Find five friends to eat chicken with Paul

Question

Paul is one of your Belgian acquaintances, and he would like you to create a program that outputs at least one of the following five strings:

12496=>14288=>15472=>14536=>14264

14264=>12496=>14288=>15472=>14536

14536=>14264=>12496=>14288=>15472

15472=>14536=>14264=>12496=>14288

14288=>15472=>14536=>14264=>12496

It represents the friends he's most proud of, and he would like to eat with them. Paul also thinks that the number 6 is perfect, and that is the only number worth using. So you cannot use any other digit than "6" in your code (0 to 5 and 7 to 9 are forbidden). Your program cannot take any inputs. The output can contain garbage before and/or after the string but should contain at least one of the above strings.

For example this is a valid output :

220frefze
f**14288=>15472=>14536=>14264=>12496**fczfe fz**15472=>14536=>14264=>12496=>14288**zfe
fzehth

I "ligthly" hinted in the question as how I expect this to be resolved, but how knows ? Maybe there is a better way... Hope you have fun.

This is code-golf : lowest score in bytes wins.

Answer 1

05AB1E, 18 17 bytes

•w[•Y·FDѨO})„=>ý

Try it online!

Explanation

•w[•                # base 214 encoding of 12496
    Y·F             # loop 2*2 times
       D            # duplicate top of stack
        Ñ           # push divisors
         ¨          # remove the last element (itself)
          O         # sum
           }        # end loop
            )       # wrap in list
             „=>ý   # join list on "=>"

In short, we calculate each number as f(n+1) = sum(divisors(f(n)) - f(n)

Answer 2

Pyke, 16 bytes

wヰw$VDlsh)J"=>

Try it here!

wヰ             -  12496
  w$            -   4
    V    )      -  repeat ^:
     D          -   duplicate(^)
      l         -     factors(^)
       s        -    sum(^)
        h       -   ^ + 1
          J     - v.join(^)
           "=>  -  "=>"

Anytime a string literal is at the end of a program, it swaps it with the token just before it, saving 1 byte in cases such as this. Pyke's factors function doesn't include the number itself nor 1. If numbers were allowed, 1 byte could be saved by replacing w$ with 4

Pyke, 21 bytes

uバ㟐㱰㣈㞸J"=>

Try it here!

Create a list of the numbers required and joins them. Not very interesting apart from the trick with the strings.

Answer 3

MATLAB, 44 bytes

['','=>@EBIJ=@>DDIJ=A@C>IJ=@A?BIJ=@>B@'-6-6]

Try it online!

I haven't found a pattern in the numbers (and it would be hard to use the pattern to anything anyways, since I can't use numbers), so I'll just go for the naive approach.

'=>@EBIJ=@>DDIJ=A@C>IJ=@A?BIJ=@>B@' is the string '12496=>14288=>15472=>14536=>14264' when 12 is added to the ASCII-values. Now, input that string, subtract 6+6, and concatenate with the empty string '' to convert it to a character array.

Answer 4

JavaScript (ES6), 57 bytes/47 (UTF-8) characters

Thanks to user5090812 for 10 B save

_=>[...'バ㟐㱰㣈㞸'].map(a=>a.charCodeAt()).join`=>`

Explanation

First we create an array and fill it with the characters in the string バ㟐㱰㣈㞸. Then we loop over the string (a has the value of the current element) and we change the character into its character code. Then we join all values in the array by =>.

Old: 67 bytes

_=>[6,6,6,6,6].map((_,a)=>`バ㟐㱰㣈㞸`.charCodeAt(a)).join`=>`

Explanation

First we create an array of length 5. Then we change the values of the array at each index for the character code of the character at that same index in the string バ㟐㱰㣈㞸, which are the numbers of all of Paul's friends in order. When we got that, we join the array together, and we use => as separator.

Usage

To use it, simply run this:

f=_=>[...'バ㟐㱰㣈㞸'].map(a=>a.charCodeAt()).join`=>`;alert(f())

Output

12496=>14288=>15472=>14536=>14264

Answer 5

Ruby, 36 bytes (26 characters)

p"バ㟐㱰㣈㞸".unpack("U*")*"=>"

Because, why not. Boring as hell.

older version - 53 bytes

p %w(jol mld oim n6b mke).map{|x|x.to_i ~-6*~-6}*'=>'

Explanation: encoding the numbers in base 25 gives the 5 six-free strings, to decode them I only have to represent the number 25 using only 6: (6-1)(6-1) => ~-6~-6

Answer 6

Perl 6, 63 59 bytes

{$/=6;$/--;join "=>",<JOL MLD OIM N6B MKE>».parse-base($/*$/)}

{join "=>",<JOL MLD OIM N6B MKE>».parse-base(--($_=6)*$_)}

Decodes the numbers from base 25, because that's the only base supported by .parse-base (2 to 36) where none of them have invalid digits.

Thanks to Neil for -3 bytes.

Perl 6, 82 75 bytes

{my \a="BXS".parse-base(6*6);join "=>",(a,{sum grep $_%%*,^$_}...^{$_==a if $++})}

{my \a="BXS".parse-base(6*6);join "=>",({$/=$_//a;sum grep $/%%*,^$/}...a)}

Decodes the number 15472 in base 36, and then generates the sequence by calculating each number as the sum of proper divisors of the previous number.

Perl 6, 69 bytes (47 characters) - noncompeting

{"{١٢۴۹6}=>{١۴۲۸۸}=>{١۵۴۷۲}=>{١۴۵۳6}=>{١۴۲6۴}"}

Doesn't use any of the forbidden ASCII digits, uses Unicode digits from the Arabic-Indic block instead (2 bytes each)! The { } string interpolations make sure they're parsed as Perl 6 number literals, and then stringified to their ASCII representations.

Okay, this is cheating - that's why I didn't use it as my main answer... :)

Answer 7

Jelly, 5 4 bytes

ȷṗȷỌ

Prints all five strings. Takes advantage of the fact that "garbage" output is allowed and buries the five strings in 10³⁰⁰³ characters of output.

How it works

ȷṗȷỌ  Main link. No arguments.

ȷ     Set the return value to 1000.
 ṗȷ   Cartesian power; form all arrays of length 1000 that consist of integers in
      [1, ..., 1000].
   Ọ  Unordinal; convert all integers to characters.

Answer 8

C, 94 84 77 Bytes

Stoopid simple. Special thanks @Neil

g(){char*m="!mnpuryzmpnttyzmqpsnyzmpqoryzmpnrp";for(;*++m;)putchar(*m-66+6);}

f(){printf("%d=>%d=>%d=>%d=>%d",'~''r'-'d','~''d'-'h','~''r'-'L','~''z'+'d','~'*'t'-'P');}

Answer 9

PHP, 73 63 60 bytes

for(;$c="|`*X6H-$*@"[$i];)echo!!$i&++$i?"=>".!!6:"",ord($c);

Run with -nr.

a little less lazy: took string as list of =>1(ascii)(ascii)
i.e.: 124,96,=>1,42,88,=>1,54,72,=>1,45,36,=>1,42,64;
print =>1 by string index, append ascii code

breakdown

            # loop through string with index $i
for(;$c="|`*X6H-$*@"[$i];)echo
            # string ascii values: 124,96,42,88,54,72,45,36,42,64
    !!$i        # true if $i>0
    &++$i       # odd if (old) $i is 0,2,4,6,8
                # -> true for 2,4,6,8
        ?"=>".!!6   # if true, print "=>1"
        :"",        # else print nothing
    ord($c);    # print ascii value

Answer 10

C++, 92 bytes

#include <cstdio>
int main(){for(char c:"UVX]ZabUXV\\abUYX[VabUXYWZabUXVZX")putchar(c-6*6);}

Answer 11

PHP, 53 bytes

<?=join('=>',unpack('v*',gzinflate('�`p������s')));

Hex dump:

00000000: 3c3f 3d6a 6f69 6e28 273d 3e27 2c75 6e70  <?=join('=>',unp
00000010: 6163 6b28 2776 2a27 2c67 7a69 6e66 6c61  ack('v*',gzinfla
00000020: 7465 2827 bb60 70c1 bcc0 e684 c50e 7300  te('.`p.......s.
00000030: 2729 2929 3b                             ')));

Output:

12496=>14288=>15472=>14536=>14264

Explanation:

Each of the five-digit integer sections is encoded as an unsigned short little endian, then concatenated together and the result is gzipped. This happens to produce a byte steam that has no offending digit characters, which is then hard-coded into a string. To extract, un-gzip the stream, unpack the two-byte shorts, interpret each as a string, and join with >=.

Answer 12

Java 8, 134 bytes

Golfed:

()->{String s="";for(String a:new String[]{"JOL","MLD","OIM","N6B","MKE"}){if(!s.isEmpty())s+=("=>");s+=Long.valueOf(a,25);}return s;}

Ungolfed, full program:

import java.util.function.*;

public class FindFiveFriendsToEatChickenWithPaul {

  public static void main(String[] args) {
    System.out.println(toString(() -> {
      String s = "";
      for (String a : new String[] { "JOL", "MLD", "OIM", "N6B", "MKE" }) {
        if (!s.isEmpty()) s += ("=>");
        s += Long.valueOf(a, 25);
      }
      return s;
    }));

  }

  private static String toString(Supplier<String> s) {
    return s.get();
  }

}

Answer 13

Batch, 191 bytes

@set/as=n=66*(66+66+66+6*6)+66/6+66/6+6,u=6/6
@call:t
@call:t
@echo %s: ==^>%
@exit/b
:t
@call:c
:c
@for /l %%i in (%u%,%u%,%n%)do @set/an-=%%i*!(%n%%%%%i)
@set/an=-n
@set s=%s% %n%

I estimate that it would take a minimum of 32 bytes to calculate each number using only 6s plus another 32 to print them all out, which is already 192 bytes, so I'm winning by computing the amicable chain. Also, I think five %s in a row is a record for me. Also, neat Batch trick: the %n% is substituted before the for loop is evaluated, so the loop calculates all the factors of n, and subtracts them from n, thus resulting in the negation of the desired result.

Answer 14

Jelly, 12 bytes

“<ọ’ÆṣÐĿj“=>

Prints the fourth string and nothing else.

Try it online!

How it works

“<ọ’ÆṣÐĿj“=>  Main link. No arguments.

“<ọ’          Yield the 1-based indices of '<' and 'ọ' in Jelly's code page, i.e.,
              [61, 222], and convert the array from base 250 to integer.
              This yields 15472.
      ÐĿ      Iteratively call the link to the left until the results are no longer
              unique and return the array of all unique results.
    Æṣ        Compute the proper digit sum of the previous value (initially 15472).
        j“=>  Join, separating by the string "=>".

Answer 15

Python 2, 78 72 bytes

print''.join(chr(ord(x)-6-6)for x in'=@>DDIJ=A@C>IJ=@A?BIJ=@>B@IJ=>@EB')

Edit - Thanks to Stewie Griffin for saving 6 bytes!

Also, another solution would be to output all the possible permutations. OP says garbage is fine.

from itertools import permutations
print str(list(permutations(''.join(str(x)+'.'for x in range(int('9'*5)).replace(',','').replace('\'','') 
# also 9 and 5 need to be converted using ord and chr

There is too much redundancy in converting from int or list to str. I guess this would be easier in some esoteric languages, but I know none of them.