Presentation on theme: "The Angel Wins!! SASMS, Nov 13, 2006 David Rhee. 1. Conway’s Angel Problem An angel of power p and a devil plays a game on a infinitely big chessboard."— Presentation transcript:
1. Conway’s Angel Problem An angel of power p and a devil plays a game on a infinitely big chessboard. The angel occupies a square. On angel's turn, she jumps to a different square that is at most p squares away. On devil's turn, he destroys one square. They alternate their turn. The angel wins by surviving indefinitely, and the devil wins by trapping the angel.
2. History The problem first appeared in the book Winning Ways by Berlekamp, Conway, and Guy, in 1982. Conway came up with some partial result such as : If the angel has power 1, then the devil can win. (1982) If the angel never moves down, then the devil can win. (1982) If the angel always increases the distance from the origin, then the devil can win. (1996)
2. History The angel seems to be advantaged, but the devil can never make a mistake. In 3D case, the angel wins if p≥13. Solved in late 2006 by Péter Gács (high p), Brian Bowditch (p≥4), Oddvar Kloster (p≥2), and András Máthé (p≥2), independently. If the angel is not allowed to travel more than squares south of a square that she already visited, the devil wins.
3. Definitions Norm Way of measuring length. Use for this talk. Angel can move at most p units in this norm. Devil wins if angel is trapped in a finite ball in this norm.
3. Definitions Origin : Angel’s starting square Nice Devil : Devil that never destroys squares that the angel already visited or could have visited. Journey : A sequence of squares that are moves of an angel Strategy : A function from the set of all finite journeys to the set of squares Block : A set of destroyed squares that are connected Block Not a block
3. Definitions 4. The squares used in this path form a journey, and we call this journey reduced and write rdj(J). 2. For each, draw an arrow to the first square,, where d(, )≤p. 3. Find a path, starting at, and ending at. Reduced Journey 1. Consider a journey, J= (,,…, ). (0,0)(1,2)(4,3)(4,1)(5,-1)(6,0)(6,-2)(4,-5)(0,0)(1,2) (4,1)(6,-2)(4,-5)
4. Nice Devil Proposition If devil has a winning strategy against the angel with power p, then the nice devil also has a winning strategy against the same angel. Proof Let S to be a strategy where devil can trap angel of power p. Then, define
4. Nice Devil Claim If J is a legal journey against the strategy T, then rdj(J) is a legal journey against S.
5. Runner We solve the problem for high powered angel by constructing a method to defeat the nice devil. She visits at most 2p neighboring squares in one move. She does not leave B(A;p), where A is the starting position for the turn. She uses left-hand rule. She paints all the walls she touch. She paints at most 2p walls per turn. The devil destroys all squares left of y-axis.
5. Runner Property If the runner paints a same wall twice, then the first wall to be painted twice is the first wall ever painted. Proof Suppose that e is the first wall to be painted twice. If the claim is false, then there exists a wall, f, that was painted right before e was painted for the first time. But the devil is nice, so he can’t do much in that area. Then f has to be painted right before e is painted the second time.
5. Runner Proposition If p>9000, the runner never goes back to the origin. Proof After t turns, the devil destroyed at most t squares, creating at most 4t new walls. Since the runner paints at least p walls each turn, she painted at least tp walls. Therefore, she painted at least tp–4t walls on the y-axis. She can only go up while doing this, so she makes at least tp–4t move going up. The runner can go down at most t times (the devil made at most t squares), so she is at least tp–5t north of the origin after t turns. tp–5t=t(p-5)>0 so the runner never goes back to the origin.
5. Runner Corollary If p>9000, then the runner wins against the nice devil. Proof Suppose that the nice devil wins. Then, he can trap the runner in a finite ball. There are only so many walls to paint in the ball, so she has to repaint at some point. Then the first wall to be repainted is the first wall she ever painted. In order to do that, she has to return to the origin. But, by the previous proposition, this does not happen. Contradiction!
6. Power 2 We modify the runner’s strategy little bit. The skinny runner can make diagonal moves and is thin enough to go through tiny gaps.
6. Power 2 Proposition The skinny runner never reaches x-axis again. Proof Suppose that this happened after t turns. Let a be the number of walls painted by angel so far. Then a is at least 2t-1, since angel paints at least 2 walls for the t-1 turns and it takes at least another painting to get to the x-axis. Therefore, a≥2t-1 –(1). Let N-9000 be bigger than the y-coordinate of any of the angel and the devil’s activity. Now, consider A = (set of squares that are above the x-axis, below y=N, and right of x=-1).
6. Power 2 Let all the squares outside of A be restored. We ask the skinny runner to run until she repaints a wall, which has to be the first wall she painted. Let l be the length of the cycle she painted. Clearly, l≥a+2N+5 –(2) from the diagram. The squares inside the cycle forms a block. There are at most N+t destroyed squares in there, which provides at most 2(N+t)+2 walls. This block has at least l walls, so 2(N+t)+2≥l –(3).
6. Power 2 2N+2t+2 ≥ l – (3) ≥ a+2N+5 – (2) ≥ 2t-1+2N+5 – (1) ≥ 2N+2t+4 This is a contradiction, therefore the skinny runner does not come back to the x-axis.
Bibliography Angel Problem. (2007, August 12). In Wikipedia, The Free Encyclopedia. Retrieved August 21, 2007, from http://en.wikipedia.org/wiki/Angel_problem http://en.wikipedia.org/wiki/Angel_problem Máthé, A. (2007). The angel of power 2 wins [electronic version]. Combinatorics, Probability and Computing, 16, 363–374. Retrieved on August 21, 2007, from http://amathe.dyn.elte.hu/letolt.php?angel-mathe.pdf http://amathe.dyn.elte.hu/letolt.php?angel-mathe.pdf
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