# 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

# Jelly, 6 bytes (non-competing)

D*L$S=  Try it online! D Get the digits of the input *L$     Raise each element to power of its length
S    Sum
=   Equals input?


# APL (Dyalog Unicode), 8 bytesSBCS

Tacit prefix function taking the input as a string.

⍎≡1⊥⍎¨*≢


Try it online!

⍎ 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)

# 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


Try it online

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


Not the sortest but my take.

# Python 2.7: 59 60 chars

a=input();(0,1)[sum(int(i)**len(str(a))for i in str(a))==a]

• It's supposed to output true or false (also needs to take input). – Timtech Jan 10 '14 at 11:46
• @Timtech or 1 and 0 – Eduard Florinescu Jan 10 '14 at 14:05
• @EduardFlorinescu Still needs to take input. – Iszi Jan 10 '14 at 14:12
• @lszi now should work ;) – Eduard Florinescu Jan 10 '14 at 15:13
• For converting boolean value to integer int(BoolExpr) is shorter than (0,1)[BoolExpr]. +(BoolExpr) even shorter. – AMK Jan 11 '14 at 18:24

# Pyth, 13 characters

JwqvJsm^vdlJJ


Explanation:

Jw                J=input()
^vdlJ      eval(d)^len(J)
m^vdlJJ     map each character in J to eval(d)^len(J)
qvJsm^vdlJJ     print(eval(J)==sum(map each character in J to eval(d)^len(J)))


# JavaScript, 56 characters

n=prompt();n.split('').reduce((a,i)=>a+i**n.length,0)==n

This makes use of the exponentiation operator, so you have to be running a modern browser for this to work.

# C, 252220225 111 bytes

int f(char *a){for(int i=0;i<strlen(a);i++){r+=((int)a[i]);for(int j=0;j<strlen(a);j++)r*=r;}return r==(int)a;}


Returns 0 if false and 1 if true. Thanks to @DrMcMoylex for saving many bytes and explaining stuff.

• Welcome to the site! Some tips for you: 1) You have a lot of unnecessary whitespace. You could remove most of the newlines and spaces/indentation and it would still run fine. 2) A function is allowed, so you could return '1' or '0' instead of printing it. That would save a lot of bytes. 3) Since you're only doing one thing in each 'for' loop, you could join the inner code with the loop itself. For example: for(int i=0;i<strlen(a);r+=a[i++]) – DJMcMayhem Nov 8 '16 at 17:56
• @DrMcMoylex How would you know the output if I used return? There's nothing printed in the terminal... – Aryaman Nov 8 '16 at 18:48
• You could write a small wrapper to print the return value of the function – DJMcMayhem Nov 8 '16 at 18:51
• @DrMcMoylex Using printf instead of making a new function uses less bytes, I believe. Thanks for the welcome, btw! – Aryaman Nov 8 '16 at 19:03
• No that's not what I meant. I meant you don't need to print anything. Returning is a valid form of output. You could print the return value for testing, but your submission can just be a function. Then you won't need to parse input or print the output. For example int f(char *a){for(int i=0;i<strlen(a);i++){r+=((int)a[i]);for(int j=0;j<strlen(a);j++)r*=r;}return r==(int)a;} – DJMcMayhem Nov 9 '16 at 19:15

# Perl, 27 +3 = 30 bytes

Run with -F. Older versions of Perl might require you to run with -nF instead, if -F does not imply -n.

grep$;+=$_**$#F,@F;say$_==$ Prints 1 if narcissistic, prints nothing otherwise. (thanks to @Dada for byte-count correction, and for -2 bytes) • grep$;+=$_**$#F,@F;say$_==$ to save two bytes ($; instead of $a, and grep...,@F instead of grep{...}@F). However, note that -F counts as 3 bytes (-, F and a space). – Dada Nov 11 '16 at 18:27

# Clojure, 85 bytes

#(= n(int(reduce +(vec(map#(Math/pow(Character/digit % 10)(count(str n)))(str n))))))


Usage is like so:

(#(...) {number})


Ungolfed (with commentary):

(defn narcissistic [n]
; The function is altered a bit, to improve readability.
; The double arrow means that a result of a function will get "chained"
; onto the next function as the last argument:
; (->> 1 (* 2) (+ 3)) -> (->> (* 2 1) (+ 3)) -> (+ 3 (* 2 1))
(->> n
; Converts it to a string, for the next function
; 153 -> "153"
str
; Converts the string to an array of characters,
; which is then raised to the powers equal to the length of the number:
; 153 -> (1.0 125.0 27.0)
(map (#(Math/pow (Character/digit % 10) (count (str n)))))
; Converts the array to a vector (reducing only works with vectors)
; (1.0 125.0 27.0) -> [1.0 125.0 27.0]
vec
; Reduces the vector by adding them
; [1.0 125.0 27.0] -> 153.0
(reduce +)
; Turns that into an integer
; 153.0 -> 153
int
; Checks if that's equal to the original n
; 153 = 153 -> true
(= 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)


# Befunge 98, 58 bytes

1-00p&:a\v
00g1+00p>:a%\a/:!kv
\:9kv00gk:00gk*+>
@.!-$<  Try it Online! I'm sure this can be golfed further. I will take another look at it and add an explanation later... # Add++, 16 bytes L,BDdbLdXBcBsA=  Try it online! # K (ngn/k), 22 bytes {x~+/*/'(#:r)#'r:10\x}  Try it online! • there's no need for a : after a monadic verb, #:r -> #r – ngn Nov 5 '18 at 17:53 # Brachylog, 10 bytes lg;?↔z^ᵐ+?  Try it online! 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 '19 at 6:30 # R - 216 Here is a long-ish R attempt: fnNar = function(x = NULL) { if(is.null(x)) x = readline('Enter a positive integer: ') digits = as.double(unlist(strsplit(as.character(x), split = ''))) exponent = length(digits) return(x == sum(digits ^ exponent)) } # test the function fnNar() # test the function fnNar(153) fnNar(1634) fnNar(101)  ## Python 2.7 - 57 chars i=input() printsum(map(lambda x:int(x)**len(i),i))==i  There is a shorter Python answer, but I might as well toss in my contribution. i=input()  i is set to input() surrounded by backticks (which is surprisingly hard to type through SE's markdown interpreter). Surrounding x with backticks is equivalent to str(x). [backtick]input()[backtick] saves two characters over raw_input() in any case where we can assume the input is an int, which we're allowed to do: Error checking for text strings or other invalid inputs is not required. Once i is a string containing the user's input, the next line is run. I'll explain this one from the inside out: map(lambda x:int(x)**len(i),i)  map is a function in Python that takes a function and an iterable as arguments and returns a list of each item in the iterable after having the function applied to it. Here I'm defining an anonymous function lambda x which converts x to a string and raises it to the power of the length of i. This map will return a list of each character in the string i raised to the correct power, and even nicely converts it to an int for us. sum(map(lambda x:int(x)**len(i),i))  Here I take the sum of each value in the list returned from the map. If this sum is equal to the original input, we have a narcissistic number. To check this, we either have to convert this sum to a string or the input to an int. int() is two more characters than two backticks, so we convert this to a string the same way we did with the input. printsum(map(lambda x:int(x)**len(i),i))==i  Compare it to i and print the result, and we're done. ## Racket 115 bytes (let*((l(number->string n))(g(string-length l))(m(for/sum((i l))(expt(string->number(string i))g))))(if(= m n)1 0))  Ungolfed: (define (f n) (let* ((l (number->string n)) (g (string-length l)) (m (for/sum ((i l)) (expt (string->number(string i)) g)))) (if (= m n) 1 0)))  Testing: (f 153) (f 1634) (f 123) (f 654)  Output: 1 1 0 0  # Jelly, 7 bytes (non-competing) D*DL$S⁼


Try it online!

Or alternatively (uses 2 chains):

Dµ*LS⁼Ḍ


Explanation of the first code:

 D* DL$S⁼ Main link (monadic). Arguments: z ⁸ (implicit) z D List of digits of z ⁸ (implicit) z D List of digits of z L Length of z$    Last two links as a monad
*        Exponentiation with base x and exponent y
S   Sum of z
⁸ (implicit) z
⁼  Check if x equals y


# C#, 103 Bytes

Golfed:

bool N(int n){double a=0;foreach(var d in n+""){a+=Math.Pow(int.Parse(d+""),n+"".Length);}return a==n;}


Ungolfed:

public bool N(int n)
{
double a = 0;

foreach (var digit in n.ToString())
{
a += Math.Pow(int.Parse(digit + ""), n.ToString().Length);
}

return a == n;
}


Testing:

Console.WriteLine(new NarcissisticNumber().N(153));
True

Console.WriteLine(new NarcissisticNumber().N(1634));
True


## Clojure, 110 bytes

(fn[](let[s(read-line)p #(Integer/parseInt(str %))](=(p s)(reduce + 0(map #(int(Math/pow(p %)(count s)))s)))))


Reads in the user input, maps over the digits, raising each to a power equal to the number of digits in the number, then checks that the sum of the digits equals the number itself.

Ungolfed (and neatened up):

(defn narcissistic? []
parse #(Integer/parseInt (str %))
pow #(int (Math/pow % (count n-str)))
powd-digits (map #(pow (parse %)) n-str)]
(= (parse n-str) (reduce + 0 powd-digits))))


# JavaScript ES6, 50 bytes

v=>[...s=v+""].map(x=>v-=Math.pow(x,s.length))&&!v


Convert the number to a string and iterate through the digits, subtract the power computation from the original number use the ! operator to invert the logic so that a 0 result returns true and non-zero returns false.

# k, 24 bytes

{x=+/*/(#$x)#,"I"$'$x}  ## R, 173 Bytes a=readline() b=as.numeric(strsplit(as.character(a),"")[]) x=(b)^length(b) if(sum(x)==as.numeric(paste(b,sep="",collapse=""))){ print("true") } else{ print("false") }  # Factor, 90 bytes [ read [ length ] [ >array [ 1string ] map [ 10 base> ] map ] bi [ swap ^ ] with map sum ]  More readable and explained: [ read ! input [ length ] ! a function which gets the length [ >array [ 1string ] map [ 10 base> ] map ] ! another which turns a number into an array bi ! apply both to the string input [ swap ^ ] with map sum ! raise each digit to the length power and sum ]  ## Pyke, 8 bytes (noncompeting) YQlL^sq  Try it here!  Ql - len(str(input)) L^ - map(^ ** V) Y - digits(input) s - sum(^) q - ^ == input  # JavaScript (ES6), 56 bytes eval(${n=prompt()}0.split.join(**${n.length}+))==n  # Perl 6, 30 bytes {$_==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.

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


# 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.

# 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


Try it online!

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