Solving the Min Phone Calls Needed to Share Clues Problem

  • Thread starter Swimmingly!
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In summary, the conversation discusses the minimum number of phone calls needed for n detectives to obtain all n clues. An algorithm is proposed using pairs of detectives and correcting for odd numbers. A list of values for P(n) is given, with a suggestion to look at powers of 2 for a more efficient algorithm.
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Swimmingly!
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Homework Statement


We have n detectives. At the start each has a single unique clue.
The aim is for all the n detectives to obtain all the n clues.
If detective A knows clue 1 and 33; and B knows, 1, 2 and 4. If A phone calls B they share their clues. i.e. A will know 1, 2, 4 and 33 and B the same.

Q: Whats the minimum number of phone calls for all detectives to know all clues?

Homework Equations


P(n)= minimum number of phone calls for n detectives.

The Attempt at a Solution


I represent this with a polyhedra with n vertices, and with links representing calls.
A simple strategy is for A to call (n-1) people, then go in reverse order and call (n-2) people.
P(n)≤ (n-1)+(n-2) =2n-3

For a value under I use a bit of a guess of something that grows fast. Like this. So if we have 8 we get to 2 knowing all with 1+2+4 calls. 1+2+4=8-1 Therefore P(n)>n-1. (n≠2)

How to solve:
Find an algorithm for round numbers (2^m or even 2m). Introduce corrections for non round numbers.
Prove the algorithm is the best with symmetry arguments mainly.
This works I think but it looks a bit extensive.

One algorithm I use is basically:
Divide the people into n/2 pairs, with the left L and the right R person of pair p.
Link L of pair p with R of pair p+1.
Link L of p with R of p+2.
Repeat: Link L of p with R of p+k. Repeat until all know all.

For odd numbers an easy fix that's not necessarily the optimal is this: Take A away to make it an even group. Link A with a random element. Run even algorithm. Link A again with a random element.

A list of values P(n) for n, some may be wrong:
n____2_3_4_5_6_8__10
P(n)._1_3_4_5_9_12_16
 
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  • #2


Swimmingly! said:
A list of values P(n) for n, some may be wrong:
n 2 3 4 5 6 7 8 9 10
P(n) 1 3 4 5 9 12 16
I couldn't figure out that list until I 'quoted' it. That revealed gaps at 7 and 9.
The suggestion to look at powers of 2 is a good one. That reveals an alogorithm that takes n log2(n) / 2 if n is a power of 2, though I haven't proved it's optimal. Fits your P(8) = 12.
 

What is the "Min Phone Calls Needed to Share Clues Problem"?

The "Min Phone Calls Needed to Share Clues Problem" is a mathematical problem that involves determining the minimum number of phone calls needed for a group of people to share all their clues with each other.

Why is this problem important?

This problem is important because it has real-world applications in various fields such as computer science, communication networks, and social interactions. It helps us understand the complexity of information sharing and communication within a group.

What are the key steps in solving this problem?

The key steps in solving this problem are identifying the number of people in the group, determining the total number of clues that need to be shared, creating a graph to represent the connections between people and their clues, and then applying graph theory algorithms to find the minimum number of phone calls needed.

What are the challenges in solving this problem?

One of the main challenges in solving this problem is dealing with large groups of people and a high number of clues, which can result in a complex and time-consuming solution. Additionally, the problem becomes more difficult when there are restrictions on who can communicate with whom.

How can this problem be applied in real life?

This problem can be applied in various scenarios such as designing efficient communication networks, optimizing information sharing in a team or organization, or even planning social events where everyone needs to be informed about all the details. It can also be used in algorithm design and analysis in computer science.

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