Sun Mon Tue Wed Thu Fri Sat 1 2 |

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 011221121121In 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.

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.)*

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.

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.

- Prints from
*Killzone ShadowFall*from Guerrilla Games. - Prints from
*Battlefield 4*from Electronic Arts Inc. *Remake of Robert Smithson's Spiral Jetty in Minecraft*by Jan Robert Leegte.*Mirror's Edge*from Dice.*3 Levelmap parallax*by Julian van Aalderen.*Xilitla*by Rosa Menkman.*TRON grid + lightcycle*by Jelle de Graaf.- Movie with fragments from:
*Future*,*Portal 2*,*Half-Life 2*,*Take on Mars*,*Eve Online*,*Mass Effect 2*,*Borderlands 2*,*Cage*,*Alien: Isolation*.

- 8: Crane Operator Takes Breathtaking Photos of Shanghai From 2,000 Feet High
- 14: user48736353001
- 14: Sloane's gap: cultural influences in mathematics
- 24: Genome-scan for IQ discrepancy in autism: evidence for loci on chromosomes 10 and 16
- 24: The Level and Nature of Autistic Intelligence II: What about Asperger Syndrome?
- 28: How Spacetime is built by Quantum Entanglement: New Insight into Unification of General Relativity and Quantum Mechanics
- 28: Tomography from Entanglement

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