Skip to main content
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
added 20 characters in body
Source Link
Nandakumar R
  • 6k
  • 3
  • 7
  • 20

Background: Given a convex region C. One can define a graph corresponding to a planar arrangement of non overlapping congruent copies of C - each unit C is a node and an edge connects it to another node iff the two C units share some finite length of boundary. Such a graph is necessarily planar. As is well known, the average degree of such a planar graph can at most be 6.

Given any triangle, the intuitive (parallelogram based) tilings of the plane with it seem to correspond to graphs (defined above) with average degree = 3 or 4 - there appears a gap between these values and the maximum average degree of a planar graph (6).

Question: Can one have an infinite planar arrangement (not necessarily a tiling) with copies of some triangle where the average degree of the corresponding graph is between 4 and 6?

Note: With squares, it is easy to form a tiling with the corresponding graph having degree 6 at every node - the highest possible average degree - or 5 at each node.

Generalization: Given a convex 2D shape C, not necessarily one that tiles. Suppose one is also somehow given the planar arrangement(s) with infinite copies of C that maximizes the average degree of the corresponding graph. How good are these arrangements at achieving max packing density? And analogous questions may be raised in 3D.

Further thought: If one alters the above definition of the graph corresponding to an infinite non-overlapping layout with congruent copies of some convex C as: "nodes are C units as before but an edge connects two units if they even touch at a point", then, how can we upper bound the average degree of the graph for any convex C? (Note: as per this new definition, the graph won't be planar).

Background: Given a convex region C. One can define a graph corresponding to a planar arrangement of non overlapping congruent copies of C - each unit C is a node and an edge connects it to another node iff the two C units share some finite length of boundary. Such a graph is necessarily planar. As is well known, the average degree of such a planar graph can at most be 6.

Given any triangle, the intuitive (parallelogram based) tilings of the plane with it seem to correspond to graphs (defined above) with average degree = 3 or 4 - there appears a gap between these values and the maximum average degree of a planar graph (6).

Question: Can one have an infinite planar arrangement (not necessarily a tiling) with copies of some triangle where the average degree of the corresponding graph is between 4 and 6?

Note: With squares, it is easy to form a tiling with the corresponding graph having degree 6 at every node - the highest possible average degree.

Generalization: Given a convex 2D shape C, not necessarily one that tiles. Suppose one is also somehow given the planar arrangement(s) with infinite copies of C that maximizes the average degree of the corresponding graph. How good are these arrangements at achieving max packing density? And analogous questions may be raised in 3D.

Further thought: If one alters the above definition of the graph corresponding to an infinite non-overlapping layout with congruent copies of some convex C as: "nodes are C units as before but an edge connects two units if they even touch at a point", then, how can we upper bound the average degree of the graph for any convex C? (Note: as per this new definition, the graph won't be planar).

Background: Given a convex region C. One can define a graph corresponding to a planar arrangement of non overlapping congruent copies of C - each unit C is a node and an edge connects it to another node iff the two C units share some finite length of boundary. Such a graph is necessarily planar. As is well known, the average degree of such a planar graph can at most be 6.

Given any triangle, the intuitive (parallelogram based) tilings of the plane with it seem to correspond to graphs (defined above) with average degree = 3 or 4 - there appears a gap between these values and the maximum average degree of a planar graph (6).

Question: Can one have an infinite planar arrangement (not necessarily a tiling) with copies of some triangle where the average degree of the corresponding graph is between 4 and 6?

Note: With squares, it is easy to form a tiling with the corresponding graph having degree 6 at every node - the highest possible average degree - or 5 at each node.

Generalization: Given a convex 2D shape C, not necessarily one that tiles. Suppose one is also somehow given the planar arrangement(s) with infinite copies of C that maximizes the average degree of the corresponding graph. How good are these arrangements at achieving max packing density? And analogous questions may be raised in 3D.

Further thought: If one alters the above definition of the graph corresponding to an infinite non-overlapping layout with congruent copies of some convex C as: "nodes are C units as before but an edge connects two units if they even touch at a point", then, how can we upper bound the average degree of the graph for any convex C? (Note: as per this new definition, the graph won't be planar).

Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
added 446 characters in body
Source Link
Nandakumar R
  • 6k
  • 3
  • 7
  • 20

Background: Given a convex region C. One can define a graph corresponding to a planar arrangement of non overlapping congruent copies of C - each unit C is a node and an edge connects it to another node iff the two C units share some finite length of boundary. Such a graph is necessarily planar. As is well known, the average degree of such a planar graph can at most be 6.

Given any triangle, the intuitive (parallelogram based) tilings of the plane with it seem to correspond to graphs (defined above) with average degree = 3 or 4 - there appears a gap between these values and the maximum average degree of a planar graph (6).

Question: Can one have an infinite planar arrangement (not necessarily a tiling) with copies of some triangle where the average degree of the corresponding graph is between 4 and 6?

Note: With squares, it is easy to form a tiling with the corresponding graph having degree 6 at every node - the highest possible average degree.

Generalization: Given a convex 2D shape C, not necessarily one that tiles. Suppose one is also somehow given the planar arrangement(s) with infinite copies of C that maximizes the average degree of the corresponding graph. How good are these arrangements at achieving max packing density? And analogous questions may be raised in 3D.

Further thought: If one alters the above definition of the graph corresponding to an infinite non-overlapping layout with congruent copies of some convex C as: "nodes are C units as before but an edge connects two units if they even touch at a point", then, how can we upper bound the average degree of the graph for any convex C? (Note: as per this new definition, the graph won't be planar).

Background: Given a convex region C. One can define a graph corresponding to a planar arrangement of copies of C - each unit C is a node and an edge connects it to another node iff the two C units share some finite length of boundary. Such a graph is necessarily planar. As is well known, the average degree of a planar graph can at most be 6.

Given any triangle, the intuitive (parallelogram based) tilings of the plane with it seem to correspond to graphs (defined above) with average degree = 3 or 4 - there appears a gap between these values and the maximum average degree of a planar graph (6).

Question: Can one have an infinite planar arrangement (not necessarily a tiling) with some triangle where the average degree of the corresponding graph is between 4 and 6?

Note: With squares, it is easy to form a tiling with the corresponding graph having degree 6 at every node - the highest possible average degree.

Generalization: Given a convex 2D shape C, not necessarily one that tiles. Suppose one is also somehow given the planar arrangement(s) with infinite copies of C that maximizes the average degree of the corresponding graph. How good are these arrangements at achieving max packing density? And analogous questions may be raised in 3D.

Background: Given a convex region C. One can define a graph corresponding to a planar arrangement of non overlapping congruent copies of C - each unit C is a node and an edge connects it to another node iff the two C units share some finite length of boundary. Such a graph is necessarily planar. As is well known, the average degree of such a planar graph can at most be 6.

Given any triangle, the intuitive (parallelogram based) tilings of the plane with it seem to correspond to graphs (defined above) with average degree = 3 or 4 - there appears a gap between these values and the maximum average degree of a planar graph (6).

Question: Can one have an infinite planar arrangement (not necessarily a tiling) with copies of some triangle where the average degree of the corresponding graph is between 4 and 6?

Note: With squares, it is easy to form a tiling with the corresponding graph having degree 6 at every node - the highest possible average degree.

Generalization: Given a convex 2D shape C, not necessarily one that tiles. Suppose one is also somehow given the planar arrangement(s) with infinite copies of C that maximizes the average degree of the corresponding graph. How good are these arrangements at achieving max packing density? And analogous questions may be raised in 3D.

Further thought: If one alters the above definition of the graph corresponding to an infinite non-overlapping layout with congruent copies of some convex C as: "nodes are C units as before but an edge connects two units if they even touch at a point", then, how can we upper bound the average degree of the graph for any convex C? (Note: as per this new definition, the graph won't be planar).

Bumped by Community user
Bumped by Community user
added 46 characters in body
Source Link
Nandakumar R
  • 6k
  • 3
  • 7
  • 20
Loading
problem with the use of the word 'any' addressed.
Source Link
Nandakumar R
  • 6k
  • 3
  • 7
  • 20
Loading
removed capitals from title, removed cryptic abbreviations
Source Link
YCor
  • 63.9k
  • 5
  • 187
  • 286
Loading
Source Link
Nandakumar R
  • 6k
  • 3
  • 7
  • 20
Loading