Elegant Solution for the Snooker ball Problem

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In summary, the conversation discusses a problem involving a set of snooker balls, with one ball having a non-standard weight. The goal is to use a scale balance to find the non-standard ball with the minimum number of trials. Propositions and their solutions are presented, including a formula and proof for identifying a non-standard ball of known relative weight and a solution for identifying a non-standard ball of unknown relative weight. An approach to proofs of solutions is also discussed, involving dividing the set into three subsets and using the concept of "weight type." The conversation also briefly mentions the problem of two non-standard balls with equal weights.
  • #1
gautam89
Snooker balls Problems – An Approach to Elegant Solutions:

Problem Definition: In a set S of snooker balls, all of the same standard weight, there is one ball of non-standard weight.

Using a Scale Balance, find the non-standard ball with minimum number of trials- Provide a formula and a proof.
---------------------------------------------------
Proposition 1. Set S contains One non-standard ball of known relative weight -Identify non-standard :
Solution: For Quantity = 3 (to power n), or less;
Min weighing required = n.
Proposition 2: Set S contains One non-standard ball- of unknown relative weight; find the ball and identify its relative (heavier or lighter) weight ?:
Solution: For Qty = ( 3(power n) + 3(n-1) + 3 (n-2) … 3(2) + 3(1) ) or less;
Min weighing required = n +1.
---------------------------------------------------------
An approach to proofs of solutions:

Definition 1: (Ternary partition)
Divide the set S into three subsets A, B, and C, where
1. Qty(S) = Qt(A)+Qty(B) +Qty(C), and
2. Qty(A) = Qty(B) >= 1/3 x Qty(S), and
3. Qty(C) = Qty(A) - i; where i = 0,1 or 2.

Proof For Proposition 1: Weigh A with B. This will identify which of the subsets A, B, or C has the non-standard (with known relative weight) ball. – Recursively use n-1 weighings to find the non-standard ball (of known relative weight).

Definition 2 : (“Weight type”)
1. A ball is of “ heavier weight type” when it can NOT be lighter ball.
2. A ball is of “ lighter weight type” when it can NOT be heavier ball.
3. A subset of balls is called of a specific “weight type” if all the balls in that subset is of that “weight type”

Proof For proposition 2:
1. Divide the set S into its Ternary subsets A, B, C (Definition 1), and weigh set A & B.
2. If A and B are Not equal, then using the “weight type” concept and Proposition 1, one can “cross-weigh” subsets of A, B and C n times, to find the non-standard ball from A or B, and identify if it is heavier or lighter ball
3. If subsets A and B are equal then the non-standard ball is in C, and again one can use “cross weighings” and Proposition 1, n times to find and identify the relative weight of the non-standard ball in subset C.

What about problem when there are TWO non-standard balls (of equal weights)...

Gautam Pandya
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  • #2
gautam89 said:
Snooker balls Problems – An Approach to Elegant Solutions:

Problem Definition: In a set S of snooker balls, all of the same standard weight, there is one ball of non-standard weight.

Using a Scale Balance, find the non-standard ball with minimum number of trials- Provide a formula and a proof.
---------------------------------------------------
Proposition 1. Set S contains One non-standard ball of known relative weight -Identify non-standard :
Solution: For Quantity = 3 (to power n), or less;
Min weighing required = n.
Proposition 2: Set S contains One non-standard ball- of unknown relative weight; find the ball and identify its relative (heavier or lighter) weight ?:
Solution: For Qty = ( 3(power n) + 3(n-1) + 3 (n-2) … 3(2) + 3(1) ) or less;
Min weighing required = n +1.
---------------------------------------------------------
An approach to proofs of solutions:

Definition 1: (Ternary partition)
Divide the set S into three subsets A, B, and C, where
1. Qty(S) = Qt(A)+Qty(B) +Qty(C), and
2. Qty(A) = Qty(B) >= 1/3 x Qty(S), and
3. Qty(C) = Qty(A) - i; where i = 0,1 or 2.

Proof For Proposition 1: Weigh A with B. This will identify which of the subsets A, B, or C has the non-standard (with known relative weight) ball. – Recursively use n-1 weighings to find the non-standard ball (of known relative weight).

Definition 2 : (“Weight type”)
1. A ball is of “ heavier weight type” when it can NOT be lighter ball.
2. A ball is of “ lighter weight type” when it can NOT be heavier ball.
3. A subset of balls is called of a specific “weight type” if all the balls in that subset is of that “weight type”

Proof For proposition 2:
1. Divide the set S into its Ternary subsets A, B, C (Definition 1), and weigh set A & B.
2. If A and B are Not equal, then using the “weight type” concept and Proposition 1, one can “cross-weigh” subsets of A, B and C n times, to find the non-standard ball from A or B, and identify if it is heavier or lighter ball
3. If subsets A and B are equal then the non-standard ball is in C, and again one can use “cross weighings” and Proposition 1, n times to find and identify the relative weight of the non-standard ball in subset C.

What about problem when there are TWO non-standard balls (of equal weights)...

Gautam Pandya
---------------------
Notes on Elegant Proof of Snooker Ball problem
Note 1: Exceptions to the Ternary partitions:
For Qty(S) = 7, Ternary partition is 7 = 2 + 2 + 3
For Qty(S) = 10. Ternary partistion is 10 = 3 + 3 + 4

Note 2: let p = MIN ((3(power n-1), Qty(C)). Cross weighing p balls from sets A, B and C, and using Proposition 1, One can easily prove proposition 2.
Gautam Pandya
 
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  • #3
gautam89 said:
Notes on Elegant Proof of Snooker Ball problem
Note 1: Exceptions to the Ternary partitions:
For Qty(S) = 7, Ternary partition is 7 = 2 + 2 + 3
For Qty(S) = 10. Ternary partistion is 10 = 3 + 3 + 4

Note 2: let p = MIN ((3(power n-1), Qty(C)).

Note 3: If A and B are of different weights, put aside p number of balls from the higher weight set, transfer p number of balls from lower weight to higher weight set, and transfer p balls from the set C (of standard weights) to the lower weight set.

Gautam Pandya

Note 4: Weigh the modified sets A and B. If they weigh opposite to their earlier weight types then the non-standard ball must be one of the p balls transferred from the lower weight type. If modified sets are equal then the non-standard ball is one of the p balls put aside from the higher weight type. In these two cases, Recurssively one can find the non-standard ball and identify its weight type.

Note 5: If modified sets weigh the same way as the original sets A and B, then non-standard ball is amongs the "remaining" balls in A and B. In this case, take
p = 3(power n-2), and repeat steps in Notes 3 and 4 above, and recurssively prove the proposition 2.
Note 6: If A = B the repeat steps then A and B has all standard balls and repeat steps in Note 3 and 4 above for Set C.
 

Related to Elegant Solution for the Snooker ball Problem

What is the Snooker ball Problem?

The Snooker ball Problem, also known as the Packing Problem, is a mathematical problem that involves finding the most efficient way to pack a certain number of identical spheres into a container.

Why is the Snooker ball Problem important?

The Snooker ball Problem has practical applications in fields such as physics, chemistry, and engineering. It can also help improve our understanding of how objects pack together in real-world scenarios.

What is an "Elegant Solution" for the Snooker ball Problem?

An Elegant Solution for the Snooker ball Problem is a solution that minimizes the wasted space between the packed spheres, resulting in a more efficient and visually pleasing arrangement.

How is the Snooker ball Problem solved?

The Snooker ball Problem is typically solved using mathematical and computer algorithms. These algorithms use techniques such as sphere packing, optimization, and Monte Carlo simulations to find the most efficient packing arrangement.

What are the potential limitations of an Elegant Solution for the Snooker ball Problem?

An Elegant Solution for the Snooker ball Problem may not always be possible for certain arrangements or numbers of spheres. Additionally, the solution may not take into account real-world constraints or variations in the size or shape of the spheres.

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