1 4 9 5 2 6 3 24 20 16 15 13 10 7 11 12 14 17 19 26 21 25 22 18 23 8
I really liked this puzzle. I think that your approach has to do with
pairs of triples. When one triple is used it forces another tripple.
1-4-9 in the sieries forces that 19 be between 14 and 19. It is much
easier to assemble the Hamiltonian path when you assemble it in pieces that
are forced as you create the next step.
Scot Ellison
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Hi Ed,
a very nice Hamiltonian graph indeed!
Solution was easy to find; starting with 2-16, using symmetry and
deleting all impassible (crossed etc.) edges, I came to this beautiful
symmetrical solution:
2-16-15-13-10-7-3-6-1-8-23-18-22-25-11-12-14-17-20-24-21-26-19-4-9-5-2
Regards,
Berend Jan van der Zwaag
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I am a longtime reader of Mathpuzzle.com. Let me just say I love
your site. It's always fascinating and my friends and I have spent
many hours discussing the interesting puzzles and subjects you post
there. Thank you!
I have a solution to your F26 graph. Here is the sequence of
vertices:
1 6 3 7 10 13 15 16 2 5 9 4 19 26 21 24 20 17 14 12 11 25 22 18 23 8
Here is my guess as to your simple approach: "don't cross any lines".
That reduces the possibilities for vertices 16 and 11 enough to make
it pretty simple, especially with the implicit (for me, anyway)
assumption that the solution is radially symmetric. I wonder if a
non-symmetric solution is possible??
-Tom Maciukenas, Portland Oregon
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Hi,
I found a solution fairly quickly. I drew the graph on my whiteboard
and started to delete edges.
First, I tried deleting the 3 symmetric crossing edges (13-14, 3-24, and
5-22) and 1 edge each from 11 and 16 (I chose 11-25 and 2-16, to leave a
planar embedding.)
From here, things just cascade to a solution. If I have deleted an edge
from a vertex, then the other two edges are "in". If a vertiex has two
edges "in", then the other edge is "out" and I delete it.
I was left with
1 8 10 13 15 16 20 24 21 25 22 18 23
26 19 17 14 12 11 7 3 6 2 5 9 4 1
-jim boyce
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Hi Ed,
I have attached three copies of my solution to your
F26 puzzle. I arbitrarily decided to use all of the
lines inside the inner loop except for the two
smallest, and all six lines connecting the inner and
outer loops. All of the lines constructing the inner
loop were forced, leaving each outer vertex connected
to its opposite. The solution works just as well if
the outer loop connections are reversed.
-Matthew Bjorge
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Dear Ed,
just tried to figure out your approch, but can't
get it till now.
Here are two solutions
first "a sailor-ship":
1,6,2,16,15,18,23,8,10,13,14,17,20,
24,3,7,11,12,9,5,22,25,21,26,19,4,1
second nice simetry:
1,8,10,13,15,16,20,24,21,25,22,18,23,
26,19,17,14,12,11,7,3,6,2,5,9,4,1
Many greetings
Erwin Eichner
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Ed,
Neat little puzzle. Rotational symmetry is beautiful.
Clint Weaver
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