It is possible to do that with only O(n) invications, i answered it already
to this group...
2009/4/1 DragonZsnakE
> ( n cards are there initially )
>
> we pick out the first card in random it takes n-1 comparisons to figure out
> its Equivalence card -n-1 steps
> Two cards have been elimin
( n cards are there initially )
we pick out the first card in random it takes n-1 comparisons to figure out
its Equivalence card -n-1 steps
Two cards have been eliminated ( this leaves us with 2 and n-2 cards)
we pick out the 2nd card in random it takes n-3 comparisons to figure out
its Equiv
