2的幂和

31

挑战

给定一个整数输入x其中1 <= x <= 255，返回时总结给出了两种力量的结果x。

例子

给定输入：

您的程序应输出：

64 16 4 2

输入：

输出：

128 64 32 16

输入：

输出：

输入：

输出：

如果总和中没有确定的2的幂，则输出可能包含零。

例如，输入65可以输出0 64 0 0 0 0 0 1。

计分

这是代码高尔夫球，因此每种语言中最短的答案将获胜。

code-golf binary code-golf sequence integer chess code-golf number arithmetic matrix code-golf code-golf combinatorics grid set-partitions code-golf array-manipulation graph-theory code-golf number code-golf string decision-problem code-golf matrix cellular-automata 3d code-challenge restricted-source printable-ascii code-golf board-game code-golf geometry grid code-golf word-puzzle code-golf matrix sorting code-golf code-golf string decision-problem code-golf matrix cellular-automata code-golf decision-problem code-golf math number arithmetic restricted-source code-golf code-golf number integer matrix code-golf date code-golf matrix code-golf sequence combinatorics chemistry code-golf array-manipulation popularity-contest code-golf code-golf natural-language code-golf number integer sorting substitution code-golf string number date encode code-golf decision-problem code-golf string subsequence code-golf string alphabet code-golf

— 怪异的Gengar
source

5

列表是否必须从高到低排序？

— 阿达姆（Adám）

2

我们可以输出一些冗余零吗？

— 乔纳森·艾伦

4

RE：“从高到低排序”为什么要添加一个限制，该限制不属于挑战范围，并且会使大多数现有答案无效？（还有little-endian呢？！）+由于集合没有任何顺序，它使我的Python答案无效。

— 乔纳森·艾伦

5

@JonathanAllan我已取消限制。下次发布其他问题时，我会记住这一点-我对此还很陌生。:)

— SpookyGengar

6

我认为您可能想声明任何2的幂只能使用一次。否则，可能会有人对输入输出3“1 1 1”

— 黑色猫头鹰凯

Answers:

38

JavaScript (ES6), 28 bytes

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

— Arnauld
source

9

You're the only person in the whole world who can make me upvote JavaScript answers!

— sergiol

4

@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.

— Shaggy

@Shaggy Because Arnauld seems the only person to do such Javascript solutions. His answers are pure genius!

— sergiol

3

@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. ^^

— Arnauld

@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].

— Arnauld

12

Pure Bash, 20

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

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

— Digital Trauma
source

12

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]

— Jonathan Allan
source

11

Jelly, 6 bytes

BUT’2*

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}

$X=\sum_{i=1}^n x_i\cdot 2^{n-i}$ The indices

i

$i$ such that

x_{i} = 0

$x_i=0$ obviously have no contribution so we're only interested in finding those such that

x_{i} = 1

$x_i=1$ . Since subtracting

i

$i$ from

n

$n$ is not convenient (the powers of two all have exponents of the form

n - i

$n-i$ , where

i

$i$ is any index of a

1

$1$ ), 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

$2$ to those powers.

— Mr. Xcoder
source

1

"ASCII-only" Sneaky ’ there...

— Erik the Outgolfer

1

@EriktheOutgolfer I guess BUT2*H would work though.

— Mr. Xcoder

1

Pretty impressive that this works with an input of 302231454903657293676544.

— Michael Karas

9

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!

— Jonathan Allan
source

8

APL (Dyalog Extended), 7 bytes^SBCS

Anonymous tacit prefix function. Requires 0-based indexing (⎕IO←0).

2*⍸⍢⌽⍤⊤

2 two
* raised to the power of
⍸ the ɩndices where true
⍢ while
⌽ reversed
⍤ of
⊤ the binary representation

— Adám
source

8

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.

— lirtosiast
source

2

Huh! Neat little language, and all the characters are actually printable.

— LegionMammal978

And now I can't get the Peter Gabriel song out of my mind...

— Digital Trauma

8

05AB1E, 3 bytes

Ýo&

Port of @JonathanAllan's Jelly answer, so make sure to upvote him!

Contains zeros (including -loads of- trailing zeros).

Try it online or verify all test cases.

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)

— Kevin Cruijssen
source

1

... what?! Never honestly seen bitwise and used in osabie. Nice one.

— Magic Octopus Urn

@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. :)

— Kevin Cruijssen

8

Catholicon, 3 bytes

ṫĊŻ

Explanation:

ṫ       Decompose         into the largest values where:
 Ċ               the input
  Ż       the bit count is truthy (equal to one)

— Okx
source

Interesting! Get TIO'd :D

— Jonathan Allan

Works with 302231454903657293676544. Nice.

— Michael Karas

7

Wolfram Language (Mathematica), 17 bytes

#~NumberExpand~2&

Mathematica strikes again.

— lirtosiast
source

This also works with input of 302231454903657293676544.

— Michael Karas

7

R, 27 23 bytes

bitwAnd(scan(),2^(7:0))

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.

— digEmAll
source

1

23 bytes with bitwise and.

— Kirill L.

@KirillL.: brilliant !

— digEmAll

7

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)

— Embodiment of Ignorance
source

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

— polfosol ఠ_ఠ

@polfosol The output may contain zeros

— Jo King

2

@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.

— polfosol ఠ_ఠ

1

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

— Embodiment of Ignorance

6

C# (Visual C# Interactive Compiler), 38 bytes

x=>{for(int y=8;y-->0;Print(x&1<<y));}

— dana
source

aw, close :P

— ASCII-only

1

Fails for inputs 1, 2, 4, 8, 16, etc. (the x>y should be x>=y instead).

— Kevin Cruijssen

1

@ASCIIOnly - I'm telling you, the range operator is going to be sweet :)

— dana

@ASCII-only Mean while, you can use the flag /u:System.Linq.Enumerable and try this for 31 bytes

— Embodiment of Ignorance

@EmbodimentofIgnorance sure. but i'd prefer not to list language as "C# /u:System.Linq.Enumerable" :P

— ASCII-only

5

C (gcc), 39 bytes

f(n){for(;n;n&=n-1)printf("%d ",n&-n);}

— nwellnhof
source

5

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

— Jackson
source

Also works with input of 302231454903657293676544

— Michael Karas

5

Haskell, 29 bytes

(mapM(\n->[0,2^n])[7,6..0]!!)

— xnor
source

5

Ruby, 25 bytes

->n{8.times{|i|p n&2**i}}

— Reinstate Monica iamnotmaynard
source

5

C (clang), 133 110 63 58 bytes

58-byte solution thanks to @ceilingcat.

x=256;main(y){for(scanf("%d",&y);x/=2;)printf("%d ",y&x);}

— a stone arachnid
source

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.)

— Peter Cordes

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

— Peter Cordes

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

— Peter Cordes

44 bytes

— ceilingcat

4

MATL, 5 bytes

BPfqW

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]

— Luis Mendo
source

4

Perl 6, 16 12 bytes

-4 bytes thanks to Jonathan Allan

*+&2**all ^8

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

— Jo King
source

4

Japt, 8 5 bytes

Æ&2pX

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

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

— Shaggy
source

1

Nice one. You can do N&2pU with -mf to avoid the 0s

— Oliver

4

05AB1E, 9 bytes

Ýoʒ›}æʒOQ

This is also correct for 6-bytes, but it doesn't complete in time on TIO for 86:

05AB1E, 6 bytes

ÝoæʒOQ

— Magic Octopus Urn
source

1

Both your answers output an empty set for 15, instead of [1,2,4,8]

— Kevin Cruijssen

1

@KevinCruijssen needed 2**0, nice catch. Ý over L.

— Magic Octopus Urn

1

Ah, I know the feeling. Also had L instead of Ý at first in my answer.

— Kevin Cruijssen

4

Julia 0.6, 13 bytes

n->n&2.^(0:7)

— Kirill L.
source

4

CJam, 12 bytes

{:T{2\#T&}%}

— Esolanging Fruit
source

4

K (oK), 19 16 bytes

-3 bytes thanks to ngn!

{*/x#2}'&|(8#2)\

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)

— Galen Ivanov
source

you can omit the { x}

— ngn

@ngn Thanks! I tried it and it worked, but I wasn't sure it is acceptible.

— Galen Ivanov

4

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

Try it online! or Test every input!

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

— ASCII-only
source

4

PHP, 41 39 bytes

for($c=256;$c>>=1;)echo$argv[1]&$c,' ';

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

— 640KB
source

4

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

— t-clausen.dk
source

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

— MickyT

@MickyT that worked like a charm. Thanks a lot

— t-clausen.dk

3

Python 2, 43 40 bytes

f=lambda n,p=1:n/p*[1]and f(n,p*2)+[p&n]

— ovs
source

1

@JonathanAllan this definitely helped. Thanks for notifying me.

— ovs

1

...and the restriction has been lifted, so -1 byte :)

— Jonathan Allan

3

C# (Visual C# Interactive Compiler), 33 bytes

n=>{for(;n>0;n&=n-1)Print(n&-n);}

Port of @Arnauld's JavaScript (ES6) answer, so make sure to upvote him!

Explanation:

n=>{            // Method with integer parameter and no return-type
  for(;n>0      //  Continue looping as long as `n` is larger than 0:
      ;         //    After every iteration:
       n&=n-1)  //     Bitwise-AND `n` by `n-1`
    Print(      //   Print with trailing newline:
      n&-n);}   //    `n` bitwise-AND `-n`

— Kevin Cruijssen
source

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