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Convert from base 10 to base 2 without built-in base conversions

Code Golf Asked on October 27, 2021

Background:

You have been given an assignment to convert base 10 numbers to base 2 without using any premade base conversion functions. You can’t use any imported libraries either.

Problem:

Convert an input string from base 10 (decimal) to base 2 (binary). You may not use any premade base conversion code/functions/methods, or imported libraries. Since this is , the shortest answer in bytes will win.

Input will be anything from -32768 to 32767 (include sign byte handling in your code)

32 Answers

PostgreSQL, 184 bytes

WITH RECURSIVE t(a,b,c)AS(SELECT abs($1),0,0 UNION ALL SELECT a/2,a%2,1from t where a>0)select case when $1<0then '-'else ''end||string_agg(b::text,''order by a)from t where c>0OR $1=0

137 bytes if only handling positive numbers

WITH RECURSIVE t(a,b,c)AS(SELECT $1,0,0 UNION ALL SELECT a/2,a%2,1from t where a>0)select string_agg(b::text,''order by a)from t where c>0

Input is given as an integer query parameter (using the extended query protocol or by wrapping in a function) and output is given as a string.

Answered by Tim Anderson on October 27, 2021

Excel, 91 90 81 74

Trailing parens already discounted.

Compatibility note: CONCAT() was available only after later versions of 2016. It replaced CONCATENATE(). Tested in Excel online.

Formulae

  • A1 - Input
  • B1 - =A1+4^8*(A1<0) (13) - Convert to 16-bit unsigned equivalent.
  • B2 - =1+INT(LOG(B1,2)) (15) - Number of binary digits + 1

Code (46):

Convert B1 to binary.

=IF(A1=0,0,CONCAT(--ISODD(B1/2^(B2-SEQUENCE(B2)))))

Works in the full 16-bit range.

Answered by Calculuswhiz on October 27, 2021

Clojure 129 bytes#

Golfed version:

(defn p[n](loop[m 32768](if(> m 0)(do(if(= 0(bit-and m n))(print"0")(print"1"))(recur(unsigned-bit-shift-right m 1)))(println))))

Ungolfed:

(defn print-as-binary [n]
  (loop [mask  32768]
    (if (> mask 0)
      (do
        (if (= 0 (bit-and mask n))
          (print "0")
          (print "1"))
        (recur (unsigned-bit-shift-right mask 1)))
      (println))))

Answered by Bob Jarvis - Reinstate Monica on October 27, 2021

Perl 5, 26 bytes

Produces an additional leading 0 which can be avoided with two additional bytes (appending }{)

$=($_&1).$;redo if$_>>=1

Try it online!

Answered by Dom Hastings on October 27, 2021

///, 124 bytes

/~/////9/8*~8/7*~7/6*~6/5*~5/4*~4/3*~3/2*~2/1*~1/0*~*0/9*~0~/*/>Oⅼ~ⅼ>/ⅼ~ⅼO/Oⅼ~Oⅼⅼ/ⅼO~Oⅼ/_ⅼ~_~/>O/>~>~

Try it online!

Could be considered cheating, Outputs and O instead of 1 and 0, due to the mechanics of ///.

Uses examples from the wiki.

Answered by nph on October 27, 2021

APL(NARS), 17 chars, 34 bytes

{2∣⌊⍵÷2*(⍺-1)..0}

It is a copy and modify of Adam answer https://codegolf.stackexchange.com/a/90107 in the way one can add the parameter for the bits lenght, and ⎕IO for this function (here is ⎕IO=1) should have no importance...

  f←{2∣⌊⍵÷2*(⍺-1)..0}
  16 f 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 
  32 f 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 
  32 f ¯1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
  16 f ¯1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
  64 f ¯12345678
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 1 0 1 0 1 1 0 0 1 0 

it seeme easy handle the number of bits in this way (I cheked that last result should be right)

Answered by user58988 on October 27, 2021

><>,  34  33 bytes

|!/?=0:,2-}:%2:+***"  @"(0:
n{~{

Try it online!

Answered by Emigna on October 27, 2021

PowerShell, 43 bytes

param($n)15..0|%{$r+='01'[($n-shr$_)%2]}
$r

Try it online!

Answered by mazzy on October 27, 2021

PowerShell, 59 87 82 70 bytes

+28 bytes for supporting negative numbers.
-12 bytes thanks to @ASCII-only

param($d)$m=$d-lt0;while($d){$n="01"[$d%2]+$n;$d=($d-$d%2)/2}'-'*$m+$n

Try it online!

Adapted from this code. Takes input through a commandline parameter -d.

Answered by GMills on October 27, 2021

MATL, 15 17 bytes

t0<?16Ww+]`2&t]x

Try it on MATL Online

TIO

(+2 bytes removing leading 0 for negative numbers, sign bit should be the first bit.)

Output on MATL Online should be read bottom-up (MSB is at the bottom).

The main part is pretty simple: `2&t = while the value is greater than 0, divide by 2 and accumulate the remainders.

Handling negative numbers and giving them 2's complement representation was the tricky part. In the end I went with the "subtract from $ 2^N $" method of getting a number's two's complement. Since we're only required to handle values upto -32768, for negative numbers the code creates $ 2^{16} = 65536 $ with 16W, adds the input to that (eg. 65536 + (-42)), which gives something MATLAB sees as a positive number but represents the input's signed binary representation in 16-bit form.

Answered by sundar - Remember Monica on October 27, 2021

Kotlin, 82 bytes

{s:String->{var i=s.toInt()
var r=""
(0..15).map{r="${i and 1}$r"
i=i shr 1}
r}()}

Try it online!

Answered by JohnWells on October 27, 2021

C (gcc), 50 43 bytes

-7 bytes thanks to ceilingcat.

f(n){printf("-%d"+(~n?n&&f(n/2),1:0),n&1);}

Try it online!

Answered by gastropner on October 27, 2021

Small Basic, 133 bytes

A script that inputs from and outputs to the TextWindow console.

n=TextWindow.Read()
While n>0
c=c+1
x[c]=Math.Remainder(n,2)
n=Math.Floor(n/2)
EndWhile
For i=0To c-1
TextWindow.Write(x[c-i])
EndFor

Try it at SmallBasic.com Requires Silverlight and thus must be run in IE.

I/O is taken/given from the black console.

-22 bytes thanks to @Neil

Answered by Taylor Raine on October 27, 2021

Java 10, 80 71 57 bytes

n->{var r="";for(n=n<0?-n:n;n>1;n/=2)r=n%2+r;return n+r;}

-9 bytes due to a rule in the comments.. Negative base-10 inputs may return the positive/absolute base-2 value as output apparently.

Explanation:

Try it online.

n->{                   // Method with integer parameter and String return-type
  var r="";            //  Result-String, starting empty
  for(n=n<0?-n:n;      //  Transform the input to its absolute value
      n>1;             //  Loop as long as `n` is still >= 2:
      n/=2)            //    After every iteration: integer-divide `n` by 2
    r=n%2+r;           //   Prepend `n` modulo-2 to the result
  return n             //  Return the last `n`
         +r;           //  appended with the result

Answered by Kevin Cruijssen on October 27, 2021

Haskell, 66 bytes

c 0=0
c n=c(div n 2)*10+mod n 2
b('-':r)='-':b r
b r=show.c.read$r

Call with b "-1023", add main=interact b for a complete program or try it on Ideon.

c performs the conversion for positive integers.
b r=show.c.read$r converts a string to a number, applies c and converts back to string.
b('-':r)='-':b r strips a possible leading - and re-appends it to the result.

Answered by Laikoni on October 27, 2021

Dyalog APL, 11 bytes

2|⌊⎕÷2*⌽⍳16

2| The division remainder when halved of
the rounded down value of
the input
÷ divided by each of
2* two to the power of each of
⍳16 {0, 1, 2, ..., 15}

Requires ⎕IO←0 which is default on many systems.

TryAPL online!

Answered by Adám on October 27, 2021

Turing Machine Code, 272 bytes

As usual, I'm using the rule table syntax defined here. You can test it on that site or, alternatively, using this java implementation.

A lot of the code is copied from my decimal-to-hex converter here.

0 * * l B
B * * l C
C * 0 r D
D * * r E
E * * r A
A _ * l 1
A * * r *
1 0 9 l 1
1 1 0 l 2
1 2 1 l 2
1 3 2 l 2
1 4 3 l 2
1 5 4 l 2
1 6 5 l 2
1 7 6 l 2
1 8 7 l 2
1 9 8 l 2
1 _ * r Y
Y * * * X
X * _ r X
X _ _ * halt
2 * * l 2
2 _ _ l 3
3 * 1 r 4
3 1 0 l 3
4 * * r 4
4 _ _ r A

Counts down from the input in base 10 while counting up from 0 in base 2. On decrementing zero, it erases the input block and terminates.

Answered by SuperJedi224 on October 27, 2021

Apps Script + Google Sheets, 147 144 121 bytes

Script

function j(decNumb){var str='';do{str=String(decNumb%2)+str;decNumb=decNumb/2|0;}while(decNumb>=1);return parseInt(str);}

Sheet

=j(b1)

Modified version of this script by ZygD.

Answered by weatherman115 on October 27, 2021

Bash, 44

f=b+=n/2**e%2*10**e,2**e++/n?f=b:f;echo $[f]

Pass an input value to the script through the environment variable n. The decimal representation of the binary result cannot exceed LONG_MAX.

This should also be compatible with ksh93 and zsh if b and e are initialized to 0 and proper arithmetic expansion is used.

Answered by ormaaj on October 27, 2021

C# - 104

string p(int d){var r="";long i=1;while(r.Length<=64){var g=d&i;r=(g!=0)? "1"+r:"0"+r;i=i<<1;}return r;}

This method will convert decimal to binary up to 64 bits.

When executed the above method in Linqpad - rr = p(-32768); rr.Dump();

Output: 01111111111111111111111111111111111111111111111111000000000000000

Answered by Rajesh on October 27, 2021

Python 3.x: 65 characters

b=lambda n:n<2 and'01'[n]or b(n//2)+b(n%2);print(b(int(input())))

Answered by dan04 on October 27, 2021

Brainf*ck, 98 77

Obviously this isn't for the purpose of winning but what would a competition be if it didnt have a brainfk solution

++++[>++++<-]>>,<[->>++<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>[-]++++++[->++++++++<]>.[-]>[-<<<+>>>]<<<<]

Since brainfk can only handle 8bit integers and no negatives I guess it doesn't fully abide by the rules but hey I was never in it to win it.

This actually does work for 16-bit input if your interpreter supports

I even got it to output in ascii values

Here is the annotated code:

++[>++++<-]                       preload 8 onto cell 1
>>,<                                input into cell 2
[-                                  iterate over cell 1
    >>++<                               put 2 in cell 3
    [->-[>+>>]>[+[-<+>]>+>>]<<<<<]      division algorithm: converts {n d} into {0 d_minus_n%d n%d n/d}
    >[-]++++++[->++++++++<]>           clears cell 4 and puts 48(ascii of 0) into cell 5
    .[-]                                output n%2 and clear it (the bit)
    >[-<<<+>>>]                         bring n/2 into cell 2 (to be used for division in next iteration)
<<<<]                               end iterate

Shorter algorithm (77):

+>,>-<[>>[->]++[-<+]-<-]++++++++[->++++++<]>+[->+>+>+>+>+>+>+>+<<<<<<<<]>[.>]

This one can only handle 8bit integers.

The algorithm works by using a binary counter which is actually very short (one increment is >[->]++[-<+]-<- which then lays out the bits. The issue is that it is difficult to print out all of the bits

That last algorithm can be adapted to fit any number of bits at the expense of bytes. To be able to deal with N bit integers, it requires 53+3*N bytes to encode.

examples:

(1 bit) +>,>-<[>>[->]++[-<+]-<-]++++++++[->++++++<]>+[->+<]>[.>]
(2 bit) +>,>-<[>>[->]++[-<+]-<-]++++++++[->++++++<]>+[->+>+<<]>[.>]
(3 bit) +>,>-<[>>[->]++[-<+]-<-]++++++++[->++++++<]>+[->+>+>+<<<]>[.>]
etc

Answered by ASKASK on October 27, 2021

GolfScript - 17 bytes

~{.1&2/}16*;]-1%

Not too much more verbose than the built-in ~2base.

Answered by primo on October 27, 2021

Python - 61 60 characters

x=input();print"".join("01"[x>>i&1]for i in range(15,-1,-1))

Answered by C0deH4cker on October 27, 2021

C, 55 chars

Prints an extra leading zero (for the sake of 2 bytes).
Recursion within printf reverses the print order, so the algorithm extracts bits right-to-left but prints left-to-right.

EDIT: Saved a char by using putchar instead of printf.

f(x){(x*=x<0?-printf("-"):1)&&f(x/2);putchar(48+x%2);}

Answered by ugoren on October 27, 2021

Mandatory APL answer - 21 22

"01"[1+2|⌊⎕÷2⋆⊖0,⍳15]

Examples:

      "01"[1+2|⌊⎕÷2⋆⊖0,⍳15]
⎕: 0
0000000000000000
      "01"[1+2|⌊⎕÷2⋆⊖0,⍳15]
⎕: 13
0000000000001101
      "01"[1+2|⌊⎕÷2⋆⊖0,⍳15]
⎕: 9999
0010011100001111
      "01"[1+2|⌊⎕÷2⋆⊖0,⍳15]
⎕: -3
1111111111111101
      "01"[1+2|⌊⎕÷2⋆⊖0,⍳15]
⎕: 32767
0111111111111111

Answered by mniip on October 27, 2021

Javascript - 56 48 and 36 28 characters

  • Does not works with negative numbers.

Thanks to @Blender for shaving 8 characters.

This form takes input and shows output, 48 characters:

x=prompt();for(a="";x;x=~~(x/2))a=x%2+a;alert(a)

If just an instruction that puts in a variable a the binary form of a variable x is needed (and you don't bother in destroying the x value as a side-effect), here it is with 28 characters:

for(a="";x;x=~~(x/2))a=x%2+a

Answered by Victor Stafusa on October 27, 2021

JavaScript, 46

for(x=prompt(o='');x;x>>>=1)o=(x&1)+o;alert(o)

Answered by copy on October 27, 2021

Smalltalk (Smalltalk/X), 63/78

the first version creates an intermediate string (78):

t:=Number readFrom:Stdin.
((15to:1by:-1)collect:[:i|$0+(t>>i&1)]as:String)print

actually, there is no need to create the string; just output the chars (63):

t:=Number readFrom:Stdin.
15to:1by:-1 do:[:i|($0+(t>>i&1))print]

mhmh - is there a shorter way to read to a number?

Answered by blabla999 on October 27, 2021

Perl, 44

This is my first Perl program ever, so please forgive me if this can be easily golfed down further. Edit: Thank you @primo for taking 7 chars away from my answer.

$x=<>;do{@s=($x&1,@s)}while($x>>=1);print@s

$x=<>;do{push@s,$x&1}while($x>>=1);print reverse@s

The logic is essentially the same as my previous C solution.

Also, uses 64 bits.

Answered by user12205 on October 27, 2021

Javascript 59

o='';i=parseInt(prompt());do{o=(i&1)+o}while(i>>=1)alert(o)

Answered by Michael M. on October 27, 2021

C, 81

char b[17];i=15;main(x){scanf("%d",&x);while(i+1)b[i--]=(x&1)+48,x>>=1;puts(b);}

The output has strictly 16 bits (including padding zeros)

Answered by user12205 on October 27, 2021

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