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0 pointsargv_empty9y ago0 comments

It sounds like you're modeling the state space wrong. In a concurrent system, the state is not just one thread's (or one processor's, or one machine's) internal data (including program counter). That would be like saying a (single-threaded) program's state is the contents of the registers and topmost stack frame. The state in a concurrent system is the combination of every thread's internal state. And what actually happens is a sequence of such arrangements. One arrangement is not realized without replacing the previous one.

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Const-me9y ago

I’m aware the state you were talking includes state of all threads.

When you combine several sequences of state transitions (one sequence per thread), you don’t get another sequence of state transitions. When two CPU cores perform two state transitions at the same time, you typically cannot determine whichever of those transitions happened first.

We might have different definitions what’s sequence. I mean this: https://en.wikipedia.org/wiki/Sequence As you see, global order is required for bunch of elements (in this case, state transitions) to form a sequence. And in concurrent and especially parallel programming, there’s no global order for those transitions. Hence, those transitions don’t form a sequence.

argv_emptyOP9y ago

No, we just have different notions of system state. If we have a pair of, say, Turing machine heads on one tape that aren't necessarily moving in lockstep, I say that the system state is the tape contents, the two heads' locations, and the two heads' states. A transition for this system can be either or both heads moving either direction, overwriting the symbol they're pointing at, and changing its own internal state. The state {tape="000111000", head0loc=0, head0state=2, head1loc=4, head1state=6} might transition to {tape="000101000", head0loc=0, head0state=2, head1loc=5, head1state=2}, or it might transition to {tape="100111000", head0loc=1, head0state=2, head1loc=4, head1state=6}, or to {tape="100101000", head0loc=1, head0state=2, head1loc=5, head1state=2}. In a given run of the system, one of those is what actually happens. Then another happens after that, and so on. So we get a sequence of them.

A transition is not one machine head changing its position and state. {head0loc=0, head0state=2}->{head0loc=1, head0state=2} is not a transition because that does not identify a pair of whole-system states.

Const-me9y ago

> one of those is what actually happens

No it’s not. If the heads aren’t moving in lockstep, what actually happens is you’ll extremely rarely have a moment of time when both locations are whole numbers.

At one moment you have { head1loc=0, head2loc=4.17 }, head #1 reached the cell position on the tape, head #2 is still moving. After a while, you’ll have { head1loc=0.872, head2loc=5}, head #1 is on its way, head #2 reached the cell position.

Looks like for a Turing machine with two asynchronously moving heads, there’s no usable concept of “state” at all. That’s very close to what happens when doing actual parallel programming on real hardware.

Good luck applying your idea of computation as a sequence of state transitions to that.

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Const-me9y ago

I’m aware the state you were talking includes state of all threads.

argv_emptyOP9y ago

Const-me9y ago

> one of those is what actually happens

No it’s not. If the heads aren’t moving in lockstep, what actually happens is you’ll extremely rarely have a moment of time when both locations are whole numbers.

Good luck applying your idea of computation as a sequence of state transitions to that.

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