%20%27value%27-%3E%0A%20%20%7B%0A%20%20%20%20%27bestStem%27%20%3D%20stems.where%7B%3D%3D%5B%5D%20%7C%7C%20.last%20%3C%20value%7D.max%7B.length%7D%0A%20%0A%20%20%20%20stems.push%28%20bestStem%20+%20%5Bvalue%5D%20%29%0A%20%20%7D%0A%20%0A%20%20return%20stems.max%7B.length%7D%0A%7D%0A%20%0A%5B3%2C2%2C6%2C4%2C5%2C1%5D.lis.trace%0A%5B0%2C8%2C4%2C12%2C2%2C10%2C6%2C14%2C1%2C9%2C5%2C13%2C3%2C11%2C7%2C15%5D.lis.trace]]
<
{
'stems' = Number.Array.mutableArray[ [] ]
Line 2,976 ⟶ 3,140:
[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</
{{out}}
<pre>
Line 2,985 ⟶ 3,149:
=={{header|Tcl}}==
{{works with|Tcl|8.6}}
<
proc longestIncreasingSubsequence {sequence} {
Line 3,005 ⟶ 3,169:
# Pick the longest subsequence; -stride requires Tcl 8.6
return [lindex [lsort -stride 2 -index 0 $subseq] end]
}</
Demonstrating:
<
puts [longestIncreasingSubsequence {0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15}]</
{{out}}
<pre>
Line 3,016 ⟶ 3,180:
=={{header|VBScript}}==
<syntaxhighlight lang="vb">
Function LIS(arr)
n = UBound(arr)
Line 3,054 ⟶ 3,218:
WScript.StdOut.WriteLine LIS(Array(3,2,6,4,5,1))
WScript.StdOut.WriteLine LIS(Array(0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15))
</syntaxhighlight>
{{Out}}
Line 3,064 ⟶ 3,228:
=={{header|Wren}}==
{{trans|Kotlin}}
<
var n = x.count
if (n == 0) return []
Line 3,100 ⟶ 3,264:
[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]
]
lists.each { |l| System.print(longestIncreasingSubsequence.call(l)) }</
{{out}}
Line 3,109 ⟶ 3,273:
=={{header|zkl}}==
<
piles:=L();
backPtr:='wrap(np){ return(np-1,if(np) piles[np-1].len()-1 else -1) }; // maybe (-1,-1)
Line 3,125 ⟶ 3,289:
do{ n,p=piles[p][n]; r.write(n); p,n=p; }while(p!=-1);
r.reverse()
}</
<
T(0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15),"foobar")){
s:=longestSequence(ns);
println(s.len(),": ",s," from ",ns);
}</
{{out}}
<pre>
| |||