1
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Task is to output this text:

ABCDEFGHIJKLMNOPQRSTUVWXYZA
A 00000000000000000000000 Z
B                         Y
C  101010101010101010101  X
D                         W
E                         V
F   2101210121012101210   U
G                         T
H                         S
I                         R
J    12321012321012321    Q
K                         P
L                         O
M     234321012343210     N
N     123432101234321     M
O                         L
P                         K
Q    23210123210123210    J
R                         I
S                         H
T                         G
U   1210121012101210121   F
V                         E
W                         D
X  010101010101010101010  C
Y                         B
Z 00000000000000000000000 A
ABCDEFGHIJKLMNOPQRSTUVWXYZA

Rules

  • shortest code wins, but you are encouraged to post answers even if your language is among more verbose ones
  • trailing newline allowed,
  • lowercase letters allowed
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5
  • 4
    \$\begingroup\$ Hi there! I've downvoted this challenge as I'm really struggling to see any sort of pattern to the middle digits, leading most answers to be like the Bubblegum answer and do just straight up compression. I recommend you post future challenges to the Sandbox where you can get meaningful feedback before posting them to main. \$\endgroup\$ Commented Feb 22, 2017 at 19:15
  • 1
    \$\begingroup\$ @AdmBorkBork He posted it to sandbox. Some people said there wasn't much interesting about it though. \$\endgroup\$ Commented Feb 22, 2017 at 19:17
  • \$\begingroup\$ @AdmBorkBork Thank you for taking your time to comment. There definitely is pattern in middle numbers. Perhaps it's unusable in miningful way.@LliwTelracs It was one person. Perhaps I should have trusted his judgement :-) \$\endgroup\$
    – user65167
    Commented Feb 22, 2017 at 19:18
  • 3
    \$\begingroup\$ Looks like there is some pattern to me. Let's try to beat the gum! \$\endgroup\$ Commented Feb 22, 2017 at 19:26
  • \$\begingroup\$ @AdmBorkBork I think for the inside, it cycles. For example, the outermost ones cycles [0], next cycles [0,1] in 0101 fashion, next cycles [0,1,2] in 0121 fashion, so on so forth until the middle one cycles [0,1,2,3,4] in 0123-4-3210 fashion. Ascending and descending? \$\endgroup\$
    – Thunda
    Commented Feb 26, 2017 at 5:47

4 Answers 4

3
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SOGL 0.8.2, 121 120 bytes

Z A+p→**1"PļΔķ≠ƨν↕KzøjQ¶°³²y⁹ν⅔E    "§768╗@ū℮^‰Y√ψ⅜εrυd┘mZΣlL⅝6׀KTφυZŗ4M7E¤Δ≠⅓∫⁽mΜ‽κuΘ\λ`ω¬ρDDz⅞":e─░χ′⁷‘’«n{;³ZWO≤oZ±WoI}*

Explanation:

Z                                 push the uppercase alphabet
  A+                              append "A"
    p                             output in a new line
     →*                           define function "*" with the above
       *                          call the function "*"
        1                         push 1 (counter)
         "...‘                    push the middle lines (without the alphabet letters) joined
              ’«n                 convert to an array with each item 25 chars long
                 {            }   for each line
                  ;³              put the counter ontop on stack and triplicate it
                    ZW            get the counterth letter of the alphabet
                      O           output in newline
                       ≤          put the current array item ontop of stack
                        o         append
                         Z±W      get the counterth letter of the reversed alphabet
                            o     append
                             I    increase counter
                               *  call the function "*"
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3
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Bubblegum, 186 bytes

00000000: 7592 d70d c330 0c44 ff6f 0a8e e092 05dc  u....0.D.o......
00000010: 2df7 def6 1f24 3263 0429 3e09 2281 7b38  -....$2c.)>.".{8
00000020: 1284 1884 519c a459 6e8a b2aa 9bb6 eb87  ....Q..Yn.......
00000030: 719a 9775 db8f 0081 38f7 470e 84c2 ce8e  q..u....8.G.....
00000040: 48c4 756e aec8 8698 fa56 2494 2d48 6df4  H.un.....V$.-Hm.
00000050: 6c8d 9f67 d519 19f5 4dc8 291b 6128 1b50  l..g....M.).a(.P
00000060: 9cc9 f57c edf3 8ea7 d8a3 a4be 0e15 652d  ...|..........e-
00000070: 6acd 9eff b8ea 6956 ad41 2357 c34f a85a  j.....iV.A#W.O.Z
00000080: 8d96 d6ac d051 56a2 7fb5 fb1e 411b 1618  .....QV.....A...
00000090: a8cf 60a4 2cc7 4459 8659 27f8 fb24 aba6  ..`.,.DY.Y'..$..
000000a0: 58a8 2fc1 4a59 8c4d e46e 95ec 1411 76ea  X./.JY.M.n....v.
000000b0: 0b71 d0dd b57b cd77 fe09                 .q...{.w..

Try It Online!

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1
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NO!, 34067 bytes

OP asked for verbose. I give you verbose.

I would say the NO! isn't old enough to compete but the length of the code says that it's not competing

As the code is too long for SE, here is a link to the actual code: actual code. I chose a TIO page because I can't be bothered to find anything else.

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3
  • \$\begingroup\$ NO! Shouting is not the answer :) \$\endgroup\$ Commented Apr 3, 2017 at 1:59
  • \$\begingroup\$ @AnthonyPham NO! says YES! SHOUTING IS THE ANSWER! (No rudeness intended) \$\endgroup\$ Commented Apr 3, 2017 at 2:00
  • \$\begingroup\$ I think that's the only TIO link I have had to scroll up to see the first line of... \$\endgroup\$ Commented Apr 17, 2017 at 17:39
0
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PHP, 294 Bytes

$d=[0=>[0,23],2=>["01",21],5=>[2101,19],9=>[123210,17],12=>[23432101,15],13=>[12343210,15],16=>[232101,17],20=>[1210,19],23=>[10,21],25=>[0,23]];echo$j=join(range(A,Z)),"A\n";for($i=0;$i<26;$i++)echo$j[$i].str_pad($d[$i]?str_pad("",$d[$i][1],$d[$i][0]):"",25," ",2).$j[-$i-1]."\n";echo$j."A\n";

Expanded

$d=[
0=>[0,23],
2=>["01",21],
5=>[2101,19],
9=>[123210,17],
12=>[23432101,15],
13=>[12343210,15],
16=>[232101,17],
20=>[1210,19],
23=>[10,21],
25=>[0,23]];
echo$j=join(range(A,Z)),"A\n";
for($i=0;$i<26;$i++)
echo$j[$i].str_pad($d[$i]?str_pad("",$d[$i][1],$d[$i][0]):"",25," ",2).$j[-$i-1]."\n";
echo$j."A\n";
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