Read upon \"Moss's Algorithm of Counting\" before attempting Counting the Number of tokens in a stream (Not Distinct) It is trivial to see that if there are m tokens in the stream, then [logzm] many bits suffice to keep track of the number of tokens. Now consider the following randomized algorithm. Probabilistic Counting: Let X 0. d) For this part, we consider an alternate (and somewhat more elegant) way of modifying the basic estimator1 to achieve better estimates. Suppose you modify the given algorithm as follows - you increment X with probability (1 +1a)x I for some a > 0 (a = 1 in the above algorithm). What should the algorithm return now? Determine the value of a that you need to choose in order to find an estimate Y such that IY ml 5 em with probability at least 9/10? Disclaimer: The solution to the above problem can be found on the internet with a little effort. But I need an answer with good and legit explanation. [1] Basic estimator: Let Y [0,1] (h is an idealized hash func) While (stream is non-empty) Let i be the next element/token Y ( min { Y, h(i)} 1 Return ; 1