Add All Permutations of a Number

Question

You have to make something that takes in one input from a default I/O method (as an integer), and prints out the sum of all the permutations of that number (not necessarily unique)

For example:

10 would return 11 because 10 can have 2 permutations (10 and 01), and the sum of those two numbers would be 11

202 would have 6 permutations (202, 220, 022, 022, 202, 220), so the sum would be 888.

The input is positive, and the input can be taken as any type, but your program's output must be a number.

Standard loopholes are not allowed, and as usual, since this is code golf, the shortest answer in bytes wins!

Note: Peter Taylor pointed out a simplification of the challenge here. If you are reading this, you may want to think about it first to see if you can get the same (or shorter) solution. — user202729, Oct 20 '18 at 3:16

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

8

Japt `-x`, 1 byte

Takes input as a string.

á

Try it

Explanation

á     :Get permutations
      :Implicitly reduce by addition and output

edited Jun 17 '20 at 9:04

Community♦

1

answered Oct 20 '18 at 8:34

Shaggy

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\$\begingroup\$ Doesn't the -x count as an extra byte? \$\endgroup\$ – Zacharý Oct 21 '18 at 2:12
3

\$\begingroup\$ @Zacharý No, here's the meta. \$\endgroup\$ – u-ndefined Oct 21 '18 at 2:15

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xnor · Accepted Answer · 2018-10-20 05:17:34Z

6

Python 2, 66 bytes

n=s=0
P=1
for c in input():s+=int(c);n+=1;P*=n
print 10**n/9*s*P/n

Try it online!

Takes input as a string.

Python 2, 70 bytes

f=lambda k,P=1,n=0,s=0:f(k/10,P*-~n,n+1,s+k%10)if k else 10**n/9*s*P/n

Try it online!

An arithmetic method. The base case is hard to deal with because the /n causes division by zero for the inital value n=0.

answered Oct 20 '18 at 5:17

xnor

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Okx · Accepted Answer · 2018-10-20 08:03:28Z

5

05AB1E, 2 bytes

œO

Try it online!

answered Oct 20 '18 at 8:03

Okx

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\$\begingroup\$ What is the O? I know œ means permutations, but what does O mean? If you answer this, and the program follows the rules, I will accept this answer, because it has the smallest number of bytes. \$\endgroup\$ – MilkyWay90 Oct 20 '18 at 14:33
3

\$\begingroup\$ @MilkyWay90 If œ is permutations, O ought to be sum. It's customary to wait at least a week before accepting an answer. \$\endgroup\$ – Dennis Oct 20 '18 at 15:28

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Esolanging Fruit · Accepted Answer · 2019-03-23 03:23:24Z

5

CJam, 10 6 bytes

{m!1b}

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Explanation

{    }  Anonymous block taking input as string on stack
 m!     Find all permutations
   1b   Implicitly convert each to integer and sum

edited Mar 23 '19 at 3:23

answered Oct 20 '18 at 7:04

Esolanging Fruit

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\$\begingroup\$ Can you add an explanation so I can decide whether to upvote or not? \$\endgroup\$ – MilkyWay90 Mar 23 '19 at 3:19
\$\begingroup\$ @MilkyWay90 Added. \$\endgroup\$ – Esolanging Fruit Mar 23 '19 at 3:23
\$\begingroup\$ Okay! This seems like a trivial answer, so I will refrain from upvoting. However, still a great answer! Also, are CJam builtins always 2 bytes or is it just a few? \$\endgroup\$ – MilkyWay90 Mar 23 '19 at 3:26
1

\$\begingroup\$ @MilkyWay90 Most builtins in CJam are one byte, but there are two groups of two-byte builtins (starting with e and m) for "extra" operations and "math" operations respectively. 1b is actually two different operations: 1 pushes the number 1, and b performs base conversion. 1b means "convert from base 1" but it has the effect of summing an array because b doesn't check that the array elements are actually valid digits. Normally one would take the sum with :+ (reduce with addition), but 1b has the advantage that it works on empty arrays and implicitly converts to integer. \$\endgroup\$ – Esolanging Fruit Mar 23 '19 at 3:37
\$\begingroup\$ Ah, I see. That does seem to be worth an upvote \$\endgroup\$ – MilkyWay90 Mar 23 '19 at 3:41

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Jo King · Accepted Answer · 2018-10-20 01:59:17Z

4

Perl 6, 30 bytes

*.comb.permutations>>.join.sum

Try it online!

It would be nice if this was just *.permutations.sum but Perl 6 doesn't treat strings as lists of characters.

Explanation

*.comb.permutations>>.join.sum
*.comb                           # Convert to list of digits
      .permutations              # Get all permutations of the list
                   >>.join       # Join all lists of digits
                          .sum   # Get the sum of all numbers

answered Oct 20 '18 at 1:59

Jo King♦

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\$\begingroup\$ Nice job! I look forward to receiving more responses to this question! Since I am not familiar with Perl, the explanation was very useful! \$\endgroup\$ – MilkyWay90 Oct 20 '18 at 2:03
1

\$\begingroup\$ @MilkyWay90 The name of the language is “Perl 6”. It is a sister language to “Perl 5”. Normally when people say “Perl” they mean “Perl 5”. \$\endgroup\$ – Brad Gilbert b2gills Oct 20 '18 at 20:16

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Dennis · Accepted Answer · 2018-10-20 13:57:25Z

3

Python 2, 63 62 bytes

f=lambda n,k=1,s=0:n>9and k*f(n/10,k+1,s+n%10)or 10**k/9*(s+n)

Try it online!

edited Oct 20 '18 at 13:57

answered Oct 20 '18 at 13:21

Dennis

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Kirill L. · Accepted Answer · 2018-10-20 17:36:39Z

3

R, 59 57 bytes

sum(gamma(n<-nchar(x<-scan()))*x%/%10^(0:n)%%10)*10^n%/%9

Try it online!

Mathematical approach suggested by Peter Taylor and decribed by Arnauld. -2 bytes by J. Doe.

edited Oct 20 '18 at 17:36

answered Oct 20 '18 at 13:22

Kirill L.

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\$\begingroup\$ Trivial inline golf for 57 \$\endgroup\$ – J.Doe Oct 20 '18 at 15:33

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Galen Ivanov · Accepted Answer · 2018-10-20 08:07:55Z

J, 18 bytes

1#.(A.~i.@!@#)&.":

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Explanation:

                ":  - convert to string
              &.    - do the verbs in () then convert back to number
       i.           - make a list from 0 to
         @!         - factorial 
           @#       - of the number of digits in the input        
    A.~             - find all permutations, using the list above as permutation index
1#.                 - find the sum by base-1 convertion

Alternative:

Using the formula suggested by Peter Taylor:

J, 30 bytes

1#.("."0*9%~!@<:@#*_1+10^#)@":

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

2

JavaScript (ES6), 53 bytes

Recursive version, inspired by @tsh.

f=(n,s=i=0,p='')=>n?(i++||1)*f(n/10|0,s+n%10,p+1):s*p

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JavaScript (ES6), 61 60 bytes

Saved 1 byte thanks to @l4m2

n=>![...n+(x=s=p='')].map((d,i)=>(p+=1,x=x*i||1,s-=d))-s*p*x

Try it online!

How?

This is based on the formula suggested by Peter Taylor in the sandbox:

$$\left(\sum_{i=1}^n{a_i}\right)\frac{10^n-1}{9}(n-1)!$$

where \$a_i\$ is the \$i\$^th digit of the input number and \$n\$ is the total number of digits.

The result of the expression \$(10^n-1)/9\$ is a number consisting of the digit \$1\$ repeated \$n\$ times, which is what is computed in \$p\$. The factorial is stored in \$x\$ and the opposite of the sum is stored in \$s\$.

edited Jun 17 '20 at 9:04

Community♦

1

answered Oct 20 '18 at 4:50

Arnauld

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\$\begingroup\$ n=>![...n+(x=s=p='')].map((d,i)=>(p+=1,x=x*i||1,s-=d))-s*p*x \$\endgroup\$ – l4m2 Oct 20 '18 at 8:33
\$\begingroup\$ f=(n,s=i=!n,p='')=>n?++i*f(n/10|0,s+n%10,p+1):s*p/i; if input > 0 then change !n to 0. \$\endgroup\$ – tsh Oct 20 '18 at 12:24

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nosewings · Accepted Answer · 2018-10-21 00:44:04Z

2

Haskell, 62 bytes

import Data.List
f::Int->Int
f=sum.map read.permutations.show

Unfortunately, f must be annotated with a type to avoid ambiguity.

answered Oct 21 '18 at 0:44

nosewings

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1

\$\begingroup\$ Welcome to PPCG and Haskell golfing in particular! Your code works fine on TIO without the type annotations: Try it online!. Also anonymous functions are allowed, so you can drop the f=. \$\endgroup\$ – Laikoni Oct 21 '18 at 0:59

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Zacharý · Accepted Answer · 2018-10-21 02:03:43Z

MY, 12 bytes

⍞℘⌊Σ88ǵ'ƒ⇥(↵

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Some of MY's stupid builtin decisions are coming in handy.

⍞℘⌊Σ88ǵ'ƒ⇥(↵
⍞             Input
  ℘                's permutations
   ⌊                                as integers.
    Σ                                           Sum of that
     88ǵ'ƒ⇥(                                               to standard base 10 form
             ↵                                                                      output.

Chas Brown · Accepted Answer · 2018-10-20 03:41:54Z

1

Python 2, 76 bytes

s=map(int,`input()`);n=len(s);t=sum(s)*int('1'*n)
while~-n:n-=1;t*=n
print t

Try it online!

answered Oct 20 '18 at 3:41

Chas Brown

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Kirill L. · Accepted Answer · 2018-10-20 07:24:25Z

1

Ruby `-nl`, 41 bytes

p$_.chars.permutation.sum{|x|x.join.to_i}

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Full program taking input from stdin.

Ruby, 45 bytes

->n{n.digits.permutation.sum{|x|x.join.to_i}}

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Function taking input as integer. digits can be shortened to chars if input is acceptable as string, and chars can be completely removed if input is an array of characters.

edited Oct 20 '18 at 7:24

answered Oct 20 '18 at 7:12

Kirill L.

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Jo King · Accepted Answer · 2018-10-20 07:26:04Z

1

Python 2, 76 bytes

lambda x:sum(map(int,x))*int('1'*len(x))*reduce(int.__mul__,range(1,len(x)))

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xnor's answer is shorter, but I thought I'd have a go at a one-liner

answered Oct 20 '18 at 7:26

Jo King♦

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Erik the Outgolfer · Accepted Answer · 2018-10-20 08:39:17Z

1

Jelly, 5 bytes

DŒ!ḌS

Try it online!

answered Oct 20 '18 at 8:39

Erik the Outgolfer

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Neil · Accepted Answer · 2018-10-20 10:26:18Z

Charcoal, 12 bytes

Ｉ↨χＥθΠＥθ∨μΣθ

Try it online! Link is to verbose version of code. Explanation: Based on Peter Taylor's formula. ΠＥθμ doesn't quite calculate \$ (n - 1)! \$ as Charcoal's indices range from \$ 0 \$ to \$ n - 1 \$ so ∨ is used to replace the \$ 0 \$ with the sum of digits of the input, thus implicitly multiplying the two. The outer mapping then creates an array of the length of the original input with that product as elements which is then passed to base 10 conversion which effectively multiplies the product by the appropriate repunit.

    θ           Input
   Ｅ            Map over characters
       θ        Input
      Ｅ         Map over characters
         μ      Current index
        ∨       Logical Or
           θ    Input
          Σ     Digital sum
     Π          Product
  χ             Predefined variable 10
 ↨              Base conversion
Ｉ               Cast to string
                Implicitly print

Logern · Accepted Answer · 2018-10-20 13:57:06Z

1

C (gcc), 89 71 bytes

g(x,s,p,f,i){for(s=x%10,p=f=i=1;x/=10;s+=x%10)p=p*10+1,f*=i++;x=s*p*f;}

Try it online!

-18 bytes thanks to @nwellnhof

edited Oct 20 '18 at 13:57

answered Oct 20 '18 at 4:06

Logern

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\$\begingroup\$ 68 bytes \$\endgroup\$ – ceilingcat Oct 22 '18 at 23:51

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

1

APL(NARS), 25 chars, 50 bytes

{+/10⊥⍉⍎¨w[110 1‼↑⍴w←⍕⍵]}

test:

  f←{+/10⊥⍉⍎¨w[110 1‼↑⍴w←⍕⍵]}
  f¨202 1024 505
888 46662 2220

110 1‼k would return as built in all the permutations of 1..k .

edited Jun 17 '20 at 9:04

Community♦

1

answered Oct 20 '18 at 17:25

user58988

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jrook · Accepted Answer · 2018-10-20 18:04:26Z

1

Scala (74 bytes)

def%(i:Int)=s"$i".indices.permutations.map(_.map(i+"").mkString.toInt).sum

Try it online

Since permutation on digits will drop the repeated digits, we have to permute on indices.

edited Oct 20 '18 at 18:04

answered Oct 20 '18 at 8:15

jrook

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u-ndefined · Accepted Answer · 2018-10-21 01:17:41Z

1

Pyth, 5 4 bytes

sv.p

Try it online! Takes input as a string.

Explanation:

sv.pQ - Full program. Implicit Q added.

    Q - Input
  .p  - All permutations of it
 v    - Evaluate each (list of strings into integers)
s     - Sum the list

edited Oct 21 '18 at 1:17

answered Oct 20 '18 at 12:00

u-ndefined

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u-ndefined · Accepted Answer · 2018-10-21 04:02:28Z

1

Pyt, 4 bytes

ǰᒆƖƩ

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Explanation:

      Implicit input
ǰ     Join (becomes string)
 ᒆ    All permutations of it
  Ɩ   List of strings to int
   Ʃ  Sum the array

answered Oct 21 '18 at 4:02

u-ndefined

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Add All Permutations of a Number

21 Answers 21

Japt `-x`, 1 byte

Explanation

Python 2, 66 bytes

Python 2, 70 bytes

05AB1E, 2 bytes

CJam, 10 6 bytes

Explanation

Perl 6, 30 bytes

Explanation

Python 2, 63 62 bytes

R, 59 57 bytes

J, 18 bytes

Explanation:

Alternative:

J, 30 bytes

JavaScript (ES6), 53 bytes

JavaScript (ES6), 61 60 bytes

How?

Haskell, 62 bytes

MY, 12 bytes

Python 2, 76 bytes

Ruby `-nl`, 41 bytes

Ruby, 45 bytes

Python 2, 76 bytes

Jelly, 5 bytes

Charcoal, 12 bytes

C (gcc), 89 71 bytes

APL(NARS), 25 chars, 50 bytes

Scala (74 bytes)

Pyth, 5 4 bytes

Pyt, 4 bytes

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Add All Permutations of a Number

21 Answers 21

Japt -x, 1 byte

Explanation

Python 2, 66 bytes

Python 2, 70 bytes

05AB1E, 2 bytes

CJam, 10 6 bytes

Explanation

Perl 6, 30 bytes

Explanation

Python 2, 63 62 bytes

R, 59 57 bytes

J, 18 bytes

Explanation:

Alternative:

J, 30 bytes

JavaScript (ES6), 53 bytes

JavaScript (ES6), 61 60 bytes

How?

Haskell, 62 bytes

MY, 12 bytes

Python 2, 76 bytes

Ruby -nl, 41 bytes

Ruby, 45 bytes

Python 2, 76 bytes

Jelly, 5 bytes

Charcoal, 12 bytes

C (gcc), 89 71 bytes

APL(NARS), 25 chars, 50 bytes

Scala (74 bytes)

Pyth, 5 4 bytes

Pyt, 4 bytes

Your Answer

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Japt `-x`, 1 byte

Ruby `-nl`, 41 bytes