Longest increasing subsequence
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.
360 Assembly
<lang 360asm>* Longest increasing subsequence 04/03/2017 LNGINSQ CSECT
USING LNGINSQ,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) save previous context
ST R13,4(R15) link backward
ST R15,8(R13) link forward
LR R13,R15 set addressability
LA R6,1 i=1
DO WHILE=(CH,R6,LE,=H'2') do i=1 to 2
IF CH,R6,EQ,=H'1' THEN if i=1 then
MVC N,=AL2((A2-A1)/2) n=hbound(a1)
MVC AA(64),A1 a=a1
ELSE , else
MVC N,=AL2((AA-A2)/2) n=hbound(a2)
MVC AA(64),A2 a=a2
ENDIF , endif
MVC PG,=CL80': ' init buffer
LA R2,AA-2 @a
LH R3,N n
BAL R14,PRINT print a
MVC LL,=H'0' l=0
SR R7,R7 j=0
DO WHILE=(CH,R7,LE,N) do j=0 to n
MVC LO,=H'1' lo=1
MVC HI,LL hi=l
LH R4,LO lo
DO WHILE=(CH,R4,LE,HI) do while lo<=hi
LH R1,LO lo
AH R1,HI lo+hi
SRA R1,1 /2
STH R1,MIDDLE middle=(lo+hi)/2
SLA R1,1 *2
LH R1,MM(R1) m(middle+1)
SLA R1,1 *2
LH R3,AA(R1) r3=a(m(middle+1)+1)
LR R1,R7 j
SLA R1,1 *2
LH R4,AA(R1) r4=a(j+1)
LH R2,MIDDLE middle
IF CR,R3,LT,R4 THEN if a(m(middle+1)+1)<a(j+1) then
LA R2,1(R2) middle+1
STH R2,LO lo=middle+1
ELSE , else
BCTR R2,0 middle-1
STH R2,HI hi=middle-1
ENDIF , endif
LH R4,LO lo
ENDDO , end /*while*/
LH R10,LO newl=lo
LR R1,R10 newl
SLA R1,1 *2
LH R3,MM-2(R1) m(newl)
LR R1,R7 j
SLA R1,1 *2
STH R3,PP(R1) p(j+1)=m(newl)
LR R1,R10 newl
SLA R1,1 *2
STH R7,MM(R1) m(newl+1)=j
IF CH,R10,GT,LL if newl>l then
STH R10,LL l=newl
ENDIF , endif
LA R7,1(R7) j++
ENDDO , enddo j
LH R1,LL l
SLA R1,1 *2
LH R10,MM(R1) k=m(l+1)
LH R7,LL j=l
DO WHILE=(CH,R7,GE,=H'1') do j=l to 1 by -1
LR R1,R10 k
SLA R1,1 *2
LH R2,AA(R1) a(k+1)
LR R1,R7 j
SLA R1,1 *2
STH R2,SS-2(R1) s(j)=a(k+1)
LR R1,R10 k
SLA R1,1 *2
LH R10,PP(R1) k=p(k+1)
BCTR R7,0 j--
ENDDO , enddo j
MVC PG,=CL80'> ' init buffer
LA R2,SS-2 @s
LH R3,LL l
BAL R14,PRINT print a
LA R6,1(R6) i++
ENDDO , enddo i
L R13,4(0,R13) restore previous savearea pointer
LM R14,R12,12(R13) restore previous context
XR R15,R15 rc=0
BR R14 exit
PRINT LA R10,PG ---- print subroutine
LA R10,2(R10) pgi=2
LA R7,1 j=1
DO WHILE=(CR,R7,LE,R3) do j=1 to nx
LR R1,R7 j
SLA R1,1 *2
LH R1,0(R2,R1) x(j)
XDECO R1,XDEC edit x(j)
MVC 0(3,R10),XDEC+9 output x(j)
LA R10,3(R10) pgi+=3
LA R7,1(R7) j++
ENDDO , enddo j
XPRNT PG,L'PG print buffer
BR R14 ---- return
A1 DC H'3',H'2',H'6',H'4',H'5',H'1' A2 DC H'0',H'8',H'4',H'12',H'2',H'10',H'6',H'14'
DC H'1',H'9',H'5',H'13',H'3',H'11',H'7',H'15'
AA DS 32H a(32) PP DS 32H p(32) MM DS 32H m(32) SS DS 32H s(32) N DS H n LL DS H l LO DS H lo HI DS H hi MIDDLE DS H middle PG DS CL80 buffer XDEC DS CL12 temp for xdeco
YREGS
END LNGINSQ</lang>
- Output:
: 3 2 6 4 5 1 > 2 4 5 : 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 > 0 2 6 9 11 15
AutoHotkey
<lang AutoHotkey>Lists := [[3,2,6,4,5,1], [0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]]
for k, v in Lists { D := LIS(v) MsgBox, % D[D.I].seq }
LIS(L) { D := [] for i, v in L { D[i, "Length"] := 1, D[i, "Seq"] := v, D[i, "Val"] := v Loop, % i - 1 { if(D[A_Index].Val < v && D[A_Index].Length + 1 > D[i].Length) { D[i].Length := D[A_Index].Length + 1 D[i].Seq := D[A_Index].Seq ", " v if (D[i].Length > MaxLength) MaxLength := D[i].Length, D.I := i } } } return, D }</lang> Output:
3, 4, 5 0, 4, 6, 9, 13, 15
C
Using an array that doubles as linked list (more like reversed trees really). O(n) memory and O(n2) runtime. <lang c>#include <stdio.h>
- include <stdlib.h>
struct node { int val, len; struct node *next; };
void lis(int *v, int len) { int i; struct node *p, *n = calloc(len, sizeof *n); for (i = 0; i < len; i++) n[i].val = v[i];
for (i = len; i--; ) { // find longest chain that can follow n[i] for (p = n + i; p++ < n + len; ) { if (p->val > n[i].val && p->len >= n[i].len) { n[i].next = p; n[i].len = p->len + 1; } } }
// find longest chain for (i = 0, p = n; i < len; i++) if (n[i].len > p->len) p = n + i;
do printf(" %d", p->val); while ((p = p->next)); putchar('\n');
free(n); }
int main(void) { int x[] = { 3, 2, 6, 4, 5, 1 }; int y[] = { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 };
lis(x, sizeof(x) / sizeof(int)); lis(y, sizeof(y) / sizeof(int)); return 0; }</lang>
- Output:
3 4 5 0 4 6 9 13 15
C++
Patience sorting <lang cpp>#include <iostream>
- include <vector>
- include <tr1/memory>
- include <algorithm>
- include <iterator>
template <typename E> struct Node {
E value; std::tr1::shared_ptr<Node<E> > pointer;
};
template <class E> struct node_ptr_less {
bool operator()(const std::tr1::shared_ptr<Node<E> > &node1,
const std::tr1::shared_ptr<Node<E> > &node2) const {
return node1->value < node2->value; }
};
template <typename E>
std::vector<E> lis(const std::vector<E> &n) {
typedef std::tr1::shared_ptr<Node<E> > NodePtr;
std::vector<NodePtr> pileTops;
// sort into piles
for (typename std::vector<E>::const_iterator it = n.begin(); it != n.end(); it++) {
NodePtr node(new Node<E>());
node->value = *it;
typename std::vector<NodePtr>::iterator j =
std::lower_bound(pileTops.begin(), pileTops.end(), node, node_ptr_less<E>());
if (j != pileTops.begin())
node->pointer = *(j-1);
if (j != pileTops.end())
*j = node;
else
pileTops.push_back(node);
}
// extract LIS from piles
std::vector<E> result;
for (NodePtr node = pileTops.back(); node != NULL; node = node->pointer)
result.push_back(node->value);
std::reverse(result.begin(), result.end());
return result;
}
int main() {
int arr1[] = {3,2,6,4,5,1};
std::vector<int> vec1(arr1, arr1 + sizeof(arr1)/sizeof(*arr1));
std::vector<int> result1 = lis(vec1);
std::copy(result1.begin(), result1.end(), std::ostream_iterator<int>(std::cout, ", "));
std::cout << std::endl;
int arr2[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15};
std::vector<int> vec2(arr2, arr2 + sizeof(arr2)/sizeof(*arr2));
std::vector<int> result2 = lis(vec2);
std::copy(result2.begin(), result2.end(), std::ostream_iterator<int>(std::cout, ", "));
std::cout << std::endl;
return 0;
}</lang>
- Output:
2, 4, 5, 0, 2, 6, 9, 11, 15,
C#
Recursive
<lang csharp>using System; using System.Collections; using System.Collections.Generic; using System.Linq;
public static class LIS {
public static IEnumerable<T> FindRec<T>(IList<T> values, IComparer<T> comparer = null) =>
values == null ? throw new ArgumentNullException() :
FindRecImpl(values, Sequence<T>.Empty, 0, comparer ?? Comparer<T>.Default).Reverse();
private static Sequence<T> FindRecImpl<T>(IList<T> values, Sequence<T> current, int index, IComparer<T> comparer) {
if (index == values.Count) return current;
if (current.Length > 0 && comparer.Compare(values[index], current.Value) <= 0)
return FindRecImpl(values, current, index + 1, comparer);
return Max(
FindRecImpl(values, current, index + 1, comparer),
FindRecImpl(values, current + values[index], index + 1, comparer)
);
}
private static Sequence<T> Max<T>(Sequence<T> a, Sequence<T> b) => a.Length < b.Length ? b : a;
class Sequence<T> : IEnumerable<T>
{
public static readonly Sequence<T> Empty = new Sequence<T>(default(T), null);
public Sequence(T value, Sequence<T> tail)
{
Value = value;
Tail = tail;
Length = tail == null ? 0 : tail.Length + 1;
}
public T Value { get; }
public Sequence<T> Tail { get; }
public int Length { get; }
public static Sequence<T> operator +(Sequence<T> s, T value) => new Sequence<T>(value, s);
public IEnumerator<T> GetEnumerator()
{
for (var s = this; s.Length > 0; s = s.Tail) yield return s.Value;
}
IEnumerator IEnumerable.GetEnumerator() => GetEnumerator(); }
}</lang>
Patience sorting
<lang csharp>public static class LIS {
public static T[] Find<T>(IList<T> values, IComparer<T> comparer = null) {
if (values == null) throw new ArgumentNullException();
if (comparer == null) comparer = Comparer<T>.Default;
var pileTops = new List<T>();
var pileAssignments = new int[values.Count];
for (int i = 0; i < values.Count; i++) {
T element = values[i];
int pile = pileTops.BinarySearch(element, comparer);
if (pile < 0) pile = ~pile;
if (pile == pileTops.Count) pileTops.Add(element);
else pileTops[pile] = element;
pileAssignments[i] = pile;
}
T[] result = new T[pileTops.Count];
for (int i = pileAssignments.Length - 1, p = pileTops.Count - 1; p >= 0; i--) {
if (pileAssignments[i] == p) result[p--] = values[i];
}
return result;
}
}</lang>
Clojure
Implementation using the Patience Sort approach. The elements (newelem) put on a pile combine the "card" with a reference to the top of the previous stack, as per the algorithm. The combination is done using cons, so what gets put on a pile is a list -- a descending subsequence.
<lang Clojure>(defn place [piles card]
(let [[les gts] (->> piles (split-with #(<= (ffirst %) card)))
newelem (cons card (->> les last first))
modpile (cons newelem (first gts))]
(concat les (cons modpile (rest gts)))))
(defn a-longest [cards]
(let [piles (reduce place '() cards)] (->> piles last first reverse)))
(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]))</lang>
- Output:
<lang>(2 4 5) (0 2 6 9 11 15)</lang>
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 ((not-found t))
(loop
while not-found
for pile in piles
and prev = nil then pile
and i from 0
do (when (<= item (caar pile))
(setf (elt piles i) (push (cons item (car prev)) (elt piles i)) not-found nil)))
(if not-found
(append piles (list (list (cons item (caar (last piles))))))
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)
Common Lisp: Using the Patience Sort approach (alternative)
This is a different version of the code above. <lang lisp>(defun insert-item (item piles)
(multiple-value-bind
(i prev)
(do* ((prev nil (car x))
(x piles (cdr x)) (i 0 (1+ i))) ((or (null x) (<= item (caaar x))) (values i prev)))
(if (= i (length piles))
(append piles (list (list (cons item (caar (last piles)))))) (progn (push (cons item (car prev)) (elt piles i)) piles))))
(defun longest-inc-seq (input)
(do* ((piles nil (insert-item (car x) piles))
(x input (cdr x)))
((null x) (reverse (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~%" (longest-inc-seq l)))</lang>
- Output:
(2 4 5) (0 2 6 9 11 15)
D
Simple Version
Uses the second powerSet function from the Power Set Task. <lang d>import std.stdio, std.algorithm, power_set2;
T[] lis(T)(T[] items) pure nothrow {
//return items.powerSet.filter!isSorted.max!q{ a.length };
return items
.powerSet
.filter!isSorted
.minPos!q{ a.length > b.length }
.front;
}
void main() {
[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;
}</lang>
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Patience sorting
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)) out(result) {
assert(result.length <= items.length); assert(result.isSorted); assert(result.all!(x => items.canFind(x)));
} body {
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.lis.writeln;
}</lang> The output is the same.
Faster Version
With some more optimizations. <lang d>import std.stdio, std.algorithm, std.range, std.array;
T[] lis(T)(in T[] items) pure nothrow if (__traits(compiles, T.init < T.init)) out(result) {
assert(result.length <= items.length); assert(result.isSorted); assert(result.all!(x => items.canFind(x)));
} body {
if (items.empty)
return null;
static struct Node {
T value;
Node* pointer;
}
Node*[] pileTops;
auto nodes = minimallyInitializedArray!(Node[])(items.length);
// Sort into piles.
foreach (idx, x; items) {
auto node = &nodes[idx];
node.value = x;
immutable i = pileTops.length -
pileTops.assumeSorted!q{a.value < b.value}
.upperBound(node)
.length;
if (i != 0)
node.pointer = pileTops[i - 1];
if (i != pileTops.length)
pileTops[i] = node;
else
pileTops ~= node;
}
// Extract LIS from nodes.
size_t count = 0;
for (auto n = pileTops[$ - 1]; n != null; n = n.pointer)
count++;
auto result = minimallyInitializedArray!(T[])(count);
for (auto n = pileTops[$ - 1]; n != null; n = n.pointer)
result[--count] = n.value;
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> The output is the same.
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 ]
Elixir
Naive version
very slow <lang elixir>defmodule Longest_increasing_subsequence do
# Naive implementation def lis(l) do (for ss <- combos(l), ss == Enum.sort(ss), do: ss) |> Enum.max_by(fn ss -> length(ss) end) end defp combos(l) do Enum.reduce(1..length(l), [[]], fn k, acc -> acc ++ (combos(k, l)) end) end defp combos(1, l), do: (for x <- l, do: [x]) defp combos(k, l) when k == length(l), do: [l] defp combos(k, [h|t]) do (for subcombos <- combos(k-1, t), do: [h | subcombos]) ++ combos(k, t) end
end
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])</lang>
- Output:
[3, 4, 5] [0, 4, 6, 9, 13, 15]
Patience sort version
<lang elixir>defmodule Longest_increasing_subsequence do
# Patience sort implementation
def patience_lis(l), do: patience_lis(l, [])
defp patience_lis([h | t], []), do: patience_lis(t, [[{h,[]}]])
defp patience_lis([h | t], stacks), do: patience_lis(t, place_in_stack(h, stacks, []))
defp patience_lis([], []), do: []
defp patience_lis([], stacks), do: get_previous(stacks) |> recover_lis |> Enum.reverse
defp place_in_stack(e, [stack = [{h,_} | _] | tstacks], prevstacks) when h > e do
prevstacks ++ [[{e, get_previous(prevstacks)} | stack] | tstacks]
end
defp place_in_stack(e, [stack | tstacks], prevstacks) do
place_in_stack(e, tstacks, prevstacks ++ [stack])
end
defp place_in_stack(e, [], prevstacks) do
prevstacks ++ [[{e, get_previous(prevstacks)}]]
end
defp get_previous(stack = [_|_]), do: hd(List.last(stack))
defp get_previous([]), do: []
defp recover_lis({e, prev}), do: [e | recover_lis(prev)]
defp recover_lis([]), do: []
end
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])</lang>
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Erlang
Both implementations:
- Naive version
- Patience sort version.
Function combos is copied from panduwana blog.
Function maxBy is copied from Hynek -Pichi- Vychodil's answer.
<lang erlang> -module(longest_increasing_subsequence).
-export([test_naive/0, test_patience/0]).
% ************************************************** % Interface to test the implementation % **************************************************
test_naive() ->
test_gen(fun lis/1).
test_patience() ->
test_gen(fun patience_lis/1).
test_gen(F) ->
show_result(F([3,2,6,4,5,1])), show_result(F([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15])).
show_result(Res) ->
io:format("~w\n", [Res]).
% **************************************************
% ************************************************** % Naive implementation % **************************************************
lis(L) ->
maxBy(
fun(SS) -> length(SS) end,
[ lists:usort(SS)
|| SS <- combos(L),
SS == lists:sort(SS)]
).
% **************************************************
% ************************************************** % Patience sort implementation % **************************************************
patience_lis(L) ->
patience_lis(L, []).
patience_lis([H | T], Stacks) ->
NStacks =
case Stacks of
[] ->
[[{H,[]}]];
_ ->
place_in_stack(H, Stacks, [])
end,
patience_lis(T, NStacks);
patience_lis([], Stacks) ->
case Stacks of
[] ->
[];
[_|_] ->
lists:reverse( recover_lis( get_previous(Stacks) ) )
end.
place_in_stack(E, [Stack = [{H,_} | _] | TStacks], PrevStacks) when H > E ->
PrevStacks ++ [[{E, get_previous(PrevStacks)} | Stack] | TStacks];
place_in_stack(E, [Stack = [{H,_} | _] | TStacks], PrevStacks) when H =< E ->
place_in_stack(E, TStacks, PrevStacks ++ [Stack]);
place_in_stack(E, [], PrevStacks)->
PrevStacks ++ [[{E, get_previous(PrevStacks)}]].
get_previous(Stack = [_|_]) ->
hd(lists:last(Stack));
get_previous([]) ->
[].
recover_lis({E,Prev}) ->
[E|recover_lis(Prev)];
recover_lis([]) ->
[].
% **************************************************
% ************************************************** % Copied from http://stackoverflow.com/a/4762387/4162959 % **************************************************
maxBy(F, L) ->
element(
2,
lists:max([ {F(X), X} || X <- L])
).
% **************************************************
% ************************************************** % Copied from https://panduwana.wordpress.com/2010/04/21/combination-in-erlang/ % **************************************************
combos(L) ->
lists:foldl(
fun(K, Acc) -> Acc++(combos(K, L)) end,
[[]],
lists:seq(1, length(L))
).
combos(1, L) ->
[[X] || X <- L];
combos(K, L) when K == length(L) ->
[L];
combos(K, [H|T]) ->
[[H | Subcombos]
|| Subcombos <- combos(K-1, T)]
++ (combos(K, T)).
% ************************************************** </lang>
Output naive:
[3,4,5] [0,4,6,9,13,15]
Output patience:
[2,4,5] [0,2,6,9,11,15]
Go
Patience sorting <lang go>package main
import (
"fmt" "sort"
)
type Node struct {
val int back *Node
}
func lis (n []int) (result []int) {
var pileTops []*Node
// sort into piles
for _, x := range n {
j := sort.Search(len(pileTops), func (i int) bool { return pileTops[i].val >= x })
node := &Node{ x, nil }
if j != 0 { node.back = pileTops[j-1] }
if j != len(pileTops) {
pileTops[j] = node
} else {
pileTops = append(pileTops, node)
}
}
if len(pileTops) == 0 { return []int{} }
for node := pileTops[len(pileTops)-1]; node != nil; node = node.back {
result = append(result, node.val)
}
// reverse
for i := 0; i < len(result)/2; i++ {
result[i], result[len(result)-i-1] = result[len(result)-i-1], result[i]
}
return
}
func main() {
for _, d := range [][]int{{3, 2, 6, 4, 5, 1},
{0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}} {
fmt.Printf("an L.I.S. of %v is %v\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]
Haskell
Naive implementation
<lang Haskell>import Data.Ord ( comparing ) import Data.List ( maximumBy, subsequences ) import Data.List.Ordered ( isSorted, nub )
lis :: Ord a => [a] -> [a] lis = maximumBy (comparing length) . map nub . filter isSorted . subsequences -- longest <-- unique <-- increasing <-- all
main = do
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]</lang>
- Output:
[2,4,5] [0,2,6,9,11,15] [1]
Patience sorting
<lang Haskell>{-# LANGUAGE FlexibleContexts, UnicodeSyntax #-}
module Main (main, lis) where
import Control.Monad.ST ( ST, runST ) import Control.Monad ( (>>=), (=<<), foldM ) import Data.Array.ST ( Ix, STArray, readArray, writeArray, newArray ) import Data.Array.MArray ( MArray )
infix 4 ≡
(≡) :: Eq α ⇒ α → α → Bool (≡) = (==)
(∘) = (.)
lis ∷ Ord α ⇒ [α] → [α]
lis xs = runST $ do
let lxs = length xs pileTops ← newSTArray (min 1 lxs , lxs) [] i ← foldM (stack pileTops) 0 xs readArray pileTops i >>= return ∘ reverse
stack ∷ (Integral ι, Ord ε, Ix ι, MArray α [ε] μ)
⇒ α ι [ε] → ι → ε → μ ι
stack piles i x = do
j ← bsearch piles x i
writeArray piles j ∘ (x:) =<< if j ≡ 1 then return []
else readArray piles (j-1)
return $ if j ≡ i+1 then i+1 else i
bsearch ∷ (Integral ι, Ord ε, Ix ι, MArray α [ε] μ)
⇒ α ι [ε] → ε → ι → μ ι
bsearch piles x = go 1
where go lo hi | lo > hi = return lo
| otherwise =
do (y:_) ← readArray piles mid
if y < x then go (succ mid) hi
else go lo (pred mid)
where mid = (lo + hi) `div` 2
newSTArray ∷ Ix ι ⇒ (ι,ι) → ε → ST σ (STArray σ ι ε) newSTArray = newArray
main ∷ IO ()
main = do
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]</lang>
- Output:
[2,4,5] [0,2,6,9,11,15] [1]
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 ->
J
These examples are simple enough for brute force to be reasonable:
<lang j>increasing=: (-: /:~)@#~"1 #:@i.@^~&2@# longestinc=: ] #~ [: (#~ ([: (= >./) +/"1)) #:@I.@increasing</lang>
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.
Example use:
<lang j>
longestinc 3,2,6,4,5,1
2 4 5 3 4 5
longestinc 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15
0 2 6 9 11 15 0 2 6 9 13 15 0 4 6 9 11 15 0 4 6 9 13 15</lang>
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.size() == 0 ? null : 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]
JavaScript
<lang javascript>
var _ = require('underscore'); function findIndex(input){ var len = input.length; var maxSeqEndingHere = _.range(len).map(function(){return 1;}); for(var i=0; i<len; i++) for(var j=i-1;j>=0;j--) if(input[i] > input[j] && maxSeqEndingHere[j] >= maxSeqEndingHere[i]) maxSeqEndingHere[i] = maxSeqEndingHere[j]+1; return maxSeqEndingHere; }
function findSequence(input, result){ var maxValue = Math.max.apply(null, result); var maxIndex = result.indexOf(Math.max.apply(Math, result)); var output = []; output.push(input[maxIndex]); for(var i = maxIndex ; i >= 0; i--){ if(maxValue==0)break; if(input[maxIndex] > input[i] && result[i] == maxValue-1){ output.push(input[i]); maxValue--; } } output.reverse(); return output; }
var x = [0, 7, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15];
var y = [3, 2, 6, 4, 5, 1];
var result = findIndex(x); var final = findSequence(x, result); console.log(final);
var result1 = findIndex(y); var final1 = findSequence(y, result1); console.log(final1); </lang>
- Output:
[ 0, 2, 6, 9, 11, 15 ] [ 2, 4, 5 ]
jq
Use the patience sorting method to find a longest (strictly) increasing subsequence.
Generic functions:
Recent versions of jq have functions that obviate the need for the two generic functions defined in this subsection. <lang jq>def until(cond; update):
def _until: if cond then . else (update | _until) end; try _until catch if .== "break" then empty else . end;
- binary search for insertion point
def bsearch(target):
. as $in
| [0, length-1] # [low, high]
| until(.[0] > .[1];
.[0] as $low | .[1] as $high
| ($low + ($high - $low) / 2 | floor) as $mid
| if $in[$mid] >= target
then .[1] = $mid - 1
else .[0] = $mid + 1
end )
| .[0];</lang>
lis: <lang jq>def lis:
# Helper function: # given a stream, produce an array of the items in reverse order: def reverse(stream): reduce stream as $i ([]; [$i] + .);
# put the items into increasing piles using the structure:
# NODE = {"val": value, "back": NODE}
reduce .[] as $x
( []; # array of NODE
# binary search for the appropriate pile
(map(.val) | bsearch($x)) as $i
| setpath([$i];
{"val": $x,
"back": (if $i > 0 then .[$i-1] else null end) })
)
| .[length - 1]
| reverse( recurse(.back) | .val ) ; </lang>
Examples: <lang jq>( [3,2,6,4,5,1],
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]
) | lis</lang>
- Output:
<lang sh>$ jq -c -n -f lis.jq [2,4,5] [0,2,6,9,11,15] </lang>
Julia
<lang julia> function lis(arr::Vector)
if length(arr) == 0 return copy(arr) end
L = Vector{typeof(arr)}(length(arr))
L[1] = [arr[1]]
for i in 2:length(arr)
nextL = []
for j in 1:i
if arr[j] < arr[i] && length(L[j]) ≥ length(nextL)
nextL = L[j]
end
end
L[i] = vcat(nextL, arr[i])
end
return L[indmax(length.(L))]
end
@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])</lang>
- Output:
lis([3, 2, 6, 4, 5, 1]) = [2, 4, 5] lis([0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]) = [0, 2, 6, 9, 11, 15]
Kotlin
Uses the algorithm in the Wikipedia L.I.S. article: <lang scala>// version 1.1.0
fun longestIncreasingSubsequence(x: IntArray): IntArray =
when (x.size) {
0 -> IntArray(0)
1 -> x
else -> {
val n = x.size
val p = IntArray(n)
val m = IntArray(n + 1)
var len = 0
for (i in 0 until n) {
var lo = 1
var hi = len
while (lo <= hi) {
val mid = Math.ceil((lo + hi) / 2.0).toInt()
if (x[m[mid]] < x[i]) lo = mid + 1
else hi = mid - 1
}
val newLen = lo
p[i] = m[newLen - 1]
m[newLen] = i
if (newLen > len) len = newLen
}
val s = IntArray(len)
var k = m[len]
for (i in len - 1 downTo 0) {
s[i] = x[k]
k = p[k]
}
s
}
}
fun main(args: Array<String>) {
val lists = listOf(
intArrayOf(3, 2, 6, 4, 5, 1),
intArrayOf(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15)
)
lists.forEach { println(longestIncreasingSubsequence(it).asList()) }
}</lang>
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Lua
<lang lua>function buildLIS(seq)
local piles = { { {table.remove(seq, 1), nil} } }
while #seq>0 do
local x=table.remove(seq, 1)
for j=1,#piles do
if piles[j][#piles[j]][1]>x then
table.insert(piles[j], {x, (piles[j-1] and #piles[j-1])})
break
elseif j==#piles then
table.insert(piles, Template:X,)
end
end
end
local t={}
table.insert(t, piles[#piles][1][1])
local p=piles[#piles][1][2]
for i=#piles-1,1,-1 do
table.insert(t, piles[i][p][1])
p=piles[i][p][2]
end
table.sort(t)
print(unpack(t))
end
buildLIS({3,2,6,4,5,1}) buildLIS({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
Mathematica
Although undocumented, Mathematica has the function LongestAscendingSequence which exactly does what the Task asks for: <lang Mathematica>LongestAscendingSequence/@{{3,2,6,4,5,1},{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}}
Nirod
<lang nimrod>proc longestIncreasingSubsequence[T](d: seq[T]): seq[T] =
var l = newSeq[seq[T]]()
for i in 0 .. <d.len:
var x = newSeq[T]()
for j in 0 .. <i:
if l[j][l[j].high] < d[i] and l[j].len > x.len:
x = l[j]
l.add x & @[d[i]]
result = @[]
for x in l:
if x.len > result.len:
result = x
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)</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]
Objective-C
Patience sorting <lang objc>#import <Foundation/Foundation.h>
@interface Node : NSObject { @public
id val; Node *back;
} @end
@implementation Node @end
@interface NSArray (LIS) - (NSArray *)longestIncreasingSubsequenceWithComparator:(NSComparator)comparator; @end
@implementation NSArray (LIS) - (NSArray *)longestIncreasingSubsequenceWithComparator:(NSComparator)comparator {
NSMutableArray *pileTops = [[NSMutableArray alloc] init];
// sort into piles
for (id x in self) {
Node *node = [[Node alloc] init];
node->val = x;
int i = [pileTops indexOfObject:node
inSortedRange:NSMakeRange(0, [pileTops count])
options:NSBinarySearchingInsertionIndex|NSBinarySearchingFirstEqual
usingComparator:^NSComparisonResult(Node *node1, Node *node2) {
return comparator(node1->val, node2->val);
}];
if (i != 0)
node->back = pileTops[i-1];
pileTops[i] = node;
}
// follow pointers from last node
NSMutableArray *result = [[NSMutableArray alloc] init];
for (Node *node = [pileTops lastObject]; node; node = node->back)
[result addObject:node->val];
return [[result reverseObjectEnumerator] allObjects];
} @end
int main(int argc, const char *argv[]) {
@autoreleasepool {
for (NSArray *d in @[@[@3, @2, @6, @4, @5, @1],
@[@0, @8, @4, @12, @2, @10, @6, @14, @1, @9, @5, @13, @3, @11, @7, @15]])
NSLog(@"an L.I.S. of %@ is %@", d,
[d longestIncreasingSubsequenceWithComparator:^NSComparisonResult(id obj1, id obj2) {
return [obj1 compare:obj2];
}]);
}
return 0;
}</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
)
OCaml
Naïve implementation
<lang OCaml>let longest l = List.fold_left (fun acc x -> if List.length acc < List.length x
then x
else acc) [] l
let subsequences d l =
let rec check_subsequences acc = function
| x::s -> check_subsequences (if (List.hd (List.rev x)) < d
then x::acc
else acc) s
| [] -> acc
in check_subsequences [] l
let lis d =
let rec lis' l = function | x::s -> lis' ((longest (subsequences x l)@[x])::l) s | [] -> longest l in lis' [] d
let _ =
let sequences = [[3; 2; 6; 4; 5; 1]; [0; 8; 4; 12; 2; 10; 6; 14; 1; 9; 5; 13; 3; 11; 7; 15]]
in
List.map (fun x -> print_endline (String.concat " " (List.map string_of_int
(lis x)))) sequences</lang>
- Output:
3 4 5 0 4 6 9 13 15
Patience sorting
<lang ocaml>let lis cmp list =
let pile_tops = Array.make (List.length list) [] in
let bsearch_piles x len =
let rec aux lo hi =
if lo > hi then
lo
else
let mid = (lo + hi) / 2 in
if cmp (List.hd pile_tops.(mid)) x < 0 then
aux (mid+1) hi
else
aux lo (mid-1)
in
aux 0 (len-1)
in
let f len x =
let i = bsearch_piles x len in
pile_tops.(i) <- x :: if i = 0 then [] else pile_tops.(i-1);
if i = len then len+1 else len
in
let len = List.fold_left f 0 list in
List.rev pile_tops.(len-1)</lang>
Usage:
# lis compare [3; 2; 6; 4; 5; 1];; - : int list = [2; 4; 5] # 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]
Perl
Dynamic programming
<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
Patience sorting
<lang perl>sub lis {
my @pileTops;
# sort into piles
foreach my $x (@_) {
# binary search my $low = 0, $high = $#pileTops; while ($low <= $high) { my $mid = int(($low + $high) / 2); if ($pileTops[$mid]{val} >= $x) { $high = $mid - 1; } else { $low = $mid + 1; } } my $i = $low; my $node = {val => $x};
$node->{back} = $pileTops[$i-1] if $i != 0;
$pileTops[$i] = $node;
}
my @result;
for (my $node = $pileTops[-1]; $node; $node = $node->{back}) {
push @result, $node->{val};
}
return reverse @result;
}
foreach my $r ([3, 2, 6, 4, 5, 1], [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]) {
my @d = @$r; my @lis = lis(@d); print "an L.I.S. of [@d] is [@lis]\n";
}</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 6
Dynamic programming
Straight-forward implementation of the algorithm described in the video.
<lang perl6>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] < @l[$j] + 1 {
@l[$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 perl6>sub lis(@deck is copy) {
my @S = [@deck.shift() => Nil].item;
for @deck -> $card {
with first { @S[$_][*-1].key > $card }, ^@S -> $i {
@S[$i].push: $card => @S[$i-1][*-1] // Nil
} else {
@S.push: [ $card => @S[*-1][*-1] // Nil ].item
}
}
reverse map *.key, (
@S[*-1][*-1], *.value ...^ !*.defined
)
}
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]
Phix
Using the Wikipedia algorithm (converted to 1-based indexing) <lang Phix>function lis(sequence X, integer N = length(X))
sequence P = repeat(0,N)
sequence M = repeat(0,N)
integer len = 0
for i=1 to N do
integer lo = 1
integer hi = len
while lo<=hi do
integer mid = ceil((lo+hi)/2)
if X[M[mid]]<X[i] then
lo = mid + 1
else
hi = mid - 1
end if
end while
if lo>1 then
P[i] = M[lo-1]
end if
M[lo] = i
if lo>len then len = lo end if
end for
sequence res = repeat(0,len)
if len>0 then
integer k = M[len]
for i=len to 1 by -1 do
res[i] = X[k]
k = P[k]
end for
end if
return res
end function
constant tests = {{3, 2, 6, 4, 5, 1},
{0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}}
for i=1 to length(tests) do
?lis(tests[i])
end for</lang>
- Output:
{2,4,5}
{0,2,6,9,11,15}
PHP
Patience sorting <lang php><?php class Node {
public $val; public $back = NULL;
}
function lis($n) {
$pileTops = array();
// sort into piles
foreach ($n as $x) {
// binary search
$low = 0; $high = count($pileTops)-1;
while ($low <= $high) {
$mid = (int)(($low + $high) / 2);
if ($pileTops[$mid]->val >= $x)
$high = $mid - 1;
else
$low = $mid + 1;
}
$i = $low;
$node = new Node();
$node->val = $x;
if ($i != 0)
$node->back = $pileTops[$i-1];
$pileTops[$i] = $node;
}
$result = array();
for ($node = count($pileTops) ? $pileTops[count($pileTops)-1] : NULL;
$node != NULL; $node = $node->back)
$result[] = $node->val;
return array_reverse($result);
}
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))); ?></lang>
- Output:
Array
(
[0] => 2
[1] => 4
[2] => 5
)
Array
(
[0] => 0
[1] => 2
[2] => 6
[3] => 9
[4] => 11
[5] => 15
)
PicoLisp
Adapted patience sorting approach: <lang PicoLisp>(de longinc (Lst)
(let (D NIL R NIL)
(for I Lst
(cond
((< I (last D))
(for (Y . X) D
(T (> X I) (set (nth D Y) I)) ) )
((< I (car R))
(set R I)
(when D (set (cdr R) (last D))) )
(T (when R (queue 'D (car R)))
(push 'R I) ) ) )
(flip R) ) )</lang>
Original recursive glutton: <lang PicoLisp>(de glutton (L)
(let N (pop 'L)
(maxi length
(recur (N L)
(ifn L
(list (list N))
(mapcan
'((R)
(if (> (car R) N)
(list (cons N R) R)
(list (list N) R) ) )
(recurse (car L) (cdr L)) ) ) ) ) ) )
(test (2 4 5)
(glutton (3 2 6 4 5 1)))
(test (2 6 9 11 15)
(glutton (8 4 12 2 10 6 14 1 9 5 13 3 11 7 15)))
(test (-31 0 83 782)
(glutton (4 65 2 -31 0 99 83 782 1)) )</lang>
PowerShell
<lang PowerShell>function Get-LongestSubsequence ( [int[]]$A )
{
If ( $A.Count -lt 2 ) { return $A }
# Start with an "empty" pile
# (We will only store the top value in each "pile".)
$Pile = @( [int]::MaxValue )
$Last = 0
# Hashtable to hold the back pointers
$BP = @{}
# For each number in the orginal sequence...
ForEach ( $N in $A )
{
# Find the first pile with a value greater than N
$i = 0..$Last | Where { $N -lt $Pile[$_] } | Select -First 1
# Place N on the pile
$Pile[$i] = $N
# Set the back pointer for this value to the value of the previous pile
$BP["$N"] = $Pile[$i-1]
# If this is the previously empty pile, add a new empty pile
If ( $i -eq $Last )
{
$Pile += @( [int]::MaxValue )
$Last++
}
}
# Ignore the empty pile
$Last--
# Start with the value of the last pile
$N = $Pile[$Last]
$S = @( $N )
# Add the remainder of the values by walking through the back pointers
ForEach ( $i in $Last..1 )
{
$S += ( $N = $BP["$N"] )
}
# Return the series (reversed into the correct order)
return $S[$Last..0]
}</lang>
<lang PowerShell>( Get-LongestSubsequence 3, 2, 6, 4, 5, 1 ) -join ', ' ( Get-LongestSubsequence 0, 8, 4, 12, 2, 10, 6, 16, 14, 1, 9, 5, 13, 3, 11, 7, 15 ) -join ', '</lang>
- Output:
2, 4, 5 0, 2, 6, 9, 11, 15
Prolog
Works with SWI-Prolog version 6.4.1
Naïve implementation.
<lang prolog>lis(In, Out) :-
% we ask Prolog to find the longest sequence
aggregate(max(N,Is), (one_is(In, [], Is), length(Is, N)), max(_, Res)),
reverse(Res, Out).
% we describe the way to find increasing subsequence
one_is([], Current, Current).
one_is([H | T], Current, Final) :-
( Current = [], one_is(T, [H], Final));
( Current = [H1 | _], H1 < H, one_is(T, [H | Current], Final));
one_is(T, Current, Final).
</lang>
Prolog finds the first longest subsequence
?- lis([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15], Out). Out = [0,4,6,9,13,15]. ?- lis([3,2,6,4,5,1], Out). Out = [3,4,5].
Python
Python: O(nlogn) Method from Wikipedia's LIS Article[1]
<lang python>def longest_increasing_subsequence(X):
"""Returns the Longest Increasing Subsequence in the Given List/Array"""
N = len(X)
P = [0] * N
M = [0] * (N+1)
L = 0
for i in range(N):
lo = 1
hi = L
while lo <= hi:
mid = (lo+hi)//2
if (X[M[mid]] < X[i]):
lo = mid+1
else:
hi = mid-1
newL = lo
P[i] = M[newL-1]
M[newL] = i
if (newL > L):
L = newL
S = []
k = M[L]
for i in range(L-1, -1, -1):
S.append(X[k])
k = P[k]
return S[::-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, longest_increasing_subsequence(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]
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 from functools import total_ordering from bisect import bisect_left
@total_ordering class Node(namedtuple('Node_', 'val back')):
def __iter__(self):
while self is not None:
yield self.val
self = self.back
def __lt__(self, other):
return self.val < other.val
def __eq__(self, other):
return self.val == other.val
def lis(d):
"""Return one of the L.I.S. of list d using patience sorting."""
if not d:
return []
pileTops = []
for di in d:
j = bisect_left(pileTops, Node(di, None))
new_node = Node(di, pileTops[j-1] if j > 0 else None)
if j == len(pileTops):
pileTops.append(new_node)
else:
pileTops[j] = new_node
return list(pileTops[-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]
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. <lang Racket>#lang racket/base (require data/gvector)
(define (gvector-last gv)
(gvector-ref gv (sub1 (gvector-count gv))))
(define (lis-patience-sort input-list)
(let ([piles (gvector)])
(for ([item (in-list input-list)])
(insert-item! piles item))
(reverse (gvector-last piles))))
(define (insert-item! piles item)
(if (zero? (gvector-count piles))
(gvector-add! piles (cons item '()))
(cond
[(not (<= item (car (gvector-last piles))))
(gvector-add! piles (cons item (gvector-last piles)))]
[(<= item (car (gvector-ref piles 0)))
(gvector-set! piles 0 (cons item '()))]
[else (let loop ([first 1] [last (sub1 (gvector-count piles))])
(if (= first last)
(gvector-set! piles first (cons item (gvector-ref piles (sub1 first))))
(let ([middle (quotient (+ first last) 2)])
(if (<= item (car (gvector-ref piles middle)))
(loop first middle)
(loop (add1 middle) last)))))])))</lang>
- Output:
'(2 4 5) '(0 2 6 9 11 15)
Ring
<lang ring>
- Project : Longest increasing subsequence
- Date : 2017/11/23
- Author : Gal Zsolt (~ CalmoSoft ~)
- Email : <calmosoft@gmail.com>
tests = [[3, 2, 6, 4, 5, 1], [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]] res = [] for x=1 to len(tests)
lis(tests[x]) showarray(res)
end
func lis(X)
N = len(X)
P = list(N)
M = list(N)
for nr = 1 to len(P)
P[nr] = 0
next
for nr = 1 to len(M)
P[nr] = 0
next
len = 0
for i=1 to N
lo = 1
hi = len
while lo <= hi
mid = floor((lo+hi)/2)
if X[M[mid]]<X[i]
lo = mid + 1
else
hi = mid - 1
ok
end
if lo>1
P[i] = M[lo-1]
ok
M[lo] = i
if lo>len
len = lo
ok
next
res = list(len)
if len>0
k = M[len]
for i=len to 1 step -1
res[i] = X[k]
k = P[k]
next
ok
return res
func showarray(vect)
see "{"
svect = ""
for n = 1 to len(vect)
svect = svect + vect[n] + ", "
next
svect = left(svect, len(svect) - 2)
see svect
see "}" + nl
</lang> Output:
{2, 4, 5}
{0, 2, 6, 9, 11, 15}
Ruby
Patience sorting <lang ruby>Node = Struct.new(:val, :back)
def lis(n)
pileTops = []
# sort into piles
for x in n
# binary search
low, high = 0, pileTops.size-1
while low <= high
mid = low + (high - low) / 2
if pileTops[mid].val >= x
high = mid - 1
else
low = mid + 1
end
end
i = low
node = Node.new(x)
node.back = pileTops[i-1] if i > 0
pileTops[i] = node
end
result = []
node = pileTops.last
while node
result.unshift(node.val)
node = node.back
end
result
end
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])</lang>
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Rust
<lang Rust>fn lower_bound<T: PartialOrd>(list: &Vec<T>, value: &T) -> usize {
if list.is_empty() {
return 0;
}
let mut lower = 0usize;
let mut upper = list.len();
while lower != upper {
let middle = lower + upper >> 1;
if list[middle] < *value {
lower = middle + 1;
} else {
upper = middle;
}
}
return lower;
}
fn lis<T: PartialOrd + Copy>(list: &Vec<T>) -> Vec<T> {
if list.is_empty() {
return Vec::new();
}
let mut subseq: Vec<T> = Vec::new();
subseq.push(*list.first().unwrap());
for i in list[1..].iter() {
if *i <= *subseq.last().unwrap() {
let index = lower_bound(&subseq, i);
subseq[index] = *i;
} else {
subseq.push(*i);
}
}
return subseq;
}
fn main() {
let list = vec![3, 2, 6, 4, 5, 1];
println!("{:?}", lis(&list));
let list = vec![0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15];
println!("{:?}", lis(&list));
}</lang>
- Output:
[1, 4, 5] [0, 1, 3, 7, 11, 15]
Scala
Patience sorting
- Output:
See it in running in your browser by ScalaFiddle (JavaScript) or by Scastie (JVM).
<lang Scala>object LongestIncreasingSubsequence extends App {
val tests = Map(
"3,2,6,4,5,1" -> Seq("2,4,5", "3,4,5"),
"0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15" -> Seq("0,2,6,9,11,15", "0,2,6,9,13,15", "0,4,6,9,13,15", "0,4,6,9,11,15")
)
def lis(l: Array[Int]): Seq[Array[Int]] =
if (l.length < 2) Seq(l)
else {
def increasing(done: Array[Int], remaining: Array[Int]): Seq[Array[Int]] =
if (remaining.isEmpty) Seq(done)
else
(if (remaining.head > done.last)
increasing(done :+ remaining.head, remaining.tail)
else Nil) ++ increasing(done, remaining.tail) // all increasing combinations
val all = (1 to l.length)
.flatMap(i => increasing(l take i takeRight 1, l.drop(i + 1)))
.sortBy(-_.length)
all.takeWhile(_.length == all.head.length) // longest of all increasing combinations
}
def asInts(s: String): Array[Int] = s split "," map (_.toInt)
assert(tests forall {
case (given, expect) =>
val allLongests: Seq[Array[Int]] = lis(asInts(given))
println(
s"$given has ${allLongests.length} longest increasing subsequences, e.g. ${
allLongests.last.mkString(",")}")
allLongests.forall(lis => expect.contains(lis.mkString(",")))
})
}</lang>
- Output:
3,2,6,4,5,1 has 2 longest increasing subsequences, e.g. 2,4,5 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15 has 4 longest increasing subsequences, e.g. 0,2,6,9,11,15
Brute force solution
<lang Scala>def powerset[A](s: List[A]) = (0 to s.size).map(s.combinations(_)).reduce(_++_) 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)(_>_)</lang>
Scheme
Patience sorting <lang scheme>(define (lis less? lst)
(define pile-tops (make-vector (length lst))) (define (bsearch-piles x len) (let aux ((lo 0)
(hi (- len 1)))
(if (> lo hi)
lo (let ((mid (quotient (+ lo hi) 2))) (if (less? (car (vector-ref pile-tops mid)) x) (aux (+ mid 1) hi) (aux lo (- mid 1)))))))
(let aux ((len 0)
(lst lst))
(if (null? lst)
(reverse (vector-ref pile-tops (- len 1))) (let* ((x (car lst)) (i (bsearch-piles x len))) (vector-set! pile-tops i (cons x (if (= i 0) '() (vector-ref pile-tops (- i 1))))) (aux (if (= i len) (+ len 1) len) (cdr lst))))))
(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)</lang>
- Output:
(2 4 5) (0 2 6 9 11 15)
Sidef
Dynamic programming: <lang ruby>func lis(a) {
var l = a.len.of { [] }
l[0] << a[0]
for i in (1..a.end) {
for j in ^i {
if ((a[j] < a[i]) && (l[i].len < l[j].len+1)) {
l[i] = [l[j]...]
}
}
l[i] << a[i]
}
l.max_by { .len }
}
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>)</lang>
Patience sorting: <lang ruby>func lis(deck) {
var pileTops = []
deck.each { |x|
var low = 0;
var high = pileTops.end
while (low <= high) {
var mid = ((low + high) // 2)
if (pileTops[mid]{:val} >= x) {
high = mid-1
} else {
low = mid+1
}
}
var i = low
var node = Hash(val => x)
node{:back} = pileTops[i-1] if (i != 0)
pileTops[i] = node
}
var result = []
for (var node = pileTops[-1]; node; node = node{:back}) {
result << node{:val}
}
result.reverse
}
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>)</lang>
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Standard ML
Patience sorting
<lang sml>fun lis cmp n =
let
val pile_tops = DynamicArray.array (length n, [])
fun bsearch_piles x =
let
fun aux (lo, hi) =
if lo > hi then
lo
else
let
val mid = (lo + hi) div 2
in
if cmp (hd (DynamicArray.sub (pile_tops, mid)), x) = LESS then
aux (mid+1, hi)
else
aux (lo, mid-1)
end
in
aux (0, DynamicArray.bound pile_tops)
end
fun f x =
let
val i = bsearch_piles x
in
DynamicArray.update (pile_tops, i,
x :: (if i = 0 then [] else DynamicArray.sub (pile_tops, i-1)))
end in app f n; rev (DynamicArray.sub (pile_tops, DynamicArray.bound pile_tops)) end</lang>
Usage:
- lis Int.compare [3, 2, 6, 4, 5, 1]; val it = [2,4,5] : int list - lis Int.compare [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]; val it = [0,2,6,9,11,15] : int list
Swym
Based on the Python video solution. Interpreter at [[2]] <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
VBScript
<lang vb> Function LIS(arr) n = UBound(arr) Dim p() ReDim p(n) Dim m() ReDim m(n) l = 0 For i = 0 To n lo = 1 hi = l Do While lo <= hi middle = Int((lo+hi)/2) If arr(m(middle)) < arr(i) Then lo = middle + 1 Else hi = middle - 1 End If Loop newl = lo p(i) = m(newl-1) m(newl) = i If newL > l Then l = newl End If Next Dim s() ReDim s(l) k = m(l) For i = l-1 To 0 Step - 1 s(i) = arr(k) k = p(k) Next LIS = Join(s,",") End Function
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)) </lang>
- Output:
2,4,5, 0,2,6,9,11,15,
zkl
<lang zkl>fcn longestSequence(ns){ // based on Patience sorting
piles:=L();
backPtr:='wrap(np){ return(np-1,if(np) piles[np-1].len()-1 else -1) }; // maybe (-1,-1)
foreach n in (ns){ newPile:=True; // create list of sorted lists
foreach e,p in (piles.enumerate()){
if(n<p[-1][0]){ p.del(1,-1) // only need the first and last elements .append(T(n,backPtr(e))); newPile=False; break; }
}
if(newPile) piles.append(L(T(n,backPtr(piles.len()))));
}
reg r=L(),p=-1,n=0;
do{ n,p=piles[p][n]; r.write(n); p,n=p; }while(p!=-1);
r.reverse()
}</lang> <lang zkl>foreach ns in (T(T(1),T(3,2,6,4,5,1),T(4,65,2,-31,0,99,83,782,1), 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);
}</lang>
- Output:
1: L(1) from L(1)
3: L(2,4,5) from L(3,2,6,4,5,1)
4: L(-31,0,83,782) from L(4,65,2,-31,0,99,83,782,1)
6: L(0,1,3,9,11,15) from L(0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15)
4: L("f","o","o","r") from foobar
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