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# mar12 - MIT OpenCourseWare http/ocw.mit.edu 6.055J 2.038J...

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MIT OpenCourseWare http://ocw.mit.edu 6.055J / 2.038J The Art of Approximation in Science and Engineering Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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Chapter 6 Box models and conservation 6.1 Cube solitaire Here is a game of solitaire that illustrates the theme of this chapter. The following 1 0 0 0 0 0 0 0 cube starts in the configuration in the margin; the goal is to make all vertices be multiples of three simultaneously. The moves are all of the same form: Pick any edge and increment its two vertices by one. For example, if I pick the bottom edge of the front face, then the bottom edge of the back face, the configuration becomes the first one in this series, then the second one: 2 1 0 0 0 0 0 0 2 1 0 0 1 1 0 0 Alas, neither configuration wins the game. Can I win the cube game? If I can win, what is a sequence of moves ends in all vertices being multiples of 3? If I cannot win, how can that negative result be proved? Brute force – trying lots of possibilities – looks overwhelming. Each move requires choosing one of 12 edges, so there are 12 10 sequences of ten moves. That number is an overestimate because the order of the moves does not affect the final state. I could push that line of reasoning by figuring out how many possibilities there are, and how to list and check them if the number is not too large. But that approach is specific to this problem and unlikely to generalize to other problems. Instead of that specific approach, make the generic observation that this problem is 1 0 0 0 difficult because each move offers many choices. The problem would be simpler with fewer edges: for example, if the cube were a square. Can this square be turned into one where the four vertices are multiples of 3? This problem is not the original problem, but solving it might teach me enough to solve the cube. This hope motivates
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