A very long Terza Rima

Question

Description

Output the rhyme scheme for a very long Terza Rima.

Input

None.

Output

ABA
BCB
CDC
DED
EFE
FGF
GHG
HIH
IJI
JKJ
KLK
LML
MNM
NON
OPO
PQP
QRQ
RSR
STS
TUT
UVU
VWV
WXW
XYX
YZY

Rules

You can pick between separating stanzas with whitespace or newlines, so either:

ABA BCB...

OR

ABA
BCB
...

A single trailing whitespace allowed per line and one trailing newline allowed.

Output can be either uppercase or lowercase.

This is code-golf, so the shortest code in bytes for each language wins.

According to en.wikipedia.org/wiki/Terza_rima your ending is wrong. It should end with either Z or ZZ. — Chris, Dec 18, 2017 at 18:28
Can there be additional output beyond the rhyme scheme? This might save me a few bytes. — NK1406, Dec 18, 2017 at 23:02

Nnnes · Accepted Answer · 2017-12-18 17:45:25Z

2

Ruby, 32 bytes

puts (?A..?Y).map{|c|c+c.next+c}

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If a list of strings is valid output:

Ruby, 31 bytes

->{(?A..?Y).map{|c|c+c.next+c}}

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answered Dec 18, 2017 at 17:45

Nnnes

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Befunge, 26 bytes

:"A"+:,:1+,,1+:64*`#@_55+,

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edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 20, 2017 at 15:52

James Holderness

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pacman256 · Accepted Answer · 2024-01-09 23:14:18Z

2

Vyxal 3 `j`, 4 bytes

moᵛṅ

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same as vyxal 2 but the byte lost from not having vectorized mirror is saved from the one byte capital alphabet builtin

answered Jan 9 at 23:14

pacman256

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pxeger · Accepted Answer · 2024-01-21 19:15:23Z

2

Zsh, 40 bytes

eval '<<<$[[##36]'{13366..45358..1333}\]

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answered Jan 21 at 19:15

pxeger

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Martin Ender · Accepted Answer · 2017-12-18 13:51:11Z

1

Retina, 24 bytes


ABA
;{:G`
T`_p`p`[^Y]..

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answered Dec 18, 2017 at 13:51

Martin Ender

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\$\begingroup\$ 21 bytes but suffers from trailing newlines... \$\endgroup\$
– Neil
Dec 18, 2017 at 14:57
\$\begingroup\$ @Neil Yeah, you don't even need the `.... Getting rid of those trailing newlines is really annoying though. I'm hoping to make printing in loops a bit more flexible and convenient in Retina 1.0. \$\endgroup\$
– Martin Ender
Dec 18, 2017 at 15:13

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hyper-neutrino · Accepted Answer · 2017-12-18 14:19:59Z

1

brainfuck, 73 bytes

+++++++++++++[>+++++>+++++<<-]>>+<<<<+++++[>++>+++++<<-]>>[>.>.+<.+<-<.>]

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answered Dec 18, 2017 at 14:19

hyper-neutrino♦

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Nelson · Accepted Answer · 2017-12-18 14:30:09Z

1

T-SQL 189 154 bytes

declare @ char(26)='ABCDEFGHIJKLMNOPQRSTUVWXYZ',@i int=1,@@ char while(@i<26)begin  set @@=substring(@,@i,1)print @@+substring(@,@i+1,1)+@@ set @i=@i+1 end

edited Dec 18, 2017 at 14:30

answered Dec 18, 2017 at 14:23

Nelson

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Nahuel Fouilleul · Accepted Answer · 2017-12-18 14:40:35Z

1

Perl 5, 27 24 bytes

say$\=$_++,$_.$\for A..Y

previous solution

say$_,chr 1+ord,$_ for A..Y

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edited Dec 18, 2017 at 14:40

answered Dec 18, 2017 at 14:07

Nahuel Fouilleul

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0xffcourse · Accepted Answer · 2017-12-18 14:41:06Z

1

Recursiva, 13 bytes

{B25"PpZ~}(;}

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answered Dec 18, 2017 at 14:41

0xffcourse

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Conor O'Brien · Accepted Answer · 2017-12-18 15:00:46Z

J (repl), 18 bytes

u:65+(,>:,])"+i.25

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Can be made into a function for +4 bytes:

(u:65+(,>:,])"+i.25)"_

Explanation

u:65+(,>:,])"+i.25
              i.25     range [0, 25)
     (     )"+         on each number:
      ,  ,               create a list composed of the number,
       >:                the number + 1
          ]              and the number
                       this gives the 2D list [0 1 0] [1 2 1] ...
  65+                  add 65 to each element
u:                     convert to characters

Neil · Accepted Answer · 2017-12-18 15:33:14Z

1

Batch, 99 bytes

@set s=ABCDEFGHIJKLMNOPQRSTUVWXYZ
:l
@echo %s:~0,2%%s:~0,1%
@set s=%s:~1%
@if not %s%==Z goto l

Only 10 bytes shorter than a literal...

answered Dec 18, 2017 at 15:33

Neil

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Sanchises · Accepted Answer · 2017-12-18 16:00:53Z

1

MATL, 12 11 10 bytes

1 byte off thanks to Luis Mendo, by using 5B (5 in binary) to push a [1 0 1] pattern.

66:90!5B-c

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edited Dec 18, 2017 at 16:00

answered Dec 18, 2017 at 14:15

Sanchises

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ovs · Accepted Answer · 2017-12-18 16:21:39Z

1

Kotlin, 42 bytes

{('A'..'Y').map{println("$it${it+1}$it")}}

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or

{('A'..'Y').map{println(""+it+(it+1)+it)}}

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answered Dec 18, 2017 at 16:21

ovs

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Socratic Phoenix · Accepted Answer · 2017-12-18 16:52:45Z

1

Shnap, 35 bytes

for c:'A'..'Z'println(""+c+(c+1)+c)

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answered Dec 18, 2017 at 16:52

Socratic Phoenix

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FlipTack · Accepted Answer · 2017-12-18 17:44:44Z

1

Pushy, 9 bytes

Z1Z25:QKh

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Explanation:

Z1Z          \ Make the stack [0, 1, 0]
   25:       \ 25 times do:
      Q      \    Print the stack, as indexes into the uppercase alphabet
       Kh    \    Increment all values

answered Dec 18, 2017 at 17:44

FlipTack

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DJMcMayhem · Accepted Answer · 2017-12-18 18:18:46Z

1

V, 16 bytes

¬azòóˆˆ/±²±\r²òd

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Hexdump:

00000000: ac61 7af2 f388 882f b1b2 b15c 72b2 f264  .az..../...\r..d

Recursive Regular Expressions for the win!

answered Dec 18, 2017 at 18:18

DJMcMayhem

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caird coinheringaahin g · Accepted Answer · 2017-12-18 21:22:00Z

Add++, 46 44 bytes

L,"ABCDEFGHIJKLMNOPQRSTUVWXYZ"dBREpBR2DBcEJn

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Who needs all these "palindromise" functions?

2 bytes saved thanks to DLosc!

How it works (old version)

L,     - Create a nilary lambda function.
  "…"  - Push the uppercase alphabet;     STACK = ['ABC…XYZ']
  d    - Duplicate;                       STACK = ['ABC…XYZ' 'ABC…XYZ']
  BR   - Reverse;                         STACK = ['ABC…XYZ' 'ZYX…CBA']
  2D   - Duplicate from below;            STACK = ['ABC…XYZ' 'ZYX…CBA' 'ABC…XYZ']
  Ep   - Dequeue each;                    STACK = ['BCD…XYZ' 'YXW…CBA' 'BCD…XYZ']
  $BR$ - Reverse the middle string;       STACK = ['BCD…XYZ' 'ABC…WXY' 'BCD…XYZ']
  Bc   - Zip;                             STACK = [['A' 'B' 'A'] … ['Y' 'Z' 'Y']]
  EJ   - Join each;                       STACK = ['ABA' … 'YZY']
  n    - Join with newlines               STACK = ['ABA…YZY']

Οurous · Accepted Answer · 2017-12-18 22:26:57Z

1

Clean, 60 bytes

import StdEnv
t=foldr(+++)""[{c,c+one,c,' '}\\c<-['a'..'y']]

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If a list of Char ([Char]) is okay instead of a String ({#Char}) (they look the same when printed bare), then we can save a few bytes and use:

Clean, 53 bytes

import StdEnv
t=flatlines[[c,c+one,c]\\c<-['a'..'y']]

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If a list of String ([{#Char}]) representing the lines is acceptable, then we can save even more and use:

Clean, 45 bytes

import StdEnv
t=[{#c,c+one,c}\\c<-['a'..'y']]

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edited Dec 18, 2017 at 22:26

answered Dec 18, 2017 at 22:14

Οurous

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ATaco · Accepted Answer · 2017-12-18 22:53:50Z

1

Funky, 38 bytes

fori=65i<90i++print("%c%c%c"%{i i+1i})

Naïve solution, not very exciting, but it works.

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answered Dec 18, 2017 at 22:53

ATaco

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Japt, 11 6 bytes

Returns an array of lines.

;BäÈiZ

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Explanation

;B is the uppercase alphabet.
äÈ gets each pair of consecutive characters in a string and passes them through a function.
iZ prepends the current element to the first character of the current element.

Alternative, 6 bytes

;Bã mê

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Explanation

ã splits the alphabet string into an array of consecutive character pairs.
m maps over the array.
ê palindromises the current element.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 18, 2017 at 17:40

Shaggy

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\$\begingroup\$ Array of lines is allowed. \$\endgroup\$
– LiefdeWen
Dec 19, 2017 at 7:16

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score 1 · Accepted Answer · 2017-12-19 18:27:57Z

1

Brain-Flak classic, 160 150 118+2=120 bytes with -A flag

((((((()()()){}){}){}){}){}())<>(((((()()()()())){}){}){}){({}[])<>([{}])[({}())][({}[])]({}())[((()()()()()){})]{}<>}

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The separator is a newline.

-8 bytes thanks to DJ

-32 bytes thanks to DJ

edited Dec 19, 2017 at 18:27

answered Dec 18, 2017 at 17:18

user63187

\$\begingroup\$ The first 50 bytes are rather redundant. in general, rather than doing (...)(({}){}) you can do ((...){}) to save 4 bytes. And of course, this can be nested several times. To push 'a' I would do ((((((()()()){}){}){}){}){}()) instead. \$\endgroup\$
– DJMcMayhem
Dec 19, 2017 at 18:05
\$\begingroup\$ You can also use brain-flak.github.io/integer to autogolf large integers for you. \$\endgroup\$
– DJMcMayhem
Dec 19, 2017 at 18:05
\$\begingroup\$ And minus 32 bytes \$\endgroup\$
– DJMcMayhem
Dec 19, 2017 at 18:08

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brhfl · Accepted Answer · 2017-12-19 18:31:45Z

1

dc, 24 bytes

65[ddP1+dPrPdAP90>M]dsMx

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Pretty straightforward. Start with ASCII value for A, duplicate it a couple of times, print it, increment it and duplicate this, print it, swap the top-of-stack, print (the unincremented number), print a line feed, keep going as long as we haven't made it to Z (90).

answered Dec 19, 2017 at 18:31

brhfl

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Max00355 · Accepted Answer · 2017-12-19 20:04:15Z

1

Python 3, 55 bytes

' '.join(map(lambda x:"%c"*3%(x-1,x,x-1),range(66,91)))

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edited Dec 19, 2017 at 20:04

answered Dec 19, 2017 at 19:49

Max00355

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\$\begingroup\$ -2 : ' '.join(map(lambda x:"%c"*3%(x,x+1,x),range(65,90))) \$\endgroup\$
– Jakque
May 22, 2021 at 9:59

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DELETE_ME · Accepted Answer · 2017-12-23 14:53:14Z

Cubically, 34 bytes

FRF'-2(+345@+33-5@-33+5@-33200@1)6

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It was hard... At least it's doable. (without too much hardcoding)

Other Rubik's algorithm may lead to shorter sequence.

Some alternatives I tried (which is, unfortunately, longer):

35 bytes: Try it online!

36 bytes: Try it online!

Explanation:


FRF'      Rubik's cube algorithm. This algorithm has the property of make face
          value 1 equal to 10. (character of code of `\n`)
-2        Subtract face value 2 (which is 25) from the notepad.
(    )6   While notepad is nonzero:
+345        Add 90 to notepad.
@           Print notepad value.
+33-5       Add 1 to notepad.
@           Print notepad value.
-33+5       Subtract 1 from notepad.
@           Print notepad value.
-33200      Subtract 89 from notepad.
@1          Print face value 1 (which is 10, '\n')

Conor O'Brien · Accepted Answer · 2018-01-02 00:39:47Z

Premier, 18 bytes

:| s |ns:'Z~*AUc
A

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Explanation

The second line just defines the data for the program. So, starting wtih A on the stack...

:| s |ns:'Z~*AUc
:|   |n            print the following format:
  .                   TOS
   s                  TOS + 1
    .                 TOS
                   this prints out ABA, for example
       s:'Z~*      If the next char is not 'Z'
             A        append it to the data
              Uc   print a newline

This repeats until the next character is Z, at which point there is no data left, and the program terminates.

Endenite · Accepted Answer · 2018-01-02 15:22:54Z

vim, 48 28 keystrokes

:h<_EnterjjYZZPqqy2llpi ESCl@qq@qD

My first attempt at golfing in vim

Explanation

:h<_EnterjjYZZP                    Insert the alphabet by copying it from the help manual
qq                                  Define a macro `q`:
 y2l                                 Copy the first two characters (relative to the cursor)
 lp                                  Paste after the copied characters (ABCDE -> ABABCDE)
 i Esc                             Insert a space (ABABCDE -> ABA BCDE)
 l@q                                 Run the macro again with the cursor after the space
q@q                                 End the macro defenition and run it
D                                   Remove the last Z

Giuseppe · Accepted Answer · 2018-04-20 20:22:36Z

SNOBOL4 (CSNOBOL4), 83 82 81 79 72 bytes

R	&UCASE POS(I) LEN(1) . L @I LEN(1) . R	:F(END)
	OUTPUT =L R L	:(R)
END

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I think I'm finally getting the hang of @!

SNOBOL variables are initialized automatically to "" which is also equivalent to 0. So the first time the interpreter sees line R, the line is:

&UCASE POS(0) LEN(1) . L @I LEN(1) . R  :F(END)

which finds a LENgth 1 string starting at POSition 0 (the beginning of the string), assigning the matched string to L. Then we assign the position of that match to I using @, which has been incremented by 1 because of the LEN(1) . L. The next character is then matched and assign it to R, and output L R L. This continues until I is 25, since then LEN(1) . R fails to match, as L would be Z.

Dust · Accepted Answer · 2018-04-21 14:15:31Z

1

Go, 56 Bytes

func n(){for i:=65;i<90;i++{Printf("%c%c%c\n",i,i+1,i)}}

Ungolfed and readable:

func n() {
    for i := 65; i < 90; i++ {
        Printf("%c%c%c\n", i, i + 1, i)
    }
}

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answered Apr 21, 2018 at 14:15

Dust

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NK1406 · Accepted Answer · 2018-05-12 13:15:22Z

1

PHP, 72 bytes

-2 bytes because yay assumed values!

-15 bytes thanks to Francisco Hahn

<?php $x=ABA;for($i=0;$i<75;$i++){if($i%3==0){echo$x." ";}$t=$x[$i%3];$x[$i%3]=++$t;}?>

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Takes the string "ABA" through a for loop which loops through each character and increments it. Displays the current value of the string every 3 times.

edited May 12, 2018 at 13:15

answered Dec 19, 2017 at 21:46

NK1406

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\$\begingroup\$ $x=ABA;for(;$i<75;$i++){echo($i%3)?"":$x." ";$t=$x[$i%3];$x[$i%3]=++$t;} 72 \$\endgroup\$
– Francisco Hahn
May 8, 2018 at 13:42

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ollien · Accepted Answer · 2018-05-12 21:02:42Z

1

Go, 60 56 bytes

func t(){for i:=65;i<90;i++{Printf("%c%c%c\n",i,i+1,i)}}

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edited May 12, 2018 at 21:02

answered May 12, 2018 at 14:20

ollien

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A very long Terza Rima

Description

Input

Output

Rules

90 Answers 90

Ruby, 32 bytes

Ruby, 31 bytes

Befunge, 26 bytes

Vyxal 3 j, 4 bytes

Zsh, 40 bytes

Retina, 24 bytes

brainfuck, 73 bytes

T-SQL 189 154 bytes

Perl 5, 27 24 bytes

Recursiva, 13 bytes

J (repl), 18 bytes

Explanation

Batch, 99 bytes

MATL, 12 11 10 bytes

Kotlin, 42 bytes

Shnap, 35 bytes

Pushy, 9 bytes

V, 16 bytes

Add++, 46 44 bytes

How it works (old version)

Clean, 60 bytes

Clean, 53 bytes

Clean, 45 bytes

Funky, 38 bytes

Japt, 11 6 bytes

Explanation

Alternative, 6 bytes

Explanation

Brain-Flak classic, 160 150 118+2=120 bytes with -A flag

dc, 24 bytes

Python 3, 55 bytes

Cubically, 34 bytes

Premier, 18 bytes

Explanation

vim, 48 28 keystrokes

Explanation

SNOBOL4 (CSNOBOL4), 83 82 81 79 72 bytes

Go, 56 Bytes

PHP, 72 bytes

Go, 60 56 bytes

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