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Celebrity Problem - Medium


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 celebrity in minimum number of questions.

This is a very interesting problem. The answer, ie the least number of questions required, changes based on what's given. Are we guaranteed there's one celebrity in the party? Does every non-celeb know every other non-celeb? ( not very ideal, but under a best case scenario. )

Once again, the rules
  • There is at most one celebrity at the party
  • Everyone knows the celebrity.
  • The celebrity knows no one.
  • We can only ask the question "Does A know B?"

When we ask the question, we can get a "yes" or "no". Each conveying information about both the participants in the question. Let's see those outcomes...

If the Answer is "Yes", we know A is definitely not a celebrity. But B is a candidate for a celebrity.

If the Answer is "No", we know both A and B are both candidates or both are Non-Celebrities.

Turns out, we can do this in linear time. We love those kinds, don't we?

Here we simply assume the first guy is a celeb. Each time you will ask a question
Does my assumed-celeb know the current-guy ?
 There are two outcomes,

  • If the answer is "Yes", we know our assumption was wrong,we can eliminate our assumption. But the next-guy could be a celeb. So we make him our new assumption.
  • If the answer is "No", our assumption is holding strong, we can safely eliminate the current-guy.


 Evidently, every time you ask the question, one guy gets eliminated from candidacy. So we'll need to ask (n - 1) questions to get one strong candidate.

At this point, if we were assured there was one celeb we can end the algorithm here. If there's a possibility, there was no celebrity, we'll have to perform one more scan to eliminate the end guy as well.

Theoretically, that's (n - 1) questions. So we need to ask a minimum of 2(n-1) questions to figure out the celebrity. I'll attempt to prove the lower bound some other time, will be handy for my Algorithm class.


import random
N = 10
#prepare testcase
R = range(N)
rand_celeb = random.randint(0,N-1)
m = {}
for i in xrange(N):
if i != rand_celeb:
m[i] = [random.randint(0,N-1) for j in xrange(10)] + [rand_celeb]
else:
m[i] = []
def knows(i,j): #i knows j?
return j in m[i]
def find_celeb(n):
celeb = 0
for i in xrange(1,n):
if not knows(i,celeb):
celeb = i
for i in xrange(n):
if i != celeb and knows(celeb,i):
return None
return celeb
print R
print "Actual Celeb : ",rand_celeb
print "Found Celeb : ",find_celeb(len(R))
view raw Single Celebrity Problem hosted with ❤ by GitHub
Let's extend this a bit, assume there were multiple celebrities at the party (or None). (A celeb doesn't know another celeb.) Although, I don't claim this is the most efficient algorithm, it certainly works. The first part of the solution remains the same, we find a celebrity, any one will do. Notice the original algorithm will do this for you. We include one more pass at the end, asking the question "Does X know Celebrity?". If the answer is "No", we KNOW that X is another celebrity. Here's the code.

import random
N = 100
#prepare testcase
R = range(N)
rand_celeb = list(set(random.randint(0,N-1) for i in xrange(5)))
m = {}
for i in xrange(N):
if i not in rand_celeb:
m[i] = [random.randint(0,N-1) for j in xrange(10)] +rand_celeb
else:
m[i] = []
def knows(i,j): #i knows j?
return j in m[i]
def find_celebs(n):
celeb = 0
for i in xrange(1,n):
if not knows(i,celeb):
celeb = i
celebs = []
for i in xrange(n):
if i != celeb and knows(celeb,i):
return None
if not knows(i,celeb):
celebs.append(i)
return celebs
print R
print "Actual Celeb : ",rand_celeb
print "Found Celeb : ",find_celebs(len(R))
view raw Multiple Celebrity Problem hosted with ❤ by GitHub
Lovely question, Thanks to CodeBunk for posting it.

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