So in Question 8 part 2 you had multiple cycles through a graph. You started on all "A" nodes and you had to find the first time at which you were entirely on "Z" nodes.
The intended solution was seemingly to calculate the length of the cycle (as in, until you were on a "Z" node) for each starting node separately. Once you'd done this, you could find the LCM of these cycle lengths to find the overall period of the cycle (~10^13).
This solution might not work if your cycles weren't constant length -- for example imagine when walking your graph you found your i_th Z-node after {6, 7, 6, 7, 6, 7, 6, 7 ...} steps; or perhaps {6, 7, 7, 7, ...}. In the test data, this didn't occur - it was always {n, n, n, n, ...}.
I can give another example perhaps - consider you're asked to find the last 3 digits of 2^n for n >= 1. You start off calculating: 002, 004, 008, ... eventually you get to 2^103 which ends in 008. This means that the cycle length would be {103, 101, 101, 101, 101, ...} since it'll never get back to 002 or 004. Solving this is a bit more difficult than the constant cycle lengths, since it's not a simple LCM.
Is there any theorem that puts limits on how irregular the cycles can be in a general case of tape length L with N options corresponding to N outputs from each graph node?
What has to be constant is the cycle length for (node, tape instruction #) pairs, because all the state to determine every future step is based on the current node and tape position; if both are the same, the path will be the same as before. I think, at least without advanced math, the only thing to rely on for "true" loops is identical pairs of (instr #, node), not just instr # or tape steps or node individually.
I haven't spent that much time thinking about it, but my guess is that there may or may not be:
* a "tail" -- some initial part of the path before you get into a true cycle (like the 002, 004 in the example above)
* an inner repeating cycle before you reach a node you've seen before.
In the case given |t| = 0 and |c| = 1, but it's easy to construct a more complex example with nodes (A, B, C, D), edges (A->B, B->C, C->D, D->B), and 'ending nodes' being B and D. In this case left and right paths go to the same node. This case would have a tail of length 1, and then the inner cycles would be of length {2, 1, 2, 1 ...}.
As a result, valid 'ending states' (Z-nodes) for this graph would be after {1, 3, 4, 6, 7, 9, 10, ...} steps.
I can see the case of an aperiodic first cycle length, but given that each vertex in the graph created by the input -> output nodes of the traversal path has only one outgoing edge, is it possible for the cyclic graph from a1 to some z1 to have an inconsistent length? And is there ever going to be some z2 for which the aperiodic cycle length from a1 to either z1 or z2 results in a better answer to the problem given that the periodic cycle length of a1 to z1 is shorter than the cycle length of a1 to z2? Please forgive me if I am not using the correct graph theory terms.
The intended solution was seemingly to calculate the length of the cycle (as in, until you were on a "Z" node) for each starting node separately. Once you'd done this, you could find the LCM of these cycle lengths to find the overall period of the cycle (~10^13).
This solution might not work if your cycles weren't constant length -- for example imagine when walking your graph you found your i_th Z-node after {6, 7, 6, 7, 6, 7, 6, 7 ...} steps; or perhaps {6, 7, 7, 7, ...}. In the test data, this didn't occur - it was always {n, n, n, n, ...}.
I can give another example perhaps - consider you're asked to find the last 3 digits of 2^n for n >= 1. You start off calculating: 002, 004, 008, ... eventually you get to 2^103 which ends in 008. This means that the cycle length would be {103, 101, 101, 101, 101, ...} since it'll never get back to 002 or 004. Solving this is a bit more difficult than the constant cycle lengths, since it's not a simple LCM.