[seqfan] Re: Self avoiding polygons

[John Mason]

“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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