12 replies to this topic

[bonanova]

I think you've collectively touched all the bases.

 

Spoiler for your combined summary, if I may

An empty slot acts like a slot filled with samples of lesser weight.

That the lesser weight is zero does not matter.

Balance then follows from symmetry:

 

Unbalanced (not symmetric)

One filled = one empty 1, 11

 

Balanced (symmetric)
All filled = all empty 0, 12
Two filled (180^o) = two empty 2, 10
Three filled (120^o) = three empty 3, 9
Four filled (90^o) = four empty 4, 8
Six filled (60^o) = six empty 6

Then the Aha! descended: B + B = B.

Specifically, the two and three cases can be merged.

Thus,

 

Balanced (B+B)
Five filled [e.g. 12:00, 3:00, 4:00, 8:00, 9:00] = five empty. 5, 7

 

Nice!

[CaptainEd]

And Nice Puzzle to you, again and forever!

[araver]

One final take on the "special" cases missed in the first solve and TSLF's correct scenarios.

Spoiler for

5 is  balanced on (12,3,4,8,9) too.add 2 opposing, 7 is balanced  on (12,3,4,8,9,1,7) or (12,3,4,8,9,11,5), I get what you mean,thanks

TSLF, I'm glad you responded about the 5 and 7. It made me realize that I didn't fully understand my naive belief that "complementation" should preserve balance. As your example points out, the solution for 7 is NOT the complement of a solution for 5. So I wonder whether, for a larger centrifuge, of size N, the presence of a solution for M samples guarantees the existence of a solution for N-M samples.

Spoiler for but how to prove that it decomposes into periodic atomic patterns?

TBD
In the example at hand, the 7 pattern decomposes into (12,4,8), (1,7), and (3,9). But that bears no obvious relationship to the complement of 5 patterns like  (12,4,8),(1,7) or (12,4,8),(3,9)

Spoiler for so...

images are clearer.Reusing past established "facts":

• Fact: All p-balanced scenarios have a complement of (n-p)-balanced scenarios
• Corollary: All 5=balanced scenarios ARE complements of 7-balanced scenarios (and viceversa).
• Fact: All p-balanced scenarios where p is not prime and gcd(p,n)=1 can (must) be decomposed into balanced non-overlapping groups.
• Fact: All p-balanced scenarios where p is prime are composed of equally distant samples, are atomic, and gcd(p,n)=p.
• Corollary: All p-balanced scenarios where p is not prime and gcd(p,n)=1 can (must) be decomposed into balanced non-overlapping atomic groups.

TSLF's solutions for n=12 and p=5 / 7 were:
*5 is  balanced on (12,3,4,8,9)
*7 is balanced  on (12,3,4,8,9,1,7) or (12,3,4,8,9,11,5),

All the above balanced scenarios can be decomposed into atomic groups (5=2+3),(7=2+2+3). And all have a complement scenario which is also balanced and can be split into atomic groups.
TSLF's solutions and complements below:

And for the more general case:

Spoiler for Not sure,but

I dare conjecture based on the above that for any n which is a multiple of 6 (or equivalently have 2 and 3 as divisors), all cases except for 1 and n-1 samples can be decomposed into atomic non-overlapping groups of 2 or 3 samples. This seems intuitive because any number between 2 and n-2 can be written as a sum of 2s and/or 3s. So F(n)=n-2 for gcd(6,n)=6.

For a more general n, it is still tricky and I can't pinpoint to a way of ensuring non-overlapping. F(N)<=n-2 for sure.