Strobogrammatic Numbers [duplicate]

Question

Strobogrammatic Numbers

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Definition

A number which is rotationally symmetrical, i.e., it'll appear the same when rotated by 180 deg in the plane of your screen. The following figure illustrates it better,

(source: w3resource.com)

Task

Given a number as the input, determine if it's strobogrammatic or not.

Examples

• Truthy

1
8
0
69
96
69169
1001
666999
888888
101010101

• Falsey

2
3
4
5
7
666
969
1000
88881888
969696969

Rules

• The number is guaranteed to be less than a billion.
• We are considering 1 in it's roman numeral format for the sake of the challenge.
• Input can be taken as number, or an array of chars, or as string.
• Output would be a truthy/falsey value. However, you are allowed to flip the output values.
• This is a code-golf, so fewest bytes will win!

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Sandbox link

Answer 1

Perl 5 -pa, 48 34 bytes

@Arnauld had a better idea in his JavaScript answer than my first one in Perl, so I stole it.

y/0-9/01aaaa9a86/;$_="@F"==reverse

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Answer 2

05AB1E, 12 11 bytes

•yΘOñ•I<èRQ

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-1 byte thanks to Razetime's suggestion

Takes the number as a list of digits

Explained

•yΘOñ•       # Push the number 1000090860
      I<     # Decrement the input list
        èR   # And get each number when rotated 180 degrees, reversing it
          Q  # Check if that equals the input

Answer 3

Ruby -n, 30 bytes

p$_.==$_.reverse.tr'92-7','69'

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Answer 4

Node.js 11.6.0,  34  33 bytes

Expects an array of digits. Returns false for strobogrammatic, or true otherwise.

n=>n.some(c=>n.pop()-c**66%79%63)

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How?

Taking the precision errors on large integers into account¹, we look for a function:

$$f(x)=(x^p\bmod m_0)\bmod m_1$$

such that:

$$f(0)=0,\:f(1)=1,\:f(6)=9,\:f(8)=8,\:f(9)=6$$

and:

$$f(x)>9\text{ for }x\in\{2,3,4,5,7\}$$

_{1: These errors actually depend on the JS engine. This is guaranteed to work with Node 11.6.0.}

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JavaScript (ES6), 38 bytes

Expects an array of digits. Returns false for strobogrammatic, or true otherwise.

n=>n.some(c=>n.pop()!="01----9-86"[c])

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Answer 5

Python 3, 53 bytes

lambda n:["01....9.86"[int(c)]for c in n[::-1]]==[*n]

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