Dutch / Nederlands

# Diary, May 2015

 ```Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 ```

## Sunday, May 3, 2015

### Outdoor book market

This morning I went to the outdoor book market in Tuindorp 't Lansink in Hengelo. I went there by bike. I returned with two maps and four books. The two maps are about the design of a road between between Enschede and Hengelo that I used to get there. I bought the at 11:08 for € 6.00. At 11:30, I bought the book John Lennon: De mens, de mythe, de muziek (Dutch translation of John Lennon: The Man, the Myth, the Music) by Tim Riley, ISBN:9789026324406, for € 7.50. At 11:46, I bought the book De Ontdekking van de Hemel by Harry Mulisch, ISBN:9023435605 for € 1.00. At 12:09, I bought another two books for € 1.00, namely: Andy Warhol by Victor Bockris, ISBN:9789067661201 and Anaïs Nin: Gemaskerd, ontmaskerd by Élisabeth Barillé, ISBN:9789067661577. At home I discovered, I already owned another edition of the Bockris book.

## Tuesday, May 5, 2015

### Six pentagons

I still on and off thinking about the Four colour theorem. I have mostly thought about the question whether the two tree graph idea could lead to a solution. Lately, I have been thinking about a way to reduce the graphs to be resolved to those that are 'round' in the sends that they are not peanut shaped. If a graph is peanut shaped, you could cut it into two 'round' graphs that touch each other. As these are smaller than the whole graph, it means you can use some induction argument. Of course, a graph could have more than one constricion, but it seems that there must always be at least one who has a 'round' graph on one side. I guess that the proof that because the two smaller parts can be coloured, the whole graph can be coloured as well, might be rather complicated. At least it made me think about some shapes again that can be reduced, and I discovered that an area that exists of one pentagon surrounded by five pentagons can be reduced to a single pentagon. I also looked at other combinations of six pentagons, and I found one, inside a hexagon, that seems not to be reducable. I had somehow hoped that every configuration of six pentagons was reducable.

## Friday, May 8, 2015

### Cycles in transition graph

I looked at non-overlapping cycles in the transition graph related to the the four colour theorem. This resulted in the following table. The first column gives the length of the cycle. The second column gives the total number of walks starting from a point to itself. A total of 665 different walks. The third column gives the number of truly distinct cycles, ignoring walking direction, starting point, and use of 1 and 2. A total of 35 cycles. Behind that the distinct cycles are given.

``` 2:   1 1 00
3:   2 1 111
4:   6 2 0102 1212
5:  10 1 01221
6:  28 4 010101 011022 011211 012012
7:  42 2 0110121 0112212
8:  76 5 01011021 01120122 01122021 01210212 11221122
9:  84 5 010112211 010121121 011011011 011201121 112112112
10: 160 8 0101101011 0101120121 0101202021 0110110212 0110120112 0112102122 0112201122 1121122122
11:   0 0
12: 256 6 010110110211 010112201121 010120220121 011201120112 011201211021 011221121121
```
In the list of 665 walks there are no walks that appear inside another walk. I do not know what can be done with this, but it is a fact that all possible sequences are made out of these walks, were it is possible that some shorter walks occur inside longer walks.

## Sunday, May 10, 2015

### Palindrome dates

Today is a palindrome date when written in the (M)M/(D)D/(YYY)Y format: 5/10/2015. When the (M)M/(D)D/YY format is used, this is also a palindrome date: 5/10/15 and this is the case for all dates up to and including 5/19/15 and also the date 5/1/15.

## Tuesday, May 12, 2015

### Book

At 12:02:58, I bought the book De Myte van Sisyfus (Dutch translation of The Myth of Sisyphus) by Albert Camus from bookshop Broekhuis for € 7.50. I had seen this book last week at the new arrivals setion, but I could no longer find it. It appeared they had placed it under novels, while I had been looking in the philosophy section.

## Monday, May 18, 2015

### Proof with Kempe chains?

In the past week, I have been rereading parts of the book Four Colors Suffice and thinking about Kempe chains for solving the four colour theorem. For proving the four colour theorem it is sufficient to look at all 2-connected (planar) cubic graphs that are free of digons, triangles, and squares. A partial cubic graph, is a cubic graph that has some open edges on the outside. The partial cubic graphs can be ordered by the number of vertices that they contain. Every (complete) cubic graph can be reached by 'closing' one of the partial cubic graphs with three outgoing edges, by joining these three edges. Furthermore, every partial cubic graph of size n+1 can be constructed by adding a vertex to that connects either one or two of the open edges of a partial cubic graph of size n.

Every face colouring of a partial cubic graph is related to a three colouring of the edges, say with α, β, and γ (or a, b, and c, as I used in my first entry). Around each vertex the edge colours are either labeled in a clockwise or anti-clockwise manner with respect to the alphabetical order of the letters. If we assign values 1 to all clockwise and 2 to all anti-clockwise vertices, a value of 0, 1, or 2 can be assigned to each (open and closed) face by adding the values of the vertices modulo 3. For the closed (inside) faces of a partial cubic graph these should be zero. With the open (outside) faces of a partial cubic graph a cyclic sequence can be associated. With each partial cubic graphs it is possible to assign a set of sequences based on all possible face colourings (as I descibed before). It is possible that more than one face colouring of the partial cubic graph leads to the same sequence.

If a partial cubic graph of size n+1 can be constructed by adding a vertex to a partial cubic graph of size n that is connected with one open edge, then a set of sequences can be constructed by expanding the sequences of partial cubic graph of size n. Every sequence of the partial cubic graph of size n results in two sequences of the partial cubic graph of size n+1 by either inserting a 1 or a 2 at the corresponding location and adding that value (modulo 3) to its neighbours. If a partial cubic graph of size n+1 can be constructed by adding a vertex to two open edges, the set of sequences can be constructed by contracting the sequences of the partial cubic graph of size n. For every sequence that has a value 1 or 2 at the corresponding location, that value needs to be removed and substracted (modulo 3) from its neighbours. If all sequences have a 0 at the corresponding location, it is not possible to colour the partial cubic graph of size n+1 and the four colour theorem would be disproven. If we can proof that every set of sequences of all partial planar graphs of size n have at least one value unequal to 0 at every possible location, then the four colour theorem can be proven from this. (If I am not mistaken, as I reasoned before, every (complete) cubic graph can be constructed from a sequence of partial cubic graphs in increasing size such that first n/2 expansions are made and next n/2 contractions are made.)

A kempe chain is made out of a combination of two edge colours. There are thus three kinds of Kempe chains: αβ, αγ, and βγ. For a partial graph the Kempe chains form cycles (assuming that it has a correct face colouring). This is because the three edges attaced to each vertex have three distinct edge colours (either assigned in a clockwise or anti-clockwise manner). This causes each outer edge to be connected with two other outer edges through two Kempe chains marked with different combinations of the edge colours. Now suppose that there is set of sequences for a partial cubic graph of size n, such that at some location all the sequences contain the value 0. Take one of the sequences and select one of the face colourings associated with this sequence, there must be at least one, and determine the Kempe chains for this colouring. Now the edges on the sides of the open face at which the 0 is located are either connected with two, three or four Kempe chains. In case there are four connected, then pick one, and exchange the values 1 and 2 on the Kempe chain resulting in another face colouring of the partial cubic graph, which has a sequence associated with it that has a value different than 0 at the given location. In case there are three connected, one of them must connect the two edges on the side of the open face. Now exchange the values 1 and 2 on one of the other two the Kempe chains resulting in another colouring, which has a sequence associated with a value different than 0 at the location. In case there are only two Kempe chains connected to the two edges on the side of the face, exchanging the 1 and 2 values of the vertices on either of them will result in a new face colouring, but there will still be 0 at the given location. However, because every Kempe chain in a partial cubic graph traverses a distinct subset of vertices (assuming that it does not contain a digon), it means that the two edges must now be connected to three Kempe chains. If this reasoning is correct, it proves the four colour theorem, because another flip of one of the Kempe chains not connecting the two edges, will result in a sequence with a value different from 0. Thus all cases lead to a contradiction, from which it must follow that all sets of sequences of all partial cubic graphs contain at least one value different from 0 at every location.

(Remark: during the day, I have been editing this description to improve its readability.)

## Tuesday, May 19, 2015

### Resolving the details

What I call Kempe chains yesterday are more often called 'Kempe chains on edges'. When I searched for that term, I found a website about the Four color theorem by Mario Stefanutti, where he talks about impasses and resolving them with swapping along Kempe chains on edges, much like what I am using. So, it seems that some of my ideas are not original. He also gives a counter example of a graph where his method does not work. There are some important difference between his method and my proof strategy.

Since I wrote the above yesterday morning, I have continued thinking about it. One important observations is that both the edges on the side of an open face with the value 0 assigned to it, must have the same edge color, lets say α, because otherwise the vertex closing the open face could be assigned a value 1 or 2. Which means that the possible Kempe chains (on edges) that are connected to this node, must be αβ and αγ chains. Another observation is that Kempe chains of a certain type cannot cross within a planar cubic graph.

Lets look at the case where there are only two Kempe chains connected to the two edges. We look to the vertices attached to the neighbour edges on both sides. It seems impossible that both of the Kempe chains pass through one of these vertices, because then one of them must end at the outside edge (which is a contradiction). It is possible that one of the Kempe chains passes through both these vertices, but there is always one Kempe chain that does not pass through both. If one of the Kempe chains passes through one of the vertices, the other Kempe chain will need to be reversed, causing this Kempe chain to be broken into two pieces, where one piece leaves through the neighbour outside edge.

Now when both Kempe chains do not pass through either of the vertices at the neighbour edges, there is a problem. Because there is a counter example, where alternating reversing the Kempe chains will result in a cycle never to resolve the issue. But there is an escape. We simply have to 'pull' one of the Kempe chains to one of the vertices. Lets say the Kempe path leaving from the edge to the right of the open face at the location towards the right neighbour is an αβ Kempe path. Now there must be some vertices along the open face between the left edge and it neighbour. The vertex closest to the neighbour edge that is included in the αβ Kempe path, must have a γ edge colour on the edge on the right. If the edge on the left has the colour α we can reverse the Kempe path βγ going through this vertex, which causes the αβ Kempe path to extend towards the neighbour edge. This process can be repeated until the vertex at the neighbour edge is included. (It is possible that during this process the other Kempe path is broken, but that would be no problem, as that would result in a faster way in three Kempe path to be connected to the two edges.) This proves that it is aways possible in a finite number of Kempe path reversals to create a sequences that has a value different than 0 at the problem location.

## Thursday, May 21, 2015

### Old Kempe proof after all?

I started writing a paper, thinking that I had found a truely elagant way of proving the four colour theorem based on the ideas that I developed earlier this week. But when I biked home yesterday, I discovered that I had overlooked an important fact, which destroyed the simple proof. But then I got another idea, but after investigating it a little I discovered that it might also not work. Earlier this week, I wrote to some people, why I thought that the proof I had in mind, was different from all other 'simple' proof attempts, like the first one from Alfred Kempe, namely: that the proof used an incremental procedure, instead of fixing a problem in a steady state, and that the problem it has to fix is localized. The first is actually not true and even if the second is true, it might still be true that the proof still belongs to the same catergory of proofs that are never going to work in the sense that you encounter problem cases (no matter how few they are) that can not be resolved. The reason might be that these proofs are based on a limited representation of an 'object' that does not capture all the complexity that is needed to resolve all cases. This exercise did give me some new ideas that I feel are worth pondering about. I gained some deeper understanding of the problem, and maybe that is worth all the effort.

## Saturday, May 23, 2015

### Book

At 11:55:12, I bought the book Proeven van Liefde (Dutch translation of Essays in Love) by Alain de Botton from bookshop Broekhuis for € 6.95.

## Sunday, May 24, 2015

### Volunteer

In the past year, I have become a regular vistor at a local meeting for people with average or above-average intelligence with a autism spectrum disorder. At first I has some doubts whether I would be allowed to attend because I do not have an official diagnoses. But I very much feel at home and feel that I always have meaningful conversations with other people attending, and even building some friendships with people. Now they are looking for new volunteers for leading the activities and welcoming (new) people. I feel that I would be able to do this and have a meaningful input. But one of the requirements is that one should have an authism spectrum disorder. Asperger's Syndrome is one of the most common forms in the spectrum for individuals with an above average intelligence, but I feel that I do not perfectly fit with it based on the people I have met that have gotten a diagnoses. I get the impression that I have no problem with information processing, I do not suffer from face and/or emotion blindness, and I do not suffer from sensory overload. I get tired from attending meetings and interacting with people a lot. This week, I discovered that I scored rahter low on the psychopathy scale, meaning that I find it hard to be dishonest and to lie. I have some problem dealing with non-verbal communication or if people are saying something else than they mean. I do know that I am quite strong introvert and that my performance-IQ is probably significantly higher than my verbal-IQ. I also know that my interests are often different from my colleagues. I do not want to go for an official diagnoses, because for the rest, I am just functioning okay, and there no reason to get a diagnoses, to which some costs are attached.

### MySample: Right border marker

I still edit my website as a collection of HTML files, which I edit with the MySample editor that I have been adapting through the years for this purpose. I always like my files not to extend beyond the eightieth column, but that is not always easy. This afternoon, I though about adding a red line in the syntax highlighting just at the that column. It was not so difficult to implement. Just some statements at the right place. It really looks great and works nice as you can see on the image on the right.

Since the last release of the MySample, I also have made some additions to the script language. I have introduced a type for points. I have added a method for closing editors (both text and image) so that you can actually use it for batch processing large ranges of images. And there is some support to read text files. It is now also possible to just the key 1 to 9 to when pointing at point in an image file and read those points from the script. I just extend the language when I feel the need. So, the set of available methods is probably not very balanced.

## Monday, May 25, 2015

### Rijksmuseum Twenthe

This afternoon, I went to the Rijksmuseum Twenthe with Conny. We went to see the Sublime Landscapes in Gaming exhibition. We first walked through the The Forest Dweller exhibition to get there. Works we saw are:
Next we walked through the Metamorphoses. Ovid in contemporary art. We were not very impressed by this. And finally we walked (in reverse order) through The House of Heek exhibition. To my surprise, I discovered that it included three works by Peter Strucyken: Structuur II A, Structuur, and Structuur XXXXV, all from 1967. Left of the works was the work Cirkel en zeshoek in overgang by Ad Dekkers. The painting Falaises près de Pourville from 1882 by Claude Monet was also included.

## Saturday, May 30, 2015

### Book

At 13:07:45, I bought the book Eja Siepman van den Berg bij bookshop Broekhuis for € 7.50. The book contains pictures of the statues by Eja Siepman van den Berg and commentaries (in Dutch and English) by Hans Sizoo and Peter Struycken.

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