hi, I am studying Josephus problem the recurrence relation for same is
J(1) = 1; J(2n) = 2J(n) - 1, for n >= 1 j(2n + 1) = 2J(n) + 1, for n >= 1 how J(2n) = 2J(n) on R.H.S side how 2 which is called k is derived.. its is given in book that (concrete mathematics) J(2n) = newnumber(J(n)), where newnumber( k) = 2k-1. what is k..? -- You received this message because you are subscribed to the Google Groups "google-codejam" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-code?hl=en.
