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

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
– devRicher, Commented Jan 6, 2017 at 21:51

Pete Arden · Accepted Answer · 2016-11-09 17:35:28Z

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

Carcigenicate · Accepted Answer · 2016-11-10 00:18:02Z

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? [] (let [n-str (read-line) 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))))

Grax32 · Accepted Answer · 2016-11-10 18:49:49Z

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.

Paul Kerrigan · Accepted Answer · 2016-11-11 13:53:43Z

0

k, 24 bytes

{x=+/*/(#$x)#,"I"$'$x}

answered Nov 11, 2016 at 13:53

Paul Kerrigan

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brendanjhowell · Accepted Answer · 2016-11-11 16:22:45Z

R, 173 Bytes

a=readline() b=as.numeric(strsplit(as.character(a),"")[[1]]) x=(b)^length(b) if(sum(x)==as.numeric(paste(b,sep="",collapse=""))){ print("true") } else{ print("false") }

Community · Accepted Answer · 2020-06-17 09:04:33Z

Factor, 90 bytes

[ read [ length ] [ >array [ 1string ] map [ 10 base> ] map ] bi [ swap ^ ] with map sum ]

Pyke, 8 bytes (noncompeting)

YQ`lL^sq

Try it here!

 Q`l - len(str(input)) L^ - map(^ ** V) Y - digits(input) s - sum(^) q - ^ == input

George Reith · Accepted Answer · 2017-01-06 18:37:26Z

0

JavaScript (ES6), 56 bytes

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

answered Jan 6, 2017 at 18:37

George Reith

2,5341 gold badge15 silver badges16 bronze badges

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Sean · Accepted Answer · 2017-01-06 19:54:21Z

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.

\$\begingroup\$ 25 bytes \$\endgroup\$

Jo King
– Jo King

2018-11-07 05:42:06 +00:00
Commented Nov 7, 2018 at 5:42 — Jo King
– Jo King, Commented Nov 7, 2018 at 5:42

FlipTack · Accepted Answer · 2017-01-06 20:01:02Z

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)

caird coinheringaahing · Accepted Answer · 2017-05-21 14:08:52Z

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!

\$\begingroup\$ You could use strings as input and save a byte. \$\endgroup\$

Jonathan Frech
– Jonathan Frech

2017-09-19 06:49:01 +00:00
Commented Sep 19, 2017 at 6:49 — Jonathan Frech
– Jonathan Frech, Commented Sep 19, 2017 at 6:49

Xcali · Accepted Answer · 2017-09-14 13:28:49Z

0

Perl 5, 24 bytes

21 bytes of code + 3 for -pF flags

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

Try it online!

answered Sep 14, 2017 at 13:28

Xcali

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Eric · Accepted Answer · 2017-09-18 17:55:40Z

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))}

Community · Accepted Answer · 2020-06-17 09:04:33Z

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)

\$\begingroup\$ Is it necessary to convert c into number? \$\endgroup\$

l4m2
– l4m2

2018-01-05 04:14:19 +00:00
Commented Jan 5, 2018 at 4:14 — l4m2
– l4m2, Commented Jan 5, 2018 at 4:14

l4m2 · Accepted Answer · 2018-01-05 04:08:39Z

0

JavaScript 7, 41 Bytes

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

answered Jan 5, 2018 at 4:08

l4m2

32.5k2 gold badges27 silver badges116 bronze badges

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mudkip201 · Accepted Answer · 2018-01-28 21:56:05Z

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!

Esolanging Fruit · Accepted Answer · 2018-01-29 05:48:46Z

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)

Community · Accepted Answer · 2020-06-17 09:04:33Z

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>

jrtapsell · Accepted Answer · 2018-01-29 11:35:17Z

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

Brian H. · Accepted Answer · 2018-01-29 11:53:02Z

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

Axuary · Accepted Answer · 2021-05-29 15:04:13Z

0

Excel, 44 bytes

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

Link to Spreadsheet

answered May 29, 2021 at 15:04

Axuary

3,5767 silver badges11 bronze badges

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alephalpha · Accepted Answer · 2022-02-25 10:21:57Z

0

Pari/GP, 30 bytes

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

Try it online!

edited Feb 25, 2022 at 10:21

answered Nov 4, 2018 at 6:34

alephalpha

51.9k7 gold badges75 silver badges197 bronze badges

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emanresu A · Accepted Answer · 2022-09-04 06:39:25Z

0

Pyth, 10 bytes

qsm^sdl`Q`

Verify the test cases.

edited Sep 4, 2022 at 6:39

emanresu A

46.2k5 gold badges112 silver badges257 bronze badges

answered Sep 1, 2017 at 18:13

Mr. Xcoder

42.9k9 gold badges87 silver badges221 bronze badges

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Stack Exchange Network

Test a number for narcissism

83 Answers 83

C#, 103 Bytes

Clojure, 110 bytes

JavaScript ES6, 50 bytes

k, 24 bytes

R, 173 Bytes

Factor, 90 bytes

Pyke, 8 bytes (noncompeting)

JavaScript (ES6), 56 bytes

Perl 6, 30 bytes

Pushy, 9 bytes

Python 2, 44 bytes

Perl 5, 24 bytes

Swift 3.2: 107 characters

JavaScript 46 bytes

JavaScript 7, 41 Bytes

Pyt, 11 bytes

CJam, 12 bytes

JavaScript (ES7), 43 38 bytes

Try It

Kotlin, 60 bytes

Beautified

Test

TIO

Javascript (ES6) 46 bytes - Not competing

Excel, 44 bytes

Pari/GP, 30 bytes

Pyth, 10 bytes

Linked

Hot Network Questions

Test a number for narcissism

83 Answers 83

C#, 103 Bytes

Clojure, 110 bytes

JavaScript ES6, 50 bytes

k, 24 bytes

R, 173 Bytes

Factor, 90 bytes

Pyke, 8 bytes (noncompeting)

JavaScript (ES6), 56 bytes

Perl 6, 30 bytes

Pushy, 9 bytes

Python 2, 44 bytes

Perl 5, 24 bytes

Swift 3.2: 107 characters

JavaScript 46 bytes

JavaScript 7, 41 Bytes

Pyt, 11 bytes

CJam, 12 bytes

JavaScript (ES7), 43 38 bytes

Try It

Kotlin, 60 bytes

Beautified

Test

TIO

Javascript (ES6) 46 bytes - Not competing

Excel, 44 bytes

Pari/GP, 30 bytes

Pyth, 10 bytes

Linked

Related

Hot Network Questions