Rotational Average

Question

Given an input integer n >= 10, output the average of all deduplicated rotations of the integer.

For example, for input 123, the rotations are 123 (no rotation), 231 (one rotation) and 312 (two rotations). The average of those is (123 + 231 + 312) / 3 or 222.

As another example, take 4928. The rotations are 4928, 9284, 2849, and 8492. Taking the average of those four numbers equals 6388.25.

For another example, for input 445445, the deduplicated rotations are 445445, 454454, and 544544, so the output is 481481.

For input 777, there is only one deduplicated rotation, so the output is 777.

Rules

If applicable, you can assume that the input/output will fit in your language's native Integer type.
The input and output can be given by any convenient method.
Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
Standard loopholes are forbidden.
This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Do you have an example where the deduplication actually changes the output ? In your 445445 example, each 3 unique rotation happens twice, so leaving them doesn't change the output. — user2956892, Apr 16, 2018 at 15:26
@Kaldo No, I wasn't able to (manually) come up with one, but that doesn't mean one doesn't exist, so I left the deduplication rules in place. — AdmBorkBork, Apr 16, 2018 at 15:28
@Kaldo Let d be the number of digits of n and k the smallest positive integer such that rotating n k digits to the left reproduces n. Take q and 0 ≤ r < k such that d = qk + r. Rotating n both d and qk digits to the left must yield n, so r = 0. This means each unique rotation occurs q times, so deduplicating the rotations isn't needed to compute the average. — Dennis, Apr 16, 2018 at 15:38

Dennis · Accepted Answer · 2018-04-16 15:39:41Z

11

Python 3, 38 36 bytes

lambda*n:10**len(n)//9*sum(n)/len(n)

Takes the digits as separate arguments. Thanks to @Rod for suggesting Python 3, saving 2 bytes.

Try it online!

edited Apr 16, 2018 at 15:39

answered Apr 16, 2018 at 15:28

Dennis

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\$\begingroup\$ An additional byte could be saved by switching back to Python 2 and taking an array of floats. \$\endgroup\$
– Dennis
Apr 16, 2018 at 15:55
\$\begingroup\$ You don't need to take the digits as separate arguments, a regular old list does just fine \$\endgroup\$
– Asone Tuhid
Apr 16, 2018 at 20:55

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H.PWiz · Accepted Answer · 2018-04-16 19:49:06Z

9

APL (Dyalog), 9 bytes

10⊥≢⍴+/÷≢

A monadic function taking a vector of digits as an argument.

Try it online!

I take the average of the digits +/÷≢, then repeat it by the length of the input ≢⍴, and finally convert from base 10.

Conceptially, I am taking the sum of the rotations (without carrying):

 4  2  9  8
 2  9  8  4
 9  8  4  2
+8  4  2  9
 -----------
 23 23 23 23

This is just 4+2+9+8 repeated 4 times. Then converting from base 10 (which does the carrying for me) and dividing by the length. Although I divide by the length earlier on because it is equivilent and saves bytes.

edited Apr 16, 2018 at 19:49

answered Apr 16, 2018 at 17:25

H.PWiz

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1

\$\begingroup\$ That's genial :D \$\endgroup\$
– Leo
Apr 17, 2018 at 22:12
\$\begingroup\$ @Leo FWIW the answers that multiply the average by a rep digit are doing essentially the same thing \$\endgroup\$
– H.PWiz
Apr 18, 2018 at 6:38

Add a comment |

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

3

Java 10, 163 137 76 72 71 bytes

n->(Math.pow(10,n.size())-1)/9*n.stream().mapToInt(i->i).sum()/n.size()

-36 bytes thanks to @Nevay.
-61 bytes thanks to @OlivierGrégoire by creating a port of @Dennis' Python 3 answer.
-1 bytes by taking the input as a List of digits instead of String.

Explanation:

Try it online.

n->                                 // Method with String parameter and double return-type
  (Math.pow(10,n.size())-1)/9       //  Repunits the same length as the input-size
  *n.stream().mapToInt(i->i).sum()  //  multiplied by the sum of digits
  /n.size()                         //  divided by the input-size

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 16, 2018 at 15:22

Kevin Cruijssen

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1

\$\begingroup\$ 151 bytes: n->{var s=new java.util.HashSet();var r=0d;for(int l=n.length(),x;l-->0;)if(s.add(x=new Integer(n=n.substring(1)+n.charAt(0))))r+=x;return r/s.size();}, stream approach with 137 bytes: n->java.util.stream.IntStream.range(0,n.length()).map(i->new Integer(n.substring(i)+n.substring(0,i))).distinct().average().getAsDouble() \$\endgroup\$
– Nevay
Apr 16, 2018 at 19:57
1

\$\begingroup\$ Use orElse(0) instead of getAsDouble(). \$\endgroup\$
– Olivier Grégoire
Apr 17, 2018 at 7:43
\$\begingroup\$ 69 bytes, based on others' solution. Round using (int) for 5 mote bytes, if needed. \$\endgroup\$
– Olivier Grégoire
Apr 17, 2018 at 7:54
\$\begingroup\$ You don't need to cast as a double: Math.pow already takes care of that. That'll spare you 3 bytes. \$\endgroup\$
– Olivier Grégoire
Apr 17, 2018 at 9:56
\$\begingroup\$ @OlivierGrégoire It will give incorrect results if I do that. The cast to int is used so we can integer-divide by 9 and multiply by the sum of digits. Only then it should a double to get the average. If I remove both (int) and *.1 it will for example output 6388.888... instead of 6388.25 for the input 4928. And if I cast the entire thing or just the .pow to an int instead, it will output 6388. \$\endgroup\$
– Kevin Cruijssen
Apr 17, 2018 at 11:12

| Show 2 more comments

Fyr · Accepted Answer · 2018-04-17 15:47:01Z

3

Husk, 5 bytes

d´MKA

Try it online!

Explanation

d´MKA
    A  Take the average of the digits
 ´MK   Replace each element of the original list with the average
d      Join the list to get a number

Husk, 7 bytes

A§modṙŀ

Try it online!

Explanation

A§modṙŀ
      ŀ  Take the range [1..length(input)]
 §m  ṙ   Rotate the input by each element of the range
   od    Convert each list of digits to a number
A        Take the average of the list

edited Apr 17, 2018 at 15:47

answered Apr 16, 2018 at 15:37

Fyr

6113 silver badges4 bronze badges

1

\$\begingroup\$ For a puzzle, there is at least one 5 byte solution \$\endgroup\$
– H.PWiz
Apr 16, 2018 at 16:53
\$\begingroup\$ @H.PWiz I can't figure it out, could you give out a hint? :P \$\endgroup\$
– Leo
Apr 17, 2018 at 12:01
\$\begingroup\$ @Leo There is no ṙ or ŀ, and the first character (on the left) is not A \$\endgroup\$
– H.PWiz
Apr 17, 2018 at 12:21

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Giuseppe · Accepted Answer · 2018-05-06 21:31:09Z

3

R, 84 73 64 bytes

function(D,K=sum(D|1))mean(array(D,K+1:0)[1:K,1:K]%*%10^(K:1-1))

Try it online!

Input as list of digits.

Thanks to MickyT for shaving off 11 bytes! 8 bytes shaved by Dennis' proof that deduplication is unnecessary.

edited May 6, 2018 at 21:31

answered Apr 16, 2018 at 15:21

Giuseppe

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\$\begingroup\$ With a slightly different way to rotate the number for 73 \$\endgroup\$
– MickyT
Apr 16, 2018 at 21:13
\$\begingroup\$ @MickyT aaahhh clever use of recycling! \$\endgroup\$
– Giuseppe
Apr 17, 2018 at 16:23
\$\begingroup\$ @MickyT array(D,K+1:0) is shorter than matrix(D,K+1,K) by a byte. \$\endgroup\$
– Giuseppe
May 6, 2018 at 21:31

Add a comment |

tybocopperkettle · Accepted Answer · 2022-09-28 22:38:17Z

3

Vyxal, 3 bytes

Ṗ⌊ṁ

-2 bytes thanks to @lyxal

Try it online!

Explanation

Ṗ   # Permutations
 ⌊  # Convert each to an integer
  ṁ # Mean

edited Sep 28, 2022 at 22:38

answered Sep 20, 2022 at 20:49

tybocopperkettle

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1

\$\begingroup\$ Try it Online! for 4 bytes, because ṅ vectorises when there's sublists \$\endgroup\$
– lyxal
Sep 27, 2022 at 14:23
\$\begingroup\$ @Lyxal Oh yeah I just realised that a few days ago, thanks \$\endgroup\$
– tybocopperkettle
Sep 28, 2022 at 7:13
1

\$\begingroup\$ Try it Online! for 3 bytes \$\endgroup\$
– lyxal
Sep 28, 2022 at 11:02

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user2956892 · Accepted Answer · 2018-04-16 15:28:33Z

2

05AB1E, 9 bytes

vÀD}\OIg/

Try it online!

answered Apr 16, 2018 at 15:28

user2956892

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\$\begingroup\$ v with no y, interesting. \$\endgroup\$
– Magic Octopus Urn
Apr 17, 2018 at 15:01
\$\begingroup\$ gFÀD})¨Osg/ was where I was thinking. \$\endgroup\$
– Magic Octopus Urn
Apr 17, 2018 at 15:02
\$\begingroup\$ Do you know how to use the .æ = pop a compute permutations by function, usage: .æ<FUNC>} command yet? I don't either, but it seems fit to this. \$\endgroup\$
– Magic Octopus Urn
Apr 17, 2018 at 15:02
\$\begingroup\$ @MagicOctopusUrn v without y is the shortest solution I could find to perform the rotation g(input) times. I'm checking the .æ, doesn't seem like it's registering <FUNC>} \$\endgroup\$
– user2956892
Apr 17, 2018 at 15:56

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recursive · Accepted Answer · 2018-04-16 15:44:50Z

2

Stax, 6 bytes

ñJä⌠╤►

Run and debug it

This program takes a quote-delimited string as input, and expresses the average as a reduced fraction. e.g. 777/1 It's not necessary to de-duplicate the rotations. It never changes the result.

Unpacked, ungolfed, and commented, it looks like this.

:)  get all character rotations
{em convert strings back to integers
:V  mean - integer inputs means result will be rational

Run this one

edited Apr 16, 2018 at 15:44

answered Apr 16, 2018 at 15:36

recursive

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Dennis · Accepted Answer · 2018-04-16 16:09:05Z

2

Jelly, 6 5 bytes

ṙJḌÆm

Try it online!

How it works

ṙJḌÆm  Main link. Argument: A (digit array)

 J     Yield the indices of A, i.e., [1, ..., len(A)].
ṙ      Rotate A 1, ..., and len(A) units to the left, yielding a 2D array.
  Ḍ    Convert each rotation from decimal to integer.
   Æm  Take the arithmetic mean.

edited Apr 16, 2018 at 16:09

answered Apr 16, 2018 at 15:13

Dennis

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Phil H · Accepted Answer · 2018-04-16 20:26:42Z

2

Perl 6, 15 bytes

{.sum/$_*1 x$_}

Try it online!

The average is the digit average applied to each decimal position, so digit average times 111.... 1 x $_ produces a string of 1s which get coerced to strings by the multiply.

Takes a list of digits as input. A sequence would require a .cache before the sum, and a number or string input would need a .comb.

answered Apr 16, 2018 at 20:26

Phil H

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Ton Hospel · Accepted Answer · 2018-04-17 10:02:28Z

2

Perl 5 `-lpF`, 24 22 bytes

#!/usr/bin/perl -lpF
$_=1x@F/s/./+$&/g*eval

Try it online!

Doing it as a list of digits is only 1 byte shorter and feels like cheating:

#!/usr/bin/perl -p
$;+=$_}{$_=1x$./$.*$

Try it online!

edited Apr 17, 2018 at 10:02

answered Apr 16, 2018 at 19:47

Ton Hospel

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\$\begingroup\$ what's usrt? :P \$\endgroup\$
– ASCII-only
Apr 17, 2018 at 1:40
\$\begingroup\$ @ASCII-only Wait,you don't have your executables in the /usrt directory ? Anyways, fixed. Thanks \$\endgroup\$
– Ton Hospel
Apr 17, 2018 at 6:49

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G B · Accepted Answer · 2018-04-17 10:19:49Z

2

Ruby, 60 bytes

->n{a=b=c=0.0;a,b,c,n=a+n%10,b*10+1,c+1,n/10while n>0;a*b/c}

Try it online!

answered Apr 17, 2018 at 10:19

G B

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alephalpha · Accepted Answer · 2018-04-17 11:42:17Z

2

Wolfram Language (Mathematica), 22 bytes

FromDigits[0#+Mean@#]&

Try it online!

answered Apr 17, 2018 at 11:42

alephalpha

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

2

Ruby, 33 bytes

->n{10**(k=n.size)/9*n.sum*1.0/k}

Try it online!

A port of Dennis's Python 3 answer

Here's my lame attempt (74 bytes)

->s{a=(1..s.size).map{|i|s.chars.rotate(i).join.to_i}|[];a.sum*1.0/a.size}

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 17, 2018 at 15:22

Asone Tuhid

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

1

JavaScript (Node.js), 43 bytes

x=>eval(x.join`+`)*'1'.repeat(n=x.length)/n

Try it online!

Can we take input as a list of digits? – Dennis♦ 7 mins ago

@Dennis Sure, that's fine. Go save some bytes. :p – AdmBorkBork 3 mins ago

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 16, 2018 at 15:27

l4m2

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

1

Japt, 8 bytes

Takes input as an array of single digit strings.

xpUÊ)÷UÊ

Try it

Explanation

             :Implicit input of array U
 pUÊ         :Repeat each string length of U times
x   )        :Reduce by addition
     ÷UÊ     :Divide by the length of U

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 16, 2018 at 16:23

Shaggy

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

APL (Dyalog Unicode), 21 14 bytes^SBCS

+/≢÷⍨⍳∘≢(⍎⌽)¨⊂

Try it online!

Tacit prefix function. Takes input as a string.

Thanks to Adám for an enlightening 7 byte save.

How?

+/≢÷⍨⍳∘≢(⍎⌽)¨⊂ ⍝ Main fn, example argument '123'
              ⊂ ⍝ Enclose the argument (turns it into a scalar)
             ¨  ⍝ Use each of the left arguments to
         ( ⌽)   ⍝ Rotate, then
          ⍎     ⍝ Convert strings into numbers
      ⍳∘≢       ⍝ Tally (≢) the argument, then index (⍳) from 1. 
                ⍝ Returns 1 2 3 for a 3 digit argument, and rotates the argument 1, 2, then 3 times.
     ⍨          ⍝ Use the result as left argument for
    ÷           ⍝ Divide
   ≢            ⍝ By the number of rotations
+/              ⍝ And sum the results

\$\begingroup\$ :| still can't get used to APL comments \$\endgroup\$
– ASCII-only
Apr 17, 2018 at 1:40 — ASCII-only, Apr 17, 2018 at 1:40

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

1

Python 2, 83 77 bytes

def f(a):b={int(`a`[i:]+`a`[:i])for i in range(len(`a`))};print sum(b)/len(b)

EDIT: -6 bytes thanks to @ovs

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 16, 2018 at 17:07

AvahW

5434 silver badges7 bronze badges

\$\begingroup\$ 35, Dennis' solution, 38, not Dennis' solution \$\endgroup\$
– ASCII-only
Apr 17, 2018 at 1:32

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Neil · Accepted Answer · 2018-04-17 00:08:13Z

1

Charcoal, 11 bytes

Ｉ∕×ΣθＩ⭆θ1Ｌθ

Try it online! Link is to verbose version of code. Explanation:

    θ  θ  θ Input as a string
   Σ        Sum of digits
      ⭆ 1   Replace each character with the literal `1`
     Ｉ      Cast to integer
  ×         Multiply
         Ｌ  Length
 ∕          Divide
Ｉ           Cast to string
            Implicitly print

answered Apr 17, 2018 at 0:08

Neil

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Galen Ivanov · Accepted Answer · 2018-04-17 07:27:47Z

1

J, 10 bytes

10#.#$+/%#

This is a port of H.PWiz's great APL solution to J.

Takes a list of digits as an argument.

Explanation:

+/%# the average of the digits (divide % the sum of the digits +/ by their number #)

#$creates a list of copies of the average according to the number of digits

10#. convert form base 10

Try it online!

answered Apr 17, 2018 at 7:27

Galen Ivanov

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alephalpha · Accepted Answer · 2018-04-17 11:40:01Z

1

Pari/GP, 34 bytes

Takes input as a list of digits.

n->fromdigits([vecsum(n)|x<-n])/#n

Try it online!

answered Apr 17, 2018 at 11:40

alephalpha

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HatsuPointerKun · Accepted Answer · 2018-05-19 17:00:36Z

1

C++, 218 208 bytes

-10 bytes thanks to Zacharý

#include<set>
#include<cmath>
float a(int m){std::set<int>a;int n=m,t,c=0;for(;n>0;n/=10)++c;for(;n<c;++n){a.insert(m);t=m%10;m=m/10+std::pow(10.f,c-1)*t;}int s=0;for(int v:a){s+=v;}return float(s)/a.size();}

And, to test :

int main() {
    printf("%f\n%f\n%f\n%f\n",a(123),a(4928),a(445445),a(777));
}

edited May 19, 2018 at 17:00

answered May 6, 2018 at 20:50

HatsuPointerKun

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1

\$\begingroup\$ You don't need spaces between #include and <, and you can remove the {} around both ++c; and s+=v;. You might be able to move the int s=0 to the beginning with your other variables. \$\endgroup\$
– Adalynn
May 10, 2018 at 19:08
1

\$\begingroup\$ Also, I don't think you need the n=0 in the second for loop, as it should have reached 0 by then. m/=10;m+=std::pow(10.f,c-1)*t; => m=m/10+std::pow(10.f,c-1)*t. And wouldn't using int instead of auto work? \$\endgroup\$
– Adalynn
May 10, 2018 at 19:12
\$\begingroup\$ You can still move int s=0; with your other variables, and do you need the braces around the s+=v;? \$\endgroup\$
– Adalynn
May 23, 2018 at 12:37
\$\begingroup\$ 168 bytes \$\endgroup\$
– ceilingcat
Jun 16, 2018 at 0:21
\$\begingroup\$ n>0 => n might work. \$\endgroup\$
– Adalynn
Jun 30, 2018 at 21:25

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Kevin Cruijssen · Accepted Answer · 2022-09-28 10:57:59Z

1

05AB1E, 3 bytes

œÅA

Port of @tybocopperkettle's Vyxal answer, so make sure to upvote him/her as well!

Try it online or verify all test cases.

Explanation:

œ    # Get all permutations of the (implicit) input
     #  e.g. 123 → ["123","132","213","231","312","321"]
 ÅA  # Take the average of this list
     #  → 222.0
     # (which is output implicitly as result)

answered Sep 28, 2022 at 10:57

Kevin Cruijssen

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The Thonnu · Accepted Answer · 2023-02-01 15:19:48Z

1

Thunno `E`, \$ 9 \log_{256}(96) \approx \$ 7.41 bytes

DLzP.ZDAA

Attempt This Online!

Port of Vyxal.

Explanation

DLzP.ZDAA  # Implicit input. The E flag keeps it as a string.
DL         # Duplicate and get the length of the input string.
  zP       # Permutations of the input with the same length.
    .ZD    # Converted to integers from lists of characters.
       AA  # Get the average of this list of integers.
           # Implicit output.

answered Feb 1 at 15:19

The Thonnu

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Yousername · Accepted Answer · 2023-02-01 17:28:21Z

1

TI-Basic, 10 bytes

mean(Ans)int(1/9₁₀^(dim(Ans

Takes input in Ans as a list. Port of Dennis's Python 3 answer.

answered Feb 1 at 17:28

Yousername

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RK. · Accepted Answer · 2018-04-16 15:24:18Z

0

Pyth, 12 bytes

csms.<zdlzlz

Probably improvable.

c         lz     Divide by the length of the input:
 s               Reduce by +
  m     lz       Map over d = [0 ... the length of the input]:
   s.<zd         Shift the input d characters to the left and cast to int

Try it here!

answered Apr 16, 2018 at 15:24

RK.

4972 silver badges11 bronze badges

\$\begingroup\$ There's a built-in average function o. If you do that and do I/O as lists of digits, you can get it down to 8 bytes. \$\endgroup\$
– user48543
Apr 16, 2018 at 15:30
\$\begingroup\$ Ah, I was taking "If applicable, you can assume that the input/output will fit in your language's native Integer type." to mean that it had to be an integer if taken in as Q. \$\endgroup\$
– RK.
Apr 16, 2018 at 16:21

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Bolce Bussiere · Accepted Answer · 2018-04-17 01:41:34Z

0

J, 23 Bytes

(+/%#)".~.(|."0 1~i.@#)

Takes input as a string

Explanation

          (|."0 1~i.@#)  | All rotations
        ~.               | Deduplicate
      ".                 | Convert each to int
(+/%#)                   | Average

answered Apr 17, 2018 at 1:41

Bolce Bussiere

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

0

Matlab, 65 bytes

c=num2str(n);l=nnz(c);mean(str2num(c(mod((1:l)+(0:l-1)'-1,l)+1)))

Gonna work on this, pretty sure it can be done better.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 17, 2018 at 13:52

Leander Moesinger

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NikoNyrh · Accepted Answer · 2018-04-18 10:14:21Z

0

Clojure, 139 bytes

#(let[n(count %)G(set(apply map list(for[i(range n)](take n(drop i(cycle %))))))](/(apply +(for[g G](read-string(apply str g))))(count G)))

Quite sub-optimal language features for converting sequences of characters to integers.

answered Apr 18, 2018 at 10:14

NikoNyrh

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

dc, 37 bytes

This is a full program, reading input and printing the output:

?1sd[O~rzsadO<x+ldO*1+sd]dsxxOkld*la/p

It works by separating the number into its digits, and multiplying the mean of the digits by the appropriate length repdigit (which is built up in d as we go).

?                               # read input
 1sd                            # initialize d=1
    [                   ]dsxx   # define and execute recursive macro x:
     O~r                        #   remainder and quotient of /10
        zsa                     #   a = number of digits
           dO<x                 #   recurse if needed
               +ldO*1+sd        #   increment repdigit d
                             Ok         # after executing x, set precision 
                               ld*la/   # multiply by repdigit; divide by a
                                     p  # print the result

Rotational Average

Rules

32 Answers 32

Python 3, 38 36 bytes

APL (Dyalog), 9 bytes

Java 10, 163 137 76 72 71 bytes

Husk, 5 bytes

Explanation

Husk, 7 bytes

Explanation

R, 84 73 64 bytes

Vyxal, 3 bytes

05AB1E, 9 bytes

Stax, 6 bytes

Jelly, 6 5 bytes

How it works

Perl 6, 15 bytes

Perl 5 -lpF, 24 22 bytes

Ruby, 60 bytes

Wolfram Language (Mathematica), 22 bytes

Ruby, 33 bytes

Here's my lame attempt (74 bytes)

JavaScript (Node.js), 43 bytes

Japt, 8 bytes

Explanation

APL (Dyalog Unicode), 21 14 bytesSBCS

How?

Python 2, 83 77 bytes

Charcoal, 11 bytes

J, 10 bytes

Explanation:

Pari/GP, 34 bytes

C++, 218 208 bytes

05AB1E, 3 bytes

Thunno E, \$ 9 \log_{256}(96) \approx \$ 7.41 bytes

Explanation

TI-Basic, 10 bytes

Pyth, 12 bytes

J, 23 Bytes

Explanation

Matlab, 65 bytes

Clojure, 139 bytes

dc, 37 bytes

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Perl 5 `-lpF`, 24 22 bytes

APL (Dyalog Unicode), 21 14 bytes^SBCS

Thunno `E`, \$ 9 \log_{256}(96) \approx \$ 7.41 bytes

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