Code Golf: Musical Notes

问题:

The challenge

The shortest code by character count, that will output musical notation based on user input.

Input will be composed of a series of letters and numbers - letters will represent the name of the note and the number will represent the length of the note. A note is made of 4 vertical columns. The note's head will be a capital O, stem, if present will be 3 lines tall, made from the pipe character |, and the flag(s) will be made from backward slash \.

Valid note lengths are none, 1/4 of a note, 1/8 of a note, 1/16 of a note and 1/32 of a note.

       |    |\    |\    |\
       |    |     |\    |\
       |    |     |     |\
 O    O    O     O     O
 1   1/4  1/8  1/16   1/32

Notes are places on the Staff, according to their note name:

  ----

D ----
C     
B ----
A     
G ----
F     
E ----

All input can be assumed to be valid and without errors - Each note separated with a white space on a single line, with at least one valid note.

Test cases

Input:
    B B/4 B/8 B/16 B/32 G/4 D/8 C/16 D B/16
Output:
                              |\               
    --------------------------|---|\--------
          |   |\  |\  |\      |   |\      |\
    ------|---|---|\--|\-----O----|--O----|\
          |   |   |   |\  |      O        | 
    -O---O---O---O---O----|--------------O--
                          |                 
    ---------------------O------------------

    ----------------------------------------

---

Input:
    E/4 F/8 G/16 A/32 E/4 F/8 G/16 A/32 
Output:

    --------------------------------

    --------------|\--------------|\
              |\  |\          |\  |\ 
    ------|\--|\--|\------|\--|\--|\
      |   |   |  O    |   |   |  O  
    --|---|--O--------|---|--O------
      |  O            |  O          
    -O---------------O--------------

---

Input:
    C E/32 B/8 A/4 B F/32 B C/16
Output:

    ------------------------------|\
              |\                  |\
    ----------|---|---------------|-
     O        |   |              O   
    ---------O----|--O----|\-O------
          |\     O        |\        
    ------|\--------------|\--------
          |\             O           
    -----O--------------------------

Code count includes input/output (i.e full program).

回答1:

Golfscript (112 characters)

' '%:A;10,{):y;A{2/.0~|1=~:r;0=0=5\- 7%
4y@--:q'  '' O'if-4q&!q*r*{16q/r<'|\\'
'| 'if}'  'if+{.32=y~&{;45}*}%}%n}%

回答2:

Perl, 126 characters (115/122 with switches)

Perl in 239 226 218 216 183 180 178 172 157 142 136 133 129 128 126 chars

This 126 character solution in Perl is the result of a lengthy collaboration between myself and A. Rex.

@o=($/)x10;$/=$";map{m[/];$p=4+(5-ord)%7;
$_.=--$p?!($p&~3)*$'?16<$p*$'?"  |\\":"  | ":$/x4:" O  ",
$|--&&y@ @-@for@o}<>;print@o

A. Rex also proposes a solution to run with the perl -ap switch. With 111(!) characters in this solution plus 4 strokes for the extra command-line switch, this solution has a total score of 115.

$\="$:
"x5;$p=4+(5-ord)%7,s#..##,$\=~s#(.)\K$#--$p?
$_*!($p&~3)?"$1|".(16<$p*$_?"\\":$1).$1:$1x4:O.$1x3#gemfor@F

The first newline in this solution is significant.

Or 122 characters embedding the switches in the shebang line:

#!perl -ap
$\="$:
"x5;$p=4+(5-ord)%7,s#..##,$\=~s#(.)\K$#--$p?$_*!($p&~3)?"$1|".(16<$p*$_?
"\\":$1).$1:$1x4:O.$1x3#gemfor@F

(first two newlines are significant).

Half-notes can be supported with an additional 12 chars:

@o=($/)x10;$/=$";map{m[/];$p=4+(5-ord)%7;
$_.=--$p?!($p&~3)*$'?16<$p*$'?"  |\\":"  | ":$/x4:$'>2?" @  ":" O  ",
$|--&&y@ @-@for@o}<>;print@o

回答3:

LilyPond - 244 bytes

Technically speaking, this doesn't adhere to the output specification, as the output is a nicely engraved PDF rather than a poor ASCII text substitute, but I figured the problem was just crying out for a LilyPond solution. In fact, you can remove the "\autoBeamOff\cadenzaOn\stemUp" to make it look even more nicely formatted. You can also add "\midi{}" after the "\layout{}" to get a MIDI file to listen to.

o=#(open-file"o""w")p=#ly:string-substitute
#(format o"~(~a"(p"2'1""2"(p"4'1""4"(p"6'1""6"(p"8'1""8"(p"/""'"(p"C""c'"(p"D""d'"(p" ""/1"(p"
"" "(ly:gulp-file"M")))))))))))#(close-port o)\score{{\autoBeamOff\cadenzaOn\stemUp\include"o"}\layout{}}

Usage: lilypond thisfile.ly

Notes:

1. The input must be in a file named "M" in the same directory as the program.
2. The input file must end in a newline. (Or save 9 bytes by having it end in a space.)
3. The output is a PDF named "thisfile.pdf", where "thisfile.ly" is the name of the program.
4. I tested this with LilyPond 2.12.2; other versions might not work.

I haven't done much in LilyPond, so I'm not sure this is the best way to do this, since it has to convert the input to LilyPond format, write it to an auxiliary file, and then read it in. I currently can't get the built-in LilyPond parser/evaluator to work. :(

Now working on an ASCII-output solution.... :)

回答4:

C89 (186 characters)

#define P,putchar(
N[99];*n=N;y;e=45;main(q){for(;scanf(" %c/%d",n,n+1)>0;n
+=2);for(;y<11;q=y-(75-*n++)%7 P+q-4?e:79)P*n&&q<4&q>0?
124:e)P*n++/4>>q&&q?92:e))*n||(e^=13,n=N,y++P+10))P+e);}

Half-note support (+7 characters)

#define P,putchar(
N[99];*n=N;y;e=45;main(q){for(;scanf(" %c/%d",n,n+1)>0;n
+=2);for(;y<11;q=y-(75-*n++)%7 P+q-4?e:v<4?79:64)P*n&&q<4&q>0?
124:e)P*n++/4>>q&&q?92:e))*n||(e^=13,n=N,y++P+10))P+e);}

回答5:

Python 178 characters

The 167 was a false alarm, I forgot to suppress the stems on the whole notes.

R=raw_input().split()
for y in range(10):
 r=""
 for x in R:o=y-(5-ord(x[0]))%7;b=" -"[y&1]+"O\|";r+=b[0]+b[o==3]+b[-(-1<o<3and''<x[1:])]+b[2*(-1<o<":862".find(x[-1]))]
 print r

Python 167 characters (Broken)

No room for the evil eye in this one, although there are 2 filler characters in there, so I added a smiley. This technique takes advantage of the uniqueness of the last character of the note lengths, so lucky for me that there are no 1/2 notes or 1/64 notes

R=raw_input().split()
for y in range(10):
 r=""
 for x in R:o=y-(5-ord(x[0]))%7;b=" -"[y&1]+"O\|";r+=b[0]+b[o==3]+b[-(-1<o<3)]+b[2*(-1<o<":862".find(x[-1]))]
 print r

Python 186 characters <<o>>

Python uses the <<o>> evil eye operator to great effect here. The find() method returns -1 if the item is not found, so that is why D doesn't need to appear in the notes.

R=raw_input().split()
for y in range(10):
 r=""
 for x in R:o='CBAGFE'.find(x[0])+4;B=" -"[y%2];r+=B+(B,'O')[o==y]+(x[2:]and
y+4>o>y and"|"+(B,'\\')[int(x[2:])<<o>>6+y>0]or B*2)
 print r

11 extra bytes gives a version with half notes

R=raw_input().split()
for y in range(10):
 r=""
 for x in R:t='CBAGFE'.find(x[0])+4;l=x[2:];B=" -"[y%2];r+=B+(B,'@O'[l
in'2'])[t==y]+(l and y+4>t>y and"|"+(B,'\\')[int(l)>>(6+y-t)>0]or B*2)
 print r

$ echo B B/2 B/4 B/8 B/16 B/32 G/4 D/8 C/16 D B/16| python notes.py 
                              |\            
------------------------------|---|\--------
      |   |   |\  |\  |\      |   |\      |\
------|---|---|---|\--|\-----@----|--O----|\
      |   |   |   |   |\  |      @        | 
-O---O---@---@---@---@----|--------------@--
                          |                 
-------------------------@------------------

--------------------------------------------

回答6:

159 Ruby chars

n=gets.split;9.downto(0){|p|m='- '[p%2,1];n.each{|t|r=(t[0]-62)%7;g=t[2..-1]
print m+(r==p ?'O'+m*2:p>=r&&g&&p<r+4?m+'|'+(g.to_i>1<<-p+r+5?'\\':m):m*3)}
puts}

回答7:

Ruby 136

n=gets;10.times{|y|puts (b=' -'[y&1,1])+n.split.map{|t|r=y-(5-t[0])%7
(r==3?'O':b)+(t[1]&&0<=r&&r<3?'|'<<(r<t[2,2].to_i/8?92:b):b+b)}*b}

Ruby 139 (Tweet)

n=gets;10.times{|y|puts (b=' -'[y&1,1])+n.split.map{|t|r=y-(5-t[0])%7
(r==3?'O':b)+(t[1]&&0<=r&&r<3?'|'<<(r<141>>(t[-1]&7)&3?92:b):b+b)}*b}

Ruby 143

n=gets.split;10.times{|y|puts (b=' -'[y&1,1])+n.map{|t|r=y-(5-t[0])%7;m=t[-1]
(r==3?'O':b)+(m<65&&0<=r&&r<3?'|'<<(r<141>>(m&7)&3?92:b):b+b)}*b}

Ruby 148

Here is another way to calculate the flags,
where m=ord(last character), #flags=1+m&3-(1&m/4)

and another way #flags=141>>(m&7)&3, that saves one more byte

n=gets.split;10.times{|y|b=' -'[y&1,1];n.each{|t|r=y-(5-t[0])%7;m=t[-1]
print b+(r==3?'O':b)+(m<65&&0<=r&&r<3?'|'<<(r<141>>(m&7)&3?92:b):b+b)}
puts}

Ruby 181

First try is a transliteration of my Python solution

n=gets.split;10.times{|y|r="";n.each{|x|o=y-(5-x[0])%7
r+=(b=" -"[y&1,1]+"O\\|")[0,1]+b[o==3?1:0,1]+b[-1<o&&o<3&&x[-1]<64?3:0,1]+b[-1<o&&o<(":862".index(x[-1]).to_i)?2:0,1]}
puts r}

回答8:

F#, 458 chars

Reasonably short, and still mostly readable:

let s=Array.init 10(fun _->new System.Text.StringBuilder())
System.Console.ReadLine().Split([|' '|])
|>Array.iter(fun n->
for i in 0..9 do s.[i].Append(if i%2=1 then"----"else"    ")
let l=s.[0].Length
let i=68-int n.[0]+if n.[0]>'D'then 7 else 0
s.[i+3].[l-3]<-'O'
if n.Length>1 then
 for j in i..i+2 do s.[j].[l-2]<-'|'
 for j in i..i-1+(match n.[2]with|'4'->0|'8'->1|'1'->2|_->3)do s.[j].[l-1]<-'\\')
for x in s do printfn"%s"(x.ToString())

With brief commentary:

// create 10 stringbuilders that represent each line of output
let s=Array.init 10(fun _->new System.Text.StringBuilder())
System.Console.ReadLine().Split([|' '|])
// for each note on the input line
|>Array.iter(fun n->
// write the staff
for i in 0..9 do s.[i].Append(if i%2=1 then"----"else"    ")
// write note (math so that 'i+3' is which stringbuilder should hold the 'O')
let l=s.[0].Length
let i=68-int n.[0]+if n.[0]>'D'then 7 else 0
s.[i+3].[l-3]<-'O'
// if partial note
if n.Length>1 then
 // write the bar
 for j in i..i+2 do s.[j].[l-2]<-'|'
 // write the tails if necessary
 for j in i..i-1+(match n.[2]with|'4'->0|'8'->1|'1'->2|_->3)do s.[j].[l-1]<-'\\')
// print output
for x in s do printfn"%s"(x.ToString())

回答9:

C 196 characters <<o>>

Borrowing a few ideas off strager. Interesting features include the n+++1 "triple +" operator and the <<o>> "evil eye" operator

#define P,putchar
N[99];*n=N;y;b;main(o){for(;scanf(" %c/%d",n,n+1)>0;n+=2);for(;y<11;)
n=*n?n:(y++P(10),N)P(b=y&1?32:45)P((o=10-(*n+++1)%7-y)?b:79)P(0<o&o<4&&*n?'|':b)
P(*n++<<o>>6&&0<o&o<4?92:b);}

回答10:

168 characters in Perl 5.10

My original solution was 276 characters, but lots and lots of tweaking reduced it by more than 100 characters!

$_=<>;
y#481E-GA-D62 #0-9#d;
s#.(/(.))?#$"x(7+$&).O.$"x($k=10).($1?"|":$")x3 .$"x(10-$2)."\\"x$2.$"x(9-$&)#ge;
s#(..)*?\K (.)#-$2#g;
print$/while--$k,s#.{$k}\K.#!print$&#ge

If you have a minor suggestion that improves this, please feel free to just edit my code.

回答11:

Lua, 307 Characters

b,s,o="\\",io.read("*l"),io.write for i=1,10 do for n,l in
s:gmatch("(%a)/?(%d*)")do x=n:byte() w=(x<69 and 72 or 79)-x
l=tonumber(l)or 1 d=i%2>0 and" "or"-"o(d..(i==w and"O"or
d)..(l>3 and i<w and i+4>w and"|"or d)..(l>7 and i==w-3
and b or l>15 and i==w-2 and b or l>31 and i==w-1 and b or
d))end o"\n"end

回答12:

C -- 293 characters

Still needs more compression, and it takes the args on the command line instead of reading them...

i,j,k,l;main(c,v)char **v;{char*t;l=4*(c-1)+2;t=malloc(10*l)+1;for(i=0;i<10;i
++){t[i*l-1]='\n';for(j=0;j<l;j++)t[i*l+j]=i&1?'-':' ';}t[10*l-1]=0;i=1;while
(--c){j='G'-**++v;if(j<3)j+=7;t[j*l+i++]='O';if(*++*v){t[--j*l+i]='|';t[--j*l
+i]='|';t[--j*l+i]='|';if(*++*v!='4'){t[j++*l+i+1]='\\';if(**v!='8'){t[j++*l+
i+1]='\\';if(**v!='1'){t[j++*l+i+1]='\\';}}}}i+=3;}puts(t);}

edit: fixed the E

edit: down to 293 characters, including the newlines...

#define X t[--j*l+i]='|'
#define Y t[j++*l+i+1]=92
i,j,k,l;main(c,v)char**v;{char*t;l=4*(c-1)+2;t=malloc(10*l)+1;for(i=10;i;)t[--i*
l-1]=10,memset(t+i*l,i&1?45:32,l-1);t[10*l-1]=0;for(i=1;--c;i+=3)j=71-**++v,j<3?
j+=7:0,t[j*l+i++]=79,*++*v?X,X,X,*++*v-52?Y,**v-56?Y,**v-49?Y:0:0:0:0;puts(t);}