Connectedness of a Network

Here are the records from a busy telephone system with one million users:

The telephone number of the caller and the called number in record $n$ are $\operatorname{Caller}(n) = S_{2n-1}$ and $\operatorname{Called}(n) = S_{2n}$ where $S_{1,2,3,\dots}$ come from the “Lagged Fibonacci Generator”:
For $1 \le k \le 55$, $S_k = [100003 – 200003k + 300007k^3] \pmod{1000000}$.
For $56 \le k$, $S_k = [S_{k-24} + S_{k-55}] \pmod{1000000}$.
If $\operatorname{Caller}(n) = \operatorname{Called}(n)$ then the user is assumed to have misdialled and the call fails; otherwise the call is successful.
From the start of the records, we say that any pair of users $X$ and $Y$ are friends if $X$ calls $Y$ or vice-versa. Similarly, $X$ is a friend of a friend of $Z$ if $X$ is a friend of $Y$ and $Y$ is a friend of $Z$; and so on for longer chains.
The Prime Minister’s phone number is $524287$. After how many successful calls, not counting misdials, will $99\%$ of the users (including the PM) be a friend, or a friend of a friend etc., of the Prime Minister?

This problem appears to be inspired by Project Euler problem 186. Project Euler is a series of challenging mathematical problems that require careful thought and computational programming to solve.

The solution to this problem involves Graph Theory and algorithms designed for network analysis. One way to solve this problem is using Union-Find data structures.

To solve the problem, we could use an array of size 1,000,000 for the parent node of each member, and another array for the size of each group. Initially, everyone is their own parent, and the size of each group is 1. For each connection, we join the two clusters together, and add the sizes of the two groups.

If one of the numbers is the Prime Minister’s (524287), we also want to track the size of their group. When their group reaches 99% of 1,000,000, we stop and return the number of successful calls.

We also need to watch out for self-connections and ignore them, as they do not contribute to the solution.

Here’s an pseudo-code example of how this could be approached:

# Initialize parent array to each node being its own parent
parent = []
for i in range(1, 1000001):
parent[i] = i

# Initialize size array to each group having size 1
size = [1] * 1000001

# Initialize the telephone number generation process
S = []
for k in range(1, 56):
S.append((100003 – 200003*k + 300007*k**3) % 1000000)

for k in range(56, 1000001):
S.append((S[k-25] + S[k-56]) % 1000000)

# Define a findRoot function to find the group a number belongs to
def findRoot(x):
if parent[x] != x:
parent[x] = findRoot(parent[x])
return parent[x]

PM_Num = 524287
successfulCalls = 0
currCall = 1

# Loop until the PM’s group has 99% of the users
while size[findRoot(PM_Num)] < 990000: caller = S[2*currCall - 2] called = S[2*currCall - 1] if caller != called: successfulCalls += 1 rootOfCaller = findRoot(caller) rootOfCalled = findRoot(called) if rootOfCaller != rootOfCalled: if size[rootOfCaller] < size[rootOfCalled]: parent[rootOfCaller] = rootOfCalled size[rootOfCalled] += size[rootOfCaller] else: parent[rootOfCalled] = rootOfCaller size[rootOfCaller] += size[rootOfCalled] currCall += 1 return successfulCalls ``` This Python code would answer how many successful calls are needed until 99% of the users are connected to the Prime Minister. However, the specific answer would require running the code with an efficient programming environment.

More Answers:
Maximum Product of Parts
Triangles Containing the Origin
Number Mind

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