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 code-golf. Program with lowest byte-count wins.
61 Answers 61
Vim, 29 bytes
:h<_↵↵↵y$ZZ26P0qqa↵♥βjllq25@q
Where ↵ represents the Return key, ♥ the Escape key, and β the Backspace key.
-
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– DJMcMayhem2016年08月21日 12:41:09 +00:00Commented Aug 21, 2016 at 12:41
-
9\$\begingroup\$ I still think you should use
␛instead of♥. And␈instead ofβ. That's what these Unicode chars were made for. utf8icons.com/subsets/control-pictures \$\endgroup\$mbomb007– mbomb0072016年08月23日 20:36:23 +00:00Commented Aug 23, 2016 at 20:36
Python 2, 65 bytes
i=1
a=bytearray(range(65,91))*26
while a:print a[:i];a=a[i:];i+=2
-
1\$\begingroup\$ I've changed the header to Python 2 because the code would not work in Python 3. \$\endgroup\$Leaky Nun– Leaky Nun2016年08月21日 11:49:47 +00:00Commented Aug 21, 2016 at 11:49
Pyramid Scheme, 1492 bytes
^ ^
/ \ / \
/set\ /do \
^-----^ ^-----^
/a\ ^- /!\ /[\
--- ^- ^--- ^---^
^- /=\ ^- / \
^- ^---^ -^ /out\
^- ^- /a\ ^- -----^
^- / \ ---/[\ / \
-^ /51 \ ^---^ /chr\
^- ----- ^- / \ -----^
-^ ^- /do \ / \
^- ^- ^-----^ /10 \
-^ /[\ -^ ^- -----
^- ^---^ -^ -^
-^ / \ -^ /c\ ^-
^- /set\ /[\--- -^
^- ^-----^---^ ^-
-^ /a\ /+\ -^ -^
^- --- ^---^ -^ -^
-^ /a\ /2\ / \ -^
-^ --- ---/set\ -^
-^ ^-----^ ^-
-^ /c\ /a\-^
-^ --- --- -^
-^ /[\
-^ ^---^
-^ ^- -^
/]\ / \ -^
^---^ /set\ -^
^- -^ ^-----^ -^
/ \ ^- /c\ /-\ -^
/set\ ^- --- ^---^ -^
^-----^- /c\ /1\ -^
/b\ / \ --- --- /]\
--- / -1\ ^---^
----- ^- / \
/ \ /out\
/set\-----^
^-----^ -^
/b\ /?\ -^
--- ^---^ / \
/!\ -^ /chr\
^--- /+\-----^
/=\ ^---^ /+\
^---^ /b\ /1\ ^---^
/b\ / \--- ---/ \ /b\
---/25 \ /65 \---
----- -----
Why? Why not?
Jelly, 10 bytes
26RḤ’RØAṁY
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.
-
\$\begingroup\$ Was "Double" called "Unhalve" back then? Also,
ṁis great!! [waiting to say congrats for 100k rep] \$\endgroup\$Erik the Outgolfer– Erik the Outgolfer2016年09月28日 11:14:52 +00:00Commented Sep 28, 2016 at 11:14 -
\$\begingroup\$ It's just a mnemonic.
His halve andḤis its inverse (unhalve). \$\endgroup\$Dennis– Dennis2016年09月28日 12:15:18 +00:00Commented Sep 28, 2016 at 12:15 -
\$\begingroup\$ I just think of
/2or*2, so it's "Halve" or "Double". That's why I was confused. \$\endgroup\$Erik the Outgolfer– Erik the Outgolfer2016年09月28日 12:17:00 +00:00Commented Sep 28, 2016 at 12:17 -
\$\begingroup\$ Also 10 bytes:
27Ḷ²IRØAṁY\$\endgroup\$Leaky Nun– Leaky Nun2017年04月30日 04:32:37 +00:00Commented Apr 30, 2017 at 4:32 -
\$\begingroup\$ Also 10 bytes:
51Rm2RØAṁY\$\endgroup\$2017年04月30日 05:01:20 +00:00Commented Apr 30, 2017 at 5:01
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.
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}}
-
\$\begingroup\$ Mathematica has a built-in for alphabet triangles? \$\endgroup\$Buffer Over Read– Buffer Over Read2016年08月21日 18:35:00 +00:00Commented Aug 21, 2016 at 18:35
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.
-
\$\begingroup\$ 62:
i;main(j){while(i<676)printf("\n%c"+(j*j^i++||!j++),i%26+65);}\$\endgroup\$xsot– xsot2016年08月21日 12:37:00 +00:00Commented Aug 21, 2016 at 12:37 -
\$\begingroup\$ @xsot Thanks but I'm afraid that leading newlines are not allowed. \$\endgroup\$Leaky Nun– Leaky Nun2016年08月21日 12:47:41 +00:00Commented Aug 21, 2016 at 12:47
-
1\$\begingroup\$ But there's no leading newline? \$\endgroup\$xsot– xsot2016年08月21日 12:50:23 +00:00Commented Aug 21, 2016 at 12:50
-
\$\begingroup\$ 60:
i;main(j){for(;j<27;j*j^++i||puts("",j++))putchar(i%26+65);}\$\endgroup\$xsot– xsot2016年08月21日 13:00:41 +00:00Commented Aug 21, 2016 at 13:00 -
\$\begingroup\$ @immibis I guess I should post it then. \$\endgroup\$xsot– xsot2016年08月22日 23:57:47 +00:00Commented Aug 22, 2016 at 23:57
Brachylog, 37 bytes
26~l<:1aLc~j[@A:I],L~@nw
=:2%1,.#@l?,
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 STDOUTPredicate 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– Leaky Nun2016年08月21日 12:30:13 +00:00Commented Aug 21, 2016 at 12:30
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\$\begingroup\$ @LeakyNun Would classify that as string manipulation more than ASCII art imo \$\endgroup\$Fatalize– Fatalize2016年08月21日 12:33:11 +00:00Commented Aug 21, 2016 at 12:33
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\$\begingroup\$ Should I add that to the tag? \$\endgroup\$Leaky Nun– Leaky Nun2016年08月21日 12:43:44 +00:00Commented Aug 21, 2016 at 12:43
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\$\begingroup\$ Nice use of the fact that the last line is the only line which ends in
Zwhich is because26is square-free. \$\endgroup\$Leaky Nun– Leaky Nun2016年08月21日 12:44:43 +00:00Commented Aug 21, 2016 at 12:44
Perl, (削除) 42 (削除ここまで) (削除) 41 (削除ここまで) 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
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:
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.
-
1\$\begingroup\$ The column names look like a built-in alphabet to me. Use the first 26 column names. \$\endgroup\$mbomb007– mbomb0072016年08月23日 20:32:05 +00:00Commented Aug 23, 2016 at 20:32
-
1\$\begingroup\$ @mbomb007 So what? I think it would be harder to use them instead of strings. \$\endgroup\$Erik the Outgolfer– Erik the Outgolfer2016年09月28日 11:18:25 +00:00Commented Sep 28, 2016 at 11:18
-
\$\begingroup\$ @EriktheGolfer So? My point was that there is a builtin alphabet. \$\endgroup\$mbomb007– mbomb0072016年09月28日 13:26:43 +00:00Commented Sep 28, 2016 at 13:26
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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– Erik the Outgolfer2016年09月28日 13:29:27 +00:00Commented Sep 28, 2016 at 13:29
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\$\begingroup\$ @EriktheGolfer It's a suggestion. Idk how many bytes it'd be. \$\endgroup\$mbomb007– mbomb0072016年09月28日 13:30:56 +00:00Commented Sep 28, 2016 at 13:30
Husk, 10 bytes
Cİ1ṠṁK..."AZ
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
JavaScript (ES6), 77 (削除) 82 88 (削除ここまで)
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)
-
\$\begingroup\$ \o/ an ES6 answer I can run in my browser \$\endgroup\$Downgoat– Downgoat2016年08月21日 19:53:32 +00:00Commented Aug 21, 2016 at 19:53
-
\$\begingroup\$ I stole your .toString(36) logic...now you have to beat 80 bytes!! \$\endgroup\$applejacks01– applejacks012016年08月22日 12:19:52 +00:00Commented Aug 22, 2016 at 12:19
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\$\begingroup\$ Ahh I concede, I can't think of any way to beat this with my library. Thanks for the challenge! \$\endgroup\$applejacks01– applejacks012016年08月22日 15:57:28 +00:00Commented Aug 22, 2016 at 15:57
-
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++)
-
1\$\begingroup\$ Using the spread operator could save you 3 bytes:
[...'ABCDEFGHIJKLMNOPQRSTUVWXYZ']instead of'ABCDEFGHIJKLMNOPQRSTUVWXYZ'.split``. \$\endgroup\$insertusernamehere– insertusernamehere2016年08月21日 18:46:09 +00:00Commented Aug 21, 2016 at 18:46
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:]}}
MATLAB, (削除) 112 109 95 79 (削除ここまで) 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 (削除) 14 (削除ここまで) 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
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
-
\$\begingroup\$ Bravo. I thought I knew what XQuery was. Turns out I had no idea. \$\endgroup\$Jordan– Jordan2016年08月22日 04:25:52 +00:00Commented Aug 22, 2016 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– Michael Kay2016年08月22日 21:24:44 +00:00Commented Aug 22, 2016 at 21:24
Ruby, 46 bytes
26.times{|i|puts ([*?A..?Z]*26)[i*i,i*2+1]*""}
See it on ideone: http://ideone.com/3hGLB0
05AB1E (alternate) 15 bytes
×ばつ52μ1⁄4D3⁄4,3ドル⁄41⁄4FÀ
Explanation:
×ばつ # push a string containing a-za-z
52μ # Loop the rest of the program until counter = 52
1⁄4 # increment counter (it's 0 initially)
D # Duplicate the alpha string on the stack
3⁄4£ # Replace alpha with alpha[0..counter]
, # Pop the substring and print it
3⁄41⁄4FÀ # rotate the alpha string left counter++ times.
R, (削除) 120 (削除ここまで) (削除) 115 (削除ここまで) 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 !
-
1\$\begingroup\$ arf, sorry I didn't see that earlier, but since you're only using
aonce 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– plannapus2016年08月23日 13:52:19 +00:00Commented Aug 23, 2016 at 13:52 -
\$\begingroup\$ @plannapus Silly me ! Thanks again ! I also integrated the
b=seqpart in the main body, so it's even less readable ! \$\endgroup\$Rudier– Rudier2016年08月23日 13:58:14 +00:00Commented Aug 23, 2016 at 13:58
R, (削除) 81 73 65 (削除ここまで) 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="")
MY-BASIC, 72 bytes
Anonymous function that takes no input and outputs to the console.
For J=1 To 51
For K=1 To J
I=I Mod 26+1
Print Chr(64+I)
Next
Print;
Next
brainfuck, 242 bytes
++++++++[->++++++++<]>+>++++++++++>++++++++++++++++++++++++++>++++++++++++++++++++++++++>+>>+>+<<<<<[->>[-<<<<.+>>>-[>>>>[-]]>>>>[-<<<<++++++++++++++++++++++++++<<<-------------------------->>>>>>>>>>>]<<<<+<<<]<<<.>>>>>++[-<+<+>>]<[->+<]<<<]
Ungolfed:
Initial Cell Values:
0 1 2 3 4 5 6 7 8
0 65 10 26 26 1 0 1 1
++++++++
[
->++++++++<
]
>+
>++++++++++
>++++++++++++++++++++++++++
>++++++++++++++++++++++++++
>+
>>+
>+
<<<<<
[
->>
[
-<<<<.+>>>-
[
>>>>[-]
]
>>>>
[
-<<<<++++++++++++++++++++++++++
<<<--------------------------
>>>>>>>
>>>>
]
<<<<
+<<<
]
<<<.>>>
>>++
[
-<+<+>>
]
<
[
->+<
]
<<<
]
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%%
05AB1E, (削除) 18 (削除ここまで) 17 bytes
×ばつN>n£NnF¦},
Explanation
26F # for N in range(0, 26) do
×ばつ # the alphabet repeated 26 times
N>n£ # take (N+1)^2 letters
NnF¦} # throw away the first N^2 letters
, # print with newline
Rexx, (削除) 74 (削除ここまで) 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
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
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.
Dyalog APL, 18 bytes
-
\$\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– Michael Kay2016年08月21日 23:06:45 +00:00Commented Aug 21, 2016 at 23:06
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1\$\begingroup\$ @MichaelKay When can APL characters be counted as 1 byte each? \$\endgroup\$Dennis– Dennis2016年08月22日 00:09:50 +00:00Commented Aug 22, 2016 at 0:09
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\$\begingroup\$ Will you explain this, or shall I? \$\endgroup\$Adám– Adám2016年08月23日 11:26:06 +00:00Commented Aug 23, 2016 at 11:26
Explore related questions
See similar questions with these tags.
modfunction. May make one myself if I have time. \$\endgroup\$