Welcome to Lucy and Lily, a game of geometry and Galois theory. This tutorial will explain how to use, play, and win the game. To read more of this text, use the scroll bar to the right of the window. 1. Tour of the Buttons PENTAGONS: Each of the two windows has a pentagon in it. If you click on the outer part of the pentagon, it will move into the position indicated by its white shadow. The inner parts cannot move. The big pentagon on the left is linked to the big one on the right. A move on one of the pentagons forces a move on the other. The linking is color coded. Try clicking the pentagons a bit. NUMBERED BUTTONS: The numbered buttons are labelled 2,4,10,50. If you click BUTTON 2, the pentagons will move as if you had made 2 random clicks. Similar statements apply to the other buttons. You can push these buttons repeatedly. Try pushing some of the numbered buttons. RESET BUTTON: The reset button returns the pentagons to their original positions at the center of the playing board. Try the reset button. HIDE BUTTONS: If you click on the hide button it will cover up two of the screens with a blue panel. If you click the same button again, it will remove the panel. Try the hide buttons. 2. Using the viewport To avoid distractions, click on the right hide button so as to cover up the two right windows. Everything said about the screens on the left applies equally to the ones on the right. The big window at the upper left is called the playing board. The small window at the lower left is called the viewport. The viewport records in miniature the objects in the playing board. The small cyan square in the viewport surrounds the visible portion of the playing board. By clicking on the viewport, you can recenter the cyan square and thereby recenter the playing board as well. To practive navigating, press BUTTON 50 until the pentagon disappears from the playing board. Use the viewport, together with clicking, to bring the pentagon back onto the playing board. 3. How to Play There are 2 games you can play: the single game and the double game. To play the single game, click on one of the HIDE buttons to cover up two of the windows. Now press one of the numbered buttons. The goal of the game is to reunite the two separated pieces. To play the double game, uncover all windows, press one of the numbered buttons, and try to reunite both pairs of pieces simultaneously. At first it might look as if the double game was much harder than the single game. In fact, the opposite is true. The single game can be extremely difficult, if one first makes a lot of random moves (for instance by pushing BUTTON 50 several times in a row.) There is no winning strategy that I know of. On the other hand, it turns out that there is a swift winning stragety for the double game. In fact, there are many such strategies. I will explain one of them. 4. How to Win For the purposes of this explanation, call the outer pentagonal ring on the left Lucy1. Call the inner pentagon Lily1. Let Lucy2 and Lily2 be the corresponding pieces on the right. Why the names? Lucy and Lily are my daughters. Here is the strategy: Step 1: Move Lucy1 as directly as possible towards Lily1 until they touch. Step 2: Move Lucy2 as directly as possible towards Lily2 until they touch. Step 3: Move Lucy1 as directly as possible towards Lily1 until they touch. Step 4: Move Lucy2 as directly as possible towards Lily2 until they touch. And so forth. After a finite number of steps, you will have won. Try it for yourself, it works like a charm. 5. The underlying mathematics Why does the winning strategy work? Here is the explanation. First of all, when you move Lucy1 as directly as possible towards Lily1, you do so by clicking on nonadjacent colors. The coloring is set up so that nonadjacent colors on Lucy1 are adjacent on Lucy2. Thus, when Lucy1 is moving as directly as possible, Lucy2 is meandering along. The same goes with the roles of the pieces reversed. Thus, if you complete two steps of the strategy, you have decreased the combined distance from Lucy1/Lucy2 to Lily1/Lily2. I will explain below that, given any number N, there are only finitely many positions of Lucy1 and Lily2 such that the combined distance from Lucy1/Lucy2 to Lily1/Lily2 is less than N. Thus, the strategy above keeps decreasing this finite number and eventually winds up at 0. Now for the explanation of why there are only finitely many positions such that the combined distance is small. Let w= cos(2 Pi/5) + i sin(2 Pi/5) be the usual 5th root of unity. Let R be the ring Z[w]. That is, R is the set of numbers of the form a0+ a1*w^1+ a2*w^2+ a3*w^3+ a4*w^4 Here a0,a1,a2,a3,a4 are all integers. There is a map f: R ---> R defined by the formula: f(a0+ a1*w^1+ a2*w^2+ a3*w^3+ a4*w^4)= a0+ a1*w^2+ a2*w^4+ a3*w^6+ a4*w^8. The number on the right is back in R because w^6=w and w^8=w^3. Technically, f is the restriction to R of one of the Galois automorphisms of the Q[w]. If Lucy1 and Lucy2 are properly scaled then the following is true: Lucy1 is always centered at a point r in R and at the same time Lucy2 is centered at the point f(r). One should really think of Lucy1 and Lucy2 as the shadows of a 4 dimensional object, LUCY=(Lucy1,Lucy2), which is centered at the point (r,f(r)). Similarly, the combined object LILY=(Lily1,Lily2), is centered at the origin (0,0). It is a famous theorem of number theory, due to Dirichlet, that the set {(r,f(r)), r in R} (that is, the graph of the function f, restricted to R) is a 4-dimensional lattice. That is, this graph looks just like an ordinary grid in the plane, except that it is 4 dimensional. In other words, LUCY moves along a 4 dimensional grid in search of LILY and you, the player, see each of two planar projections. There are only finitely many grid points inside any ball. This explains the claim above, concerning the finiteness of the positions of Lucy1/Lucy2 whose combined distance to Lily1/Lily2 is small.