A little demonstration why trying to calculate _all_ path is a dangerous thing.

let Nx be the graph of a x dimensional cube where the edges are bidirectionally connections of 2^x nodes representing the corners.

  101--111
  /:   /|
001--011|
 |100·|110          N3
 |·   |/
000--010

[download]

When calculating paths between extremal corners you will see:

N2 (the square) has 4 corners and 2 such paths

N3 (the normal cube) has 8 corners and 18 paths (e.g. 000->001->011->010->110->100->101->111 )

N4 has 16 corners and already 6432 paths

N5 has 32 corners and the program to calculate all paths is still running and will likely never finish!

I really doubt that such a general question is a clever approach. (for comparison the number of _shortest_ paths is "only" x! i.e. 5!=120 for N5)

IMHO not only a clear case of an XY question but already XYZ, any underlying question is either far more trivially answered¹ or is simply not solvable.

This approach to precalculate all path will potentially damage the hardware.

Cheers Rolf

¹) e.g. can a node K be part of a path from B to E?

In reply to Re: Short Circuiting DFS Graph Traversal by LanX
in thread Short Circuiting DFS Graph Traversal by Limbic~Region

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