So…
After I’ve slept over it.
The Fibonacci works well, because the algorithm creates a base case with a ‘0’ MC.
The schema again:
| C | - | B |
| - | 0 | A |
| - | D | - |
| C | -D | B |
| -A | 0 | A |
| -B | D | -C |
I was just doodling last night and just kind of figured it out to make it work, but now that I look at it, it is obvious why it had to be this way.
Starting in order of the Fib numbers…
Ok, Ok, let’s do it properly.
| F3 | -F4 | F2 |
| -F1 | 0 | F1 |
| -F2 | F4 | -F3 |
Let us build it with a few simple steps.
So, three principles to the arrangement:
0. We place a zero in the middle.
1. We are considering the sides – the lines that do not go through the center. In order to bring each one of them to 0 using the Fib numbers, we simply take advantage of the fact that, by definition, we will get to 0 if we take any Fib number (2 and above) and remove the previous two. Meaning that we need two consecutive Fib numbers with one sign (positive or negative) and the next Fib number with the opposite sign.
1.1 We have 4 Fib numbers and we need them to cover 4 sides. The way to do it is to use the fact that 4 consecutive numbers form 2 overlapping subsets of 3 consecutive numbers each – F1 to F3 and F2 to F4.
Given that F2 and F3 are part of the two subsets, we place them in neighboring corners. Now F3 will form one of the subsets with F2 on their shared side and the other subset on the other side. Similar for F2.
| F3 | - | F2 |
| - | - | - |
| - | - | - |
1.2 From the subsets we know that F2 is the middle value in subset 1 and the lowest value in subset 2. Correspondingly, F3 will be the highest value in subset one and the middle value in subset 2. This means that the top side (in this rotation), that already contains F2 and F3, will be part of the two subsets.
This is where we note that at this stage we have four sides with a total of four copies of the two subsets. For now, we will forget about the sign of the numbers and focus on the values.
Once we have the F2 and F3 placed, the rest is almost automatic. We have one choice to make, which is completely arbitrary, because it gets cancelled later on by the symmetries of the MS.
We need to place either F1 or F4 between F2 and F3 that are already on the board. Let’s choose F4. This means that the newly formed side is the larger subset.
We copy each of the values across the center, resulting in another copy of the subset, just in reverse order looking in the same direction (from left to right in this case). Only two cells remain and we fill them in with the value we did not choose the last time. In this case – F1.
We get:
| F3 | F4 | F2 |
| F1 | 0 | F1 |
| F2 | F4 | F3 |
1.2 We pick a side and start multiplying by -1. The rule that governs the multiplication is what we mentioned earlier – in each of the subsets, the two smaller values have the same sign, which is also opposite to the sign of the largest value.
At the top we simply convert F4 to -F4. So F2 and F3 remain positive. This means that on the right F1 will have the same sign as the F2 and F3 will become -F3. On the left F3 is the highest value, so F1 and F2 there will be multiplied by -1. This also automatically takes care of the bottom row.
| F3 | -F4 | F2 |
| -F1 | 0 | F1 |
| -F2 | F4 | -F3 |
1.3 Sanity check – across the center each Fib number stands opposite its negative, so along with the zero at the center, so each diagonal results in a zero.
1.4 Versions – the only notable thing so far is that if we had chosen to place F1 at the top, instead of F4, we would have gotten a schema that is flipped along the F2 diagonal. It is not immediately obvious to me why F2 and not F3. Needs some looking into.
| -F3 | F1 | F2 |
| F4 | 0 | -F4 |
| -F2 | -F1 | F3 |
—
Consequence of this whole thing. We are not using the fact that the numbers are part of the Fibonacci sequence. We are merely using the property that numbers in the Fibonacci sequence have – each equals the sum of the previous two. Which means that we can also freely use the Lucas sequence or any other that shares the same building principle.
Next:
– What about Tribonacci? Can we extend the algorithm to cover 4×4 MS? There are some nice complications that are visible from the start – we will need longer Fib subsets, but we still have only four sides. Also, we have outer sides and inner sides. Delight!