BrainDen.com - Brain Teasers

## Question

In solving a detective case Y-San went to her lab to use her centrifuge.
It has twelve slots, spaced as the hours on a clock. She knows the
importance of filling the slots in a manner that maintains proper
balance, at its high operational speed.

So, she wonders aloud, just loud enough for you to overhear, just how
many different numbers of samples could she process at once and still
maintain that balance?

She of course deduced the answer as quickly as her words were formed,
But she'd love to hear your thoughts on the matter as well.

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2 samples (12 6)

3 samples (12 4 8)
4 samples (12,3 6 9)
6 samples (12 2 4 6 8 10)
8 samples (all but 12 3 6 9)
9 samples (all but 12 4 8)
10 samples (all but 12,6)
12 samples (all 12)

put water samples all unused slots and get 1, 5, 7, and 11, but I suspect that's not the point of this puzzle

Edited by CaptainEd

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n-phi(n) for n geq 2 if having it spin empty is not considered a valid way. Plus 1 if no samples is considered a case of succesful processing rather than a waste of energy.

For n=12 that is 12-4=8. Unfortunately that is just fitting the data with a formula, not really an answer.

For 12 exactly, only 1, 5,7 and 11 samples can't be balanced.

Assume that q samples are balanced then the sample distribution could be decomposed into groups, each with at least 2 samples such that each group could be processed balancefully on its own. Decomposition may not be unique.

An atomic /not decomposable further) such group would have the distance among the samples constant. So the cardinal of the group would be a divisor of n to ensure the equal distance.

Being an atomic group kinda makes that cardinal prime othwrwise it can be decomposed into equal balanced groups

That being said The integration step is tricky or my algebra is very rusty. The solution - the number of q's between 1 and n exist such that there exists a balance position of q samples on n equally distanced placeholders. The number is clearly smaller or equal to the formula above. However, constructing a sample distribution for each q not relatively prime to n is not easy to describe.

Induction should work for n prime(trivial) and for q being a divisor of n (equally space them. Done)

plus the fact that if gcd(q,n)=p then p is balanced as a divisor of n. If q=p*r with r relatively prime with n by gcd definition then one needs to acknowledge that r groups each of p samples can be placed in r balanced layers each shifted with 1 position clockwise then the layer below and projected without overlapping since the length of consecutive samples is less than n/p the distance between the sanples in each group so shfting by 1 the distance can't be covered unless q=n in which case they still wouldn't manage to overlap/collide, just fit neatly.

All in all a great image is better than the thousand characters above.

Edited by araver

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Araver, I guess that means we don't have it right. So I've got a question for you

a set of samples into two sets of different sizes, each of which is balanced? I don't think that's what you were saying, and I don't feel confident about the physics. But If we superimposed and rotated a 2-sample (12,6) with a 3-sample(1,5,9) to create (12 1 5 6 9), we would naturally get 5 and (via complementation) 7.

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Araver, I guess that means we don't have it right. So I've got a question for you

a set of samples into two sets of different sizes, each of which is balanced? I don't think that's what you were saying, and I don't feel confident about the physics. But If we superimposed and rotated a 2-sample (12,6) with a 3-sample(1,5,9) to create (12 1 5 6 9), we would naturally get 5 and (via complementation) 7.

5 is balanced on (12,3,4,8,9) too.

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Araver, I guess that means we don't have it right. So I've got a question for you

a set of samples into two sets of different sizes, each of which is balanced? I don't think that's what you were saying, and I don't feel confident about the physics. But If we superimposed and rotated a 2-sample (12,6) with a 3-sample(1,5,9) to create (12 1 5 6 9), we would naturally get 5 and (via complementation) 7.

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

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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.

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Araver, I guess that means we don't have it right. So I've got a question for you

a set of samples into two sets of different sizes, each of which is balanced? I don't think that's what you were saying, and I don't feel confident about the physics. But If we superimposed and rotated a 2-sample (12,6) with a 3-sample(1,5,9) to create (12 1 5 6 9), we would naturally get 5 and (via complementation) 7. Writing on the phone, I didn't recheck the sentences I wrote.

Yes, my first impression on composition/decomposition was using (3D) layers of a cake on top of each other:

If all layers/groups of samples are balanced AND one could shift them and project them without overlapping, maybe one would get another balanced set.

That is true for all SAME-groups regardless of overlapping (because of symmetry), but sometime during writing I became convinced that one cannot get both balance and avoid overlapping if groups are different.

Logic miss:

*Composition of balanced SAME groups preserves balance (all n-phi(n) above have easy to construct balanced atomic or SAME groups)

*Composition of balanced DIFFERENT groups doesn't ALWAYS preserve balance even if they don't overlapped(EDITED)

The missing link is the fact that if groups are different, there MAY be a shift such that composition is balanced. EDIT: Or not For 12, there is such a shift for both 5 and 7 according to TSLF. For 1 and 11 there is not (in general 1 and n-1 are never balanced). EDIT for n>=2

In general if p samples can be balanced the negative image of n-p samples are also balanced. EDIT: Less sure after I wrote this. Anyway, that only needs to be true for gcd(n,p)=1, the missing case above. In which case gcd(n,n-p)=1.

The general question still remains ....

And I don't think it's easy to construct these special cases using decomposition ...

Edited by araver

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Assuming all samples have the same mass, and the slots in the centrifuge are spaced evenly, and all the same distance from the hub,
then
* the ALL pattern (all samples are full) has the COG at the hub
* if a pattern P of samples has its COG at the hub, then
* the pattern ALL - P must have its COG at the hub, as the COG of ALL must be the weighted average of the COGs of P and ALL-P.

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)

Edited by CaptainEd

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I think you've collectively touched all the bases.

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 (180o) = two empty 2, 10
Three filled (120o) = three empty 3, 9
Four filled (90o) = four empty 4, 8
Six filled (60o) = 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!

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

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.

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)

And for the more general case:

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.

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: ## Join the conversation

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