# Sum of Powers of 2

The Challenge

Given an integer input x where 1 <= x <= 255, return the results of powers of two that when summed give x.

Examples

Given the input:

86


64 16 4 2


Input:

240


Output:

128 64 32 16


Input:

1


Output:

1


Input:

64


Output:

64


The output may contain zeros if the certain power of two is not present in the sum.

For example, input 65 may output 0 64 0 0 0 0 0 1.

Scoring

This is , so the shortest answer in each language wins.

• Does the list have to be sorted highest to lowest?
Commented Jan 28, 2019 at 20:56
• May we output some redundant zeros? Commented Jan 28, 2019 at 20:58
• RE: "sorted highest to lowest" why add a restriction that was not part of the challenge and invalidates most existing answers? (Also what about little-endian?!) + it invalidates my Python answer since sets do not have any order. Commented Jan 28, 2019 at 21:38
• @JonathanAllan I've removed the restriction. I'll keep that in mind next time I post another question - I'm still fairly new to this. :) Commented Jan 28, 2019 at 22:28
• I think you might want to state that any power of two may only be used once. Otherwise somebody could output "1 1 1" for the input 3. Commented Jan 29, 2019 at 10:58

# JavaScript (ES6), 28 bytes

f=n=>n?[...f(n&~-n),n&-n]:[]


Try it online!

• You're the only person in the whole world who can make me upvote JavaScript answers! Commented Jan 29, 2019 at 0:05
• @sergiol, why wouldn't you normally upvote a JS solution? A good solution is a good solution regardless of the language used or who posted it. Commented Jan 29, 2019 at 13:23
• @Shaggy Because Arnauld seems the only person to do such Javascript solutions. His answers are pure genius! Commented Jan 29, 2019 at 15:11
• @sergiol Thanks for the compliment, but that's not quite true. I'm regularly outgolfed by more clever answers -- and that's what this site is all about. ^^ Commented Jan 29, 2019 at 15:18
• @Oliver I'm not sure. It seems like leading zeros (before 128) are forbidden. Otherwise, another possible variant is f=n=>n&&f(n&~-n)+[,n&-n]. Commented Jan 29, 2019 at 19:47

# Jelly, 4 bytes

-2 since we may output zeros in place of unused powers of 2 :)

Ḷ2*&


Try it online!

### How?

Ḷ2*& - Link: integer, n         e.g. 10
Ḷ    - lowered range of n            [  0,  1,  2,  3,  4,  5,  6,  7,  8,  9]
2*  - two to the power of           [  1,  2,  4,  8, 16, 32, 64,128,256,512]
& - bit-wise and                  [  0,  2,  0,  8,  0,  0,  0,  0,  0,  0]


# Jelly, 6 bytes

BUT’2*


Try it online!

### Explanation

BUT here is an explanation (note: I had assumed that we may only output the powers of 2 themselves and nothing else):

BUT’2* – Monadic link. Takes a number N as input. Example: 86
B      – Convert N to binary.                              [1, 0, 1, 0, 1, 1, 0]
U     – Reverse.                                          [0, 1, 1, 0, 1, 0, 1]
T    – Truthy indices.                                   [2, 3, 5, 7]
’   – Decrement.                                        [1, 2, 4, 6]
2* – Raise 2 to that power.                            [2, 4, 16, 64]


"Proof" that it works correctly. The standard representation of an integer $$\ X\$$ in base 2 is a list $$\\{x_1, x_2, x_3,\cdots, x_n\}\$$, where $$\x_i\in\{0,1\},\:\forall\:\: i\in\overline{1,n}\$$, such that: $$X=\sum_{i=1}^n x_i\cdot 2^{n-i}$$ The indices $$\i\$$ such that $$\x_i=0\$$ obviously have no contribution so we're only interested in finding those such that $$\x_i=1\$$. Since subtracting $$\i\$$ from $$\n\$$ is not convenient (the powers of two all have exponents of the form $$\n-i\$$, where $$\i\$$ is any index of a $$\1\$$), instead of finding the truthy indices in this list we reverse it and then find them "backwards" with UT. Now that we've found the correct indices all we have to do is raise $$\2\$$ to those powers.

• "ASCII-only" Sneaky ’ there... Commented Jan 28, 2019 at 21:05
• @EriktheOutgolfer I guess BUT2*H would work though. Commented Jan 28, 2019 at 21:08
• Pretty impressive that this works with an input of 302231454903657293676544. Commented Feb 3, 2019 at 13:47

# Pure Bash, 20

echo $[2**{7..0}&$1]


Try it online!

### Explanation

          {7..0}     # Brace expansion: 7 6 5 4 3 2 1 0
2**{7..0}     # Brace expansion: 128 64 32 16 8 4 2 1
2**{7..0}&$1 # Brace expansion: 128&n 64&n 32&n 16&n 8&n 4&n 2&n 1&n (Bitwise AND)$[2**{7..0}&$1] # Arithmetic expansion echo$[2**{7..0}&$1] # and output  # Python, 35 bytes lambda n:[n&2**i for i in range(8)]  Little-endian with zeros at unused powers of 2. Try it online! # Catholicon, 3 bytes ṫĊŻ  Try it online! Explanation: ṫ Decompose into the largest values where: Ċ the input Ż the bit count is truthy (equal to one)  • Interesting! Get TIO'd :D Commented Jan 28, 2019 at 22:25 • Works with 302231454903657293676544. Nice. Commented Feb 3, 2019 at 13:54 # APL (Dyalog Extended), 7 bytesSBCS Anonymous tacit prefix function. Requires 0-based indexing (⎕IO←0). 2*⍸⍢⌽⍤⊤  Try it online! 2 two * raised to the power of ⍸ the ɩndices where true ⍢ while ⌽ reversed ⍤ of ⊤ the binary representation # Sledgehammer 0.2, 3 bytes ⡔⡸⢣  Decompresses into {intLiteral[2],call[NumberExpand,2]}. Sledgehammer is a compressor for Wolfram Language code using Braille as a code page. The actual size of the above is 2.75 bytes, but due to current rules on meta, padding to the nearest byte is counted in code size. • Huh! Neat little language, and all the characters are actually printable. Commented Jan 29, 2019 at 15:41 • And now I can't get the Peter Gabriel song out of my mind... Commented Jan 31, 2019 at 22:14 # 05AB1E, 3 bytes Ýo&  Port of @JonathanAllan's Jelly answer, so make sure to upvote him! Contains zeros (including -loads of- trailing zeros). Explanation: Ý # Create a list in the range [0, (implicit) input] # i.e. 15 → [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] # i.e. 16 → [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16] o # Take 2 to the power of each value # → [1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768] # → [1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536] & # Bitwise-AND each value with the (implicit) input # 15 → [1,2,4,8,0,0,0,0,0,0,0,0,0,0,0,0] # 16 → [0,0,0,0,16,0,0,0,0,0,0,0,0,0,0,0,0] # (and output the result implicitly)  • ... what?! Never honestly seen bitwise and used in osabie. Nice one. Commented Jan 29, 2019 at 15:40 • @MagicOctopusUrn I indeed also don't use it very often. Can't even find any other answer I've used & in. xD I have used Bitwise-XOR a couple of times, like here or here and Bitwise-NOT once here (which I later removed again after golfing further..). I do use Bitwise-AND, XOR, OR, NOT, SHIFT, etc. pretty often in Java, but in 05AB1E not so much. :) Commented Jan 29, 2019 at 15:55 # Wolfram Language (Mathematica), 17 bytes #~NumberExpand~2&  Try it online! Mathematica strikes again. • This also works with input of 302231454903657293676544. Commented Feb 3, 2019 at 13:56 # R, 27 23 bytes bitwAnd(scan(),2^(7:0))  Try it online! ### Unrolled code and explanation : A = scan() # get input number A from stdin # e.g. A = 65 bitwAnd( A , 2^(7:0)) # bitwise AND between all powers of 2 : 2^7 ... 2^0 and A # and implicitly print the result # e.g. B = bitwAnd(65, c(128,64,32,16,8,4,2,1)) = c(0,64,0,0,0,0,0,1)  • 4 bytes thanks to @Kirill L. • 23 bytes with bitwise and. Commented Jan 29, 2019 at 9:53 • @KirillL.: brilliant ! Commented Jan 29, 2019 at 13:16 # C# (Visual C# Interactive Compiler), 29 bytes Contains 5 unprintable characters. n=>"€@ ".Select(a=>a&n)  # Explanation //Lambda taking one parameter 'n' n=> //String with ASCII characters 128, 64, 32, 16, 8, 4, 2, and 1 "€@ " //Iterate through all the chars of the above string and transform them to .Select(a=> //A bitwise AND operation between the integer value of the current char and the input value a&n)  Try it online! • But we need to get rid of zeros, like n=>new int[8].Select((j,i)=>1<<i&n).Where(i=>i!=0) The part before Where is five bytes shorter btw Commented Jan 29, 2019 at 8:49 • @polfosol The output may contain zeros – Jo King Commented Jan 29, 2019 at 9:06 • @JoKing Still, n=>new int[8].Select((j,i)=>1<<i&n) is 35 bytes long and we won't need additional flags and text encodings. Commented Jan 29, 2019 at 13:02 • Using ascii characters 0-7 should be shorter, eg n=>"INSERT ASCII HERE".Select(a=>1<<a&n) But I'm on a mobile device that can't display or type unprintables, so I'll have to wait till I get home to update the answer Commented Jan 29, 2019 at 16:34 # C (gcc), 39 bytes f(n){for(;n;n&=n-1)printf("%d ",n&-n);}  Try it online! • 37 bytes – c-- Commented Jul 18, 2022 at 2:46 # C# (Visual C# Interactive Compiler), 38 bytes x=>{for(int y=8;y-->0;Print(x&1<<y));}  Try it online! • aw, close :P Commented Jan 29, 2019 at 7:45 • Fails for inputs 1, 2, 4, 8, 16, etc. (the x>y should be x>=y instead). Commented Jan 29, 2019 at 9:37 • @ASCIIOnly - I'm telling you, the range operator is going to be sweet :) – dana Commented Jan 29, 2019 at 13:43 • @ASCII-only Mean while, you can use the flag /u:System.Linq.Enumerable and try this for 31 bytes Commented Jan 29, 2019 at 23:18 • @EmbodimentofIgnorance sure. but i'd prefer not to list language as "C# /u:System.Linq.Enumerable" :P Commented Jan 30, 2019 at 4:54 # Perl 6, 16 12 bytes -4 bytes thanks to Jonathan Allan *+&2**all ^8  Try it online! Returns an All Junction with 8 elements. This is a rather non-standard way of returning, but generally, Junctions can act as ordered (at least until autothreading is implemented) lists and it is possible to extract the values from one. ### Explanation: *+& # Bitwise AND the input with 2** # 2 raised to the power of all ^8 # All of the range 0 to 7  # 05AB1E, 7 bytes 2вRƶ<oò  explanation: 2в convert input to binary array R reverse array ƶ< multiply each item by it's index and subtract 1 oò 2^item then round down  Try it online! • Also works with input of 302231454903657293676544 Commented Feb 3, 2019 at 14:00 # Julia 0.6, 13 bytes n->n&2.^(0:7)  Try it online! # Haskell, 29 bytes (mapM(\n->[0,2^n])[7,6..0]!!)  Try it online! # Ruby, 25 bytes ->n{8.times{|i|p n&2**i}}  Try it online! # C (clang), 13311063 58 bytes 58-byte solution thanks to @ceilingcat. x=256;main(y){for(scanf("%d",&y);x/=2;)printf("%d ",y&x);}  Try it online! • In C89, you can declare like main(){} and the return type defaults to int. Same for variables at global scope. Also, at least on normal implementations like clang, printf and scanf work without prototypes. You get warnings of course, but it's still valid C89 (maybe) or at least K&R C for them to be implicitly declared. The types of the C objects you pass as args defines how they're passed, so a char* and int* will Just Work without truncating pointers to 32-bit on x86-64 or anything. (Default argument promotions happen, same as for variadic functions which they are anyway.) Commented Jan 30, 2019 at 3:46 • Or is this aiming to be valid C11 with no undefined behaviour? If so, proudly proclaim it. :) And BTW, writing a function that takes an output array as an arg would probably be smaller. Anyway, see Tips for golfing in C Commented Jan 30, 2019 at 3:48 • You can use bitwise & to check if a bit is set. Like y&(1<<x)&&printf("%d ",1<<x);. Or to not skip zeros, just printf("%d ", y&(1<<x)). Or instead of counting bit positions, use x=256 and x>>=1 to shift the mask. main(y){int x=256;for(scanf("%d",&y);x>>=1;)printf("%d ",y&x);} 63 bytes Try it online! clang will even compile that with -std=c11 Commented Jan 30, 2019 at 4:11 • 44 bytes Commented Feb 9, 2019 at 1:25 # APL (Dyalog Extended), 6 bytes 2*⍸⌽⊤⎕  2 to the * power of the ⍸ indices of the true bits of the ⌽ reversed ⊤ binary encoding of the ⎕ input Try it online! # MATL, 5 bytes BPfqW  Try it online! # Explanation Consider input 86 as an example. B % Implicit input. Convert to binary (highest to lowest digits) % STACK: [1 0 1 0 1 1 0] P % Flip % STACK: [0 1 1 0 1 0 1] f % Find: indices of nonzeros (1-based) % STACK: [2 3 5 7] q % Subtract 1, element-wise % STACK: [1 2 4 6] W % Exponential with base 2, element-wise. Implicit display % STACK: [2 4 16 64]  # Japt, 8 5 bytes Æ&2pX  Try it Æ&2pX :Implicit input of integer U Æ :Map each X in the range [0,U) & : Bitwise AND of U with 2pX : 2 to the power of X  ## Alternative Suggested by Oliver to avoid the 0s in the output using the -mf flag. N&2pU  Try it N&2pU :Implicitly map each U in the range [0,input) N :The (singleton) array of inputs & :Bitwise AND with 2pX :2 to the power of U :Implicitly filter and output  • Nice one. You can do N&2pU with -mf to avoid the 0s Commented Jan 29, 2019 at 2:54 # APL(NARS) 18 chars, 36 bytes {(⍵⊤⍨8⍴2)/⌽1,2*⍳7}  test:  f←{(⍵⊤⍨8⍴2)/⌽1,2*⍳7} f 86 64 16 4 2 f 240 128 64 32 16 f 1 1 f 64 64 f 3 2 1  # 05AB1E, 9 bytes Ýoʒ›}æʒOQ  Try it online! This is also correct for 6-bytes, but it doesn't complete in time on TIO for 86: # 05AB1E, 6 bytes ÝoæʒOQ  Try it online! • Both your answers output an empty set for 15, instead of [1,2,4,8] Commented Jan 29, 2019 at 7:14 • @KevinCruijssen needed 2**0, nice catch. Ý over L. Commented Jan 29, 2019 at 15:29 • Ah, I know the feeling. Also had L instead of Ý at first in my answer. Commented Jan 29, 2019 at 15:31 # CJam, 12 bytes {:T{2\#T&}%}  Try it online! # K (oK), 19 16 bytes -3 bytes thanks to ngn! {*/x#2}'&|(8#2)\  Try it online! oK does not have power operator, that's why I need a helper function {*/x#2} (copy 2 x times and reduce the resulting list by multiplication) • – ngn Commented Jan 31, 2019 at 12:49 • @ngn Thanks! I tried it and it worked, but I wasn't sure it is acceptible. Commented Jan 31, 2019 at 15:25 # Alchemist, 125 bytes _->In_x+128a+m m+x+a->m+b m+0x+a->n+a m+0a->o+Out_b+Out_" " n+b->n+x+c n+0b+a->n+c n+0a->p o+b->o+c o+0b->p p+2c->p+a p+0c->m  ### Explanation _->In_x+128a+m # Initialize universe with input, 128a (value to compare to) and m (state) m+x+a->m+b # If c has been halved, subtract min(a, x) from a and x and put its value into b m+0x+a->n+a # If x < a, continue to state n m+0a->o+Out_b+Out_" " # Else print and continue to state o n+b->n+x+c # Add min(a, x) (i.e. x) back to x, and add it to c (we're collecting a back into c) n+0b+a->n+c # Then, add the rest of a to c n+0a->p # Then, go to state p o+b->o+c # Add min(a, x) (i.e. a) to c - x _is_ greater than a and so contains it in its binary representation, so we're not adding back to x o+0b->p # Then, go to state p p+2c->p+a # Halve c into a p+0c->m # Then go to state m  # PHP, 41 39 bytes for($c=256;$c>>=1;)echo$argv[1]&$c,' ';  Try it online! Or 38 with no fun >>= operator and PHP 5.6+: for($x=8;$x--;)echo$argv[1]&2**$x,' ';  Or 36 with little-endian ("0 2 4 0 16 0 64 0") output: while($x<8)echo$argv[1]&2**$x++,' ';


Really I just wanted to use the >>= operator, so I'm sticking with the 39.

Tests:

$php pow2.php 86 0 64 0 16 0 4 2 0$php pow2.php 240
128 64 32 16 0 0 0 0

$php pow2.php 1 0 0 0 0 0 0 0 1$php pow2.php 64
0 64 0 0 0 0 0 0

\$php pow2.php 65
0 64 0 0 0 0 0 1


# TSQL, 43 39 bytes

Can't find a shorter fancy solution, so here is a standard loop. -4 bytes thanks to MickyT and KirillL

DECLARE @y int=255

,@ int=128s:PRINT @y&@ SET @/=2IF @>0GOTO s


Try it out

• using the bitwise and (&) you can save a few with the following ,@ int=128s:print @y&@ set @/=2IF @>0GOTO s. This is hinted by @KirillL for the R answer Commented Jan 31, 2019 at 19:49
• @MickyT that worked like a charm. Thanks a lot Commented Feb 1, 2019 at 0:02