*The four colour conjecture is a fascinating mathematical problem —because it is extremely easy to understand what it says. So it has a huge watcher-base. Also it is infuriatingly difficult to prove, which is puzzling (if it is a fact, which is not quite 100% certain). The present author has been fascinated by it for nearly 70 years. It is extraordinarily difficult to see how a simple proof could be put together, and there are even some arguments which appear to suggest that it might be invalid. *

*It has been allegedly resolved by Appel and Haken and others using “computer proofs” which, however, leave a slight doubt in the mind of sceptical critics. This blog is the second in a short series which offers an elementary way of starting to analyse the problem —one which might eventually lead towards a proof (or even, improbably, a disproof).*

* In the previous blog (Part 1) we started to look at the process of constructing a fully saturated, fully triangualised network —the field of the inquiry. This blog is Part 2 of that process. It is surely essential to have a clear, precise concept of the field of configurations (networks) one is claiming can be four-coloured. *

*To recap, an origin O is chosen of order 5. (Any such network necessarily contains many nodes of order 5.) *

*O is surrounded by a pentagon, a “circuit” of nodes, comprised of five original nodes and the five links joining them together. Next, five new nodes can be added immediately, each one being the vertex of a triangle whose base is one of the five sides of the original pentagon. Then these “new nodes” can be (a) either connected together by five new links, or (b) they can be connected together by strings of “newer” nodes as shown in Diagram E —thus forming a second, now longer, circuit of O. *

*Notice that no pair of two nodes on the first circuit can be linked directly, because that would entrap one of the five nodes with only 3 links (order 3) —a precondition ruled out by the decision to focus on saturated networks..*

*It is also impossible to add links which join pairs of nodes on the *second* circuit (of both type (a) and (b)) for the same reason. *

*Following this initial set-up, a succession of new circuits of nodes moving ever further from O can be constructed. They grow initially and may later pass through countless phases of growth and temporary contraction. Such a simple concentric network will eventually end, with a final circuit of nodes. *

* This is not the whole story, After the second circuit has been completed, later circuits may contain extra looping links between pairs of nodes —provided at least five nodes are involved in the “loop” thus created. *

*Such a “loop” creates an empty space which can be, in effect, the start-circuit of a fresh set of concentric sub-circuits, i.e. a sequence of sub-circuits independent of the initial ones. Standing back and taking a three-dimensional view of this overall structure each concentric set of sub-circuits can be treated as a “limb”, the set of all such limbs forming a tree. A large standard circuit (after the 2*^{nd}*) may sprout a number of looping links, each one potentially becoming the start of a new independent concentric “limb”. From such simple beginnings an immense field of complex, sophisticated networks may be conceptualised.*

*But it is, in the last analysis, the individual limb, and particularly the “closing down” of each individual limb, which represents the essence of the problem. A hyper complex network may have a mass of limbs like a dense thornbush, but it is, in the end, the way in which the four-colouring process successfully copes (if it can) with “a typical limb’s completion” which holds the key to the problem.*

*>>>>An interesting hypothesis presents itself at this point: that the four-colour problem can be tamed by adopting a suitable protocol for colouring each node of each new circuit as it presents itself in the construction of the network. Such a protocol may be said to be a way of <<colouring as you go>>.*

*The simplest way to complete a typical simple, linear limb, is to introduce a terminator node, T which links to all the nodes on the final circuit, and thereby ‘completes’ the limb. (See Diagram E.)*

* [This is not the only way a typical limb can be completed, as we see presently.]*

*The first condition is that every node on the final circuit of a limb must have at least two links to earlier nodes and two links to its adjacent nodes on the circuit. This means that a circuit composed of the kind of new nodes encountered in option (b) of the second circuit cannot qualify. This condition is necessary because these nodes on the final circuit of a limb must each have five links after it has been linked to the final “terminator” node T,*

*It means that there are two kinds of circuits and sub circuits in a typical network, (i) those that are *completable*, (ii) those that are *incompletable*. The second circuit of type (a) was, trivially, completable. The second circuit (b) was incompletable.*

*We turn now to focus on the final circuit or sub-circuit of a limb.*

*Clearly the nodes on the final circuit must all be colourable using only three colours: because a different fourth colour will be needed to colour the terminator T.*

* Yes. A colouring protocol can be conceptualised which guarantees that every successive circuit working outwards from O has only three colours, one of which is a minority colour. (There are relatively few pf these on the circuit.)*

* Let’s call the colour of O red. Those of the nodes on the first circuit (a pentagon) may be taken to be blue, yellow, blue, yellow and green. (See Diagram F). It is convenient to call red and green the ‘Robinhood’ colours, and blue and yellow the ‘Ukrainean’ colours. (These colour-pairs will be called ‘paired colours’.) The 1*^{st}* circuit has, say, alternate Ukrainean colours and a singleton green.*

* [In effect we are *calling* the colour chosen for O “red”. Likewise we are calling the first singleton colour “green”, and the four remaining colours “blue” and “yellow”.] *

* The next circuit (the 2*^{nd}*) may be a circuit with many nodes, some of which are initially only of power 3. Such a circuit cannot be completed by means of a terminator node, because some of its nodes will end-up as being of power 4. They can however be coloured alternately using the Robinhood colours, but at least one Ukrainean coloured node will be needed to cope with those 2*^{nd}* circuit nodes which are linked directly to the green node of the 1*^{st}* circuit. (See Diagram G.)*

* This pattern can clearly be repeated again and again until we reach the final circuit which will be colourable using only three colours. Therefore the network can be completed by using the remaining colour for the terminator node T..*

* So here we have a colouring protocol which is an example of “colouring the network as you go”. The unspoken premiss here is that the colour decisions of the protocol can be made in a way which stay in place.*

* Another way of completing a limb, however, emerges. The final circuit may be completed by attaching all its nodes to a “terminating triangle” of three differently coloured nodes. See DIAGRAM H.*

* Now it is fairly easy to see that completion by this device cannot occur if only three colours are present on the final circuit.*

* It follows that the hypothesis that the colouring can be done in a stable way “as you go” is invalid. This is an important negative result which removes what would have been a pleasantly simplifying approach.*

**CHRISTOPHER ORMELL 1 ^{st} NOVEMBER 2023**

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