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16/03/2014, 19:20
... child, in which case we have (wi+1,1 , wi+1,2 ) = (ui,1 , ui,2 ), xi+1,1 = xi,1 , xi+1,2 = xi,2 , and yi+1 = vi,1 yvi,2 Otherwise, i+ 1 will be the right child of i , and we have (wi+1,1 , wi+1,2 ... b15 a5 In this section, we prove that the string z has no 2decomposition.7 It helps to visualize strings in MIX as closed curves in a plane If w is a string in MIX, by plotting the coordinates ... c17 is a substring of ui,1 Since (ui,1 , ui,2 ) is long, (wi+1,1 , wi+1,2 ) = (ui,1 , ui,2 ) Therefore, the condition (†) holds with i + in place of i Case ui,1 does not contain c Then (ui,1...