Oh! thanks! On May 9, 7:16 pm, Paul Smith <[email protected]> wrote: > Let x = E[n=3]. > > We have an equation of the form x = 2 + x/3. > > Solve for x. > > Paul Smith > > [email protected] > > On Mon, May 9, 2011 at 3:10 PM, Eagle <[email protected]> wrote: > > @Pedro, > > > In your last two steps, > > >> E[n=3] = 2 + 1/3 * E[n=3] > > >> E[n=3] = 3 > > > there is something wrong. How are you eliminating E[n=3] on the right > > hand side of the equation? > > > Eagle > > > On May 9, 12:28 am, Pedro Osório <[email protected]> wrote: > >> 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 > > athttp://groups.google.com/group/google-code?hl=en.
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