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Question

I'm a bit stumped by a conjecture proposed in an Algorithm Analysis class I'm taking. Here is the question.

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Assume that we have a hash table structured as a vector of lists, resolving collisions by sequential search of buckets. Without making any assumptions on the hash function used, show the following, where n is the number of items in the table and b is the number of buckets (or slots) in the table implementation. e(k) denotes the expected number of buckets of size k and p(k) denotes the probability a given bucket has exactly k items.

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I interpret the above description as basically describing a situation where n items are inserted into any one of b buckets. A bucket can contain between n and 0 items. It can be empty or it can contain all of the n items. We can make no assumptions about the p(k) function that denotes the probability that any one item will be inserted into any of the buckets.

The conjecture that I need to prove is.....

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Show that e(k) = bp(k)

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That is, show that the expected number of buckets of size k is equal to the probability that any given bucket has k items multiplied by b buckets.

I'd like to start this problem by saying that the probability that any one bucket receives an item is simply p(1) = 1/b. But this doesn't work since I'd have to assume that the p(k) function distributes items into the buckets evenly, which it may not necessarily do.

I'm not exactly sure how to set this problem up with so little information. Help would be much appreciated.

Answer 1

Well... For me, you just have to say that the probability for one bucket to have k elements is independent of the probability that any other bucket has k elements. Therefore

e(k) = probability of bucket 1 having k elements
     + probability of bucket 2 having k elements
     + ...
     + probability of bucket b having k elements
     = b*p(k)

I am not sure that is it what you are waiting for. The result seems pretty straightforward to me. Sorry if it did not help.

Answer 2

I think this is a great spot to use linearity of expectation.

Let's define some new random variables. For any bucket i and capacity k, let X _{i, k} be 1 if bucket i has k elements in it and 0 otherwise. This means that the total number of buckets with capacity k is given by

X _{1, k} + X _{2, k} + ... + X _{n, k}

Therefore, e(k) is given by

e(k) = E[X _{1, k} + X _{2, k} + ... + X _{n, k} ]

via linearity of expectation, we have

e(k) = E[X _{1, k} ] + E[X _{2, k} ] + ... + E[X _{n, k} ]

Now, let's look at any one of the X _{i, k} variables. Notice that X _{i, k} is 1 if bucket i has exactly k elements and is 0 otherwise.

Therefore:

E[X _{i, k} ] = 0 · Pr[bucket i doesn't have k elements] + 1 · Pr[bucket i does have k elements]

= Pr[bucket i does have k elements]

And hey! We know what that probability is - it's p(k). Therefore, E[X _{i, k} ] = p(k), so we have that

e(k) = E[X _{1, k} ] + E[X _{2, k} ] + ... + E[X _{n, k} ]

= p(k) + p(k) + ... + p(k)

= np(k)

and we're done!

Hope this helps!