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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.
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.
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.
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.
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.
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.)
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.
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.
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.
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.
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.
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.
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.
This months interesting links
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