Talk:Euclidean rhythm

Anon 9/4/26: `So, there's a pretty weird issue here. The step-through of the algorithm presented shows the connection to the Euclidean algorithm nicely, but does not actually generate the rhythms given in the examples section. For instance, distributing 3 beats over 8 time steps gives x..x.x.. when using the approach shown in the "Summary of Algorithm" section, while the equivalent example is x..x..x.

I read Toussaint's original paper and it looks like he uses a different algorithm/approach entirely. It seems like he uses the Bjorklund algorithm, which actually doesn't utilize the Euclidean algorithm or the sequence of r numbers. Would it make sense to include some information about that, or perhaps a summary of Bjorklund, or some other note basically saying that the Summary of Algorithm section doesn't produce the canonical results in this article?` — Preceding unsigned comment added by ~2026-47920-50 (talk) 02:58, 5 September 2026 (UTC)

I did some more work on this and I'm actually wrong here (I will note that this is not original research per se: I'm doing a personal project related to this, using the article as a base, and writing up results coming from that project as I find them). The current visual step through of the algorithm does reflect what Toussaint outlines in the paper, but it discusses it in such vague terms as to be hard to follow.

Here's what the paper says:

"Bjorklund’s algorithm will be described simply by using one of his examples. Consider a sequence with

n = 13 and k = 5. Since 13 − 5 = 8, we start by considering a sequence consisting of 5 one’s followed by

8 zero’s which should be thought of as 13 sequences of one bit each:

[1 1 1 1 1 0 0 0 0 0 0 0 0]

We begin moving zero’s by placing a zero after each one, to produce five sequences of two bits each, with

three zero’s remaining:

[10] [10] [10] [10] [10] [0] [0] [0]

Next we distribute the three remaining zeros in a similar manner, by placing a [0] sequence after each [10]

sequence to obtain:

[100] [100] [100] [10] [10]

Now we have three sequences of three bits each, and a remainder of two sequences of two bits each.

Therefore we continue in the same manner, by placing a [10] sequence after each [100] sequence to obtain:

[10010] [10010] [100]

The process stops when the remainder consists of only one sequence (in this case the sequence [100]), or

we run out of zero’s. The final sequence is thus the concatenation of [10010], [10010], and [100]:

[1 0 0 1 0 1 0 0 1 0 1 0 0]

Note that one could proceed one step further in this process by inserting [100] into [10010] [10010]. How-

ever, Bjorklund argues that since the sequence is cyclic it does not matter (hence his stopping rule). Bjork-

lund [5] shows that the final sequence may be computed from the initial sequence using O(n) arithmetic

operations in the worst case."

So this does follow the example given, but the phrasing is difficult to follow because it lays out the xs and .s in a justified grid when Bjorklund's algorithm uses strings in an array.

Notably, Bjorklund's algorithm can use the Euclidean algorithm to determine how many items from the end to put on the strings at the beginning of the array at any given step. The sequence of r values from the Euclidean algorithm in the case of 5 distributed over 13 is [5, 3, 2, 1, 0] and Bjorklund's distributes 5 [.] items first, then 3 [.] items, then 2 [x.] items, then 1 [x..] item, then combines.

But this took reading the original paper and some experimentation to figure out on my part.

Would it make sense to rewrite the section using Bjorklund's as the clear example ,and highlighting the connection to the output of the Euclidean algorithm alongside a more functional algorithm for deriving a canonical rhythm? ~2026-47920-50 (talk) 21:53, 5 September 2026 (UTC)

Requested move

Euclidean Rhythm → Euclidean rhythm – Unnecesary capitalisation Rob Kam (talk) 18:51, 7 November 2014 (UTC)

• Support sources don't capitalise it either. Stickee (talk) 07:47, 8 November 2014 (UTC)
• Move done as uncontroversial request. Dekimasuよ! 23:45, 9 November 2014 (UTC)