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Unformatted text preview: CSE 21 Winter 2008 Final Exam Solutions Roy Liu University of California at San Diego 1 a By guessing that g ( n ) = αr n 1 + βr n 2 , we get 0 = r 2 5 r + 6 r ∈ { 2 , 3 } . Applying initial conditions, we get that g (0) = 0 = α + β g (1) = 1 = α · 2 + β · 3 α = 1 β = 1 g ( n ) = 2 n + 3 n . b By guessing that g ( n ) = αr n + βnr n , we get 0 = r 2 4 r + 4 r = 2 . Applying initial conditions, we get that g (0) = 0 = α g (1) = 1 = α · 2 + β · 2 α = 0 β = 1 / 2 g ( n ) = n 2 n 1 . 2 Let D be the random variable counting the number of defective light bulbs chosen. We have E [ D ] = 3 summationdisplay i =0 i * Pr [ D = i ] = 0 * parenleftbigg 4 parenrightbiggparenleftbigg 6 3 parenrightbiggslashbiggparenleftbigg 10 3 parenrightbigg + 1 * parenleftbigg 4 1 parenrightbiggparenleftbigg 6 2 parenrightbiggslashbiggparenleftbigg 10 3 parenrightbigg + 2 * parenleftbigg 4 2 parenrightbiggparenleftbigg 6 1 parenrightbiggslashbiggparenleftbigg 10 3 parenrightbigg + 3 * parenleftbigg 4 3 parenrightbiggparenleftbigg 6 parenrightbiggslashbiggparenleftbigg 10 3 parenrightbigg ....
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 Spring '08
 Graham
 Graph Theory, Kruskal's algorithm, Prim's algorithm

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