Range, Reverse, Sum!

Question

Given a positive integer n as input, output the reversed range sum of n.

A reversed range sum is created by making an inclusive range up to n, starting with 1 and including n, reversing each of the numbers inside, and summing it.

Example:

Here is what would happen for an input of 10:

Range: [1,2,3,4,5,6,7,8,9,10]

Reverse: [1,2,3,4,5,6,7,8,9,01] (1-char numbers reversed are themselves, 10 reversed is 01 or 1)

Sum: 46

Numbers with 3+ digits are reversed the same way numbers with 2 digits are. For example, 1234 would become 4321.

Test cases:

Input -> Output

10 -> 46
5 -> 15
21 -> 519
58 -> 2350
75 -> 3147
999 -> 454545

Complete text cases to input of 999 can be found here, thanks very much to @fireflame241.

Answer 1

05AB1E, 3 bytes

Code

LíO

Uses the 05AB1E encoding. Try it online!

Explanation

L       # Range
 í      # Reverse
  O     # Sum

Answer 2

Bash + GNU utils, 24

seq $1|rev|paste -sd+|bc

Try it online.

Explanation

seq $1                    # range
      |rev                # reverse (each line)
          |paste -sd+|bc  # sum

Answer 3

Perl 6, 20 bytes

{(1..$_)».flip.sum}

Test it

Expanded:

{
   ( 1 .. $_ )\  # Range
   ».flip        # flip each value in the Range (possibly in parallel)
   .sum          # sum up the list
}

Answer 4

JavaScript (ES6), 42 bytes

f=n=>n&&+[...n+""].reverse().join``+f(n-1)

My favorite doubly-recursive solution is unfortunately 3 bytes longer:

f=n=>n&&+(g=x=>x?x%10+g(x/10|0):"")(n)+f(n-1)

Answer 5

Retina, 41 36 35 bytes

.+
$*
1
1$`¶
1+
$.&
%O^$`.

.+
$*
1

Try it online! Link includes test cases. Edit: Saved 5 bytes thanks to @FryAmTheEggman. Saved 1 byte thanks to @PunPun1000. Explanation:

.+
$*

Convert to unary.

1
1$`¶

Create a range from 1 to n.

1+
$.&

Convert back to decimal.

%O^$`.

Reverse each number.

.+
$*

Convert back to unary.

1

Sum and convert back to decimal.

Answer 6

Jelly, 4 bytes

Ṛ€ḌS

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How?

Ṛ€ḌS - Link: n
Ṛ€   - reverse for €ach (in implicit range)
  Ḍ  - convert from decimal list (vectorises)
   S - sum

Answer 7

Haskell, 34 bytes

\n->sum$read.reverse.show<$>[1..n]

Simple and straightforward.

Answer 8

C (gcc), 63 bytes

f(n){int t=0,c=n;for(;c;c/=10)t=t*10+c%10;return n?t+f(n-1):0;}

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Answer 9

cQuents, 4 bytes

;\r$

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Explanation

       Implicit input n.
;      Series mode. Outputs the sum of the sequence from 1 to n.
 \r$   Each item in the sequence equals:
 \r    String reverse of
   $                     current index (1-based)

Answer 10

Python 2, 38 bytes

Can't compute higher terms than the recursion limit:

f=lambda x:x and int(`x`[::-1])+f(x-1)

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Answer 11

Brachylog, 4 bytes

⟦↔ᵐ+

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Explanation

⟦↔ᵐ+
⟦        range from 0 to input
 ↔ᵐ      map reverse
   +     sum

Answer 12

Röda, 56 41 36 bytes

15 bytes saved thanks to @fergusq

{seq 1,_|parseInteger`$_`[::-1]|sum}

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This is an anonymous function that takes an integer from the input stream and outputs an integer to the output stream.

Explanation

{seq 1,_|parseInteger`$_`[::-1]|sum} Anonymous function
 seq 1,_                             Create a sequence 1 2 3 .. input and push each value to the stream
        |                            For each value in the stream:
                     `$_`             Cast it into a string
                         [::-1]       And reverse it
         parseInteger                 And parse the resulting string as an integer, while pushing the value to the stream
                               |sum  Sum all the values in the stream

Answer 13

C# (.NET Core), 103 97 bytes

using System.Linq;r=>new int[r+1].Select((_,n)=>int.Parse(string.Concat((n+"").Reverse()))).Sum()

Try it online!

TIO link outputs all the results from 1 to 999, so feel free to check my work.

I expected this to be a bit shorter, but it turns out Reverse() returns an IEnumerable<char> instead of another string so I had to add some extra to turn it back into a string so I could parse it to an int. Maybe there's a shorter way to go from IEnumerable<char> to int correctly.

Of minor note, this also uses the functions Range() Reverse() and Sum() all in order.

-6 bytes thanks to TheLethalCoder

Answer 14

Ruby, 56, 52, 41, 39 bytes

->n{(1..n).sum{|i|i.to_s.reverse.to_i}}

Ruby, 34 bytes (if lambda param is a string)

->n{(1..n).sum{|i|i.reverse.to_i}}

Thanks to @Unihedron for the second solution.

Answer 15

Mathematica, 47 bytes

Tr[FromDigits@*Reverse/@IntegerDigits@Range@#]&

Try it online! (in order to work on mathics we need to replace "Tr" with "Total")

Answer 16

Python 2, 50 47 bytes

-3 bytes thanks to officialaimm!

lambda n:sum(int(`i+1`[::-1])for i in range(n))

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Answer 17

Japt, 7 5 bytes

-2 bytes thanks to @Shaggy.

õs xw

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Explanation

õs xw  Implicit input of integer U
õs     Create range [1,U] and map to strings
    w  Reverse each string
   x   Sum the array, implicitly converting to numbers.

Old solution, 7 bytes

Keeping this since it's a really cool use of z2.

õs z2 x

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Explanation

õs z2 x  Implicit input of integer U
õs       Create range [1,U] and map to strings
   z2    Rotate the array 180°, reversing strings
      x  Sum the array, implicitly converting back to integers

Answer 18

Charcoal, 14 13 bytes

-1 byte thanks to Carlos Alejo

Ｉ∕…·⁰Ｎ«⁺ιＩ⮌Ｉκ

Try it online! Link is to verbose version.

Explanation

Ｉ                  Cast
  ∕     «           Reduce
   …·⁰Ｎ            Inclusive range from 0 to input as number
         ⁺          Plus
          ι         i
           Ｉ⮌Ｉκ   Cast(Reverse(Cast(k)))

Answer 19

C++, 146 bytes

#include<string>
using namespace std;int r(int i){int v=0,j=0;for(;j<=i;++j){auto t=to_string(j);reverse(t.begin(),t.end());v+=stoi(t);}return v;}

Answer 20

Husk, 7 6 3 bytes

ṁ↔ḣ

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Ungolfed/Explanation

  ḣ  -- With the list [1..N] ..
ṁ    -- .. do the following with each element and sum the values:
 ↔   --    reverse it

Answer 21

Perl 5, 29 27 22 + 1 (-p) = 23 bytes

map$\+=reverse,1..$_}{

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Answer 22

Magneson, 102 bytes

That's not very visible, so here's a scaled up version (Note: Won't actually run, and still isn't very pretty)

Magneson operates by parsing an image and evaluating commands from the colours of the pixels it reads. So stepping through the image for this challenge, we have:

• R: 0, G: 1, B: 1 is an integer assignment command, which takes a string for the variable name and the value to assign. We'll use this to store the sum total.
• R: 0, G: 1, B: 0 is a prebuilt string with the value VAR_1 (Note: This is only while we're asking for a string; the colour code has a separate function when used elsewhere).
• R: 3, G: 0, B: 0 is a raw number. Magneson handles standard numbers by requiring the Red component to be exactly 3, and then forms a number by using the blue value directly plus the green value multiplied by 256. In this case, we're just getting the number 0.
• R: 0, G: 1, B: 1 is another integer assignment command. This time, we're storing an iteration variable, to keep track of which number we're on
• R: 0, G: 1, B: 1 is a prebuilt string with the value VAR_2 (Once more, only when we need a string)
• R: 3, G: 0, B: 0 is the number 0, once more. Onto the interesting bits now.
• R: 1, G: 0, B: 0 indicates the start of a loop. This takes a number and loops the following snippet of code that many times.
• R: 2, G: 0, B: 0 is the STDIN function, or at least it is when we need a number. This reads a line of input from the console and turns it into a number, since we asked for a number.
• R: 0, G: 8, B: 0 starts off our looping code, and it is an additive command. This adds a number to an integer variable, and so takes a string for the variable name, and the number to add.
• R: 0, G: 1, B: 1 is the prebuilt string for VAR_2, which is our iteration variable.
• R: 3, G: 0, B: 1 is a raw number, but this time it's the number 1.
• R: 0, G: 8, B: 0 is another addition command.
• R: 0, G: 1, B: 0 is the string for VAR_1, which is our sum total.
• R: 0, G: 3, B: 0 is a function that reverses a string. In the context of asking for a number, it then converts the reversed string to a number.
• R: 0, G: 2, B: 1 is an integer retrieval command, and will retrieve the number stored in a provided variable. In the context of asking for a string (such as from the reverse command), it converts the number to a string.
• R: 0, G: 1, B: 1 is the name VAR_2; our iteration variable.
• R: 1, G: 0, B: 1 is the marker to end the loop, and go back to the start of the loop if the criteria isn't met (so if we need to keep looping). Otherwise, proceed onwards.
• R: 0, G: 0, B: 1 is a very simple println command, and takes a string.
• R: 0, G: 2, B: 1 retrieves an integer from a variable

• R: 0, G: 1, B: 0 is the name of our sum total variable, VAR_1

  All in all, the program:

• Assigns the value 0 to VAR_1 and VAR_2
• Loops from 0 to a number provided in STDIN
  • Adds one to VAR_2
  • Adds the integer value of reversing VAR_2 to VAR_1
• Prints the contents of VAR_1

Answer 23

APL (Dyalog), 10 7 bytes

3 bytes golfed thanks to @Adám by converting to a tradfn from a train

+/⍎⌽⍕⍳⎕

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⎕          Input (example input: 10)
⍳          Range; 1 2 3 4 5 6 7 8 9 10
⍕          Stringify; '1 2 3 4 5 6 7 8 9 10'
⌽          Reverse; '01 9 8 7 6 5 4 3 2 1'
⍎          Evaluate; 1 9 8 7 6 5 4 3 2 1
+/         Sum; 46

Answer 24

CJam, 12 bytes

ri){sW%i}%:+

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-1 thanks to Business Cat.

Explanation:

ri){sW%i}%:+
r            Get token
 i           To integer
  )          Increment
   {sW%i}    Push {sW%i}
    s         To string
     W        Push -1
      %       Step
       i      To integer
         %   Map
          :+ Map/reduce by Add

Answer 25

Java 8, 97 bytes

IntStream.range(1,n+1).map(i->Integer.valueOf(new StringBuffer(""+i).reverse().toString())).sum()

---

EDIT

As per the comment of Kevin Cruijssen, I would like to improve my answer.

Java 8, 103 bytes

n->java.util.stream.LongStream.range(1,n+1).map(i->new Long(new StringBuffer(""+i).reverse()+"")).sum()

Answer 26

RProgN 2, 8 bytes

{Ø.in}S+

Explained

{Ø.in}S+
{    }S # Create a stack in range 0 through the implicit input, using the function defined
 Ø.     # Append nothing, stringifying the number
   i    # Reverse the string
    n   # Convert back to a number
       +# Get the sum of the stack, and output implicitly.

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Answer 27

Pyth, 8 6 bytes

^{-2 bytes thanks to FryAmTheEggman!}

sms_`h

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Answer 28

Tcl, 66 bytes

time {incr s [regsub ^0+ [string rev $n] ""];incr n -1} $n
puts $s

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Answer 29

Neim, 4 bytes

Δ)

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Explanation

Δ )              for each element 1 to n (outputs list)
                reverse 
                sum 

Answer 30

Excel VBA, _{80 78} 49 Bytes

Anonymous VBE Immediate Window function that takes input expected type Integer from range [A1] determines all of the values that fall within the range 1:[A1], and outputs the sum of the reversed values to the VBE immediate window

For i=1To[A1]:s=s+Val(StrReverse(Str(i))):Next:?s

Answer 31

C (gcc), 71 bytes

q(n,x,p){p=n?q(n/10,x*10+n%10):x;}f(w,a,e){for(a=0;w;)a+=q(w--,0);e=a;}

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Answer 32

TXR Lisp: 62 56 bytes:

(opip(range 1)(mapcar[chain tostring reverse toint])sum)

Interactive:

1> (opip(range 1)(mapcar[chain tostring reverse toint])sum)
#<intrinsic fun: 0 param + variadic>
2> [*1 10]
46
3> [*1 999]
454545

The following 44 byte expression, inspired by the Pari/GP solution, is possible; however, it requires the sum macro to be defined:

1> (defmacro sum (var from to expr)
      (with-gensyms (accum)
        ^(for ((,var ,from) (,accum 0))
              ((<= ,var ,to) ,accum)
              ((inc ,accum ,expr) (inc ,var)))))
sum
2> (do sum x 1 @1 (toint(reverse(tostring x))))
#<interpreted fun: lambda (#:arg-01-0171 . #:rest-0170)>
3> [*2 10]
46
4> [*2 999]
454545

Answer 33

Ruby, 38 35 bytes

Similar to the previous Ruby solution, but a full program and also shorter (for less than an hour)!

p (?1..gets).sum{|x|x.reverse.to_i}

Shortened by 3 bytes thanks to user akostadinov

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Explanation

p             # Inspect and print. Written as infix notation to avoid using ()
(?1..gets)    # All strings from "1" to "input", based on successive string format
  .sum{|x|    # Enumerable -> Map to value -> Sum by value
      x.reverse.to_i
  }

Answer 34

Gaia, 7 bytes

@…)¦v¦Σ

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Range, Reverse, and sum!

@             # push an input. stack: n
 …            # generate range. stack: [0...n-1]
  )¦          # map over the list with increment. stack: [1...n]
    v¦        # map over the list with reverse. stack: [1...n], but all digitally reversed
      Σ       # sum the list; output TOS.

Answer 35

Gol><>, 26 bytes

&IFLPWaSD$|~rlMFa*+|&+&|&h

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Explanation

&I                      &h < register init, read "n"; print register (h = n;)
  FLP                  |   < For loop, do "n" times for each x in [1] to [n]
     WaSD$|                < modpow "x" to digits: [123] [3 12] [3 2 1] [3 2 1 0]
           ~rlMFa*+|       < Ditch zero, reverse stack, build the number back
                    &+&    < register += final x

Answer 36

Pyke, 7 bytes

SF`_b)s

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S       -   range(1, input)
 F   )  -  for i in ^:
  `     -     str(i)
   _    -    reversed(^)
    b   -   int(^)
     s  - sum(^)

Answer 37

MATL, 5 bytes

:VPUs

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Answer 38

Python 3, 50 bytes

lambda n:sum(int(str(n)[::-1])for n in range(n))

Quite close to idiomatic Python code, and honestly I don't really see much room for improvement.

Answer 39

Pyth - 7 bytes

sms_`dS

Explanation

sms_`dSQ  Q added implicitly
s         Sum of
 m        map
  s_`d    integer representation of reversal of string representation
          to
      SQ  Range from 1 to input

Answer 40

Pyt, 3 bytes

ř₫Ʃ

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The characters perform the following operations: range, reverse, sum.

Answer 41

Jelly, 5 bytes

RDUḌS

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R      ' [R]ange
 D     ' Integer -> [D]ecimal
  U    ' [U]pend (vectorized reverse?)
   Ḍ   ' [Ḍ]ecimal -> Integer
    S  ' [S]um

Answer 42

MY, 11 bytes

iC4ǵ'ƒ⇹(Σ↵

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How?

• input as integer
• i [1 ... pop()] inclusive
• C4ǵ' push the string '\x4C' (The reverse command (⌽), which works on numbers)
• ƒ as a function
• ⇹ mapped over the argument (pushes a function)
• ( apply the function
• Σ sum
• ↵ output with a new line

MY is capable of something, woohoo!

Answer 43

Ly, 31 bytes

ns[l1-s]pr[s>lSrJs>l<p<p]>>r&+u

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Answer 44

Javascript,89 77 72 bytes

saved 12 bytes thanks to @Justin Mariner

saved 5 bytes thanks to @Neil

n=>-eval(`-'${[...[...Array(n+1).keys()].join`'-'`].reverse().join``}'`)

Explanation:

[...Array(n+1).keys()] creates an array from 0 to n.

.reverse(‌).join`` reverses every value.

.join`'-'` and eval(`-'{stuff}'`) subtracts the values in order, producing the negative of the answer. - negates the negative to a positive.

Answer 45

Pari/GP, 43 bytes

n->sum(i=1,n,fromdigits(Vecrev(digits(i))))

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Answer 46

QBIC, 19 bytes

[:|_F!a$|p=p+!A!}?p

Explanation

[:|     FOR a = 1 to n (read from cmd line)
_F   |      FLIP and assign to A$
  !a$        a string cast of our loop counter
p=p+!A!    increment p with A$ cast back to number
}           NEXT
?p         Print the result

Answer 47

Swift 4, 63 61 bytes

{v in(1...v).reduce(0,{$0+Int(String("\($1)".reversed()))!})}

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---

Explanation

• {v in} - Creates an anonymous (lambda-like) function with a parameter v.

• (1...v) - Inclusive range from 1 to v (although it kind of breaks the rules of logic).

• Int(String("\($1)".reversed())) - Reverses the Number by converting it to a String, returning ReverseCollection<String>. That cannot be casted directly to an integer (Swift borked types!), so we have to cast it to String again, before making the conversion to an Integer.

• reduce(0,{$0+...}) - Sums the mapped range.

Answer 48

LOGO, 33 bytes

[reduce "+ map "reverse iseq 1 ?]

There is no "Try it online!" link because all online LOGO interpreter does not support template-list.

That is a template-list (equivalent of lambda function in other languages).

Usage:

pr invoke [reduce[?+reverse ?2]iseq 1 ?] 10

(invoke calls the function, pr prints the result)

prints 46.

Explanation:

LOGO stores numbers as words, therefore apply reverse on a number reverse the digits in that number.

reduce "f [a b c d e] (where [a b c d e] is a list) will calculate f(a, f(b, f(c, f(d, e)))). So reduce "+ list will calculate sum of values of list.

Answer 49

Swift 3, 98 bytes

func f(a:Int)->Int{return (0...a).map{Int(String(String($0).characters.reversed()))!}.reduce(0,+)}

Answer 50

PowerShell, 53 bytes

(1.."$args"|%{-join"$_"["$_".length..0]})-join'+'|iex

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Literal translation of the specification. Creates a range with .., reverses each by using string manipulation and -joining them back into a single number, then -joins each number together with a + and piping that to iex (short for Invoke-Expression and similar to eval).

Answer 51

GolfScript, 15 bytes

~),{`-1%~}%{+}*

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Explanation:

~),{`-1%~}%{+}* Input all at once
~               Eval
 )              Increment
  ,             Exclusive range
   {     }      Push block
    `            Repr
     -1          -1
       %         Every xth element
        ~        Eval
          %     Map
           { }  Push block
            +    Add
              * Reduce

Answer 52

R, 89 bytes

sum(strtoi(sapply(strsplit(paste(1:scan()),''),function(x)paste(rev(x),collapse='')),10))

Reads from stdin; returns the value.

sum(                                          #compute the sum
 strtoi(                                      #convert string to int
  sapply(                                     #iterate over
   strsplit(                                  #this where each element is the list of characters in the number
    paste(1:scan()),                          #convert range to character
          ""),                                #split on each character
   function(x)paste(rev(x), collapse="")),    #reverse the list and concatenate digits
         10))                                 #as base 10

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Answer 53

PHP, 39+1 bytes

while($i++<$argn)$s+=strrev($i);echo$s;

Run as pipe with -nR.

Answer 54

Rust, 110 bytes

|x:i64|(1..x+1).map(|v|format!("{}",v).chars().rev().collect::<String>().parse::<i64>().unwrap()).sum::<i64>()

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I like Rust, but it is not a very efficient language for golfing!

Answer 55

Common Lisp, 62 bytes

(loop as i to(read)sum(parse-integer(reverse(format()"~d"i))))

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Answer 56

REXX, 46 bytes

arg n
s=0
do i=1 to n
  s=s+reverse(i)
  end
say s

Answer 57

Swift 3, 83 68 characters

let f={n in Array(1...n).reduce(0){$0+Int(String("($1)".characters.reversed()))!}}

{(1...$0).reduce(0){$0+Int(String("\($1)".characters.reversed()))!}}

This would probably be shortened by Swift 4's new String and Character handling.

Answer 58

k, 13 bytes

+/.:'|:'$1+!:

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         1+!: /create list 1 2 3 ... x
        $     /turn each number into a string
     |:'      /reverse each string
  .:'         /eval each string
+/            /sum

Answer 59

Java 8, 84 bytes

Lambda from int to long (e.g. IntToLongFunction). Adaptation of coder-croc's solution.

n->{long s=0;while(n>0)s+=new Long(new StringBuffer(""+n--).reverse()+"");return s;}

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Answer 60

Recursiva, 10 bytes

smBa"I_Va"

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Answer 61

J, 18 Bytes

+/(".@|.@":)"0>:i.

Explanation:

+/                 | Sum of
  (".@|.@":)       | Reverse
            "0     | Each member of
              >:i. | Integers from 1 to n inclusive