Current question to explore: can we (sometimes) derive 5cdc from o6c4c?

I have experimentally found, that sometimes we can take o6c4c solution, find weights for the cycles and produce some nowhere–zero 5–flow.

Experiments also show that we can always construct some (3,3)–flow parity–pair–cover (which always leads to 5cdc) from any nowhere–zero 5–flow.

We also have another connection between 6c4c and 5cdc — through Petersen colouring.

And it is also known, that there is no obvious oriented version of Petersen colouring.

But it looks like we could have 2 different connections between 6c4c and 5cdc (more on that later).

So the plan is following:

• take some graph, e. g. 18g1 (more on that notation later);
• generate all of its Petersen colourings, at first in the format of poor and rich edges, and then just any possible normal/Petersen colouring/mapping;
• take also Petersen graph - generate all possible 6c4c solutions (there is just 1) and all possible 5cdc solutions (about 10 for 96555 configuration and about 30 for 86655 configuration);
• now we combine all of 18g1 Petersen colourings with all possible 6c4c–5cdc pairs coming from Petersen graph;
• after that we take some o6c4c solution for 18g1;
• we find all compatible nowhere–zero 5–flows for it;
• we try somehow to find all (3,3)–flow parity–pair–cover, compatible with these nz5–flows;
• we find all 5cdc from these (3,3)–flow parity–pair–covers;
• …
• PROFIT!