24
\$\begingroup\$

Task

Your task is to print this exact text:

A
BCD
EFGHI
JKLMNOP
QRSTUVWXY
ZABCDEFGHIJ
KLMNOPQRSTUVW
XYZABCDEFGHIJKL
MNOPQRSTUVWXYZABC
DEFGHIJKLMNOPQRSTUV
WXYZABCDEFGHIJKLMNOPQ
RSTUVWXYZABCDEFGHIJKLMN
OPQRSTUVWXYZABCDEFGHIJKLM
NOPQRSTUVWXYZABCDEFGHIJKLMN
OPQRSTUVWXYZABCDEFGHIJKLMNOPQ
RSTUVWXYZABCDEFGHIJKLMNOPQRSTUV
WXYZABCDEFGHIJKLMNOPQRSTUVWXYZABC
DEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKL
MNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVW
XYZABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJ
KLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXY
ZABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOP
QRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHI
JKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZABCD
EFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZA
BCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZ

Specs

  • You may do it in all-lowercase instead of all-uppercase.
  • Trailing newlines at the end of the triangle is allowed.
  • Trailing spaces after each line is allowed.
  • You must print to STDOUT instead of outputting an array of strings.

Scoring

This is . Program with lowest byte-count wins.

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  • 1
    \$\begingroup\$ What do you mean by "strikes again"? Was there another challenge you made like this? \$\endgroup\$ – haykam Aug 21 '16 at 13:46
  • \$\begingroup\$ @Peanut codegolf.stackexchange.com/questions/87496/alphabet-triangle \$\endgroup\$ – Beta Decay Aug 21 '16 at 13:58
  • 1
    \$\begingroup\$ Seems fairly trivial do we really need (another) alphabet challenge? \$\endgroup\$ – Rohan Jhunjhunwala Aug 21 '16 at 21:44
  • 2
    \$\begingroup\$ It is a good challenge, but I think we have outstripped saturation of these alphabet challenges, nothing personal. \$\endgroup\$ – Rohan Jhunjhunwala Aug 21 '16 at 21:45
  • \$\begingroup\$ Actually looking for an alphabet challenge that the letter at a position cannot be calculated by simple expressions from its coordinates involving the mod function. May make one myself if I have time. \$\endgroup\$ – Weijun Zhou Jan 31 '18 at 20:17

54 Answers 54

39
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Vim, 29 bytes

:h<_↵↵↵y$ZZ26P0qqa↵♥βjllq25@q

Where represents the Return key, the Escape key, and β the Backspace key.

enter image description here

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  • 3
    \$\begingroup\$ How do you always beat me to the vim answer on these? Aargh +1 anyway, I can't not upvote vim! :) \$\endgroup\$ – DJMcMayhem Aug 21 '16 at 12:41
  • 7
    \$\begingroup\$ I still think you should use instead of . And instead of β. That's what these Unicode chars were made for. utf8icons.com/subsets/control-pictures \$\endgroup\$ – mbomb007 Aug 23 '16 at 20:36
9
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Python 2, 65 bytes

i=1
a=bytearray(range(65,91))*26
while a:print a[:i];a=a[i:];i+=2
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  • 1
    \$\begingroup\$ I've changed the header to Python 2 because the code would not work in Python 3. \$\endgroup\$ – Leaky Nun Aug 21 '16 at 11:49
7
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Jelly, 10 bytes

26RḤ’RØAṁY

Try it online!

How it works

26RḤ’RØAṁY  Main link. No Arguments.

26          Set the return value to 26.
  R         Range; yield [1, 2, ..., 25, 26].
   Ḥ        Unhalve; yield [2, 4, ..., 50, 52].
    ’       Decrement; yield [1, 3, ..., 49, 51].
     R      Range; yield [[1], [1, 2, 3], ..., [1, ..., 49], [1, ..., 51]].
      ØA    Yield the uppercase alphabet.
        ṁ   Mold the alphabet as the array of ranges. This repeats the uppercase
            letters over an over again, until all integers in the range arrays
            have been replaced with letters.
         Y  Join, separating by linefeeds.
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  • \$\begingroup\$ Was "Double" called "Unhalve" back then? Also, is great!! [waiting to say congrats for 100k rep] \$\endgroup\$ – Erik the Outgolfer Sep 28 '16 at 11:14
  • \$\begingroup\$ It's just a mnemonic. H is halve and is its inverse (unhalve). \$\endgroup\$ – Dennis Sep 28 '16 at 12:15
  • \$\begingroup\$ I just think of /2 or *2, so it's "Halve" or "Double". That's why I was confused. \$\endgroup\$ – Erik the Outgolfer Sep 28 '16 at 12:17
  • \$\begingroup\$ Also 10 bytes: 27Ḷ²IRØAṁY \$\endgroup\$ – Leaky Nun Apr 30 '17 at 4:32
  • \$\begingroup\$ Also 10 bytes: 51Rm2RØAṁY \$\endgroup\$ – HyperNeutrino Apr 30 '17 at 5:01
7
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VBA Excel (80 bytes, 1742 bytes)


Excel, 1742 bytes

Inspired by the ugoren's creative answer, I managed to find an Excel formula to create the pattern as shown in the OP.

=MID(REPT("ABCDEFGHIJKLMNOPQRSTUVWXYZ",26),(ROW()-1)^2+1,2*ROW()-1)

Paste this formula in cell A1, then drag all over range A1:A26.

The length of the formula is 67 bytes but you have to replicate it 26 times, so it's equal to 67*26=1742 bytes. Here is the output:

enter image description here


Excel VBA, 80 bytes

Now it's possible we integrate Excel with VBA to automate the process and to save many bytes since VBA is built into most Microsoft Office applications, including Excel. Write and run the following code in the Immediate Window (use combination keys CTRL+G to display it in Visual Basic Editor):

[A1:A26]="=MID(REPT(""ABCDEFGHIJKLMNOPQRSTUVWXYZ"",26),(ROW()-1)^2+1,2*ROW()-1)"

The program works by printing the Excel formula above to the range A1:A26. Unfortunately, both Excel and VBA have no built-in alphabet.

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  • \$\begingroup\$ The column names look like a built-in alphabet to me. Use the first 26 column names. \$\endgroup\$ – mbomb007 Aug 23 '16 at 20:32
  • 1
    \$\begingroup\$ @mbomb007 So what? I think it would be harder to use them instead of strings. \$\endgroup\$ – Erik the Outgolfer Sep 28 '16 at 11:18
  • \$\begingroup\$ @EriktheGolfer So? My point was that there is a builtin alphabet. \$\endgroup\$ – mbomb007 Sep 28 '16 at 13:26
  • 1
    \$\begingroup\$ @mbomb007 You said "Use the first 26 column names", which I perceived as "Use the first 26 column names instead of what you currently use", that's why I replied. \$\endgroup\$ – Erik the Outgolfer Sep 28 '16 at 13:29
  • \$\begingroup\$ @EriktheGolfer It's a suggestion. Idk how many bytes it'd be. \$\endgroup\$ – mbomb007 Sep 28 '16 at 13:30
5
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Haskell, 67 bytes

_#53=[]
s#i=take i s:drop i s#(i+2)
mapM putStrLn$cycle['A'..'Z']#1

A simple recursion over the length i of the line. In each step the next i chars are taken from an infinite repetition of the alphabet.

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4
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Mathematica, 90 bytes

StringRiffle[Flatten[Alphabet[]&~Array~26]~Internal`PartitionRagged~Range[1,51,2],"
",""]&

Anonymous function. Takes no input and returns a string as output. Golfing suggestions welcome. An example of what Internal`PartitionRagged does:

In[1]:= Internal`PartitionRagged[{2, 3, 5, 7, 11, 13}, {2, 3, 1}]               

Out[1]= {{2, 3}, {5, 7, 11}, {13}}
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  • \$\begingroup\$ Mathematica has a built-in for alphabet triangles? \$\endgroup\$ – Buffer Over Read Aug 21 '16 at 18:35
4
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C, 79 bytes

main(i,j){for(i=0,j=1;i<676;i++){putchar(i%26+65);if(j*j==i+1){puts("");j++;}}}

My first answer in C \o/

Golfing suggestions are more than welcome.

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  • \$\begingroup\$ 62: i;main(j){while(i<676)printf("\n%c"+(j*j^i++||!j++),i%26+65);} \$\endgroup\$ – xsot Aug 21 '16 at 12:37
  • \$\begingroup\$ @xsot Thanks but I'm afraid that leading newlines are not allowed. \$\endgroup\$ – Leaky Nun Aug 21 '16 at 12:47
  • 1
    \$\begingroup\$ But there's no leading newline? \$\endgroup\$ – xsot Aug 21 '16 at 12:50
  • \$\begingroup\$ 60: i;main(j){for(;j<27;j*j^++i||puts("",j++))putchar(i%26+65);} \$\endgroup\$ – xsot Aug 21 '16 at 13:00
  • \$\begingroup\$ @immibis I guess I should post it then. \$\endgroup\$ – xsot Aug 22 '16 at 23:57
4
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Brachylog, 37 bytes

26~l<:1aLc~j[@A:I],L~@nw
=:2%1,.#@l?,

Try it online!

Explanation

  • Main predicate:

    26~l         Let there be a list of 26 elements
    <            This list is an ascending list of integers
    :1aL         Apply predicate 1 to that list ; the resulting list of strings is L
    c            Concatenate the list of strings into one big string
    ~j[@A:I],    That big string is the result of juxataposing the alphabet I times to itself
    L~@n         Create a string which when splitted on line breaks results in L
    w            Write that string to STDOUT
    
  • Predicate 1: used to generate variable strings of odd lengths.

    =            Assign a value to the Input
    :2%1,        That value must be odd
    .#@l?,       Output is a string of length Input
    
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  • \$\begingroup\$ Finally you do an ascii-art challenge \$\endgroup\$ – Leaky Nun Aug 21 '16 at 12:30
  • \$\begingroup\$ @LeakyNun Would classify that as string manipulation more than ASCII art imo \$\endgroup\$ – Fatalize Aug 21 '16 at 12:33
  • \$\begingroup\$ Should I add that to the tag? \$\endgroup\$ – Leaky Nun Aug 21 '16 at 12:43
  • \$\begingroup\$ Nice use of the fact that the last line is the only line which ends in Z which is because 26 is square-free. \$\endgroup\$ – Leaky Nun Aug 21 '16 at 12:44
3
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Pyth, 12 bytes

jtc*G26*Rd26

Try it online!

\$\endgroup\$
3
\$\begingroup\$

JavaScript (ES6), 77 82 88

EcmaScript 6 required just to save 1 byte using a template string literal for newline.

for(i=r=l=o='';l+52;r++||(r=l-=2,o+=`
`))o+=(i++%26+10).toString(36);alert(o)

Less golfed

for(i = r = l = o = '';
    l + 52;
    r++ || (r = l -= 2, o += `\n`))
  o += (i++ % 26 + 10).toString(36);
alert(o);

Test

for(i=r=l=o='';l+52;r++||(r=l-=2,o+=`
`))o+=(i++%26+10).toString(36);alert(o)

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  • \$\begingroup\$ \o/ an ES6 answer I can run in my browser \$\endgroup\$ – Downgoat Aug 21 '16 at 19:53
  • \$\begingroup\$ I stole your .toString(36) logic...now you have to beat 80 bytes!! \$\endgroup\$ – applejacks01 Aug 22 '16 at 12:19
  • \$\begingroup\$ Ahh I concede, I can't think of any way to beat this with my library. Thanks for the challenge! \$\endgroup\$ – applejacks01 Aug 22 '16 at 15:57
3
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Perl, 42 41 39 bytes

perl -E '@b=(A..Z)x26;say splice@b,0,$#a+=2for@b'

Just the code:

@b=(A..Z)x26;say splice@b,0,$#a+=2for@b

An obvious shorter version unfortunately triggers an internal perl problem (Use of freed value in iteration):

say splice@b,0,$#a+=2for@b=(A..Z)x26
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2
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JavaScript, 129 bytes

z=1,i=0,g=a=>{b=0,c="";while(a+a-1>b){c+='ABCDEFGHIJKLMNOPQRSTUVWXYZ'.split``[i>26?i=0:i++];b++}console.log(c)};while(z<26)g(z++)
\$\endgroup\$
  • 1
    \$\begingroup\$ Using the spread operator could save you 3 bytes: [...'ABCDEFGHIJKLMNOPQRSTUVWXYZ'] instead of 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'.split``. \$\endgroup\$ – insertusernamehere Aug 21 '16 at 18:46
2
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Pyke, 14 bytes

G26*WDoh<
Ko>D

Try it here!

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2
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Go, 133 bytes

package main
import S "strings"
func main(){s:=S.Repeat("ABCDEFGHIJKLMNOPQRSTUVXXYZ",26)
for i:=1;s!="";i+=2{println(s[:i]);s=s[i:]}}
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2
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MATLAB, 112 109 95 79 77 bytes

This will also work with Octave, you can try online here.

a=['' 65:90 65:90]';j=0;for i=1:2:52;b=circshift(a,j);j=j-i;disp(b(1:i)');end

So after some major changes, I've saved a further 14 32 bytes. This one is getting to be much more like the length I would have expected from MATLAB. I've left the old version below as it is substantially different.

a=['' 65:90 65:90]';        %Create 2 copies of the upper case alphabet
j=0;                        %Initialise cumulative sum
for i=1:2:52;               %For each line length
    b=circshift(a,j);       %Rotate the alphabet string by the cumulative offset
    j=j-i;                  %Update cumulative negative sum of offsets.
    disp(
         b(1:i)'            %Extract first 'line length' characters from rotated alphabet.
               );           %Display next line (disp adds newline automatically)
end

Original version:

a=['' repmat(65:90,1,26)];c=cumsum(1:2:51);disp(cell2mat(arrayfun(@(s,f)[a(s:f) 10],[1,c(1:25)+1],c,'Un',0)))

Wow that one ended up being longer than I thought it would. I'll see if I can't knock a few bytes off it.

An ungolfed version to explain:

a=['' repmat(65:90,1,26)]; %Create 26 copies of the upper case alphabet
c=cumsum(1:2:51);          %Calculate the end index of each row in the above array, by cumulatively summing the length of each line
disp(
     cell2mat(
              arrayfun(@(s,f)
                             [a(s:f) 10], %Extract the part of the alphabet and insert a new line.
                                         [1,c(1:25)+1],c, %start index is the previous end of line plus 1. End index is as calculated by cumsum.
                       'Un',0 %The returned values are not all the same length
                       )   %Do this for each line
              )            %Convert back to a matrix now new lines inserted
     )                     %And display it

Acknowledgements

  • 3 bytes saved - thanks @LuisMendo
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2
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XPath 3.0 (and XQuery 3.0), 84 bytes

codepoints-to-string((0 to 25)!(subsequence(((1 to 26)!(65 to 90)),.*.+1,2*.+1),10))

Explanation:

(1 to 26)!(65 to 90) is the alphabet 26 times

(0 to 25)!(subsequence(XX, start, len),10) takes 26 subsequences of this, each followed by newline

subsequence(X, .*.+1, 2*.+1) takes successive subsequences with start position and length: (1, 1), (2, 3), (5, 5), (10, 9) etc.

codepoints-to-string() turns Unicode codepoints into characters

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  • \$\begingroup\$ Bravo. I thought I knew what XQuery was. Turns out I had no idea. \$\endgroup\$ – Jordan Aug 22 '16 at 4:25
  • \$\begingroup\$ I doubt that many of the posts here tell you much about the language they are written in. \$\endgroup\$ – Michael Kay Aug 22 '16 at 21:24
2
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Ruby, 46 bytes

26.times{|i|puts ([*?A..?Z]*26)[i*i,i*2+1]*""}

See it on ideone: http://ideone.com/3hGLB0

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2
\$\begingroup\$

05AB1E (alternate) 15 bytes

A2×52µ¼D¾£,¾¼FÀ

Try it online!

Explanation:

A2×              # push a string containing a-za-z
   52µ           # Loop the rest of the program until counter = 52
      ¼          # increment counter (it's 0 initially)
       D         # Duplicate the alpha string on the stack
        ¾£       # Replace alpha with alpha[0..counter]
          ,      # Pop the substring and print it
           ¾¼FÀ  # rotate the alpha string left counter++ times.
\$\endgroup\$
2
\$\begingroup\$

R, 120 115 111 bytes

v=c();for(i in 1:26){v=c(v,c(rep(LETTERS,26)[(sum((b=seq(1,51,2))[1:i-1])+1):sum(b[1:i])],"\n"))};cat(v,sep="")

Ungolfed :

a=rep(LETTERS,26)
b=seq(1,51,2)
v=vector()

for(i in 1:26)
    {
    v=c(v,c(a[(sum(b[1:i-1])+1):sum(b[1:i])],"\n"))
    }

cat(v,sep="")

Basically, b is the vector of the odd numbers between 1 and 51, thus giving the length of each line. Obviously, the sum function sums the numbers of this vector, and gives the starting and ending indexes.

-5 bytes thanks to @plannapus !
-4 bytes thanks to @plannapus !

\$\endgroup\$
  • 1
    \$\begingroup\$ arf, sorry I didn't see that earlier, but since you're only using a once you don't actually need to define it, meaning you can shave a couple more bytes: b=seq(1,51,2);v=c();for(i in 1:26){v=c(v,c(rep(LETTERS,26)[(sum(b[1:i-1])+1):sum(b[1:i])],"\n"))};cat(v,sep="") works. \$\endgroup\$ – plannapus Aug 23 '16 at 13:52
  • \$\begingroup\$ @plannapus Silly me ! Thanks again ! I also integrated the b=seq part in the main body, so it's even less readable ! \$\endgroup\$ – Frédéric Aug 23 '16 at 13:58
2
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R, 81 73 65 63 bytes

A simple for loop approach. Repeat the alphabet 26 times and loop through a sliding index range that is calculated using (i^2-2*i+2):i^2.

for(i in 1:26)cat(rep(LETTERS,26)[(i^2-2*i+2):i^2],"\n",sep="")
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2
\$\begingroup\$

Husk, 10 bytes

Cİ1ṠṁK…"AZ

Try it online!

Explanation

Cİ1ṠṁK…"AZ
      …"AZ    Get the alphabet
   ṠṁK        Replace each letter with the whole alphabet
C             Cut the resulting string into lines with lengths
 İ1            equal to the list of odd numbers
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1
\$\begingroup\$

Batch, 123 bytes

@set s=
@for /l %%i in (1,2,51)do @call set s=%%s%%ABCDEFGHIJKLMNOPQRSTUVWXYZ&call echo %%s:~0,%%i%%&call set s=%%s:~%%i%%
\$\endgroup\$
1
\$\begingroup\$

05AB1E, 18 17 bytes

26FA26×N>n£NnF¦},

Explanation

26F                 # for N in range(0, 26) do
   A26×             # the alphabet repeated 26 times
       N>n£         # take (N+1)^2 letters
           NnF¦}    # throw away the first N^2 letters
                ,   # print with newline

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Rexx, 74 72 bytes

i=1;m=1;do 26;say substr(copies(xrange('A','Z'),26),i,m);i=i+m;m=m+2;end

Ungolfed:

i=1
m=1
do 26
  say substr(copies(xrange('A','Z'),26),i,m)
  i=i+m
  m=m+2
end
\$\endgroup\$
1
\$\begingroup\$

TSQL, 129 bytes

USE MASTER in the beginning of the script is to ensure that the query is run in the master database which is default for many users(not counting bytes for that).

Golfed:

USE MASTER

SELECT SUBSTRING(REPLICATE('ABCDEFGHIJKLMNOPQRSTUVWXYZ',26),number*number+1,number*2+1)FROM spt_values WHERE number<26and'P'=type

Ungolfed:

USE MASTER

SELECT SUBSTRING(REPLICATE('ABCDEFGHIJKLMNOPQRSTUVWXYZ',26),number*number+1,number*2+1)
FROM spt_values
WHERE number<26and'P'=type

Fiddle

Fiddle for older version using xml path

\$\endgroup\$
1
\$\begingroup\$

PowerShell, 68 bytes

$a=-join(65..90|%{[char]$_})*26;26..1|%{$a=$a.Insert($_*$_,"`n")};$a

The section before the first semicolon produces a string containing 26 copies of the uppercase alphabet. The next section injects linebreaks at the index of each square number (working backward so I don't have to account for the shifting). Finally, the $a at the end just shoves that string variable onto PowerShell's equivalent of STDOUT.

\$\endgroup\$
1
\$\begingroup\$

Dyalog APL, 18 bytes

↑2{⍺↓⍵⍴⎕a}/×⍨0,⍳26

\$\endgroup\$
  • \$\begingroup\$ 18 bytes or 18 characters? To encode these 18 characters as 18 bytes, you need a custom character encoding; and we could compress any of the solutions if we choose our character encoding carefully enough. \$\endgroup\$ – Michael Kay Aug 21 '16 at 23:06
  • 1
    \$\begingroup\$ @MichaelKay When can APL characters be counted as 1 byte each? \$\endgroup\$ – Dennis Aug 22 '16 at 0:09
  • \$\begingroup\$ Will you explain this, or shall I? \$\endgroup\$ – Adám Aug 23 '16 at 11:26
1
\$\begingroup\$

C, 60 bytes

i;main(j){for(;j<27;j*j^++i||puts("",j++))putchar(i%26+65);}
\$\endgroup\$
  • \$\begingroup\$ puts only takes one argument. (Some undefined behaviour is permitted in codegolf normally but this is a bit too far outside the usual lanes) \$\endgroup\$ – M.M Aug 23 '16 at 1:13
  • \$\begingroup\$ @M.M Undefined behaviour is exploited all the time. The rule is that a submission is valid as long as it works in some compiler, otherwise we would have to explicitly rule out a long list of exceptions. This code works in gcc so it's a valid submission. \$\endgroup\$ – xsot Aug 23 '16 at 1:38
1
\$\begingroup\$

C++, 111 bytes

void a(){int c=65,i,j;for(i=0;i<26;i++){for(j=0;j<=2*i;j++){std::cout<<(char)c;c++;if(c==91)c=65;}std::cout<<'\n';}}

First try at one of these. Uses an int "c" to record which letter it needs to print at any given time. Once "c" passes 90 ('Z') it gets reset to 65 ('A'). Prints the pyramid using for loops.

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  • \$\begingroup\$ Nice answer! You could do if(c<92)c=65 to take one byte off, and you might also be able to do int a() instead of void a(), but I'm not positive if that works without the return. Other than that, I think you need to include #include <iostream> in your byte count. \$\endgroup\$ – DJMcMayhem Aug 23 '16 at 6:05
  • \$\begingroup\$ I believe you meant if(c>90)c=65, but thanks for the suggestion, it's a good idea. Also, I guess I'll include it, thanks. \$\endgroup\$ – limelier Aug 26 '16 at 8:02
1
\$\begingroup\$

PHP, 76 69 bytes

for(;$i<26&&$a.=join(range(A,Z));)echo substr($a,$i**2,1+2*$i++)."
";
  • Create 26 alphabet (more than enough) and contactenate them in $a
  • loop for i < 26
  • display $a substring start i^2, end 2*i+1
\$\endgroup\$

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