Puzzles with numbers and things

[snookersfun]

My orang-utan has been scratching his head, something he likes to do a lot, but is unfortunately still not convinced.

I think the first part of your answer showed that there must be A set of at least 46 round robin matches where the socks thrown into the audience are all of the same colour. Is that right?

In the 2nd part, are you arguing here that if we take that set of at least 46 matches (which could, of course, be ANY combination of different matches), my orang utan must be able to find among them a set of 3 matches such that only 3 players are involved?

yes and yes? But obviously the APE didn't like my arguing... be back in a sec., this computer just got repossesed by a child...

[Vidas]

Snookersfun, - you work so quickly!

i tried slightly different way...
ok, 17 players.
X plays 16 matches.
he plays at least 6 matches(16/3>5)
wearing socks of the same color(green).
X-A X-B X-C X-D X-E X-F
if any two of A...F will play each other in green socks then ok.
If not, they all (A..F) should play each other in pink/ginger socks.
and we need to find (among those 6 people) group of three players all playing each other either in pink or in ginger socks. ...and this part of the problem is similar to problem #10-2 (with smacks and pots of jam).

[davis_greatest]

Good answer

I can't fault much of that, Vidas! Well done

You missed out the tricky bit though - showing that 70p + 3 x 10p = £1, so we have enough money to go to the matches.

I might let you off and give you the point - please fill in the missing details though to show what happens when down to 6 players and 2 colours of socks...

[snookersfun]

sounds good Vidas!

[davis_greatest]

sounds good Vidas!

I'll think about the scoring... I think the whole point will be going to Vidas though. However, snookersfun, your answer has inspired the next question. I shall wait till this evening though before posting it...

[snookersfun]

that's more than OK with me!
Vidas' answer was quite more elegant and complete

[Vidas]

I might let you off and give you the point - please fill in the missing details though to show what happens when down to 6 players and 2 colours of socks...

Ok...
6 players (A-F) have to wear only pink or ginger socks.
A plays 5 people B,C,D,E,F.
5/2 > 2, so 3 or more players have to wear same color of socks during matches with A.(for example A-B, A-C, A-D matches).
Matches B-C, B-D, C-D: if all have to wear the other color - you should go to these three matches...
if not - at some match they wear the same color as (A-B, A-C, A-D) - and we get group of three again (them and A).

[davis_greatest]

Congratulations, Vidas!

That completes the proof! Described almost as elegantly as my orang-utan explained it. Well done!

SO HERE IS THE SCOREBOARD AFTER ROUND 16

snookersfun……………………….…..7
robert602…………………………………4
abextra……………………………..…...3½
Vidas……………………………………….2½

(scoreboard adds up to number of rounds+1 since 2 members each got a point for round 14)

Next question to follow in around 3 hours.

[davis_greatest]

Round 17 - Orang-utan goes ape!

In round 16, snookersfun wanted to prove that with 17 players playing each other in round robin matches, if we take ANY 46 of those matches it would ALWAYS be possible to find a set of 3 matches involving a total of only 3 players.

My orang-utan did not believe that snookersfun would ever succeed in proving this - the reason being, my orang-utan believes, that it is NOT always possible. In fact, just after rating this thread by clicking on the orange box to the right of "Rating" near the top of this screen, my orang-utan has found that, with these 17 players, he CAN choose 46 round robin matches in which there exists NO set of 3 matches involving only 3 players.

Here is an example, supposing that we chose only 5 matches, involving the 17 players (A to Q). Suppose we choose:

A v B
A v C
B v D
C v D
E v Q

From this set of 5 matches, it is impossible to find 3 matches involving only 3 players. However, this is only 5 matches. I mentioned above that my orang-utan can find 46 matches from which it is impossible to choose 3 matches involving only 3 players. Not only this, but being so clever, he has found the greatest possible number of matches from which it is impossible to choose 3 matches involving only 3 players.

How many matches has he found, and how did he do this?

PS Being rather busy, my orang-utan has not yet proved that the number he has found is the maximum possible. However, you will be awarded the point if you can find a number, at least as big as my orang-utan's number, and describe how you will choose this number of matches to make it impossible to choose 3 matches from these that involve only 3 players. Clear?

[snookersfun]

lol, I'll sit this one out

[Vidas]

Er.. 72 matches?

[Obligation]

I'll plump with 142.

[davis_greatest]

Let's hear Vidas's go with 72 and Obligation's go with 142...

On second thoughts, given that there are only 136 round robin matches altogether (16+15+14+...+3+2+1) = 17 x 16/2, I don't think that Obligation is going to be able to come up with 142

So, Vidas, let's hear you explain how to choose 72 matches - with your proof that there will not be any group of 3 matches from these that involve only 3 players in total.

[snookersfun]

I got to those 72 as well, but small concern here, are we now on the way to proove, that you could NOT have found a group of three matches, such that only three players are in it (because somehow, I got the feeling, that the remaining group, also doesn't have any)?

[Vidas]

It depends on how to pick 72 matches. Lets try...

A B.. Q
1 2.. 17

Players with odd numbers meet players with even, and vice versa.

1 plays 2, 4, ... 16
2 plays 3, 5, ..15, 17
...
15 - 16
16 - 17.
Total: 8+8+7+7+6+6+5+5+4+4+3+3+2+2+1+1 = 72
It can be nicely represented by tournament table - it looks like chessboard.
If we pick any three players, those who are both odd or both even,
won't meet in those 72 matches.