# Test a number for narcissism

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):

13 + 53 + 33 = 1 + 125 + 27 = 153

1634:

14 + 64 + 34 + 44 = 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 list, does not qualify.

OEIS A005188

• Is it ok if I output True if it's such a number, but anything else (in this case the number itself) if not? – devRicher Jan 6 '17 at 21:51

## APL (15)

∆≡⍕+/(⍎¨∆)*⍴∆←⍞

Outputs 1 if true and 0 if false.

Explanation:

• ∆←⍞: read a line (as characters), store in
• (⍎¨∆)*⍴∆: evaluate each character in and raise it to the power ⍴∆
• ∆≡⍕+/: see if the input equals the string representation of the sum of these
• what did i just read – Jbwilliams1 Feb 18 '14 at 20:13
• @LagWagon God's language – tomsmeding Jul 8 '14 at 8:00

# Brachylog, 10 bytes

lg;?↔z^ᵐ+?

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The predicate succeeds if the input is an Armstrong number and fails if it is not. If run as a full program, success prints true. and failure prints false.

lg;?↔z     A list of pairs [a digit of input, the length of input].
^ᵐ   A list of numbers where each is a digit of the input raised to the power of its length.
+  The sum of those numbers.
? Attempt to unify that sum with the input.
• Originally, this was nine bytes, but I realized that if I gave it a number more than nine digits long it would stop acting correctly, as previously I hadn't put g in there and it worked by virtue of single-digit lengths being length-1 sequences of themselves. – Unrelated String Mar 1 at 6:30

# Tcl, 77 bytes

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

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# Retina, 118 bytes

Byte count assumes ISO 8859-1 encoding.

.*
$0$0¶$0 (?<=^\d*)\d x (?=.*$)
¶
0¶

\d+
$* ¶(1+) ¶$1 $1 {^x ? }s(?<=x.*)1(?=1* (1+))$1
1*

1¶1
11
^(1*)¶+\1\b

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Explanation

.*                          # Copy number 3 times. For Length, Unary, and Digits
$0$0¶$0 (?<=^\d*)\d # Convert first copy to x's (Length) x (?=.*$)                     # Split up digits of last copy, each on their own line
¶
0¶                          # Remove zeros, because they leave blank lines

\d+                         # Convert to unary
$* ¶(1+) # Duplicate each separated digit ¶$1 $1 {^x ? # While x's exist, remove an x ... }s(?<=x.*)1(?=1* (1+)) # and multiply each value by the digit (nth power)$1
1*                         # Remove original digits

1¶1                         # Remove lines between digits
11
^(1*)¶+\1\b                 # Match if values are equal

# Ly, 51 bytes

ns>lS0sp11[ppl:Isp>l<ysp>l<sp>,^<lsy,=!]>&+s<<l=fp

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# Pari/GP, 35 bytes

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

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# K (ngn/k), 22 bytes

{x~+/*/'(#:r)#'r:10\x}

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• there's no need for a : after a monadic verb, #:r -> #r – ngn Nov 5 '18 at 17:53

# 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

# 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) }
}

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# APL (Dyalog Unicode), 8 bytesSBCS

Tacit prefix function taking the input as a string.

⍎≡1⊥⍎¨*≢

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is the evaluated argument

identical to

1⊥ the sum (lit. convert from base-1) of

⍎¨ the evaluated characters (i.e the digits)

* raised to the power of

the tally of characters (i.e. digits)

# 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>

# CJam, 12 bytes

{_Ab_,f#:+=}

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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
=    and compare
_               to the original: 1 (true)

# 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?

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L,BDdbLdXBcBsA=

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# JavaScript 7, 41 Bytes

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

# 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)

• Is it necessary to convert c into number? – l4m2 Jan 5 '18 at 4:14

## Swift 3.2: 107 characters

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

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# Japt, 149 7 bytes

¶ì_xpZÊ

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## Explanation

Implicit input of integer U.

ì_

Convert U to an array of digits (ì), pass it through a function and convert back to an integer after.

xpZÊ

Reduce by addition (x), raising each element to the power (p) of the length (Ê) of the array in the process.

Check if the result is strictly equal to U.

• I think ¥U¬®n pUlÃx would work for 11 bytes ;) – Oliver May 18 '17 at 17:39

## R, 71696656 48

Reduced by 8 bytes thanks to @Giuseppe! The idea was to perform the integer division before the modulo operation.

i=nchar(a<-scan()):0;a==sum((a%/%10^i%%10)^i[1])

(3-year) old version with corresponding explanation:

i=nchar(a<-scan()):1;a==sum(((a%%10^i)%/%10^(i-1))^i[1])

a<-scan() takes a number (integer, real,...) as input (say 153 for the example).
i becomes a vector containing 3 to 1 (the number of characters of a being 3).
%% is vectorized so a%%10^i means a modulo 1000, 100 and 10: it therefore gives 153, 53, 3.
(a%%10^i)%/%10^(i-1) is the integer division of that vector by 100, 10, 1: therefore, 1, 5, 3.
We elevate that with the first element of i which is the number of characters (here digits) of a, i. e. 3, thus giving a vector containing 1, 125, 27 that we sum and compares to a.

• Does the integer division always round down? Otherwise, you could run into problems with e.g. 370 (a narcissistic number) turning into 4,7,0 (which would return false) or 270 (non-narcissistic) turning into 3,7,0 (returning true). – Iszi Nov 14 '13 at 20:40
• Integer division doesn't round... The integer division of 370 by 100 is 3 with the remainder of 70 and not 3.70. – plannapus Nov 15 '13 at 7:40
• 48 bytes...somebody bumped this to the homepage! – Giuseppe Sep 1 '17 at 19:55

# R, 53 bytes

sum(scan(t=gsub("(.)","\\1 ",x<-scan()))^nchar(x))==x

The gsub regex inserts spaces in between characters, so that the scan function will be able to read the number into a vector of digits.

• +1 i would have never thought of doing that, it's brilliant. – plannapus Nov 15 '13 at 16:26

# Pyth (non-competing), 10 bytes

qsm^sdlQ

Verify the test cases.

# 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 ins)==s

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

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echo $[ printf "+%c**${#1}" $(fold -w1<<<$1) == $1 ] Try it online! Takes input as parameter, outputs 1 if is narcissist, 0 if not. fold prints each digit one per line, printf adds +<digit>**<length> forming the arithmetic expression which is evaluated and compared to the original input. # Befunge 98, 58 bytes 1-00p&:a\v 00g1+00p>:a%\a/:!kv \:9kv00gk:00gk*+> @.!-$<

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I'm sure this can be golfed further. I will take another look at it and add an explanation later...

# 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^=#

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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)

{$_==sum .comb.map(* **.comb)} Pretty straightforward. The chars method would be more typically used to get the number of characters in the input string, but comb returns those characters as a list, which evaluates to the number of characters in numeric context, and saves us a byte. # Jelly, 6 bytes (non-competing) D*L$S=

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D        Get the digits of the input
*L$Raise each element to power of its length S Sum = Equals input? # JavaScript (ES6), 56 bytes eval(${n=prompt()}0.split.join(**\${n.length}+`))==n

Python, 90 Bytes

a,z=input(),[]
for x in list(a):z.append(int(x)**len(a))
print(1 if sum(z)==int(a) else 0)