Help me with my polyrhythms

Question

I'm a musician, and I need more polyrhythms in my life!

A polyrhythm occurs in music (and in nature) when two events (claps, notes, fireflies flashing etc.) are occurring at two different regular intervals. The two kinds of event happen a different number of times in the same interval.

If I tap with my left hand twice, and with my right hand 3 times, in the same space of time, it looks a little bit like this:

  ------
R . . .
L .  .  

The hyphens at the top denote the length of the polyrthmic pattern, which is the lowest common multiple or 2 and 3. This can be understood as the point at which the pattern repeats.

There's also a 'metarhythm', which is the pattern produced when either hand is tapping:

  ------
R . . .
L .  .  
M . ...

This is a simple, and very common polyrhythm, with a ratio of 3:2.

Let's just say I don't want to do a simple polyrhythm that I can work out in my head, so I need something to work it out for me. I could do it long-form on paper, or...

---

Rules:

• Write some code to generate and display a polyrhythm diagram, as described above.
• Any old language, try for the fewest bytes.
• Your code takes two arguments:
  • Number of taps with the Left hand (positive integer)
  • Number of taps with the right hand (positive integer)
• It will work out the length, which is the lowest common multiple for the two arguments.
• The top line will consist of two whitespace characters followed by hyphens displaying the length (length * '-')
• The second and third lines will show the pattern for the right and left hands:
  • It will start with an R or L, do denote which hand it is, followed by a space.
  • The interval for that hand is the length divided by it's argument.
  • The taps will start at the third character, denoted by any character you choose. From then on it will display the same character 'interval' characters apart.
  • It will not be longer than the length line.
• The fourth line is the metarhythm:
  • It will start with an upper case M, followed by a space.
  • From the third character onwards, it will show a character (any character you choose) in every position where there's a tap on either the right or the left hand.
• Trailing whitespace is irrelevant.

---

Test cases:

r = 3, l = 2

  ------
R . . .
L .  .  
M . ...

r = 4, l = 3

  ------------
R .  .  .  .    
L .   .   .    
M .  .. . ..

r = 4, l = 5

  --------------------
R .    .    .    .                     
L .   .   .   .   .      
M .   ..  . . .  ..

r = 4, l = 7

  ----------------------------
R .      .      .      .      
L .   .   .   .   .   .   .   
M .   .  ..   . . .   ..  .

r = 4, l = 8

  --------
R . . . . 
L ........
M ........

---

Happy golfing!

Answer 1

JavaScript (ES6), 131 bytes

Outputs 0 as the tap character.

r=>l=>`  ${g=n=>n?s.replace(/./g,(_,x)=>[,a=x%(k/r),x%=k/l,a*x][n]&&' '):++k%l|k%r?'-'+g():`-
`,s=g(k=0)}R ${g(1)}L ${g(2)}M `+g(3)

Try it online!

How?

We use the same helper function \$g()\$ for two different purposes.

When \$g()\$ is called with no argument or an argument equal to \$0\$, it recursively builds the hyphen string of length \$k=\operatorname{lcm}(l,r)\$ with a trailing linefeed:

g = _ => ++k % l | k % r ? '-' + g() : `-\n`

This string is saved in \$s\$.

When \$g()\$ is called with \$1 \le n \le 3\$, it generates a tap string by replacing each hyphen at position \$x\$ in \$s\$ with either a space or \$0\$:

g = n => s.replace(/./g, (_, x) => [, a = x % (k / r), x %= k / l, a * x][n] && ' ')

Answer 2

Java 11, 226 234 233 219 bytes

String h(int r,int l,int m){var s="";for(;m>0;)s+=m%r*(m--%l)<1?'.':32;return s;}

r->l->{int a=r,b=l,m;for(;b>0;b=a%b,a=m)m=b;m=r*l/a;return"  "+repeat("-",m)+"\nR "+h(m/r,m+1,m)+"\nL "+h(m/l,m+1,m)+"\nM "+h(m/r,m/l,m);}

Kind of lengthy; too bad Java does not have an lcm() function. Try it online here (TIO does not have Java 11 yet, so this uses a helper method instead of String.repeat()).

My initial version took the interval between taps instead of the number of taps. Fixed now. Thanks to Kevin Cruijssen for golfing 1 byte.

Ungolfed:

String h(int r, int l, int m) { // helper function returning a line of metarhythm; parameters are: tap interval (right hand), tap interval (left hand), length
    var s = ""; // start with an empty String
    for(; m > 0; ) // repeat until the length is reached
        s += m % r * (m-- % l) < 1 ? '.' : 32; // if at least one of the hands taps, add a dot, otherwise add a space (ASCII code 32 is ' ')
    return s; // return the constructed line
}

r -> l -> { // lambda taking two integers in currying syntax and returning a String
    int a = r, b = l, m; // duplicate the inputs
    for(; b > 0; b = a % b, a = m) // calculate the GCD of r,l using Euclid's algorithm:
        m=b; // swap and replace one of the inputs by the remainder of their division; stop once it hits zero
    m = r * l / a; // calculate the length: LCM of r,l using a=GCD(r,l)
    return // build and return the output:
    "  " + "-".repeat(m) // first line, m dashes preceded by two spaces
    + "\nR " + h(m / r, m + 1, m) // second line, create the right-hand rhythm; by setting l = m + 1 for a metarhythm, we ensure there will be no left-hand taps
    + "\nL " + h(m / l, m + 1, m) // third line, create the left-hand rhythm the same way; also note that we pass the tap interval instead of the number of taps
    + "\nM " + h(m / r, m / l, m); // fourth line, create  the actual metarhythm
}

Answer 3

Python 2, 187 185 183 174 166 156 148 147 145 bytes

Uses - as the tap character

a,b=r,l=input()
while b:a,b=b,a%b
w=r*l/a
for x,y,z in zip(' RLM',(w,r,l,r),(w,r,l,l)):print x,''.join('- '[i%(w/y)!=0<i%(w/z)]for i in range(w))

Try it online!

---

Saved:

• -2 bytes, thanks to Jonathan Frech

Answer 4

Perl 6, 85 80 78 bytes

-2 bytes thanks to Jo King.

{«'  'R L M»Z~'-'x($!=[lcm] @_),|(@_.=map:{' '~(0~' 'x$!/$_-1)x$_}),[~|] @_}

Try it online!

Returns a list of four lines.

Answer 5

Python 2, 185 228 223 234 249 bytes

def f(r,l):
     c='.';d=' ';M,R,L=[r*l*[d]for _ in d*3]
     for i in range(r*l):
      if i%r<1:L[i]=M[i]=c
      if i%l<1:R[i]=M[i]=c
      if r<R.count(c)and l<L.count(c):R[i]=L[i]=M[i]=d;break
     print d,i*'-','\nR',''.join(R),'\nL',''.join(L),'\nM',''.join(M)

Try it online!

Answer 6

Jelly, 32 bytes

æl/Ḷ%Ɱµa/ṭ=0ị⁾. Z”-;ⱮZ“ RLM”żK€Y

Try it online!

Takes input as a list [L,R].

æl/       Get LCM of this list.
   Ḷ      Range [0..LCM-1]
    %Ɱ    Modulo by-each-right (implicitly the input, [L,R]):
           [[0%L ... (LCM-1)%L], [0%R ... (LCM-1)%R]]
µ         Take this pair of lists, and:
 a/ṭ      Append their pairwise AND to the pair.
    =0    Is zero? Now we have a result like:
              [[1 0 0 1 0 0 1 0 0 1 0 0 1 0 0]
               [1 0 0 0 0 1 0 0 0 0 1 0 0 0 0]
               [1 0 0 1 0 1 1 0 0 1 1 0 1 0 0]]

ị⁾.       Convert this into dots and spaces.
Z”-;ⱮZ    Transpose, prepend a dash to each, transpose. Now we have
              ['---------------'
               '.  .  .  .  .  '
               '.    .    .    '
               '.  . ..  .. .  ']

“ RLM”ż       zip(' RLM', this)
       K€     Join each by spaces.
         Y    Join the whole thing by newlines.

Answer 7

C (gcc), 204 bytes

p(s){printf(s);}
g(a,b){a=b?g(b,a%b):a;}
h(r,l,m){for(;m;)p(m%r*(m--%l)?" ":".");}
f(r,l,m,i){m=r*l/g(r,l);p("  ");for(i=m;i-->0;)p("-");p("\nR ");h(m/r,m+1,m);p("\nL ");h(m/l,m+1,m);p("\nM ");h(m/r,m/l,m);}

Port of my Java answer. Call with f(number_of_right_hand_taps, number_of_left_hand_taps). Try it online here.

Answer 8

Pyth, 53 bytes

j.b+NYc"  L R M "2++*\-J/*FQiFQKm*d+N*\ t/JdQsmeSd.TK

Definitely room to golf. Will do so when I have time.
Try it here

Explanation

j.b+NYc"  L R M "2++*\-J/*FQiFQKm*d+N*\ t/JdQsmeSd.TK
                       J/*FQiFQ                        Get the LCM.
                    *\-                                Take that many '-'s.
                               Km*d+N*\ t/dJQ          Fill in the taps.
                                             smeSd.TK  Get the metarhythm.
                  ++                                   Append them all.
      c"  L R M "2                                     Get the prefixes.
 .b+NY                                                 Prepend the prefixes.
j                                                      Join with newlines.

Answer 9

C# (Visual C# Interactive Compiler), 254 bytes

---

Golfed Try it online!

(r,l)=>{int s=l>r?l:r,S=s;while(S%l>0|S%r>0)S+=s;string q(int a){return"".PadRight(S/a,'.').Replace(".",".".PadRight(a,' '));}string R=q(S/r),L=q(S/l),M="";s=S;while(S-->0)M=(R[S]+L[S]>64?".":" ")+M;return"  ".PadRight(s+2,'-')+$"\nR {R}\nL {L}\nM {M}";}

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Ungolfed

( r, l ) => {
    int
        s = l > r ? l : r,
        S = s;

    while( S % l > 0 | S % r > 0 )
        S += s;

    string q( int a ) {
        return "".PadRight( S / a, '.' ).Replace( ".", ".".PadRight( a, ' ' ) );
    }

    string
        R = q( S / r ),
        L = q( S / l ),
        M = "";

    s = S;

    while( S-- > 0 )
        M = ( R[ S ] + L[ S ] > 64 ? "." : " " ) + M;

    return "  ".PadRight( s + 2, '-') + $"\nR {R}\nL {L}\nM {M}";
}

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Full code

Func<Int32, Int32, String> f = ( r, l ) => {
    int
        s = l > r ? l : r,
        S = s;

    while( S % l > 0 | S % r > 0 )
        S += s;

    string q( int a ) {
        return "".PadRight( S / a, '.' ).Replace( ".", ".".PadRight( a, ' ' ) );
    }

    string
        R = q( S / r ),
        L = q( S / l ),
        M = "";

    s = S;

    while( S-- > 0 )
        M = ( R[ S ] + L[ S ] > 64 ? "." : " " ) + M;

    return "  ".PadRight( s + 2, '-') + $"\nR {R}\nL {L}\nM {M}";
};

Int32[][]
    testCases = new Int32[][] {
        new []{ 3, 2 },
        new []{ 4, 3 },
        new []{ 4, 5 },
        new []{ 4, 7 },
        new []{ 4, 8 },
    };

foreach( Int32[] testCase in testCases ) {
    Console.Write( $" Input: R: {testCase[0]}, L: {testCase[1]}\nOutput:\n{f(testCase[0], testCase[1])}" );
    Console.WriteLine("\n");
}

Console.ReadLine();

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Releases

• v1.0 - 254 bytes - Initial solution.

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Notes

• None

Answer 10

Charcoal, 52 bytes

≔θζＷ﹪ζη≧⁺θζζ↙≔⮌Ｅζ⟦¬﹪×ιθζ¬﹪×ιηζ⟧ζＦζ⊞ι⌈ι↓Ｅζ⭆ι§ .λ←↓RLM

Try it online! Link is to verbose version of code. Explanation:

≔θζＷ﹪ζη≧⁺θζ

Calculate the LCM of the inputs by taking the first multiple of R that's divisible by L.

ζ↙

Print the LCM, which automatically outputs the necessary row of -s. Then move to print the rhythm from right to left.

≔⮌Ｅζ⟦¬﹪×ιθζ¬﹪×ιηζ⟧ζ

Loop over the numbers from the LCM down to 0 and create an array of lists representing the beats of the right and left hands.

Ｆζ⊞ι⌈ι

Loop over the beats and add the metarhythm.

↓Ｅζ⭆ι§ .λ

Print the reversed beats downwards, but as this is an array they end up leftwards.

←↓RLM

Print the header.

Answer 11

Ruby, 130 126 bytes

->*a{puts"  "+?-*s=a[0].lcm(a[1])
r,l=a.map!{|e|(?.+' '*(s/e-1))*e}
[?R,?L,?M].zip(a<<r.gsub(/ /){l[$`.size]}){|e|puts e*" "}}

Try it online!

Answer 12

Python 2, 117 bytes

a,b=input();n=a
while n%b:n+=a
for i in-1,1,2,3:print'_RLM '[i],''.join(' -'[i%2>>m*a%n|i/2>>m*b%n]for m in range(n))

Try it online!

Answer 13

Pyth, 49 bytes

J/*FQiFQjC+c2" RLM    "ms@L" -"!M++0d*Fdm%Ld/LJQJ

Expects input in the form [r,l]. Uses - to display taps. Try it online here, or verify all test cases at once here.

J/*FQiFQjC+c2" RLM    "ms@L" -"!M++0d*Fdm%Ld/LJQJ   Implicit: Q=eval(input())
 /*FQiFQ                                            Compute LCM: (a*b)/(GCD(a,b))
J                                                   Store in J
                                        m       J   Map d in [0-LCM) using:
                                            /LJQ      Get number of beats between taps for each hand
                                         %Ld          Take d mod each of the above
                                                    This gives a pair for each beat, with 0 indicating a tap
                       m                            Map d in the above using:
                                     *Fd              Multiply each pair (effecively an AND)
                                 ++0d                 Prepend 0 and the original pair
                               !M                     NOT each element
                        s@L" -"                       Map [false, true] to [' ', '-'], concatenate strings
                                                    This gives each column of the output
           c2" RLM    "                             [' RLM','    ']
          +                                         Prepend the above to the rest of the output
         C                                          Transpose
        j                                           Join on newlines, implicit print

Answer 14

R, 161 149 146 bytes

function(a,b){l=numbers::LCM(a,b)
d=c(0,' ')
cat('  ',strrep('-',l),'\nR ',d[(x<-l:1%%a>0)+1],'\nL ',d[(y<-l:1%%b>0)+1],'\nM ',d[(x&y)+1],sep='')}

Try it online!

I definitely feel like there's room for improvement here, but I tried a few different approaches and this is the only one that stuck. Getting rid of the internal function definition would make me quite happy, and I tried a bunch of restructures of the cat() to make it happen. Nevermind, as soon as I posted I realised what I could do. Still definitely some efficiency savings to be found.

There are other LCM functions in libraries with shorter names, but TIO has numbers and I considered that more valuable at this point.

Answer 15

C++ (gcc), 197 bytes

int f(int a,int b){std::string t="  ",l="\nL ",r="\nR ",m="\nM ";int c=-1,o,p=0;for(;++p%a||p%b;);for(;o=++c<p;t+="-")l+=a*c%p&&++o?" ":".",r+=b*c%p&&++o?" ":".",m+=o-3?".":" ";std::cout<<t+l+r+m;}

Try it online!