The problem asks for the minimum number of calls required in a gossip scheme that completes in the minimal time ( T(n) ), where ( T(n) = \lceil \log_2 n \rceil ) for even ( n ) and ( T(n) = \lceil \log_2 n \rceil + 1 ) for odd ( n ).

Key Observations:

1. Minimal Total Calls (Unconstrained Time): The minimal total number of calls to disseminate all information, regardless of time...

Yes. If you insist on completing gossiping in the minimum number of rounds T(n) (i.e., T(n) = ceil(log2 n) for even n, and T(n) = ceil(log2 n)+1 for odd n), then the minimum possible number of calls is

Cmin(n) = ceil( n · ceil(log2 n) / 2 ).

Why this is the right answer:

• Lower bound: In any round, a node can participate in at most one call, and a node’s knowledge can at most double each time ...

Yes.
If one insists on finishing the gossiping process in the minimum possible number T(n) of rounds

  T(n) = ⌈log₂ n⌉ (n even)
  T(n) = ⌈log₂ n⌉ + 1 (n odd)

then the minimum possible number of calls is known and can be written in closed form.

Notation
Let C(n) be the smallest number of calls that allow all n people to know every rumour in exactly T(n) rou...

Minimum Calls in Minimal Time Gossip Problem

The question asks about the minimum number of calls needed in a gossip scheme that completes in the minimum possible time T(n), where:

• T(n) = ⌈log₂ n⌉ for even n
• T(n) = ⌈log₂ n⌉ + 1 for odd n

This is an interesting variation of the gossip problem that involves a trade-off between minimizing calls and minimizing time.

Analysis

First, let's ...

<think> I need to understand the question. The user is asking about the minimum number of calls in a gossip scheme that achieves the minimum time, which is T(n) = ceil(log2 n) for even n and ceil(log2 n) + 1 for odd n.

First, recall the gossip problem: n nodes, each with a unique piece of info, and in each round, pairs call each other and exchange all info they know. The goal is to minimize the n...

Of course. This is an excellent question that gets to the heart of the trade-off between the number of calls and the time (number of rounds) in the gossip problem.

Yes, there are definitive results for the minimum number of calls required to achieve the minimum possible time, . You've correctly identified that minimizing calls and minimizing time are conflicting goals. The call schem...

Let

 C(n , t) := minimum possible number of calls (edges) of a telephone–gossip scheme on n persons that finishes in at most t parallel rounds.
The question is to determine C(n , T(n)) where

 T(n)= ⌈log2 n⌉ if n is even
 T(n)= ⌈log2 n⌉+1 if n is odd ,

i.e. the minimum number of calls that still allows one to finish in the minimum possible number of rounds.

T...

Below is a brief summary of what is known about the trade‐off

“minimum number of calls C(n,r) vs. number of rounds r”

and in particular what happens at the time‐optimal value

r=T(n)=⌈log₂n⌉ (if n even),
r=T(n)=⌈log₂n⌉+1 (if n odd).

1. A simple lower bound (“doubling argument”)

– In each two‐way call a node’s total information can at most double.
– Hence in order...

This is an excellent question about the gossip problem with time constraints. Let me provide a comprehensive answer based on the current state of knowledge in this area.

The Minimum Calls Problem with Time Constraint

The problem you're asking about is to find the minimum number of calls needed to complete gossiping on nodes within rounds (time units).

Known Results

1. *...