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| kelvSYC... |
 Posted: Sat Oct 31, 2009 12:12 pm |
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One question that I want to ask is the number of ways that you can
arrange the hexes on a Settlers of Catan board (not counting harbors or
number tokens), not counting rotations.
First, I have 19!/3!3!4!4!4! from the composition of the hexes. Which
is roughly 2.44 x 10^11. If I were to not count rotations, I'd divide
this by 6 to get roughly 4.07 x 10^10.
Is this correct? |
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| James Dow Allen... |
| Posted: Sat Oct 31, 2009 12:13 pm |
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On Nov 1, 1:12 am, kelvSYC <no.em... at (no spam) here.com> wrote:
[quote]One question that I want to ask is the number of ways that you can
arrange the hexes on a Settlers of Catan board (not counting harbors or
number tokens), not counting rotations.
First, I have 19!/3!3!4!4!4! from the composition of the hexes. Which
is roughly 2.44 x 10^11. If I were to not count rotations, I'd divide
this by 6 to get roughly 4.07 x 10^10.
Is this correct?
[/quote]
Thank you for your very thoughtful post! The people reading this ng
are dying of boredom, and are now delighted to Google around and
find out what a Settlers of Catan board is (and how to *not*
count its harbors). An inconsiderate poster would have spoiled this
fun by including a description.
Although I wasn't *quite* bored enough to Google for Settlers of
Catan, I'll bet your enumeration is incorrect because you ignore
a key fact about such counting. Rather than just tell you what
I mean, I'll return the courtesy you showed us and just mention
that Augustin Cauchy studied such enumeration problems. Study
all of his works (  . Come back if you still have questions.
Glad to help.
James Dow Allen |
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| Jake Wildstrom... |
| Posted: Sun Nov 01, 2009 12:31 pm |
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In article <ken.pledger-F05F4D.08471502112009 at (no spam) news.eternal-september.org>,
Ken Pledger <ken.pledger at (no spam) mcs.vuw.ac.nz> wrote:
[quote]In article <311020091212591362%no.email at (no spam) here.com>,
kelvSYC <no.email at (no spam) here.com> wrote:
One question that I want to ask is the number of ways that you can
arrange the hexes on a Settlers of Catan board (not counting harbors or
number tokens), not counting rotations.
First, I have 19!/3!3!4!4!4! from the composition of the hexes. Which
is roughly 2.44 x 10^11. If I were to not count rotations, I'd divide
this by 6 to get roughly 4.07 x 10^10.
[/quote]
This is right, although in part due to dumb luck. You'd be sunk if any
configuration had a nontrivial rotation onto itself; as far as I can
tell none of them do.
This problem would become significant if you wanted to also disallow
reflections; you couldn't get this answer simply by dividing your
above answer by 2. There _are_ Settlers boards which map onto
themselves under reflection, and this problem would involve
identification of which boards those are; then you could apply of Burnside's
Lemma.
[quote]You'll need to explain carefully what a Settlers of Catan board
looks like, for the sake of us ignorant mathematicians.
[/quote]
Hey, if SIAM can assume we know, so can he.
(I received a membership-promoting poster recently from SIAM featuring
images of what I can only surmise were events at conferences. There
was a picture of a poster session, a picture of a couple of people
networking, and a picture of a Settlers of Catan game in progress)
-Jake |
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| Ken Pledger... |
| Posted: Sun Nov 01, 2009 2:47 pm |
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In article <311020091212591362%no.email at (no spam) here.com>,
kelvSYC <no.email at (no spam) here.com> wrote:
[quote]One question that I want to ask is the number of ways that you can
arrange the hexes on a Settlers of Catan board (not counting harbors or
number tokens), not counting rotations.
First, I have 19!/3!3!4!4!4! from the composition of the hexes. Which
is roughly 2.44 x 10^11. If I were to not count rotations, I'd divide
this by 6 to get roughly 4.07 x 10^10.
Is this correct?
[/quote]
You'll need to explain carefully what a Settlers of Catan board
looks like, for the sake of us ignorant mathematicians. :-)
Ken Pledger. |
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