On Oct 2, 7:20 am, eSKay <[email protected]> wrote:
> What exactly do you prove here?
>
The first player always has a winning move.
> You just make some statements, which should be proved. shouldn't it?
>
> Or am I missing something??
>
It's a 2-player game that's deterministic, zero-sum,
perfect information, finite, and without ties. So a
winning strategy exists for one of the players.
The proof doesn't indicate what the winning play is
for any given N, just that a winning play always exists
for the first player.
--
Geoff
> On Oct 2, 7:08 am, saltycookie <[email protected]> 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 <[email protected]> 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 !- Hide quoted text -
>
> - Show quoted text -
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