A new proof for the number of lozenge tilings of quartered hexagons

Abstract

It has been proven that the lozenge tilings of a quartered hexagon on the triangular lattice are enumerated by a simple product formula. In this paper we give a new proof for the tiling formula by using Kuo's graphical condensation. Our result generalizes a Proctor's theorem on enumeration of plane partitions contained in a "maximal staircase".

Lozenge Tilings of a Halved Hexagon with an Array of Triangles Removed from the Boundary

“…It is worth noticing that the author gave another proof for the equalities (3.1), (3.2), and (3.3) in [18] by using Kuo condensation. We can also prove (3.4) by using the same method.…”

A q-enumeration of lozenge tilings of a hexagon with three dents

“…[YYZ,YZ,Ku06,Sp,Ci15,Fu] for various aspects and generalizations of Kuo condensation; and e.g. [CK,CL,CF14,CF15,KW,La15a,La15b,La15c,La14,LMNT,Zh] for recent applications of the method.…”

Problems in the Enumeration of Tilings

“…Jockusch and Propp [41] introduced the "quartered Aztec diamonds" as quarters of an Aztec diamond divided by two zigzag cuts passing the center (see Figure 2.6). These regions have been re-investigated and generalized in [52][53][54][55]. These papers showed that one could transform a "quartered Aztec rectangle" (a natural generalization of the quartered Aztec diamond) into a quartered hexagons using certain local graph transformations.…”