PPan Galactic Division
Rich SchwartzSeptember 27, 2018
The purpose of these notes is to explain Peter Doyle and Cecil Qiu’s proof [2]that “division by N ” is possible on a set-theoretic level. To state the resultformally, let S × N stand for S × { , ..., ( N − } . Here S is an arbitrary set. Theorem 1.1.
Let A and B be sets. If there is an injective map from A × N to B × N then there is an injective map from A to B . The significance of Doyle and Qiu’s proof is that it produces an effective(i.e. canonical) injection from A → B in a simple and straightforward way.In particular, their proof avoids using the Well-Ordering Principle or anyother form of the Axiom of Choice. For comparison, in 1949 Tarski [3] gavean effective proof, but Tarski’s proof is complicated. Conway and Doyle [1]sought a simpler proof, but found nothing as simple as the proof here. Formore on the long and tangled history of this problem, see [2].Theorem 1.1 is an immediate consequence of induction and the followingresult. Theorem 1.2.
Let N be any positive integer. If there is an injection from A × N to A × N then there is an injection from A × ( N − to B × ( N − . The proof is the same regardless of the choice of N . Following Doyle andQiu, I’ll give the proof when N = 4, so that it looks like a game of cards.Doyle and Qiu call their game Pan Galactic Division, because they contendthat a definite fraction of intelligent civilizations in the universe will have hitupon their canonical algorithm. 1 a r X i v : . [ m a t h . C O ] A p r The Rules of the Game
Think of B × B and the suits spades, hearts, diamonds, clubs are justother names for 3 , , ,
0. Think of A as a set of players, each staring at 4spots in front of them, named spades, hearts, diamonds, clubs in orderfrom left to right.Enough of the cards in the deck are dealt out to the players so that eachplayer has a hand of 4 cards, placed into the 4 spots. (The suit on a cardneed not match the suit which names the spot it is in.) The injective mapfrom A × B × Shape Up and
Ship Out . (1) Shape Up: Each player having at least one spade arranges to havea spade as the leftmost card. This is done by swapping the leftmost spadewith the leftmost card, if such a swap is necessary. Figures 1 and 2 show anexample, before and after the player shapes up.Figure 1: Before shaping up: The Chicken of spades is the leftmost spade.Figure 2: After shaping up: The Chicken of spades swaps with Ape of clubs.2 ames of the Hands:
All players shape up simultaneously. Once that isdone, any hand which has a spade in it will also have a spade in the leftmostspot. In this case, the name of this leftmost spade names the hand. In ourexample, the name of the hand is “Chicken”, because the Chicken of spadesoccupies the leftmost position. (Since the sets A and B might be very large,we can’t expect our cards to be as in a traditional deck.) The hands withoutspades are not named. (2) Ship Out : After a player shapes up, any named hand in her handwill have a spade on the left. Say that a bad spade is a spade in a namedhand which is not the leftmost spade. In our example above, the 2 of spadesis a bad spade. A hand can have up to three bad spades.In the Ship Out round, each player with a bad spade swaps their left-most bad spade with another card in the game, according to the followingrule. Recall that the spots in front of the player are named spades, hearts,diamonds, clubs, from left to right. The player asks for the card which hasthe same suit as the name of the spot occupied by the leftmost bad spade,and the same name as the name of the hand. There is never any conflict:two players will not ask for the same card.Let’s continue with our example above. In our example, there is one badspade, namely the 2 of spades. Even though this card lies in the rightmostspot, it is the leftmost bad spade, because it is the only bad spade. The 2of spades occupies the clubs spot, according to our scheme, and it lies in theChicken hand. Therefore, the player swaps the 2 of spades with the Chickenof clubs.The Chicken of clubs either is part of another player’s hand, or elsewherein the player’s own hand, or else is still in the deck; the distinction does notmatter. Figure 3 shows our hand after the player ships out.Figure 3: Two of spades is traded for the Chicken of clubs.3 Playing the Game
The game is now played indefinitely, with the rounds alternating: shape up,ship out, shape up, ship out, . . . Notice that a hand never loses its leftmostspade card, because the Ship Out rule will never require anyone to call for aspade. So, any player shapes up at most once.The quality of a named hand is assessed according to how many cardshave names which match the name of the hand and suits which match thename of the spot they occupy. Let’s consider our example. Before shippingout, our player has 1-of-a-kind, so to speak, because the Chicken of spades isher only card which has the same name as the hand (by definition) and is inthe correct position. After shipping out, our player has acquired the Chickenof clubs and placed it into the clubs position. Thus, now she has 2-of-a-kind.Notice that our player has improved her hand by shipping out.As our example suggests, a player always improves her hand by shippingout a spade. A player’s hand can have a spade shipped in by having one ofher cards called away. For instance, in the example below, our player losesthe ape of clubs and acquires the bolt of spades. The bolt of spade is a badspade in the diamonds position, as shown in Figure 4.Figure 4: The bolt of spades occupies the diamonds position.In the next round, our player ships out the bolt of spades and gets theChicken of diamonds, as shown in Figure 5. In this way, she has improvedher hand to 3-of-a-kind. So, any time a player participates in a ShippingOut round, either actively or passively, her hand improves after at most twomore rounds of the game. 4igure 5: The bolt of spades is traded for the Chicken of diamonds.
From the discussion above, we conclude that during this indefinite game, aplayer’s hand can change at most 8 times. Moreover, once a player’s handhas stopped changing, the player has no bad spades. In other words, oncethe hand has stabilized, it has no spades in the 3 non-left spots.We imagine that the game proceeds indefinitely. There is a well-definednon-spade card associated to each pair (player, non-left spot): We simplywait until the player’s hand stabilizes - and it will stabilize - and then wesee what card is occupying the relevant spot. This gives us an injective mapfrom A × B × References [1] Peter G. Doyle and John Horton Conway. Division by three,1994, arXiv:math/0605779 [math.LO]. http://arxiv.org/abs/math/0605779 .[2] Peter G. Doyle and Cecil Qiu. Division by four, 2015, arXiv:1504.01402[math.LO]. http://arxiv.org/abs/1504.01402 .[3] A. Tarski. Cancellation laws in the arithmetic of cardinals.