[seqfan] Re: eulerian orientations of hypercube graphs
hv at crypt.org
hv at crypt.org
Sat Oct 29 12:49:29 CEST 2022
Ah, I see at least how to take advantage of the indistinguishability.
For d=4, for example, pick an initial square face ABCD. Let "AB"
represent the edge going out from A, and in to B.
We pick A as out first vertex, and assign AB, AC as its two out-edges,
with EA, IA as its two in-edges. AB, AC now represent a pair of
indistinguishable dimensions, as do AE, AI, and we will multiply
the results by 4C2 = 6.
For the first pair, we can now try setting a) BD and CD, b) BD and DC,
and c) DB and DC: we now count a + 2b + c, since (BD + DC) is
symmetric with (DB + CD). For each combination, we similarly set
3 selections of EM, IM for the second pair. In this case we end up
testing 9 sets out of 16; the same strategy for d=6 tests 16 sets
out of 64 (so not the 36x improvement I had hoped for).
That suggests to me that the insight from d=4 might come by classifying
the structures of the cube faces.
:Ah, that certainly suggests brute force won't get far even with d=6.
:Since all directions look alike to the first vertex, we can choose any
:arrangement of the 6 edges and multiply the results by 20.
:Now the three dimensions we choose for "out" edges from that vertex
:are indistinguishable, as are the other three, which ought to reduce
:the problem another 36-fold - but I don't see how to take advantage
:of that, and it would still leave at least 4 x 10^22 to count.
:Maybe if the 2970 solutions for d=4 could be classified it might
:yield some insight.
:Zach DeStefano <zachdestefano at gmail.com> wrote:
::I found some bounds for this problem.
::According to "Bounds on the Number of Eulerian Orientations" by A.
::Schrijver (1983), the number of Eulerian orientations of a 2k-regular graph
:on n vertices is between (2^-k (2k choose k))^n and sqrt((2k choose k)^n).
::>From this we have that the value for d = 6 is between 2.9 x 10^25 and 4.3 x
::10^41 and the value for d = 8 is between 1.2 x 10^164 and 1.5 x 10^236.
::On Fri, Oct 28, 2022 at 12:51 PM Hans Havermann <gladhobo at bell.net> wrote:
::> Is it only a coincidence that A054759(4) is also 2970?
::> > On Oct 28, 2022, at 10:27 AM, hv at crypt.org wrote:
::> > I can confirm d=4 gives 2970.
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