Print the digital root

Question

^{This is different from My Word can beat up your Word as it is less complex and only requires you to calculate it, and not compare them.}

To find the digital root, take all of the digits of a number, add them, and repeat until you get a one-digit number. For example, if the number was 12345, you would add 1, 2, 3, 4, and 5, getting 15. You would then add 1 and 5, giving you 6.

Your task

Given an integer N (0 <= N <= 10000) through STDIN, print the digital root of N.

Test cases

1 -> 1
45 -> 9
341 -> 8
6801 -> 6
59613 -> 6
495106 -> 7

Remember, this is code-golf, so the code with the smallest number of bytes wins.

• nimi: Maybe a subtask of this challenge.
• AdmBorkBork: Very closely related to this challenge ... maybe close enough for a dupe.
• Ton Hospel: Please be more precise when saying number. In particular. must input 0 be supported?
• Martin Ender: @TimmyD I think that this one is the much cleaner challenge without adding letter to integer conversion, computing the function for two values and including the literal STALEMATE. It might be better to close the other one as a dupe of this.
• Erik the Outgolfer: @MartinEnder I retracted my close vote, I think it's unfair to close a good challenge as a dupe of another more complex challenge.

Answer 1

Pyke, 1 byte

s

Try it here!

Takes the digital root of the input

Answer 2

Jelly, (削除) 7 5 4 (削除ここまで) 3 bytes

ḃ9Ṫ

TryItOnline! or all test cases

How?

The digital root is known to obey the formula (n-1)%9+1.
This is the same as the last digit in bijective base 9
(and due to implementation that 0ḃ9=[] and []Ṫ=0 this handles the edge-case of zero).

ḃ9Ṫ - Main link: n
ḃ9 - convert to bijective base 9 digits (a list)
 Ṫ - tail (get the last digit)

Answer 3

JavaScript (ES6), (削除) 16 (削除ここまで) 10 bytes

n=>--n%9+1

Test cases

let f =
n=>--n%9+1
console.log(f(1)); // -> 1
console.log(f(45)); // -> 9
console.log(f(341)); // -> 8
console.log(f(6801)); // -> 6
console.log(f(59613)); // -> 6
console.log(f(495106)); // -> 7

Answer 4

Python, (削除) 16 (削除ここまで) 20 bytes

+4 bytes to handle edge case of zero.

lambda n:n and~-n%9+1

repl.it

• Karl Napf: Wow. This is so easy it can be ported to any language. You can even ~-input()%9+1
• Emigna: Doesn't work for 0 unfortunately.
• Jonathan Allan: @KarlNapf Wouldn't that need a print?
• Karl Napf: @JonathanAllan Ah, possibly. I just tested it in the REPL environment and that did it.
• Jonathan Allan: @ the anonymous user who attempted an edit - it would have actually broken the code (made an input of 0 result in 9 rather than 0, which is what is catered for by the n and part of the code) furthermore it would have counted as 19 bytes not 13 (since the print and the space must be counted).

Answer 5

MATL, 3 bytes

9X\

Try it online!

A lot of (now deleted answers) tried using modulo 9 to get the result. This is a great shortcut, but unfortunately does not work for multiples of 9. MATL has a function for modulo on the interval [1, n]. Using this modulo, we have 1 % 3 == 1, 2 % 3 == 2, 3 % 3 == 3, 4 % 3 == 1, etc. This answer simply takes the input modulo nine using this custom modulo.

Answer 6

Mathematica, (削除) 27 (削除ここまで) 11 bytes

Mod[#,9,1]&

Mathematica's Mod takes a third parameter as an offset of the resulting range of the modulo. This avoids decrementing the input and incrementing the output.

Answer 7

Julia, 12 bytes

!n=mod1(n,9)

or

n->mod1(n,9)

mod1 is an alternative to mod which maps to the range [1, n] instead of [0, n).

Answer 8

PHP, 15 Bytes

<?=--$argn%9+1;

Previous version PHP, 55 Bytes

$n=$argn;while($n>9)$n=array_sum(Str_split($n));echo$n;

• CT14.IT: Exactly how I did it!
• Jörg Hülsermann: @CT14.IT I can delete this post if you wish. Your deleted post ws 1 minute earlier and you have only forgot the while loop
• CT14.IT: Nah the deleted answer was wrong because I didn't read the question properly to start with, I didnt attempt to sum the generated number
• Crypto: You can add the trick of other answers <?=--$argv[1]%9+1?>

Answer 9

Retina, 7 bytes

{`.
*
.

Try it online!

I see lots of mathematical solutions, but in Retina the straightforward approach seems to be the best one.

Explanation

{` makes the whole program run in a loop until the string doesn't change anymore. The loop consists of two stages:

.
*

Convert each digit to unary.

.

Count the number of characters (=convert the unary number to decimal).

This works because converting each digit to unary with no separator between digits creates a single unary number which is equal to the sum of all digits.

Answer 10

R, (削除) 72 67 (削除ここまで) 29 bytes

Edit: Thanks to @rturnbull for shaving off two bytes.

n=scan();`if`(n%%9|!n,n%%9,9)

• rturnbull: I recently learned that ifelse can be replaced by `if`, with identical behavior, which saves you a couple of bytes.
• Billywob: @rturnbull I was always wondering how ` if ` worked. Could you give an example or maybe add it to Tips for golfing in ?
• rturnbull: The simplest way to understand it is that it's a non-vectorized ifelse. In this case, `if`(n%%9|!n,n%%9,9) provides identical behavior to the code you've posted. As far as I can tell, this behavior is undocumented! I'll add a comment to the tips thread.

Answer 11

Husk, 4 bytes

Recursive method (4 bytes) (Credit to @ovs)

ωoΣd

 # implicit parameter 0 (refers to the last parameter)
 d # list of digits in base 10
 Σd # sum of list
 oΣd # compose the two functions
ωoΣd # repeat until the function returns the same result as the previous one

Try it online!

Modulus method (4 bytes)

→%9←

 # implicit parameter 0 (refers to the last parameter)
 ← # previous number (n - 1)
 %9← # modulus by 9
→%9← # next number (n + 1)

Try it online!

• ovs: ωoΣd works for the recursive version
• scpchicken: cool ill change it soon

Answer 12

Haskell, (削除) 35 (削除ここまで) 34 bytes

until(<10)$sum.map(read.pure).show

Try it on Ideone.

Explanation:

until(<10)$sum.map(read.pure).show
 show -- convert int to string
 map( ). -- turn each char (digit) into
 pure -- a string 
 read. -- and then a number
 sum. -- sum up the list of numbers
until(<10)$ -- repeat until the result is < 10

Answer 13

Perl, 15 bytes

Includes +2 for -lp

Give input on STDIN

root.pl <<< 123

root.pl

#!/usr/bin/perl -lp
$_&&=~-$_%9+1

This is the boring solution that has already been given in many languages, but at least this version supports 0 too

More interesting doing real repeated additions (though in another order) is in fact only 1 byte longer:

#!/usr/bin/perl -p
s%%$_+=chop%reg

Answer 14

Java 7, 63 bytes

int f(int n){int s=0;for(;n>0;n/=10)s+=n%10;return s>9?f(s):s;}

Recursive function which just gets digits with mod/div. Nothing fancy.

Cheap port

of Jonathan Allan's would be a measly 28 bytes:

int f(int n){return~-n%9+1;}

Answer 15

Labyrinth, 8 bytes

?(_9%)!@

using the equation (n-1)%9+1:

• ? reads the input as decimal and pushes it to the stack
• ( decrements the top of the stack
• _ pushes a zero onto the top of the stack
• 9 push the top of the stack popped times 10 the digit (in this case, 9)
• % pops y, pops x, pushes x%y
• ) increments the top of the stack
• ! pops the top of the stack and out puts it as a decimal string
• @ terminates the program

Answer 16

Hexagony, (削除) 19 (削除ここまで) 15 bytes

.?<9{(/>!@!/)%' 

More Readable:

 . ? < 
 9 { ( /
> ! @ ! / 
 ) % ' .
 . . . 

Try it online!

-3 bytes by taking a different approach, making the 0 edge case trivial.
-1 byte by fixing 0 edge case bug

Using the formula ((n-1) mod 9) + 1 like a lot of other solutions aswell.

Answer 17

Japt -h, (削除) 6 (削除ここまで) 4 bytes

Æ=ìx

Try it

Æ=ìx :Implicit input of integer U
Æ :Map the range [0,U)
 = : Reassign to U
 ì : Convert to digit array
 x : Reduce by addition
 :Implicit output of last element

Answer 18

Brachylog, 9 bytes

#0|@e+:0&

Try it online!

Explanation

#0 Input = Output = a digit
 | OR
 @e Split the input into a list of digits
 + Sum
 :0& Call this predicate recursively

Alternative approach, 11 bytes

:I:{@e+}i#0

This one uses the meta-predicate i - Iterate to call I times the predicate {@e+} on the input. This will try values of I from 0 to infinity until one makes it so that the output of i is a single digit which makes #0 true.

Answer 19

05AB1E, 6 bytes

[SODg#

Try it online!

Explanation

[ # infinite loop
 S # split into digits
 O # sum digits
 Dg# # if length == 1: break

Answer 20

Ruby, 12 bytes

->n{~-n%9+1}

• Ton Hospel: 19 ? Shouldn't that be 9 ?
• TuxCrafting: @TonHospel Yes, stupid error :P

Answer 21

JavaScript (ES6), (削除) 41 (削除ここまで) 38 bytes

Saved 3 bytes, thanks to Bassdrop Cumberwubwubwub

Takes and returns a string.

f=s=>s[1]?f(''+eval([...s].join`+`)):s

Test cases

f=s=>s[1]?f(''+eval([...s].join`+`)):s
console.log(f("1")); // -> 1
console.log(f("45")); // -> 9
console.log(f("341")); // -> 8
console.log(f("6801")); // -> 6
console.log(f("59613")); // -> 6
console.log(f("495106")); // -> 7

• Bassdrop Cumberwubwubwub: You can change s.split`` to [...s]

Answer 22

CJam, (削除) 19 (削除ここまで) 13 bytes

r{:~:+_s9円>}g

Interpreter

Explanation:

r{:~:+_s9円>}g Code
r Get token
 {:~:+_s9円>} Block: :~:+_s9円>
 ~ Eval
 : Map
 + Add
 : Map
 _ Duplicate
 s Convert to string
 \ Swap
 9 9
 > Greater than
 g Do while (pop)

Thanks to 8478 (Martin Ender) for -6 bytes.

---

CJam, 6 bytes

ri(9%)

Suggested by 8478 (Martin Ender). Interpreter

I was thinking about it, but Martin just got it before me. Explanation:

ri(9%) Code
r Get token
 i Convert to integer
 ( Decrement
 9 9
 % Modulo
 ) Increment

• Martin Ender: Single-command map and reduce can both be written with prefix :, so you can do :~:+. It also doesn't hurt to run the block at least once so you can use a g loop instead of a w loop.
• Erik the Outgolfer: @MartinEnder r{_,1>}{:~:+`}w works, but I don't know how on earth am I supposed to use g here.
• Martin Ender: E.g. like this: r{:~:+_s9円>}g (of course the closed form solution ri(9%) is much shorter.
• Erik the Outgolfer: @MartinEnder Oh gawd, for real now, I'm such a beginner...
• ThePlasmaRailgun: The second one doesn't work on multiples of 9

Answer 23

Factor, 24

Smart, mathy answer.

[ neg bitnot 9 mod 1 + ]

63 for dumb iterative solution:

[ [ dup 9 > ] [ number>string >array [ 48 - ] map sum ] while ]

Answer 24

J, 8 bytes

**1+9|<:

Uses the formula d(n) = ((n-1) mod 9) + 1 with the case d(0) = 0.

Usage

 f =: **1+9|<:
 (,.f"0) 0 1 45 341 6801 59613 495106
 0 0
 1 1
 45 9
 341 8
 6801 6
 59613 6
495106 7

Explanation

**1+9|<: Input: integer n
 <: Decrement n
 9| Take it modulo 9
 1+ Add 1 to it
* Sign(n) = 0 if n = 0, else 1
 * Multiply and return

Answer 25

Pyth - (削除) 7 (削除ここまで) (削除) 4 (削除ここまで) (削除) 6 (削除ここまで) 7 bytes

Not the best one, but still beats a decent amount of answers:

|ejQ9 9

Like the previous version, but handling also cases of multiples of 9, using logical or.

---

This version fails the 45 testcase:

ejQ9

Explanation:

 jQ9 -> converting the input to base 9
e -> taking the last digit

Try it here

Try the previous version here!

---

Previous solutions:

&Qh%tQ9

Explanation:

 tQ -> tail: Q-1
 %tQ9 -> Modulo: (Q-1)%9
 h%tQ9 -> head: (Q-1)%9+1
&Qh%tQ9 -> Logical 'and' - takes the first null value. If Q is 0 - returns zero, otherwise returns the (Q-1)%9+1 expression result

You're invited to try it here!

• Dennis: Your 4-byte version fails test case 45.
• xnor: Won't this give 0 for multiples of 9?
• Yotam Salmon: Yeah, I just noticed it. Will do some fixing there. Apparently, jQ9 doesn't act like Jelly's ḃ9 :-P
• Toby Speight: Does this give the correct answer when input is 0?
• Yotam Salmon: The original solution does. And darn, I was 15, where did you come from now :P

Answer 26

APL (Dyalog), (削除) 15 (削除ここまで) 9 bytes bytes

×ばつ1+9|-∘1

Try it online!

Answer 27

COBOL (GNU), 131 bytes

PROGRAM-ID.A.DATA DIVISION.
	WORKING-STORAGE SECTION.
	01 N PIC 9(9).
PROCEDURE DIVISION.ACCEPT N.COMPUTE N=FUNCTION REM(N- 1,9)+ 1

Try it online!

Explanation :

PROGRAM-ID.A.DATA DIVISION.
 * The first line is necessary and can't be removed
 WORKING-STORAGE SECTION.
 * sadly so is this (accept input / output)
 01 N PIC 9(9).
 * This accepts input N that is upto 999,999,999 (output is via same)
 PROCEDURE DIVISION.
 * Time to start working on program itself.
 ACCEPT N.
 * Accepts the input from stdin.
 COMPUTE N = FUNCTION REM(N - 1, 9) + 1
 * compute n such that it is mod of n - 1 and 9 and then add one (standard forumla of --N%9+1

Output is via display N.

• ceilingcat: Suggest LOCAL-STORAGE instead of WORKING-STORAGE

Answer 28

05AB1E, 3 bytes

ΔSO

Try it online or verify all test cases.

Or a minor alternative:

Δ1ö

Try it online or verify all test cases.

Explanation:

Δ # Loop until the result no longer changes,
 # using the (implicit) input-integer in the first iteration
 S # Convert it to a list of digits
 O # Sum those together
 # (after which the resulting digit is output implicitly)
Δ # Loop until the result no longer changes,
 # using the (implicit) input-integer in the first iteration
 1ö # Convert it from base-10 to base-1
 # (which is a non-vectorizing way to sum an integer's digits in 05AB1E)
 # (after which the resulting digit is output implicitly)

Answer 29

><> (Fish), 36 bytes

>0$:1(?v:&a%+&a,:1%-20.
^v?(a:~<
n<;

Try it online! - Interactive Version

Expects input as a number pushed onto the stack in advance.

Explanation

There are 2 basic loops, one for summing the digits and one for checking if the number is less than 10 yet.

 :1(?v

Checks if the number is less than 1, if not all the digits are summed.

 :&a%+&a,:1%

Add the modulo to the accumulator, and then push the division again. We subtract the number %1 to make it a integer.

 20.

Skip the first 3 letters of the bottom row.

^v?(a:~<
n<;

If the number is less than 10 (a is 10 in fish, it uses hexadecimal), print it and exit. Otherwise, go back to the start of the program with the number pushed to the stack.

>0$

Push the accumulator to the second from first spot on the stack.

• Emigna: You can save 20 bytes by performing the operations on the input and erroring out: :a(?n:a%$a,:1%-+

Answer 30

Python 2, (削除) 54 (削除ここまで) 51 bytes

i=input()
while~-len(i):i=`sum(map(int,i))`
print i 

Thanks to Oliver and Karl Napf for helping me save 3 bytes

• Oliver Ni: You can change while len(i)>1 to while~-len(i) to save one byte.
• Karl Napf: I think you can omit the ticks around input() and force the input the be enclosed in quotes to save 2 bytes.
• Erik the Outgolfer: @KarlNapf I don't think you can do this when the input is an integer.
• Daniel: @EriktheGolfer, the op said that the input can be taken as a string