Aztec diamond problem: Spider move

This article will introduce the amazing Aztec diamond model. I think this is a question that can be entered into the "Mathematical book". Its Expression is simple and elementary, but it is closely related to many of the most profound and wonderful questions in the counting combination, of course, it also has an amazing solution. I can't help but write it out and share it with you.

So what is the Aztec diamond model? Very simple: Stack $2, 4, \ cdots, 2n $ squares in sequence, and then perform an axial reflection on $ x $, the resulting image is named $ N $ level Aztec diamond map $ az_n $; A domino cover in an Aztec diamond map is to cover each square of $ az_n $ with some $1 \ times2 $ dominoes without overlap, as shown in: ($ n = 4 $ in the example)

Our question is: $ N $ Aztec diamond map $ az_n $ how many different dominoes are covered?

Domino coverage is essentially a perfect match of a plan. Consider each square in a new map $ g_n $ and $ az_n $ as the vertex in $ g_n $, the two vertices in $ g_n $ are adjacent only when their corresponding square has a public edge in $ az_n $. In the case of $ n = 4 $, $ G_4 $ is like: (the square in the shadow is called the cell space. You can see a total of $ n ^ 2 $ cells. The spider movement to be introduced later is the transformation defined in the cell cavity)

Now we have transformed the problem into finding the number of perfect match for a plan. The most basic idea to solve this problem is the weight function.

Set $ G $ to a simple plot. $ G $ each edge of $ e $ has a weight value $ W (e) $, $ W (E) $ is usually a variable, such as $ X, Y, Z, W $. For every perfect match of $ G $ M $, define the weight of $ M $ W (m) $ as the product of the weight values of all edges in $ M $: \ [W (m) = \ prod _ {e \ In m} W (E), \]

Then define the Weight Function of $ G $ M (g) $ for the sum of all perfectly matched weights of $ G $:

\ [M (G) = \ sum _ {m} W (M). \]

If the weight of each edge of $ G $ is 1, $ M (g) $ is the number of perfect matches of $ G $. However, in general, the edge weight $ W (e) $ is an uncertain element, so $ M (g) $ is a multivariate function that contains uncertain elements. However, as long as the expression with the weight function $ M (g) $ is obtained, the number of perfect matches for $ G $ is obtained by assigning all the variables to 1.

It is of course a good thing to be able to obtain the weight function, because the weight function contains a lot of information about the graph and can help us calculate many other values of interest. For example, if you specify an edge $ e $, how many matches do not contain $ e $? Therefore, you only need to set the weight of edge $ e $ to 0, and keep the other edge weight to 1, which can be substituted into the weight function.

It seems that finding the weight function of an image is more complicated than directly calculating the number of perfect matches. Why should we leave the distance? What are the mysteries of the Weight Function method?

This is the key: For a complex graph, we want to convert it into a simple graph through some "operations" or "transformations", such as deleting some vertices and edges, or replace a part with a new graph. Of course, these transformations usually change the number of perfectly matched graphs,The weight function maintains some immutability or similarity before and after the transformation.In this way, we can find the recursive relationship satisfied by the weight function and then calculate it.

The Aztec diamond model is composed of N. elkies, G. kuperberg, M. larsen and J. propp was proposed in his paper in 1992. This is a wonderful article, but it is also difficult to understand (it requires a deep understanding of algebra ). In that article, they gave a total of four solutions, both of which are very "advanced ". Today, there are no more than a dozen solutions for counting the Aztec diamond model. However, the most amazing solution is undoubtedly the method we will introduce called "Spider mobile. Although it was later found to be equivalent to the fourth solution Domino shuffling algorithm in the papers of elkies and others, the expression of spider movement is more concise and elegant. Roughly speaking: The Domino shuffling algorithm is used to reverse the transformation steps of spider movement. They have their own functions: We use spider movement to count and Domino shuffling algorithm to generate random matching.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~

We need two simple but extraordinary latencies:

Theorem: vertex splitting

: Set the neighbor of the vertex $ V \ in G $ N (v) $ to be divided into two non-intersecting sets $ N (v) = H \ cup K $, here, $ h $ and $ K $ allow an empty set. Let's make a simple transformation to the figure $ G $: split two new vertices from $ V ', v'' $, and use $ V' $ "instead of" $ V $ to connect to the vertices in $ h $, use $ v'' $ "instead of" $ V $ to connect to the vertex in $ K $, $ V $ to $ V ', when the edge weight between v'' $ is defined as 1, the new image $ G' has the same weight function as the source image: $ M (G) = M (G ') $.

The proof is very simple. You can try it on your own.

In the Journey to the West, Sun Wukong has a stunt, pulling out a handful of hair, blowing a breath, shouting a change, these hairs become many small Sun Wukong, and then playing monsters at the tables, this is Sun Wukong's separation technique. Our idea here is to make $ V $ connect $ H, K $, when the two bodies are separated for them, it only needs to connect $ V' and v'' $.

The vertex splitting theorem is usually a preprocessing step, which has little influence on the graph. The core part is the second theorem: Spider movement.

2: Spider Movement

Suppose a part of the figure $ G $ is shown in:

The weights of the four edges in the center cell are $ W, X, Y, Z $, and the four edges $ A, B, C, the weights of the four edges connected by d $ are 1.

Now we delete the center cell and connect $ a, B, c, d $ to a new cell, and the specified edge weight is (note $ \ Delta = WZ + XY $)

\ [W' = \ frac {z} {\ Delta}, \ quad Z' = \ frac {w} {\ Delta }, \ quad x' =\frac {y }{\ Delta}, \ quad y' =\frac {x }{\ Delta }. \]

The relationship between the new image $ G' $ and the weight function of the source image $ G $ is as follows:
\ [M (G) = \ Delta M (G'). \]

Proof: Set $ s =\{ a, B, c, d \}$ according to the matching method of $ S $ in $ G $, all perfect matches of $ G $ can be divided into three categories:

The four vertices in I. $ S $ match the four vertices in the cell cavity one by one;

The four vertices in II. $ S $ match the external vertices;

Iii. $ S $ has two vertices matching the vertex in the cell cavity, and the other two matching the external vertex.

It is recommended that you understand how these three situations correspond to the matching of $ G' $, because they are important for understanding the following Domino shuffling algorithms.

For each matching $ M $ of class I, set the factor product of the remaining edge in $ M $ to $ q $, $ \ Pi (m) = Q $ because the four edges inside $ M $ are 1. After deleting the inner cell, we can define the matching of $ A, B, C, and D $ in two ways, that is, $ \ {AB, CD \} $ or $ \ {ad, BC \} $. The weight of the former is $ w'z' Q $, and the latter is $ X 'y' Q $, and the sum is $ (w'z' + x 'y ') = Q/\ Delta $.

In short, for each matching of class I, we can convert it into two matches in $ G' $, and the sum of the transformed matching weights is the original $1/\ Delta $.

For matching in Class II, because the four vertices in the cell cavity are paired with each other, there are two possible methods: $ wzq $ and $ xyq $, among them, $ q $ is still the product of the weights that match the rest. However, after deleting the cell, these two types may become one with the weight of $ q $.

In short, the matching of Class II can be combined into a match in $ G' $ by pairing them, and the Transformed Weight is also the original $1/\ Delta $.

Set the vertex $ \ {, B \} $ matches the vertex in the cell (the other three possible scenarios are $ \ {B, c \ }$, $ \ {CD \ }$, $ \ {, d \}$, analysis is similar, and $ \{ a, c \}$ and $ \{ B, d \}$ are not possible ). Therefore, the weight of $ M $ is $ ZQ $. After deleting the cell space, there is only one way to define the matching in $ G' $ without affecting the matching of other parts, the weight is $ w'q $, therefore, the transformed weight is still the original $ \ frac {w'q} {ZQ }=\ frac {1} {\ Delta} $.

Based on the above three cases, we get $ M (G) = \ Delta M (G') $.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~

Solutions to Aztec diamond problems:

Let's give a solution to the Aztec diamond problem: Set the Weight Function of $ g_n $ to $ W (g_n) $ and place it at an angle of 45 degrees, this clearly shows the $ n ^ 2 $ cell. The weight of each cell edge is the same as that above, and the clockwise direction is $ W, X, Z, y $.

First, use the vertex split repeatedly in the $ n ^ 2 $ cell. In the following figure, the newly added edge weight is 1:

Then, use $ n ^ 2 $ to move the second image in each cell cavity to obtain the third image:

The Weight Function of the first and second images is the same, both of which are $ W (g_n) $; the Weight Function of the third figure after $ n ^ 2 $ spider movement is $ \ dfrac {W (g_n) }{\ Delta ^ {n ^ 2 }}$.

We can use another angle to calculate the Weight Function of the third image. This is also the whole proof of the peak and return, the dark flowers and flowers! Note that the matching of the third graph is restricted to death on the boundary. The actual free part is an internal low-level graph $ g _ {n-1} $, so we can strip the outer ring loop!

The Edge Weight rules of this subgraph are $ w/\ delta, X/\ delta, Z/\ delta, Y/\ Delta $, therefore, the weight function of this subgraph is $ \ dfrac {W (G _ {n-1}) }{\ Delta ^ {n (n-1 )}} $ (because $ g _ {n-1} $ contains $ n (n-1) $ edge, half of the number of vertices in $ g _ {n-1} $ ). In this way, we use two methods to obtain the Weight Function of the third image, that is, \ [\ frac {W (g_n )} {\ Delta ^ {n ^ 2 }}=\ frac {W (G _ {n-1}) }{\ Delta ^ {n (n-1) }}, \]

That is, \ [W (g_n) = \ Delta ^ NW (G _ {n-1}), \]

Note $ n = 1 $ W (G_1) = \ Delta $, and $ W (g_n) = \ Delta ^ {\ frac {n (n + 1 )} {2 }}$. In particular, $ W = x = y = z = 1 $, we get the $ N $ level Aztec diamond map $ ad_n $. The number of all dominoes covered is $2 ^ {\ Frac. {n (n + 1 )} {2 }}$.

How is it? Is the entire process very exciting? Spider mobility is similar to the $ Y-\ Delta $ transform in an electrical network. Similar graph transformation techniques are commonly used in precisely solvable lattice models.

Although the counting of the Aztec diamond model has been completed so far, it is just the beginning of a wonderful story. We haven't talked about how to turn spider movement into a random matching shuffling algorithm, and other brilliant solutions to the Aztec diamond model. There is also a very interesting phenomenon of "frozen Arctic Ocean, wait. This may be left in future articles.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~

References

1. Generalized Domino-shuffling, James Propp, 2000.

2. altenating sign matrices and Domino tilings, elkies, kuperberg. 1991.

3. Random Domino tilings and the Arctic Circle theorem, jockush, Propp

Aztec diamond problem: Spider move