[Guest]

problem 1) hard.

27 people want to play in groups of three, where everyone plays each day, and no one is paired with the same person more than once. if possible, this should take 13 days.

problem 2) insane.

after you find the thirteen days, construct triangles with each number on a point of the triangle for each pairing. with these triangles, form hexagons such that the central point is all the same number, and each outer adjacent number differs by 3. how many such hexagons can you construct?

example:

    _________

   /\8    12/\

  /11\     /9 \

 /    \   /    \

/15___1\1/1____7\

\18   1/1\1    4/

 \    /   \    /

  \ 3/     \5 /

   \/6_____2\/

yes, I know I'm evil.

Edited September 27, 2010 by phillip1882

[Guest]

problem 1) hard.

27 people want to play in groups of three, where everyone plays each day, and no one is paired with the same person more than once. if possible, this should take 13 days.

problem 2) insane.

after you find the thirteen days, construct triangles with each number on a point of the triangle for each pairing. with these triangles, form hexagons such that the central point is all the same number, and each outer adjacent number differs by 3. how many such hexagons can you construct?

example:

    _________

   /\8    12/\

  /11\     /9 \

 /    \   /    \

/15___1\1/1____7\

\18   1/1\1    4/

 \    /   \    /

  \ 3/     \5 /

   \/6_____2\/

yes, I know I'm evil.

cool puzzle, I will get on it as soon as I get home.

[Guest]

If they're groups of 3 - no-one is PAIRED

[bonanova]

If they're groups of 3 - no-one is PAIRED

Each person plays in one threesome each of the 13 days,

playing with all 26 other players, exactly once, without repeats.

[bonanova]

Here's the first dent in this hideous problem.

Name the players A-Z and /.

There are 101 threesomes that satisfy the no-pair repeated rule.

Three players, in this case they are A, B and C, play 13 times.

The others all play 11 times.

Two tasks remain:

1. Schedule 9 or fewer threesomes per day, over 13 days, so that no player plays twice in a day.
2. These are the triangles. Start making hexagons.
   .

But first, some sleep.

Re-reading the OP,

Everyone plays each day?

That would require 119 = 13x9 threesomes.

[Guest]

that would require 13x9 = 117 threesomes.

yes. there are 3 C 27 = 2925 possible threesomes.

[bonanova]

that would require 13x9 = 117 threesomes.

yes. there are 3 C 27 = 2925 possible threesomes.

Not that many that satisfy the OP:

Here are two threesomes: ABC and ABD. Both pair A with B.

" ... and no one is paired with the same person more than once."

[Guest]

bonanova here's the problem with that logic.

you can do 16 people in groups of 4 over 5 days, each group unique.

but under your constuction you would have...

AAAA BBB CCC DDD

BEHK EFG EFG EFG

CFIL HIJ IJH JHI

DGJM KLM MKL LMK

only 13 unique pairs.

Edited September 28, 2010 by phillip1882

[bonanova]

bonanova here's the problem with that logic.

you can do 16 people in groups of 4 over 5 days, each group unique.

but under your constuction you would have...

AAAA BBB CCC DDD

BEHK EFG EFG EFG

CFIL HIJ IJH JHI

DGJM KLM MKL LMK

[/code]

only 13 unique pairs.


If you credit me with using logic, it might be more than I deserve. 


The OP asks for [color=#000000]unique [color=#ff0000]pairings[/color][/color]. The word 'pair' means 'two'. 


Consider poker hands:

QQQ is not a pair of Queens. AA is a pair of Aces.


To requir[color=#000000]e [/color][color=#ff0000][color=#000000]unique [/color]threesomes[/color], you might say:

no one[color=#000000] is [/color][color=#000000]grouped with the same [/color][color=#ff0000][color=#000000]person[/color]s[/color] more than once.


In golfing parlance, ABCD is a [color=#ff0000]foursome[/color].

[Guest]

fine, fine unique groupings. you know what i mean, that's what's important.

[Guest]

1. Schedule 9 or fewer threesomes per day, over 13 days, so that no player plays twice in a day.

Here you go: http://img571.imageshack.us/img571/8769/golfthingy.jpg

Edited September 28, 2010 by MaybeMyNameIsStine

[Guest]

cngrats on solving part one.

now on to part 2. anyone dares?

[superprismatic]

cngrats on solving part one.

now on to part 2. anyone dares?

I don't understand part 2 at all. Please rephrase it.

It doesn't seem to have anything to do with part 1.

Also, it seems that a countably infinite number of

such hexagons can be made from your example just by

adding any integer to every number there. The properties

you mentioned (all the same in the center; outer adjacent

numbers differ by 3) are preserved by doing this.

I've been trying for quite a while to make some sense

of it, without success. I suspect that I would love the

problem if I could only understand it.

[Guest]

In my solution, i used the letters from A-Z and /, because that's what bonanova did, and bonanova's smart. But to solve part 2, you need to substitute the letters with numbers 1-28. Then you can make a lot of triangles (each threesome makes a triangle) and solve part 2.

...or, that's how I understood it

[Guest]

Bonus question: How many of the groups form a three-letter word at my schedule

(if you solve this, you'll win a free walking trip to Middelfart in Denmark. If you happen to live somewhere where an ocean will get in the way, you should be even more eager to win, because then you'd have a nice opportunity to test your lungs! I'm very gracious today)

EDIT: spelling

Edited September 28, 2010 by MaybeMyNameIsStine

[bonanova]

I don't understand part 2 at all. Please rephrase it.

It doesn't seem to have anything to do with part 1.

Also, it seems that a countably infinite number of

such hexagons can be made from your example just by

adding any integer to every number there. The properties

you mentioned (all the same in the center; outer adjacent

numbers differ by 3) are preserved by doing this.

I've been trying for quite a while to make some sense

of it, without success. I suspect that I would love the

problem if I could only understand it.

SP, I'm kind of with you on this puzzle.

But I don't think you can use just any combination of numbers [or letters] at the corners.

[Perhaps only] the 117 'pairings' that solved the puzzle.

But take that only as a guess.

[superprismatic]

SP, I'm kind of with you on this puzzle.

But I don't think you can use just any combination of numbers [or letters] at the corners.

[Perhaps only] the 117 'pairings' that solved the puzzle.

But take that only as a guess.

Bonanova,

The many mappings to numbers make things pretty large. I'm a bit puzzled as to

why Phillip1882 didn't frame the second part with nonagons instead of hexagons.

But, it's his puzzle, so I can't complain!

Edit: removed dumb statement.

Edited September 28, 2010 by superprismatic

[Guest]

right the 117 pairings that solved the puzzle. i did call the puzzle insane for a reason.

but note that each number is only used 13 times. therefore you really only have 13 C 6 = 1716 possibilities for a single number, you just need to multiply however many you get by 13 to get a rough estimate of how many total you have.

[Guest]

The part 1 can be solve by the expression:

AlluniqueCombination = 27 c 3 - 36x1 - 36x2 ...36xn

until alluniqueCombination > 0, this happen whit n = 13, 13 days

27c3 -36x1-36x3..36x13 = 177. 36xn its all the combination already used on the previous days.

Edited September 30, 2010 by dreamcrash

[Guest]

The part 1 can be solve by the expression:

We have 27 c 3 of combination , total combination = 27 c 3 = 2925

In each day we have 27 pair combination + 9 threesome combination , so we have 36 different combinations

In day 2 we can't use the 36 combination of day 1, and in day 3 we can use the 72 pervious combination of day 2 and 1,

So to find the expression the all the unique combination we have :

day 1 day 2 day 3 day n

AlluniqueCombination = 27 c 3 - 36x1 - 36x2 ...36xn while alluniqueCombination > 0,

this happen whit n = 13, 13 days

27c3 -36x1-36x3..36x13 = 177.

Edited September 30, 2010 by dreamcrash

[Guest]

Part 2, Not very confidence.

.....__________

.../e\E ......d/D\

../...\......./.....\

./.....\...../.......\

/F__A\A/A___c\

\f....A/A\A....C/

.\....../....\...../

..\..../.......\.b/

...\G/g____B\/

The restrains are :

Only have 27 numbers ,All diferente numbers. b = B +3, c = C+ 3 , ect.

So:

For letter

A : 27 numbers;

B : 24 ( Excluding the number of A, the number A-3 because this made b = A-3 +3 , and the number 27 because b = 27 + 3= 30.

b : 1 (only the number B+3)

C: 21 (Excluding the number of A , A-3, B, b ,B-3, and 27)

D : 18

E : 15

F : 12

G : 9

Solution : (27 x 24 x 21 x 18 x 15 x 12 x 9) ? .excluindo the fact that you can permute the triangules its self.

Edited September 30, 2010 by dreamcrash

[Guest]

dreamcrash, after day 1, you eliminate the following.

assume day 1 is the following.

1 2 3

4 5 6

...

25 26 27

then you eliminate...

1 2 4

1 2 5

1 2 6

...

1 2 27

1 4 3

1 5 3

...

1 27 3

4 2 3

5 2 3

...

27 2 3

4 5 1

...

27 23 24

and of course the original orientations.

for a total of 657 possibilities eliminated.