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Note that this is not the same as Print the alphabet four times.
This task is to write a program to generate four copies of each letter of the English alphabet, one letter per line, on standard output:
A
A
A
A
B
B
B
B
etc.
The output should include newlines after each letter.
Lowercase letters and/or extra whitespace are acceptable.
The solution must be a complete program.
This is code-golf so answers will be scored in bytes with a lower score being the goal.
⍪4/⎕A
Matrix format (⍪) of 4-replication (4/) of alphabet (⎕A).
for i in range(104):print chr(i/4+65)
As i goes from 0 to 104, it is divided by four and added to the ASCII value for A, and the resulting character is printed.
3/4 is 0, while 3.0/4.0 is 0.75; range() generates integers.
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Dec 13, 2013 at 17:43
for i in range(104):print(chr(i//4+65))
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write(rep(LETTERS,e=4),1)
Former version with 30 bytes:
cat(rep(LETTERS,e=4),sep="\n")
1 instead of "" to specify stdout in write as well, which saves another byte.
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I submit this, an uncompetitively long answer, simply because I don't see a C submission yet. And that makes me sad. :-/
LATER: Props to @moala for doing a "/4" int version of this, saving 13 chars!
float i;main(){while(i<26)printf("%c\n",65+(int)i),i+=.25;}
4#u:65+i.26 1
I'm still pretty shaky with J, so this could probably be improved
#) instead of division like so: 4#u:65+i.26 1. Also, ~ swaps a function's arguments, so if you ever find yourself doing (expression) F value, you can replace that with value F~ expression to save a character.
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puts ([*?A..?Z]*4).sort
All credit to @manatwork -- upvote his comment, not this. :)
puts [*?A..?Z].map{|i|[i]*4} or puts ([*?A..?Z]*4).sort, so the letters get ordered as in the example.
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Dec 13, 2013 at 19:02
puts (?A..?Z).map{|i|[i]*4} is a character shorter. You can call map directly on a Range, so you don't need the splat in this case.
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Dec 13, 2013 at 19:08
Golfed code:
[char[]](65..90*4)|Sort
Walkthrough:
[char[]](...) takes an array of objects and converts them to ASCII characters.
65..90 are the ASCII codes for A-Z.
*4 repeats the series 4 times.
|Sort sorts the output.
Note:
If you want this written to a file, just throw >, followed by a file name, at the end.
x a=a++a
main=putStr$['A'..'Z']>>=x.x.(:"\n")
putStr$['A'..'Z']>>=("golf">>).(:"\n") saves 8 bits
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1+::'g`#@_4/'A+,a,
Works by storing a number and ending when it reaches 104. Prints out the corresponding character of the alphabet for the number divided by 4, followed by a newline. But if I need not add a newline after each letter, then it is 16 chars:
1+::'g`#@_4/'A+,
Can be reduced if I can print more characters (ie all of them four times)(7 6 chars, even works in Befunge 93):
1+:4/,
With newline:
1+:4/,a,
46:
i;main(){while(i<104)printf("%c\n",65+i++/4);}
44:
i=260;main(j){for(;(j=i++>>2)<91;puts(&j));}
44 too:
i=260;main(j){while(j=i++>>2,j<91)puts(&j);}
Thanks to @marinus, 43:
i=260;main(j){while(j=i++/4,j<91)puts(&j);}
Should I add a bounty for getting to 42? :)
Java: 56
for(int i=0;i<104;)System.out.println((char)(i++/4+65));
edit: changed from 'print' to 'println'
The solution must be complete enough to execute. I should be able to invoke an implementation of the language, paste the code from the answer, and get results, without typing any additional code. i think your solution violates this condition
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Jul 20, 2016 at 11:58
+++[[-<+>>++<]+>]<<<<<<+[->,++++[->.<<<.>>]>+<<]
Prints in lowercase, separated by carriage returns. Uses wrapping 8 bit cells as well as cells left of the origin, though you can prepend a > to counter the latter.
26,{65+...}%+n*
printf %s\\n {A..Z}{,,,}
BrainF* ,79 60
+++++++++++++[->++>+>+++++<<<]>>---<[->>>++++[-<.<.>>]<+<<]
+++++++++++++[>+>+++++>++<<<-]>--->>[<.<.>.<.>.<.>.<.>+>-]
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Dec 14, 2013 at 7:32
'h 0 [do] [i] 4 / 'A + emit cr [loop]
4ú*SÖi
Explanation
4ú*SÖi
4 * Do 4 times
ú Create string of alphabet in lowercase
S Sort it
Ö Switch Case
i Push each character of string
4 Bytes with lowercase and no newline:
4ú*S
In hex:
B409BA160189D7B96800F6C1037502FE05CD21E2F5C3400A24
This is a complete MS-DOS .COM program. Copy the byte sequence to the file with .com extension and run it from DOSBox
Disassembly:
00: B4 09 mov ah,0x09 ;INT 21h "Write string to STDOUT" function
02: BA 16 01 mov dx,0x116 ;Address of the string s ('$'-terminated)
05: 89 D7 mov di,dx ;Because there's no way to dereference address in DX
07: B9 68 00 mov cx,104 ;CX=26*4
_0000000A:
0A: F6 C1 03 test cl,0x03 ;When lower two bits are zero...
0D: 75 02 jne _00000011 ;...do not skip the next instruction
0F: FE 05 inc b,[di] ;*s++
_00000011:
11: CD 21 int 21 ;Print the string
13: E2 F5 loop _0000000A ;Until --CX==0
15: C3 retn
16: 40 db 0x40 ;s[0], starts with 'A'-1
17: 0A db 0x0A ;'\n'
18: 24 db '$' ;Terminator required by the print function
Binary:
00000000: 400a 24b8 1109 8bd6 b168 d0c8 102c cd21 @.$......h...,.!
00000010: e2f8 c3 ...
Build with xxd -r and test in DOSBox or your favorite DOS VM.
Unassembled:
40 0A 24 DB '@', 0AH, '$' ; the letter + NL + '$' string terminator
B8 0911 MOV AX, 0911H ; AH = DOS print string; AL = 0001 0001 bit pattern
8B D6 MOV DX, SI ; DX to output string
B1 68 MOV CL, 26*4 ; Loop 4 times for each letter in alphabet
ALOOP:
D0 C8 ROR AL, 1 ; roll repeating bit pattern to the right
10 2C ADC BYTE PTR[SI], CH ; add rightmost bit to ASCII character
CD 21 INT 21H ; print letter to console
E2 F8 LOOP ALOOP ; continue looping
C3 RET ; return to DOS
Notes:
For most flavors of DOS, we know the values of many CPU registers at the start of execution and we know SI is 0100H, and CH is 0. So by placing the output string in the beginning of the program, [SI] will already be pointing to it. These bytes ('@',0AH,'$') decode to two benign instructions (INC AX, OR AH, [SI]), which have no effect on the program.
AL contains the bit pattern of 0001 0001 which is bit rolled right each time and the rightmost bit added to the ASCII value of the output char in memory.
j, 5 byteskAdds
Look ma, no Unicode!
kA # Uppercase output
dd # Quadruple
s # Sorted
# (j flag) join by newlines
Adds here is what is adding everything 4 times ;)
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ØAx4Y
Explanation:
ØAx4Y Main link
ØA “ABCDEFGHIJKLMNOPQRSTUVWXYZ”
4 4
x Repeat each element of x y times
Y Join x with newlines
.........
-1@''4#'.Q.A;
AWK, 48
Lets try it with AWK...
END{s=65;for(i=104;i--;s+=0==i%4)printf"%c\n",s}
As suggested by manatwork we can get rid of 2 chars
AWK, 46 (Edit)
END{for(i=104;i--;s+=0==i%4)printf"%c\n",s+65}
AWK,40 (editing MarkReed's code)
END{for(;i<104;){printf"%c\n",i++/4+65}}
END{for(i=104;i--;s+=0==i%4)printf"%c\n",s+65}.
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Dec 14, 2013 at 20:01
END block means it requires an input stream (even if it's /dev/null) to work. Does that modify the char count? Anyway, BEGIN{for(;++i<104;){printf"%c\n",i/4+65}} is 5 chars shorter.
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Dec 15, 2013 at 2:36
END pattern's requirement for an input stream...
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Dec 15, 2013 at 22:00
65..90|%{,[char]$_*4}
A slightly different approach to Iszi's. And shorter :-)
C# LINQ 115 Bytes110 Bytes
Enumerable.Range(65, 26).SelectMany(i => Enumerable.Repeat(i,4))
.ToList().ForEach(i=> Console.WriteLine((char)i));
A4×{S»
Explanation:
A # Push 'abcdefghijklmnopqrstuvwxyz'
4× # Repeat four times
{ # Sort
S # Split into list
» # Join by newlines
# Implicit print
Without newlines, 4 bytes
A4×{
With help from alephalpha and Mark S.
Print/@{#,#,#,#}&/@Alphabet[];
Print /@ {#, #, #, #} & /@ "A"~CharacterRange~"Z";
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Dec 14, 2013 at 7:27
Print/@{#,#,#,#}&/@Alphabet[]; instead of Print/@{#,#,#,#}&/@"A"~CharacterRange~"Z";
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for(var a=259;a++<363;WriteLine((char)(a/4)));
