# Six Musical Friends

Six musicians are participating in a music festival. During each concert, some of the musicians perform, while the others listen in the audience.

What is the fewest number of concerts needed so every musician has listened (as an audience member) to every other musician perform?

Source: Aust MS Gazette, Volume 42, No. 2, in the Puzzle Corner section. See here.

• I suppose that playing in a very (VERY) large room, then sitting down and listening to the echo of the concert would be cheating? Jul 11, 2015 at 3:59

I believe I have it

Four concerts!

 Concert format: --- are playing for xxx
--- xxx
Concert |Done performing |Done listening
abc def |                |
cde abf |C               |          F
aef bdc |AE              |         DB
bdf cae |BDF             |        CAE

 My newest reasoning:
In a 1 performing for 5, 5 relationships are formed, X is heard by abcdef
In a 3 performing for 3, 9 relationships are formed
a is heard by def | b is heard by def | c is heard by def
so 3x3 is more efficient achieving the number of relatioships needed

So there are 6 people that need to be heard by 5 people, this is 30 relationships
To do this in 3 turns we would need to generate 10 relationships per concert.
Looking at the concert options there are only 0x6, 1x5, 2x4, 3x3 and symmetric concerts
which are the same production of relationships.
These produce, 0, 5, 8, and 9 respectively.
None of these are greater than 10 (Last I checked)
However, very interestingly this means that it is potentially possible that
2x4 could produce a solution, however this does not prove it possible as there could
be duplicate relationships forced.
But it does prove 4 concerts in the minimum.


Thanks to mmking for format help

• Working on formatting Jul 10, 2015 at 18:09
• Who's playing, and who's listening? Jul 10, 2015 at 18:11
• Fun fact, it doesnt matter, the activities are Identical, but I intended left performing, right listening Jul 10, 2015 at 18:11
• This HAS to be optimal. Unless we count recording all 6 of them in one performance, and then letting all 6 watch that video, reducing it to 1 performance!
– JLee
Jul 10, 2015 at 18:16
• Very nice! If you prove your solution is optimal, I will accept your answer. Jul 10, 2015 at 18:21

Unless I'm missing something, the answer is

4

abc defade bcffcd abefbe acd

• Alternatively, one listens and the rest play Jul 10, 2015 at 17:54

Performing Audience

P    A
abcd ef
cdef ab
abef cd

fbd eac
eac fbd


5 performances but I think it can done in less.

• Nice! Is this optimal?
– JLee
Jul 10, 2015 at 17:59
• Not sure haven't looked for c=4
– Bob
Jul 10, 2015 at 18:00
• Yeah, If I was a faster poster I would have had something similar to this I was think 2 v 4 and then you just need to get each to play for their partner, I had not yet stumbled upon the 3v3 idea. Well done Jul 10, 2015 at 18:00

Solutions for four have been posted.

Three is impossible, because each musician must either:

• play alone (and have all 5 listen)
• play at least twice (if they are not playing alone, they must play for musician(s) playing with them the first time).

They must also either

• be in the audience alone (and listen to all five)
• be in the audience at least twice (they must listen to the person who was in the audience with them the first time)

For everyone to listen twice and play twice, 24 'slots' are needed, but only 18 are available.

If one plays alone, then in the remaining two concerts, the 5 others must all play twice, which means they cannot listen again.

If one listens alone, then in the remaining two concerts, the 5 others must all listen twice, which means they cannot play again.

Therefore it is impossible with three concerts.

There is no solution for 4 concerts other than for them all to be 3 vs 3, because:

Nobody can listen to anybody more than twice or play for anybody more than twice, there are only 24 slots, and we've already established 24 are required.

So suppose ABCD played to EF.

A would still need to play to B and C and D, and EF cannot listen to A again, so AEF to BCD is the next concert.

B also needs to play to ACD. Now E and F have both played to C and heard C, so cannot do either again without going over 24 spaces.

So no 4 vs 2 concerts.

It does not matter in fact whether E was listening in the first concert or playing, so the 5 vs 1 concert is out as well.

OK, I will go ahead and kick things off with

6, as follows.

I'm still trying other configurations to see if this is optimal.