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 -~----------~----~----~----~------~----~------~--~---
