Last week we saw that the best strategy for Wythoff’s game is to always move to one of the blue squares in the diagram below (if you can). We found the locations of the blue squares with an iterative algorithm that builds outward from (0,0), but we haven’t yet discovered the underlying pattern in how these blue cells are arranged. Let’s go pattern hunting!

As we noticed last week, the blue cells are arranged in a symmetric “V” shape, formed from two seemingly straight lines. What are the slopes of these lines? This is the question du jour.

Looking more carefully, the upper branch of the “V” has only two types of “gaps”: consecutive blue cells are separated by either a “short” (1,2) step (a.k.a. a knight’s move) or a “long” (2,3) step. So, if we believe that this “V” branch is a perfect line, then that line’s slope should be somewhere between the slopes of these two steps, namely \(3/2=1.5\) and \(2/1=2\). Knowing more about how these steps are arranged will tell us more about the “V”.

Specifically, label these “short” and “long” steps as a and b respectively. What we care about is the ratio of bs to as. Why? For example, if there were an equal number of as and bs on average, then the slope of the line would be the same as the vector \(a+b=(3,5)\), i.e., the slope would be 5/3. If there were, say, two bs for each a in the long run, then the line would fall in the direction of \(a+2b=(5,8)\), with slope 8/5. Unfortunately, the ratio we seek is neither 1 nor 2; what is this magic number?

Let’s write down the sequence of jumps along our line. From the diagram above we see that it starts a b a b b a …, and using last week’s iterative algorithm, we can compute farther into this infinite sequence:

This sequence seems to be composed of “blocks” of (a b) and (a b b):

So for every a there are either one or two bs, meaning the ratio of bs to as is somewhere between 1 and 2. So the slope should be between 5/3 and 8/5. But what is it exactly? To answer this, we need to know how these “short” and “long” blocks are arranged.

Write A = (a b) and B = (a b b), so we can rewrite the sequence of blocks as

Hold on; this sequence looks familiar. It’s exactly the same pattern as our original jump sequence! So it seems that the pattern of blocks is exactly the same as the pattern of the letters themselves! This is a “self-generating” sequence!

In particular, if the ratio of bs to as is some number r, then the ratio of (a b b) blocks to (a b) blocks is also r. But if we have one block of (a b) and r blocks of (a b b) then we have a total of \(1+2r\) bs and \(1+r\) as, so the ratio of bs to as is \(\frac{1+2r}{1+r}=r\). This simplifies to \(1+r=r^2\), and the solution is the shiny number \(r=\frac{1+\sqrt{5}}{2}\), known as the golden ratio and denoted \(\phi\) (Greek letter phi).

Now we know that our purported line should be in the direction of \(a+b\cdot\phi = (1+2\phi,2+3\phi)\), so the slope is \((2+3\phi)/(1+2\phi)=\phi\). And since our “V” shape is symmetric, the other line should have slope \(1/\phi = \frac{\sqrt{5}-1}{2}\). Done and done!

Well, yes and no. We found the slopes of the lines in the “V”, but why do they form lines at all? And what does all this have to do with the Fibonacci numbers?