I write, therefore I am
With this variation on a famous statement by the philosopher Descartes, I would like to express that writing about what happens in my life is important to me.
ScheltemaThis morning, I visited bookshop Scheltema in Amsterdam. I mostly looked around on the top floors with books on sale and second hand books. At 11:46, I purchased a black, hardback Moleskine Daily Diary / Planner of 2015 (ISBN:9788867322626) for € 16.90. Stedelijk Museum in Amsterdam. There I went to see the exhibition Bad Thoughts - Collection Martijn and Jeannette Sanders. It includes the work Computer Structuren 2, 1970-1971 by Peter Struycken. At the end of the afternoon, I went to the opening of the exhibition RGB exit with seventeen new works by Peter Struycken in the series 'Kleurverhouding' (colour relationships). Each work consists of two square paintings on canvas with the size 60 cm. The left one is painted according to a fixed pattern on a grid of six by six squares using six colours, such that every colour occurs exactly six times. The right one is painted on a five by five square where the colours occur at the ratios 1, 2, 3, 5, 6, and 8. The pattern used for each work is unique. The hues of the six colours are placed on fixed positions on the colour wheel that are evenly spaced. The saturation and the brightness are different for each work. The idea of this works is to show that the impact of saturation and brightness have a greater impact on the colour experience than the hue. I met Peter Struycken while he was talking with Carel Blotkamp. We only briefly talked. Many people talked with him. There was a lot of interest for his works. I did talk with Zsa-Zsa Eyck and agreed that in the future I should maybe investigate the works that they still have in stock. Maybe I am going to visit this exhibition again before it ends at October 18, just to have a quiet look at all the works as it might be the last time that they can be seen together.
Wednesday, September 3, 2014
Impatiens glanduliferaYesterday, I noticed some special flowers along the road, and I took the close-up picture on the right. I thought I had found some orchid, but when at home I tried to determine it, I could not find me. This morning, I asked some colleagues if they had some idea. One of them thought it was an impatiens. After some searching, I concluded that it might be an impatiens glandulifera. Someone on twitter affirmed this.
For the love of PhysicsAt 12:51:48, I bought the book For the love of Physics by Walter Lewin, ISBN:9781451607130, from bookshop Broekhuis for € 9.95 (second hand). Chapter 15 of the book, with the title Ways of Seeing talks about his friendship with Peter Struycken. Walter Lewin helped Peter Struycken by making some suggestions for mathematical formuleas that he could use for his computer art works. Many of his art works are based on these suggestions.
Thursday, August 14, 2014
Wednesday, August 13, 2014
Third visit to Peter StruyckenI visited Peter Struycken. Besides talking about all kinds of things, we also went throught the list of his works on the page I have made and reordered the works before 1969 based on his recollections about which works were made in the same periode. We also discovered that there are still some early works missing when we looked at some pictures that were taken during the openings of some early exhibitions. I also returned the part of his personal archive that I took with me last October and took some new part home.
Thursday, August 7, 2014
Triangulation of regular polygonsThe number of ways a regular polygon with n-2 sides can be dissected into triangles (also known as triangulation) is given by the Catalan number Cn. Every pair of triangulation for a given regular polygon has a number of (internal) lines in common. I wrote a program to count these numbers and the results are given in the table below. The second column gives the Catalan number for the value given in the first column. The following columns give the number of pairs that have the number of lines in common equal to the number given above the column. The total number of pairs is equal to Cn(Cn-1)/2. The diagonal sequence 1, 5, 21, 84, 330, and so on, seems to be equal to OEISsequence A002054.
n C 0 1 2 3 4 5 6 7 ----------------------------------------------------------------- 2 2 1 3 5 5 5 4 14 34 36 21 5 42 273 308 196 84 6 132 2436 2928 1992 960 330 7 429 23391 29898 21555 11220 4455 1287 8 1430 237090 321490 244420 135080 58630 20020 5005 9 4862 2505228 3594756 2872694 1670812 773773 292292 88088 19448
Another dead-endLast Wednesday, I found an interesting question on MathOverflow related to the the Four colour problem by the user Robert Carlson (maybe Bob Carlson from the University of Colorado). The question Are all Hamiltonian planar graphs 4 colorable? Does this imply all planar graphs are colorable? talks about how a Hamiltonian cycle on the faces of a planar graph, would split that graph in two tree graphs. It talks about 'diamond switches' and states:
4-connectedThe paper A theorem on planar graphs by William Thomas Tutte, which appeared in Trans. American Math. Soc. 82: pages 99-116, states in Theorem II (on page 115): Let G be any 4-connected planar graph having at least two edges. Then G has a Hamiltonian circuit. 4-connected means that it is not possible to separate the graph into two parts by removing 3 or less vertices. The dual graph of interesting graphs with respect to the Four colour problem have only faces with three edges and all vertices have degree five or higher, meaning that they have five or more edges. Because of the triangle faces (faces with three edges connecting three vertices), every minimal set of vertices that separates the dual graph must be a cycle. If this is not the case, it means that there are two vertices on the sequence of vertices that separate the two components on the plane. These two vertices will be connected to both vertices on of the two components, and because there are only triangle faces, it means that there must be an edge between vertices in the separate components, which would be a contradition. Now suppose that the dual graph is 3-connected, then the original graph can be split in two parts that are connected with three edges. It is obvious that this graph can be reduced. This means that the dual graph of all non-reducable colouring problems must be at least 4-connected. And from Theorem II it follows that the dual graph must have a Hamiltonian cycle.
Double tree graphsSome weeks ago, I realized that all possible expansions of the 00 sequence according to the rules I described on November 23 (of the same length) must have some sequence in common. What I mean is that if you start with the sequence, choose an expansion sequence consisting of n successive locations, and applies this to the sequence 00, resulting in 2n sequences (when expanding with 1 and 2 at each location), that this collection of sequences always has a sequence in common with any other expansion sequence. Yesterday, evening, I thought about the idea, that maybe I could proof this by studying the properties of the graph you get, when you glue two tree graphs (represeting each expansion pattern) together and see if it has some special properties. With some smaller tree graphs, I got the idea that they always contain a face with three or four vertices and that such a face could be easily removed. But after some calculations, I concluded that there was no evidence for such a property. This morning, I decided to look for the smallest counter example of a combined graph with only faces with five or more vertices. That is when I made the above drawing of the dodecahedron graph and drew a line with pencil visiting all faces once, and thus spliting the graph into two tree graphs. The digits give (except for some errors) the number of vertices of the face on given side of the pencil line. The pencil line is actually a Hamiltonian cycle in the dual graph. Next, I realized that maybe the dual graph of every interesting graph with respect to the Four colour problem has a Hamiltonian cycle. According to Tutte, each 4-connected planar graph has a Hamiltonian cycle. If this is the case for all interesting (non-reducable) graphs, then it would be sufficient to proof what I wanted to proof in the first place to proof the Four colour theorem as well. That would simplify the matter a lot, except for the fact that it will problably be very difficult to proof that all expansions (with the same number of expansions) have at least one sequence in common.
Flowers in magnoliaI noticed some flowers in our magnolia. Many of them are in new branches. It also needs to be pruned again. It seems it is growing higher and higher each year. Annabel is feeding branches to her rabbit, who seem to like it very much. The little magnolia plant now has nine leaves, but the last three leaves have brown edges and they appear to be smaller than earlier leaves. bookshop Broekhuis for € 8.95. The book is about some popular math topics at the M.Sc. level.
Saturday, July 5, 2014
Toki Pona: The language of goodThis evening, I got the book Toki Pona: The language of good by Sonja Lang (formerly Sonja Elen Kisa), ISBN:9780978292300 from bookshop Broekhuis, which I had ordered, and paid € 19.99. In the evening I read through the book skipping some parts. The book is about the constructed language Toki Pona, which only has 120 words. The book contains lessons for the language, but also a kind of hieroglyphs of each word and a sign language. A few years ago, I made some attempt to learn the language. I think I am going to try again. The only negative point that I have about the book is that here and there it contains some reference to the (recently acquired) religious convictions of the authors, which I find are not appropriate for this kind of work.
GasI try keep track of our gas usage and milages. For this purpose, I always fill up the gas tank and note down the distance we travelled. Our car also has a milage indicator but it is not always very reliable. Last week, Annabel got some gas, but she did not fill up the tank, because the price was quite high. Now I don't know how much gas was used during both periods. But I can use the readings from the milage indicator to make a good guess. Annabel filled 20.02 liter of gas and noted down a distance of 586.5 Km with a milage reading of 15.6 Km/l. Yesterday, I filled 35.55 liter with a distance of 309.6 Km and a milage reading of 16.0 Km/l. The total distance traveled is 896.1 Km and the total gas is 55.57 liter, which gives an average milage of about 16.1256 Km/l. The weighted milage according to the indicator is equal to (15.6 * 586.5 + 16.0 * 309.6) / 896.1, which is: 15.7382 Km/l. This means that the reported values of the milage indicator have to be multiplied with 1.02461 to arrive at an estimation of the true usage during the given periode. This results in an estimated usage of 36.69 liter of gas during the first periode and 18.88 liter during the second periode. These add up to 55.57 liter for both periodes. Yesterday, I paid € 61.47. But taking into accound the estimated gas usage during the last periode, it means that actually only € 32.65 was used. The remained has to be added up to the amount of the first periode (€ 35.42) resulting in € 64.24. Now I can use these numbers to calculate how much Annabel has to pay for gas that she used during both periods based on the administration that we keep, assuming that she agrees with the above calculations, of course.
Trip to China 2010
-- contact -- Frans
My life as a hacker
The Art of Programming
HTML to LaTeX
eXtreme Programming Hamilton cycles