คำเตือน: นี่ไม่ใช่ความท้าทาย "เฮ้ลองวาดเค้กด้วยศิลปะ ASCII"! โปรดอ่านต่อ);

บางครั้งที่ผ่านมามันเป็นวันเกิดของฉันฉันตอนนี้ 33

ดังนั้นจึงมีประเพณีทางสังคมที่น่าอึดอัดใจนี้ประกอบด้วยการเชิญครอบครัวและเพื่อน ๆ วางเทียนที่มีหมายเลขเหมือนบนเค้กร้องเพลงและของขวัญเปิด

   33   
--------

แทนที่จะใช้ตัวเลขฉันสามารถใช้ระบบเลขฐานสองในการวางเทียนมาตรฐานได้: ฉันวางไว้ 6 ชิ้นบนเค้กและจุดสองจุด

 100001
--------

ฉันเห็นได้ว่าทั้งเลขทศนิยมและเลขฐานสองในยุคของฉันนั้นช่างน่ากลัวมาก!

ท้าทาย

ฉันต้องการที่จะรู้ว่าหมายเลขอื่น ๆ สามารถวางบนเค้กด้วยเทียนและเป็น Palindromic ทศนิยมและไบนารี

เขียนโปรแกรม / ฟังก์ชั่นเพื่อทดสอบว่าตัวเลขเป็น palindromic ทั้งทศนิยมและไบนารี แต่เดี๋ยวก่อนมีอีกมากมาย: ในระบบเลขฐานสอง, เลขศูนย์นำหน้าสำหรับการทดสอบ!

อินพุต

ทศนิยม x จำนวนที่ผมอยากจะทดสอบว่ามันเป็นวันเกิดของ palindromic ด้วย 0 <x <2 ³² -1 (ใช่คนที่อยู่ในมิติของฉันอาศัยอยู่นานมาก)

เอาท์พุต

ความจริงถ้ามันตรงตามเงื่อนไขทั้งสองนี้ Falsey:

• การแทนค่าทศนิยมของตัวเลขคือ palindrome มาตรฐาน
• การแทนเลขฐานสองของจำนวนนั้นเป็นแบบมาตรฐานและการเพิ่มศูนย์นำหน้าอาจช่วยได้

กรณีทดสอบ

1 > 1 => Truthy
6 > 110 (0110) => Truthy
9 > 1001 => Truthy
10 > 1010 (01010) => Falsey, 10 is not palindromic
12 => 1100 (001100) => Falsey, 12 is not palindromic
13 => 1101 (...01101) => Falsey, neither 13 nor 1101 are palindromic
14 => 1110 (01110) => Falsey, 14 is not palindromic
33 > 100001 => Truthy
44 > 101100 (..0101100) => Falsey, 101100 is not palindromic
1342177280 > 1010000000000000000000000000000 (00000000000000000000000000001010000000000000000000000000000) => Falsey, 1342177280 is not palindromic (but the binary representation is)
297515792 > 10001101110111011101100010000 (000010001101110111011101100010000) => Truthy

กฎระเบียบ

• ช่องโหว่มาตรฐานไม่ได้รับอนุญาต
• อนุญาตให้มีการแปลงและการทดสอบห้องสมุดในตัว
• นี่คือโค้ดกอล์ฟซึ่งเป็นรหัสที่สั้นที่สุดที่จะชนะ!

ขอให้โชคดีและในที่สุดก็สุขสันต์วันเกิด!

05AB1E , 7 ไบต์

b0Ü‚DíQ

ลองออนไลน์! หรือเป็นTest Suite

คำอธิบาย

b         # convert input to binary
 0Ü       # remove trailing zeroes
   ‚      # pair with input
    D     # duplicate
     í    # reverse each (in the copy)
      Q   # check for equality

Python 3, 59 bytes

lambda a:all(c==c[::-1]for c in[str(a),bin(a).strip('0b')])

Try it online!

-3 bytes thanks to Rod
-3 bytes thanks to Connor Johnston

JavaScript (ES6), 65 bytes

Returns 0 or 1.

n=>(g=b=>[...s=n.toString(b)].reverse().join``==s)()&g(2,n/=n&-n)

How?

The helper function g() takes an integer b as input and tests whether n is a palindrome in base b. If b is not specified, it just converts n to a string before testing it.

We get rid of the trailing zeros in the binary representation of n by isolating the least significant 1 with n&-n and dividing n by the resulting quantity.

Fun fact: it's truthy for 0 because (0/0).toString(2) equals "NaN", which is a palindrome. (But 0 is not a valid input anyway.)

Test cases

let f =

n=>(g=b=>[...s=n.toString(b)].reverse().join``==s)()&g(2,n/=n&-n)

console.log(f(1 )) // Truthy
console.log(f(6 )) // Truthy
console.log(f(9 )) // Truthy
console.log(f(10)) // Falsey
console.log(f(12)) // Falsey
console.log(f(13)) // Falsey
console.log(f(14)) // Falsey
console.log(f(33)) // Truthy
console.log(f(44)) // Falsey

Mathematica, 52 49 bytes

i=IntegerReverse;i@#==#&&!i[#,2,Range@#]~FreeQ~#&

Try it on Wolfram Sandbox

Usage

f = (i=IntegerReverse;i@#==#&&!i[#,2,Range@#]~FreeQ~#&);

f[6]

True

f /@ {9, 14, 33, 44}

{True, False, True, False}

Explanation

i=IntegerReverse;i@#==#&&!i[#,2,Range@#]~FreeQ~#&

i=IntegerReverse                                   (* Set i to the integer reversing function. *)
                 i@#==#                            (* Check whether the input reversed is equal to input. *)
                       &&                          (* Logical AND *)
                          i[#,2,Range@#]           (* Generate the binary-reversed versions of input, whose lengths *)
                                                   (* (in binary) are `{1..<input>}` *) 
                                                   (* trim or pad 0s to match length *)
                                        ~FreeQ~#   (* Check whether the result is free of the original input *)
                         !                         (* Logical NOT *)

Version with builtin PalindromeQ

PalindromeQ@#&&!IntegerReverse[#,2,Range@#]~FreeQ~#&

Pyth - 13 bytes

&_I.s.BQ`Z_I`

Test Suite.

Japt, 14 bytes

s ꬩ¢w n2 ¤ê¬

Test it online!

Explanation

 s ê¬ © ¢   w n2 ¤  ê¬
Us êq &&Us2 w n2 s2 êq   Ungolfed
                         Implicit: U = input integer
Us êq                    Convert U to a string and check if it's a palindrome.
        Us2 w            Convert U to binary and reverse. 
              n2 s2      Convert to a number, then back to binary, to remove extra 0s.
                    êq   Check if this is a palindrome.
      &&                 Return whether both of these conditions were met.

Proton, 57 bytes

a=>(b=c=>c==c[to by-1])(str(a))*b(bin(a)[2to].strip('0'))

Try it online!

APL, 27 31 Bytes

∧/(⌽≡⊢)∘⍕¨{⍵,⊂{⍵/⍨∨\⍵}⌽2⊥⍣¯1⊢⍵}

How's it work? Using 6 as the argument...

      2⊥⍣¯1⊢6 ⍝ get the bit representation
1 1 0

      ⌽2⊥⍣¯1⊢6 ⍝ reverse it (if it's a palindrome, it doesn't matter)
0 1 1

      {⍵/⍨∨\⍵}⌽2⊥⍣¯1⊢6 ⍝ drop off the trailing (now leading 0's)
1 1

      6,⊂{⍵/⍨∨\⍵}⌽2⊥⍣¯1⊢6 ⍝ enclose and concatenate the bits to the original number
┌─┬───┐
│6│1 1│
└─┴───┘

      (⌽≡⊢)∘⍕ ⍝ is a composition of
      ⍕ ⍝ convert to string and 
      (⌽≡⊢) ⍝ palindrome test

      (⌽≡⊢)∘⍕¨6,⊂{⍵/⍨∨\⍵}⌽2⊥⍣¯1⊢6 ⍝ apply it to each of the original argument and the bit representation
  1 1

      ∧/(⌽≡⊢)∘⍕¨6,⊂{⍵/⍨∨\⍵}⌽2⊥⍣¯1⊢6  ⍝ ∧/ tests for all 1's (truth)
  1

Try it on TryAPL.org

Jelly, 8 bytes

Bt0ŒḂaŒḂ

Try it online!

Brachylog, 7 bytes

↔?ḃc↔.↔

Try it online!

That's a lot of ↔'s…

Explanation

With the implicit input and output, the code is: ?↔?ḃc↔.↔.

?↔?        The Input is a palindrome
   ḃ       Convert it to the list of its digits in binary
    c      Concatenate it into an integer
     ↔     Reverse it: this causes to remove the trailing 0's
      .↔.  The resulting number is also a palindrome

APL (Dyalog Classic), 26 bytes

{≡∘⌽⍨⍕⍵,⍵,⍨(<\⊂⊢)⌽2⊥⍣¯1⊢⍵}

Explanation

                  2⊥⍣¯1⊢⍵  encode ⍵ as binary
                 ⌽         reverse
           (<\⊂⊢)          partition from first 1
      ⍵,⍵,⍨                prepend and append ⍵
    ⍕                     turn into text string
≡∘⌽⍨                       match text with its reverse (f⍨X is XfX, where f is a composed function that reverses its right argument and matches with left)

Try it online!

Perl, 53 +3 (-pal) bytes

$_=sprintf"%b",$_;s/0+$//;$_="$_/@F"eq reverse"@F/$_"

try it online

Pyt, 10 bytes

Returns [1] if true, [0] if false

ĐɓƖ₫áĐ₫=ʁ∧

Try it online!

Explanation:

              Implicit input
Đ             Duplicate input
ɓ             Get input in binary (as string)
Ɩ             Cast to integer
₫             Reverse the digits (this removes any trailing zeroes)
á             Push the stack into a list
Đ             Duplicate the list
₫             Reverse the digits of each element of the list
=             Are the two lists equal element-wise
ʁ∧            Reduce the list by bitwise AND

Retina, 72 bytes

.+
$*_;$&
+`(_+)\1
$+0
0_
_
0+;
;
+`\b(\w)((\w*)\1)?\b
$3
(\B;\B)|.*
$.1

Try it online! Link includes test cases. Works by creating a unary duplicate of the original number, but using _s so that it's not confused by e.g. an input of 11. The unary number is then converted to "binary" and trailing zeros stripped. Palindromes then get successively truncated and the last stage tests whether there is anything left.

Mathematica, 70 bytes

(P=PalindromeQ)[PadRight[#~IntegerDigits~2,#~IntegerExponent~2]]&&P@#&

Husk, 14 bytes

¤&S=↔ȯ↓=0↔ḋ⁰d⁰

Try it online!

Ungolfed/Explanation

             d⁰  -- digits of input in decimal
          ḋ⁰)    -- digits of input in binary
         ↔       --   reverse
     (↓=0        --   and drop all leading zeros
¤&               -- test the two lists if
  S=↔            --   they're equal to their own reverse

Gaia, 10 bytes

ṙṭ@ḍ2⁻Πbṭ∧

Try it online!

Explanation

Instead of checking with leading zeroes in binary, I check without without the trailing zeros.

ṙ           String representation of number
 ṭ          Is palindromic?
  @         Push input again
   ḍ        Prime factors
    2⁻      Remove all 2s
      Π     Product
       b    Convert to binary
        ṭ   Is palindromic?
         ∧  Logical and

C (gcc), 105 bytes

r(n,b){int s=0,c=n;for(;n;n/=b)s=s*b+n%b;return s==c;}
c;f(n){for(c=n;c%2<1;c/=2);return r(n,10)&r(c,2);}

Try it online!

C# (.NET Core), 130 129 179 173 + 23 bytes

a few things, thank you to Ed Marty for pointing out that i need to check for as many 0's padded in front for a palindrome. And i need to make sure that i can check up to x^³²-1.

x=>{var a=Convert.ToString(x,2);var b=x+"";Func<string,bool>p=z=>z.SequenceEqual(z.Reverse());return new int[a.Length].Select((_,z)=>p(new string('0',z)+a)).Any(z=>z)&p(b);}

Try it online!

Python 2, 56 bytes

lambda n:all(s==s[::-1]for s in(`n`,bin(n).strip("0b")))

Try it online!

Uses Python's strip method to both remove the bin(..)'s output's leading 0b and the binary number's trailing zeros (as they will always have a matching bit).

Pyth, 25 22 19 18 17 bytes

-3 6 7 8 bytes by learning the language further

Ks_.Bsz&_IzqKs_`K

Explanation:

Ks        Set K to the integer version of...
 _.BsJ    Reverse string of the binary input
&         And
 _Iz      Is the input equal to the reverse of itself?
 qKs_`K   Is K equal to int(the reverse of basically string(K))

I am sure that this can be golfed down, I'll be working on that.

Test Suite

PHP, 69+1 bytes

$q=trim(decbin($p=$argn),0);if(strrev($p)==$p&&strrev($q)==$q)echo$p;

Run as pipe with -nR
Echoes the original input for truthy / nothing for falsey

Try it online!

Octave, 68 66 bytes

@(x)all([d=num2str(x) b=deblank(['' dec2bin(x)-48])]==flip([b d]))

Try it online!

Initial offering from Octave.

We basically create an array containing the number as a decimal string and the number as a binary string with trailing 0's removed. Then we create an array with the same to strings but with the binary and decimal numbers flipped. Finally both arrays are compared and the result is either true if they match (both palindromes) or false if they don't (one or both not palindromes).

• Save 2 bytes using flip instead of fliplr.

APL2 (not Dyalog), 36 bytes

(N≡⌽N←⍕N)^∨/(((⌽B)⍳1)↓B)⍷B←(32⍴2)⊤N←

First let B be the 32-bit representation of N:

B←(32⍴2)⊤N

Then mirror B and find the position of the 1st 1:

(⌽B)⍳1

Then drop that many positions from B. This will preserve the correct number of leading 0s.

Then perform FIND and an OR-REDUCTION to see if the cropped B contains its own mirror.

Now let's look at N, the decimal. The left-most bracketed expression converts N to a character vector and checks if it MATCHes its own mirror.

Finally, an AND joins the two checks.

In APL2 I can't make a neat lambda so I wrote a one-liner and included the assignment arrow. Hope this isn't cheating.

Java 8, 105 104 bytes

n->{String t=n+n.toString(n,2).replaceAll("0*$","")+n;return t.contains(new StringBuffer(t).reverse());}

Explanation:

Try it here.

n->{                         // Method with Integer parameter and boolean return-type
  String t=n                 //  Create a String `t` starting with the input Integer
    +n.toString(n,2)         //  Append the binary representation of the input Integer,
      .replaceAll("0*$","")  //   with all trailing zeroes removed
    +n;                      //  Append the input Integer again
  return t.contains(new StringBuffer(t).reverse());
                             //  Return true if `t` is a palindrome
}                            // End of method