Why rhombus tilings are just piling boxes with decending restrictions Suppose $H$ is a hexagon with sidelength $a\times b\times c$, and it's innner angles are all $\frac{2\pi}{3}$. Then it's well known that the rhombus tilings of $H$ is in one-to-one correspondece with plane partitions of shape $a\times b$ and largest number not exceeding $c$. 
That is, if we draw a 3D graph of any tiling of $H$, it will look like piling unit boxes at the corner of the wall, but with decending conditions. This fact is quite obvious, but I still want a rigorous proof. My question is, can we give a real bijection between these two objects, rather than just saying "draw a graph and it's obvious that ..." ? 


Consider $\mathbf{R}$ is the set of rhombic tilings and $\mathbf{B}$ is the set of box pilings. That there exists a 1-1 function from $\mathbf{B}\rightarrow\mathbf{R}$ I think is quite obvious - i.e. box sides are just rhombi, therefore we have a rhombic tiling. Perhaps the more difficult part is seeing that this function is onto. Let us instead show that there is a function from $\mathbf{R}\rightarrow\mathbf{B}$, and we will have shown that there exists a bijection.
I think this becomes trivial once you look at all possible combinations at any vertex. If we show that each rhombic tiling around a vertex looks (locally) like the surface of some box pile, then we have that all vertices form a mesh of the box pile. Hence, we have found that for each rhombic tile there is a box pile.
The possible combinations are (sorry for the lack of visuals, but you can find most of these examples in the picture you provided):

*

*3 rhombi arranged top, bottom left, bottom right -> top corner of a box.

*3 rhombi arranged bottom, top left, top right -> top of a box (or floor) from below, right side of a box (or wall) from the left, left side a box (or wall) from the right.

*6 rhombi arranged as a star -> a box above, to the bottom left, to the bottom right.

*5 rhombi arranged with obtuse angle facing bottom -> tops of 3 boxes below to the bottom, left and right, and the adjacent sides of one box above.

*5 rhombi arranged with obtuse angle facing top -> tops of 3 boxes, and the sides of the two front facing boxes.

*5 rhombi arranged with obtuse angle facing lower left -> 2 boxes stacked to the left of vertex, and one box to the right.

*5 rhombi arranged with obtuse angle facing lower right -> 2 boxes stacked to the right of vertex, and one box to the left.

*5 rhombi arranged with obtuse angle facing upper left -> 1 box to the right and and right side of boxes (or wall) to the left and above.

*5 rhombi arranged with obtuse angle facing upper right -> 1 box to the left and and left side of boxes (or wall) to the right and above.

*4 rhombi arranged with obtuse angle above and below -> the floor or 4 boxes below.

*4 rhombi arranged with obtuse angle to top left, bottom right -> the left wall or 4 boxes to the left.

*4 rhombi arranged with obtuse angle to top right, bottom left -> the right wall or 4 boxes to the right.

And that is all the possible arrangements of rhombi around a vertex (2 permutations of 3, 3 permutations of 4, 6 permutations of 5, and 1 permutation of 6).
