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Re^15: Finding All Paths From a Graph From a Given Source and End Nodeby BrowserUk (Patriarch) |
on May 22, 2011 at 02:58 UTC ( [id://906110]=note: print w/replies, xml ) | Need Help?? |
Hm. That suggests that having transitioned from the A state to the B state, you cannot go backwards to the A state. But then you have two sentences that have an overriding termination (or perhaps extension) criteria:
And in your example, having transitioned from c-d, the reaction(s) can find their way back to the c state; and then leave via a different route. Your path:A->B->C->D->G->H->C->J->K So my question is, if the reaction can go from c-d once; and can find its way back to c; why can it not transition c-d a second time? Which would make the path a-b-c-d-g-h-c-d-e an elongation of the path you've identified as legitimate: A->B->C->D->E Thinking, I hope logically, about a subject I know next to nothing about, I considered the possibility that whatever enzymes are in the original mix, when the C state occures it preferentially reacts with whichever enzyme(s) are required to transition to state D. And in doing so, depletes the mix of that(those) enzymes. Which means that when the mix arrives at the C state a second time, those enzymes are no longer present so it can now react with other enzymes that cause it to transition to state J. Is that a viable, even if technically inaccurate, model of what goes on? If so, I think an efficient solution is possible. Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
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