Hi, I thought I had completely understood how pd handles the order of connections, but I can't explain what is going on in this simple case:

So, that's a beginner question. I know that I should assure the order from right to left using a [t b b], however I'm just trying to understand what's happening when I click the main bang, which is:

1. Connecting the upper delay first: Upper bang flashes first, lower bang later.
2. Connecting the lower delay first: Both bangs flash at the same time (lower bang receives a bang first), lower bang later

My thoughts:

1. Upper delay receives main bang and schedules it to 600ms later (600ms later it goes to upper bang and lower delay). Lower delay receives main bang and schedules it to 600ms later. 600ms have passed: Upper bang flashes; lower delay receives bang from upper delay thus overriding the message to the lower bang that should be flashing now (?) (because of the message from main bang) Why would that be the order? Finally 600ms later, lower bang flashes.

2. Lower delay receives main bang and schedules it to 600ms later. Upper delay receives main bang and schedules it to 600ms later. 600ms have passed: lower bang flashes. "At the same time" (but later), a bang comes from the upper delay. Upper bang flashes. 600ms later: lower bang flashes.

Could someone clarify how to think systematically about the order issue in general (and this case specifically) once for all or point me to a tutorial that handles the subject? Thanks.

@Cmaj7 Hm I'm not really sure what your question is, because you are thinking about everything correctly.

In case 1, you click the main [bang]. Because you connected the left-most [delay] first, it receives the "bang" a fraction of a second before the right [delay]. Thus, the left-most [delay] will output a "bang" a fraction of a second sooner than the right [delay]. This causes the right most delay to reset its timer and output a "bang" after another 600 ms.

In case 2, you click the main [bang]. Because you connected the right-most [delay] first, it receives the "bang" a fraction of a second before the left [delay]. Thus, the right-most [delay] will output a "bang" a fraction of a second sooner than the left [delay]. This means that the right [delay] will receive a second bang a fraction of a second after it outputs its own "bang", so it will "bang" again in 600ms.

Perhaps the only caveat I can see being important here is that if a [delay] receives a "bang", and then receives another "bang" before its "timer is up", then the internal "timer" is just reset. So for example, if you have a delay of [delay 600], and it receives a bang, only to receive another bang 300ms later, then it will output a "bang" effectively 900ms after the first "bang".

This same phenomenon is happening in your example, except the second bang--instead of arriving 300ms later as in the example in the previous paragraph--is coming in one control rate cycle, or cpu tick, later, which I believe is 1/20 of a second (someone who is knowledgable should correct me)

This is why the order in which you connect objects is so important in some circumstances. Although I've just restated what you said above so I'm not sure how to help you. All I can say is maybe read this: http://en.flossmanuals.net/pure-data/dataflow-tutorials/order-of-operations/

"Thus, the left-most [delay] will output a "bang" a fraction of a second sooner than the right [delay]. This causes the right most delay to reset its timer"

Yes! Actually you did help me . That's exactly the point of my confusion: when the lower delay receives two bangs "at the same time" but in different orders. The delay is reset in one case but not in the other. The last section of your link was very helpful: visualizing it as a tree with its rightmost branch being executed first makes it clear to think about the order. Thanks.