Counting in binary nybbles

Question

This challenge is to output to your terminal, window, canvas or screen the numbers zero to 10 inclusive. Each outputted number must be shown as a 4-bit wide nybble, so zero must show as 0000 and so on.

You may separate each number outputted with a space, comma or carriage return. Smallest solution wins but the numbers can be displayed in any order you like as long as there are no repeating numbers in your sequence.

Entries in low-level binary languages need not worry about the comma or white space separators if it is not possible to output with commas or white spaces (i.e., the standard output is limited to binary only, or your solution is for an early computer kit such as the KIM-1 which has a limited digital display).

Answer 1

MATL, 6 bytes

0:10YB

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Explanation

0:10    % Create the array [0...10]
YB      % Convert this array to a binary string where each number is 
        % placed on a new row
        % Implicitly display the result
        

Answer 2

05AB1E, 9 8 bytes

T         # push 10
 4ã       # cartesian product repeat with 4
   R      # reverse list
    T>£   # take the first 11 elements of the list
      »   # join by newline and display

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Answer 3

JavaScript, 46 bytes

for(i=15;i++<26;)alert(i.toString(2).slice(1))

Why use a padding function when you can simply add 16 to each number and slice off the first binary digit?

Answer 4

Japt, 7 bytes

GôA,_¤Å

And here I was thinking Japt was doomed to be longer than every other golfing language...

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Explanation

GôA,_¤Å  // Implicit: A = 10, G = 16
GôA      // Create the inclusive range [G...G+A].
    _    // Map each item Z to Z
     ¤   //   .toString(2)
      Å  //   .slice(1).
         // Implicit: output result of last expression

Normally commas can be removed in Japt, but this one is there because of a bug: _ normally means function(Z){Z, but for some reason the compiler thinks A_ means function(A,Z){Z.

Answer 5

Bash + GNU utils, 26

• 4 bytes saved thanks to @Dennis

seq -w 0 1010|sed /[2-9]/d

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Answer 6

Bash + Unix utilities, 29 26 bytes

dc -e2o8927II^*8/p|fold -4

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This is the same length as @DigitalTrauma/@Dennis's solution, but uses a completely different method.

Output is:

1010
0010
0110
0001
1001
0101
0100
0111
0011
1000
0000

(Any order is allowed.)

---

Pure Bash, 34 bytes

echo 0{0,1}{0,1}{0,1} 10{00,01,10}

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Output is:

0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010

Answer 7

J, 6 bytes

#:i.11

Thanks to miles for cutting it down to 6 bytes!

Answer 8

Jelly, 7 bytes

2Bṗ4ṫ6Y

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(5 bytes if trailing lines of nybbles are allowed, 2Bṗ4Y)

How?

Prints in descending order.

2Bṗ4ṫ6Y - Main link, no arguments
2B      - 2 converted to binary -> [1,0]
  ṗ4    - Cartesian 4th power -> [[1,1,1,1], [1,1,1,0], ..., [0,0,0,0]]
                            i.e.  16       , 15         ..., 0
    ṫ6  - tail from 6th item  -> [[1,0,1,0], [1,0,0,1], ..., [0,0,0,0]]
                            i.e.  10       , 9        , ..., 0
      Y - join with line feeds
        - implicit print

An alternative 7-byter is 2ṗ4Ịṫ6Y, the [1,0] is replaced with [1,2] and Ị is the "is insignificant" monad (abs(z)<=1), converting 2s to 0s.

Answer 9

Python 3.6, 36 35 bytes

i=11
while i:i-=1;print(f"{i:04b}")

-1 byte thanks to @JonathanAllan

Python 3.5 and earlier:

i=11
while i:i-=1;print("{:04b}".format(i))

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Answer 10

PHP, 33 bytes

while($i<11)printf('%04b ',$i++);

Answer 11

CJam, 12 bytes

B{G+2b1>}%N*

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Explanation

The Cartesian power approach would have been my choice, but was already taken.

So this generates numbers from 0 to 10, and for each it adds 16 and converts to binary. Adding 16 ensures that the required leading zeros are produced, together with an extra leading one which is removed.

B             e# Push 11
 {      }%    e# Map over "11", implicitly converted to the array [0 1 ... 10]
  G+          e# Add 16. This makes sure there will be 5 binary digits: a leading 1
              e# which will have to be removed and the remaining, valid digits
    2b        e# Convert to array of binary digits
      1>      e# Remove first digit
          N*  e# Join by newlines. Implicitly converts arrays to strings

Answer 12

Intel 8080 machine code, MITS Altair 8800, 10 bytes

00 00 00 00 00 00 00 00 76

Listing:

0000   00   NOP      
0001   00   NOP      
0002   00   NOP      
0003   00   NOP      
0004   00   NOP      
0005   00   NOP      
0006   00   NOP      
0007   00   NOP      
0008   00   NOP      
0009   00   NOP      
000A   76   HLT 

Uses the CPU's program counter (PC, aka instruction pointer) to count from 0 to 10 then halt. The output display is conveniently in nibble format already.

Output:

Output is displayed on Blinkenlights A3-A0.

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Answer 13

MATLAB / Octave, 13 bytes

dec2bin(0:10)

Online Demo

Answer 14

Jelly, 10, 9, 8 bytes

⁴r26BḊ€Y

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I'm not that great at jelly, so I'd be open to any tips!

This uses Emigna's first algorithm

---

Thanks to Dennis for shaving off two bytes making me tie his own answer. :P

Explanation:

      Ḋ€    # Return all but the first element of each item in the list:
⁴r26        #   [16, 17, 18, ... 26]
     B      #   Converted to binary
        Y   # And joined with newlines

Answer 15

Jelly, 8 bytes

2Ḷṗ4ḣ11Y

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How it works

2Ḷṗ4ḣ11Y  Main link.

2Ḷ        Unlength 2; yield [0, 1].
  ṗ4      Take the fourth Cartesian power.
    ḣ11   Head 11; discard all but the first eleven elements.
       Y  Join, separating by linefeeds.

Answer 16

Python 2, 38 36 bytes

n=16;exec"print bin(n)[3:];n+=1;"*11

Thanks to @DJMcMayhem for golfing off 2 bytes!

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Answer 17

SmileBASIC, 26 bytes

FOR I=0TO&HA?BIN$(I,4)NEXT

Answer 18

RProgN, 15 Bytes

~16.aL1{2B26q}:

This has been a very good modivation to add a pad function. The entirety of ]L4\-'0'\m\., more than half the code, is to pad.

_Saved 6 bytes thanks to @ETHProductions, that's the pad function cut in half.

Explained

~16.aL1{2B26q}:
~               # Zero Space Segment
 16.            # The literal number 16
    aL          # The length of the Alphabet
      1         # The literal number 1
       {     }: # For each number between 16 and 26 inclusive
        2B      # Convert to base 2
          26q   # Get the characters between 2 and 6 inclusive.

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Answer 19

Retina, 36 33 bytes

%%%%
+`(^|\b)%
0$%'¶$%`1
11!`\d+

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Explanation

%%%%

Replace the empty (non-existent) input with %%%%.

+`(^|\b)%
0$%'¶$%`1

On the first run of this stage, it will match ^% and essentially replace the text %%%% with the two lines 0%%% and 1%%%. The stage will loop until the output stops changing. On the second run, it will match \b% (since digits count as word characters and % doesn't), and replace the groups by duplicating them and adding 0 to one copy and 1 to the other: 0%%% becomes the lines 00%% and 01%% (and the same sort of thing for 1%%%). Through this loop all 16 bitstrings will be produced, linefeed separated.

11!`\d+

The first 11 matches of \d+ (a run of at least 1 digit) are retrieved. The matches are output in a linefeed-separated list.

Answer 20

Ruby, 25 bytes

11.times{|n|puts"%04b"%n}

Answer 21

BF, 121 101 bytes

,....>,.<...+.>.<-..+.-.>.<..+..>.<-.+.-..>.<.+.-.+.>.<-.+..-.>.<.+...>.<.-...>.<+.-..+.>.<.-.+.-.!0

Requires a trailing newline. Makes use of ! symbol (so, check the box that says !) with this interpreter (try it online!).

Potentially 51 bytes if each operator was considered as 4 bits

Answer 22

C#, 96 bytes

---

Golfed

()=>{for(int i=0;i++<11;)System.Console.WriteLine(System.Convert.ToString(i,2).PadLeft(4,'0'));}

---

Ungolfed

() => {
    for( int i = 0; i++ < 1; )
        System.Console.WriteLine( System.Convert.ToString( i, 2 ).PadLeft( 4, '0' ) );
}

---

Full code

using System;

namespace Namespace {
    class Program {
        static void Main( string[] args ) {
            m();

            Console.ReadLine();
        }

        static void m() {
            for( Int32 i = 0; i++ < 11; )
                Console.WriteLine(
                    Convert.ToString( i, 2 ). // Converts the number to binary code
                    PadLeft( 4, '0' ) );      // Fills the number with the missing '0's
        }
    }
}

---

Releases

• v1.0 - 96 bytes - Initial solution.

Answer 23

C 170 120 bytes

n,i,m,k[4]={0};f(){for(m=0;m<=10;m++){n=m;i=0;for(;n;i++){k[i]=n;n/=2;}for(i=4;i>0;i--)printf("%d",k[i-1]%2);puts("");}}

Ungolfed version:

void f()
{
    int n,i,m,k[4]={0};


   for(m=0;m<=10;m++)
   {
      n=m;
      i=0;

      for(;n;i++)
      {
         k[i]=n;
         n/=2;
      }  
      for(i=4;i>0;i--)
         printf("%d",k[i-1]%2);

      puts("");        
   }
}

Can definitely be shortened!?

@Ahemone Awesome idea, Thanks!

Should work now! Try it online!

Answer 24

R - 23

We can use intToBin function from the R.utils package:

R.utils::intToBin(0:10)

[1] "0000" "0001" "0010" "0011" "0100" "0101" "0110" "0111" "1000" "1001" "1010"

Answer 25

C, 75 68 69 bytes

Approach 1: 75 73 74 bytes

m;p(i){putchar(i?m&i?49:48:9);}r(){for(m=11;m--;p(4),p(2),p(1),p(0))p(8);}

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---

Approach 2: 68 69 bytes

m,n,o;f(){for(m=11;m--;)for(n=m,o=5;o--;n*=2)putchar(o?n&8?49:48:9);}

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Answer 26

C, 68 bytes

f(){puts("0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010");}

Somehow, this is the shortest C answer so far…

EDIT: I was missing 1000 somehow. Well, it’s still winning.

Answer 27

Commodore VIC-20/VC-20/C64 - 95 69 68 67 tokenized BASIC bytes used (obfuscated and minimized)

0dEfnb(x)=sgn(xandb):fOi=0to10:b=8:fOj=0to3:?rI(stR(fnb(i)),1);:b=b/2:nE:?:nE

Copy and paste the above text into WinVice, for instance. Just about fits onto 80 characters on a Commodore 64.

Here's the non-obfuscated symbolic listing for illustrative purposes:

0 DEF FN B(X)=SGN(X AND B)
1 FOR I=0 TO 10
2  LET B=8
3  FOR J=0 TO 3
4   PRINT RIGHT$(STR$(FN B(I)),1);
5   LET B=B/2
6  NEXT J
7  PRINT
8 NEXT I

Of course this may be minimised and further obfuscated. Explanation:

The DEF FN command in line zero is a simple function which accepts one numeric parameter. This will give the algebraic sign the number passed to it as a parameter, in this case is ANDed with the value B (declared elsewhere, but becomes global once declared).

Line 1 starts the loop counter I

Line 2 sets the value of the 3rd BIT.

Line 3 starts loop counter J (we want bits 0 - 3 essentially).

Line 4 gets the STR$ value (string value) of the function FN B(I), this is then passed to the RIGHT$ command as CBM BASIC auto-pads outputted numbers with a white space, so we get the last character in the string created by STR$.

Line 5 moves the BIT counter down one (like x >> 1 in C I suppose).

Line 6 moves the J loop up one.

Line 7 prints a new line (line echo PHP_EOL; in PHP).

Line 8 moves the I loop up one.

I've 'nested' the listing in the example to better show the loop and loop order.

Answer 28

Python 2, 44 bytes

for x in range(11):print bin(x)[2:].zfill(4)

This uses the zfill function which works like rjust except it always padds with 0 so you don't waste bytes on an argument.

Answer 29

Pyke, 8 bytes

TFw0+b2t

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TFw0+b2t - for i in range(10):
  w0+    -    i+16
     b2  -   bin(^)
       t -  ^[:-1]

Also 8 bytes:

TF 4@b2t

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   4@    - set_bit(4, i)

Answer 30

Pyth - 8 7 bytes

<5^_`T4

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