Test a number for narcissism

Question

A Narcissistic Number is a number which is the sum of its own digits, each raised to the power of the number of digits.

For example, take \$153\$ (3 digits):

\$1^3 + 5^3 + 3^3 = 153\$

\$1634\$:

\$1^4 + 6^4 + 3^4 + 4^4 = 1634 = 1 + 1296 + 81 + 256 = 1634\$

The Challenge:

Your code must take input from the user and output True or False depending upon whether the given number is a Narcissistic Number.

Error checking for text strings or other invalid inputs is not required. 1 or 0 for the output is acceptable. Code that simply generates a list of Narcissistic Numbers, or checks the user input against a hardcoded list, does not qualify.

OEIS A005188

Answer 1

JavaScript (ES6), 56 bytes

eval(`${n=prompt()}0`.split``.join(`**${n.length}+`))==n

Answer 2

Perl 6, 30 bytes

{$_==sum .comb.map(* **.comb)}

Pretty straightforward. The chars method would be more typically used to get the number of characters in the input string, but comb returns those characters as a list, which evaluates to the number of characters in numeric context, and saves us a byte.

Answer 3

Pushy, 9 bytes

Note that this answer is non-competing as Pushy postdates the challenge. However, I thought I'd post it because it's interestingly short for a simple stack-based language:

VsLKeS^=#

Try it online!

It works like this (how the two stacks would look for the example input is shown on the right):

    \ Implicit input, for example 1634.  [1634], []
V   \ Copy into second stack.            [1634], [1634]
s   \ Split into individual digits       [1,6,3,4], [1634]
L   \ Push stack length                  [1,6,3,4,4], [1634]
Ke  \ Raise all to this power            [1,1296,81,256], [1634]
S   \ Take sum                           [..., 1634], [1634]
^=  \ Check equality with initial input  [..., True], []
#   \ Output this boolean (as 0/1)     

Answer 4

Bash, 54 bytes

echo $[ `printf "+%c**${#1}" $(fold -w1<<<$1)` == $1 ]

Try it online!

Takes input as parameter, outputs 1 if is narcissist, 0 if not.

fold prints each digit one per line, printf adds +<digit>**<length> forming the arithmetic expression which is evaluated and compared to the original input.

Answer 5

Python 2, 44 bytes

I think this is the shortest you can get for Python. Uses a lambda rather than a full program.

lambda s:sum(int(x)**len(`s`)for x in`s`)==s

Try it online!

Based off @danmcardle's answer.

Original answer, 51 bytes

lambda s:sum(map(lambda x:int(x)**len(`s`),`s`))==s

Try it online!

Answer 6

Perl 5, 24 bytes

21 bytes of code + 3 for -pF flags

for$i(@F){$_-=$i**@F}

Try it online!

Answer 7

Swift 3.2: 107 characters

Of course Swift is absolutely not a quick language but I thought I would try:

func n(s:String){let c=s.utf8.map{Double($0)-48};print(c.reduce(0){$0+pow($1,Double(c.count))}==Double(s))}

Answer 8

JavaScript 46 bytes

F=n=>![...n].reduce((x,c)=>x-(+c)**n.length,n)

console.log(F("153") == true)
console.log(F("370") == true)
console.log(F("371") == true)
console.log(F("152") == false)
console.log(F("150") == false)

Answer 9

JavaScript 7, 41 Bytes

s=>[...s].map(x=>N+=x**s.length,N=0)|N==s

Answer 10

Pyt, 11 bytes

ĐĐḶ⌊⁺⇹ą⇹^Ʃ=

Explanation:

                      Implicit input
ĐĐ                    Duplicate the input twice
   Ḷ⌊⁺                Get number of digits in the input
      ⇹ą              Convert input to array of digits
        ⇹^Ʃ           sum the digits raised to the power of the number of digits
           =          equals input?

Try it online!

Answer 11

CJam, 12 bytes

{_Ab_,f#:+=}

Try it online!

Explanation:

{          }   anonymous block, input: 1634
   b           get digits in base
  A              10: [1 6 3 4]
     ,         take the length: 4
       #       and raise
      f          each element
    _              of the list of digits
               to that power: [1 1296 81 256]
        :      fold over
         +       addition: 1634
          =    and compare
 _               to the original: 1 (true)

Answer 12

JavaScript (ES7), 43 38 bytes

Takes input as a string.

n=>n==eval([...n,0].join`**n.length+`)

---

Try It

f=
n=>n==eval([...n,0].join`**n.length+`)
o.innerText=f(i.value="153")
oninput=_=>o.innerText=f(i.value)

<input id=i type=number><pre id=o>

Answer 13

Kotlin, 60 bytes

map{Math.pow(it-'0'+0.0,length+0.0)}.sum()==this.toInt()+0.0

Beautified

map {
    Math.pow(it - '0' + 0.0, length + 0.0)
}.sum() == this.toInt() + 0.0

Test

fun String.f() =
map{Math.pow(it-'0'+0.0,length+0.0)}.sum()==this.toInt()+0.0

fun main(args: Array<String>) {
    (0..1000).filter { it.toString().f() }.forEach { println(it) }
}

TIO

TryItOnline

Answer 14

Javascript (ES6) 46 bytes - Not competing

n=>(s=[...''+n]).forEach(v=>n-=v**s.length)|!n

Explanation:

n=>
(s=[...''+n])        // Convert input to array of chars and assign to s
  .forEach(v=>       // Loop over s (returns undefined)
    n-=v**s.length)  // reduce input by char^length
  |                  // undefined is falsy, so we OR
  !n                 // OR with negated input 
                     // returns 1 if 0, 0 otherwise

Answer 15

Tcl, 77 bytes

proc N n {expr [join [lmap d [split $n ""] {expr $d**[string le $n]}] +]==$n}

Try it online!

Answer 16

Excel, 44 bytes

=A1=SUM(MID(A1,SEQUENCE(LEN(A1)),1)^LEN(A1))

Link to Spreadsheet

Answer 17

Pari/GP, 30 bytes

n->[n-=d^#s|d<-s=digits(n)];!n

Try it online!

Answer 18

Pyth, 10 bytes

qsm^sdl`Q`

Verify the test cases.

Answer 19

Thunno 2, 6 bytes

dDl*S=

Attempt This Online!

Explanation

dDl*S=  # Implicit input     ->  153
d       # Cast to digits     ->  [1, 5, 3]
 Dl     # Push the length    ->  [1, 5, 3]  3
   *    # Exponentiate       ->  [1, 125, 27]
    S   # Sum the list       ->  153
     =  # Equals the input?  ->  1
        # Implicit output