Originally Posted by
TKHunny I would put it on an exam in a methematics class. I suppose I could be different.
Your formula is interesting. How do you propose to prove it?
This doesn't quite do it...
Four, for example:
A<==>B ==> A(b) and B(a)
B(a)<==>C ==> B(ac) and C(ab)
C(ab)<==>D ==> C(abd) and D(abc)
D(abc)<==>A(b) ==> D(abc) and A(bcd)
A(bcd)<==>B(a) ==> A(bcd) and B(abc)
It's a little irritating that A has to call B twice, but this may establish an absolute maximum number of necessary calls, one greater than the number of participants.
A<==>B ==> A(b) and B(a)
C<==>D ==> C(d) and D(c)
D(c)<==>A(b) ==> A(bcd) and D(abc)
B(a)<==>C(d) ==> B(acd) and D(abc)
One Shorter!
Maybe five participants.
A<==>B ==> A(b) and B(a)
B(a)<==>C ==> B(ac) and C(ab)
C(ab)<==>D ==> C(abd) and D(abc)
D(abc)<==>E ==> D(abce) and E(abcd)
E(abcd)<==>A(b) ==> A(bcde) and E(abcd)
A(bcde)<==>B(ac) ==> A(bcde) and B(acde)
B(acde)<==>C(abd) ==> B(acde) and C(abde)
Okay, now I have to revise the absolute maximum. I'm going with (n-1)+(n-2) = 2n-3. I could prove that.
Fine, I think we have an absolute maximum. If you can ALWAYS do it with one fewer, maybe you have something. Are you sure you cannot EVER do it in two fewer? This is mathematics. You must PROVE it.