Domino coverage of the Board: dimer lattice model, Pfaff polynomial, and kasteleyn Theorem

This time we will introduce a classic problem in the counting combo: dimer lattice model. The question is: How many different dominoes are there for a 64-square chess board? Here, the covering refers to covering the entire board without repetition or omission.

It is a possible method of overwriting (the picture is from Wikipedia ):

The answer to this question is 12988816. A very large number is definitely not counted one by one. In 1961, German physicist kasteleyn first solved this problem by referring to a conclusion in linear algebra. Next we will introduce his method.

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Step 1: dominoes overwrite COUNT = perfectly matched count

Consider the entire chessboard as a graph $ G $, and the 64 squares as the vertices of $ G $. The two vertices in $ G $ are adjacent when and only when their corresponding squares are adjacent. In this way, a simple undirected plan $ G $ is obtained, and the number of Domino overwrites in the calculation board is converted to the number of perfect matches in the calculation of $ G $.

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Step 2: Use the Pfaff polynomial to combine all the perfect matches of a graph into an expression.

This step is long and the details involved are also complicated.

Let's take a look at an example of a level 4 opposition matrix:

\ [\ Det \ begin {pmatrix} 0 & A _ {12} & A _ {13} & A _ {14} \-A _ {12} & 0 & _ {23} & A _ {24} \-A _ {13} &-A _ {23} & 0 & A _ {34} \-A _ {14} &-A _ {24} &-A _ {34} & 0 \ end {pmatrix} = (A _ {12} A _ {34}-A _ {13} A _ {24} + A _ {14} A _ {23 }) ^ 2. \] What did you find? The determining factor of this anticalled matrix can be expressed as the square of a polynomial! In addition, observe the subscripts of each monominal on the right. They are $ \ {(12), (34) \ }$, $ \ {(14), (23), respectively) \}$, $ \ {(13), (24) \}$, exactly all the matching methods of the set $ \ {1, 2, 3, 4 \} $!

This conclusion is not accidental. In fact, any even-order inverse matrix can be expressed as the square of a polynomial, which is called the Pfaff polynomial.

So what about the odd-order antisymmetric matrix? Their determinant is 0, so they are not considered.

The following describes the Pfaff polynomials.

Consider a way to pair the set $ s =\{ 1, 2, \ ldots, 2n \}$ (which is divided into $ N $ pairs) $ \ Pi $, recorded as \ [\ Pi = (I _1, J_1) (I _2, J_2) \ cdots (I _n, j_n), \ quad I _k <j_k, \ quad k = 1, \ ldots, n. \] $ \ Pi $ is a matching of the set $ S $.

For the given $ 2n $ level objection matrix $ A = (A _ {IJ}) $, define \ [A _ \ Pi = A _ {I _1j_1} A _ {I _2j_2} \ cdots A _ {I _nj_n }. \]

$ A _ \ Pi $ is the product of the elements in the upper triangle of the matrix $ A $.

We also need to define the symbol $ \ Text {sign} (\ PI) $ that matches $ \ Pi $. Mark $1, 2, \ ldots, 2n $ on the number axis, and connect them with the arc above the number axis in sequence $ (I _1, J_1) $, $ \ cdots $, $ (I _n, j_n) $, the number of all intersections between arcs is recorded as $ K $, and $ \ Text {sign} (\ PI) = (-1) ^ {k} $ is defined.

Theorem:Set $ A $ to $ 2n $ level objection matrix, then \ [\ det A = (\ sum _ {\ PI} \ Text {sign} (\ PI) A _ \ PI) ^ 2 = [\ Text {PF (a)}] ^ 2. \]

Here, $ \ Pi $ runs all the matches of $ \ {1, 2, \ ldots, 2n \} $.

Proof: according to the definition of the determinant,

\ [\ Det A = \ sum _ {\ Sigma} \ Text {sign} (\ sigma) A _ {\ Sigma }=\ sum _ {\ Sigma} \ Text {sign} (\ sigma) A _ {1 \ sigma (1 )} A _ {2 \ sigma (2)} \ cdots A _ {n \ sigma (n )}. \]

Here, $ \ Sigma $ run the replacement group $ S _ {2n} $.

In the Rotation decomposition of $ \ Sigma $, if the length of a rotation is odd, for example, set $ \ Sigma = \ sigma_1 \ sigma_2 \ cdots \ sigma_l $, where the length of $ \ sigma_1 $ is an odd number and its minimum element is smaller than that of all other elements whose length is an odd number, define $ \ Sigma '= \ sigma_1 ^ {-1} \ sigma_2 \ cdots \ sigma_l $, then $ \ Text {sign} (\ sigma) =\text {sign} (\ Sigma ') $, but \ [A _ {\ sigma_1 }=\ prod A _ {I \ sigma_1 (I )} =-\ prod A _ {\ sigma_1 (I) I} =-\ prod A _ {I \ sigma_1 ^ {-1} (I )} =-A _ {\ sigma_1 ^ {-1 }}. \]

Therefore, the sum of $ \ Sigma $ and $ \ Sigma '$ is offset. (Note that $ \ Sigma $ corresponds to exactly $ \ Sigma $) therefore, in the sum of $ \ det A $, consider the replicas Whose Rotation decomposition lengths are even.

In the $ [\ Text {pf} (a)] ^ 2 $ expansion, each item is shown in \ [\ Text {sign} (\ PI) A _ {\ PI} \ cdot \ Text {sign} (\ mu) A _ {\ Mu }, \] We need to prove that each of these items and the rest of the Rotation decomposition length are even $ \ Sigma $ corresponding to $ \ Text {sign} (\ sigma) A _ {\ Sigma} $ corresponds to each other.

First, we need to establish a pair of matching $ (\ Pi, \ mu) $ and the decomposition type are the one-to-one correspondence between the replacement of the even rotation $ \ Sigma $, then it is proved that the corresponding permission is equal.

For any pair of $ (\ Pi, \ mu) $, they convert $ s =\{ 1, 2, \ ldots, 2n \}$ into a regular Graph with a vertex degree of 2, this figure can be expressed as the sum of several loops. in a loop, the edges of $ \ Pi $ and $ \ Mu $ are staggered, therefore, the length of each loop is an even number. We map each loop to an even rotation: Find the smallest element in the loop, and then start from this element and circle it in the direction of matching $ \ Pi $, in this way, an even rotation is obtained. If this is done for each loop, a replacement with an even Rotation decomposition length is obtained. In turn, given such a replacement, it is easy to get the corresponding two matching $ (\ Pi, \ mu) $ against the above steps.

However, the most difficult part is to prove that the corresponding permission is equal.

Everything depends on the following theorem:

Theorem:If $ I $ and $ I + 1 $ are not matched in $ \ Pi $, then the $ I $ and $ I + 1 $ matches $ \ PI '$ and $ \ Pi $.

Proof of the theorem: Combine $ I $ and $ I + 1 $ and their "spouse" into a set of four vertices for $ M $, the remaining $2n-4 $ points are $ s \ backslash M $.

The intersection of two ARCs belonging to $ s \ backslash M $ is not affected by the exchange of $ I $ and $ I + 1 $, therefore, the number of intersections of $ s \ backslash M $ does not change before and after the switch.

An arc in $ s \ backslash M $ and an arc in $ M $ may be affected by switching, however, the parity between the arc in $ s \ backslash M $ and the total number of arc intersections in $ M $ is not affected by switching. Set $ XY $ to an arc in $ s \ backslash M $. If an odd vertex in $ M $ falls in the middle of $ x $ and $ y $, the arc $ XY $ and $ M $ have an odd intersection point. Similarly, if $ M $ has an even number of vertices falling between $ x $ and $ y $, the arc $ XY $ has an even number of intersections with the arc in $ M $. This fact is intuitive, but it must be carefully said and cool, so we will leave it for your own verification.

Finally, the changes of the arc point in $ M $ before and after the switch are investigated. However, $ M $ has only four vertices, and there are only three possible matches for the four vertices. You only need to verify them one by one.

To sum up, in these three cases, the number of intersection points is not affected by the exchange. In the second case, the parity of the number of intersection points is not affected by the exchange. In the third case, the number of intersection points changes by 1. In this way, we prove that the replaced symbol will change the number after the switch.

Now let's go back to the proof of the theorem. What we need to prove is that the corresponding given previously maintains equal weights. That is

\ [\ Text {sign} (\ PI) A _ {\ PI} \ cdot \ Text {sign} (\ mu) A _ {\ Mu} =\text {sign} (\ sigma) A _ {\ Sigma }. \]

Obviously we have

\ [A _ {\ PI} A _ {\ Mu} = (-1) ^ {e (\ sigma)} A _ {\ Sigma }. \ quad E (\ sigma) =\#\{ I: \ sigma (I) <I \}. \]

So as long as the certificate

\ [\ Text {sign} (\ PI) \ Text {sign} (\ mu) = (-1) ^ {e (\ sigma )} \ Text {sign} (\ sigma ). \ Quad (\ AST) \]

We do not know whether this equation is true or not, but we can explain that as long as it is true for a special replacement $ \ Sigma $, then it is true for all replicas that are associated with $ \ Sigma $.

Assume that this equation is true for a certain $ \ Sigma $, and consider exchanging $ I $ and $ I + 1 $ in $ \ Sigma $. This is equivalent to replacing $ (I, I + 1) $ in $ \ Sigma $, so $ \ Text {sign} (\ sigma) $ remains unchanged.

1. if $ I $ is not adjacent to $ I + 1 $ in the Rotation decomposition of $ \ Sigma $, that is to say, $ \ Sigma $ does not change $ I $ to $ I + 1 $ or $ I + 1 $ to $ I $, therefore, they are not a pair in both $ \ Pi $ and $ \ Mu $. After switching, $ \ Text {sign} (\ PI) $ and $ \ Text {sign} (\ mu) $ changes, so the product remains unchanged. $ E (\ sigma) $ obviously remains unchanged, so the equation remains valid.

2. if $ I $ is adjacent to $ I + 1 $ in the Rotation decomposition of $ \ Sigma $, there are two possibilities: the first one is $ (I, I + 1) $ this pair belongs to one of $ \ Pi $ and $ \ Mu $, but does not belong to the other, then $ \ Text {sign} (\ PI) $ and $ \ Text {sign} (\ mu) $ change the number of another variable, $ E (\ sigma) $ Change 1, and the equation remains valid. The second possibility is that $ (I, I + 1) $ belongs to both $ \ Pi $ and $ \ Mu $. In this case, $ (I, I + 1) $ forms a separate rotation, so $ \ Text {sign} (\ PI) $ and $ \ Text {sign} (\ mu) $ remain unchanged, $ E (\ sigma) $ remains unchanged, so the equation remains valid.

Therefore, we only need to select a replacement in each of the classes to verify that $ (\ AST) $ is valid. The simplest choice is \ [\ Sigma = (12 \ cdots K) (k + 1 \ cdots) \ cdots (\ cdots N). \]

Verify the replacement directly.

So far, we have proved the theorem.

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Step 3: Pfaff orientation of the Plan, which converts all the matches in the Pfaff polynomial into the same number

The conclusion of Pfaff polynomials inspires us to use it to calculate the number of all perfect matches for a graph $ G $.

First, an arbitrary directed graph $ \ overrightarrow {g} $ is obtained for the edge of $ G $. Write the $ \ overrightarrow {g} $ adjacent matrix $ A = (A _ {IJ }): $ \ [A _ {IJ }=\ left \ {\ begin {array} {ll} 1 & V_ I \ rightarrow V_j \-1 & V_j \ rightarrow V_ I \ 0 & \ Text {else }. \ end {array} \ right. \]

Then $ A $ is an antisymmetric matrix and \ [\ det A = [\ sum _ {\ PI} \ Text {sign} (\ PI) A _ {\ PI}] ^ 2. \]

Note that the value of $ \ Text {sign} (\ PI) A _ {\ PI} $ for each item is 1,-1, or 0; $ A _ {\ PI} $ is not 0 if and only when $ \ Pi $ gives a perfect match for the graph $ \ overrightarrow {g} $. Therefore, every perfect match in the figure $ G $ corresponds to each non-zero entry in the Pfaff polynomial. Unfortunately, these non-zero items include + 1 and-1. What you add to them is not the number of all matches. However, we can think like this: can we assign $ G $ with + 1 or-1 to $ A _ {IJ} $ through proper targeting, so that each item $ \ Text {sign} (\ PI) A _ {\ PI} $ is of the same number? If yes, the number of perfect matches required is $ \ SQRT {\ det A} $, the determining square of the corresponding adjacent matrix $ A $.

When proving the Pfaff polynomial, we used the corresponding: \ [\ Text {sign} (\ PI) A _ {\ PI} \ cdot \ Text {sign} (\ mu) A _ {\ Mu} =\text {sign} (\ sigma) A _ {\ Sigma }. \]

Therefore, you only need to make every $ \ Text {sign} (\ sigma) A _ {\ Sigma} = 1 $. Set $ \ Sigma =\sigma_1 \ cdots \ sigma_l $, then $ \ Text {sign} (\ sigma) = (-1) ^ L $, therefore, you only need to make every $ A _ {\ sigma_ I} =-1 $.

Note that each $ \ sigma_ I $ is an even loop. $ A _ {\ sigma_ I} $ is the product of the edge weight of this loop along a certain direction, therefore, as long as an odd side is positive (of course, an odd side is reversed.

Definition:In a directed graph $ G $, if the $ C $ length of a loop is an even number, and the remaining parts still have a perfect match after $ C $ is deleted, $ C $ is a good loop. If a $ G $ ), this is called a Pfaff targeting of $ G $.

It is difficult to find the Pfaff orientation of a general graph, but it is very easy to map it. This is the theorem below:

Theorem [kasteleyn ]:If $ G $ is a simple floor plan, you can set a proper orientation for the $ G $ edge so that when every side of $ G $ is walking counterclockwise (the external infinity area is not counted ), each has an odd number of edges that are consistent with the walking direction. This type of orientation is the Pfaff orientation of $ G $.

Proof: First, it indicates that such a ing exists, so that every plane of $ G $ is an odd ing. Sum up the number of faces: $ f = 0 $, then $ G $ is a tree, and any direction is Pfaff. Conclusion: A simple directed graph with $ F-1 $ is set up. For a graph with $ F $, find an edge adjacent to the external area. $ e $, delete $ e $ to get a directed graph with $ F-1 $. Therefore, we can make every plane odd and then add $ e $ back, you can give $ e $ a proper targeting to make the last aspect odd.

Next, we will show that this orientation is Pfaff. For any good loop $ C $ in $ G $, follow each plane inside $ C $ counterclockwise, and then count the total number of lines that are consistent with our walking direction.

Assume that $ C $ has a length of $ L $, and $ C $ contains $ p $ vertices, $ q $ edges, and $ r$ faces.

Assume that the $ C_ I $ Edge Orientation is counter-clockwise when walking along each side ($ I = 1, \ ldots, r$ ), $ C $ the number of counter-clockwise edges is $ C $, then the total number of counter-clockwise edges encountered is \ [\ sum _ {I = 1} ^ rc_ I = C + q. \] This is because every side inside $ C $ has been taken twice, and it is calculated every time in a clockwise direction; however, the side on $ C $ is calculated only once for those with counter-clockwise orientation.

Since each $ C_ I $ is an odd number, \ [r \ equiv C + Q \ (\ Text {mod} \ 2). \]

On the other hand, we use the Euler's Theorem to obtain the region contained in $ C $.

\ [(P + l)-(q + l) + r = 1. \]

Therefore, $ p $ is the opposite of $ C $ parity, but $ p $ is an even number, so $ C $ is an odd number, which proves the theorem.

Back to the question at the beginning of the article: calculate the number of dominoes covered by the chess board. We calculate the number of perfect matches for the general $ n \ times N $ board. First, we need to make the Pfaff orientation. The following figure shows the Pfaff targeting method for the general rectangular board:

The next step is to write out the ing matrix of this directed graph. (The order of vertex labels is the first line from left to right, the second line, and so on ). Set

\ [B = \ begin {pmatrix} 0 & 1 & 0 & \-1 & 0 & 1 & \ &-1 & 0 & 1 &\\&& & \ ddots & 1 \\& &-1 & 0 \ end {pmatrix }_{ n \ times n }. \]

Then, the adjacent matrix is $ A =\begin {pmatrix} B & I & \-I &-B & \ &-I & B &&\ \ & \ ddots & I \ &-I & (-1) ^ {n-1} B \ end {pmatrix }. $

$ A $ here is a block matrix of $ n ^ 2 $.

Change the ID of the row and column of $ A $ to obtain

\ [\ Det A = \ det \ begin {pmatrix} B &-I & \-I & B &-I & \ &-I & B &- I <\& & \ ddots &-I \\& &-I & B \ end {pmatrix }. \]

Thus \ [\ det A = \ det (B \ otimes I-I \ otimes c). \]

Where

\ [C =\begin {pmatrix} 0 & 1 & 0 & \ 1 & 0 & 1 & \\\ & 1 & 0 & 1 &\\&&&\ ddots & 1 \\& & 1 & 0 \ end {pmatrix }. \]

The rest is the feature value in linear algebra. I will not write it here. The final answer is

\ [\ SQRT {\ det A} = 2 ^ {n ^ 2/2} \ prod _ {k = 1} ^ n \ prod _ {L = 1} ^ N (\ cos ^ 2 \ frac {k \ PI} {n + 1} + \ cos ^ 2 \ frac {L \ PI} {n + 1 }) ^{ 1/4 }. \]

This is the number of perfect matches required.

Domino coverage of the Board: dimer lattice model, Pfaff polynomial, and kasteleyn Theorem