Find the Odd odds

Question

Given an unordered collection of positive integers by any reasonable input method, return all of the sub-collections that have an odd number of odd elements (i.e. have an odd total).

This is code-golf so you should aim to minimize the byte count of your program.

Since some languages only have ordered collections (lists, arrays, vectors, etc.) or don't have an unordered collection that allows duplicates, you may use ordered collections (regardless of your language choice), however you should not output any duplicate collections with different orders (e.g. [2,3] and [3,2]). You may output in whatever order you see fit.

Test cases

[2,3,7,2] -> [[3],[7],[2,3],[2,7],[2,2,3],[2,2,7]]
[2,4,6,8] -> []
[4,9]     -> [[9],[4,9]]

Answer 1

05AB1E, 6 bytes

{æÙʒOÉ

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{æÙʒOÉ
{      Sort
 æ     Powerset
  Ù    Uniqufy
   ʒ   Keep elements where
    O                      the sum
     É                             is uneven

-2 bytes thanks to @EriktheOutgolfer

Answer 2

Python 3, 93 bytes

f=lambda x,r=[[]]:x and f(x[1:],r+[y+x[:1]for y in r])or{(*sorted(y),)for y in r if sum(y)&1}

Returns a set of tuples. Most likely way too long.

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Answer 3

Pyth, 10 9 8 bytes

{f%sT2yS

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         # implicit input
       S # sort input, this way the subsets will already be sorted
      y  # all subsets
 f       # filter elements when ..
   sT    # the sum ..
  %  2   # is odd
{        # remove all duplicated elements
         # implicit output

Answer 4

Python 2, 91 bytes

r=[[]]
for n in input():r+=map([n].__add__,r)
print{tuple(sorted(y))for y in r if sum(y)&1}

Prints a set of tuples. If a set of strings is allowed, tuple(sorted(y)) can be replaced with `sorted(y)` for 86 bytes.

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Answer 5

Perl 6, 50 bytes

{.combinations.grep(*.sum!%%2).unique(:as(*.Bag))}

To filter out the same-up-to-ordering combinations I filter out duplicates by converting each to a Bag (unordered collection) before comparing. Unfortunately I couldn't find a way to accept a Bag as input that was as concise.

Answer 6

Brachylog, 11 bytes

o⊇ᵘ{+ḃt1&}ˢ

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I hoped to find a shorter solution, but here's the best I could do.

Explanation

o⊇ᵘ{+ḃt1&}ˢ    
o                                        the input, sorted
 ⊇ᵘ           Find all unique subsets of

   {    &}ˢ   Then select only those results where
    +                                          the sum
     ḃ                           the base 2 of
      t        The last digit of
       1                                               is 1.

Yeah, I could've used modulo 2 to check for oddness, but that's not an odd approach ;)

Answer 7

Mathematica 31 44 38 bytes

Among all subsets of the input set, it returns those for which the sum, Tr, is odd.

6 bytes saved thanks to alephalpha.

Select[Union@Subsets@Sort@#,OddQ@*Tr]&

---

 Select[Union@Subsets@Sort@#,OddQ@*Tr]&[{2,3,7,2}]

{{3}, {7}, {2, 3}, {2, 7}, {2, 2, 3}, {2, 2, 7}}

Answer 8

Jelly, 8 bytes

ṢŒPSḂ$ƇQ

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Bug-fix thanks to Jonathan Allan.

-1 byte by using the alias for filter thanks to Riolku.

Answer 9

PHP, 126 bytes

for(;++$i>>$argc<1;sort($t),$s&1?$r[join(_,$t)]=$t:0)for ($t=[],$j=$s=0;++$j<$argc;)$i>>$j&1?$s+=$t[]=$argv[$j]:0;print_r($r);

takes input from command line arguments; run with -nr or try it online.

breakdown

for(;++$i>>$argc<1;             # loop through subsets
    sort($t),                       # 2. sort subset
    $s&1?$r[join(_,$t)]=$t:0        # 3. if sum is odd, add subset to results
    )                               # 1. create subset:
    for ($t=[],$j=$s=0;++$j<$argc;)     # loop through elements
        $i>>$j&1?                       # if bit $j is set in $i
        $s+=$t[]=$argv[$j]:0;           # then add element to subset
print_r($r);                    # print results