Fold a List in Half

Question

We are going to fold a list of integers. The procedure to do so is as follows, If the list is of even length, make a list of half of its length where the nth item of the new list is the sum of the nth item of the old list and the nth-to-last item of the old list. For example if we had the list

[1 2 3 4 5 6 7 8]

We would fold it like so

 [8 7 6 5]
+[1 2 3 4]
__________
 [9 9 9 9]

If the list is of odd length, to fold it we first remove the middle item, fold it as if it were even and the append the middle item to the result.

For example if we had the list

[1 2 3 4 5 6 7]

We would fold it like so

 [7 6 5]
+[1 2 3]
__________
 [8 8 8]
++     [4]
__________
 [8 8 8 4]

Task

Write a program or function that takes a list of integers as input and outputs that list folded.

This is a code-golf question so answers will be scored in bytes, with fewer bytes being better.

Sample implementation

Here's an implementation in Haskell that defines a function f that performs a fold.

f(a:b@(_:_))=a+last b:f(init b)
f x=x

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Answer 1

JavaScript (ES6), 46 43 bytes

f=(a,[b,...c]=a)=>c+c?[b+c.pop(),...f(c)]:a

f=(a,[b,...c]=a)=>c+c?[b+c.pop(),...f(c)]:a

console.log(f([1,2,3,4,5,6,7,8]).join(', ')) // 9, 9, 9, 9  ✓
console.log(f([1,2,3,4,5,6,7]).join(', '))   // 8, 8, 8, 4  ✓

.as-console-wrapper{max-height:100%!important}

Saved 3 bytes with inspiration from Asaf.

Answer 2

Perl, 42 38 chars

sub f{@a=map{$+pop}splice@,0,@/2;@a,@}

sub f{(map{$_+pop}splice@_,0,@_/2),@_} 

Try for example like so:

perl -e 'my @input=(1..9); sub f{(map{$_+pop}splice@_,0,@_/2),@_}  print join(",",f(@input));

Answer 3

Pyth, 18 17 13 bytes

V.Tc2Q aYsN;Y

My original approach was

WtQ aY+.)Q.(Q0;+Y

-1 byte thanks to Mr. Xcoder

-4 bytes thanks to FryAmTheEggman

Answer 4

C++17, 75 73 71 bytes

As unnamed lambda, accepting a container like vector or list, returns via modifying the input:

[](auto&L){for(auto a=L.begin(),b=L.end();a<--b;L.pop_back())*a+++=*b;}

Using the well known 'goes-to' operator <-- and the triple plus +++

Ungolfed and example:

#include<iostream>
#include<vector>

using namespace std;

auto f=
[](auto&L){
 for(
  auto a=L.begin(),b=L.end();
  a<--b;
  L.pop_back()
 )
 *a+++=*b;
}
;

void test(auto L) {
 for(auto x:L)cout << x << ", ";
 cout << endl;
 f(L);
 for(auto x:L)cout << x << ", ";
 cout << endl << endl;
}

int main() { 
 vector<int> A = {1,2,3,4,5,6,7,8}, B = {1,2,3,4,5,6,7};
 test(A);
 test(B);
}

Answer 5

Java 8, 93 bytes

Double digits! This is a lambda that takes an int[] and returns an int[].

l->{int n=l.length,i=0;for(;i<n/2;)l[i]+=l[n-++i];return java.util.Arrays.copyOf(l,n/2+n%2);}

Ungolfed lambda

l -> {
    int n = l.length, i = 0;
    for (; i < n / 2; )
        l[i] += l[n - ++i];
    return java.util.Arrays.copyOf(l, n / 2 + n % 2);
}

Quite straightforward. It folds the second half in place onto the first half of the input and returns a copy of just the first half.

Surprisingly, the array copy in the return statement seems to be the cheapest way to handle the final element quirk for odd-length inputs.

Answer 6

PHP, 67 bytes

function($l){while($l)$o[]=array_shift($l)+array_pop($l);return$o;}

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Answer 7

Perl 6, 28 bytes

{.[^*/2]Z+.[$_-1...$_/2,$_]}

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Explanation:

{                          }   # Anonymous code block
 .[^*/2]                       # The first half of the list
        Z+                     # Zip added to
          .[$_-1...$_/2        # The other half of the list
                       ,$_]    # And zero for the middle element

Answer 8

J, 22 bytes

({.+/@,:|.@}.)~>.@-:@#

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Answer 9

APL (Dyalog Unicode), 21 bytes^SBCS

-3 bytes thanks to @Adám.

(⌊2÷⍨≢)(↑{+⌿↑⍺⍵}∘⌽↓)⊢

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Explanation:

(⌊2÷⍨≢)(↑{+⌿↑⍺⍵}∘⌽↓)⊢  ⍝ Monadic function train
(⌊2÷⍨≢)                  ⍝ Left portion:
     ≢                   ⍝ Take the length of the input...
  2÷⍨                    ⍝ Divide it by two...
 ⌊                       ⍝ And floor it. This gives our midpoint index. Call it "X"
                      ⊢  ⍝ Right portion: return the original input. Call it "Y"
       (↑{+⌿↑⍺⍵}∘⌽↓)   ⍝ Midddle portion: takes X and Y as arguments
        ↑           ↓    ⍝ Take and drop Y by X. Essentially splits Y in half
                         ⍝ Presents the two halves to the next function
                 ∘⌽     ⍝ Reverse the second half
         {+⌿↑⍺⍵}       ⍝ Final function, takes first half and reversed second half
              ⍺⍵        ⍝ Construct a nested list of first and second halves...
             ↑          ⍝ ...and "mix" them into a matrix. Has the nice property that
                        ⍝ it will pad the first half with a zero if needed.
          +⌿           ⍝ Sum the matrix along the columns, return resulting vector

Answer 10

Ruby, 34 bytes

f=->l{l[1]?[l.shift+l.pop]+f[l]:l}

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Answer 11

Common Lisp, 106 bytes

(lambda(l)(setf(values a b)(floor(length l)2))`(,@(#1=subseq(mapcar'+ l(reverse l))0 a),@(#1#l a(+ a b))))

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Answer 12

Java (JDK), 102 95 bytes

i->{int l=i.length,o=l/2+l%2,r[]=new int[o];for(;o-->0;)r[o]=i[o]+(o<l+~o?i[l+~o]:0);return r;}

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-7 Thanks to Kevin Cruijssen

Answer 13

jq, 58 bytes

[reverse[:length/2],.[:length/2|floor]]|transpose|map(add)

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Answer 14

Scala, 65 bytes

l=>l take(l.size+1)/2 zipAll(l.reverse take l.size/2,0,0)map(_+_)

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Answer 15

k9, 12 bytes

+/++(;|)@'2^

Version 2021.09.20.

Cut in two, reverse second list, pad with zeroes, add.

Answer 16

Thunno, \$ 10 \log_{256}(96) \approx \$ 8.23 bytes

2APAur0Tz+

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Sort-of port of Adnan's 05AB1E answer

Explanation:

2APAur0Tz+  # Implicit input
2AP         # Split in half
   Au       # Dump onto stack
     r      # Reverse top item
      0T    # Append a 0
        z+  # Zipped addition

Thunno's normal addition (+) will provide a nested list of each element of the first list plus each element of the second list.

Thunno's zipped addition (z+) zips the two lists together, discarding items if they are not of equal length.

So, we must append a 0 to the top list (since it will be the shorter one if the input contained an odd number of items) and then use zipped addition.

Answer 17

C# (Visual C# Interactive Compiler), 71 bytes

x=>{int i=0,l=x.Length;for(;i<l/2;)x[i]+=x[l-++i];return x[..(l-l/2)];}

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-2 bytes thanks to @ceilingcat!

Answer 18

JavaScript (Node.js), 41 bytes

f=([b,...c])=>1/b?[b+~~c.pop(),...f(c)]:c

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So many ways to reach 41

Answer 19

Japt, 10 bytes

There's gotta be a 9 here, somewhere.

òUÊ/2 oÔyx

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òUÊ/2 oÔyx     :Implicit input of array U
ò              :Partitions of length
 UÊ            :  Length of U
   /2          :  Divided by 2
      o        :Modify last element
       Ô       :  Reverse
        y      :Transpose
         x     :Reduce each by addition

Also 10 bytes

íUÔoUÊz¹mx

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íUÔoUÊz¹mx     :Implicit input of array U
í              :Interleave with
 UÔ            :  Reverse U (mutates the original)
   o           :  Pop this many items (also mutates the original)
    UÊ         :    Length of U
      z        :    Floor divided by 2
       ¹       :End interleave
        m      :Map
         x     :  Reduce by addition

Answer 20

Nibbles, 6.5 bytes

/`\~_!:\@0@+

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/   Fold
`\~  split into two parts
_     input (row from STDIN as a list of integers)
!    zip
:     join
\      reverse
@       the second part
0      0
@     the first part
+     with addition

Answer 21

Julia 1.0, 69 68 bytes

!s=(~=length;~s%2>0&&insert!(s,(1+~s)÷2,0);(s+reverse(s))[1:~s÷2])

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-1 byte thanks to MarcMush: replace ==1 with >0

Answer 22

JavaScript (Node.js), 62 bytes

a=>a.map((e,i)=>e+a[c=(b=a.length)-i-1]*(c!=i)).slice(0,++b/2)

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