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anonymous
 2 years ago
For a theoretical graph, the nth cube Q_n is a simple graph whose verices are the 2^n points (x_1...x_n) in R^n. So that for each x_i=0 or 1, and whose two vertices adjacent if thet agree in exactly n1 coordinates.
show that if n>=2 the Q_n has a hamiltonian cycle.
anonymous
 2 years ago
For a theoretical graph, the nth cube Q_n is a simple graph whose verices are the 2^n points (x_1...x_n) in R^n. So that for each x_i=0 or 1, and whose two vertices adjacent if thet agree in exactly n1 coordinates. show that if n>=2 the Q_n has a hamiltonian cycle.

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0• ok. so i am going to use induction and i hasve my base step be N=2 so i have a 2 dimensional cube with 2^2=4 vertives in R^2. with coordinates (00) (01) (10)(11)... which is really a two dimensional square with those coordinates as vertices. i can easily see it is a hamiltonian cycle.true.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i can even visualize a 3d cube whose eight vertices in R3whose edges are named: (000)(001)(010)(100)(011)(101)(110)(111) create a cube. which creates a hamiltonian cycyle. dw:1383750838900:dw

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0however when i get to the n and n+1 i can't think of it... i don't know how to PROVE there is a hamiltonian cycle ... unless there is some formula that i can find to guarantee the edge numbers i need....
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