Puzzles with numbers and things

[davis_greatest]

Well, the answer is definitely not greater than n. The question is whether it is ever less than n, except when n=2 (when the answer is 1).

Let's think...

[elvaago]

If n is 2 then the answer is 1.
If n is 3 then the answer is 3. A to B, A to C, B to C.
If n is 4, then the answer is 6. A to B, A to C, A to D, B to C, B to D, C to D.
If n is 5, then the answer is 10. A to B, A to C, A to D, A to E, B to C, B to D, B to E, C to D, C to E, D to E.

So if you look at this, I'm going to throw otu the formula:
With n people and n games, the number of required phonecalls is (n-1)+(n-2)+(n-3) etc. I know there's a proper formula for this, but I've forgotten how it goes.

So in the above examples:
n = 2 -> n-1 = 1
n = 3 -> (n-1) + (n-2) = 2 + 1 = 3
n = 4 -> (n-1) + (n-2) + (n-3) = 3 + 2 + 1 = 6
n = 5 -> (n-1) + (n-2) + (n-3) + (n-4) = 4 + 3 + 2 + 1 = 10

Maybe someone smarter can give the exact formula for this?

[davis_greatest]

I had been assuming that someone could not speak to two or more other people at the same time. Is that the case, however, or are people allowed to be on more than one call at once?

[davis_greatest]

If n is 2 then the answer is 1.
If n is 3 then the answer is 3. A to B, A to C, B to C.
If n is 4, then the answer is 6. A to B, A to C, A to D, B to C, B to D, C to D.
If n is 5, then the answer is 10. A to B, A to C, A to D, A to E, B to C, B to D, B to E, C to D, C to E, D to E.

So if you look at this, I'm going to throw otu the formula:
With n people and n games, the number of required phonecalls is (n-1)+(n-2)+(n-3) etc. I know there's a proper formula for this, but I've forgotten how it goes.

So in the above examples:
n = 2 -> n-1 = 1
n = 3 -> (n-1) + (n-2) = 2 + 1 = 3
n = 4 -> (n-1) + (n-2) + (n-3) = 3 + 2 + 1 = 6
n = 5 -> (n-1) + (n-2) + (n-3) + (n-4) = 4 + 3 + 2 + 1 = 10

Maybe someone smarter can give the exact formula for this?

elvaago, you won't need this many calls! Remember that it's not necessary for each pair of people to speak. Once I know the score from your match, I can tell it to robert602 on the phone - you don't need to speak to him on the phone directly.

PS Your formula would be n(n-1)/2. But it is too big.

[elvaago]

I never claimed to be Stephen Hawking. ;-)

maybe I really should let the smart people solve these problems!

[snookersfun]

wouldn't that be like players playing each other n(n-1)/2

[elvaago]

Wait, in the situation of 3 people.
A calls B, A tells B their score, B tells A their score. B calls C, B tells C their score and A's score. C tells B their score. But A still needs to know C's score, so in the case of 3, 3 calls are still required.

With 4 people.
A calls B, A and B are fully updated. B calls C, B and C are fully updated, but A still needs to know C's score. C calls D, then C and D are fully updated. If D then calls A, A knows everything, but B still doesn't know D's score. So B and D still need to talk or B and A need to talk again. So with 4 people I guess you need 5 calls.

Now with 5 people.
A-B, B-C, C-D, D-E, E-A. Then you need B-D, C-E. That's 7.

I'm going to throw in the towel for real now. :-)

[davis_greatest]

Well, the answer is definitely not greater than n. The question is whether it is ever less than n, except when n=2 (when the answer is 1).

Let's think...

I take this back! In fact, the answer is bigger than n (if n>4)! Think I've solved it now but I'll give the others more time (that's also my excuse to let me check it )

[elvaago]

Please Pm me the answer, I won't tell anyone else. :-)

[Robert602]

I take this back! In fact, the answer is bigger than n (if n>4)! Think I've solved it now but I'll give the others more time (that's also my excuse to let me check it )

If you do have the answer, already, then I've just got to say I'm impressed. Nobody in my year at college was able to do this one, and a few of them are pretty tuned in. So, kudos .

You were indeed right to assume 1-to-1 conversations only, the original question was a bit clearer about that before I snookerfied it.

[snookersfun]

is it smaller than 2n-3?

[davis_greatest]

is it smaller than 2n-3?

Yeah, but notalot!

... assuming n>2 (for n=2 it's one phone call)

[davis_greatest]

Yeah, but notalot!

... assuming n>2 (for n=2 it's one phone call)

I meant assuming n>3 (for n=2 it's one call and for n=3 you need 3 calls). Once n>=4 though it's easier to see the pattern.

[snookersfun]

are you sure?

n-1 phone calls will totally inform 2 fans (the two last calling), leaving n-2 fans, which still need to be informed somehow. How can you do that in less than n-2 calls?

[davis_greatest]

As sure as I can be while trying to type an answer while on the phone to work about something else and typing an email on my other computer about something else entirely Definitely if n>=4, it's easy to find a way to spread the scores with 2n-4 calls.
If n=2 though, you need 1 call, and with n=3 you need 3.