[seqfan] Re: Self avoiding polygons

[Allan Wechsler]

John Mason, I believe the assertion to be true. Proof sketch: rotate the
given 2n+2-edge polygon into "portrait" orientation, so that its height is
2 or more. Now position a horizontal line above the polygon, and lower it
until it contacts the northernmost boundary. This northernmost boundary
will consist of some number of disjoint horizontal line segments; pick any
one of them, and say it consists of k contiguous collinear edges. The row
of cells immediately below that segment forms a k-by-1 rectangle, which can
be removed to form a predecessor polygon with 2n edges.

On Sat, Feb 20, 2021 at 1:06 PM John Mason <masonmilan33 at gmail.com> wrote:

> “This is not my sort of area, but my immediate thought is that a square
> would
> not be generated by that procedure, since all its antecedents would have
> perimeter >= that of the square.”
>
> My description of the procedure was probably imprecise. What I meant was
> that any contiguous segment could be “shifted”. So, take a domino:
> O
> O
> It can then generate:
> Polygon 1, generated by extending the bottom border down:
> O
> O
> O
> Polygon 2, generated by extending a length 1 segment right:
> OO
> O
> Polygon 3, generated by extending a length 2 segment right:
> OO
> OO
>
> john
>
>
> From: hv at crypt.org
> Sent: 20 February 2021 18:58
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: Self avoiding polygons
>
> John Mason <masonmilan33 at gmail.com> wrote:
> :Hi seqfans.
> :I would be grateful if anyone could tell me of an existing proof of the
> following assertion, relative to self-avoiding polygons on the square
> lattice (aka the boundaries of "profane" polyominoes):
> :
> :Premise: one way to build such a polygon of perimeter 2n+2 is to take a
> polygon of perimeter 2n, find within that polygon a segment of length s,
> and "shift" a component of that segment, of integral length, "outwards" by
> one unit, but only if such a manoeuvre does not touch, even at a corner,
> any existing part of the polygon.
> :So for example:
> :OOO
> :OOO
> :OOO
> :Can generate (among others):
> :OOO
> :OOOO
> :OOOO
> :
> :Assertion: any self-avoiding polygon on the square lattice, of size 2n+2,
> for n > 2, may be generated from some polygon of size 2n by using the
> above-described procedure.
>
> This is not my sort of area, but my immediate thought is that a square
> would
> not be generated by that procedure, since all its antecedents would have
> perimeter >= that of the square.
>
> Hugo
>
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