Implement the Thanos sorting algorithm

Question

The sorting algorithm goes like this:

While the list is not sorted, snap half of all items (remove them from the list). Continue until the list is sorted or only one item remains (which is sorted by default). This sorting algorithm may give different results based on implementation.

The item removal procedure is up to the implementation to decide, but the list should be half as long as before after one pass of the item removal procedure. Your algorithm may decide to remove either the first half or the list, the last half of the list, all odd items, all even items, one at a time until the list is half as long, or any not mentioned.

The input list can contain an arbitrary amount of items (within reason, let’s say up to 1000 items), not only perfectly divisible lists of 2^n items. You will have to either remove (n+1)/2 or (n-1)/2 items if the list is odd, either hardcoded or decided randomly during runtime. Decide for yourself: what would Thanos do if the universe contained an odd amount of living things?

The list is sorted if no item is smaller than any previous item. Duplicates may occur in the input, and may occur in the output.

Your program should take in an array of integers (via stdin or as parameters, either individual items or an array parameter), and return the sorted array (or print it to stdout).

Examples:

// A sorted list remains sorted
[1, 2, 3, 4, 5] -> [1, 2, 3, 4, 5]

// A list with duplicates may keep duplicates in the result
[1, 2, 3, 4, 3] -> [1, 3, 3] // Removing every second item
[1, 2, 3, 4, 3] -> [3, 4, 3] -> [4, 3] -> [3] // Removing the first half
[1, 2, 3, 4, 3] -> [1, 2] // Removing the last half

[1, 2, 4, 3, 5] could give different results:

// Removing every second item:
[1, 2, 4, 3, 5] -> [1, 4, 5]

or:

// Removing the first half of the list
[1, 2, 4, 3, 5] -> [3, 5] // With (n+1)/2 items removed
[1, 2, 4, 3, 5] -> [4, 3, 5] -> [3, 5] // With (n-1)/2 items removed

or:

// Removing the last half of the list
[1, 2, 4, 3, 5] -> [1, 2] // With (n+1)/2 items removed
[1, 2, 4, 3, 5] -> [1, 2, 4] // With (n-1)/2 items removed

or:

// Taking random items away until half (in this case (n-1)/2) of the items remain
[1, 2, 4, 3, 5] -> [1, 4, 3] -> [4, 3] -> [4]

Answer 1

Python 3, 38 42 39 bytes

q=lambda t:t>sorted(t)and q(t[::2])or t

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-3 bytes thanks to @JoKing and @Quuxplusone

Answer 2

R, 41 bytes

x=scan();while(any(x-sort(x)))x=x[!0:1];x

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Answer 3

Python 2, 39 bytes

f=lambda l:l>sorted(l)and f(l[::2])or l

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Answer 4

Brachylog (v2), 6 bytes

≤₁|ḍt↰

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This is a function submission. Input from the left, output to the right, as usual. (The TIO link uses a command-line argument that automatically wraps the function into a full program, so that you can see it in action.)

Explanation

≤₁|ḍt↰
≤₁       Assert that {the input} is sorted {and output it}
  |      Handler for exceptions (e.g. assertion failures):
   ḍ     Split the list into two halves (as evenly as possible)
    t    Take the last (i.e. second) half
     ↰   Recurse {and output the result of the recursion}

Bonus round

≤₁|⊇ᵇlᵍḍhtṛ↰

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The snap's meant to be random, isn't it? Here's a version of the program that choses the surviving elements randomly (while ensuring that half survive at each round).

≤₁|⊇ᵇlᵍḍhtṛ↰
≤₁            Assert that {the input} is sorted {and output it}
  |           Handler for exceptions (e.g. assertion failures):
   ⊇ᵇ         Find all subsets of the input (preserving order)
     lᵍ       Group them by length
       ḍht    Find the group with median length:
         t      last element of
        h       first
       ḍ        half (split so that the first half is larger)
          ṛ   Pick a random subset from that group
           ↰  Recurse

This would be rather shorter if we could reorder the elements, but whyever would a sorting algorithm want to do that?

Answer 5

Perl 6, 30 bytes

$!={[<=]($_)??$_!!.[^*/2].&$!}

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Recursive function that removes the second half of the list until the list is sorted.

Explanation:

$!={                         }    # Assign the function to $!
    [<=]($_)??                    # If the input is sorted
              $_                  # Return the input
                !!                # Else
                  .[^*/2]         # Take the first half of the list (rounding up)
                         .&$!     # And apply the function again

Answer 6

C# (Visual C# Interactive Compiler), 76 bytes

g=>{while(!g.OrderBy(x=>x).SequenceEqual(g))g=g.Take(g.Count()/2);return g;}

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Answer 7

Java 10, 106 97 bytes

L->{for(;;L=L.subList(0,L.size()/2)){int p=1<<31,f=1;for(int i:L)f=p>(p=i)?0:f;if(f>0)return L;}}

-9 bytes thanks to @OlivierGrégoire.

Try it online.

Only leave the first halve of the list every iteration, and removes \$\frac{n+1}{2}\$ items if the list-size is odd.

Explanation:

L->{               // Method with Integer-list as both parameter and return-type
  for(;;           //  Loop indefinitely:
      L=L.subList(0,L.size()/2)){
                   //    After every iteration: only leave halve the numbers in the list
    int p=1<<31,   //   Previous integer, starting at -2147483648
        f=1;       //   Flag-integer, starting at 1
    for(int i:L)   //   Inner loop over the integer in the list:
      f=p>(p=i)?   //    If `a>b` in a pair of integers `a,b`:
         0         //     Set the flag to 0
        :          //    Else (`a<=b`):
         f;        //     Leave the flag the same
    if(f>0)        //   If the flag is still 1 after the loop:
      return L;}}  //    Return the list as result

Answer 8

Wolfram Language (Mathematica), 30 bytes

#//.x_/;Sort@x!=x:>x[[;;;;2]]&

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@Doorknob saved 12 bytes

Answer 9

JavaScript (ES6),  49  48 bytes

Saved 1 byte thanks to @tsh

Removes every 2nd element.

f=a=>a.some(p=c=>p>(p=c))?f(a.filter(_=>a^=1)):a

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Answer 10

Ruby, 37 bytes

->r{r=r[0,r.size/2]while r.sort!=r;r}

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Answer 11

05AB1E, 8 7 bytes

[Ð{Q#ιн

-1 byte thanks to @Emigna.

Removes all odd 0-indexed items every iteration, so removes \$\frac{n-1}{2}\$ items if the list-size is odd.

Try it online or verify some more test cases (or verify those test cases with step-by-step for each iteration).

7 bytes alternative by @Grimy:

ΔÐ{Ê>äн

Removes the last \$\frac{n}{2}\$ items (or \$\frac{n-1}{2}\$ items if the list-size is odd) every iteration.

Try it online or verify some more test cases (or verify those test cases with step-by-step for each iteration).

Explanation:

[        # Start an infinite loop:
 Ð       #  Triplicate the list (which is the implicit input-list in the first iteration)
  {Q     #  Sort a copy, and check if they are equal
    #    #   If it is: Stop the infinite loop (and output the result implicitly)
  ι      #  Uninterweave: halve the list into two parts; first containing all even-indexed
         #  items, second containing all odd-indexed items (0-indexed)
         #   i.e. [4,5,2,8,1] → [[4,2,1],[5,8]]
   н     #  And only leave the first part

Δ        # Loop until the result no longer changes:
 Ð       #  Triplicate the list (which is the implicit input-list in the first iteration)
  {Ê     #  Sort a copy, and check if they are NOT equal (1 if truthy; 0 if falsey)
    >    #  Increase this by 1 (so 1 if the list is sorted; 2 if it isn't sorted)
     ä   #  Split the list in that many parts
      н  #  And only leave the first part
         # (and output the result implicitly after it no longer changes)

Answer 12

TI-BASIC (TI-84), 47 42 45 44 bytes

-1 byte thanks to @SolomonUcko!

Ans→L1:Ans→L2:SortA(L1:While max(L1≠Ans:iPart(.5dim(Ans→dim(L2:L2→L1:SortA(L1:End:Ans

Input list is in Ans.
Output is in Ans and is implicitly printed out.

Explanation:

Ans→L1                  ;store the input into two lists
Ans→L2
SortA(L1                ;sort the first list
                        ; two lists are needed because "SortA(" edits the list it sorts
While max(L1≠Ans        ;loop until both lists are strictly equal
iPart(.5dim(Ans→dim(L2  ;remove the latter half of the second list
                        ; removes (n+1)/2 elements if list has an odd length
L2→L1                   ;store the new list into the first list (updates "Ans")
SortA(L1                ;sort the first list
End
Ans                     ;implicitly output the list when the loop ends

---

Note: TI-BASIC is a tokenized language. Character count does not equal byte count.

Answer 13

Jelly, 7 bytes

m2$⁻Ṣ$¿

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Answer 14

Haskell, 57 55 bytes (thanks to ASCII-only)

f x|or$zipWith(>)x$tail x=f$take(div(length x)2)x|1>0=x

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---

Original Code:

f x|or$zipWith(>)x(tail x)=f(take(div(length x)2)x)|1>0=x

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---

Ungolfed:

f xs | sorted xs = f (halve xs)
     | otherwise = xs

sorted xs = or (zipWith (>) xs (tail xs))

halve xs = take (length xs `div` 2) xs

Answer 15

Gaia, 9 8 bytes

eo₌⟨2%⟩↻

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Explanation:

e		| eval input as a list
       ↻	| until
 o		| the list is sorted
  ₌		| push additional copy
   ⟨2%⟩  	| and take every 2nd element

Answer 16

Julia 1.0, 30 bytes

-x=x>sort(x) ? -x[1:2:end] : x

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Takes every second element of the array if not sorted.

Answer 17

Octave, 49 bytes

l=input('');while(~issorted(l))l=l(1:2:end);end;l

Try it online! This was a journey where more boring is better. Note the two much more interesting entries below:

50 bytes

function l=q(l)if(~issorted(l))l=q(l(1:2:end));end

Try it online! Instead of the uninteresting imperative solution, we can do a recursive solution, for only one additional byte.

53 bytes

f(f=@(g)@(l){l,@()g(g)(l(1:2:end))}{2-issorted(l)}())

Try it online! Yep. A recursive anonymous function, thanks to @ceilingcat's brilliant answer on my tips question. An anonymous function that returns a recursive anonymous function by defining itself in its argument list. I like anonymous functions. Mmmmm.

Answer 18

K (oK), 22 20 bytes

Solution:

{(*2 0N#x;x)x~x@<x}/

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Iterate over the input until it's sorted... if it's not sorted take first n/2 items.

{(*2 0N#x;x)x~x@<x}/ / the solution
{                 }/ / lambda that iterates
                <x   / indices that sort x ascending (<)
              x@     / apply (@) these indices back to x
            x~       / matches (~) x? returns 0 or 1
 (       ; )         / 2-item list which we index into
          x          / original input (ie if list was sorted)
       #x            / reshape (#) x
   2 0N              / as 2 rows
  *                  / take the first one      

Edits:

• -2 bytes thanks to ngn

Answer 19

MATL, 11 bytes

tv`1L)ttS-a

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This works by removing every second item.

Explanation

t      % Take a row vector as input (implicit). Duplicate
v      % Vertically concatenate the two copies of the row vector. When read with
       % linear indexing (down, then across), this effectively repeats each entry
`      % Do...while
  1L)  %   Keep only odd-indexed entries (1-based, linear indexing)
  t    %   Duplicate. This will leave a copy for the next iteration
  tS   %   Duplicate, sort
  -a   %   True if the two arrays differ in any entry
       % End (implicit). A new iteration starts if the top of the stack is true
       % Display (implicit). Prints the array that is left on the stack

Answer 20

Java (JDK), 102 bytes

n->{for(;n.stream().reduce((1<<31)+0d,(a,b)->a==.5|b<a?.5:b)==.5;n=n.subList(0,n.size()/2));return n;}

There's already a C# answer, so I decided to try my hand on a Java answer.

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Answer 21

Husk, 6 5 bytes

1 byte saved thanks to Zgarb

ΩΛ<Ċ2

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Explanation

ΩΛ<Ċ2
Ω         Repeat until
 Λ<         all adjacent pairs are sorted (which means the list is sorted)
   Ċ2         drop every second element from the list

Answer 22

C++ (gcc), 103 bytes

I can't comment, but I improved the version from movatica by reducing the includes, and using auto.

-2 Bytes: ceilingcat
-2 Bytes: ASCII-only

#include<regex>
auto f(auto l){while(!std::is_sorted(l.begin(),l.end()))l.resize(l.size()/2);return l;}

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Answer 23

Crystal, 59 50 bytes

With Array#sort (59 50 bytes):

def f(a);while a!=a.sort;a.pop a.size//2;end;a;end

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Without Array#sort (102 93 bytes):

def f(a);while a.map_with_index{|e,i|e>a.fetch i+1,Int32::MAX}.any?;a.pop a.size//2;end;a;end

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Answer 24

Japt, 10 9 bytes

£=ëÃæÈeXñ

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£=ëÃæÈeXñ     :Implicit input of array U
£             :Map
 =            :  Reassign to U
  ë           :  Every second element, starting with the first
   Ã          :End map
    æ         :Get the first element that returns true
     È        :When passed through the following function as X
      e       :  Test for equality with
       Xñ     :  X sorted

Answer 25

VDM-SL, 99 bytes

f(i)==if forall x in set inds i&x=1or i(x-1)<=i(x) then i else f([i(y)|y in set inds i&y mod 2=0]) 

Never submitted in vdm before, so not certain on language specific rules. So I've submitted as a function definition which takes a seq of int and returns a seq of int

A full program to run might look like this:

functions
f:seq of int +>seq of int
f(i)==if forall x in set inds i&x=1or i(x-1)<=i(x) then i else f([i(y)|y in set inds i&y mod 2=0]) 

Answer 26

Pyth, 10 bytes

.W!SIHhc2Z

Try it online here. This removes the second half on each iteration, rounding down. To change it to remove the first half, rounding up, change the h to e.

.W!SIHhc2ZQ   Q=eval(input())
              Trailing Q inferred
  !SIH        Condition function - input variable is H
   SIH          Is H invariant under sorting?
  !             Logical not
      hc2Z    Iteration function - input variable is Z
       c2Z      Split Z into 2 halves, breaking ties to the left
      h         Take the first half
.W        Q   With initial value Q, execute iteration function while condition function is true

Answer 27

C++ (gcc), 139 137 116 bytes

-2 bytes thanx to ceilingcat, -21 bytes thanx to PeterZuger

#include<regex>
auto f(std::vector<int>l){while(!std::is_sorted(l.begin(),l.end()))l.resize(-~l.size()/2);return l;}

Resize the vector to its first half until it's sorted.

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Answer 28

Julia 1.0, 33 bytes

-a=issorted(a) ? a : -a[1:end÷2]

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Answer 29

Husk, 7 bytes

ΩS=Oo←½

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removes last half.

Answer 30

Jelly, 6 bytes

m⁻Ṣ$¿2

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Very similar to Erik's existing answer, but I came up with this independently, and Erik's no longer active.

How it works

m⁻Ṣ$¿2 - Main link. Takes a list L on the left
    ¿  - Loop while the condition is false, updating L each time:
   $   -   Condition:
  Ṣ    -     L sorted
 ⁻     -     L ≠ L sorted
m    2 -   Body: Take every second element of L