# Is this a Major Scale (or equivalent)?

Sandbox

The major scale (or Ionian scale) is one of the most commonly used musical scales, especially in Western music. It is one of the diatonic scales. Like many musical scales, it is made up of seven notes: the eighth duplicates the first at double its frequency so that it is called a higher octave of the same note.

The seven musical notes are:

C, D, E, F, G, A, B, C (repeated for example purposes)

A major scale is a diatonic scale. Take the previous succession of notes as a major scale (Actually, It is the scale C Major). The sequence of intervals between the notes of a major scale is:

whole, whole, half, whole, whole, whole, half

where "whole" stands for a whole tone (a red u-shaped curve in the figure), and "half" stands for a semitone (a red broken line in the figure). In this case, from C to D exist a whole tone, from D to E exist a whole tone, from E to F exist half tone, etc...

We have 2 components that affects the tone distance between notes. These are the Sharp symbol (♯) and the flat symbol (♭).

The Sharp symbol (♯) adds half tone to the note. Example. From C to D we mentioned that exists a whole tone, if we use C♯ instead C then from C♯ to D exists half tone.

The Flat symbol (♭) do the opposite of the Sharp symbol, it subtract half tone from the note. Example: From D to E we mentioned that exists a whole tone, if we use Db instead D then from Db to E exists a tone and a half.

By default, from Note to Note exist a whole tone except for E to F and B to C in where just half tone exists.

Note in some cases using enharmonic pitches can create an equivalent to a Major Scale. An example of this is C#, D#, E#, F#, G#, A#, B#, C# where E# and B# are enharmonic but the scale follows the sequence of a Major Scale.

## Challenge

Given a scale, output a truthy value if it is a Major Scale or equivalent, otherwise output a falsey value.

Rules

• Standard I/O method allowed
• Standard rules apply
• You don't need to take in consideration the 8th note. Assume the input will only consist of 7 notes
• Assume double flat (♭♭), double sharp (♯♯) or natural sign (♮) don't exist

## Test cases

C, D, E, F, G, A, B                 => true
C#, D#, E#, F#, G#, A#, B#          => true
Db, Eb, F, Gb, Ab, Bb, C            => true
D, E, Gb, G, A, Cb, C#              => true
Eb, E#, G, G#, Bb, B#, D            => true
-----------------------------------------------
C, D#, E, F, G, A, B                => false
Db, Eb, F, Gb, Ab, B, C             => false
G#, E, F, A, B, D#, C               => false
C#, C#, E#, F#, G#, A#, B#          => false
Eb, E#, Gb, G#, Bb, B#, D           => false

• So, E# is equal to F, and Cb equal to B? – Abigail Nov 8 '18 at 14:13
• and Cx (or C##) = D – SaggingRufus Nov 8 '18 at 18:32
• Btw, Pentatonic scales do not have one of each letter :v – Luis felipe De jesus Munoz Nov 8 '18 at 19:48
• @Neil Chromatic scales do not have unique letters and I'm sure there is a type of scale that doesnt follow an ascending order – Luis felipe De jesus Munoz Nov 8 '18 at 21:38
• I'm going to have to upvote this because @Neil downvoted it thank you very much – David Conrad Nov 9 '18 at 18:03

# Perl 6, 766563 59 bytes

-4 bytes thanks to Phil H

{221222==[~] (.skip Z-$_)X%12}o*>>.&{13*.ord+>3+?/\#/-?/b/}  Try it online! ### Explanation *>>.&{ ... } # Map notes to integers 13*.ord # 13 * ASCII code: A=845 B=858 C=871 D=884 E=897 F=910 G=923 +>3 # Right shift by 3: A=105 B=107 C=108 D=110 E=112 F=113 G=115 # Subtracting 105 would yield A=0 B=2 C=3 D=5 E=7 F=8 G=10 # but isn't necessary because we only need differences +?/\#/ # Add 1 for '#' -?/b/ # Subtract 1 for 'b' { }o # Compose with block (.skip Z-$_)        # Pairwise difference
X%12    # modulo 12
[~]  # Join
221222==     # Equals 221222

• If you're going to do a pairwise difference and modulo 12, you don't need to subtract 105; it's just an offset. -4 chars: tio.run/… – Phil H Nov 8 '18 at 16:46
• @PhilH Yes, of course. Thanks! – nwellnhof Nov 8 '18 at 16:53
• That's a really clever way of mapping the notes to their relative values, +1 from me! – Sok Nov 9 '18 at 15:44

# Node.js v10.9.0, 787671 69 bytes

a=>!a.some(n=>(a-(a=~([x,y]=Buffer(n),x/.6)-~y%61)+48)%12-2+!i--,i=3)


Try it online!

### How?

Each note $$\n\$$ is converted to a negative number in $$\[-118,-71]\$$ with:

[x, y] = Buffer(n) // split n into two ASCII codes x and y
~(x / .6)          // base value, using the ASCII code of the 1st character
- ~y % 61          // +36 if the 2nd character is a '#' (ASCII code 35)
// +38 if the 2nd character is a 'b' (ASCII code 98)
// +1  if the 2nd character is undefined


Which gives:

  n   | x  | x / 0.6 | ~(x / 0.6) | -~y % 61 | sum
------+----+---------+------------+----------+------
"Ab" | 65 | 108.333 |    -109    |    38    |  -71
"A"  | 65 | 108.333 |    -109    |     1    | -108
"A#" | 65 | 108.333 |    -109    |    36    |  -73
"Bb" | 66 | 110.000 |    -111    |    38    |  -73
"B"  | 66 | 110.000 |    -111    |     1    | -110
"B#" | 66 | 110.000 |    -111    |    36    |  -75
"Cb" | 67 | 111.667 |    -112    |    38    |  -74
"C"  | 67 | 111.667 |    -112    |     1    | -111
"C#" | 67 | 111.667 |    -112    |    36    |  -76
"Db" | 68 | 113.333 |    -114    |    38    |  -76
"D"  | 68 | 113.333 |    -114    |     1    | -113
"D#" | 68 | 113.333 |    -114    |    36    |  -78
"Eb" | 69 | 115.000 |    -116    |    38    |  -78
"E"  | 69 | 115.000 |    -116    |     1    | -115
"E#" | 69 | 115.000 |    -116    |    36    |  -80
"Fb" | 70 | 116.667 |    -117    |    38    |  -79
"F"  | 70 | 116.667 |    -117    |     1    | -116
"F#" | 70 | 116.667 |    -117    |    36    |  -81
"Gb" | 71 | 118.333 |    -119    |    38    |  -81
"G"  | 71 | 118.333 |    -119    |     1    | -118
"G#" | 71 | 118.333 |    -119    |    36    |  -83


We compute the pairwise differences modulo $$\12\$$ between these values.

The lowest possible difference between 2 notes is $$\-47\$$, so it's enough to add $$\4\times12=48\$$ before applying the modulo to make sure that we get a positive result.

Because we apply a modulo $$\12\$$, the offset produced by a '#' is actually $$\36 \bmod 12 = 0\$$ semitone, while the offset produced by a 'b' is $$\38 \bmod 12 = 2\$$ semitones.

We are reusing the input variable $$\a\$$ to store the previous value, so the first iteration just generates $$\\text{NaN}\$$.

For a major scale, we should get $$\[ \text{NaN}, 2, 2, 1, 2, 2, 2 ]\$$.

We use the counter $$\i\$$ to compare the 4th value with $$\1\$$ rather than $$\2\$$.

• Great approach, much more interesting than my answer – Skidsdev Nov 19 '18 at 15:49

# JavaScript (Node.js), 150131 125 bytes

l=>(l=l.map(x=>'C0D0EF0G0A0B'.search(x)+(x=='#'|-(x=='b')))).slice(1).map((n,i)=>(b=n-l[i])<0?2:b)+""=='2,2,1,2,2,2'


Try it online!

-19 bytes thanks to Luis felipe
-6 bytes thanks to Shaggy

## Ungolfed:

function isMajor(l) {
// Get tone index of each entry
let array = l.map(function (x) {
// Use this to get indices of each note, using 0s as spacers for sharp keys
let tones = 'C0D0EF0G0A0B';
// Get the index of the letter component. EG D = 2, F = 5
let tone = tones.search(x);
// Add 1 semitone if note is sharp
// Use bool to number coercion to make this shorter
tone += x == '#' | -(x=='b');
});
// Calculate deltas
let deltas = array.slice(1).map(function (n,i) {
// If delta is negative, replace it with 2
// This accounts for octaves
if (n - array[i] < 0) return 2;
// Otherwise return the delta
return n - array[i];
});
// Pseudo array-comparison
return deltas+"" == '2,2,1,2,2,2';
}

• [...'C0D0EF0G0A0B'] instead of 'C0D0EF0G0A0B'.split('') and +"" instead of .toString() to save some bytes – Luis felipe De jesus Munoz Nov 8 '18 at 14:36
• x=='#'|-(x=='b') instead of x=='#'?1:(x=='b'?-1:0) save some bytes too – Luis felipe De jesus Munoz Nov 8 '18 at 14:39
• @LuisfelipeDejesusMunoz Oh nice thanks! I can't believe I forgot about array expansion and adding an empty string – Skidsdev Nov 8 '18 at 14:43
• "If delta is negative, replace it with 2" sounds wrong. I think you need to take the difference modulo 12. – nwellnhof Nov 8 '18 at 15:58
• @nwellnhof In my tests, all major scales either had the correct deltas to begin with, or, if they spanned an octave, had one delta at -10 rather than 2. Replacing negative deltas fixes that. I don't think -10 % 12 == 2. Although come to think of it this might fail in some cases... – Skidsdev Nov 8 '18 at 16:33

# Dart, 198 197 196 189 bytes

f(l){var i=0,j='',k,n=l.map((m){k=m.runes.first*2-130;k-=k>3?k>9?2:1:0;return m.length<2?k:m=='#'?k+1:m=='b'?k-1:k;}).toList();for(;++i<7;j+='${(n[i]-n[i-1])%12}');return'221222'==j;}  Try it online! Loose port of the old Perl 6 answer https://codegolf.stackexchange.com/a/175522/64722 f(l){ var i=0,j='',k, n=l.map((m){ k=m.runes.first*2-130; k-=k>3?k>9?2:1:0; return m.length<2?k:m=='#'?k+1:m=='b'?k-1:k; }).toList(); for(;++i<7;j+='${(n[i]-n[i-1])%12}');
return'221222'==j;
}

• -1 byte by using ternary operators for #/b
• -1 byte by using ifs instead of ternaries for the scale shifts
• -7 bytes thanks to @Kevin Cruijssen

Old version :

# Dart, 210 bytes

f(l){var i=0,k=0,n={'C':0,'D':2,'E':4,'F':5,'G':7,'A':9,'B':11,'b':-1,'#':1},j='',y=[0,0];for(;++i<7;j+='${(y-y)%12}')for(k=0;k<2;k++)y[k]=n[l[i-k]]+(l[i-k].length>1?n[l[i-k]]:0);return'221222'==j;}  Try it online! Ungolfed: f(l){ var i=0,k=0,n={'C':0,'D':2,'E':4,'F':5,'G':7,'A':9,'B':11,'b':-1,'#':1},j='',y=[0,0]; for(;++i<7;j+='${(y-y)%12}')
for(k=0;k<2;k++)
y[k]=n[l[i-k]]+(l[i-k].length>1?n[l[i-k]]:0);

return'221222'==j;
}


A whole step is 2, a quarter is 1. Mod 12 in case you jump to a higher octave. Iterates through all notes and computes the difference between the ith note and the i-1th note. Concatenates the result and should expect 221222 (2 whole, 1 half, 3 wholes).

• -2 bytes by not assigning 0 to k
• -4 bytes by using j as a String and not a List
• -6 bytes thanks to @Kevin Cruijssen by removing unnecessary clutter in loops
• I don't know Dart, but parts are similar as Java. Therefore: changing i=1 to i=0 can reduce a byte by changing for(;i<7;i++) to for(;++i<7;). In addition, the brackets {} can be removed around that loop, by putting the j+=... inside the third part of the loop: for(;++i<7;j+='\${(y-y)%12}'). And one last thing is changing return j=='221222'; to return'221222'==j; to get rid of the space. -6 (210 bytes) after these modifications. – Kevin Cruijssen Nov 8 '18 at 15:02
• Thanks, didn't know about those tricks for the loops – Elcan Nov 8 '18 at 15:28
• Np. In your new 196-bytes version you can also golf it to 189 bytes by changing if(k>9)k--;if(k>3)k--; to k-=k>3?k>9?2:1:0; and k+=m.length<2?0:m=='#'?1:m=='b'?-1:0;return k; to return m.length<2?k:m=='#'?k+1:m=='b'?k-1:k;. :) – Kevin Cruijssen Nov 8 '18 at 17:49
• Damn, I still have a lot to learn it seems, thanks ! – Elcan Nov 8 '18 at 19:00
• Well, I've been golfing for 2.5 years now, and even I get tips on things to golf all the time. :) It's pretty easy to miss something yourself initially, and with time you think about different ways to golf. :) Tips for golfing in <all languages> might be interesting to read through if you haven't yet. And some of the Tips for golfing in Java might also be applicable in Dart, since the golfs I did in your answers were based on my Java knowledge, since this is the first time I see Dart. ;) – Kevin Cruijssen Nov 8 '18 at 20:38

# C (gcc), -DA=a[i] + 183 = 191 bytes

f(int*a){char s,b,h=0,i=0,j=0,d;for(;A;A==35?b[i-h++-1]++:A^98?(b[i-h]=A*13>>3):b[i-h++-1]--,i++);for(;j<7;d=(b[j]-b[j-1])%12,d=d<0?d+12:d,s[j++-1]=d+48);a=!strcmp(s,"221222");}


Try it online!

Based on the Perl answer.

Takes input as a wide string.

Ungolfed:

int f(int *a){
char s, b;
int h, i, j;
h = 0;
for(i = 0; a[i] != NULL; i++){
if(a[i] == '#'){
b[i-h-1] += 1;
h++;
}
else if(a[i] == 'b'){
b[i-1-h] -= 1;
h++;
}
else{
b[i-h] = (a[i] * 13) >> 3;
}
}
for(j = 1; j < 7; j++){
int d = (b[j] - b[j-1]) % 12;
d = d < 0? d + 12: d;
s[j-1] = d + '0';
}
return strcmp(s, "221222") == 0;
}


# Python 3, 175136134114 112 bytes

def f(t):r=[ord(x)//.6+ord(x[1:]or'"')%13-8for x in t];return[(y-x)%12for x,y in zip(r,r[1:])]==[2,2,1,2,2,2]


Try it online!

An one-liner Python 3 implementation.

Thanks to @Arnauld for idea of calculate tones using division and modulo.
Thanks to @Jo King for -39 bytes.

# [Python] 269 202 bytes

Improvements from Jo King:

p=lambda z:"A BC D EF G".index(z)+"b #".index(z[1:]or' ')-1
def d(i,j):f=abs(p(i)-p(j));return min(f,12-f)
q=input().replace(' ','').split(',')
print([d(q[i],q[i+1])for i in range(6)]==[2,2,1,2,2,2])


Try it!

Ungolfed, with test driver:

tone = "A BC D EF G"   # tones in "piano" layout
adj = "b #"            # accidentals

def note_pos(note):
if len(note) == 1:
note += ' '
n,a = note

def note_diff(i, j):
x, y = note_pos(i), note_pos(j)
diff = abs(x-y)
return min(diff, 12-diff)

def is_scale(str):
seq = str.replace(' ','').split(',')
div = [note_diff(seq[i], seq[i+1]) for i in (0,1,2,3,4,5)]
return div == [2,2,1,2,2,2]

case = [
("C, D, E, F, G, A, B", True),
("C#, D#, E#, F#, G#, A#, B#", True),
("Db, Eb, F, Gb, Ab, Bb, C", True),
("D, E, Gb, G, A, Cb, C#", True),
("Eb, E#, G, G#, Bb, B#, D", True),

("C, D#, E, F, G, A, B", False),
("Db, Eb, F, Gb, Ab, B, C", False),
("G#, E, F, A, B, D#, C", False),
("C#, C#, E#, F#, G#, A#, B#", False),
("Eb, E#, Gb, G#, Bb, B#, D", False),
]

for test, result in case:
print(test + ' '*(30-len(test)), result, '\t',
"valid" if is_scale(test) == result else "ERROR")

• Yes, I see the white space -- still inculcated with too much PEP-8, I'm afraid. I apparently missed something; is an execution link required here? – Prune Nov 9 '18 at 2:03
• Though, if you want the link, 202 bytes with some quick golfing. You could definitely golf some more by changing to a different input format – Jo King Nov 9 '18 at 2:28
• Ah ... I'm too used to Python returning the final expression as the process value. Thanks for the pointers and hints. – Prune Nov 9 '18 at 2:32
• You can get 156 bytes if you switch to a function taking a list of strings. Also, TIO has an auto formatter in the link section that you can use – Jo King Nov 9 '18 at 2:49
• @JoKing, you're perfectly welcome to edit this answer or post your own; commenting with a link separates the improvements by one level. – Prune Nov 9 '18 at 16:40

# Ruby, 109 bytes

->s{(0..11).any?{|t|s.map{|n|(w="Cef;DXg<E=Fhi>G j8A d9B:")[(w.index(""<<n.sum%107)/2-t)%12]}*''=='CfDX<=h'}}


Try it online!

# [Wolfram Language (Mathematica) + Music package], 114 bytes

I love music and found this interesting, but I was out playing real golf when this code golf opportunity came down the pike so my submission is a little tardy.

I figured I'd try this a totally different way, utilizing some actual music knowledge. It turns out the music package of Mathematica knows the fundamental frequency of the named notes. First I convert the input string into sequence of named notes. Next, I take the ratios of each successive note and double any that are less than 2 (to account for octave shift). Then I compare these ratios to the ratios of the Ionian scale which has roughly a 6% frequency difference between half notes and 12% between full notes.

More than half of the bytes spent here are to convert the input into named symbols.

.06{2,2,1,2,2,2}+1==Round[Ratios[Symbol[#~~"0"]&/@StringReplace[# ,{"b"->"flat","#"->"sharp"}]]/.x_/;x<1->2x,.01]&
`

Try it online!