Putting the Why in Wythoff

Last week we drew dots at multiples of the vectors \(v=(\phi,\phi^2)\) and \(w=(\phi^2,\phi)\), and we colored grid-cells green or yellow according to whether they contain a dot or not. Our remaining task was to prove these three facts:

  • Endgame condition: The cell (0,0) is green.
  • Greens lose: From a green cell, there are no other green cells accessible with a single Wythoff’s game move.
  • Yellows win: From any yellow cell, it is possible to move to a green cell with one Wythoff’s game move.
Our guess at the solution to Wythoff's game
Our guess at the solution to Wythoff's game: multiples of vectors v and w (dotted) and their containing cells (dark green).

The “Endgame” condition is easy: the dot at \(0\cdot v = (0,0)\) makes cell (0,0) green. One condition down, two to go!

The “Greens lose” condition asks us to show that no two green cells are in the same row, column, or (slope 1) diagonal—otherwise, you could get from one to the other with a Wythoff move. In fact, we will prove something stronger:

Covering Theorem: Every row, column, and diagonal contains exactly one green cell.

Pictorially, this says that each thick line in the diagrams below will hit one and only one green cell:

Covering the columns and diagonals
All columns (left), diagonals (right), and rows (not pictured) contain exactly one green cell. (Some of these green cells are beyond the image boundaries.) Because the green cell locations are symmetric through the line y=x, the diagram for the rows is redundant and therefore omitted.

It is not difficult to describe why each diagonal is covered, i.e. has exactly one green cell. In the diagram on the right, the central diagonal line has equation \(y-x=0\); the one above it has equation \(y-x=1\); the next is \(y-x=2\); and so on. And because \(\phi^2-\phi=1\), the vector v jumps from one diagonal line to the next at each step. Vector w similarly jumps from line to line in the other direction.

By contrast, I find the fact that each column is covered quite surprising. It says that the upper and lower branches of the “V” perfectly complement each other, each filling exactly the columns that the other skips! Before we prove this fact, let’s see why it helps us.

If we assume the covering theorem for now, we can finish the three conditions needed for our proof. We already saw that (0,0) is green, and the covering theorem certainly tells us that each row, column, and diagonal has at most one green cell, so the “Greens lose” condition holds. We just have to prove the “Yellows win” property: that you can always move to a green cell from any yellow cell. The idea is simple: if your yellow cell is above the “V”[1] then move down to the unique green cell in your column (which may be on either the upper or lower “V” branch). If you are below the “V”, move left to the green cell in your row. Finally, if you are inside the “V”, move diagonally. In this way, you can always move to a green cell with one Wythoff move.

Proving the "Yellows win" condition
All yellow cells can reach a green cell with one Wythoff move. Yellow cells above the "V" can move down to a green cell; yellow cells below the "V" can move left to a green cell; yellow cells inside the "V" can move diagonally to a green cell.

With this third and last property verified, we’re finally done with our proof! …except that we still need to prove the mysterious covering theorem. We’ll discuss this next week. Come back then for the exciting QED!


  1. i.e., its lower-left corner is above the line \(y=\phi\cdot x\) []

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