# Binary representation of a number is palindrome or not?

Write a full program to find whether the binary representation of a number is palindrome or not?

Sample Input
5

Sample Output
YES


Print YES if binary representation is palindrome and NO otherwise.

• What should be the output when it's not a palindrome? Feb 8, 2011 at 15:20
• @dogbert It should be 'NO' without the quotes. Feb 8, 2011 at 15:22
• How do you know it's a palindrome? Because the values from the first nonzero to the end of the "string" are palindromic? This smells really bad to me, as a challenge. Feb 9, 2011 at 23:40
• Much as I <3 gnibbler's answer, it's not actually the shortest solution, and any question tagged [code-golf] should pick the shortest solution as the winner. Feb 10, 2011 at 3:48
• Input is given how?
– Joey
Mar 2, 2011 at 8:34

## Python - 46 chars

n=bin(input())[2:]
print'YNEOS'[n!=n[::-1]::2]

• Wow. What does [n!=n[::-1]::2] do? Feb 8, 2011 at 21:38
• @Dogbert, n[::-1] is a slice. The start and end indexs are empty, so it means the whole string. The stepsize is -1, so when you see [::-1] it is a short way to reverse a string/list etc. So n!=n[::-1] is True (ie 1) when n is not a palindrome. Therefore when n is a palindrome, you get 'YNEOS'[0::2] - start at 0 and take every 2nd character. When n is not a palindrome you get 'YNEOS'[1::2] - start at 1 and take every second character :) Feb 8, 2011 at 22:22
• I think people are voting for the slice trick :), rightly so. :P +1 Feb 9, 2011 at 6:08

## Golfscript -- 22 chars

~2base.-1%="YES""NO"if


# Vyxal, 85 bitsv1, 10.625 bytes

-6 bits thanks to @lyxal

-3 bits thanks to @the-thonnu

bḂ≠∨ȯno½iN


Try it Online!

b            # binary of input
Ḃ≠          # is not equal to its reverse (essentially: is not a palindrome?)
∨ȯno    # compressed string yesno
½   # split into two strings of equal length
i  # index that inequality (0/1) into that
N # case swapped to uppercase

• Try it Online! for 11.375 bytes May 10, 2023 at 5:16
• Or Try it Online! for 11.125 bytes May 10, 2023 at 5:17
• 10.625 bytes May 11, 2023 at 18:33

## Ruby, 41 39

$><<%w(YES NO)[(n="%b"%$*)<=>n.reverse]


Thanks to Michael Kohl's "%b"%gets trick.

• Very nice, I like this a lot! +1 for using the spaceship in a creative way :-) Mar 6, 2011 at 17:23

C 84 81 74 Characters

r;main(v,x){for(scanf("%d",&v),x=v;v;v/=2)r=r*2|v&1;puts(r-x?"NO":"YES");}


It does not use any function like string reverse.

• Couldn't you save a few characters changing r<<=1 into r*=2, v>>=1 into v/=2 and {} into ;?
– user56228
Sep 23, 2016 at 13:20
• @paxdiablo Indeed. Changed. Thanks a lot. Sep 24, 2016 at 7:48
• r*=2,r|=v&1 -> r=r*2|v&1 (-2) Sep 24, 2016 at 13:00
• and moving that term to the body of the loop saves another byte. Sep 24, 2016 at 13:01

## Javascript - 79 77 chars

alert((a=(prompt()*1).toString(2))-a.split("").reverse().join("")?"NO":"YES")


prompt()*1 : Quick trick to convert string to number.

.toString(2) : That's how you convert to binary in javascript.

a.split("").reverse().join("") : There is no native support to reverse string, so you have to convert string to array and array to string.

("[part1]" - "[part 2]")?"YES":"NO" : - is a replacement for != to save 1 char.

• Excellent explanation. Mar 3, 2011 at 1:44

# PHP - 41

<?=strrev($n=decbin(cat))==$n?@YES:@NO;


Test:

php 713.php <<< 5
YES
php 713.php <<< 6
NO

• If you're going to use shell calls to get the input, might as well use m4 instead of cat to save one. There's also pg and dd (which writes some bytes to stderr).
– Nabb
Feb 8, 2011 at 15:34
• Have You tried that on Windows? ;) Sep 24, 2016 at 13:19

# Perl, 45 characters

$_=sprintf'%b',shift; print reverse==$_?YES:NO


Ruby, 43 characters

puts((n="%b"%gets)==n.reverse ? "YES":"NO")

• Save 2: puts (n="%b"%gets)==n.reverse ? :YES: :NO Mar 2, 2011 at 16:15

('NO','YES')[($a=[Convert]::ToString("$input",2))-eq-join$a[64..0]]  # 05AB1E, 17 12 bytes ‘NO…Ü‘#EbÂQè  -5 bytes thanks to Adnan. Try it online! • Hey nice! I tried to golf it a bit and came to 12 bytes ‘NO…Ü‘#EbÂQè :). Sep 25, 2016 at 19:14 • Great! I still don't know how to use/make compressed strings. Also, I didn't know the function bin() existed Sep 25, 2016 at 19:47 • There is actually a detailed example here, if you are interested :). Sep 25, 2016 at 20:05 # JavaScript, 75 bytes 69 bytes alert((a=[...(+prompt()).toString(2)]).some(x=>x-a.pop())?"NO":"YES")  Try it online! A different approach from HoLyVieR's solution Explanation +prompt() : Converts string to number [...s] : converts a string s to an array of chars .some((x) => x - a.pop()) : checks whether there is one element in the array that does not equal the last element of the array Edit 69 chars, thanks to the suggestions by Shaggy • 69 bytes, or 55 bytes as a function. May 10, 2023 at 22:47 • 67 bytes, or 53 bytes as a function. May 10, 2023 at 23:11 • @Shaggy, thanks for the first one, I'll add it, as for the second one, I think it makes more sense to be suggested in HoLyVieR's answer May 11, 2023 at 12:35 # Arturo, 47 bytes print(=reverse<=as.binary do arg\0)?->'YES->'NO  ## Python (51) n=bin(input())[2:] print'YES'if n==n[::-1]else'NO'  • You can ['NO','YES'][n==n[::-1]] Nov 3, 2016 at 15:35 # Perl (73) No string reverse: print f(split//,sprintf'%b',shift); sub f{@_<=1?YES:shift!=pop()?NO:f(@_)}  # Perl (127) This one constructs all palindromes up to 2^32. sub f{ my($x,$l)=@_;$l+=2,f(($x<<$_)+1+(1<<$l-1),$l)?return 1:0 for 1..15-$l/2;$x-$ARGV[0]?0:1 } print f(0,1)+f(0,0)+f(1,1)?YES:NO  ## Bash, 55 chars C=dc<<<$1\ 2op;[ $C = rev<<<$C ]&&echo YES||echo NO

• Well, technically that's bash and dc and rev :-)
– user56228
Sep 23, 2016 at 13:26

## J - 33 characters

13 : ';(]-:|.)#:y{''YES'';''NO'''


## J: 24

((-:|.)#:x){2 3$'NO YES'  eg:  ((-:|.)#:5){2 3$'NO YES'
YES
((-:|.)#:12){2 3$'NO YES' NO ((-:|.)#:125){2 3$'NO YES'
NO
((-:|.)#:63){2 3$'NO YES' YES  ## Haskell (79) 0?k=n;n?k=div n 2?(nmod2+k*2);f x|x==x?0="YES"|True="No";main=interact$f.read

• Don't forget: In Haskell, this will work with really big numbers. Feb 8, 2011 at 21:13
• Ahm, that's actually 79 characters. ;-) Feb 8, 2011 at 22:49

## C (77 bytes)

r,t;main(n){for(t=n=atoi(gets(&n));n;r*=2,r|=n%2,n/=2);puts(r-t?"NO":"YES");}


TEST

## Pyth, 18 bytes

%2>"YNEOS"!qJ.BQ_J


Also 18 bytes:

@,"NO""YES"qJ.BQ_J


# PHP, not competing

I wanted to do it without using strings at all.

iterative solution, 78 bytes

for($x=log($n=$argv[1],2);$i<$x&($n>>$i^$n>>$x-$i^1);$i++);echo$i<$x/2?NO:YES;  recursive solution, 113 bytes function p($n,$x=0){return$n<2?$n:is_pal(($n&(1<<$x=log($n,2)/2)-1)^$n>>$x+!is_int($x));}echo p($argv[1])?YES:NO;


If n is a binary palindrome, the upper half xor the lower half is also a binary palindrome and vice versa.

a port of the excellent C answer from fR0DDY, 58 bytes

for($x=2*$v=$argv[1];$x/=2;$r=$r*2|$x&1);echo$r-$v?NO:YES;  a binary reverse. Columbus´ egg. # Jelly, 12 bytes (non-competing) BṚ⁼Bị“YES“NO  Try it online! Explanation: BṚ⁼Bị“YES“NO Main link. Arguments: z. B Binary representation of z. Ṛ Reversed. B Binary representation of z. ⁼ Check if x is equal to y. “YES“NO [['Y', 'E', 'S'], ['N', 'O']] ị xth element of y (1-indexed).  Before printing, Python's str function is mapped through a list, and then the elements are concatenated, so you see YES or NO. # Wolfram Language (Mathematica), 46 bytes If[PalindromeQ@IntegerDigits[#,2],"YES","NO"]&  Try it online! # Arn, 16 bytes ±î«Áýãf©¯tvf–É‰2  Try it! # Explained Unpacked: "YES"^(=\|:;b)||"NO. Man that yes/no required output really killed my byte count  "YES" String ^ Repeated ( Begin expression \ Fold with... = ...equality |: Bifurcate _ Variable initialized to STDIN; implied ;b Binary representation ) End expression || Boolean OR "NO String, ending quote implied  This works because empty strings are falsey. # Pyth, 16 bytes ?q_K.BQK"YES""NO  Try it online! ?q_K.BQK"YES""NO K.BQ // Assign binary input to K ?q_K K // Evaluate whether K and reversed K are equal "YES""NO // Ternary output.  # APL (Dyalog Unicode), 23 bytes (SBCS) 'NO' 'YES'⊃⍨≡∘⌽⍨2⊥⍣¯1⊢⎕  -3 bytes from Bubbler, after fitting the question requirements. Instead of performing an if-else like below, this program uses APL's representation of true as 1 and false as 0 to select from an array, where 'NO' is in the 0th index and 'YES' is in the first index. Try it online! # APL (Dyalog Unicode), 26 bytes (SBCS) {(⌽≡⊢)2(⊥⍣¯1)⍵:'YES'⋄'NO'}  ## Explanation {(⌽≡⊢)2(⊥⍣¯1)⍵:'YES'⋄'NO'} { } function wrapper ⍵ take the right argument 2(⊥⍣¯1) convert to base 2, and split into -1 groups(gets all digits) (⌽≡⊢) reverse equals right expression? :'YES'⋄'NO'} if the above is true, display 'YES', otherwise 'NO'  Try it online! • @petStorm Right, somehow I didn't notice that either. Adding ⎕ at the end solves the problem. Aug 28, 2020 at 0:24 • You can write a full program with test cases on TIO like this. Some golfing gives 24 bytes (which is the same length as tacit). Aug 28, 2020 at 0:32 • Oh, sorry about that. I'll fix it soon. Thanks for the help! Aug 28, 2020 at 1:58 • 23 bytes with (⌽≡⊢) → ≡∘⌽⍨ (because ≡∘⌽⍨x → x≡∘⌽x → x≡⌽x) Aug 28, 2020 at 2:38 • A second commutation. Got it. Aug 28, 2020 at 2:39 # Retina, 80 78 bytes Byte count assumes ISO 8859-1 encoding. .+$*
+(1+)\1
${1}0 01 1 ^((.)*?).??((?<-2>.)*$)
$1¶$3
O\$^.(?=.*¶)

^(.*)¶\1


Try it online

Convert to unary. Convert that to binary. Cut the number in half and remove a middle digit if there is one. Reverse the first half. Match if both halves are equal.

# Japt, 13 bytes

¢è¬?yu:"NO


Try it