I agree with you. If we remove 1 from the initial status, the situation seems rather complex. However, it's a very interesting problem.
On Oct 2, 12:41 pm, nikhil <nikhilgar...@gmail.com> wrote: > Good job !! > just to add , i think the exact winning strategy would be based on > distribution of factors of numbers and that would depend on > distribution of primes... > So it occurs to me that there might not be a "closed form" winning > strategy possible ! > > Still thanks for the proof ! > > cheers > > - > nikhil > Every single person has a slim shady lurking. > > On Oct 2, 7:08 am, saltycookie <saltycoo...@gmail.com> wrote: > > > Here is a proof. Unfortunately, the proof is not constructive.The > > secret of winning is "1", which is a fator of every integer. > > > If the first player(player A) can win by removing a number between 2 > > to n, then our hypothesis holds. Or else, A can't win by removing any > > number between 2 to n. We denote the situation after removing number i > > from [1, n] by S(n, i), then for i = 2...n, S(n, i) is a winning > > situation. A can then remove number 1 at the first step. No matter > > what B removes in the next step, he will leave a situation S(n, i)(i > > is the number B removes), which is a winning situation for the next > > player(A). > > > On 10月1日, 上午2时53分, nikhil <nikhilgar...@gmail.com> wrote: > > > > we have all the numbers written from 1- n. 2 players play > > > alternatively. At any turn , a player removes a number and along with > > > all its divisors present in the list. Player to remove last number > > > wins. > > > > so given initial number n and player who is starting first , we are to > > > find who wins if both play optimum. > > > > NOW , i have found that the the player who starts ALWAYS wins. Can > > > anyone prove this or still better come up with a real strategy ! > > > > cheers > > > - > > > nikhil > > > Every single person has a slim shady lurking ! --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
