Alphabet triangle

Question

You are to print this exact text:

A
ABA
ABCBA
ABCDCBA
ABCDEDCBA
ABCDEFEDCBA
ABCDEFGFEDCBA
ABCDEFGHGFEDCBA
ABCDEFGHIHGFEDCBA
ABCDEFGHIJIHGFEDCBA
ABCDEFGHIJKJIHGFEDCBA
ABCDEFGHIJKLKJIHGFEDCBA
ABCDEFGHIJKLMLKJIHGFEDCBA
ABCDEFGHIJKLMNMLKJIHGFEDCBA
ABCDEFGHIJKLMNONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWXWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWXYXWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWXYZYXWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWXYXWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWXWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNONMLKJIHGFEDCBA
ABCDEFGHIJKLMNMLKJIHGFEDCBA
ABCDEFGHIJKLMLKJIHGFEDCBA
ABCDEFGHIJKLKJIHGFEDCBA
ABCDEFGHIJKJIHGFEDCBA
ABCDEFGHIJIHGFEDCBA
ABCDEFGHIHGFEDCBA
ABCDEFGHGFEDCBA
ABCDEFGFEDCBA
ABCDEFEDCBA
ABCDEDCBA
ABCDCBA
ABCBA
ABA
A

Specs

Extra trailing newlines are allowed at the end of the output.
Extra trailing spaces (U+0020) are allowed at the end of each line, including the extra trailing newlines.
You can use all lowercase instead of all uppercase, but you cannot print partially lowercase partially uppercase.
You can return the text as a function output instead of printing it in a full program.

Scoring

Since this is a triangle, and a triangle has 3 sides, and 3 is a small number, your code should be small in terms of byte-count.

"Since a triangle has 3 sides and 3 is a small number, so your code should be small in terms of byte count." seems legitimate — Rohan Jhunjhunwala, Commented Aug 4, 2016 at 22:00
Squares of numbers composed only of 1 seems related: 1*1 = 1 ~= A , 11*11 = 121 ~= ABA , 111*111 = 12321 ~= ABCBA ... — Caridorc, Commented Aug 5, 2016 at 13:28

Mitch Schwartz · Accepted Answer · 2018-07-17 20:12:43Z

3

brainfuck, 79 bytes

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

Formatted:

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

Try it online

edited Jul 17, 2018 at 20:12

answered Jul 17, 2018 at 19:26

Mitch Schwartz

5,8891 gold badge27 silver badges26 bronze badges

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Dennis · Accepted Answer · 2016-08-03 08:12:05Z

Sesos, 27 bytes

0000000: a85ab9 ac5daa f8b1c7 785f9b b961f7 665c1b 73ccfc  .Z..]....x_..a.f\.s..
0000015: c01ecb 987303                                     ....s.

Try it online! Check Debug to see the generated SBIN code.

Sesos assembly

The binary file above has been generated by assembling the following SASM code.

add 65, rwd 1, add 10, rwd 2, add 25
jmp
    fwd 3
    jmp
        put, fwd 1
    jnz
    rwd 1
    jmp
        fwd 1, add 1, fwd 1, add 1, rwd 2, sub 1
    jnz
    fwd 2
    jmp
        rwd 2, add 1, fwd 2, sub 1
    jnz
    rwd 1, add 1, rwd 2
    jmp
        put, rwd 1
    jnz
    rwd 1, sub 1
jnz
fwd 3
jmp
    jmp
        put, fwd 1
    jnz
    rwd 1, get, rwd 1
    jmp
        put, rwd 1
    jnz
    fwd 2
; jnz (implicit)

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

2

///, 208 bytes

/'/\/\///+/BA
AB'-/DC+CD'&/FE-EF')/HG&GH'=/JI)IJ'|/LK=KL'7/NM|MN'1/PO7OP'q/RQ1QR'p/TSqST'6/VUpUV'[/XW6WX/A
A+C+C-E-E&G&G)I)I=K=K|M|M7O7O1Q1QqSqSpUpU6W6W[Y[YZY[Y[W6W6UpUpSqSqQ1Q1O7O7M|M|K=K=I)I)G&G&E-E-C+C+A
A

Not winning of course, but here it is, predating Martin Ender...

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 3, 2016 at 9:17

Erik the Outgolfer

40.6k5 gold badges45 silver badges122 bronze badges

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izabera · Accepted Answer · 2016-08-03 09:42:22Z

2

bash, 80 bytes

printf -v a %s {A..Z};for i in {1..26} {25..1};{ echo ${a::i}`rev<<<${a::i-1}`;}

edited Aug 3, 2016 at 9:42

answered Aug 3, 2016 at 9:14

izabera

9174 silver badges15 bronze badges

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

2

Javascript (ES5), 105 bytes

Beating all other ES6 answers using some good ol' vanilla JS looping!

for(a="",n=g=1;n++<52;n<27?g++:g--){for(z=x=0;z++<g*2-1;z<g?x++:x--)a+=String.fromCharCode(65+x);a+="\n"}a

I could save 1 byte by using ES6 and changing

a+="\n"

to

a+=`
`

but nah.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 5, 2016 at 13:18

Bassdrop Cumberwubwubwub

5,9492 gold badges24 silver badges50 bronze badges

\$\begingroup\$ Since you don't have console.log anywhere I imagine this is supposed to be the contents of a function. If you wrapped it in the ()=>{} that you need to actually execute your code and tried it, it would return undefined. It needs to be return a. If you had done everything so that it can actually be run (including the ()=>{} and return ) it would be 119 bytes. tl;dr this code doesn't work. \$\endgroup\$
– Pandacoder
Commented Aug 5, 2016 at 18:37
\$\begingroup\$ Nice. Also (10+x).toString(36) is shorter than String.fromCharCode(65+x) and like @Pandacoder said, you should put something like alert(a) at the end to make it valid. \$\endgroup\$
– user81655
Commented Aug 14, 2016 at 10:11

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TheLethalCoder · Accepted Answer · 2016-08-05 16:03:44Z

2

C#, 289 215 206 190 179 171 159 157 156 153 147 bytes

(n,i,s)=>{s=new string[51];for(i=0;i<26;){for(n=65;n-65<i;)s[i]+=(char)n++;for(;n>64;)s[i]+=(char)n--;s[50-i]=s[i++];}return string.Join("\n",s);};

Saved 3 bytes thanks to VisualMelon

Saved 6 bytes thanks to mdfst13

Formatted version:

(n,i,s)=>
{
    s=new string[51];

    for(i=0;i<26;)
    {
        for(n=65;n-65<i;)
            s[i]+=(char)n++;

        for(;n>64;)
            s[i] += (char)n--;

        s[50-i]=s[i++];
    }

    return string.Join("\n",s);
};

156 byte version (when the irrelevant whitespace is removed) of counting backwards, uses a similar concept to the above method but does so in reverse.

(n, i, s) =>
{
    s = new string[i = 51];
    for (; --i > 24;)
    {
        for (n = 65; i + n < 116;)
            s[i] += (char)n++;

        for (n = 50 - i + 64; n > 64;)
            s[i] += (char)n--;

        s[50 - i] = s[i];
    }

    return string.Join("\n", s);
};

edited Aug 5, 2016 at 16:03

answered Aug 3, 2016 at 13:00

TheLethalCoder

7,23927 silver badges43 bronze badges

\$\begingroup\$ Feel free to share any improvements to my code if you find any \$\endgroup\$
– TheLethalCoder
Commented Aug 4, 2016 at 14:03
\$\begingroup\$ An explanation would be nice! You can use a for loop instead of a while, saving you the ; on the statement prior. The last statement in the outer loop can be put inside the last inner for loop (s[...]=s[i]+=) (need to juggle the i++ a bit, can certainly be done with only one extra byte). You've also got 51-i-1 in there, which can be reduced ;) I would be looking into reducing the two inner loops into one, and counting i backwards (so you assign it to 51 inside the array constructor). \$\endgroup\$
– VisualMelon
Commented Aug 4, 2016 at 17:51
\$\begingroup\$ @VisualMelon I've been looking at it for ages and I never saw the 51 - I - 1 haha, and I can't put the last statement in the last inner loop as it doesn't get called on the first pass but didn't know I could chain the assignment like that so that's cool! And I've been looking into combining the loops, getting there \$\endgroup\$
– TheLethalCoder
Commented Aug 5, 2016 at 8:06
\$\begingroup\$ @VisualMelon I reversed the outer loop and ended up adding 3 bytes when I got it working, will see if I can golf it down though \$\endgroup\$
– TheLethalCoder
Commented Aug 5, 2016 at 9:39

Add a comment |

ETHproductions · Accepted Answer · 2016-10-25 22:21:06Z

2

JavaScript (ES6), 89 bytes

R=(n,s=10,c=s.toString(36))=>n?c+R(n-1,s+1)+c:c
C=(n=0,r=R(n))=>n<25?r+`
`+C(n+1)+`
`+r:r

Not the shortest, but I wanted to try my hand at recursion. Call with C().

answered Oct 25, 2016 at 22:21

ETHproductions

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u-ndefined · Accepted Answer · 2019-06-14 09:05:29Z

2

Pepe, 94 bytes

rEeEeEEeEeREeEeeeeeEREEEEEEerEEReeeReeEREEEeEREEEeEREEeeREEEEEreererEEReeeReeEReReREEeEreeReee

Try it online!

answered Jun 14, 2019 at 9:05

u-ndefined

2,32311 silver badges21 bronze badges

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Helena · Accepted Answer · 2019-06-14 17:23:04Z

brainfuck, 96 bytes

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

Try it online!

; Initialize memory as: 0 (25) 0 "\n" "A" 0
+++++[->+++++>>++>+++++++++++++<<<<]>

[-                          ; Repeat 25 times
    >>>[.>]                 ; print all the characters on the band
    <[->+>+<<]>+>[-<<+>>]   ; copy the last character and increment
    <<<[.<]                 ; print all but the last two characters backwards
<]
>>>
[                           ; while there is still characters left
    [.>]                    ; print all the characters
    <[-]                    ; remove last character
    <[.<]                   ; print all characters backwards
>>]

Dom Hastings · Accepted Answer · 2019-06-17 19:36:25Z

2

Perl 5 with `-M5.010`, 54 bytes

for$@(A..Z){@;=($_.$/,@;);s/$*/$*$@$*/;say;$*=$@}say@;

Try it online!

Saved bytes thanks to @Xcali!

edited Jun 17, 2019 at 19:36

answered Aug 3, 2016 at 11:31

Dom Hastings

24.4k4 gold badges56 silver badges93 bronze badges

\$\begingroup\$ Found a method which takes 56 bytes \$\endgroup\$
– Xcali
Commented Jun 14, 2019 at 22:58
\$\begingroup\$ Thanks @Xcali! With you suggestion managed to get it down to 54. And some added obfuscation for entertainment... \$\endgroup\$
– Dom Hastings
Commented Jun 17, 2019 at 19:37

Add a comment |

Aaroneous Miller · Accepted Answer · 2021-07-20 14:51:08Z

2

Vyxal `j`, 8 bytes

kA¦ømƛøm

Try it Online!

Explanation:

kA        # Uppercase ALPHABET
  ¦       # Prefixes (A, AB, ABC, ...)
   øm     # Mirror without duplicating middle
     ƛ    # For each:
      øm  #   Mirror without duplicating middle

answered Jul 20, 2021 at 14:51

Aaroneous Miller

9,6901 gold badge24 silver badges39 bronze badges

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MarcMush · Accepted Answer · 2021-07-21 11:30:56Z

2

Julia 0.4, 84 bytes

!()=(- =join;a=-char(65:90);b=[a[1:x+1]*a[x:~0:1]for x=0:25];vcat(b,b[25:~0:1])-"
")

Try it online!

edited Jul 21, 2021 at 11:30

MarcMush

6,63513 silver badges17 bronze badges

answered Aug 4, 2016 at 0:39

Mama Fun Roll

9,9281 gold badge36 silver badges46 bronze badges

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

1

JavaScript (ES6), 141

for(a=65,q=[],z=[];a<91;z.push(a++))q.push(String.fromCharCode(...z,a,...[...z].reverse()))
a=q.pop(),alert([...q,a,...q.reverse()].join`\n`)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 3, 2016 at 10:25

edc65

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

1

Retina, 90 88 86 76 bytes

I'm sure it can be made shorter, maybe by using non-verbose replacements and simply outputting at the end. (I tried. I don't think I can do it shorter that way.)

:`
A
A
ABA
{:`

(.)([^Z]\1)
$1$2$2
}T1`__L`L`(?<=(.)[^Z])\1
+:`(.).\1|^A$
$1

Try it online

Thanks to LeakyNun for 14 bytes off

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 3, 2016 at 15:39

mbomb007

23.5k7 gold badges63 silver badges135 bronze badges

\$\begingroup\$ @LeakyNun Thanks, idk why I thought I needed that part... \$\endgroup\$
– mbomb007
Commented Aug 3, 2016 at 16:21
\$\begingroup\$ Match the second stage with ^ and insert only AB. \$\endgroup\$
– Martin Ender
Commented Aug 4, 2016 at 14:20

Add a comment |

Alexander · Accepted Answer · 2016-08-03 18:18:19Z

Swift 2.2, 157 142 Bytes

let t={(l:Int)in[Int](1..<l)+[Int](l.stride(to:0,by:-1))}
for x in(t(n).map{t($0).map{"\(UnicodeScalar($0+64))"}.reduce(""){$0+$1}}){print(x)}

ungolfed:

let makeTriangleArray = {(limit: Int) -> [Int] in
    return [Int](1 ..< limit) + [Int](limit.stride(to: 0, by: -1))
}

let n = 26
let lines = makeTriangleArray(n).map{
    makeTriangleArray($0).map{String(UnicodeScalar($0 + 64))}
                         .reduce(""){ $0 + $1}
}

for line in lines {
    print(line)
}

157 bytes:

let t={(l:Int)in[Int](1..<l)+[Int](l.stride(to:0,by:-1))}
print(t(n).map{t($0).map{"\(UnicodeScalar($0+64))"}.joinWithSeparator("")}.joinWithSeparator("\n"))

ungolfed:

let makeTriangleArray = {(limit: Int) -> [Int] in // "called t" above
    return [Int](1 ..< limit) + [Int](limit.stride(to: 0, by: -1))
}

let n = 26
let s = makeTriangleArray(n).map{
    makeTriangleArray($0).map{String(UnicodeScalar($0 + 64))}
                         .joinWithSeparator("")
}.joinWithSeparator("\n")
print(s)

Community · Accepted Answer · 2017-04-13 12:39:03Z

1

Jelly, 16 bytes

ŒRAạ
25ÇÇ€‘ịØAj⁷

This uses the same method in my Python answer.

Try it online.

Explanation

ŒRAạ  Define a helper link. Input: n
ŒR    Create the range [-abs(n), , ..., 0, ..., abs(n)]
  A   Take the absolute value of each
   ạ  Return the absolute difference between each value and n

25ÇÇ€‘ịØAj⁷  Main link. No arguments
25           The constant 25
  Ç          Call the helper link on 25 to get [0, 1, ..., 25, ..., 1, 0]
   Ç€        Call the helper link on each value in the previous
     ‘       Increment every value
       ØA    The alphabet (the string 'A..Z')
      ị      Select the char at each index in the alphabet
         j⁷  Join them using newlines and return

edited Apr 13, 2017 at 12:39

CommunityBot

1

answered Aug 3, 2016 at 18:53

miles

17k2 gold badges30 silver badges94 bronze badges

\$\begingroup\$ This is extremely close to this answer \$\endgroup\$
– Leaky Nun
Commented Aug 3, 2016 at 18:59
\$\begingroup\$ @LeakyNun It generates the indices into the alphabet using a different method. Since there is only one output the intermediate values only finally become identical after the increment. \$\endgroup\$
– miles
Commented Aug 3, 2016 at 19:19

Add a comment |

Neil · Accepted Answer · 2016-08-04 10:04:30Z

1

Batch, 226 224 bytes

@set z=ABCDEFGHIJKLMNOPQRSTUVWXYZ
@set a=YXWVUTSRQPONMLKJIHGFEDCBA
@for /l %%i in (0,1,25)do @call echo %%z:~0,%%i%%%%z:~%%i,1%%%%a:~-%%i,%%i%%
@for /l %%i in (24,-1,0)do @call echo %%z:~0,%%i%%%%z:~%%i,1%%%%a:~-%%i,%%i%%

Edit: Saved 2 bytes thanks to @EʀɪᴋᴛʜᴇGᴏʟғᴇʀ.

edited Aug 4, 2016 at 10:04

answered Aug 3, 2016 at 8:00

Neil

177k12 gold badges74 silver badges281 bronze badges

\$\begingroup\$ I think you can do )do, but not any other space (except in(), maybe... \$\endgroup\$
– Erik the Outgolfer
Commented Aug 3, 2016 at 9:20
\$\begingroup\$ @EʀɪᴋᴛʜᴇGᴏʟғᴇʀ Oddly enough I have done this before, just the once: codegolf.stackexchange.com/a/68936/17602 \$\endgroup\$
– Neil
Commented Aug 4, 2016 at 10:04

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

1

QBasic, 151 122 bytes

Really just for fun (via QB64) - it doesn't golf well

Taking some ABS() inspiration from Joffan's VBA answer but applying it differently:

FOR i=-25 TO 25
FOR n=0 TO 25-abs(i)
s$=s$+chr$(n+65)
IF n THEN t$=t$+chr$(90-abs(i)-abs(n))
NEXT
? s$+t$
s$=""
t$=""
NEXT

Previously:

FOR i=0 TO 25
p(i)
NEXT
FOR i=0 to 25
p(25-i)
NEXT
SUB p(i)
FOR n=0 TO i
s$=s$+chr$(n+65)
IF n THEN t$=t$+chr$(65+i-n)
NEXT
? s$+t$
s$=""
t$=""
END SUB

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 3, 2016 at 14:41

Chris H

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downrep_nation · Accepted Answer · 2016-08-04 12:07:09Z

1

C#, 162 bytes

void p(){for(int i=1;i<54;i++){var a=Enumerable.Range(65,i>27?54-i:i).Select(n=>(char)n);Console.WriteLine(new string(a.Concat(a.Reverse().Skip(1)).ToArray()));}}

Try it online!

answered Aug 4, 2016 at 12:07

downrep_nation

1,1926 silver badges12 bronze badges

1

\$\begingroup\$ I was wondering what the linq version would look like, you need System.Console.. though right? and shouldn't you also include using System.Linq in the byte count although I'm unsure about that \$\endgroup\$
– TheLethalCoder
Commented Aug 4, 2016 at 12:15
\$\begingroup\$ Change your for loop to for(int i=0;++i<54;) to save some bytes \$\endgroup\$
– TheLethalCoder
Commented Aug 4, 2016 at 12:25

Add a comment |

MonkeyZeus · Accepted Answer · 2016-08-04 15:41:41Z

HTML, 1412 bytes

<pre>A
ABA
ABCBA
ABCDCBA
ABCDEDCBA
ABCDEFEDCBA
ABCDEFGFEDCBA
ABCDEFGHGFEDCBA
ABCDEFGHIHGFEDCBA
ABCDEFGHIJIHGFEDCBA
ABCDEFGHIJKJIHGFEDCBA
ABCDEFGHIJKLKJIHGFEDCBA
ABCDEFGHIJKLMLKJIHGFEDCBA
ABCDEFGHIJKLMNMLKJIHGFEDCBA
ABCDEFGHIJKLMNONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWXWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWXYXWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWXYZYXWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWXYXWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWXWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVWVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUVUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTUTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSTSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRSRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQRQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPQPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNOPONMLKJIHGFEDCBA
ABCDEFGHIJKLMNONMLKJIHGFEDCBA
ABCDEFGHIJKLMNMLKJIHGFEDCBA
ABCDEFGHIJKLMLKJIHGFEDCBA
ABCDEFGHIJKLKJIHGFEDCBA
ABCDEFGHIJKJIHGFEDCBA
ABCDEFGHIJIHGFEDCBA
ABCDEFGHIHGFEDCBA
ABCDEFGHGFEDCBA
ABCDEFGFEDCBA
ABCDEFEDCBA
ABCDEDCBA
ABCDCBA
ABCBA
ABA
A</pre>

Actually, with native HTML it's impossible to do better than this — Richard Hamilton, Commented Aug 5, 2016 at 17:36
@LeakyNun You must be on a whole 'nother level. What is – – – – – – and how do I make it work? — MonkeyZeus, Commented Aug 5, 2016 at 17:44
A possible interpretation of @LeakyNun's comment: codegolf.stackexchange.com/a/88984 — manatwork, Commented Aug 6, 2016 at 10:56

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

1

R, 95 bytes

L=LETTERS;cat("A\n");for(i in 2:50)cat(L[1:(26-abs(i-26))],L[(25-abs(i-26)):1],'\n');cat("A\n")

Ungolfed:

L=LETTERS; #Took idea from @Frederic, I apologize if I shouldn't (first time posting)
cat("A\n")
for(i in 2:50)
{
    cat(L[1:(26-abs(i-26))],L[(25-abs(i-26)):1],'\n')
}
cat("A\n")

Basically, the formula in the cat function in the for loop is a transformation on the absolute value function, mapping n -> n for Letters 2 through 26 and then mapping Letters 27 -> 25, 28 -> 24, ..., 50 -> 2. I don't think it would work if I included the 'A' in this formula, which is why I printed it out separately.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 4, 2016 at 17:41

Sol

111 bronze badge

\$\begingroup\$ 100 bytes and actually valid. \$\endgroup\$
– Giuseppe
Commented Jan 15, 2018 at 17:57

Add a comment |

Crypto · Accepted Answer · 2016-08-05 06:01:01Z

1

PHP, 80 83 bytes

Making half of the triangle then print it in reverse did the trick.

for($a="A";$a<"Z";$b.=$a++)$c.=$b.$a.strrev($b)."
";echo"$c{$b}Z".strrev($c.$b);

edited Aug 5, 2016 at 6:01

answered Aug 4, 2016 at 6:17

Crypto

8824 silver badges8 bronze badges

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

1

MATLAB, 53 Bytes

MATLAB is not made for strings, so this needs some looping. I haven't found a faster way to create a list 1 2 3 4 3 2 1, than [1:4,3:-1:1], which seems really verbose.

Anyway, it's only beaten by 11/55 other submissions, and most of those are golfing languages.

for i=[1:26,25:-1:1]
disp(['',64+[1:i,i-1:-1:1]])
end

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 5, 2016 at 11:39

Stewie Griffin

46.5k13 gold badges132 silver badges295 bronze badges

1

\$\begingroup\$ Just spent an entertaining fifteen minutes proving you right that this is indeed the shortest - even inline definition of an anonymous function (Octave-only) is still longer. \$\endgroup\$
– Sanchises
Commented Jul 25, 2017 at 11:42

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

1

R, 71 69 bytes

L=LETTERS;for(i in c(1:26,25:1))cat(c(L[1:i],L[i:1][-1]),"\n",sep="")

See here on an online interpreter.

Edit: moved LETTERS outside the loop in order to avoid the curly brackets.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 5, 2016 at 11:44

plannapus

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\$\begingroup\$ 59 bytes \$\endgroup\$
– Giuseppe
Commented Jan 15, 2018 at 17:59
\$\begingroup\$ oh I didn't notice that my suggestion was identical to this one. \$\endgroup\$
– Giuseppe
Commented Jan 15, 2018 at 17:59

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Pandacoder · Accepted Answer · 2016-08-05 18:50:11Z

1

Javascript (ES6), 113 bytes

i=>{for(i=0;i<51;i++,console.log(L))for(j=s=i<26?i:50-i,L="";j>=0;)k=String.fromCharCode(65+j),L=k+(j---s?L+k:L)}

Thanks to manatwork for pointing me to a tip written by William Barbosa to reduce by a byte.

edited Aug 5, 2016 at 18:50

answered Aug 5, 2016 at 18:39

Pandacoder

1906 bronze badges

\$\begingroup\$ As William Barbosa wrote in his tip, is shorter to declare an unused dummy parameter. \$\endgroup\$
– manatwork
Commented Aug 5, 2016 at 18:45

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user61264user61264 · Accepted Answer · 2016-10-27 07:40:50Z

1

Rust, 120 bytes

fn main(){for j in -25i8..26{let k=25-j.abs();for i in -k..k+1{print!("{}",(65+k-i.abs())as u8 as char);}println!("");}}

My first attempt at codegolfing in Rust... codegolfing in anything, really. A straightforward attempt, just using for loops and abs().

fn main() {
    for j in -25i8..26 {
        let k=25-j.abs();
        for i in -k..k+1{
            print!("{}",(65+k-i.abs())as u8 as char);
        }
        println!("");
    }
}

answered Oct 27, 2016 at 7:40

user61264

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Jörg Hülsermann · Accepted Answer · 2017-05-01 02:33:39Z

1

PHP, 78 Bytes

foreach(range(A,Z)as$r)$o.="\n$t".strrev($t.=$r);echo$o.substr(strrev($o),51);

edited May 1, 2017 at 2:33

answered Oct 26, 2016 at 11:38

Jörg Hülsermann

13.3k3 gold badges15 silver badges36 bronze badges

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Spiffo · Accepted Answer · 2017-09-10 19:36:50Z

1

Python, 156 bytes

Not impressive, but I'm proud of myself because I'm a noob:

def p(k):
    x=k-1
    while x>-k:
        print(chr(64+(k-abs(x))),end="")
        x-=1
q=25
while q>-27:
    p(26-abs(q))
    print()
    q-=1

answered Sep 10, 2017 at 19:36

Spiffo

111 bronze badge

1

\$\begingroup\$ Welcome to PPCG! You can save a bunch of bytes by making your indentation single spaces. You can also use tricks like print();q-=1. \$\endgroup\$
– Stephen
Commented Sep 10, 2017 at 20:26

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

1

SOGL V0.12, 10 bytes

øZ{⁴Κ}¹╬,k

Try it Here!

Explanation:

ø           push an empty string
 Z{  }      for each character in the alphabet do
   ⁴          duplicate the item before the top item
    Κ         reverse add to that
              or simpler - on each iteration creates a duplicate of ToS but with the current character appended.
      ¹     wrap the stack in an array
       ╬,   quad-palindromize without spacing to a square
         k  remove the 1st line

The last k could be removed if the question allowed leading newlines.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 14, 2017 at 18:12

dzaima

20.3k2 gold badges41 silver badges75 bronze badges

\$\begingroup\$ Your interpreter site runs fine half of the time and half of the time, the run button seems to do nothing! \$\endgroup\$
– 0xffcourse
Commented Sep 15, 2017 at 3:43
1

\$\begingroup\$ @officialaimm ..huh. It's probably that Processing.js hasn't yet downloaded/transpiled the source. When I get the time I'll make the run button blank out while its loading \$\endgroup\$
– dzaima
Commented Sep 15, 2017 at 5:09
\$\begingroup\$ ah, makes sense. \$\endgroup\$
– 0xffcourse
Commented Sep 15, 2017 at 5:59

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jrtapsell · Accepted Answer · 2018-01-30 13:57:00Z

1

Kotlin, 99 93 88 bytes

(('A'..'Y')+'Z'.downTo('A')).map{println((('A'..it-1)+it.downTo('A')).joinToString(""))}

That's still longer than I expected but I doubt I'll be able to golf it further.

Factorisation doesn't save the day, best I could do is 93 bytes :

fun r(a:Char,b:Char)=(a..b-1)+b.downTo(a)
r('A','Z').map{println(r('A',it).joinToString(""))}

edited Jan 30, 2018 at 13:57

jrtapsell

1,0037 silver badges15 bronze badges

answered Aug 4, 2016 at 10:03

Aaron

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Alphabet triangle

Specs

Scoring

96 Answers 96

brainfuck, 79 bytes

Sesos, 27 bytes

Sesos assembly

///, 208 bytes

bash, 80 bytes

Javascript (ES5), 105 bytes

C#, 289 215 206 190 179 171 159 157 156 153 147 bytes

JavaScript (ES6), 89 bytes

Pepe, 94 bytes

brainfuck, 96 bytes

Perl 5 with -M5.010, 54 bytes

Vyxal j, 8 bytes

Julia 0.4, 84 bytes

JavaScript (ES6), 141

Retina, 90 88 86 76 bytes

Swift 2.2, 157 142 Bytes

Jelly, 16 bytes

Explanation

Batch, 226 224 bytes

QBasic, 151 122 bytes

C#, 162 bytes

HTML, 1412 bytes

R, 95 bytes

PHP, 80 83 bytes

MATLAB, 53 Bytes

R, 71 69 bytes

Javascript (ES6), 113 bytes

Rust, 120 bytes

PHP, 78 Bytes

Python, 156 bytes

SOGL V0.12, 10 bytes

Kotlin, 99 93 88 bytes

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