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 code-golf, the shortest answer in bytes will win.
Input will be anything from -32768 to 32767 (include sign byte handling in your code)
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
Trailing parens already discounted.
Compatibility note: CONCAT()
was available only after later versions of 2016. It replaced CONCATENATE()
. Tested in Excel online.
A1
- InputB1
- =A1+4^8*(A1<0)
(13) - Convert to 16-bit unsigned equivalent.B2
- =1+INT(LOG(B1,2))
(15) - Number of binary digits + 1Convert 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
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
Produces an additional leading 0
which can be avoided with two additional bytes (appending }{
)
$=($_&1).$;redo if$_>>=1
Answered by Dom Hastings on October 27, 2021
/~/////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/>~>~
Could be considered cheating, Outputs ⅼ
and O
instead of 1
and 0
, due to the mechanics of ///
.
Uses examples from the wiki.
Answered by nph on October 27, 2021
{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
Answered by Emigna on October 27, 2021
Answered by mazzy on October 27, 2021
+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
Adapted from this code. Takes input through a commandline parameter -d
.
Answered by GMills on October 27, 2021
t0<?16Ww+]`2&t]x
(+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
{s:String->{var i=s.toInt()
var r=""
(0..15).map{r="${i and 1}$r"
i=i shr 1}
r}()}
Answered by JohnWells on October 27, 2021
Answered by gastropner on October 27, 2021
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
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:
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
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
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.
Answered by Adám on October 27, 2021
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
function j(decNumb){var str='';do{str=String(decNumb%2)+str;decNumb=decNumb/2|0;}while(decNumb>=1);return parseInt(str);}
=j(b1)
Modified version of this script by ZygD.
Answered by weatherman115 on October 27, 2021
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
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
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
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
~{.1&2/}16*;]-1%
Not too much more verbose than the built-in ~2base
.
Answered by primo on October 27, 2021
x=input();print"".join("01"[x>>i&1]for i in range(15,-1,-1))
Answered by C0deH4cker on October 27, 2021
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
"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
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
for(x=prompt(o='');x;x>>>=1)o=(x&1)+o;alert(o)
Answered by copy on October 27, 2021
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
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
o='';i=parseInt(prompt());do{o=(i&1)+o}while(i>>=1)alert(o)
Answered by Michael M. on October 27, 2021
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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