I'm looking for a well done patch for midi Euclidean rhythms generation, with the possibility to vary steps, pulses and offset.
Thanks for any advice.

[I use Purr Data on Ubuntu 18.04]

a.

very interesting. Two completely different ways: a matricial approach and an apparently simple and effective algorithm.
Ok, the graphic part is missing (with clocks and inscribed geometries ...) to "visualize" the rhythms, but the engine that generates the Euclidean rhythms works perfectly.
I have tested both patches, they are great.
Bye,

a.

@atux Can you please repost the links to the two patches you tested? I'm particularly interested in the arithmetic of post #7 but can't make sense of the text representation of @Stutter's patch (and all the hurleur links are broken for me).

Ok, I see that I was preceded.... :D
I add @kyro's patch with matrices: euclidean.pd

I tested @Stutter's patch as subpatch.
For example, with the triad: 3,0,8 the rhythm is obtained:

TUM ta ta TUM ta ta TUM ta

Bye,
a.

Mike Moreno has made these Euclidean rhythm things based on the Stutter patch I think...
https://github.com/MikeMorenoDSP/mian-

Thanks, at the link above the patch that interests me is:
eucgui-help.pd

But I don't know how to use the [midirealtimein] object, there is also no help file: connecting an external keyboard and pressing some keys nothing happens, not even when printing with [print( from outlets.
What should happen?

(I specify that [notein] object, for example, works correctly with my midi keyboard.)

Anyway I tested the patch by putting a metro and works fine.

[I use Purr Data on Ubuntu 18.04]

a.

Also @noDSP made this a while ago and I imagine the control values are the same as those you should receive on the left outlet of [midirealtimein]
midirealtimeout.pd

David.

[1] Is there a base rotation for Euclidean rhythms?
[2] For arbitrary E(k,n), what is the offset between these algorithms?

If you follow the algorithm in http://cgm.cs.mcgill.ca/~godfried/publications/banff.pdf, E(5,13) = 1001010010100, But if you use the much simpler one from @Stutter you get 1001001010010.

If you flip stutter's result three digits to the left you get banff's. It's a matter of a [- 3] somewhere.