Numbers with Rotational Symmetry

Question

Given an integer, output a truthy value if it is the same upside-down (rotated 180°) or a falsy value otherwise.

0, 1, and 8 have rotational symmetry. 6 becomes 9 and vice versa.

Sequence of numbers producing truthy results: OEIS A000787

0, 1, 8, 11, 69, 88, 96, 101, 111, 181, 609, 619, 689, 808, 818, 888, 906, 916, 986, 1001, 1111, 1691, 1881, 1961, 6009, 6119, 6699, 6889, 6969, 8008, 8118, 8698, 8888, 8968, 9006, 9116, 9696, 9886, 9966, 10001, 10101, 10801, 11011, 11111, 11811, 16091, ...

This question is inspired by my own reputation at the time of posting: 6009.

Answer 1

05AB1E, 22 16 15 14 bytes

Code:

Â23457ð«-69‡Q

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---

Previous code:

Â69‡Q¹¹„vd•ÃQ*

To figure out if the string is rotational symmetric, we just need to transliterate 69 with 96, reverse the string and check if they are equal. The other thing we need to know is if the number only contains the digits 0, 1, 8, 6 and 9. So that's exactly what we are going to do:

                     # Bifurcate the input, which pushes input and input[::-1]
 69Â                  # Bifurcate 69, which pushes 69 and 96.
    ‡                 # Transliterate 69 with 96 in the input, e.g. 1299 becomes 1266.
     Q                # Check for equality.
      ¹¹              # Push input again twice.
        „vd•          # Compressed version for 10869.

The „vd• part actually converts the string vd from base 190 to base 10. You can try this out here.

            Ã        # Keep the characters of the second string in the first string.
                       667788 would become 6688 (since 7 is not in 10869).
             Q       # Check for equality.
              *      # Multiply the top two numbers, which actually does an AND function.

Uses CP-1252 encoding. Try it online!

Answer 2

Python 2, 50 bytes

lambda n:`n`==`map('01xxxx9x86'.find,`n`)`[-2::-3]

The method '01xxxx9x86'.find takes a digit character to it's upside-down number, with any unflippable digit giving -1. This function is mapped to the reversed number string, producing a list of digits.

This is converted to a string with the [1::3] trick, except it's reversed instead by doing [-2::-3] (thanks to Dennis for this, saving 4 bytes), and compared to the original number string. Any -1's from unflippable digits will misalign the conversion, making it fail.

---

56 bytes:

lambda n:`n`[::-1]==`n`.translate('01xxxx9x86______'*16)

Checks if the number string reversed is the same as it with the upside-down replacements. Digits that can't be flipped are replaced with 'x' to always give the wrong answer.

The replacement is done with translate on a string of 256 chars, replacing the corresponding ASCII values. Only the 10 values 48 to 57 matter, but I padded to length 16 to make the total length be 256. I wonder if there's a shorter way.

Some other approaches (lengths 59, 60, 60):

lambda n:set(zip(`n`,`n`[::-1]))<=set(zip('01896','01869'))
r=lambda s:set(zip(s,s[::-1]));lambda n:r(`n`)<=r('9018106')
lambda n:all(a+b in'001188969'for a,b in zip(`n`,`n`[::-1]))

Answer 3

Ruby, 54 46 bytes

 ->a{(b=a.to_s).tr('1-9','1w-z9x86').reverse==b}

I don't know, is anonymous functions like that allowed or not

Basically same idea as Python2 answer. If input is not integer, act bad (i.e. aba gives true)

Answer 4

JavaScript (ES6), 56 bytes

n=>n==[...''+n].reverse().map(c=>'0100009086'[c]).join``

Answer 5

Perl, 29 26 bytes

Includes +1 for -p

Run with the input on STDIN:

rotation.pl <<< 69

rotation.pl:

#!/usr/bin/perl -p
$_=reverse==y/962-7/69a/r

Answer 6

Jelly, 16 15 bytes

,ȷ9+90860¤Dị/⁼Ṛ

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How it works

,ȷ9+90860¤Dị/⁼Ṛ  Main link. Argument: n (integer)

         ¤       Evaluate the three links to the left as a niladic chain:
 ȷ9              Yield 1000000000.
    90860        Yield 90860.
   +             Add to yield 1000090860.
,                Pair; yield [n, 1000090860].
          D      Convert both integers to base 10.
           ị/    Reduce by index, i.e., for each digit of n, retrieve the element
                 of [1, 0, 0, 0, 0, 9, 0, 8, 6, 0] at that index (1-based).
              Ṛ  Yield the digits of n in reversed order.
             ⁼   Test for equality.

Answer 7

Retina, 57 49 bytes

8 bytes saved thanks to @Martin Büttner.

+`^(([018])(.*)\2|9(.*)6|6(.*)9)$
$3$4$5
^[018]?$

• Apply reduction as such: 1610880191 -> 61088019 -> 108801 -> 0880 -> 88 -> (empty).
• Returns 1 if only 0, 1, 8, or (empty) is left.
• Returns 0 otherwise.

Try it online!

Answer 8

Retina, 40 38 33 bytes

$
;$_
T`92-7`69`;.+
+`(.);\1
;
^;

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Explanation

This uses a completely different approach from the other Retina answer. Instead of removing all symmetric parts, we simply perform the transformation of reversing the string and swapping 6 and 9s and then compare for equality. To make sure that no non-symmetric digits appear we also turn them into 9s in one half.

$
;$_

We duplicate the input by matching the end of the string and inserting ; followed by the entire input.

T`92-7`69`;.+

This performs a character transliteration only on the second half by matching it with ;.+. The two transliteration sets expand to:

9234567
6999999

Because 2-7 denotes a range and the target set is padded with the last character to match the length of the source set. Hence, the stage swaps 6 and 9 and also turns all of 23457 into 9s.

+`(.);\1
;

Repeatedly (+) remove a pair of identical characters around the ;. This will either continue until only the ; is left or until the two characters around the ; are no longer identical, which would mean that the strings aren't the reverse of each other.

^;

Check whether the first character is ; and print 0 or 1 accordingly.

Answer 9

sh, 40 33 bytes

[ `rev<<<$1|tr 6923457 96` = $1 ]

Input via command line argument, output via exit code. Generate all testcases:

for i in `seq 0 10000`
do
  if [ `rev<<<$i|tr 6923457 96` = $i ]
  then
    echo $i
  fi
done

Answer 10

TSQL, 122 bytes

I am new to code golf, so not quite sure how to count the characters. Counting it as 1 here since the number used in this case is 8

This will return 1 when the reversed value match and nothing when it doesn't match:

SELECT 1FROM(values(8))x(x)WHERE(x)NOT like'%[23457]%'and REPLACE(REPLACE(x,6,4),9,6)=REVERSE(REPLACE(REPLACE(x,9,4),9,6))

Human readable:

SELECT 1
FROM
  (values(808))x(x)
WHERE(x)
  NOT like'%[23457]%'
  and REPLACE(REPLACE(x,6,4),9,6)=REVERSE(REPLACE(REPLACE(x,9,4),9,6))

Answer 11

GNU sed, 84 bytes

(including +1 for -r flag)

:
s/^6(.*)9$/\1/
s/^9(.*)6$/\1/
s/^([081])(.*)\1$/\2/
t
s/^[081]$//
s/.+/0/
s/^$/1/

If the line ends with a rotation of its beginning character, trim both ends by one. Repeat until there's no match. Account for a single symmetric character, then if anything remains, the input was not symmetric, and we return false; otherwise return true.

Answer 12

Pyth - 21 bytes

&!-z+`180K`69qzX_z_KK

Test Suite.

Answer 13

Pyth, 17 bytes

!-FmC_B`d,QC\

Test it in the Pyth Compiler.

How it works

!-FmC_B`d,QC\  (implicit) Store the input in Q.

            \  Yield the Unicode character with code point 1068901.
           C    Compute its code point.
         ,Q     Pair the input and 1068901.
   m            Map; for each d in [Q, 1068901]:
       `d         Yield the string representation of d.
     _B           Yield the pair of `d` and `d` reversed.
    C             Zip. For the second string, this gives
                  ['11', '00', '69', '88', '96', '00', '11'].
 -F             Fold by difference, i.e., removes all pairs in the second list
                from the first list.
!               Logically negate the result, returning True iff the list is empty.

Answer 14

Visual Basic for Applications, 150 111 bytes

Usable in console or as an UDF.

Function c(b)
c=(Not b Like"*[!01869]*")And b=Replace(Replace(Replace(StrReverse(b),9,7),6,9),7,6)
End Function

Improved by taking advantage of implicit type conversions and by doing three steps swapping instead of two steps on each side of the equation. Count includes Function and End Function statements.

Answer 15

C, 82 bytes

char*s="0100009086";f(x){int y=0,z=x;for(;x;x/=10)y=y*10+s[x%10]-'0';return z==y;}

Expanded

char *s = "0100009086";
int f(int x)
{
    int y=0, z=x;
    for(; x; x/=10)
        y = y*10 + s[x%10]-'0';
    return z==y;
}

Explanation

We reverse the digits of x using modulo-10 arithmetic, replacing 6 and 9 by their reflections as we go. We replace the rotationally-asymmetric digits by zeros (note that we can handle symmetric fives and/or twos by simply changing the replacement table s). If the new number is equal to the original (saved in 'z'), then it is rotationally symmetric.

Test program

#include <stdio.h>
int main()
{
    int i;
    for(i=0;  i <= 16091; ++i)
        if (f(i))
            printf("%d\n", i);
}

This prints the list of symmetric numbers given in the question.

Answer 16

MATL, 25 21 22 bytes

j69801VmAGtP69VtPXE=vA

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Explanation

j       % Explicitly grab input as string
69801   % Number literal
V       % Convert to a string
mA      % Check if all input chars are members of this list
G       % Explicitly grab the input
tP      % Duplicate and flip it
69      % Number literal
V       % Convert to string ('69')
tP      % Duplicate and flip it ('96')
XE      % Replace all '6' with '9', and '9' with '6'
=       % Check that this substituted string is equal to the original
v       % Vertically concatenate all items on the stack
A       % Ensure everything is true
        % Implicitly print boolean result

Answer 17

Seriously, 23 bytes

,$;`"01xxxx9x86"í`MRεj=

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This is essentially a port of xnor's Python 2 solution.

Explanation:

,$;`"01xxxx9x86"í`MRεj=
,$;                      push two copies of str(input)
   `"01xxxx9x86"í`M      map: get each digit's rotated digit (or x if not rotatable)
                   Rεj   reverse and join on empty string to make a string
                      =  compare equality with original input

Answer 18

Kotlin, 69 bytes

This takes the number, converts it to a string, rotates it, and then compares it with the original as a string for equality. Non-rotatable digits are simply converted to 0

{i->"$i".map{"0100009086"[it-'0']}.joinToString("").reversed()=="$i"}

Test it Here!