โ Scribed by Aviezri S. Fraenkel
No coin nor oath required. For personal study only.
The game of the title is played on the system of five orbits drawn in the figure . Each system consists of four vertices and five or six edges. On distinct vertices of each system, either one electron (dime) and one positron (penny) or two electrons and two positrons are placed. A typical arrangement is indicated in the figure (-for electron, + for positron).
Two physicists play the game by moving alternately. Each physicist at his turn selects a particle and moves it to an adjacent vertex JA along a directed edge, provided that either u is unoccupied or u is occupied by a particle of the opposite type. In the latter case, of course, both particles get annihilated, as physicists tell us particles and antiparticles do.
The object of the game is to make the last move. Thus the physicist who first finds himself without particles is the loser, the other the winner. If there is no last move, the outcome is defined to be a tie.
Remarks. (i) The two upper systems of orbits are identical, and so are the three lower systems. Both systems differ only in that the former has one additional edge. Although this is a very simple case, it may not be immediately obvious that for the initial position indicated in the figure, for example, there is precisely one winning move which enables the physicist making it to win, no matter how his opponent plays. (Which one is it?) All other moves lead either to the 0 c 1977, A.
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