Lacings continued
Pfoertner, Hugo
Hugo.Pfoertner at muc.mtu.de
Fri Dec 13 15:30:07 CET 2002
-----Original Message-----
Von: Antti Karttunen [mailto:karttu at megabaud.fi]
Gesendet am: 13 December, 2002 15:08
An: njas at research.att.com
Cc: Pfoertner, Hugo; seqfan at ext.jussieu.fr;
math-fun at mailman.xmission.com
Betreff: Re: Lacings continued
<sequence definition snipped>
>
>Dear Neil,
>
>Because you have counted 132546 as a solution in A078602, I reckon that
your condition
>"and cannot pass though three consecutive and adjacent eyelets that are in
a line"
>means that "cannot pass _in order_ three consecutive and adjacent eyelets
on
>the same side."
If we want to count "real world" lacings things like the "132546" config
should be
excluded from being counted. I would like another condition that applies to
all
lacings I have seen:
From each eyelet there is at least one path to the opposite side. This
condition
should guarantee that there is a closing force on the eylet and not just a
force in
tangential direction as it is the case with eyelets 3 and 4 in the 1322546
config.
>I think here it is a question of simply counting of permutations with some
forbidden
>subsequences. I.e. in the case a(3) = 21, we have permutations of [1..6],
but
>with the first and the last element fixed as 1 and 6, so actually counting
the
>permutations of [2,3,4,5], but discarding the cases where we would have a
>either an increasing [i,i+1,i+2] or decreasing subsequence [i+2,i+1,i]
>(where i+2 is either <= n or i > n) e.g. from 4! = 24 permutations of
>[1,2,3,4,5,6] with 1 and 6 fixed we have to further discard the three cases
>[1,2,3,4,5,6] (violates the condition on both sides), [1,2,3,5,4,6] (on the
left side)
>and [1,3,2,4,5,6] (on the right side), so we are left with 24-3 = 21
solutions.
>
>Reasoning on these lines should easily (?) lead to a formula for A078602.
>At least it's now very easy to write a program in a declarative language
>like Prolog or Haskell to count the solutions.
>
>Then regarding Hugo's point about whether the lace enters each hole
>(Btw, some of his pictures resemble my son's attempts at this difficult
art...)
>from the outside or inside, doesn't this just add an extra factor
>of 2^(2n) to the count?
That would be the case, if all combinations were allowed. I want to exclude
inside/outside changes if the lace connects two eyeholes on the same side.
This complicates things.
>Regarding which lace goes underneath or over which other, it seems more
>complex. I don't know whether thinking in terms of braids actually
clarifies
>or confuses this issue:
>
>http://mathworld.wolfram.com/BraidGroup.html
>and
>http://www.brown.edu/Students/OHJC/topology/index.html
>and
>http://math.ucr.edu/home/baez/braids.html
>
>(at least one could borrow some notation from there?)
I'll try to learn something from the links.
>
>
>Terveempänä,
>
>Antti Karttunen
Best Regards,
Hugo Pfoertner
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