Print this maximally long URL

Question

Task

This one is simple. We want to compress a URL, but don't trust URL shorteners.

Write a program that prints to stdout (or a 0-argument function that returns) the following, working URL:

http://a.b.c.d.e.f.g.h.i.j.k.l.m.n.oo.pp.qqq.rrrr.ssssss.tttttttt.uuuuuuuuuuu.vvvvvvvvvvvvvvv.wwwwwwwwwwwwwwwwwwwwww.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx.yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy.zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz.me

Code golf, standard rules.

You may output uppercase or lowercase.

Leading or trailing whitespace is ok.

Context

Explanation taken from the website:

The domain is created to reach the maximum number of allowed characters (255 (really 253)) with an exponential curve in the length of the letters as you proceed through the alphabet. The formula used is "1 + 62 * (10/7)^(x-26)". To help illustrate this curve, reference the distribution on this spreadsheet. It's colorful because I like colors and wanted to do a progressive "rainbow" animation in CSS3.

Answer 1

APL (Dyalog), 41 38 bytes

-3 bytes thanks to Bubbler

2⌽'mehttp://',∊'.',¨⍨⎕A⍴¨⍨⌊1+62×.7*⍒⎕A

Try it online!

Outputs the URL with the letters capitalised. Uses the 0 indexed formula \$ \lfloor 1 + 62 \times 0.7^{25-x} \rfloor \$, since \$ (\frac{10}{7})^{x-25} = ((\frac{7}{10})^{-1})^{x-25} = (\frac{7}{10})^{25-x}\$

Explanation:

2⌽                         ⍝ Rotate by two places (moving the 'me' to the end)
  'mehttp://'              ⍝ The start string
             ,             ⍝ Followed by
              ∊               ⍝ The flattened string of
               '.'            ⍝ A period
                  ,⍨¨         ⍝ Appended to the end of each of
                     ⎕A            ⍝ The uppercase alphabet
                       ⍴¨⍨         ⍝ Where each letter is repeated by
                          ⌊               ⍝ The floor of
                           1+             ⍝ 1 plus
                             62×          ⍝ 62 times
                                .7*       ⍝ 0.7 to the power of
                                   ⍒⎕A  ⍝ The range 26 to 1

Answer 2

05AB1E, 25 24 23 21 bytes

žXAS.7Ƶ-t∞-m>×….me¬ýJ

-1 byte thanks to @Neil's analysis that *(10/7)** is the same as /.7**.
-3 bytes thanks to @Grimmy using a different formula and ingenious use of ý!

Try it online.

Explanation:

The formula used to get the correct amount of characters of the alphabet, where \$n\$ is the 1-based index of the alphabetic letter:
\$a(n) = \left\lfloor0.7^{(\sqrt{208}-n)}+1\right\rfloor\$

žX                     # Push builtin "http://"
      Ƶ-               # Push compressed 208
        t              # Take the square-root of that: 14.422...
         ∞             # Push an infinite list of positive integers: [1,2,3,...]
          -            # Subtract each from the 14.422...: [13.442...,12.442...,...]
    .7     m           # Take 0.7 to the power of each of these: [0.008...,0.011...,...]
            >          # Increase each by 1: [1.008...,1.011...,...]
  AS                   # Push the lowercase alphabet as list of characters,
             ×         # and repeat each the calculated values amount of times as string
                       # (which implicitly truncates the floats to integers, and ignores
                       #  the floats beyond the length of the alphabet)
              ….me     # Push ".me"
                  ¬    # Push its head (the "."), without popping the ".me" itself
                   ý   # Join with delimiter. Normally it will use the top value as
                       # delimiter and joins the top-1'th list. In this case however, the
                       # top-1'th item is a string, so instead it will join the entire stack
                       # together. BUT, because the stack contains a list, it will instead
                       # only join all lists on the stacks by the "." delimiter
                    J  # And finally join the three strings on the stack together
                       # (after which this result is output implicitly)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ- is 208.

Answer 3

Python 3, ^{86 \$\cdots\$ 81} 77 bytes

Saved 3 bytes thanks to mypetlion!!!
Saved 4 bytes thanks to Shieru Asakoto!!!

print(f"http://{'.'.join(chr(i+97)*int(1+.7**-i/120)for i in range(26))}.me")

Try it online!

Uses Shieru Asakoto's 0-based formula.

Answer 4

Perl 5, 61 57 55 bytes

-4 bytes thanks to @Neil

say"http://",(map{$_ x(1+62/.7**(++$p-26))."."}a..z),me

Try it online!

Answer 5

JavaScript (Node.js), 77 bytes

Based on @ShieruAsakoto's formula. Builds the URL recursively.

f=n=>n>25?".me":(n?".".padEnd(2+.7**-n/120,Buffer([97+n])):"http://a")+f(-~n)

Try it online!

Answer 6

Java (JDK), 108 bytes

v->{var s="HTTP://";for(char c=64,t;++c<91;s+=(""+c).repeat(t/=Math.pow(.7,c-90))+c+".")t=62;return s+"ME";}

Try it online!

Credits

• -4 bytes thanks to Kevin Cruijssen

Answer 7

Charcoal, 27 bytes

http://⭆β⁺×ι⊕×⁶²Ｘ·⁷⁻²⁵κ.¦me

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

http://                     Implicitly print literal string `http://`
        β                   Lowercase alphabet
       ⭆                    Map over letters and join
                      κ     Current index
                   ⁻²⁵      Subtract from 25
                Ｘ·⁷         Raise 0.7 to that power
             ×⁶²            Multiply by 62
            ⊕               Increment
          ×ι                Repeat letter that many times
         ⁺             .    Concatenate literal string `.`
                        ¦   Implicitly print
                         me Implicitly print literal string `me`

Answer 8

Jelly, 49 48 46 31 bytes (send help)

A niladic link printing the URL

26RU.7*×89ĊØa×⁾meṭj”.“http://”;

-2 bytes thanks to Kevin

-15 bytes thanks to Nick!

I never wrote anything this complex in Jelly and the tacicity isn't obvious to me yet... So this is very golfable (see 49 byte link). I would appreciate feedback and golfing tips in small chunks so that I can digest it!

You can try this online.

Answer 9

Wolfram Language (Mathematica), 71 bytes

"http://"<>Array[Table[Alphabet[][[#]],1+1.43^(#-26)62]<>"."&,26]<>"me"

Try it online!

Answer 10

C (gcc), 99 bytes

n;f(i){for(i=printf("http://");i<33;)for(n=62/pow(.7,i++-32)+2;n--;)putchar(n?89+i:46);puts("me");}

-3 or so bytes thanks to Noodle9!

-8 bytes thanks to gastropner!

Try it online!

Answer 11

JavaScript (Node.js), 93 bytes

_=>`http://${[...Array(26)].map((x,i)=>Buffer(Array(1+.7**-i/120|0).fill(97+i))).join`.`}.me`

Try it online!

When turning the formula into 0-indexed, the formula becomes \$1+62\times(\frac{10}{7})^{x-25}=1+0.0083146\times0.7^{-x}\$, and then 0.0083146 is approximated by 1/120.

Answer 12

Japt, 29 bytes

;`http://{C®p.7**T´/#xÄ +'.}´

Saved 2 bytes thanks to @Shaggy

Test it

Answer 13

R, 124 115 63 bytes

This is way better than my original approach. Thanks to @Giuseppe and @Grimmy

cat("http://");cat(strrep(letters,1+62*.7^(25:0)),"me",sep=".")

Try it online!

Answer 14

Haskell, 86 bytes

g(c,x)=replicate(floor$1+62*1.43**x)c++"."
f="http://"++(zip['a'..][-25..0]>>=g)++"me"

Try it online!

Answer 15

C++ (gcc), 126 123 bytes

-3 bytes thanks to S.S. Anne

Some acrobatics needed to avoid having to include cmath.

#import<map>
using s=std::string;s f(){s r="http://";for(float i=26,p=7456;i--;p*=.7)r+=s(1+62/p,122-i)+'.';return r+"me";}

Try it online!

Answer 16

Keg, 46 40 bytes

`3);://`0&(‡26   |⑻a+ⁿ⅍7
/⑻±Ëx/ℤ1+⑹*\.)`me

Try it online!

A port of the Python 3 answer by Noodle9.

-6 bytes due to using string compression and some other things

Answer 17

PHP, 94 bytes

for($c='a',$s='http://',$i=-26;$i++;$c++,$s.='.')for($l=62/.7**$i;$l-->=0;$s.=$c);echo$s.'me';

Try it online!

Probably still golfable, uses the formula from the question and PHP's char incrementing

EDIT: saved 3 bytes with code reorganization

EDIT2: another byte less starting from -26 and removing ($i-26)

Answer 18

Ruby, 64 bytes

Conveniently, when multiplying a string with a non-whole positive number, Ruby truncates the number to determine how many times to repeat the string.

puts"http://#{(?a..?z).map{|c|c*(1+62/0.7**(c.ord-122))}*?.}.me"

Try it online!

Answer 19

MATLAB 92 bytes

b="";for i=1:26;b=b+repelem(char(96+i),floor(1+62*(10/7).^(i-26)))+".";end;"http://"+b+"me"