Perfect match for "bra": High Temperature Expansion and vertex expansion techniques

The purpose of this article is to use an example to introduce two classic methods for accurately solving models in statistical mechanics: High Temperature Expansion and vertex expansion.

The problem is: consider a very similar figure like "bra:

Note that this graph is not a floor plan! The edges of the above two solid wires are respectively glued together with the edges of the following two solid wires. the edges of the two solid wires on the left and those on the right are also bonded separately. The dotted line is not an edges, it is only used to describe the bonding orientation. Therefore, the degree of each vertex in this graph is 3.

Imagine its stereoscopy: This is a bra that has been worn on a pretty girl, so the upper and lower sides and the Left and Right solid sides are equivalent to the tied rope.

How many different perfect matches does this image have?

The answer is $64 $. Do you want to know why?

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First, we will introduce a technique used in counting composition: high temperature expansion. It comes from the Ising Model in physics.

Set $ G $ to a finite connected graph, and the vertex is marked as $1, 2, \ cdots, N $ in sequence. Assign a variable $ \ sigma_ I \ In \ {+ 1,-1 \} $ to each vertex $ I $. Each Assignment Method $ \ Sigma =\{\ PM1, \ cdots, \ PM1 \}$ is called a state of the system, the system has a total of $2 ^ N $ possible states.

The configuration function $ Z $ of the entire system is defined

\ [Z = \ sum _ {\ Sigma} \ prod _ {(I, j) \ In e} e ^ {k \ sigma_ I \ sigma_j}. \]

Here, the product number of the inner layer is all the sides of $ G $, and the sum Number of the outer layer is all possible States $2 ^ N $, $ K $ is a normal number.

Finding a system's allocation function is very important. onsager obtained the exact solution of the allocation function of the two-dimensional Ising Model in 1944, it is regarded as the first major theoretical achievement in statistical physics, and the precise solution of the three-dimensional Ising model has not yet been solved.

Theorem:$ E ^ {k \ sigma_ I \ sigma_j} = \ cosh (k) (1 + \ sigma_ I \ sigma_j \ Tanh (k) $.

Proof of the theorem: \ [\ begin {Align} e ^ {k \ sigma_ I \ sigma_j} & =\ frac {e ^ {k \ sigma_ I \ sigma_j} + e ^ {-k \ sigma_ I \ sigma_j }}{ 2} + \ frac {e ^ {k \ sigma_ I \ sigma_j}-e ^ {-k \ sigma_ I \ sigma_j }}{ 2} \ & = \ cosh (k \ sigma_ I \ sigma_j) + \ sinh (k \ sigma_ I \ sigma_j) \\&=\ cosh (k \ sigma_ I \ sigma_j) (1 + \ Tanh (k \ sigma_ I \ sigma_j )) \\&=\ cosh (k) (1 + \ sigma_ I \ sigma_j \ Tanh (k )). \ end {Align} \]

The last equal sign here is because $ \ sigma_ I \ sigma_j = \ PM1 $, and $ \ cosh $ and $ \ Tanh $ are the even functions and odd functions respectively.

Now we can use the above theorem to simplify $ Z $ :( for simple note $ A = \ Tanh (k) $)

\ [\ Begin {Align} Z & = \ sum _ {\ Sigma} \ prod _ {(I, j) \ In e} \ cosh (k) (1 + \ sigma_ I \ sigma_j \ Tanh (k) \ & = [\ cosh (k)] ^ {| E |} \ sum _ {\ Sigma} \ prod _ {(I, j) \ In e} (1 + A \ sigma_ I \ sigma_j ). \ end {Align} \]

If the product in the above formula is expanded, it is the sum of all subgraphs $ h $ of $ G $:

\ [\ Prod _ {(I, j) \ In e} (1 + A \ sigma_ I \ sigma_j) = \ sum _ {H} a ^ {| E (h) |} \ prod _ {(I, j) \ in E (h)} \ sigma_ I \ sigma_j. \]

Therefore, \ [\ begin {Align} Z & = [\ cosh (k)] ^ {| E |} \ sum _ {\ Sigma} \ sum _ {H} a ^ {| E (h) |}\ prod _ {(I, j) \ in E (h)} \ sigma_ I \ sigma_j \\& = [\ cosh (k)] ^ {| E |} \ sum _ {H} a ^ {| E (h) |} \ sum _ {\ Sigma} \ prod _ {(I, j) \ in E (h)} \ sigma_ I \ sigma_j. \ end {Align} \]

We found \ [\ begin {Align *} \ sum _ {\ Sigma} \ prod _ {(I, j) \ in E (h )} \ sigma_ I \ sigma_j & =\ sum _ {\ Sigma} \ sigma_1 ^ {d_h (1)} \ sigma_2 ^ {d_h (2 )} \ cdots \ sigma_n ^ {d_h (n) }\\& = (\ sum _ {\ sigma_1 = \ PM1} \ sigma_1 ^ {d_h (1 )}) (\ sum _ {\ sigma_2 = \ PM1} \ sigma_2 ^ {d_h (2 )}) \ cdots (\ sum _ {\ sigma_n = \ PM1} \ sigma_n ^ {d_h (n )}). \ end {Align *} \]

Here $ d_h (I) $ indicates the degree of vertex $ I $ in the subgraph $ h $.

If $ d_h (I) $ is an odd number, then $ \ sum _ {\ sigma_ I = \ PM1} \ sigma_ I ^ {d_h (I)} = 0 $; otherwise, it is equal to 2. Therefore, we obtain the following formula for expanding the temperature:

\ [Z = 2 ^ {| v |} [\ cosh (k)] ^ {| E |} \ sum _ {\ Text {$ h $ is an Euler's subgraph of $ G $} a ^ {| E (h) | }. \]

Here, the Euler's subgraph of $ G $ is defined as a subgraph with an even number of vertices.

The high temperature expansion formula can help us calculate the number of Euler's subgraphs of $ G $. This is the theorem below:

Theorem:If $ G $ is a finite connected graph, the number of Euler's subgraphs of $ G $ is $2 ^ {| E |-| v | + 1} $.

Proof: set the number of Euler's subgraphs of $ G $ to $ N $, so that $ k \ to + \ infty $, then $ \ cosh (k) \ Sim \ dfrac {e ^ k} {2} $, $ A = \ Tanh (k) \ sim1 $, so

\ [Z \ sim2 ^ {| v |} (\ frac {e ^ k} {2}) ^ {| E |} \ cdot n. \ quad k \ to + \ infty \]

Then \ [\ sum _ {\ Sigma} \ prod _ {(I, j) \ In e} e ^ {k \ sigma_ I \ sigma_j} \] on the other side

What about it? In this formula, when all $ \ sigma_ I $ values are + 1 or-1, the value of the two items is the largest, and the sum is $2 e ^ {k | E |}$, the order of other items is strictly less than these two items, so $ Z \ SIM 2E ^ {k | E |}$, $ k \ to + \ infty $. If the two estimates are equal, $ n = 2 ^ {| E |-| v | + 1} $ is obtained.

The way we come to this conclusion seems to be twists and turns, and we were surprised by a seemingly unrelated question. This was derived by physicists. It is not surprising that mathematicians have a more direct approach (using a Tutte polynomial or linear algebra), but this technique was initially developed.

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Fisher vertex Expansion Technique

Vertex expansion technique is a graph transformation technique introduced by Fisher when solving the precise solution of the two-dimensional Ising model. One of the following versions is available:

Set $ G $ to a 3-regular graph (each vertex has a degree of 3). We perform the following operations on $ G $ to convert $ G $ into a new graph $ G' $: for each vertex $ V $ of $ G $, scale $ V $ "to a 3-ring, set the vertex of the 3-ring to replace $ V $ with the three neighbors of $ V $. Figure $ G' $ is a fisher image of $ G $.

Theorem [Fisher]:All the even subgraphs in the figure $ G $ correspond to all the perfect matches in the figure $ G' $.

Proof: Set $ p $ to an even subgraph of $ G $, so that $ P ^ \ ast $ is the complement subgraph of $ p $ in $ G $: the vertex set of $ P ^ \ ast $ is also $ V (G) $, but the edge of $ P ^ \ ast $ is composed of those that do not belong to $ p $. Because $ p $ is an even subgraph and $ G $ is a three-time graph, each vertex degree of $ p $ is 0 or 2, therefore, the degree of each vertex of $ P ^ \ ast $ is 1 or 3.

Let's take a look at $ P ^ \ ast $. There are only two possibilities for $ P ^ \ ast $ at a vertex $ V $:

The first possibility is that all three edges of the vertex $ V $ belong to $ P ^ \ ast $. Therefore, we define $ G' $ matching in this part, for example, right:

The second possibility: if only one of the three edges of the vertex $ V $ belongs to $ P ^ \ ast $, the matching of $ G' $ in this part is defined as right:

In this way, for the given $ P ^ \ ast $, we specify a match at each 3-ring of $ G' $, then, does the local match "Work Together" constitute a $ G' $ match? This requires that the matching is compatible. For example, in $ G' $, two adjacent 3-rings $ \ delta_1 $ and $ \ delta_2 $, $ v_1 \ In \ delta_1 $ and $ V_2 \ In \ delta_2 $ are connected by an edge $ e $. If $ \ delta_1 $ matches another vertex in $ v_1 $ and $ \ delta_1 $, the matching at $ \ delta_2 $ matches $ V_2 $ and $ v_1 $, so these two matching are obviously incompatible. Therefore, we need to note that edge $ e $ is either used by both $ \ delta_1 $ and $ \ delta_2 $, or not used at the same time. This is easy: If $ \ delta_1 $ uses $ e $, it indicates that the edges between the two vertices in the graph $ G $ expanded to $ \ delta_1 $ and $ \ delta_2 $ belong to $ P ^ \ ast $, therefore, $ \ delta_2 $ is also used.

The inverse ing is also obvious. For every matching of $ G' $, every 3-ring is reduced to a single vertex, and a graph with a vertex degree of 1 or 3 is obtained, then perform the complement operation to obtain an even subgraph of $ G $.

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Now we can answer the question about the number of perfect matches in the "bra" Graph: Remember this graph as $ G $. $ G $ meets the following requirements:

1. Its vertex is the union of several 3-rings without common vertices.

2. Its edges are composed of the edges in the 3-ring and the edges connecting the 3-ring.

3. Each vertex level of $ G $ is 3.

Compress each 3-ring of $ G $ into a vertex, and the resulting graph is recorded as $ h $. Then $ h $ is a 3-regular graph, therefore, the number of perfect matches of $ G $ is equal to the number of even subgraphs of $ h $. We have used the High Temperature Expansion Technique to calculate the number of even subgraphs of $ h $ as $2 ^ {| E (h) |-| V (h) | + 1} $, the following figure shows these quantities using the $ G $ parameter.

First $ | V (h) | = | V (G) |/3 $. Second, the sum of all vertex degrees in the figure $ h $ is equal to $2 | E (h) | $, while $ h $ is 3-regular, SO \ [3 | V (h) | = 2 | E (h) |, \]

That is, $ | E (h) | = | V (G) |/2 $, so the number of perfect matches for $ G $ is

\ [2 ^ {| E (h) |-| V (h) | + 1} = 2 ^ {\ frac {| V (G) |} {6} + 1 }. \]

Perfect match for "bra": High Temperature Expansion and vertex expansion techniques