In the previous section we studied elementary knots. We discovered that we use quite a lot of them in our day to day life. Different shapes of elementary knots can be found in books and tutorials. For the complex knots, the situation is quite opposite. We use and know very few knots that involve more than 2 ropes.

We will now evaluate what complex knots could be (as opposed to elementary) and use the concept of black boxes to discover their properties. A black box containing an elementary knot will be our starting point. Again, let us observe it : it has 4 ropes ends, we could say 2 entering ropes and the 2 same exiting ropes. Again, let us repeat that this knot can be of several types, +1, 0, -1. Now what if we try to add some complexity into it? Can we add an entering rope? Yes but not only! The rope HAS to exit somewhere...

Starting elementary knot	Adding one end only is impossible	Adding one rope = adding 2 ends	The knot can be unbalanced	A more natural shape appears

A first result from this short drawing session is that knots ALWAYS have an even number of rope ends (half entering in and half exiting from the knot). Thus, a natural way of sorting the knots shall be by number of ropes. The second result is also very important. A knot with an odd number of rope ends doesn't exist. But a knot with an odd number of branches exists. Its branches should be made of doubled rope ends; in some cases each of them containing one entering and one exiting end but not necessarily. Physical knots with branches containing one single end will eventually be unbalanced and not really nice looking.

Order 3 knots are knots with 3 branches (having 2 rope ends each). After numbering the rope ends we will now draw them, linking any possible input to any possible output. Doing so we will obtain a view of all the knots of order 3 and we shall analyze the result to understand properties of complex knots. Of course we will not draw the physical knots themselves, but rather the set of possible connections between inputs and outputs regardless of how the not can be done physically. Just keep in mind that our knot is a black box. We don't need to know how the ropes are interconnected inside the knot. We just care the inputs and the outputs.

The routine we will execute to cover all the cases is the following : Connect end 1 (input) with immediate possible end (output), Then the next available end (input) to the next available end (output) etc... until no end is available anymore. Then after obtaining a knot, start again but change the first possible connection that has not yet been tested and apply the same routine (first possible input with first possible output that has not yet been tested.) etc....

The first knot will be (1-2;3-4;5-6) Then change the first possible thing (1-2 becomes 1-3) and apply the same routin:. The first available end is 2 that we shall connect to 4. Then the 2 remaining ends (5 and 6) will connect and you'll obtain the second knot that will be (1-3;2-4;5-6). The third will be (1-4;2-3;5-6) The fourth will be (1-5;2-3;4-6) ... etc... ... up to (1-5;2-6;3-4), and finally (1-6;2-5;3-4).

Again, this drawing session helps us to identiy rules and behaviors. For example, you can notice 3 perfectly and yet different symmetrical knots :
(1-2;3-4;5-6) , (1-6;2-3;4-5) and (1-4;2-5;3-6).
Also, you can notice that some knots are the same but simply rotated by 1/3 of turn. There are actually 4 groups :
Group 1 : (1-3;2-4;5-6) , (1-2;3-5;4-6) , (1-5;2-6;3-4)
Group 2 : (1-4;2-3;5-6) , (1-2;3-6;4-5) , (1-6;2-5;3-4)
Group 3 : (1-5;2-3;4-6) , (1-6;2-4;3-5) , (1-3;2-6;4-5)
Group 4 : (1-3;2-5;4-6) , (1-5;2-4;3-6) , (1-4;2-6;3-5)
Eventually the total of different knots of order 3 is 7, which was difficult to predict from the beginning.

If you draw knots with doubled branches, you can start with knots of order 2, then 3 (as we just did), then 4, etc... The number of different knots depends on the axes and centers of symmetry of the knot. Often a given knot will simply be the image of a previous one by a transformation. Therefore, you shall disregard it in your counting. Counting such knots takes us back to an ancient mathematical challenge called "Tsuro tiles". The general solution of this problem is not an easy one unless we go deeper in the groups theory. We won't do so together but if we did we would end up studying a very interesting result called : The Burnside lemma. This result will just tell us the number of different knots that exist for a given order. Although this number is reasonably small for knots of orders below 6, it increases rapidly to astronomical figures.

Order 2 : 3 different knots. Those are the elementary knots.
Order 3 : 7 different knots out of 15 as drawn earlier in this section.
Order 4 : 35 different knots out of 105. The number is not so large, you may try to draw all of them. This is an excellent exercise.
Order 5 : 193 different knots out of 945. The tabulation starts to be challenging. Still doable manually, but challenging...
Order 6 : 1799 different knots out of 10,395! This is humanly doable but VERY boring and the probability or error gets significant now.
Order 7 : 19311 different knots out of 135,135. Sorting out the useful knots becomes a real nightmare.
Order 8 : 254143 different knots to be found within 2,027,025. I believe drawing them is neither wise nor worth.
etc ...

For order 10 there are by far more than 50 millions combinations. You can find a general formula in article A132102 of the Online Encyclopedia of Integer Series; which I contributed to :) OK... That's a little crazy. But don't freak out. Now let's take this one step further and simplify the whole thing.

We are now facing a very simple problem : We want to control the rope path and achieve our bondages just as we draw them. How many knots do we need to learn? Well, You don't want to follow a maths lecture containing all of the explanations. The number is 21; or let's say around 21 to a first approximation. In case you look at orientation (we shall talk about it later on) this is more like 50.

The fact is that regular knots are very few. Here is the way how to figure out how many knots we need to learn. Remind that branches are always doubled. An entering rope will all enter from (let's say) the left side of a branch. All the entering ropes ends will be the symmetric of the first entering rope, thus entering on the left hand side of a branch. There is no choice to be made here. The first entering rope will exit from the right hand side of a given branch. And just like it all the other exiting rope ends will exit from the right hand side of a branch, as they result from the symmetry of the first exiting end. So to say, the exiting ends have no choice but mimic the first one. Only the first exiting end "has a choice" to pick its exiting branch. This leaves us with very few possibilities and we shall even be able to verify this statement using a knot of order 5.

1 blue rope entering from the left : All branches are equivalent.	Symmetry	Knot is symmetrical. Ropes on the left will all be "entering" ends		Exiting branches are not equivalent. There are 5 different choices	Symmetry	Each of these 5 ropes replicates the red one and exits on the right.
	x5				x5

There are only 5 mathematically different symmetrical knots of order 5 which are defined by the 5 possible choices for the first exiting end. Following the same discussion, there are only 3 mathematically different symmetrical knots with 3 branches (of order 3), 4 of order 4, etc... The theoretical number of symmetrical knots to learn if we want to use the knots up to order 6 is 6+5+4+3+2+1 = 21. Of course you can discuss of what use a knot of order 1 would be but, As we do not go by numbers I prefer to say that the number is "approximately 21". Also, you will soon discover that several worthy knots of order 2 and higher orders exist; (and you want to learn those ones).
But what you have to understand, is that for each order, the number of symmetrical knots equals the order number itself. i.e. there are only 8 symmetrical knots out of the 2541413 possible knots of order 8; only 3 symmetrical knots out of the 7 different knots of order 3...

Of course we just did the counting here. But don't worry about the look of the knots! As we are working on symmetrical species, you guess those knots are particularly elegant!

The conclusion of this chapter is double : We established a sorting criteria and proved that knowing the physical shape of our knots is not really needed as long as the connections that result from it are known as well. Actually knowing the resulting connections is enough to draw your bondage. Bondages can be designed using Black Boxes. In the next chapter we will address the knot tabulation problem. Once a relevant tabulation is created, we will have to find physical shapes of all our knots.