The Celebrity Problem - GeeksforGeeks

Another classical problem.

In a party of N people, only one person is known to everyone. Such a person may be present in the party, if yes, (s)he doesn’t know anyone in the party. We can only ask questions like “does A know B? “. Find the stranger (celebrity) in minimum number of questions.

We can describe the problem input as an array of numbers/characters representing persons in the party. We also have a hypothetical function HaveAcquaintance(A, B) which returns true if A knows B, false otherwise. How can we solve the problem, try yourself first.

We measure the complexity in terms of calls made to HaveAcquaintance().

Graph:

We can model the solution using graphs. Initialize indegree and outdegree of every vertex as 0. If A knows B, draw a directed edge from A to B, increase indegree of B and outdegree of A by 1. Construct all possible edges of the graph for every possible pair [i, j]. We have N_C2 pairs. If celebrity is present in the party, we will have one sink node in the graph with outdegree of zero, and indegree of N-1. We can find the sink node in (N) time, but the overall complexity is O(N²) as we need to construct the graph first.

Recursion:

We can decompose the problem into combination of smaller instances. Say, if we know celebrity of N-1 persons, can we extend the solution to N? We have two possibilities, Celebrity(N-1) may know N, or N already knew Celebrity(N-1). In the former case, N will be celebrity if N doesn’t know anyone else. In the later case we need to check that Celebrity(N-1) doesn’t know N.

Solve the problem of smaller instance during divide step. On the way back, we may find a celebrity from the smaller instance. During combine stage, check whether the returned celebrity is known to everyone and he doesn’t know anyone. The recurrence of the recursive decomposition is,

T(N) = T(N-1) + O(N)

T(N) = O(N²). You may try Writing pseudo code to check your recursion skills.

Using Stack:

The graph construction takes O(N²) time, it is similar to brute force search. In case of recursion, we reduce the problem instance by not more than one, and also combine step may examine M-1 persons (M – instance size).

We have following observation based on elimination technique (Refer Polya’s How to Solve It book).

• If A knows B, then A can’t be celebrity. Discard A, and B may be celebrity.
• If A doesn’t know B, then B can’t be celebrity. Discard B, and A may be celebrity.
• Repeat above two steps till we left with only one person.
• Ensure the remained person is celebrity. (Why do we need this step?)

We can use stack to verity celebrity.

1. Push all the celebrities into a stack.
2. Pop off top two persons from the stack, discard one person based on return status of HaveAcquaintance(A, B).
3. Push the remained person onto stack.
4. Repeat step 2 and 3 until only one person remains in the stack.
5. Check the remained person in stack doesn’t have acquaintance with anyone else.

We will discard N elements utmost (Why?). If the celebrity is present in the party, we will call HaveAcquaintance() 3(N-1) times. Here is code using stack.

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