Sunday, September 03, 2006

Tiling Rectangle

The marriage theorem from the previous post provides a direct proof that the 8x8 square with two diagonal corners removed cannot be tiled by dominoes. However it is incapable in providing the number of possible tiling schemes if tiling solutions exist. The number is usually quite large.
However, in 1961, Kasteleyn and others actually found the exact solution for tiling a rectangle lattice with dominoes. The solution given by Kasteleyn involves the use of Pfaffian. Pfaffian is a special kind of matrices and less known to today's students. The problem of tiling rectangles itself is interesting for many physical and chemical problems, for example, the adsorption of diatomic molecules on a regular surface, the cell-cluster theory of liquds, the Ising problem (spin arrangement in two-dimensional lattice). Therefore the search for the exact solution and its asymptotic behaviour is not just for curiosity only but also driven by the real needs.
Here I sketch the main idea described by Kasteleyn. Anyone interested in knowing the details should read read the original paper, Physica, 1961, vol. 27, pp. 1209-25.
Consider a lattice, m x n, and m is even and n can be odd or even. (If both m and n are odd, it is obviouesly no tiling solution.) Each square at the lattice point can be labeled by {i (from 1 to m),j (from 1 to n)}. The square can also be uniquely labeled by just one number if an order is defined. For example, we can define a new label, p (from 1 to mxn )and p is related with {i,j} as follows,
  {i,j} <--> p = (j-1)m + j.
We use p numbers to describe a configuration of dominoes, i.e. tiling scheme. First we consider a trivial configuration which is a solution for lattices. It can be drawn schematically as follows,

  o-----o       o-----o       o-----o       o-----o
{1,1} {2,1}  {3,1}{4,1}  {5,1}{6,1}  {7,1}{8,1}

  o-----o       o-----o       o-----o       o-----o
{1,2} {2,2}

  o-----o       o-----o       o-----o       o-----o
{1,3} {2,3}

  o-----o       o-----o       o-----o       o-----o
{1,4} {2,4}

A configuration can be writen as a list, {{p1,p2},{p3,p4}, ... {p(nm-1), pnm}} and the above trivial configuration can be writen as, {{1,2},{3,4},...{nm-1, nm}}. However, if we exchange the order of each pair list, {pi,p{i+1)}, it still represents a same configuration. Or if we exchange of the order of child lists of the parent list, the configuration keeps too. Therefore, we need to define an order to eliminate such ambagiousness. One convention of the following can be adopted for the order,

  p1 < p2; p3 < p4; ...; p(mn-1) < p(mn); and
  p1 < p3 < p5 < ... < p(mn-1).

Such ordering relates the configuration with a Pfaffian. A Pfaffian is a number attributed to a triangular array of coefficients in the following way,

  Pf[{a(i,j)|i,j=1,...N, N even}]= Sum[p[P]Permutation[a(k1,k2)a(k3,k4)...a(k(n-1),kn)]],

where p[P] (=1 or -1) is the parity of the permutation P and the sum runs over all possible permutations.
The number of dominoes' configuration is then given by Pf[D(p;p')]. The element of the matrix D(p;p') can be defined as follows. If there is no bonding between p and p' (not covered by the same domino), the element is 0. If the two are connected by a horizonal bond, the element is set as 1 and if connected by a vertical bond, the element could be 1 or -1. More exactly, the element is given by the following equations.

  D(i,j;i+1,j)=1;
  D(i,j;i,j+1)=(-1)^i;
  D(i,j;i',j')=0, otherwise.

(If the dominoes are not symmetric, the corresponding nondegeneracy should be mulitplied.) The derivation of the first two rules needs a proof. The essential part of the proof lies in the transformation to construct any new configuration from the trivial configuration given above. 1) the sites with bond breaking and formation form polygons with alternating old and new bonds. 2) all the polygons do not overlap or cross. 3) by just moving the bonds of the polygons by just one bond distance, new configuration and old configuration can be exchanged.

Once the matrix D(p;p') is constructed, the next job is to calculate the Pfaffian. The following property can be used,

  (Pf[D])^2 = det D,

where D is the skey-symmetric matrix extended from the triangular array of elements of the Pfaffian. The matrix can be decomposed in the following way,

  D = Qm X En + Fm X Qn,

where En is the nxn unit matrix, and Q is the matrix in the following format, {{0,1,0,0,-...},{-1,0,1,0,0...},{0,-1,0,1,0},...} and F is the matrix in the following format, {{-1,0,0,0...},{0,1,0,0,...},{0,0,-1,0,...},...}. By diagonzation the above matrices product, the configuration number can be found.

  Pf[D] = Product[Product[2(Cos[k Pi/(m+1)]^2 + Cos[l Pi/(n+1)]^2)^(1/2),{l,1,n}],{k,1,m/2}].

The above result could be further simplified slightly. For a simple calculation shows that the number of ways to tile a 8x8 lattice with dominoes is 12988816.
The result is quite surprsing. Since we know the answer must be an integer, yet it is given as the products of a series of irrational numbers. Kasteleyn also looked at the situation when perodic conditions are applied to the lattice, i.e. lattice on torus. More discussion could be given on the application of this result.

Note: On the ICM 2006 in Madrid, Spain, Andrei Okounkov has been awarded a Fields medal for his study on random tiling pattern.

Wednesday, August 30, 2006

Marriage Theorem

Once upon a time, there is a planet with men and women living harmoniously together. On the planet, a man or a woman has neighbors of opposite sex only. They live so well that they only choose one of their neighbors as their husband or wife... However, is it always possible for a man to be able to find a wife or a woman to find a husband? Of course, the answer is simplely no if there are not the same number of men and women. But if there are exactly same number of men and women, what is the answer then?

Well, this marriage problem is wierd, but it is related and helpful to understand another problem, the tiling of squares with dominos.

The following is a 8x8 square with two corners removed. One may ask if the shape can be tiled with dominos. A domino is composed with two adjacent squares. If the corners are present, the answer is obviously positive. However, after removing the two corners, there are 62 squares. Can it be covered completely with 31 dominos? The question at first sight is very difficult, which is generally true for tiling problem due to large number of possible tiling schemes. If fact, if we add an additional property to the identical squares, we may turn the problem into the above marriage probem and find the answer very easily.


Domino


We may color all the squares in two colors alternatingly like a chessboard fasion. See below. If we put a domino anywhere on the board, one square of the domino must be green and the other white. Therefore, if a tiling solution exists, the number of green squares and white squares should be the same. However, we find there are 32 green squares and only 30 white squares. (The alternating pattern is an antiferromagnetic arrangement in the 2-dimensional space and there is a residual net spin though.) So the coloring scheme turns to be crucial to find the negative answer.

Coloring Scheme


However, the coloring method cannot give a definitive answer if we can color the squares alternatingly for the same number of squares. For example, in the following 8x8 square, we remove one square in green and one in white. The resulted shape is obviously tilable. (Though it is maybe difficult to write down all the tiling possibility, we can at least find a few.)


However, in the same 8x8 square, we remove two whites and two greens and obtain the pattern shown below (the black square is one of the holes). There is no way to tile the upper left corner with a domino, for which the corner is an isolated square.


There are a lot of more to say about tiling and marriage theorem. But so far, i only understand little of them.

Friday, August 18, 2006

Clouds Appreication

The Cloud Appreiciation Society,
A nice website for cool pictures of clouds. Check it out at,
http://www.cloudappreciationsociety.org/gallery/

Thursday, August 17, 2006

Amplitudes of Tetrahedral Spin Network

after Dan Christensen and Igor Khavkine;
using the following equations,

The amplitude is the real part of f(q)/g(q),
where q is a complex number,
f(q) = (q^4-q^2+1) (q^4+q^3+q^2+q^+1) (q^4-q^3+q^2-q^+1) (q^6+q^5+q^4+q^3+q^2+q^+1) (q^6-q^5+q^4-q^3+q^2-q^+1) (q^20-q^18-q^14-q^12+q^10-q^8-q^6-q^2+1)
and
g(q) = (q^8)((q^4+q^2+1)^2) ((q^4+1)^5).

The value of amplitude is colored using the modulation function,
adjust(x) = log(log(abs(x)+1)+1).



The argument of the result using the same modulation function for the color description.


Recover the negative (-Pi, 0) and positive (0, Pi) difference in the argument.


still not understand what it means..

Tuesday, August 15, 2006

Paint Opera

Sophie von Hellermann: Rosina


Makiki Kudo: Princess Yue-yang


Cecily Brown: Untitled


Barnaby Furnas: Untitled

NYTimes: Where Bel Canto Meets Paintbrush
http://www.nytimes.com/2006/08/15/arts/design/15hero.html?hp&ex=1155700800&en=3dae22a5ca19de54&ei=5094&partner=homepage

Sunday, August 13, 2006

Painting

by an unkown artist. taken on July, 2006 in an office building in Dresden, Germany.

Blue Sea


Lotous(?)

Friday, July 28, 2006

Goodstein Sequence: Power of Infinity

A postive integer number can be expressed in a recursive expotenial form form. For example, in the base of 2, the number 299 can be written as the following binary tree form.

{Plus, {Power, 2, {Plus, {Power, 2, {Plus, {Power, 2, 1}, 1}}, 0}}, {Plus, {Power, 2, {Plus, {Power, 2, {Plus, {Power, 2, 1}, 0}}, 1}}, {Plus, {Power, 2, {Plus, {Power, 2, 1}, 1}}, {Plus, {Power, 2, 1}, 1}}}}

The Goodstein sequence is generated in the following steps:
1. Start with any positive integer number and write it in the above recursive expotenial form with the base 2;
2. Replace the base 2 in the above form with 3 and then substract 1 from 1.
3. Recursively repeat the above steps to generate the sequence.

Apparently, the sequence will increase very rapidly. For example, if we start at 5, the first 61 numbers in the Goodstein sequence are:

{5, 27, 255, 467, 775, 1197, 1751, 2454, 3325, 4382, 5643, 7126, 8849, 10830, 13087, 15637, 18499, 21691, 25231, 29137, 33427, 38119, 43231, 48781, 54787, 61267, 68239, 75721, 83731, 92287, 101407, 111108, 121409, 132328, 143883,156092, 168973, 182544, 196823, 211828, 227577, 244088, 261379,279468, 298373, 318112, 338703, 360164, 382513, 405768, 429947, 455068,481149, 508208, 536263, 565332, 595433, 626584, 658803, 692108, 726517}.

We can graph the sequence as follows.
Goodstein sequence starting at 5


It seems that the sequence increases in a power law. However, after a finite number of steps, though very large, the sequence will become 0 at last. More amazingly, this cannot be proved in the Peano arithmetic. Baez showed how the infinity of ordinals can be used to prove this result, http://math.ucr.edu/home/baez/week236.html.