>>For eg: if <=4 then seems
>>to be 2*n-3
>>
>>For 4<n<=16 seems to be 2n-4

>For 8, it's 11. How does it fit with your formula?
Hi Alex

I simply cannot figure out how to arrive at 11 calls for 8 people. The best I can do is 12 calls. Neither can I figure out how you got the formula 3N/2-1 for N=2^n. To me the formula seems to be n*2^(n-1).

Anyway, here are a couple of elegant 'proofs' that I got from the sci.math groups. Each prove that 2N-4 is sufficient, but not minimal.

Math Induction:
Step 1 Prove that the solution is valid for some n>=4
Step 2 For N=N+1, have the 'new' person call any one person (1 call)
Step 3 For the N persons excluding the 'new' person, share the secret as
before. All N people now know the 'new' persons secret. (2N-4 calls) per
formula
Step 4 One from this N calls up the 'new' person (1 call)
Step 5 Total for N+1 = 1+2N-4+1 = 2N-2 = 2(N+1) -4. Hence valid for N=>
valid for N+1

Or intuitive one ...
Number the people from 1...n. Have person 1 call all people greater than 4.
(N-4) calls.
Now 1..4 share the secret in 4 calls (4 calls)
Now the person 1 calls all people greater than 4 again. (N-4 calls)
Total = N-4+4+N-4 = 2N-4.

Still no solution that it is indeed least number of calls. And no formulation in sight for my added twist.

Manoj.