RATS अनुक्रम

Question

आपका कार्य RATS अनुक्रम का nth शब्द उत्पन्न करना है, जहाँ n इनपुट है। RATS अनुक्रम को रिवर्स ऐड तब सॉर्ट अनुक्रम के रूप में भी जाना जाता है। यह क्रम भी यहां पाया जा सकता: http://oeis.org/A004000 ।

परीक्षण के मामलों:

0 > 1
1 > 2
2 > 4
3 > 8
4 > 16
5 > 77
6 > 145
7 > 668

उदाहरण के लिए, 5 के लिए आउटपुट 77 है क्योंकि 16 + 61 = 77 है। इसके बाद 77 को सॉर्ट किया जाता है।

सबसे कम सबमिशन जीतता है। यह मेरी पहली चुनौती है इसलिए मुझे उम्मीद है कि यह कोई नकल या कुछ नहीं है।

Comments

क्या इनपुट में एक इंटेगर होना चाहिए या यह एक स्ट्रिंग भी हो सकता है?

— डेनकर

@DenkerAffe क्या आप एक स्ट्रिंग के रूप में एक संख्या का मतलब है?

— justaprogrammer

@justaprogrammer Yea, इसलिए मुझे पूर्णांक के रूप में 123 के बजाय "123" मिल सकता है। मेव कुछ बाइट्स बचाएंगे।

— डेनकर

77 + 77 = 154 नहीं है? या मुझसे कोई चीज चूक रही है? संपादित करें: ओह, हाँ, मैं सॉर्ट करना भूल गया।

— डेन्हम कोटे

@DenhamCoote मुझे लगता है कि आपका मतलब है "ओह रैट एस , मैं सॉर्ट करना भूल गया!"

— मार्टिन एंडर

Answer 1

MATL , 11 12 बाइट्स

1i"tVPU+VSU

इनपुट एक है स्ट्रिंग (एकल उद्धरण के साथ) में एक पूर्णांक का प्रतिनिधित्व एकल । स्ट्रिंग इनपुट को चुनौती द्वारा अनुमति दी जाती है, और यूनिरी एक मान्य प्रारूप है ।

इसे ऑनलाइन आज़माएं!

व्याख्या

1      % push number 1 to the stack
i      % input. Will be a string of "n" ones 
"      % for loop: repeat n times (consumes string)
  t    %   duplicate
  V    %   convert to string
  P    %   reverse
  U    %   convert to number
  +    %   add
  V    %   convert to string
  S    %   sort
  U    %   convert to number
       % loop is implicitly ended
       % stack content is implicitly displayed    

Comments

मुझे नहीं पता कि मुझे कौन सा डराता है / अधिक परेशान करता है, MATL या जेली ... +1

— डाउनगैट

Answer 2

05AB1E , 6 बाइट्स

कोड:

$FDR+{

स्पष्टीकरण:

$       # Push 1 and input
 F      # For N in range(0, input)
  D     # Duplicate top of the stack
   R    # Reverse top of the stack
    +   # Add top two items
     {  # Sort top of the stack
        # Implicitly print top of the stack

यह भी 0 बाइट प्रोग्राम के साथ काम करता है ।

Comments

@ अदनान तीन दिन पहले , वास्तव में। फिर भी, अच्छी तरह से खेला ...

— दरवाज़े

@Doorknob बस समय में है

— अदनान

आप अपने स्रोत कोड को समाप्त करके 6 बाइट्स बचा सकते हैं।

— डेनिस

आप 05AB1Eपहले अग्रणी शून्य को समाप्त करके छोटा कर सकते हैं , और फिर छोड़ सकते हैं 1, जैसा कि 1E==E। फिर आप बस 5ABE, -2 बाइट्स प्राप्त करें।

— दोष

@ डेनिस का महान अवलोकन

— अदनान

Answer 3

सीजेएम, 15 बाइट्स

1ri{_sW%i+s$i}*

इसका परीक्षण यहां करें।

व्याख्या

1     e# Push 1 as the start of the sequence.
ri    e# Read input and convert to integer N.
{     e# Run this block N times...
  _s  e#   Duplicate and convert to string.
  W%  e#   Reverse string.
  i+  e#   Convert back to integer and add to previous value.
  s$  e#   Convert to string and sort.
  i   e#   Convert back to integer for the next iteration.
}*

Comments

ये सभी भाषाएँ इतनी कम कैसे हो सकती हैं

— justaprogrammer

@justaprogrammer अंतर्निहित कार्यों के लिए एकल-वर्ण नाम मदद करते हैं। ;) CJam, Pyth और Brachylog सभी गोल्फिंग भाषाएँ हैं, जो विशेष रूप से कोड गोल्फ को ध्यान में रखकर बनाई गई हैं। ( En.wikipedia.org/wiki/Code_golf#Dedicated_golfing_languages देखें । ) इसके बाद APL और J जैसी भाषाएं भी हैं जो सभी भाषाओं को गॉर्जिंग नहीं कर रही हैं, लेकिन समान रूप से प्रचलित हैं, क्योंकि डिजाइनरों ने सोचा कि यह एक अच्छा विचार होगा।

— मार्टिन एंडर

Which one do you recommend the most for winning challenges like these?

— justaprogrammer

@justaprogrammer I wouldn't pick one based on which one is winning these challenges (that would likely be Pyth or Jelly). It can be just as fun to golf in a "normal" language (especially because there might be more competition within that language). For a golfing language, it's probably more important that you enjoy using it. CJam is quite fun - it's a stack-based which makes you bend your mind a bit more than other languages, and at the same time it's quite a powerful language, that I've started to use for simple throwaway scripts outside of golf, which is a good boost to my productivity.

— Martin Ender

These languages look very interesting and I can't wait to learn one myself. I don't know what jelly is? Is that some kind of gelatine or something?

— justaprogrammer

Answer 4

Pyth, 17 13 12 bytes

uS`+vGv_GQ\1

u        Q\1    reduce range(input()) on base case of "1" (string)
   +vG          eval the string (to get a number), and add...
      v_G       the same number, reversed first and then eval'd
 S`             convert back to string and sort

Try it on the online interpreter.

Comments

What is this magic? How does this work?

— justaprogrammer

@justaprogrammer I've added an explanation. :)

— Doorknob

Huh, but how. How do you test this code?

— justaprogrammer

@justaprogrammer I've added a link to an online interpreter that you can run the code on.

— Doorknob

This is awesome, it is so short, yet so beautiful

— justaprogrammer

Answer 5

Python 2, 72

f=lambda x,n=1:x and f(x-1,int(''.join(sorted(`n+int(`n`[::-1])`))))or n

Recursive function, makes use of the Python 2 shorthand for __repr__, which will break once the function reaches very large values (an L will be appended to the number's string), I'm not certain from the spec if there is a place where we can stop, but if not changing to str() only adds 6 bytes, but then it becomes slightly shorter to output as a string, at 75 bytes:

f=lambda x,n='1':x and f(x-1,''.join(sorted(str(int(n)+int(n[::-1])))))or n

1 byte saved thanks to trichoplax on this version

Comments

Is that a surplus space before the or in the second code block?

— trichoplax

@trichoplax thanks for the catch :)

— FryAmTheEggman

Answer 6

JavaScript ES6, 70 bytes

Saved 1 byte thanks to @user81655

f=n=>n?+[...+[...''+(b=f(n-1))].reverse().join``+b+''].sort().join``:1

sigh JavaScript is really verbose. A lot (> 50%) of the code is just case to string + array function + join + cast to int. I've tried reduce, eval, and all sorts of stuff but this seems to be the shortest.

Try it online (All browsers work)

Comments

Just like mine, but better (and posted earlier). Bah!

— edc65

String Manipulation is JS is so long, you have my condolences

— MayorMonty

@user81655 cool, thanks! I would of never thought to re-order that way

— Downgoat

f=n=>n?[...+[...b=f(n-1)].reverse().join``+b+''].sort().join``:'1' if returning string allowed

— l4m2

Answer 7

Brachylog, 19 bytes

0,1 .|-1=:0&Rr+R=o.

Explanation

0,1 .               § If Input is 0, unify the Output with 1
     |              § Else
      -1=:0&R       § unify R with the output of this main predicate, with input = Input - 1
             r+R=o. § Reverse R, add it to itself and order it, then unify with the Output.

Answer 8

Haskell, 67 bytes

import Data.List
g i=i:g(sort$show$read i+read(reverse i))
(g"1"!!)

Usage example: (g"1"!!) 7-> "668".

It's a direct implementation of the definition: starting with "1", repeatedly append the reverse-add-sort result of the current element. The main function (g"1"!!) picks the ith element.

Comments

This is the most readable program under 70 bytes!

— Gaurav Agarwal

Answer 9

Julia, 77 bytes

n->(x=1;for _=1:n x=(p=parse)(join(sort(["$(x+p(reverse("$x")))"...])))end;x)

This is a lambda function that accepts an integer and returns an integer. To call it, assign it to a variable.

Ungolfed:

function f(n::Int)
    # Begin x at 1
    x = 1

    # Repeat this process n times
    for _ = 1:n
        # Add x to itself with reversed digits
        s = x + parse(reverse("$x"))

        # Refine x as this number with the digits sorted
        x = parse(join(sort(["$s"...])))
    end

    # Return x after the process (will be 1 if n was 0)
    return x
end

Answer 10

Jelly, 13 12 bytes

I'm sure this can probably be golfed, as this is my first answer in Jelly/in a tacit language.

DUḌ+ðDṢḌ Performs RATS
1Ç¡      Loops

D        Converts integer to decimal
 U       Reverses
  Ḍ      Converts back to integer
   +     Adds original and reversed
    ð    Starts new chain
     D   Converts back to decimal
      Ṣ  Sorts
       Ḍ Back to integer again

1        Uses 1 instead of input
 Ḍ       Uses line above
  ¡      For loop

EDIT: Saved 1 byte, thanks to Dennis

Answer 11

Java 1.8, 251 bytes

interface R{static void main(String[]a){int i,r,n=1,c=0,t=Byte.valueOf(a[0]);while(++c<=t){i=n;for(r=0;i!=0;i/=10){r=r*10+i%10;}n+=r;a[0]=n+"";char[]f=a[0].toCharArray();java.util.Arrays.sort(f);n=Integer.valueOf(new String(f));}System.out.print(n);}}

Expanded

interface R{
static void main(String[]args){
    int input,reversed,nextValue=1,count=0,target=Byte.valueOf(args[0]);
    while(++count<=target){
        input=nextValue;
        for(reversed=0;input!=0;input/=10){reversed=reversed*10+input%10;}
        nextValue+=reversed;
        args[0]=nextValue+"";
        char[]sortMe=args[0].toCharArray();
        java.util.Arrays.sort(sortMe);
        nextValue=Integer.valueOf(new String(sortMe));
    }
    System.out.print(nextValue);
}
}

Comments

Why do you use interface R instead of class R which is 4 bytes shorter?

— Will Sherwood

@WillSherwood because you can then omit the public modifier on main(), making it shorter overall :)

— Denham Coote

Answer 12

Seriously, 17 bytes

1,`;$R≈+$S≈`n

Try it online!

Explanation:

1,`;$R≈+$S≈`n
1              push 1
 ,`       `n   do the following n times:
   ;$R≈        reverse
       +       add
        $S≈    sort

Answer 13

Lua, 179 156 Bytes

I can't see how I could golf it more, but I'm sure there's a way. Thanks to @LeakyNun I took the time to come down on this and golf it the proper way, I could maybe still win some bytes by using another approach.

k=0z=table
for i=0,io.read()do
t={}(""..k+(""..k):reverse()):gsub("%d",function(d)t[#t+1]=d
end)z.sort(t)k=k<1 and 1or tonumber(z.concat(t,""))
end
print(k)

Ungolfed and explanations

k=0                                  
z=table                              -- z is a pointer on the table named table
                                     -- it allows me to use its functions
                                     -- while saving 4 bytes/use

for i=0,io.read()                    -- Iterate n times for the nth element
do
  t={}
  (""..a+(""..a):reverse())          -- we add k with its "reversed" value
                                     -- and convert the whole thing to a string
    :gsub(".",function(d)            -- for each character in it, use an anonymous fucntion
       t[#t+1]=d end)                -- which insert them in the array t
  z.sort(t)                          
  a=a<1 and 1 or                     -- if i==0, k=1
     tonumber(z.concat(t,""))        -- else we concat t in a string and convert it to number
end
print(k)

Comments

Well it looks like you aren't here anymore... but maybe you can refer to my java answer.

— Leaky Nun

@LeakyNun Well, I'm not participating a lot this times but still peaking at challenges from time to times, I'll try to take a look at your answer, but even without that I see some thing that can be golfed pretty easily (a=a<1 and 1or for instance).

— Katenkyo

we would be glad – I would be glad – to have you back.

— Leaky Nun

Answer 14

Brachylog 2, 11 bytes, language postdates challenge

;1{↔;?+o}ⁱ⁽

Try it online!

Explanation

;1{↔;?+o}ⁱ⁽
  {     }ⁱ  Repeatedly apply the following,
 1            starting at 1,
;         ⁽   a number of times equal to the input:
   ↔            reverse,
    ;?+         add the original input,
       o        then sort the resulting number

I'm not quite clear on what this does with zero digits, but the question doesn't state any particular handling, and they probably don't appear in the sequence anyway.

Answer 15

ES6, 79 bytes

n=>eval("r=1;while(n--)r=+[...+[...r+''].reverse().join``+r+''].sort().join``")

82 bytes without eval:

n=>[...Array(n)].reduce(r=>+[...+[...r+''].reverse().join``+r+''].sort().join``,1)

All those conversions are painful.

@edc65 I actually saved 4 bytes by switching from map to reduce this time... no doubt you'll prove me wrong again though.

Comments

for is shorter: n=>eval("for(r=1;n--)r=+[...+[...r+''].reverse().join``+r+''].sort().join``")

— Downgoat

@Doᴡɴɢᴏᴀᴛ Doesn't work for n=0, even after I've fixed the syntax errors.

— Neil

Answer 16

Python 2, 91 Bytes

Input as Integer, result is printed to the screen.

def f(n):
 t=1
 for i in range(n):t=int("".join(sorted(str(int(str(t)[::-1])+t))))
 print t

This could be a lot shorter with some recursion magic I guess, but I cant wrap my head around it yet. Gonna have a fresh look later and hopefully improve this one.

Answer 17

Python 2, 83 bytes

def f(n):
 v='1'
 for _ in v*n:v=''.join(sorted(str(int(v)+int(v[::-1]))))
 print v

Answer 18

Perl 6, 40 bytes

{(1,{[~] ($_+.flip).comb.sort}...*)[$_]} # 40

( If you want it to return an Int put a + right before [~] )

Usage:

# give it a lexical name
my &RATS = {…}

say RATS 5; # 77

# This implementation also accepts a list of indexes

# the first 10 of the sequence
say RATS ^10; # (1 2 4 8 16 77 145 668 1345 6677)

Answer 19

Mathematica 10.3, 66 61 bytes

Nest[FromDigits@Sort@IntegerDigits[#+IntegerReverse@#]&,1,#]&

Quite simple.

Answer 20

PHP, 102 Bytes

$r=[1];$i++<$argn;sort($v),$r[]=join($v))$v=str_split(bcadd(strrev($e=end($r)),$e));echo$r[$argn];

Online Version

PHP, 95 Bytes

n <= 39

for($r=[1];$i++<$argn;sort($v),$r[]=join($v))$v=str_split(strrev($e=end($r))+$e);echo$r[$argn];

Answer 21

Java, 171 167 163 160 bytes

int f(int n){int a=n>0?f(n-1):0,x=10,b[]=new int[x],c=a,d=0;for(;c>0;c/=x)d=d*x+c%x;for(a+=d;a>0;a/=x)b[a%x]++;for(;a<x;c=b[a++]-->0?c*x+--a:c);return n>0?c:1;}

Try it online!

Not the longest entry! \o/

Comments

@Katenkyo see this

— Leaky Nun

It is ok f(1)...f(20) But from f(21) result seems wrong...

— RosLuP

Loss of precision I suppose.

— Leaky Nun

Answer 22

Python 2, 70 bytes

f=lambda n,a=1:n and int(''.join(sorted(`f(n-1)+f(n-1,-1)`))[::a])or 1

Try it online!

Answer 23

Axiom, 146 bytes

c(r:String):NNI==reduce(+,[(ord(r.i)-48)*10^(#r-i) for i in 1..#r]);f(n:INT):NNI==(n<1=>1;v:=f(n-1);v:=v+c(reverse(v::String));c(sort(v::String)))

test and results [RATS sequence]

(3) -> [[i, f(i)] for i in 0..20]
   (3)
   [[0,1], [1,2], [2,4], [3,8], [4,16], [5,77], [6,145], [7,668], [8,1345],
    [9,6677], [10,13444], [11,55778], [12,133345], [13,666677], [14,1333444],
    [15,5567777], [16,12333445], [17,66666677], [18,133333444], [19,556667777],
    [20,1233334444]]

Answer 24

Tcl, 91 bytes

proc R n {join [lsort [split [expr {$n?[set v [R [expr $n-1]]]+[string rev $v]:1}] ""]] ""}

Try it online!