polyominoes, gray codes, and venn diagrams
TRANSCRIPT
Polyominoes, Gray Codes, and Venn Diagrams
Stirling Chow1 Frank Ruskey1
1Department of Computer ScienceUniversity of Victoria, CANADA
Northwest Theory Day, April 2005
Definition - polyomino
I Made from unit squaresjoined along edges.
I No holes allowed.
I Must be connected.
I Translations allowed butnot flips or rotations.
(a)
(b)
(c)
(d)
(e)
Definition - polyomino
I Made from unit squaresjoined along edges.
I No holes allowed.
I Must be connected.
I Translations allowed butnot flips or rotations.
(a)
(b)
(c)
(d)
(e)
Definition - polyomino
I Made from unit squaresjoined along edges.
I No holes allowed.
I Must be connected.
I Translations allowed butnot flips or rotations.
(a)
(b)
(c)
(d)
(e)
Definition - polyomino
I Made from unit squaresjoined along edges.
I No holes allowed.
I Must be connected.
I Translations allowed butnot flips or rotations.
(a)
(b)
(c)
(d)
(e)
Definition - convexity conditions
I Column convex if intersections with vertical lines areconnected.
I Satisfy the recurrence: an = 5an−1 − 7an−2 + 4an−3
([Polya],nice proof:[Hickerson]).
I Convex if both row and column convex.
I Gray codes: column convex only.
I Further restriction: number of cells in each column is fixed.
(a) column−convexnot row−convex
(b) row−convexnot column−convex
(c) convex
Definition - convexity conditions
I Column convex if intersections with vertical lines areconnected.
I Satisfy the recurrence: an = 5an−1 − 7an−2 + 4an−3
([Polya],nice proof:[Hickerson]).
I Convex if both row and column convex.
I Gray codes: column convex only.
I Further restriction: number of cells in each column is fixed.
(a) column−convexnot row−convex
(b) row−convexnot column−convex
(c) convex
Definition - convexity conditions
I Column convex if intersections with vertical lines areconnected.
I Satisfy the recurrence: an = 5an−1 − 7an−2 + 4an−3
([Polya],nice proof:[Hickerson]).
I Convex if both row and column convex.
I Gray codes: column convex only.
I Further restriction: number of cells in each column is fixed.
(a) column−convexnot row−convex
(b) row−convexnot column−convex
(c) convex
Definition - convexity conditions
I Column convex if intersections with vertical lines areconnected.
I Satisfy the recurrence: an = 5an−1 − 7an−2 + 4an−3
([Polya],nice proof:[Hickerson]).
I Convex if both row and column convex.
I Gray codes: column convex only.
I Further restriction: number of cells in each column is fixed.
(a) column−convexnot row−convex
(b) row−convexnot column−convex
(c) convex
Definition - convexity conditions
I Column convex if intersections with vertical lines areconnected.
I Satisfy the recurrence: an = 5an−1 − 7an−2 + 4an−3
([Polya],nice proof:[Hickerson]).
I Convex if both row and column convex.
I Gray codes: column convex only.
I Further restriction: number of cells in each column is fixed.
(a) column−convexnot row−convex
(b) row−convexnot column−convex
(c) convex
I Column counts: [a1, a2, . . . , ak ]; below is the set [1, 2, 2, 1].
I Shift limits: 〈A1,A2, . . . ,Ak−1〉, where Ai = ai + ai+1 − 1;thus [1, 2, 2, 1] → 〈2, 3, 2〉.
I Not unique [1, 3, 1, 3] → 〈3, 3, 3〉 and [2, 2, 2, 2] → 〈3, 3, 3〉.I No polyomino for 〈1, 2, 1〉.I Encode individual polyominoes as
(p1, p2, . . . , pk−1) ∈ A1 × A2 × · · · × Ak−1
(1,2,1)(1,2,0)(1,1,1)(1,1,0)(1,0,1)(1,0,0)
(0,0,0) (0,0,1) (0,1,0) (0,1,1) (0,2,0) (0,2,1)
I Column counts: [a1, a2, . . . , ak ]; below is the set [1, 2, 2, 1].
I Shift limits: 〈A1,A2, . . . ,Ak−1〉, where Ai = ai + ai+1 − 1;thus [1, 2, 2, 1] → 〈2, 3, 2〉.
I Not unique [1, 3, 1, 3] → 〈3, 3, 3〉 and [2, 2, 2, 2] → 〈3, 3, 3〉.I No polyomino for 〈1, 2, 1〉.I Encode individual polyominoes as
(p1, p2, . . . , pk−1) ∈ A1 × A2 × · · · × Ak−1
(1,2,1)(1,2,0)(1,1,1)(1,1,0)(1,0,1)(1,0,0)
(0,0,0) (0,0,1) (0,1,0) (0,1,1) (0,2,0) (0,2,1)
I Column counts: [a1, a2, . . . , ak ]; below is the set [1, 2, 2, 1].
I Shift limits: 〈A1,A2, . . . ,Ak−1〉, where Ai = ai + ai+1 − 1;thus [1, 2, 2, 1] → 〈2, 3, 2〉.
I Not unique [1, 3, 1, 3] → 〈3, 3, 3〉 and [2, 2, 2, 2] → 〈3, 3, 3〉.I No polyomino for 〈1, 2, 1〉.I Encode individual polyominoes as
(p1, p2, . . . , pk−1) ∈ A1 × A2 × · · · × Ak−1
(1,2,1)(1,2,0)(1,1,1)(1,1,0)(1,0,1)(1,0,0)
(0,0,0) (0,0,1) (0,1,0) (0,1,1) (0,2,0) (0,2,1)
I Column counts: [a1, a2, . . . , ak ]; below is the set [1, 2, 2, 1].
I Shift limits: 〈A1,A2, . . . ,Ak−1〉, where Ai = ai + ai+1 − 1;thus [1, 2, 2, 1] → 〈2, 3, 2〉.
I Not unique [1, 3, 1, 3] → 〈3, 3, 3〉 and [2, 2, 2, 2] → 〈3, 3, 3〉.I No polyomino for 〈1, 2, 1〉.I Encode individual polyominoes as
(p1, p2, . . . , pk−1) ∈ A1 × A2 × · · · × Ak−1
(1,2,1)(1,2,0)(1,1,1)(1,1,0)(1,0,1)(1,0,0)
(0,0,0) (0,0,1) (0,1,0) (0,1,1) (0,2,0) (0,2,1)
I Column counts: [a1, a2, . . . , ak ]; below is the set [1, 2, 2, 1].
I Shift limits: 〈A1,A2, . . . ,Ak−1〉, where Ai = ai + ai+1 − 1;thus [1, 2, 2, 1] → 〈2, 3, 2〉.
I Not unique [1, 3, 1, 3] → 〈3, 3, 3〉 and [2, 2, 2, 2] → 〈3, 3, 3〉.I No polyomino for 〈1, 2, 1〉.I Encode individual polyominoes as
(p1, p2, . . . , pk−1) ∈ A1 × A2 × · · · × Ak−1
(1,2,1)(1,2,0)(1,1,1)(1,1,0)(1,0,1)(1,0,0)
(0,0,0) (0,0,1) (0,1,0) (0,1,1) (0,2,0) (0,2,1)
I The change pi := pi ± 1 corresponds to a small earthquake.fault line earthquake begins two unit shift
(1,1,3,2,3,1) (1,1,3,2,3,1) (1,1,5,2,3,1)
I Can generate these using Gray codes for mixed-radix numbers(H-path in k − 1 dimensional grid graph).
I A more interesting move is a single cell move within a column.
I Then pi := pi ± 1 and pi+1 := pi+1 ∓ 1 (except atextremities).
I The change pi := pi ± 1 corresponds to a small earthquake.fault line earthquake begins two unit shift
(1,1,3,2,3,1) (1,1,3,2,3,1) (1,1,5,2,3,1)
I Can generate these using Gray codes for mixed-radix numbers(H-path in k − 1 dimensional grid graph).
I A more interesting move is a single cell move within a column.
I Then pi := pi ± 1 and pi+1 := pi+1 ∓ 1 (except atextremities).
I The change pi := pi ± 1 corresponds to a small earthquake.fault line earthquake begins two unit shift
(1,1,3,2,3,1) (1,1,3,2,3,1) (1,1,5,2,3,1)
I Can generate these using Gray codes for mixed-radix numbers(H-path in k − 1 dimensional grid graph).
I A more interesting move is a single cell move within a column.
I Then pi := pi ± 1 and pi+1 := pi+1 ∓ 1 (except atextremities).
I The change pi := pi ± 1 corresponds to a small earthquake.fault line earthquake begins two unit shift
(1,1,3,2,3,1) (1,1,3,2,3,1) (1,1,5,2,3,1)
I Can generate these using Gray codes for mixed-radix numbers(H-path in k − 1 dimensional grid graph).
I A more interesting move is a single cell move within a column.
I Then pi := pi ± 1 and pi+1 := pi+1 ∓ 1 (except atextremities).
I Define two operations (and their inverses) on A-sequences:τi : pi := pi + 1 (only at the endpoints)σi : pi := pi − 1; pi+1 := pi+1 + 1;
I The “problem” of Ai = 1.I Implies that ai , ai+1 = 1. (Since Ai = ai + ai+1 − 1)I I.e., cells i and i + 1 are frozen in place.
I If two or more Ai = 1 then underlying graph is disconnected;otherwise it is connected.
I The graph is denoted G ([a1, . . . , ak ]) or G (〈A1, . . . ,Ak−1〉).I An interesting case occurs when Ak−1 = 1 and Ai > 1,
for i = 1, 2, . . . , k − 2: the right-frozen case.
I Free = neither side frozen.
I Define two operations (and their inverses) on A-sequences:τi : pi := pi + 1 (only at the endpoints)σi : pi := pi − 1; pi+1 := pi+1 + 1;
I The “problem” of Ai = 1.I Implies that ai , ai+1 = 1. (Since Ai = ai + ai+1 − 1)I I.e., cells i and i + 1 are frozen in place.
I If two or more Ai = 1 then underlying graph is disconnected;otherwise it is connected.
I The graph is denoted G ([a1, . . . , ak ]) or G (〈A1, . . . ,Ak−1〉).I An interesting case occurs when Ak−1 = 1 and Ai > 1,
for i = 1, 2, . . . , k − 2: the right-frozen case.
I Free = neither side frozen.
I Define two operations (and their inverses) on A-sequences:τi : pi := pi + 1 (only at the endpoints)σi : pi := pi − 1; pi+1 := pi+1 + 1;
I The “problem” of Ai = 1.I Implies that ai , ai+1 = 1. (Since Ai = ai + ai+1 − 1)I I.e., cells i and i + 1 are frozen in place.
I If two or more Ai = 1 then underlying graph is disconnected;otherwise it is connected.
I The graph is denoted G ([a1, . . . , ak ]) or G (〈A1, . . . ,Ak−1〉).I An interesting case occurs when Ak−1 = 1 and Ai > 1,
for i = 1, 2, . . . , k − 2: the right-frozen case.
I Free = neither side frozen.
I Define two operations (and their inverses) on A-sequences:τi : pi := pi + 1 (only at the endpoints)σi : pi := pi − 1; pi+1 := pi+1 + 1;
I The “problem” of Ai = 1.I Implies that ai , ai+1 = 1. (Since Ai = ai + ai+1 − 1)I I.e., cells i and i + 1 are frozen in place.
I If two or more Ai = 1 then underlying graph is disconnected;otherwise it is connected.
I The graph is denoted G ([a1, . . . , ak ]) or G (〈A1, . . . ,Ak−1〉).I An interesting case occurs when Ak−1 = 1 and Ai > 1,
for i = 1, 2, . . . , k − 2: the right-frozen case.
I Free = neither side frozen.
I Define two operations (and their inverses) on A-sequences:τi : pi := pi + 1 (only at the endpoints)σi : pi := pi − 1; pi+1 := pi+1 + 1;
I The “problem” of Ai = 1.I Implies that ai , ai+1 = 1. (Since Ai = ai + ai+1 − 1)I I.e., cells i and i + 1 are frozen in place.
I If two or more Ai = 1 then underlying graph is disconnected;otherwise it is connected.
I The graph is denoted G ([a1, . . . , ak ]) or G (〈A1, . . . ,Ak−1〉).I An interesting case occurs when Ak−1 = 1 and Ai > 1,
for i = 1, 2, . . . , k − 2: the right-frozen case.
I Free = neither side frozen.
I Define two operations (and their inverses) on A-sequences:τi : pi := pi + 1 (only at the endpoints)σi : pi := pi − 1; pi+1 := pi+1 + 1;
I The “problem” of Ai = 1.I Implies that ai , ai+1 = 1. (Since Ai = ai + ai+1 − 1)I I.e., cells i and i + 1 are frozen in place.
I If two or more Ai = 1 then underlying graph is disconnected;otherwise it is connected.
I The graph is denoted G ([a1, . . . , ak ]) or G (〈A1, . . . ,Ak−1〉).I An interesting case occurs when Ak−1 = 1 and Ai > 1,
for i = 1, 2, . . . , k − 2: the right-frozen case.
I Free = neither side frozen.
I Define two operations (and their inverses) on A-sequences:τi : pi := pi + 1 (only at the endpoints)σi : pi := pi − 1; pi+1 := pi+1 + 1;
I The “problem” of Ai = 1.I Implies that ai , ai+1 = 1. (Since Ai = ai + ai+1 − 1)I I.e., cells i and i + 1 are frozen in place.
I If two or more Ai = 1 then underlying graph is disconnected;otherwise it is connected.
I The graph is denoted G ([a1, . . . , ak ]) or G (〈A1, . . . ,Ak−1〉).I An interesting case occurs when Ak−1 = 1 and Ai > 1,
for i = 1, 2, . . . , k − 2: the right-frozen case.
I Free = neither side frozen.
I Define two operations (and their inverses) on A-sequences:τi : pi := pi + 1 (only at the endpoints)σi : pi := pi − 1; pi+1 := pi+1 + 1;
I The “problem” of Ai = 1.I Implies that ai , ai+1 = 1. (Since Ai = ai + ai+1 − 1)I I.e., cells i and i + 1 are frozen in place.
I If two or more Ai = 1 then underlying graph is disconnected;otherwise it is connected.
I The graph is denoted G ([a1, . . . , ak ]) or G (〈A1, . . . ,Ak−1〉).I An interesting case occurs when Ak−1 = 1 and Ai > 1,
for i = 1, 2, . . . , k − 2: the right-frozen case.
I Free = neither side frozen.
1,1,0
2,0,0 0,1,0
2,1,0
1,0,0
0,0,0
LemmaIf k is even, then G ([a1, . . . , ak ]) isbipartite.
Proof.Define partite sets according to theparity of∑
j odd
pj (=∑
j
jpj mod 2).
Example
G ([2, 2, 1, 1]) is bipartite.
2,1
0,0
1,1
2.0 0,1
1,0
LemmaIf k > 3 is odd, then G ([a1, . . . , ak ])is bipartite iff A1 = 1 or Ak−1 = 1.
Proof.Odd cycle: τ1, σ1, . . . , σk−2, τ
−1k−1:
Example
G ([2, 2, 1]) is not bipartite.
2,1
0,0
1,1
2.0 0,1
1,0
LemmaIf k > 3 is odd, then G ([a1, . . . , ak ])is bipartite iff A1 = 1 or Ak−1 = 1.
Proof.Odd cycle: τ1, σ1, . . . , σk−2, τ
−1k−1:
0000,
Example
G ([2, 2, 1]) is not bipartite.
2,1
0,0
1,1
2.0 0,1
1,0
LemmaIf k > 3 is odd, then G ([a1, . . . , ak ])is bipartite iff A1 = 1 or Ak−1 = 1.
Proof.Odd cycle: τ1, σ1, . . . , σk−2, τ
−1k−1:
0000,1000,
Example
G ([2, 2, 1]) is not bipartite.
2,1
0,0
1,1
2.0 0,1
1,0
LemmaIf k > 3 is odd, then G ([a1, . . . , ak ])is bipartite iff A1 = 1 or Ak−1 = 1.
Proof.Odd cycle: τ1, σ1, . . . , σk−2, τ
−1k−1:
0000,1000,0100,
Example
G ([2, 2, 1]) is not bipartite.
2,1
0,0
1,1
2.0 0,1
1,0
LemmaIf k > 3 is odd, then G ([a1, . . . , ak ])is bipartite iff A1 = 1 or Ak−1 = 1.
Proof.Odd cycle: τ1, σ1, . . . , σk−2, τ
−1k−1:
0000,1000,0100,0010,
Example
G ([2, 2, 1]) is not bipartite.
2,1
0,0
1,1
2.0 0,1
1,0
LemmaIf k > 3 is odd, then G ([a1, . . . , ak ])is bipartite iff A1 = 1 or Ak−1 = 1.
Proof.Odd cycle: τ1, σ1, . . . , σk−2, τ
−1k−1:
0000,1000,0100,0010,0001,
Example
G ([2, 2, 1]) is not bipartite.
2,1
0,0
1,1
2.0 0,1
1,0
LemmaIf k > 3 is odd, then G ([a1, . . . , ak ])is bipartite iff A1 = 1 or Ak−1 = 1.
Proof.Odd cycle: τ1, σ1, . . . , σk−2, τ
−1k−1:
0000, 0000.1000,0100,0010,0001,
Example
G ([2, 2, 1]) is not bipartite.
LemmaIf k > 2 is even, then G (a) has no Hamilton path if A2i+1 is oddfor all i , unless a2 = a3 = · · · = ak−1 = 1.
Proof.Define a sign-reversing involution... If all A2i+1 are odd, thenparity difference is
d(A) =∏
j even
Aj .
Example
G (〈3, 2, 3, 2, 3〉) has no H-path.G (〈3, 1, 1, 1, 3〉) is a 3 by 3 grid and has a H-path.Conjecture: There is a H-path if k even and some A2i+1 is even.
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Proof.Convert each edge of H into a path of BC vertices in H ′. Thestructure of B,C is a B by C grid graph. Need to be careful aboutthe τ moves in H.... Need to show the existence of particular typesof H-cycles in grid graphs....
TheoremIf A has the form (N>3)
0|1(O>3N>1)∗(N)0|1E or it’s reverse, then
there is a 1-move Gray code for A
Proof.Add pairs of columns on the left...
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
0 0 0 00 0 0 0
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
? ? ? ?0 0 0 0 00 0 0 0 1? ? ? ?
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
? ? ? ?0 0 0 0 00 0 0 0 1? ? ? 1 0
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
0 1 1 0 00 0 0 0 00 0 0 0 10 0 1 1 0
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
0 1 1 0 0 ? ? ?0 0 0 0 0 0 0 0 10 0 0 0 1 1 1 1 00 0 1 1 0 ? ? ?
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
0 1 1 0 0 0 1 1 10 0 0 0 0 0 0 0 10 0 0 0 1 1 1 1 00 0 1 1 0 1 1 0 0
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
0 1 1 0 0 0 1 1 1 ? ? ?0 0 0 0 0 0 0 0 1 1 1 1 10 0 0 0 1 1 1 1 0 0 0 0 10 0 1 1 0 1 1 0 0 ? ? ?
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
0 1 1 0 0 0 1 1 1 0 0 1 10 0 0 0 0 0 0 0 1 1 1 1 10 0 0 0 1 1 1 1 0 0 0 0 10 0 1 1 0 1 1 0 0 0 1 1 0
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
0 1 1 0 0 0 1 1 1 0 0 1 1 ? ? ?0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 10 0 0 0 1 1 1 1 0 0 0 0 1 1 1 10 0 1 1 0 1 1 0 0 0 1 1 0 ? ? ?
TheoremIf G (〈A1,A2, . . . ,Ak−1〉) has a Hamilton path H and BC is evenand BC 6= 6, then G (〈B,A1,A2, . . . ,Ak−1,C 〉) has a Hamiltonpath H ′.
Example
Converting G (2, 2) into G (2, 2, 2, 2).0 0 1 10 1 0 1
0 1 1 0 0 0 1 1 1 0 0 1 1 0 0 10 0 0 0 0 0 0 0 1 1 1 1 1 1 1 10 0 0 0 1 1 1 1 0 0 0 0 1 1 1 10 0 1 1 0 1 1 0 0 0 1 1 0 0 1 1
TheoremIf A has the form (N>3)
0|1(O>3N>1)∗(N)0|1E or it’s reverse, then
there is a 1-move Gray code for A
Right-frozen case – a poset
I Operations are τ1 and σ1, σ2, . . ..
I Underlying poset G (〈A1,A2, . . . ,Ak−1, 1〉) orders p-sequences:(p1, p2, . . . , pk−1) ≤ (q1, q2, . . . , qk−1) iff
k−1∑i=j
pi ≤k−1∑i=j
qi
I The operations are the cover relations.
I G (〈2, 2, . . . , 2︸ ︷︷ ︸n
, 1〉) is M(n) (e.g., [Lindstrom][Stanley])
I Is a distributive lattice in general.
I Join irreducibles: A′1A
′2 · · ·A′
i0 · · · 0xA′j+1 · · ·A′
k−1, where0 ≤ x < A′
j := Aj − 1. There are∑
j jA′j of them.
I Is self-dual under the mappingp1 · · · pk−1 7→ (A′
1 − p1) · · · (A′k−1 − pk−1).
Right-frozen case – a poset
I Operations are τ1 and σ1, σ2, . . ..
I Underlying poset G (〈A1,A2, . . . ,Ak−1, 1〉) orders p-sequences:(p1, p2, . . . , pk−1) ≤ (q1, q2, . . . , qk−1) iff
k−1∑i=j
pi ≤k−1∑i=j
qi
I The operations are the cover relations.
I G (〈2, 2, . . . , 2︸ ︷︷ ︸n
, 1〉) is M(n) (e.g., [Lindstrom][Stanley])
I Is a distributive lattice in general.
I Join irreducibles: A′1A
′2 · · ·A′
i0 · · · 0xA′j+1 · · ·A′
k−1, where0 ≤ x < A′
j := Aj − 1. There are∑
j jA′j of them.
I Is self-dual under the mappingp1 · · · pk−1 7→ (A′
1 − p1) · · · (A′k−1 − pk−1).
Right-frozen case – a poset
I Operations are τ1 and σ1, σ2, . . ..
I Underlying poset G (〈A1,A2, . . . ,Ak−1, 1〉) orders p-sequences:(p1, p2, . . . , pk−1) ≤ (q1, q2, . . . , qk−1) iff
k−1∑i=j
pi ≤k−1∑i=j
qi
I The operations are the cover relations.
I G (〈2, 2, . . . , 2︸ ︷︷ ︸n
, 1〉) is M(n) (e.g., [Lindstrom][Stanley])
I Is a distributive lattice in general.
I Join irreducibles: A′1A
′2 · · ·A′
i0 · · · 0xA′j+1 · · ·A′
k−1, where0 ≤ x < A′
j := Aj − 1. There are∑
j jA′j of them.
I Is self-dual under the mappingp1 · · · pk−1 7→ (A′
1 − p1) · · · (A′k−1 − pk−1).
Right-frozen case – a poset
I Operations are τ1 and σ1, σ2, . . ..
I Underlying poset G (〈A1,A2, . . . ,Ak−1, 1〉) orders p-sequences:(p1, p2, . . . , pk−1) ≤ (q1, q2, . . . , qk−1) iff
k−1∑i=j
pi ≤k−1∑i=j
qi
I The operations are the cover relations.
I G (〈2, 2, . . . , 2︸ ︷︷ ︸n
, 1〉) is M(n) (e.g., [Lindstrom][Stanley])
I Is a distributive lattice in general.
I Join irreducibles: A′1A
′2 · · ·A′
i0 · · · 0xA′j+1 · · ·A′
k−1, where0 ≤ x < A′
j := Aj − 1. There are∑
j jA′j of them.
I Is self-dual under the mappingp1 · · · pk−1 7→ (A′
1 − p1) · · · (A′k−1 − pk−1).
Right-frozen case – a poset
I Operations are τ1 and σ1, σ2, . . ..
I Underlying poset G (〈A1,A2, . . . ,Ak−1, 1〉) orders p-sequences:(p1, p2, . . . , pk−1) ≤ (q1, q2, . . . , qk−1) iff
k−1∑i=j
pi ≤k−1∑i=j
qi
I The operations are the cover relations.
I G (〈2, 2, . . . , 2︸ ︷︷ ︸n
, 1〉) is M(n) (e.g., [Lindstrom][Stanley])
I Is a distributive lattice in general.
I Join irreducibles: A′1A
′2 · · ·A′
i0 · · · 0xA′j+1 · · ·A′
k−1, where0 ≤ x < A′
j := Aj − 1. There are∑
j jA′j of them.
I Is self-dual under the mappingp1 · · · pk−1 7→ (A′
1 − p1) · · · (A′k−1 − pk−1).
Right-frozen case – a poset
I Operations are τ1 and σ1, σ2, . . ..
I Underlying poset G (〈A1,A2, . . . ,Ak−1, 1〉) orders p-sequences:(p1, p2, . . . , pk−1) ≤ (q1, q2, . . . , qk−1) iff
k−1∑i=j
pi ≤k−1∑i=j
qi
I The operations are the cover relations.
I G (〈2, 2, . . . , 2︸ ︷︷ ︸n
, 1〉) is M(n) (e.g., [Lindstrom][Stanley])
I Is a distributive lattice in general.
I Join irreducibles: A′1A
′2 · · ·A′
i0 · · · 0xA′j+1 · · ·A′
k−1, where0 ≤ x < A′
j := Aj − 1. There are∑
j jA′j of them.
I Is self-dual under the mappingp1 · · · pk−1 7→ (A′
1 − p1) · · · (A′k−1 − pk−1).
Right-frozen case – a poset
I Operations are τ1 and σ1, σ2, . . ..
I Underlying poset G (〈A1,A2, . . . ,Ak−1, 1〉) orders p-sequences:(p1, p2, . . . , pk−1) ≤ (q1, q2, . . . , qk−1) iff
k−1∑i=j
pi ≤k−1∑i=j
qi
I The operations are the cover relations.
I G (〈2, 2, . . . , 2︸ ︷︷ ︸n
, 1〉) is M(n) (e.g., [Lindstrom][Stanley])
I Is a distributive lattice in general.
I Join irreducibles: A′1A
′2 · · ·A′
i0 · · · 0xA′j+1 · · ·A′
k−1, where0 ≤ x < A′
j := Aj − 1. There are∑
j jA′j of them.
I Is self-dual under the mappingp1 · · · pk−1 7→ (A′
1 − p1) · · · (A′k−1 − pk−1).
Further properties
I The rank of a p-sequence is
r(p) =k−1∑j=1
jpj .
I Rank generating function is
k−1∏j=1
1− z jAj
1− z j
I Since G is the cover graph of a distributive lattice, its prism isHamiltonian ([Pruesse & R]). Thus G 2 is Hamiltonian. Notthe same as moving two cells!
Further properties
I The rank of a p-sequence is
r(p) =k−1∑j=1
jpj .
I Rank generating function is
k−1∏j=1
1− z jAj
1− z j
I Since G is the cover graph of a distributive lattice, its prism isHamiltonian ([Pruesse & R]). Thus G 2 is Hamiltonian. Notthe same as moving two cells!
Further properties
I The rank of a p-sequence is
r(p) =k−1∑j=1
jpj .
I Rank generating function is
k−1∏j=1
1− z jAj
1− z j
I Since G is the cover graph of a distributive lattice, its prism isHamiltonian ([Pruesse & R]). Thus G 2 is Hamiltonian. Notthe same as moving two cells!
LemmaThe graph G (〈2, . . . , 2, 1〉) has a Hamilton path if and only if(n+1
2
)is even and n 6= 5.
Proof.This follows from the results of Savage, Shields, and West.
TheoremFor all n > 0 the graph G (〈2, 2, . . . , 2︸ ︷︷ ︸
n
〉) is Hamiltonian.
Proof.Induction on n with a suitably strengthened hypothesis...
LemmaThe graph G (〈2, . . . , 2, 1〉) has a Hamilton path if and only if(n+1
2
)is even and n 6= 5.
Proof.This follows from the results of Savage, Shields, and West.
TheoremFor all n > 0 the graph G (〈2, 2, . . . , 2︸ ︷︷ ︸
n
〉) is Hamiltonian.
Proof.Induction on n with a suitably strengthened hypothesis...
LemmaThe graph G (〈2, . . . , 2, 1〉) has a Hamilton path if and only if(n+1
2
)is even and n 6= 5.
Proof.This follows from the results of Savage, Shields, and West.
TheoremFor all n > 0 the graph G (〈2, 2, . . . , 2︸ ︷︷ ︸
n
〉) is Hamiltonian.
Proof.Induction on n with a suitably strengthened hypothesis...
LemmaThe graph G (〈2, . . . , 2, 1〉) has a Hamilton path if and only if(n+1
2
)is even and n 6= 5.
Proof.This follows from the results of Savage, Shields, and West.
TheoremFor all n > 0 the graph G (〈2, 2, . . . , 2︸ ︷︷ ︸
n
〉) is Hamiltonian.
Proof.Induction on n with a suitably strengthened hypothesis...
LemmaThere is no Hamilton path in G (〈A1,A2, . . . ,Ak , 1〉) if (1) has thesame parity as A1A2 · · ·Ak .
k∑j=1
j(Aj − 1) =k∑
j=1
jA′j . (1)
Proof.Note that 00 · · · 00 and A′
1A′2 · · ·A′
k0 are pendant vertices in aA1A2 · · ·Ak vertex bipartite graph G and that their distance fromeach other is precisely (1).
LemmaThere is no Hamilton path in G (〈A1,A2, . . . ,Ak , 1〉) if (1) has thesame parity as A1A2 · · ·Ak .
k∑j=1
j(Aj − 1) =k∑
j=1
jA′j . (1)
Proof.Note that 00 · · · 00 and A′
1A′2 · · ·A′
k0 are pendant vertices in aA1A2 · · ·Ak vertex bipartite graph G and that their distance fromeach other is precisely (1).
44
43
42
41
40
40
30
20
10
00
34
00
10
20
30
40
44
34
2443
33 14
34
24
43042342
42 04
14
1332
41 03032241
1231
21 02
11 00
01
0240
01
I The Poset G (〈5, 5, 1〉). Joinirreducibles in yellow.
I The subposet of join-irreducibles.
I Both are self-dual.
I Note: G (〈5, 5, 1〉) notrank-unimodal.
44
43
42
41
40
40
30
20
10
00
34
00
10
20
30
40
44
34
2443
33 14
34
24
43042342
42 04
14
1332
41 03032241
1231
21 02
11 00
01
0240
01
I The Poset G (〈5, 5, 1〉). Joinirreducibles in yellow.
I The subposet of join-irreducibles.
I Both are self-dual.
I Note: G (〈5, 5, 1〉) notrank-unimodal.
44
43
42
41
40
40
30
20
10
00
34
00
10
20
30
40
44
34
2443
33 14
34
24
43042342
42 04
14
1332
41 03032241
1231
21 02
11 00
01
0240
01
I The Poset G (〈5, 5, 1〉). Joinirreducibles in yellow.
I The subposet of join-irreducibles.
I Both are self-dual.
I Note: G (〈5, 5, 1〉) notrank-unimodal.
44
43
42
41
40
40
30
20
10
00
34
00
10
20
30
40
44
34
2443
33 14
34
24
43042342
42 04
14
1332
41 03032241
1231
21 02
11 00
01
0240
01
I The Poset G (〈5, 5, 1〉). Joinirreducibles in yellow.
I The subposet of join-irreducibles.
I Both are self-dual.
I Note: G (〈5, 5, 1〉) notrank-unimodal.
Posets - Join Irreducibles
200 001
122
000
220
221
212 022
202
201
012
022
Expanding
I G (〈3, 3, 3, 1〉).I Ends with 2 or
not.
I G (〈3, 3, 3, 3, 1〉).
Posets - Join Irreducibles
0222
0122
0222
1222
2012
2022
2122
2212
2202
2002
0002
0012
2221
2220 2201
2200 2001
00012000
0000
Expanding
I G (〈3, 3, 3, 1〉).I Ends with 2 or
not.
I G (〈3, 3, 3, 3, 1〉).
Posets - Join Irreducibles
0222
0122
0222
1222
2012
2022
2122
2212
2202
2002
0002
0012
2221
2220 2201
2200 2001
00012000
0000
Expanding
I G (〈3, 3, 3, 1〉).I Ends with 2 or
not.
I G (〈3, 3, 3, 3, 1〉).
More on unimodality
I Let c(S) be the smallest value of k for which P(S i ) is rankunimodal for all i ≥ k.
I Example What is c(2, 3)? The rank gf is(1 + z)(1 + z2 + z4) = 1 + z + z2 + z3 + z4 + z5.
I Ranks of 2, 3, 2:1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 2 2 2 1 1 1
I Ranks of 2, 3, 2, 3 are1 1 1 2 2 2 1 1 1
1 1 1 2 2 2 1 1 11 1 1 2 2 2 1 1 1
1 1 1 2 3 3 2 3 4 3 2 3 3 2 1 1 1which is not unimodal.
I Continuing, P((2, 3)5) is not unimodal, but P((2, 3)k) appearsto be unimodal for k ≥ 6 (checked up to k = 200).
More on unimodality
I Let c(S) be the smallest value of k for which P(S i ) is rankunimodal for all i ≥ k. Well defined?
I Example What is c(2, 3)? The rank gf is(1 + z)(1 + z2 + z4) = 1 + z + z2 + z3 + z4 + z5.
I Ranks of 2, 3, 2:1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 2 2 2 1 1 1
I Ranks of 2, 3, 2, 3 are1 1 1 2 2 2 1 1 1
1 1 1 2 2 2 1 1 11 1 1 2 2 2 1 1 1
1 1 1 2 3 3 2 3 4 3 2 3 3 2 1 1 1which is not unimodal.
I Continuing, P((2, 3)5) is not unimodal, but P((2, 3)k) appearsto be unimodal for k ≥ 6 (checked up to k = 200).
More on unimodality
I Let c(S) be the smallest value of k for which P(S i ) is rankunimodal for all i ≥ k. Well defined?
I Example What is c(2, 3)? The rank gf is(1 + z)(1 + z2 + z4) = 1 + z + z2 + z3 + z4 + z5.
I Ranks of 2, 3, 2:1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 2 2 2 1 1 1
I Ranks of 2, 3, 2, 3 are1 1 1 2 2 2 1 1 1
1 1 1 2 2 2 1 1 11 1 1 2 2 2 1 1 1
1 1 1 2 3 3 2 3 4 3 2 3 3 2 1 1 1which is not unimodal.
I Continuing, P((2, 3)5) is not unimodal, but P((2, 3)k) appearsto be unimodal for k ≥ 6 (checked up to k = 200).
More on unimodality
I Let c(S) be the smallest value of k for which P(S i ) is rankunimodal for all i ≥ k. Well defined?
I Example What is c(2, 3)? The rank gf is(1 + z)(1 + z2 + z4) = 1 + z + z2 + z3 + z4 + z5.
I Ranks of 2, 3, 2:1 1 1 1 1 1◦ ◦ ◦ 1 1 1 1 1 1
1 1 1 2 2 2 1 1 1
I Ranks of 2, 3, 2, 3 are1 1 1 2 2 2 1 1 1
1 1 1 2 2 2 1 1 11 1 1 2 2 2 1 1 1
1 1 1 2 3 3 2 3 4 3 2 3 3 2 1 1 1which is not unimodal.
I Continuing, P((2, 3)5) is not unimodal, but P((2, 3)k) appearsto be unimodal for k ≥ 6 (checked up to k = 200).
More on unimodality
I Let c(S) be the smallest value of k for which P(S i ) is rankunimodal for all i ≥ k. Well defined?
I Example What is c(2, 3)? The rank gf is(1 + z)(1 + z2 + z4) = 1 + z + z2 + z3 + z4 + z5.
I Ranks of 2, 3, 2:1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 2 2 2 1 1 1
I Ranks of 2, 3, 2, 3 are1 1 1 2 2 2 1 1 1◦ ◦ ◦ ◦ 1 1 1 2 2 2 1 1 1
◦ ◦ ◦ ◦ 1 1 1 2 2 2 1 1 1
1 1 1 2 3 3 2 3 4 3 2 3 3 2 1 1 1which is not unimodal.
I Continuing, P((2, 3)5) is not unimodal, but P((2, 3)k) appearsto be unimodal for k ≥ 6 (checked up to k = 200).
c(n, m)2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
2 1 6 1 7 1 5 1 7 1 5 1 5 1 5 1 5 1 5 13 7 6 8 8 8 a 8 a 9 a b a b a c c c c c4 4 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 15 6 1 3 4 6 5 6 5 6 6 7 6 7 6 7 6 7 7 76 1 1 1 3 1 1 1 3 1 3 1 3 1 3 1 3 1 3 17 3 1 1 2 4 4 4 4 4 4 5 5 5 6 5 6 5 6 58 3 1 1 3 1 3 1 1 1 3 1 1 1 1 1 1 1 1 19 4 1 1 1 3 3 3 3 4 4 4 4 4 4 4 4 4 4 510 3 1 1 1 1 1 1 3 1 1 1 3 1 1 1 3 1 3 111 3 1 3 1 1 2 4 4 4 4 4 4 4 4 4 4 4 4 412 1 1 1 1 1 3 1 3 1 3 1 3 1 3 1 1 1 3 113 4 1 3 1 3 1 2 2 3 3 3 3 3 3 4 4 4 4 414 4 1 3 3 1 3 1 3 1 3 1 3 1 1 1 3 1 1 115 1 4 4 4 1 4 4 2 4 4 4 4 4 4 4 4 4 4 416 4 1 3 3 1 1 1 3 1 3 1 3 1 3 1 3 1 3 117 4 1 3 3 1 1 3 1 3 2 3 3 3 3 3 3 3 3 318 4 1 1 1 1 1 1 3 1 3 1 3 1 3 1 3 1 3 119 4 1 4 3 1 1 1 1 1 2 3 2 3 3 3 3 4 4 420 3 1 3 3 1 3 1 1 1 3 1 3 1 3 1 3 1 3 1
Remarks about c(m, n)
I Only the 2,2 entry is proven.
I Most entries are checked out to k = 30. The numbers arehuge; i.e., the underlying poset has (nm)30 elements.
I Some conjectures:I For odd m, limn→∞ c(m, n) = cm. In particular, c3 = 16 (the
value 16 first occurs for k = 90).I For even m with an odd factor, for large enough n, if n even
then c(m, n) = 1 and if n is odd then c(m, n) = 3.I If m = 2d then c(m, n) = 1 for n ≥ 2d+1.I For m ≥ 25, c(m, 3) = 1 if 5 | m; otherwise, c(m, 3) = 4.I For m ≥ 15, c(m, 3) = 1, 3, 4 if
m = ({0}, {5, 10}, other) mod 15.
Remarks about c(m, n)
I Only the 2,2 entry is proven.
I Most entries are checked out to k = 30. The numbers arehuge; i.e., the underlying poset has (nm)30 elements.
I Some conjectures:I For odd m, limn→∞ c(m, n) = cm. In particular, c3 = 16 (the
value 16 first occurs for k = 90).I For even m with an odd factor, for large enough n, if n even
then c(m, n) = 1 and if n is odd then c(m, n) = 3.I If m = 2d then c(m, n) = 1 for n ≥ 2d+1.I For m ≥ 25, c(m, 3) = 1 if 5 | m; otherwise, c(m, 3) = 4.I For m ≥ 15, c(m, 3) = 1, 3, 4 if
m = ({0}, {5, 10}, other) mod 15.
Remarks about c(m, n)
I Only the 2,2 entry is proven.
I Most entries are checked out to k = 30. The numbers arehuge; i.e., the underlying poset has (nm)30 elements.
I Some conjectures:I For odd m, limn→∞ c(m, n) = cm. In particular, c3 = 16 (the
value 16 first occurs for k = 90).I For even m with an odd factor, for large enough n, if n even
then c(m, n) = 1 and if n is odd then c(m, n) = 3.I If m = 2d then c(m, n) = 1 for n ≥ 2d+1.I For m ≥ 25, c(m, 3) = 1 if 5 | m; otherwise, c(m, 3) = 4.I For m ≥ 15, c(m, 3) = 1, 3, 4 if
m = ({0}, {5, 10}, other) mod 15.
Remarks about c(m, n)
I Only the 2,2 entry is proven.
I Most entries are checked out to k = 30. The numbers arehuge; i.e., the underlying poset has (nm)30 elements.
I Some conjectures:I For odd m, limn→∞ c(m, n) = cm. In particular, c3 = 16 (the
value 16 first occurs for k = 90).I For even m with an odd factor, for large enough n, if n even
then c(m, n) = 1 and if n is odd then c(m, n) = 3.I If m = 2d then c(m, n) = 1 for n ≥ 2d+1.I For m ≥ 25, c(m, 3) = 1 if 5 | m; otherwise, c(m, 3) = 4.I For m ≥ 15, c(m, 3) = 1, 3, 4 if
m = ({0}, {5, 10}, other) mod 15.
Remarks about c(m, n)
I Only the 2,2 entry is proven.
I Most entries are checked out to k = 30. The numbers arehuge; i.e., the underlying poset has (nm)30 elements.
I Some conjectures:I For odd m, limn→∞ c(m, n) = cm. In particular, c3 = 16 (the
value 16 first occurs for k = 90).I For even m with an odd factor, for large enough n, if n even
then c(m, n) = 1 and if n is odd then c(m, n) = 3.I If m = 2d then c(m, n) = 1 for n ≥ 2d+1.I For m ≥ 25, c(m, 3) = 1 if 5 | m; otherwise, c(m, 3) = 4.I For m ≥ 15, c(m, 3) = 1, 3, 4 if
m = ({0}, {5, 10}, other) mod 15.
Remarks about c(m, n)
I Only the 2,2 entry is proven.
I Most entries are checked out to k = 30. The numbers arehuge; i.e., the underlying poset has (nm)30 elements.
I Some conjectures:I For odd m, limn→∞ c(m, n) = cm. In particular, c3 = 16 (the
value 16 first occurs for k = 90).I For even m with an odd factor, for large enough n, if n even
then c(m, n) = 1 and if n is odd then c(m, n) = 3.I If m = 2d then c(m, n) = 1 for n ≥ 2d+1.I For m ≥ 25, c(m, 3) = 1 if 5 | m; otherwise, c(m, 3) = 4.I For m ≥ 15, c(m, 3) = 1, 3, 4 if
m = ({0}, {5, 10}, other) mod 15.
Remarks about c(m, n)
I Only the 2,2 entry is proven.
I Most entries are checked out to k = 30. The numbers arehuge; i.e., the underlying poset has (nm)30 elements.
I Some conjectures:I For odd m, limn→∞ c(m, n) = cm. In particular, c3 = 16 (the
value 16 first occurs for k = 90).I For even m with an odd factor, for large enough n, if n even
then c(m, n) = 1 and if n is odd then c(m, n) = 3.I If m = 2d then c(m, n) = 1 for n ≥ 2d+1.I For m ≥ 25, c(m, 3) = 1 if 5 | m; otherwise, c(m, 3) = 4.I For m ≥ 15, c(m, 3) = 1, 3, 4 if
m = ({0}, {5, 10}, other) mod 15.
Remarks about c(m, n)
I Only the 2,2 entry is proven.
I Most entries are checked out to k = 30. The numbers arehuge; i.e., the underlying poset has (nm)30 elements.
I Some conjectures:I For odd m, limn→∞ c(m, n) = cm. In particular, c3 = 16 (the
value 16 first occurs for k = 90).I For even m with an odd factor, for large enough n, if n even
then c(m, n) = 1 and if n is odd then c(m, n) = 3.I If m = 2d then c(m, n) = 1 for n ≥ 2d+1.I For m ≥ 25, c(m, 3) = 1 if 5 | m; otherwise, c(m, 3) = 4.I For m ≥ 15, c(m, 3) = 1, 3, 4 if
m = ({0}, {5, 10}, other) mod 15.
Open Problems
I Is the Gray code necessary condition sufficient?
I Remove the BC 6= 6 from a previous theorem.
I Gray codes for general column-convex polyominoes?
I Gray codes for convex polyominoes?
I Hexominoes?
I Prove some of the unimodality results. Perhaps the numbersare asymptotically normal?
Open Problems
I Is the Gray code necessary condition sufficient?
I Remove the BC 6= 6 from a previous theorem.
I Gray codes for general column-convex polyominoes?
I Gray codes for convex polyominoes?
I Hexominoes?
I Prove some of the unimodality results. Perhaps the numbersare asymptotically normal?
Open Problems
I Is the Gray code necessary condition sufficient?
I Remove the BC 6= 6 from a previous theorem.
I Gray codes for general column-convex polyominoes?
I Gray codes for convex polyominoes?
I Hexominoes?
I Prove some of the unimodality results. Perhaps the numbersare asymptotically normal?
Open Problems
I Is the Gray code necessary condition sufficient?
I Remove the BC 6= 6 from a previous theorem.
I Gray codes for general column-convex polyominoes?
I Gray codes for convex polyominoes?
I Hexominoes?
I Prove some of the unimodality results. Perhaps the numbersare asymptotically normal?
Open Problems
I Is the Gray code necessary condition sufficient?
I Remove the BC 6= 6 from a previous theorem.
I Gray codes for general column-convex polyominoes?
I Gray codes for convex polyominoes?
I Hexominoes?
I Prove some of the unimodality results. Perhaps the numbersare asymptotically normal?
Open Problems
I Is the Gray code necessary condition sufficient?
I Remove the BC 6= 6 from a previous theorem.
I Gray codes for general column-convex polyominoes?
I Gray codes for convex polyominoes?
I Hexominoes?
I Prove some of the unimodality results. Perhaps the numbersare asymptotically normal?
Thanks for still being here!
And please let me know if you have seen that poset before...
And on to the Venn diagrams.
Definition - Venn diagram
I Made from simple closedcurves C1,C2, . . . ,Cn.
I Infinite intersection notallowed (usually, but nothere!).
I Let Xi denote the interioror the exterior of the curveCi and considerX1 ∩ X2 ∩ · · · ∩ Xn.
I Euler diagram if each suchintersection is connected.
I Venn diagram if Euler andno intersection is empty.
I Independent family if nointersection is empty.
Definition - Venn diagram
I Made from simple closedcurves C1,C2, . . . ,Cn.
I Infinite intersection notallowed (usually, but nothere!).
I Let Xi denote the interioror the exterior of the curveCi and considerX1 ∩ X2 ∩ · · · ∩ Xn.
I Euler diagram if each suchintersection is connected.
I Venn diagram if Euler andno intersection is empty.
I Independent family if nointersection is empty.
Definition - Venn diagram
I Made from simple closedcurves C1,C2, . . . ,Cn.
I Infinite intersection notallowed (usually, but nothere!).
I Let Xi denote the interioror the exterior of the curveCi and considerX1 ∩ X2 ∩ · · · ∩ Xn.
I Euler diagram if each suchintersection is connected.
I Venn diagram if Euler andno intersection is empty.
I Independent family if nointersection is empty.
Definition - Venn diagram
I Made from simple closedcurves C1,C2, . . . ,Cn.
I Infinite intersection notallowed (usually, but nothere!).
I Let Xi denote the interioror the exterior of the curveCi and considerX1 ∩ X2 ∩ · · · ∩ Xn.
I Euler diagram if each suchintersection is connected.
I Venn diagram if Euler andno intersection is empty.
I Independent family if nointersection is empty.
Definition - Venn diagram
I Made from simple closedcurves C1,C2, . . . ,Cn.
I Infinite intersection notallowed (usually, but nothere!).
I Let Xi denote the interioror the exterior of the curveCi and considerX1 ∩ X2 ∩ · · · ∩ Xn.
I Euler diagram if each suchintersection is connected.
I Venn diagram if Euler andno intersection is empty.
I Independent family if nointersection is empty.
Definition - Venn diagram
I Made from simple closedcurves C1,C2, . . . ,Cn.
I Infinite intersection notallowed (usually, but nothere!).
I Let Xi denote the interioror the exterior of the curveCi and considerX1 ∩ X2 ∩ · · · ∩ Xn.
I Euler diagram if each suchintersection is connected.
I Venn diagram if Euler andno intersection is empty.
I Independent family if nointersection is empty.
“polyVenn” diagrams
I Minimum area Venn diagrams.
I Curves are the boundaries of a polyomino.I Each interior/exterior intersection is a square.
I Total bounded area is 2n − 1.I Each polyomino is a 2n−1-omino.
I Introduced by Mark Thompson on a recreational math webpage.
PolyVenn diagram with congruent curves
B
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A CB
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A CB
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A CB
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Top figure colored bycardinality of the underlyingset. Red = 1, yellow = 2, etc.
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n = 7.
In a minimum bounding box
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D D
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CCCC
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A A A
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AAA
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A
A naıve way to construct a polyVenn diagram for any n.
DECECDBEBDBCAEADACABEDCB
ABC
BCDE ACDE ABDE ABCE ABCD CDE BDE BCE BCD ADE ACE ACD ABE ABD
A ABCDE
I A 5-Venn diagram with curve A highlighted.
I Area in general is 322n.
Symmetric chain decomposition of the Boolean lattice
.
ABC
BCACAB
CA
(a)
AD
ADACBCAB
CDBDABD
D
ACD
C
BCD
B
ABC
A
$\emptyset$
ABCD
(b)
$\emptyset$
D
BD
B
ABCD
BCDACDABD
CD
D
AD
ABD
ABCD
A B
BCAB
ABC BCD
ACD
(c)
BD CD
C
AC
I A partition of all 2n
subsets into chains of theform
x1 ⊂ x2 ⊂ · · · ⊂ xt
where |xi | = n − |xt−i+1|and |xi | = |xi−1 − 1|.
I Various algorithms known:
I De Bruijn, vanEbbenhorstTengbergen, andKruyswijk.
I Aigner.
.
ACE
CDE BCE
ABE
ABCE
DEADE
ABDE
ABD
ACDEBCDE
AB
ABC
ABCD
ABCDE
BD
BDE
BECD
E
AE
D
AD
C
AC
ACD
B
BC
BCD
A
CE
I Another 5-Venn diagram withcurve A highlighted.
I Based on symmetric chaindecomposition.
I Area is 2n +( nn/2
).
I Are there congruent n-polyVenns for n ≥ 6? We saw earlierthat they exist for n = 2, 3, 4, 5.
I Is there a 5-polyVenn whose curves are convex polyominoes?
I Are there minimum bounding box n-polyVenns for n ≥ 6? Wesaw earlier that they exist for n = 2, 3, 4, 5.
I Are there minimum area n-polyVenns for n ≥ 8? We sawearlier examples for n = 6, 7.
I One problem for which we have not attempted solutions is theconstruction of n-polyVenns that fill an w × h box, wherewh = 2n − 1. Of course, a necessary condition is that 2n − 1not be a Mersenne prime. For example, is there are4-polyVenn that fits in a 3× 5 rectangle or a 6-polyVenn thatfits in a 7× 9 or 3× 27 rectangle?
Thanks for coming!