Hi Ricardo, For your example regarding 3, you have the following possibilities:
1 2 3 - perfect 1 3 2 - one correct 2 1 3 - one correct 2 3 1 - wrong 3 1 2 - wrong 3 2 1 - one correct As you can see, if goro hits without holding anything, 1/6 of the time it will be sorted, 1/2 of the time there will be 2 out of place and 1/3 of the time there will be 3 out of place. This means that: E[n=3] = 1 + (1/6 * 0 + 1/2 * E[n=2] + 1/3 * E[n=3]) As you have shown, E[n=2] = 2, so: E[n=3] = 2 + 1/3 * E[n=3] E[n=3] = 3 On May 8, 7:57 pm, werneckpaiva <[email protected]> wrote: > I read too and it doesn't make any sense for me. > > I understand that this is a geometric distribution where if P(X)=p so > E(X)=1/p. So, if you have 2 numbers unsorted and Goro hits the table, > there is 0.5% chance to stay in the same position and 0.5% chance to > swap positions. But, if you have 3 unsorted elements, there are 6 > different permutations, so, P(X)=1/6 and the E(X)=6. My solution is > hold 1 number, swap the other 2; hold the sorted element and swap the > other 2 remaining, so 2 + 2 = 4 hits > > 3 1 2 > 1 3 2 > 1 2 3 > > But this doesn't seem to be the correct answer. The developers > solutions say that for 3 unsorted numbers needs only 3 hits. Anyone > knows how to explain that? > > Regards, > > Ricardo > > On May 7, 8:24 pm, SwiftCoder <[email protected]> wrote: > > > > > > > > > I looked at some of the solutions, as per that umber of hits are same > > as the count of numbers which are not at their correct sorted > > position. Is that so? -- 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.
