30
\$\begingroup\$

Write a program to play the popular English nursery rhyme.

Sheet music (This file is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license. Attribution: Helix84 at the English language Wikipedia; Blahedo at the English language Wikipedia.)

Some Wikipedia articles that may be useful:

Some guidelines for your submission:

  • Your program must use the computer's sound card. If your programming language doesn't have convenient access to audio hardware, your program must create an output file in some standard format such as WAV or MIDI.

  • Your program must actually generate its output. For example, embedding the Ogg Vorbis file from Wikipedia would not be allowed.

  • The audio quality must be acceptable. At the very least, the song should be easily recognizable. Preferably, it should sound good as well.

  • The focus should be on code size, sound quality, or both (explain which one you decided on). Elegant solutions would also be great. Have fun!

\$\endgroup\$
5
  • \$\begingroup\$ can I approximate note frequencies? \$\endgroup\$
    – Ming-Tang
    May 20, 2011 at 4:06
  • 1
    \$\begingroup\$ Why is this closed? \$\endgroup\$
    – user58826
    May 16, 2017 at 18:12
  • \$\begingroup\$ @programmer5000 I'd say the close reason describes it extremely well... \$\endgroup\$ Sep 28, 2017 at 10:58
  • 3
    \$\begingroup\$ Even after making it code golf, I'd say this is still off topic. The audio quality must be acceptable. is not an objective validity criterion. \$\endgroup\$
    – Dennis
    Sep 28, 2017 at 15:46
  • 4
    \$\begingroup\$ @Dennis I'd say that's more like "unclear" rather than "off-topic". \$\endgroup\$ Sep 28, 2017 at 17:45

19 Answers 19

30
\$\begingroup\$

QBasic (56)

A$="CCGGAAG2FFEEDDC2"
B$="GGFFEED2"
PLAY "L4"+A$+B$+B$+A$

Focus is on reminiscence :)

(Don't have a QBasic to test this though)

\$\endgroup\$
11
  • 1
    \$\begingroup\$ Works on my DOSBox installation at least, but could you modify it to play the full song? \$\endgroup\$ Jan 31, 2011 at 11:49
  • \$\begingroup\$ Done. I'll go work on my reading skills :/ \$\endgroup\$
    – Eelvex
    Jan 31, 2011 at 11:57
  • \$\begingroup\$ Hmm is this page wrong? en.wikibooks.org/wiki/QBasic/Appendix#PLAY -- >An octave begins with C and ends with B. I did half expect QBASIC to be middle C based, but that notation would suggest it's A220 based if exactly correct :) \$\endgroup\$
    – mootinator
    Jan 31, 2011 at 15:35
  • 7
    \$\begingroup\$ Wow, this brings back memories of my first programming experiences with QBasic...which involved, among other things, writing cheesy music! \$\endgroup\$ Jan 31, 2011 at 16:40
  • 3
    \$\begingroup\$ +1 for trip down memory lane! Now all I need is a DRAW command sample :) \$\endgroup\$ May 20, 2011 at 21:32
16
\$\begingroup\$

JavaScript (214 212 211 characters)

Open Safari, Opera, or Google Chrome to JavaScript Shell, then enter the code below:

for(s="",y=192e3;x=--y/4e3|0;)s+="~ "[(y%4e3>800|x%8==1)&Math.pow(2,"024579702457245702457970"[x>>1]/12)*y/31%2];open("data:audio/wav;base64,UklGRiXuAgBXQVZFZm10IBAAAAABAAEAQB8AAEAfAAABAAgAZGF0YQHuAgCA"+btoa(s))

Unminified for readability (even then it may be hard to understand):

for(s = "", y = 192E3; x = --y / 4E3 | 0;) {
  s += "~ "[(y % 4E3 > 800 | x % 8 == 1) & Math.pow(2, "024579702457245702457970"[x >> 1] / 12) * y / 31 % 2];
}
open("data:audio/wav;base64,UklGRiXuAgBXQVZFZm10IBAAAAABAAEAQB8AAEAfAAABAAgAZGF0YQHuAgCA" + btoa(s));

With several more characters, it could work on Firefox as well, but you can change the audio/wav part to at least save the WAV file.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ Works just fine in Chrome. \$\endgroup\$
    – mootinator
    Jan 31, 2011 at 23:19
  • \$\begingroup\$ @mootinator: Works for me as well. I hadn't thought of checking it in Chrome -- it had not gained WAV file support until fairly recently. code.google.com/p/chromium/issues/detail?id=23916 \$\endgroup\$ Jan 31, 2011 at 23:42
  • \$\begingroup\$ Wow! This is wonderful. \$\endgroup\$ May 19, 2011 at 22:00
  • \$\begingroup\$ what do you mean by "with several more characters"? What would those characters be? \$\endgroup\$ May 19, 2011 at 22:02
  • \$\begingroup\$ @cf_PhillipSenn: When I ran the code on Firefox, I got a non-functional QuickTime Player. I had to change the code open(...) to Audio(...).play() (8 more characters) to get it to use Firefox's working built-in audio player. \$\endgroup\$ May 20, 2011 at 2:19
10
\$\begingroup\$

C# (Length: LOL)

So, what I did here was implement support for generating a .wav file from the string used for the QBasic solution in C# (single octave, no accidentals). Emphasis was on:

  1. Avoiding unsafe code blocks
  2. Not wasting too much of my time doing it
  3. Making it relatively simple to extend

using System;
using System.Collections.Generic;
using System.Linq;
using System.Reflection;
using System.Text.RegularExpressions;
using System.IO;

namespace ConsoleApplication1
{
    public static class Extension
    {
        public static byte[] ToByteArray(this object o)
        {
            return o.GetType().GetProperties(BindingFlags.Public | BindingFlags.Instance)
                .SelectMany(x =>
                                {
                                    var value = x.GetValue(o, null);
                                    if (value.GetType().Equals(typeof (UInt16)))
                                    {
                                        return BitConverter.GetBytes((UInt16) value);
                                    }
                                    if (value.GetType().Equals(typeof (UInt32)))
                                    {
                                        return BitConverter.GetBytes((UInt32) value);
                                    }
                                    if (value.GetType().Equals(typeof(char[])))
                                    {
                                        return ((char[]) value).Select(y => Convert.ToByte(y));
                                    }
                                    if (value.GetType().Equals(typeof(byte[])))
                                    {
                                        return (byte[]) value;
                                    }
                                    throw new NotImplementedException();
                                }).ToArray();
        }
    }
    public class Wave
    {
        public readonly byte[] WavFile; 

        public Wave(string notes)
        {
            var header = new Header();
            var data = new List<Chunk>();
            var f = new Fmt(8000);
            data.Add(f);
            data.Add(new WavData(notes, f));
            var thefile = data.SelectMany(x => x.ToByteArray()).ToArray();
            header.Size = (uint)thefile.Length + 4;
            WavFile = header.ToByteArray().Concat(thefile).ToArray();
        }
        class WavData: Chunk
        {
            private static IEnumerable<byte> RenderNote(string note, int length, Fmt fmt)
            {
                double frequency;
                switch (note)
                {
                    case "A":
                        frequency = 440;
                        break;
                    case "B":
                        frequency = 493.883;
                        break;
                    case "C":
                        frequency = 261.626;
                        break;
                    case "D":
                        frequency = 293.665;
                        break;
                    case "E":
                        frequency = 329.628;
                        break;
                    case "F":
                        frequency = 349.228;
                        break;
                    case "G":
                        frequency = 391.995;
                        break;
                    default:
                        throw new NotImplementedException("Unsupported Note");
                }
                var result = new byte[fmt.SampleRate / length * 2];  // For 120BPM tempo
                for (int i = 0; i < result.Length; i++)
                {
                    double time = (i % fmt.SampleRate) / (double)fmt.SampleRate;
                    double position = time * frequency;
                    if (result.Length - i <= fmt.SampleRate / 16)
                        result[i] = 127;
                    else
                        result[i] = (byte)Math.Round((Math.Sin(position * 2 * Math.PI) + 1) * 127);
                }
                return result;
            }
            public WavData(string notes, Fmt fmt)
            {
                Samples = new byte[0];
                foreach (var note in Regex.Matches(notes, @"[A-G][1|2|4|8]?").OfType<Match>().Select(x => x.Value))
                {
                    Samples = Samples.Concat(RenderNote(note[0] + "", note.Length > 1 ? note[1] - '0' : 4, fmt)).ToArray();
                }

            }
            public override char[] Id
            {
                get { return "data".ToCharArray(); }
            }
            public override uint DataSize
            {
                get { return (uint)Samples.Length; }
            }
            public byte[] Samples { get; private set; }
        }
        class Fmt : Chunk
        {
            public Fmt(UInt32 sampleRate)
            {
                CompressionCode = 1; // Unknown/PCM
                Channels = 1;
                SampleRate = sampleRate;
                SignificantBits = 8;
            }
            public override char[] Id
            {
                get { return "fmt ".ToCharArray();}
            }
            public override uint DataSize
            {
                get { return 16; }
            }
            public UInt16 CompressionCode { get; private set; }
            public UInt16 Channels { get; private set; }
            public UInt32 SampleRate { get; private set; }
            public UInt32 AvgBytesPerSecond { get { return SampleRate*BlockAlign; } }
            public UInt16 BlockAlign { get { return (UInt16) (SignificantBits/8*Channels); } }
            public UInt16 SignificantBits { get; private set; }
        }
        class Header
        {
            public Header()
            {
                Type = "RIFF".ToCharArray();
                RiffType = "WAVE".ToCharArray();
                Size = 0;
            }

            public char[] Type { get; private set; }
            public UInt32 Size { get; set; }
            public char[] RiffType { get; private set; }
        }
        abstract class Chunk
        {
            public abstract char[] Id { get; }
            public abstract UInt32 DataSize { get; }
        }
    }
    class Program
    {
        public static void Main(string[] args)
        {
            var p1 = "CCGGAAG2";
            var p2 = "FFEEDDC2";
            var p3 = "GGFFEED2";
            var w = new Wave(p1+p2+p3+p3+p1+p2);
            using (var f = new FileStream("testfile.wav", FileMode.Create))
                f.Write(w.WavFile, 0, w.WavFile.Length);
        }
    }
}
\$\endgroup\$
3
  • \$\begingroup\$ I noticed the output wave has a very small amplitude in my answer above. Clearly I'm missing something about how an 8 bit sample works. \$\endgroup\$
    – mootinator
    Mar 9, 2011 at 15:58
  • \$\begingroup\$ Ah, fixed it. Order of operations is important. Sample output here: dl.dropbox.com/u/469380/testfile.wav \$\endgroup\$
    – mootinator
    Mar 9, 2011 at 16:17
  • \$\begingroup\$ FromDigits["LOL",36]==28101 That does not look like LOL<sub>36</sub> bytes... \$\endgroup\$ Mar 11, 2016 at 15:15
10
\$\begingroup\$

C - 520

#include <linux/fd.h>
#include <time.h>
struct timespec t,p;char*q="AAHHJJH  FFEECCA  HHFFEEC  HHFFEEC  AAHHJJH  FFEECCA";x,y,z,v,w;main(){x=open("/dev/fd0",3);for(y;q[y];y++){clock_gettime(CLOCK_MONOTONIC,&p);if(q[y]>' ')for(w=z=0;z<4e8;z+=t.tv_nsec,w++){struct floppy_raw_cmd s={0};s.flags=FD_RAW_NEED_SEEK;v=!v;s.track=v;ioctl(x,FDRAWCMD,&s);clock_gettime(CLOCK_MONOTONIC,&t);t.tv_nsec=(w+1)*5e8/pow(2.,q[y]/12.)-(t.tv_sec-p.tv_sec)*1e9-t.tv_nsec+p.tv_nsec;t.tv_sec=0;nanosleep(&t,0);}t.tv_nsec=2e8;nanosleep(&t,0);}}

Why use past century's hardware like speakers and headphones? This excellent piece of code lets you play the song on a modern piece of hardware: a floppy drive!
No special requirements:

  • An IDE floppy drive
  • Linux kernel
  • Compile with -lm
  • Make sure the program can access /dev/fd0, so either chown the device or run as superuser

Bends the rules a bit, but let's for a second consider the floppy drive a sound device, and the IDE controller an integrated sound card.

\$\endgroup\$
1
  • \$\begingroup\$ I'd say this is reasonable, floppy drives are surprisingly capable for this task youtu.be/LkqYLOa76E0 \$\endgroup\$
    – Hotkeys
    Mar 11, 2016 at 17:37
7
\$\begingroup\$

Desmos, 148 116 bytes

The viewport has been manually changed to x=0 to 20, and y=0 to 12, because the audio trace feature in Desmos is affected by the viewport (the song will become distorted if the viewport is moved or zoomed in/out). I'm not sure if that contributes to the byte count or not. For now, I'm going to assume that it doesn't affect byte count, but please let me know otherwise if this isn't the case.

A=join(F(0,7,9,7,x),F(5,4,2,0,x))
B=F(7,5,4,2,x)
F(a,b,c,d,x)=\join([a,b,c]\{x<1,1.1<x<2.1\},d\{x<2\})
join(A,B,B,A)

Try It On Desmos!

Try It On Desmos! - Prettified

I focused on both song quality and code length (basically, shortest code in which the audio still sounds good).

To play the audio, click the join(A,B,B,A) expression (it should be outlined in blue). Then press ALT+T, and after that, H, after which you will hear the audio play if everything is done correctly.

Big thanks to @hyper-neutrino for helping me find the sequence of numbers that represents the notes in chat. I was having a very hard time finding that myself; I don't know a thing about music lol.

Explanation

This answer takes advantage of one of Desmos's accessibility features: Audio Tracing, which can play any graph as a sound. By plotting horizontal line segments at specific places in the viewport, the audio tracing will be able to play that line segment as a single note for a certain duration, depending on how long that line segment is. Specifically, the viewport is split as follows:

Bottomest vs toppest lines of the graph correspond to the range of an octave... the 12 steps within this range correspond to the half-tones in the music scale. (Source)

As I am not familiar with music at all, I pretty much have no idea what this means, but hopefully that is understandable to all of you that know music.

The next part is to actually find the sequence of y-values so that it will play the song. As I mentioned above, hyper-neutrino basically did all the work and found the correct sequence of y-values which play the song (I pretty much have no idea how he found them; something about intervals and semitones?). The sequence is as follows:

0 0 7 7 9 9 7, 5 5 4 4 2 2 0, 7 7 5 5 4 4 2, 7 7 5 5 4 4 2, 0 0 7 7 9 9 7, 5 5 4 4 2 2 0

where every 7th note (the note before each comma, and the last note) is a half note, meaning it is played twice as long as the other notes, which are quarter notes. Graphically, this means that the line segment that represents a half note is twice as long as that of a quarter note.

With that out of the way, the next thing to do is to actually put all of this into Desmos. So how are we going to do this? Well, first I found that a line segment of length 1 is long enough for a note, so each quarter note will be length 1, while each half note will be length 2.

Secondly, we will need to put a short pause in between every note, or else two repeated notes will just play as one half note instead. I decided to use a 0.1 length pause in between each note.

My plan was to construct a huge list of line segments which Desmos can graph and audio trace over.

A pattern I noticed with the notes is that every 7 notes always follows a a a b b c c d (half note) pattern. To take advantage of that, I made a function F which takes in 4 notes a,b,c,d and constructs a list of line segments in accordance to that pattern. Let's take a look at it:

F(a,b,c,d,x)=\join([a,b,c]\{x<1,1.1<x<2.1\},d\{x<2\})

The function F takes first takes in 4 notes a,b,c,d, then x because we are plotting something (the function won't work without it).

Let's first take a look at the first part of the function:

[a,b,c]\{x<1,1.1<x<2.1\}

Because of the list [a,b,c], this entire expression results in a list of 3 horizontal line segments: one segment at y levels a, b, and c. Now let's take a look at the domain restriction. It splits the lines y=[a,b,c] into two parts: one where x<1 and another where 1.1<x<2.1, effectively splitting it into two line segments of length 1 with a 0.1 gap in between them (the viewport is from x=0, so x<1 is functionally the same as 0<x<1). The audio trace will first play the x<1 segment (one quarter note), then a very short pause later, play the 1.1<x<2.1 segment (another quarter note). It does this for a, then b, then c.

Finally, this list is joined with d\{x<2\}, which is a line segment of length 2 at y level d. This will play the note d for twice as long, emulating a half note.

With a full understanding of function F, we can now encode the number sequence shown earlier with F.

First note that the first 14 notes (0 0 7 7 9 9 7, 5 5 4 4 2 2 0) are repeated at the end as well. Also, the middle 14 notes (7 7 5 5 4 4 2, 7 7 5 5 4 4 2) is just 7 7 5 5 4 4 2 repeated twice.

We can encode 0 0 7 7 9 9 7 as F(0,7,9,7,x) and 5 5 4 4 2 2 0 as F(5,4,2,0,x). By joining the two resultant lists and storing it in a variable A, we can re-use these 14 notes at the end of the song.

We can encode 7 7 5 5 4 4 2 as F(7,5,4,2,x) and set it to a variable B so that we can re-use these notes.

Finally, we can join the notes like so: join(A,B,B,A) to get the final sequence of notes. Desmos will then automatically graph all the line segments in the resultant list, which we can then audio trace over.

\$\endgroup\$
1
  • \$\begingroup\$ zoomed out and was not disappointed \$\endgroup\$ May 1 at 5:27
6
\$\begingroup\$

Python (259)

import pysynth

c=('c',4)
g=('g',4)
a=('a',4)
b=('b',4)
d=('d',4)
e=('e',4)
f=('f',4)
g2=('g',2)
c2=('c',2)
d2=('d',2)

s=(c,c,g,g,a,a,g2,f,f,e,e,d,d,c2,g,g,f,f,e,e,d2,g,g,f,f,e
            ,e,d2,c,c,g,g,a,a,g2,f,f,e,e,d,d,c2)

pysynth.make_wav(s,fn="s.wav")
\$\endgroup\$
6
  • 3
    \$\begingroup\$ couldn't this be shortened to import ttls? \$\endgroup\$
    – zzzzBov
    Mar 11, 2011 at 20:38
  • \$\begingroup\$ @zzz: Um, is that a joke? \$\endgroup\$
    – John
    Mar 11, 2011 at 20:48
  • 2
    \$\begingroup\$ @zzz: -Bangs-head-on-desk- What are you saying exactly? \$\endgroup\$
    – John
    Mar 11, 2011 at 20:54
  • 1
    \$\begingroup\$ @John it's obviously a koan. you should either bow or hit him. \$\endgroup\$
    – ixtmixilix
    Sep 9, 2011 at 8:41
  • 1
    \$\begingroup\$ (very late comment) For reference, here is a version -60 bytes, because it's really golfed. \$\endgroup\$ Jun 18, 2016 at 20:50
6
\$\begingroup\$

bash + say + gunzip, 136 bytes

say, of course, being the OS X text-to-speech command. This is... dorky. Yeah, let's go with dorky.

printf '<117 bytes>'|gunzip|sh

The 117 bytes is, of course, a gzip stream containing unprintable characters. Here's an xxd dump of the script including those characters:

00000000: 7072 696e 7466 2027 1f8b 085c 305c 305c  printf '...\0\0\
00000010: 305c 305c 3002 032b 4eac 54d0 2d0b c9c8  0\0\0..+N.T.-...
00000020: cf4d 2c56 c8e7 c2ca 75cc cb4b c4ce 71cb  .M,V....u..K..q.
00000030: ccc7 c90b 4b4d 85f0 7252 530b 14f4 4ca0  ....KM..rRS...L.
00000040: c2de 8945 a979 4061 6cbc e0c4 dcc4 bc92  ...E.y@al.......
00000050: 8c44 dc02 2e89 7999 a939 685c 5c74 7723  .D....y..9h\\tw#
00000060: ec44 755c 6e2a 8f8a ee19 581b 8767 1402  .Du\n*....X..g..
00000070: 5c30 fa36 7e25 2599 025c 305c 3027 7c67  \0.6~%%..\0\0'|g
00000080: 756e 7a69 707c 7368                      unzip|sh

Explanation

The 117 bytes is the following script gzipped:

say -vThomas o
say -vThomas o
say -vAnna o
say -vAnna o
say -vFiona o
say -vFiona o
say -vVeena o
sleep .4
say -vKaren o
say -vKaren o
say -vSamantha o
say -vSamantha o
say -vDaniel o
say -vDaniel o
say -vThomas o
sleep .4
say -vVeena o
say -vVeena o
say -vKaren o
say -vKaren o
say -vSamantha o
say -vSamantha o
say -vDaniel o
sleep .4
say -vVeena o
say -vVeena o
say -vKaren o
say -vKaren o
say -vSamantha o
say -vSamantha o
say -vDaniel o
sleep .4
say -vThomas o
say -vThomas o
say -vAnna o
say -vAnna o
say -vFiona o
say -vFiona o
say -vVeena o
sleep .4
say -vKaren o
say -vKaren o
say -vSamantha o
say -vSamantha o
say -vDaniel o
say -vDaniel o
say -vThomas o

That's right, I just made a bunch of different say voices say "o." To figure out which ones, I wrote a script using aubionotes to get a quick-and-dirty estimate of each voice's pitch, then did a lot of trial and error the find ones that sound mostly right.

I considered trying to golf this manually, but there's so much repetition that I figured Zopfli would make shorter work of it, so I took the easy way out.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ I want to hear how it sounds! Perhaps include a render of it for those of us not on OSX? \$\endgroup\$
    – bb94
    Feb 2, 2021 at 22:07
5
\$\begingroup\$

C, 277 characters

#include<math.h>
a[]={0,7,9,7,5,4,2,0,7,5,4,2,7,5,4,2,0,7,9,7,5,4,2,0},i,j,f;main(){unsigned char
b[8000];f=open("/dev/dsp",1);for(i=0;i<24;i++){for(j=0;j<8000;j++){b[j]=(i%4==3
||j/400%20!=9?1+sinf(j*powf(2,a[i]/12.)):1)*127;}for(j=0;j<8000;j+=write(f,b+j,
8000-j));}close(f);}

Perl, 203 characters

open F,'>:raw','/dev/dsp';for$a(0,7,9,17,5,4,2,10,7,5,4,12,7,5,4,12,0,7,9,17,5,4
,2,10){$b=pack'C*',map 127*($a>9||$_/400%20!=9?1+sin($_*2**($a%10/12)):1),0..
7999;$b=substr$b,syswrite F,$b while length$b}

Conveniently, OSS's /dev/dsp defaults to 8kHz mono u8; all I do here is open the device and write computed samples.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Post Perl separately; you're on codegolf :) \$\endgroup\$
    – Eelvex
    Nov 2, 2011 at 3:10
  • \$\begingroup\$ 179 - j,f;main(i){char b[8000];f=open("/dev/dsp",1);for(i=24;i--;write(f,b,j))for(j=0;j<8000;j++)b[j]=(i%4&&j/400%20==9?1:1+sinf(j*powf(2,("@BDEGIG@BDEGBDEG@BDEGIG@"[i]-64)/12.)))*127;} \$\endgroup\$
    – gastropner
    Dec 30, 2017 at 12:07
5
\$\begingroup\$

C, 96 chars

main(t){for(;++t>>16<3;)putchar(t*!!(t>>9&7|!(-t>>12&7))*(96+"#d|dOE3#dOE3dOE3"[t>>13&15])>>5);}

Produces raw 8-bit unsigned mono audio data in classic bytebeat style. Recommended sample rates for playback are between 8 and 16 kHz; changing the sample rate changes the tempo and pitch.

To compile and play under Linux, save the code above as twinkle.c and run the following commands:

gcc twinkle.c -o twinkle
./twinkle | aplay

Some notes on how the code works:

  • The general trick used for bytebeat compositions like this is that putchar() takes an integer value but only prints the low eight bits of it. Thus, putchar(t), where t is an increasing counter, generates a sawtooth wave, and the frequency of the wave can be altered by multiplying t with a suitable value.

  • !!(t>>9&7|!(-t>>12&7)) produces the repeating 6+1 note pattern. Specifically, !!(t>>9&7) evaluates to 0 whenever t>>9 & 7 == 0 and to 1 otherwise. Thus, it produces a 512-sample gap in the waveform every 4096 samples, while the !(-t>>12&7) eliminates every eighth such gap.

  • 96+"#d|dOE3#dOE3dOE3"[t>>13&15] generates the melody: the ASCII code of each characters in the string plus 96 gives the relative frequency of the corresponding note. Actually, the values are the approximate frequencies in Hz of concert pitch notes in the 3rd / small octave, i.e. with A corresponding to 220. However, since the base tone with which these values are multiplied is about 64 Hz (when played at 16 kHz, or 32 Hz when played at 8 kHz), we need to scale the result down by five octaves with >>5 to get the frequency back into a reasonable range.

Ps. If you want to try this code on a JavaScript-based bytebeat player, replace [t>>13&15] with .charCodeAt(t>>13&15).

\$\endgroup\$
4
\$\begingroup\$

PowerShell: 207

Golfed code:

filter n {switch($_){C{262}D{294}E{330}F{349}G{392}A{440}}}$t="CCGGAAGFFEEDDCGGFFEEDGGFFEEDCCGGAAGFFEEDDC";1..6|%{$t[0..5]|n|%{[console]::beep($_,600)};$t[6]|n|%{[console]::beep($_,1200)};$t=$t.SubString(7)}

Ungolfed, with comments:

# Filter to define note frequencies.
filter n {switch($_){C{262}D{294}E{330}F{349}G{392}A{440}}}

# Notes for Twinkle, Twinkle, Little Star.
$t="CCGGAAGFFEEDDCGGFFEEDGGFFEEDCCGGAAGFFEEDDC"

# Run through each phrase in the song.
1..6|%{
    # Play first six notes as quarter notes.
    $t[0..5]|n|%{[console]::beep($_,600)}
    # Play seventh note as half note.
    $t[6]|n|%{[console]::beep($_,1200)}
    # Left-shift $t by 7 notes.
    $t=$t.SubString(7)
}

Not the greatest-sounding rendition of the song ever, but it works.

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  • 1
    \$\begingroup\$ (late comment) Can you do n{ instead of n {? \$\endgroup\$ Jun 18, 2016 at 20:35
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HyperCard 2.2 - 113

play harpsichord "c c g g a a gh fq f e e d d ch gq g f f e e dh gq g f f e e dh cq c g g a a gh fq f e e d d ch"

Usage: Start HyperCard, type ⌘M to open the Message Box, paste the code above, and press enter.

harpsichord may be replaced with flute or boing to get different sounds.

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  • \$\begingroup\$ (very late comment) harpsichord -> flute -6 remove the space between instrument and quotes -1 total -7 \$\endgroup\$ Jun 20, 2016 at 17:25
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Arduino, 688

int length=15;char notes[]="ccggaag ffeeddc ggffeed ggffeed ccggaag ffeeddc";int beats[]={1,1,1,1,1,1,2,1,1,1,1,1,1,2,4};int tempo=300;void playTone(int tone,int duration){for(long i=0;i<duration*1000L;i+=tone*2){digitalWrite(11,HIGH);delayMicroseconds(tone);digitalWrite(11, LOW);delayMicroseconds(tone);}}void playNote(char note, int duration){char names[]={'c','d','e','f','g','a','b','C'};int tones[]={1915,1700,1519,1432,1275,1136,1014,956};for(int i=0;i<8;i++){if(names[i]==note){playTone(tones[i], duration);}}}void setup(){pinMode(11, OUTPUT);}void loop(){for(int i=0;i<length;i++){if(notes[i]==' '){delay(beats[i]*tempo);}else{playNote(notes[i],beats[i]*tempo);}delay(tempo/2);}}

Plug in the buzzer on output 11. I concentrated mainly on quality, but also on code length.

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    \$\begingroup\$ (late comment) "ccggaag ffeeddc ggffeed ggffeed ccggaag ffeeddc " remove space? int beats[] = remove space? for (long remove space? 11, LOW remove space? note, int remove space? i], duration remove space? 11, OUTPUT remove space? \$\endgroup\$ Jun 18, 2016 at 11:27
  • 1
    \$\begingroup\$ Did you actually forget some spaces? As far as I understand, the space in ffeeddc " is for some sort of delay, which you do not need at the end, and beats[] = has no reason to have a space. Also, you should make a version concentrating on code length, just for the heck of it! \$\endgroup\$ Jun 18, 2016 at 20:33
  • \$\begingroup\$ I suppose the variables could all be one letter but I don't think it's worth the time on such an old answer. Thanks for those tips. \$\endgroup\$
    – Timtech
    Jun 18, 2016 at 21:01
  • \$\begingroup\$ Ok, feel free to post as a separate answer if you'd like. \$\endgroup\$
    – Timtech
    Jun 18, 2016 at 21:07
  • \$\begingroup\$ @Timtech As the question is closed, that can't be done unfortunately... :-( \$\endgroup\$
    – wizzwizz4
    Oct 24, 2016 at 16:00
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JavaScript, 194 177 176

Now that WebAudio is a thing, I thought it would be fun to write this. (Yes, I know ES6 and WebAudio came out after the challenge was posted.) Tested on Microsoft Edge.

a=new AudioContext,o=a.createOscillator(i=0)
o.connect(a.destination)
setInterval(_=>o.frequency.value=4*[65,73,82,87,98,110]['045432104321432104543210'[i++]]|0,1e3)
o.start(1)

Given the lack of strict output requirements, I took the liberty of detuning some frequency values to be divisible by 4. It still sounds passable, relative-pitch wise. The final notes are off by -11, -10, -9, -6, 0, and 0 cents, for c, d, e, f, g, and a respectively, giving a total window of 11 cents. Not great to the trained ear, but recognizable as the tune, which is all the challenge requires.

There are also no breaks between notes, but that wasn't a requirement either.

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Wolfram Language (Mathematica), 93 bytes

Sound[s=SoundNote;If[#>9,s[#-10,2],{s@#,s@#}]&/@Join[a={0,7,9,17,5,4,2,10},b={7,5,4,12},b,a]]
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2
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Python 317 305 301

This is my solution, using only standard python libraries:

import math,wave,struct;d=24000;r=1100.;t=wave.open("t.wav","w");t.setparams((1,2,int(r),d,"NONE",""));a=[0,7,9,7];b=[5,4,2,0];c=[7,5,4,2]
for h in[math.sin(6*[240*2**(j/12.)for j in a+b+c+c+a+b][x/1000]*(x/r))*(x%500>9 or x/1000%4>2)for x in range(d)]:t.writeframes(struct.pack('h', int(h*64000/2)))

And here it is with some more whitespace for readability:

import math,wave,struct;d=24000;r=1100.
a=[0,7,9,7];b=[5,4,2,0];c=[7,5,4,2];s=[240*2**(j/12.) for j in a+b+c+c+a+b]
z=[math.sin(6*s[int(x/1000)]*(x/r))*(x%500>10 or int(x/1000)%4>2) for x in range(d)]
t=wave.open("t.wav","w");t.setparams((1,2,int(r),d,"NONE",""))
for h in z:t.writeframes(struct.pack('h', int(h*64000./2)))
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2
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Powershell, 120 117 bytes

[Console]::beep, note labels and frequencies inspired by Iszi

 ($a='ccggaaGffeeddC')+'ggffeeD'*2+$a|% t*y|%{[console]::beep((0,110,0,65,73,82,87,98)[$_-band7]*4,600+600*($_-lt97))}

Main idea:

  • The melody is encoded in a string.
  • The notes are encoded with chars A,C,D,E,F,G.
  • Uppercase means a double duration.
  • 3 lower bits ($_-band7) of each note uses as index in the frequencies array (A->1,C->3,D->4...)
  • This script uses the reduced sampling rate for frequencies: (0,110,0,65,73,82,87,98)[$_-band7]*4 instead Iszi's (0,440,0,262,294,330,349,392)[$_-band7]. [console]::Beep is not the most accurate musical instrument, so it can slightly fake :)

Explanation: For each char from the melody string ccggaaGffeeddCggffeeDggffeeDccggaaGffeeddC, the script:

  • lookups frequensies from the array using the lower bits as index
  • calculates a duration based on char uppercase/lowercase
  • calls [console]::beep to play the note
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1
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SmileBASIC, 45 bytes

BGMPLAY"{M=CCGGAAG2FFEEDDC2}{M}[GGFFEED2]2{M}
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1
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C (MSVC + Win32), 163 bytes

#include<windows.h>
#include<math.h>
main(){for(char*s="DDKKMMK IIHHFFD KKIIHHF KKIIHHF DDKKMMK IIHHFFD ";*s;s++)*s-32&&Beep(220*pow(1.06,*s-65),s[1]-32?400:800);}

In Windows 7, Beep was rewritten to pass the beep to the default sound device for the session. This is normally the sound card, except when run under Terminal Services, in which case the beep is rendered on the client.

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0
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Raku, 115 bytes

for 'xPHPZ`jxPZ`jPZ`jxPHPZ`jx'.ords.kv ->\i,\p {print 9*(4*$_ div p%2&&2900>$_)*(3==i%4||!(1500>$_>1400))for ^3e3}

Pipe to aplay to listen.

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