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):

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

marinus · Accepted Answer · 2013-11-14 06:57:29Z

39

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

answered Nov 14 '13 at 6:57

marinus

29.8k6 gold badges62 silver badges109 bronze badges

9

\$\begingroup\$ what did i just read \$\endgroup\$ – Jbwilliams1 Feb 18 '14 at 20:13
4

\$\begingroup\$ @LagWagon God's language \$\endgroup\$ – tomsmeding Jul 8 '14 at 8:00

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Howard · Accepted Answer · 2013-11-14 05:25:51Z

GolfScript, 16 characters

~.`:s{48-s,?-}/!

Input must be given on STDIN, output is 0 or 1 indicating non-narcissistic / narcissistic number.

Explanation of the code:

~              # Evaluate the input to get a number
.              # Accumulator (initially the number itself)
`:s            # Convert number to string and assign to variable s
{              # Loop over characters of the string
  48-          # Reduce character value by 48
  s,           # Push length of input number
  ?            # Power
  -            # Subtract result from accumulator
}/
!              # Not! (i.e. iff accumulator was zero it was a narcissistic number)

I did a double-take on `` ~.` `` but it seems impossible to improve. Nice one. — Peter Taylor, Nov 20 '13 at 17:28

alephalpha · Accepted Answer · 2013-11-14 04:43:37Z

15

Mathematica, 43 chars

Tr[#^Length@#&@IntegerDigits@#]==#&@Input[]

answered Nov 14 '13 at 4:43

alephalpha

22.8k3 gold badges33 silver badges98 bronze badges

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breadbox · Accepted Answer · 2013-11-14 02:32:42Z

14

Perl, 38 characters

perl -lpe '$@=y///c;$s+=$_**$@for/./g;$_=$_==$s'

A pretty straightforward implementation.

Here's a slightly different version that fits in 35 characters:

perl -lpe '$@=y///c;$s+=$_**$@for/./g;$_-=$s'

This version outputs a false value if the input is narcissistic, otherwise it outputs a (Perl-accepted) true value. One might argue that this backwards version falls within the limits of the challenge description, but upon reflection I decided not to. I'm not that desperate to improve my score. Yet.

answered Nov 14 '13 at 2:32

breadbox

6,86324 silver badges44 bronze badges

\$\begingroup\$ “Error checking for text strings or other invalid inputs is not required.” – So why not suppose the input will be valid number, without trailing newline? echo -n 153 | perl -pe '…' will work without -l. \$\endgroup\$ – manatwork Nov 14 '13 at 12:26
\$\begingroup\$ I think so long as you define what your true and false outputs are, it should be legal \$\endgroup\$ – Cruncher Nov 14 '13 at 16:48
\$\begingroup\$ Strictly speaking, the wording of the challenge text does leave a bit of ambiguity as to what True/False or 0/1 should mean, so I'll let this one pass. A different script of equal length that returns true for narcissistic values would have the advantage, though. \$\endgroup\$ – Iszi Nov 14 '13 at 20:47
\$\begingroup\$ Same idea but shorter: perl -pe'map$s+=$_**@y,@y=/./g;$_=$_==$s' \$\endgroup\$ – msh210 Jun 19 '16 at 7:48

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rationalis · Accepted Answer · 2013-11-14 06:04:48Z

13

J, 23 chars

(".=+/@("."0^#))(1!:1)1

(1!:1)1 is keyboard input (returning a string).

". converts input to a number; "0 specifies a rank (dimension) of 0, in other words, taking each character and converting it to a number.

^ is the power function and # is the length function, thus taking each digit to the power of the length of the string (equivalently, the number of digits).

+/ is just sum, and = is comparing the sum and number.

edited Nov 14 '13 at 6:04

answered Nov 14 '13 at 3:57

rationalis

4462 silver badges8 bronze badges

2

\$\begingroup\$ "Your code must take input from the user and output True or False depending upon whether the given number is a Narcissistic Number." (emphasis mine) \$\endgroup\$ – John Dvorak Nov 14 '13 at 4:45
\$\begingroup\$ @JanDvorak My bad -- added keyboard input. \$\endgroup\$ – rationalis Nov 14 '13 at 6:08

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histocrat · Accepted Answer · 2013-11-14 15:32:32Z

12

Ruby, 34+5=39

With command-line flags

ruby -nlaF|

Run

p eval [$F,0]*"**#{~/$/}+"+"==#$_"

Outputs true or false.

answered Nov 14 '13 at 15:32

histocrat

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3

\$\begingroup\$ This may be the most Ruby flags I've ever seen in a legitimate code golf :P \$\endgroup\$ – Doorknob♦ Nov 21 '13 at 21:24

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plannapus · Accepted Answer · 2017-09-02 10:19:46Z

11

R, 71 69 66 56 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.

edited Sep 2 '17 at 10:19

answered Nov 14 '13 at 7:14

plannapus

8,58013 silver badges42 bronze badges

\$\begingroup\$ 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). \$\endgroup\$ – Iszi Nov 14 '13 at 20:40
\$\begingroup\$ Integer division doesn't round... The integer division of 370 by 100 is 3 with the remainder of 70 and not 3.70. \$\endgroup\$ – plannapus Nov 15 '13 at 7:40
1

\$\begingroup\$ 48 bytes...somebody bumped this to the homepage! \$\endgroup\$ – Giuseppe Sep 1 '17 at 19:55

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mbomb007 · Accepted Answer · 2016-11-08 15:38:57Z

9

Python 3, 56 bytes

Not very obfuscated, but a simple solution.

s = input()
print(int(s)==sum(int(c)**len(s)for c in s))

edited Nov 8 '16 at 15:38

mbomb007

18.8k5 gold badges48 silver badges120 bronze badges

answered Nov 14 '13 at 3:08

danmcardle

6954 silver badges8 bronze badges

1

\$\begingroup\$ The [ and ] are unnecessary, and you can drop the space in front of for too, so: sum(int(c)**len(s)for c in s) \$\endgroup\$ – marinus Nov 14 '13 at 7:12
\$\begingroup\$ That's awesome! Thanks for the tip. \$\endgroup\$ – danmcardle Nov 14 '13 at 12:05
1

\$\begingroup\$ You can save two characters by removing the spaces in s = input() and another one by moving this to 2.7 where print isn't a function. \$\endgroup\$ – Ben Nov 16 '13 at 17:31
\$\begingroup\$ Good point, edited. \$\endgroup\$ – danmcardle Nov 17 '13 at 5:32
\$\begingroup\$ I think you should point out that adding braces to print (hence one character more) would make this a valid Python 2.x and Python 3.x solution. \$\endgroup\$ – Martin Thoma Nov 18 '13 at 18:24

| show 6 more comments

Vlad Preda · Accepted Answer · 2013-11-14 15:12:42Z

8

PHP, 80 74 66 chars

Very straightforward PHP solution:

<?for(;$i<$l=strlen($a=$argv[1]);)$s+=pow($a[$i++],$l);echo$s==$a;

It assumes error_reporting doesn't include notices, otherwise quite a few extra characters will be needed to initialize $s=0; and $i=0.

Thx @manatwork for shortening many chars.

edited Nov 14 '13 at 15:12

answered Nov 14 '13 at 14:41

Vlad Preda

1815 bronze badges

\$\begingroup\$ Don't assign $a and $l in separate statements. <?for($i=0;$i<$l=strlen($a=$argv[1]);$i++){$s+=pow($a[$i],$l);}echo$s==$a; is shorter. \$\endgroup\$ – manatwork Nov 14 '13 at 14:58
\$\begingroup\$ As you already have a statement which generate a notice, just add another: remove the loop control variable initialization. Incrementing the loop control variable also not need to be a standalone statement. And the braces are definitely not needed: <?for(;$i<$l=strlen($a=$argv[1]);)$s+=pow($a[$i++],$l);echo$s==$a;. \$\endgroup\$ – manatwork Nov 14 '13 at 15:09
\$\begingroup\$ @manatwork: Thank you for the warm welcome to codegolf :) \$\endgroup\$ – Vlad Preda Nov 14 '13 at 15:13
\$\begingroup\$ Can be golfed to this for(;$i<$l=strlen($a=$argn);)$s+=$a[$i++]**$l;echo$s==$a; \$\endgroup\$ – Jörg Hülsermann May 17 '17 at 16:51

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manatwork · Accepted Answer · 2013-11-14 20:32:57Z

8

Dc: 48 characters

[1pq]Sr?d0rdZSz[d10/r10%lz^rSh+Lhd0!=c]dScx+=r0p

Sample run:

bash-4.1$ dc -e '[1pq]Sr?d0rdZSz[d10/r10%lz^rSh+Lhd0!=c]dScx+=r0p' <<< '153'
1

bash-4.1$ dc -e '[1pq]Sr?d0rdZSz[d10/r10%lz^rSh+Lhd0!=c]dScx+=r0p' <<< '1634'
1

bash-4.1$ dc -e '[1pq]Sr?d0rdZSz[d10/r10%lz^rSh+Lhd0!=c]dScx+=r0p' <<< '2013'
0

answered Nov 14 '13 at 20:32

manatwork

17k4 gold badges39 silver badges73 bronze badges

\$\begingroup\$ Never actually used dc, save for frantic typos in attempt to write cd \$\endgroup\$ – Stan Strum Jan 5 '18 at 1:54

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0 ' · Accepted Answer · 2016-11-11 20:56:04Z

8

K, 24 23

{x=+/xexp["I"$'a]@#a:$x}

Shaved 1 char with reordering

{x=+/{x xexp#x}"I"$'$x}

edited Nov 11 '16 at 20:56

0 '

3,3401 gold badge15 silver badges29 bronze badges

answered Nov 14 '13 at 15:34

tmartin

3,88710 silver badges16 bronze badges

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Cyoce · Accepted Answer · 2017-09-01 18:55:57Z

8

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.

edited Sep 1 '17 at 18:55

Cyoce

2,39518 silver badges27 bronze badges

answered Nov 15 '13 at 13:29

flodel

2,33511 silver badges16 bronze badges

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

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tmartin · Accepted Answer · 2013-11-14 17:28:43Z

6

Kona, 18

...

{x=+/(0$'u)^#u:$x}

answered Nov 14 '13 at 17:28

tmartin

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Rynant · Accepted Answer · 2013-11-19 19:31:10Z

6

Powershell, 75 63 62 60 58

Edit: Updated per @Iszi's comment ~~(note: this counts on $x not existing)~~

Edit: Added @Danko's changes.

[char[]]($x=$n=read-host)|%{$x-="$_*"*$n.length+1|iex};!$x

58 56 chars

If input is limited to 10 digits (includes all int32)

($x=$n=read-host)[0..9]|%{$x-="$_*"*$n.length+1|iex};!$x

edited Nov 19 '13 at 19:31

answered Nov 14 '13 at 20:49

Rynant

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\$\begingroup\$ I was wondering if someone was going to do PowerShell before I did. \$\endgroup\$ – Iszi Nov 14 '13 at 20:55
\$\begingroup\$ Save 12 characters by adding another variable $x and using += to do your summing instead of measure -sum then test $x-eq$n. \$\endgroup\$ – Iszi Nov 14 '13 at 21:03
1

\$\begingroup\$ 61 chars: ($x=$n=read-host)-split''|%{$x-=[math]::pow($_,$n.length)};!$x \$\endgroup\$ – Danko Durbić Nov 15 '13 at 9:04
1

\$\begingroup\$ @DankoDurbić, Nice! Type coercion often comes in handy with PoSh code golfing. I only get 62 though when I run '($x=$n=read-host)-split""|%{$x-=[math]::pow($_,$n.length)};!$x'.length \$\endgroup\$ – Rynant Nov 15 '13 at 15:06
1

\$\begingroup\$ @Rynant Good point. I ran your length check in PowerShell and came up with 62 as well. When running a length check similarly against the actual script, it comes up 61. This is probably because of how PowerShell handles the '' which you replaced with ''. I took the original script into Excel to double-check with =LEN("($x=$n=read-host)-split''|%{$x-=[math]::pow($_,$n.length)};!$x") and got 62 also. Of course, we could always count it manually - but who really does that? \$\endgroup\$ – Iszi Nov 15 '13 at 15:22

| show 11 more comments

user1354557 · Accepted Answer · 2013-11-14 20:24:17Z

5

Python 2.x - 51

Same concept as crazedgremlin's solution for Python 3.x:

s=input();print s==sum(int(c)**len(`s`)for c in`s`)

answered Nov 14 '13 at 20:24

user1354557

2511 silver badge6 bronze badges

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Josh · Accepted Answer · 2014-01-10 15:53:26Z

4

C - 97 93 characters

a,b;main(c){scanf("%d",&c);b=c;for(;c;c/=10)a+=pow(c%10,(int)log10(b)+1);printf("%d",a==b);}

With indentation:

a,b;
main(c) { 
  scanf("%d",&c);
  b=c;
  for(;c;c/=10)
    a+=pow(c%10,(int)log10(b)+1);
  printf("%d",a==b);
}

edited Jan 10 '14 at 15:53

answered Jan 10 '14 at 15:32

Josh

2,2311 gold badge11 silver badges22 bronze badges

2

\$\begingroup\$ You don't have to define int for global variables. \$\endgroup\$ – Konrad Borowski Jan 10 '14 at 15:49
\$\begingroup\$ Woah. You're reading the input into argc. \$\endgroup\$ – SIGSTACKFAULT Jan 6 '17 at 14:16
\$\begingroup\$ Also, shouldn't having to do -lm at compile-time count +1 byte? \$\endgroup\$ – SIGSTACKFAULT Jan 6 '17 at 14:21
\$\begingroup\$ @Blacksilver the -lm flag is not required for C89 compilers. \$\endgroup\$ – Josh Jan 9 '17 at 14:55
\$\begingroup\$ Aha. Learn a new thing every day. \$\endgroup\$ – SIGSTACKFAULT Jan 9 '17 at 14:59

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Teun Pronk · Accepted Answer · 2014-02-18 15:34:23Z

Delphi - 166

uses System.SysUtils,math;var i,r,l:integer;s:string;begin r:=0;readln(s);l:=length(s);for I:=1to l do r:=round(r+power(strtoint(s[i]),l));writeln(inttostr(r)=s);end.

With indent

uses System.SysUtils,math;
var
  i,r,l:integer;
  s:string;
begin
  r:=0;
  readln(s);
  l:=length(s);
  for I:=1to l do
    r:=round(r+power(strtoint(s[i]),l));
  writeln(inttostr(r)=s);
end.

Magic Octopus Urn · Accepted Answer · 2016-11-08 15:47:49Z

4

05AB1E, 7 bytes (Non-competing)

DSDgmOQ

Try it online!

-2 bytes thanks to @daHugLenny

edited Nov 8 '16 at 15:47

answered Nov 8 '16 at 15:09

Magic Octopus Urn

13.3k4 gold badges49 silver badges128 bronze badges

3

\$\begingroup\$ You can replace §1ô with S \$\endgroup\$ – acrolith Nov 8 '16 at 15:45

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Nathan Baum · Accepted Answer · 2013-11-15 11:22:40Z

3

Haskell 2010 - 76 characters

main=do x<-getLine;print$(==x)$show$sum$map((^length x).(+(-48)).fromEnum)x

edited Nov 15 '13 at 11:22

answered Nov 15 '13 at 3:54

Nathan Baum

512 bronze badges

1

\$\begingroup\$ You shouldn't post the number of ms to run the code, but the number of chars you used. ;) \$\endgroup\$ – user unknown Nov 15 '13 at 4:02

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manatwork · Accepted Answer · 2013-11-15 14:21:50Z

3

Awk: 40 39 characters

{for(;i<NF;)s+=$(i+++1)**NF;$0=$0==s}1

Sample run:

bash-4.1$ awk -F '' '{for(;i<NF;)s+=$(i+++1)**NF;$0=$0==s}1' <<< '153'
1

bash-4.1$ awk -F '' '{for(;i<NF;)s+=$(i+++1)**NF;$0=$0==s}1' <<< '1634'
1

bash-4.1$ awk -F '' '{for(;i<NF;)s+=$(i+++1)**NF;$0=$0==s}1' <<< '2013'
0

edited Nov 15 '13 at 14:21

answered Nov 15 '13 at 13:04

manatwork

17k4 gold badges39 silver badges73 bronze badges

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user unknown · Accepted Answer · 2013-11-16 01:56:14Z

3

Bash, 64 chars

for((a=$1;a>0;s+=(a%10)**${#1},a/=10));do :; done;echo $[s==$1]

~~a=$1;p=${#a};for((;a>0;a/=10));do s=$((s+(a%10)**p));done;echo $((s==$1))~~

edited Nov 16 '13 at 1:56

answered Nov 15 '13 at 4:00

user unknown

4,08024 silver badges28 bronze badges

1

\$\begingroup\$ You are using variable p in a single place, so no need for it. You can move the initialization of variable a into the for to spare its separate ;: for((a=$1;a>0;a/=10));do s=$[s+(a%10)**${#1}];done;echo $[s==$1]. \$\endgroup\$ – manatwork Nov 15 '13 at 11:08
1

\$\begingroup\$ By moving the evaluation into the for one more character can be shortened: for((a=$1;a>0;s+=(a%10)**${#1},a/=10));do :; done;echo $[s==$1]. \$\endgroup\$ – manatwork Nov 15 '13 at 11:16
\$\begingroup\$ Oh, curious! I tried something like that, but it didn't work. Curious what went wrong. \$\endgroup\$ – user unknown Nov 16 '13 at 1:53

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criptych · Accepted Answer · 2014-01-10 15:20:57Z

3

Lua (101 chars)

Lua isn't known for being concise, but it was fun to try anyway.

for n in io.lines()do l,s=n:len(),0 for i=1,l do d=n:byte(i)s=s+(d-48)^l end print(s==tonumber(n))end

Improvements welcome.

answered Jan 10 '14 at 15:20

criptych

1811 silver badge8 bronze badges

\$\begingroup\$ As it is not required that your program can handle and process a list of numbers, I would not use bytes to implement that functionality. Replacing the loop for n in io.lines()do [...]end with n=io.read() saves some bytes (TIO). \$\endgroup\$ – Jonathan Frech Sep 1 '17 at 19:55

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zzzzBov · Accepted Answer · 2016-11-08 15:00:05Z

3

JavaScript - 70 58 characters

for(i in a=b=prompt())b-=Math.pow(a[i],a.length)
alert(!b)

Note:

If you're testing this in your dev console on Stack Exchange, be aware that there are a number of non-standard properties added to String.prototype that will break this solution, such as String.prototype.formatUnicorn. Please be sure to test in a clean environment, such as on about:blank.

edited Nov 8 '16 at 15:00

answered Nov 15 '13 at 14:52

zzzzBov

2,4702 gold badges12 silver badges19 bronze badges

\$\begingroup\$ I count 70 characters there. \$\endgroup\$ – manatwork Nov 15 '13 at 16:10
\$\begingroup\$ @manatwork, whoops, forgot to count the newline. \$\endgroup\$ – zzzzBov Nov 15 '13 at 19:10
\$\begingroup\$ Great trick that decrementation! \$\endgroup\$ – manatwork Nov 16 '13 at 13:12
2

\$\begingroup\$ it always returns true for me, regardless of input \$\endgroup\$ – koko Jul 8 '14 at 9:27
\$\begingroup\$ @koko, I've added a note to explain why you're receiving incorrect results. \$\endgroup\$ – zzzzBov Nov 8 '16 at 15:00

| show 1 more comment

Hypino · Accepted Answer · 2016-11-10 18:24:33Z

3

Java - 84 bytes

(a,l)->{int s=0;for(byte c:a.getBytes())s+=Math.pow(c-48,l);return a.equals(""+s);};

Non-lambda version: 101 bytes:

boolean n(String a,int l){int s=0;for(byte c:a.getBytes())s+=Math.pow(c-48,l);return a.equals(""+s);}

Called like this:

interface X {
    boolean n(String a, int l);
}

static X x = (a,l)->{int s=0;for(byte c:a.getBytes())s+=Math.pow(c-48,l);return a.equals(""+s);};

public static void main(String[] args) {
    System.out.println(n("153",3));
    System.out.println(n("1634",4));
    System.out.println(n("123",3));
    System.out.println(n("654",3));
}

Returns:

true
true
false
false

edited Nov 10 '16 at 18:24

answered Nov 8 '16 at 19:14

Hypino

2211 silver badge6 bronze badges

\$\begingroup\$ You can remove the parenthesis around the lambda arguments, a,l-> works exactly the same. \$\endgroup\$ – FlipTack Jan 6 '17 at 19:50
\$\begingroup\$ I know you've answer this almost a year ago, but you can golf two bytes: (a,l)-> can be a->l-> and byte can be int: a->l->{int s=0;for(int c:a.getBytes())s+=Math.pow(c-48,l);return a.equals(""+s);} \$\endgroup\$ – Kevin Cruijssen Sep 18 '17 at 14:47

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Shaggy · Accepted Answer · 2017-09-14 10:42:15Z

3

Japt, 14 9 7 bytes

¶ì_xpZÊ

Try it online

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.

edited Sep 14 '17 at 10:42

answered May 18 '17 at 16:39

Shaggy

22.2k3 gold badges21 silver badges73 bronze badges

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

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Smetad Anarkist · Accepted Answer · 2013-11-14 08:36:47Z

2

F# - 92 chars

let n=stdin.ReadLine()
n|>Seq.map(fun x->pown(int x-48)n.Length)|>Seq.sum=int n|>printf"%b"

answered Nov 14 '13 at 8:36

Smetad Anarkist

2681 silver badge10 bronze badges

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Paul Richter · Accepted Answer · 2013-11-17 10:11:38Z

Common Lisp - 116 102 characters

(defun f(m)(labels((l(n)(if(> n 0)(+(expt(mod n 10)(ceiling(log m 10)))(l(floor n 10)))0)))(= m(l m))))

Formatted:

(defun f(m)
  (labels((l(n)
            (if(> n 0)
               (+(expt(mod n 10)(ceiling(log m 10)))
                 (l(floor n 10)))
               0)))
    (=(l m)m)))

Paul Richter · Accepted Answer · 2013-11-20 06:03:30Z

2

Smalltalk - 102 99 characters

[:n|a:=n asString collect:[:e|e digitValue]as:Array.^n=(a collect:[:each|each raisedTo:a size])sum]

At the Workspace, send value: with the number, and Print It.

edited Nov 20 '13 at 6:03

answered Nov 15 '13 at 3:12

Paul Richter

7705 silver badges6 bronze badges

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It'sNotALie. · Accepted Answer · 2014-01-11 23:40:05Z

2

C#, 117

using System.Linq;class A{int Main(string[] a){return a[0].Select(c=>c-'0'^a[0].Length).Sum()==int.Parse(a[0])?1:0;}}

answered Jan 11 '14 at 23:40

It'sNotALie.

2091 silver badge7 bronze badges

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Angs · Accepted Answer · 2016-11-08 17:10:18Z

2

Haskell, 68 66 bytes

d 0=[]
d n=mod n 10:d(div n 10)
sum.(\a->map(^length a)a).d>>=(==)

Usage:

*Main> sum.(\a->map(^length a)a).d>>=(==) $ 1634
True

edited Nov 8 '16 at 17:10

answered Nov 8 '16 at 16:49

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Test a number for narcissism

73 Answers 73

APL (15)

GolfScript, 16 characters

Mathematica, 43 chars

Perl, 38 characters

J, 23 chars

Ruby, 34+5=39

R, 71 69 66 56 48

Python 3, 56 bytes

PHP, 80 74 66 chars

Dc: 48 characters

K, 24 23

R, 53 bytes

Kona, 18

Python 2.x - 51

C - 97 93 characters

Delphi - 166

With indent

05AB1E, 7 bytes (Non-competing)

Haskell 2010 - 76 characters

Awk: 40 39 characters

Bash, 64 chars

Lua (101 chars)

JavaScript - 70 58 characters

Java - 84 bytes

Japt, 14 9 7 bytes

Explanation

F# - 92 chars

Common Lisp - 116 102 characters

Smalltalk - 102 99 characters

C#, 117

Haskell, 68 66 bytes

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